THEORY OF MACHINES AND MECHANISMS

Theory of Machines and Mechanisms, 4e Uicker et al.

© Oxford University Press 2015. All rights reserved.

Solutions Manual to accompany

THEORY OF MACHINES

AND MECHANISMS Fourth Edition

International Version

John J. Uicker, Jr. Professor Emeritus of Mechanical Engineering

University of Wisconsin – Madison

Gordon R. Pennock Associate Professor of Mechanical Engineering

Purdue University

Joseph E. Shigley Late Professor Emeritus of Mechanical Engineering

The University of Michigan



PART 1

KINEMATICS AND MECHANISMS



Chapter 1

The World of Mechanisms

1.1 Sketch at least six different examples of the use of a planar four-bar linkage in practice.

They can be found in the workshop, in domestic appliances, on vehicles, on agricultural

machines, and so on.

Since the variety is unbounded no standard solutions are shown here.

1.2 The link lengths of a planar four-bar linkage are 0.2, 0.4, 0.6 and 0.6 m. Assemble the

links in all possible combinations and sketch the four inversions of each. Do these

linkages satisfy Grashof's law? Describe each inversion by name, for example, a crank-

rocker mechanism or a drag-link mechanism.

0.2, 0.6, 0.4, 0.6s l p q ; these linkages all satisfy Grashof’s law

since 0.2 0.6 0.4 0.6 . Ans.

Drag-link mechanism Drag-link mechanism Ans.

Crank-rocker mechanism Crank-rocker mechanism Ans.

Double-rocker mechanism Crank-rocker mechanism Ans.

1.3 A crank-rocker linkage has a 250 mm frame, a 62.5 mm crank, a 225 mm coupler, and a

187.5 mm rocker. Draw the linkage and find the maximum and minimum values of the

transmission angle. Locate both toggle positions and record the corresponding crank

angles and transmission angles.



Extremum transmission angles: min 1 max 353.1 98.1 Ans.

Toggle positions: 2 2 4 440.1 59.1 228.6 90.9 Ans.

1.4 In Fig. P1.4, point C is attached to the coupler; plot its complete path.

1.5 Find the mobility of each mechanism illustrated in Fig. P1.5.



(a) 1 26, 7, 0;n j j 3 6 1 2 7 1 0 1m Ans.

(b) 1 28, 10, 0;n j j 3 8 1 2 10 1 0 1m Ans.

(c) 1 27, 9, 0;n j j 3 7 1 2 9 1 0 0m Ans.

Note that the Kutzbach criterion fails in this case; the true mobility is m=1. The

exception is due to a redundant constraint. The assumption that the rolling contact

joint does not allow links 2 and 3 to separate duplicates the constraint of the fixed

link length 2 3O O .

(d) 1 24, 3, 2;n j j 3 4 1 2 3 1 2 1m Ans.

Notice that each coaxial pair of sliding ground joints is counted as only a single

prismatic pair.

1.6 Use the Kutzbach criterion to determine the mobility of the mechanism illustrated in Fig.

P1.6.

1 25, 5, 1;n j j 3 5 1 2 5 1 1 1m Ans.

Notice that the double pin is counted as two single j1 pins.

1.7 Find a planar mechanism with a mobility of one that contains a moving quaternary link.



How many distinct variations of this mechanism can you find?

To have at least one quaternary link, a planar mechanism must have at least eight links.

The Grübler criterion then indicates that ten single-freedom joints are required for

mobility of m = 1. According to H. Alt, “Die Analyse und Synthese der achtgleidrigen

Gelenkgetriebe”, VDI-Berichte, vol. 5, 1955, pp. 81-93, there are a total of sixteen

distinct eight-link planar linkages having ten revolute joints, seven of which contain a

quaternary link. These seven are shown below: Ans.

1.8 Use the Kutzbach criterion to detemine the mobility of the planar mechanism illustrated

in Fig. P1.8. Clearly number each link and label the lower pairs (j1) and higher pairs (j2)

on Fig. P1.8.

1 25, 5, 1;n j j 3 5 1 2 5 1 1 1m Ans.



1.9 For the mechanism illustrated in Fig. P1.9, determine the number of links, the number of

lower pairs, and the number of higher pairs. Using the Kutzbach criterion determine the

mobility. Is the answer correct? Briefly explain.

1 24, 3, 2;n j j 3 4 1 2 3 1 2 1m Ans.

If it is not evident that the input shown will increment this device in the direction shown,

then consider incrementing link 3 downward. Since it seems intuitive that this

determines the position of all other links, this verifies that mobility of one is correct.

1.10 Use the Kutzbach criterion to detemine the mobility of the planar mechanism illustrated

in Fig. P1.10. Clearly number each link and label the lower pairs (j1) and higher pairs (j2)

on Fig. P1.10. Treat rolling contact to mean rolling with no slipping.

1 25, 5, 1;n j j 3 5 1 2 5 1 1 1m Ans.



1.11 For the mechanism illustrated in Fig. P1.11 treat rolling contact to mean rolling with no

slipping. Determine the number of links, the number of lower pairs, and the number of

higher pairs. Using the Kutzbach criterion determine the mobility. Is the answer correct?

Briefly explain.

1 27, 8, 1;n j j 3 7 1 2 8 1 1 1m Ans.

This result appears to be correct. If all parts remain assembled and within the limits of

travel of the joints shown, then it appears that when any one member is locked the total

system becomes a structure.

1.12 Does the Kutzbach criterion provide the correct result for the planar mechanism

illustrated in Fig. P1.12? Briefly explain why or why not.

1 24, 2, 3;n j j 3 4 1 2 2 1 3 2m Ans.

This result appears to be correct. If any part except the wheel is moved, other parts are

required to follow. However, after all other parts are in a set position, the wheel is still

able to rotate because of slipping against the frame at A.



1.13 The mobility of the mechanism illustrated in Fig. P1.13 is m = 1. Use the Kutzbach

criterion to determine the number of lower pairs and the number of higher pairs. Is the

wheel rolling without slipping, or rolling and slipping, at point A on the wall?

Suppose that we identify the number of constraints at A by the symbol k. Then if we

account for all links and all other joints as follows, the Kutzbach criterion gives

1 25; 4; 1; 1;kn j j j 3 5 1 2 4 1 1 1 3 ;m k k

Therefore, to have mobility of 1m , we must have 2k constraints at A. The wheel

must be rolling without slipping. Ans.

1.14 Devise a practical working model of the drag-link mechanism.



1.15 Find the time ratio of the linkage of Problem 1.3.

From the values of 2 and 4 we find 188.5 and 171.5 .

Then, from Eq. (1.5), 1.099Q . Ans.

1.16 Plot the complete coupler curve of the Roberts' mechanism illustrated in Fig. 1.24b. Use

AB = CD = AD = 62.5 mm and BC = 31.25 mm.

1.17 If the crank of Fig. 1.11 is turned 25 revolutions counterclockwise, how far and in what

direction does the carriage move?

Screw and carriage move by (25 rev)/(6 rev/mm) = 4.17 mm to the right.

Carriage moves (7 rev)/(18 rev/mm) = 3.57 mm to the left with respect to the screw.

Net motion of carriage = 25/6 – 25/7 = 25/42 = 0.59 mm to the right. Ans.

More in-depth study of such devices is covered in Chapter 9.



1.18 Show how the mechanism of Fig. 1.15b can be used to generate a sine wave.

With the length and angle of crank 2 designated as R and 2, respectively, the horizontal

motion of link 4 is 4 2 2cos sin 90x R R .

1.19 Devise a crank-and-rocker linkage, as in Fig. 1.14c, having a rocker angle of 60. The

rocker length is to be 0.50 m.



1.20 A crank-rocker four-bar linkage is required to have a time ratio Q = 1.2. The rocker is to

have a length of 62.5 mm and oscillate through a total angle of 60 . Determine a

suitable set of link lengths for the remaining three links of the four-bar linkage.

Following the procedure of Example 1.4, the required time ratio gives

1801.2

180Q

and, therefore, we must have 16.36 .

Then, with the X-line chosen at 30°, the drawing shown below (dimensioned 10 times

size) gives measured distances of 4 2 1 108.5 mmO OR r ,

2 2 3 2 160.5 mmB OR r r , and

1 2 3 2 111 mmB OR r r . From these we get one possible solution, which has link lengths

1 2 3 4111 mm, 24.8 mm, 135.8 mm, 62.5 mmr r r r Ans.

111 mm

160.5 mm



Chapter 2

Position and Displacement

2.1 Describe and sketch the locus of a point A that moves according to the equations

cos 2xAR at t , sin 2

yA

R at t , and 0zAR .

The locus is the spiral shown. Ans.

2.2 Find the position difference to point P from point Q on the curve 2 16y x x , where 2x

PR and 4xQ

R .

2

2 2 16 10y

PR ; ˆ ˆ2 10P R i j

2

4 4 16 4y

QR ; ˆ ˆ4 4Q R i j

ˆ ˆ2 14 14.142 98.1PQ P Q R R R i j Ans.



2.3 The path of a moving point is defined by the equation 22 28y x .

Find the position difference to point P from point Q if 4xPR and

3xQ

R .

2

2 4 28 4y

PR ; ˆ ˆ4 4P R i j

2

2 3 28 10y

QR ; ˆ ˆ3 10Q R i j

ˆ ˆ7 14 15.652 63.4PQ P Q R R R i j Ans.

2.4 The path of a moving point P is defined by the equation 360 / 3y x . What is the displacement of the point if its motion

begins when 0xPR and ends when 3x

PR ?

3

0 60 0 / 3 60y

PR ; ˆ0 60P R j

3

3 60 3 / 3 51y

PR ;

ˆ ˆ3 3 51P R i j

ˆ ˆ(3) (0) 3 9 9.487 71.57P P P R R R i j Ans.

2.5 If point A moves on the locus of Problem 2.1, find its displacement from t = 1.5 to t = 2.

ˆ ˆ ˆ1.5 1.5 cos3 1.5 sin3 1.5A a a a R i j i

ˆ ˆ ˆ2.0 2.0 cos4 2.0 sin 4 2.0A a a a R i j i

ˆ2.0 1.5 3.5A A A a ΔR R R i Ans.

2.6 The position of a point is given by the equation 2100 j te R . What is the path of the

point? Determine the displacement of the point from t = 0.10 to t = 0.60.

The point moves in a circle of radius 100 with center at the origin. Ans.

0.628 ˆ ˆ0.10 100 80.902 58.779je R i j

3.770 ˆ ˆ0.60 100 80.902 58.779je R i j

ˆ ˆ0.60 0.10 161.804 117.557 200.0 216 ΔR R R i j Ans.



2.7 The equation 2 /104 j tt e R defines the position of a point. In which direction is

the position vector rotating? Where is the point located when t = 0? What is the next

value t can have if the orientation of the position vector is to be the same as it is when t =

0? What is the displacement from the first position of the point to the second?

Since the polar angle for the position vector is

/10t , then /d dt is negative and therefore

the position vector is rotating clockwise. Ans.

2 00 0 4 4 0je R Ans.

The position vector will next have the same

direction when /10 2t , that is, when t=20. Ans.

2 220 20 4 404 0je R

20 0 400 0 R R R Ans.

2.8 The location of a point is defined by the equation 2 / 304 2 j tt e R , where t is time in

seconds. Motion of the point is initiated when t = 0. What is the displacement during the

first 3 s? Find the change in angular orientation of the position vector during the same

time interval.

0 ˆ0 0 2 2 0 2je R i

9/ 30 ˆ ˆ3 12 2 14 54 8.229 11.326je R i j

ˆ ˆ3 0 6.229 11.326 12.926 61.19 ΔR R R i j Ans.

54 0 54 ccw Ans.



2.9 Link 2 in Fig. P2.9 rotates according to the equation / 4t . Block 3 slides outward

on link 2 according to the equation 2 2r t . What is the absolute displacement

3PR from t = 1 to t = 3? What is the apparent displacement 3/ 2PR ?

3

2 / 42j j t

P re t e R

3

ˆ ˆ1 3 45 2.121 2.121P R i j

3

ˆ ˆ3 11 135 7.778 7.778P R i j

3 3 3

ˆ ˆ3 1 9.899 5.657 11.402 150.26P P P ΔR R R i j Ans.

3

0 2

/ 2 2ˆ2j

P re t R i

3 / 2 2

ˆ1 3P R i

3 /2 2

ˆ3 11P R i

3 3 3/2 /2 /2 2

ˆ3 1 8P P P ΔR R R i Ans.

2.10 A wheel with center at O rolls without slipping so that its center is displaced 250 mm to

the right. What is the displacement of point P on the periphery during this interval?

Since the wheel rolls without slipping,

O POR R .

/

250 mm /150 mm 1.667 rad 95.51O POR R

For POR ,

270 95.51 174.49 ˆ ˆ150 mm 174.49 149.3 14.4 mmPO

R i j

ˆ ˆ ˆ ˆ250 149.3 14.4 150 mm

P O PO PO

ΔR ΔR R R

i i j j

6 inPO

R ˆ ˆ100.7 164.4 mm 192.8 mm 58.51P R i j Ans.

2.11 A point Q moves from A to B along link 3 whereas link 2 rotates from 2 30 to

2 120 . Find the absolute displacement of Q.

3

ˆ ˆ0.3 m 30 259.8 150.0 mmQ R i j

3

ˆ ˆ0.3 m 120 150.0 259.8 mmQ R i j

3 3 3

ˆ ˆ409.8 109.8 mmQ Q Q ΔR R R i j

5 /3ˆ600.0 mmQ BA ΔR R i



2 4

0.3 mAO BO

R R ;4 2

0.6 mBA O O

R R 5 3 5 /3Q Q Q ΔR ΔR ΔR

5

ˆ ˆ190.2 109.8 mm 219.6 mm 30Q ΔR i j Ans.

2.12 The linkage is driven by moving the sliding block 2. Write the loop-closure equation.

Solve analytically for the position of sliding block 4. Check the result graphically for the

position where 45 .

The loop-closure equation is

A B AB R R R Ans.

/12 jj

A B AB

j

B AB

R e R R e

R R e

500 mm, 15AB

R

Taking the imaginary components of this, we get

sin15 sinA ABR R

sin sin 45500 mm 1365 mm

sin15 sin15A ABR R

Ans.

2.13 The offset slider-crank mechanism is driven by the rotating crank 2. Write the loop-

closure equation. Solve for the position of the slider 4 as a function of 2 .

20 mm

AOR , 50 mm

BAR , and 140 mm

CBR

C A BA CB R R R R

32/ 2 jjj

C A BA CBR R e R e R e

Taking real and imaginary parts,

2 3cos cosC BA CBR R R and 2 30 sin sinA BA CBR R R

and, solving simultaneously, we get

1 23

sinsin A BA

CB

R R

R

with 390 90



22

2 2

2

2 2 2

cos sin

50cos 19 200 1 000sin 2 500sin

C BA CB A BAR R R R R

Ans.

2.14 For the mechanism illustrated in Fig. P2.14, define a set of vectors that is suitable for a

complete kinematic analysis of the mechanism. Label and show the sense and orientation

of each vector on Fig. P2.14. Write the vector loop equation(s) for the mechanism.

Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown

variables, and any constraints. If you have identified constraints then write the constraint

equation(s).

One suitable set of two vector loop equations is

? ? ?

2 3 4 5 1

R R R R R 0 and ? 1 2 3

2 3 44 24 22

I C C C

R R R R R 0 Ans.

The angle 2 is a reasonable input. Three constraint equations are required.

44 4 (C1) 24 4 (C2) 22 2 (C3) Ans.

There are four unknowns 3 4 5 24, , , and R .



2.15 Assume rolling with no slip between pinion 5 and rack 4 in the mechanism illustrated in

Fig. P2.15. Define a set of vectors that is suitable for a complete kinematic analysis of

the mechanism. Label and show the sense and orientation of each vector on Fig. P2.15.

Write the vector loop equation(s) for the mechanism. Identify suitable input(s) for the

mechanism. Identify the known quantities, the unknown variables, and any constraints.

If you have identified constraints then write the constraint equation(s).

One suitable set of vectors is as shown. The vector loop equation is

? ?

2 3 6 4 15 1 R R R R R R 0ÖI Ö Ö ÖÖ ÖÖ ÖÖ

with 5 5 6R Ans.

The angle 2 is a suitable input. Ans.

There are two unknown variables, 3

and R6. Ans.



2.16 For the geared five-bar mechanism illustrated in Fig. P2.16, there is rolling with no

slipping between gears 2 and 5. Define a set of vectors that is suitable for a complete

kinematic analysis of the mechanism. Label and show the sense and orientation of each

vector on Fig. P2.16. Write the vector loop equation(s) for the mechanism. Identify

suitable input(s) for the mechanism. Identify the known quantities, the unknown


equation(s).


? C

2 3 4 5 1 R R R R R 0ÖI Ö Ö? Ö ÖÖ

with 2 2 5 5 0 Ans.

The angle 2 is a suitable input.



2.17 For the mechanism illustrated in Fig. P2.17, gear 3 is pinned to link 4 at point B, and is

rolling without slipping on the semi-circular ground link 1. The radius of the semi-

circular ground link is 1 and the radius of gear 3 is 3. Define a set of vectors that is

suitable for a complete kinematic analysis of the mechanism shown. Label and show the

sense and orientation of each vector in Fig. P2.17. Write the vector loop equation(s) for

the mechanism. Identify suitable input(s) for the mechanism. Identify the known

quantities, the unknown variables, and any constraints. If you have identified constraints

then write the constraint equation(s).


?

2 4 5 1 R R R R 0ÖI Ö Ö? ÖÖ

with 2 2 3 3R Ans.

The angle 2 is a suitable input.





of each vector in Fig. P1.6. Write the vector loop equation(s) for the mechanism.



equation(s).

One set of vectors suitable for a kinematic analysis of the mechanism is shown below.

The corresponding vector loop equations are

? ?

1 2 3 13

I

a

R R R R 0 and ? ? ?

3 4 15 13

C

b

R R R R 0 Ans.

with the constraint equation 13 13 constant.a bR R Ans.








equation(s).

One set of vectors suitable for a kinematic analysis of the mechanism is shown here.

The corresponding set of vector loop equations is

? ?

1 2 4 5

I

R R R R 0 and ? ?

1 22 3 35

C

R R R R 0 Ans.

with the constraint equation 22 2 . Ans.








equation(s).

One set of vectors suitable for a complete kinematic analysis of this mechanism is as

shown.


?

1 3 32 2

I

R R R R 0 and ? ? 1 2 ?

11 4 34 33 32 2

C C I

R R R R R R 0 Ans.

with the two constraint equations

34 4 90° C1 and 33 4 180° C2. Ans.








equation(s).

One set of vectors suitable for a kinematic analysis of the mechanism is shown.


1 11 2 3

I

R R R R 0 and 1 ? ?

11 2 34 4 9

I C

R R R R R 0 Ans.

with the constraint equation

34 3 C1 Ans.

However, these equations do not analyze the angular displacement of the small wheel,

body 5. In order to do this, we might consider the apparent angular displacement as seen

by an observer fixed on vector 9 and viewing the point of contact between bodies 5 and

1. The non-slip condition would provide the constraint

1 1/9 5 5/9

1 1 9 5 5 9

5 5 1 5 9

5 5 9 9

0

0R

.Ans

where 5 is the radius of wheel 5 and 5 is the angular displacement of body 5.

2.22 Write a calculator program to find the sum of any number of two-dimensional vectors

expressed in mixed rectangular or polar forms. The result should be obtainable in either

form with the magnitude and angle of the polar form having only positive values.

Because the variety of makes and models of calculators is vast and no standards exist for

programming them, no solution is shown here.



2.23 Write a computer program to plot the coupler curve of any crank-rocker or double-crank

form of the four-bar linkage. The program should accept four link lengths and either

rectangular or polar coordinates of the coupler point relative to the coupler.

Again the variety of programming languages makes it difficult to provide a standard

solution. However, one version, written in ANSI/ISO FORTRAN 77, is supplied here as

an example. There are also no universally accepted standards for programming graphics.

Therefore the Tektronix PLOT10 subroutine library, for display on Tektronix 4010 series

displays, is chosen as an older but somewhat recognized alternative. The symbols in the

program correspond to the notation shown in Figure 2.19 of the text. The required input

data are:

X5, Y5, -1R1, R2, R3, R4,

R5, θ5, 1

The program can be verified using the data of Example 2.7 and checking the results

against those of Table 2.3.

PROGRAM CCURVE

C

C A FORTRAN 77 PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK-ROCKER

C OR DOUBLE-CRANK FOUR-BAR LINKAGE, GIVEN ITS DIMESNIONS.

C ORIGINALLY WRITTEN USING SUBROUTINES FROM TEKTRONIX PLOT10 FOR

C DISPLAY ON 4010 SERIES DISPLAYS.

C REF:J.J.UICKER,JR, G.R.PENNOCK, & J.E.SHIGLEY, ‘THEORY OF MACHINES

C AND MECHANISMS,’ FOURTH EDITION, OXFORD UNIVERSITY PRESS, 2009.

C EXAMPLE 2.6

C

C WRITTEN BY: JOHN J. UICKER, JR.

C ON: 01 JANUARY 1980

C

C READ IN THE DIMENSIONS OF THE LINKAGE.

READ(5,1000)R1,R2,R3,R4,X5,Y5,IFORM

1000 FORMAT(6F10.0,I2)

C

C FIND R5 AND ALPHA.

IF(IFORM.LE.0)THEN

R5=SQRT(X5*X5+Y5*Y5)

ALPHA=ATAN2(Y5,X5)

ELSE

R5=X5

ALPHA=Y5/57.29578

END IF

Y5=AMAX1(0.0,R5*SIN(ALPHA))

C

C INITIALIZE FOR PLOTTING AT 120 CHARACTERS PER SECOND.

CALL INITT(1200)

C

C SET THE WINDOW FOR THE PLOTTING AREA.

CALL DWINDO(-R2,R1+R2+R4,-R4,R4+R4+Y5)

C

C CYCLE THROUGH ONE CRANK ROTATION IN FIVE DEGREE INCREMENTS.

TH2=0.0

DTH2=5.0/57.29578

IPEN=-1

DO 2 I=1,73



CTH2=COS(TH2)

STH2=SIN(TH2)

C

C CALCULATE THE TRANSMISSION ANGLE.

CGAM=(R3*R3+R4*R4-R1*R1-R2*R2+2.0*R1*R2*CTH2)/(2.0*R3*R4)

IF(ABS(CGAM).GT.0.99)THEN

CALL MOVABS(100,100)

CALL ANMODE

WRITE(7,1001)

1001 FORMAT(//’ *** THE TRANSMISSION ANGLE IS TOO SMALL. ***’)

GO TO 1

END IF

SGAM=SQRT(1.0-CGAM*CGAM)

GAM=ATAN2(SGAM,CGAM)

C

C CALCULATE THETA 3.

STH3=-R2*STH2+R4*SIN(GAM)

CTH3=R3+R1-R2*CTH2-R4*COS(GAM)

TH3=2.0*ATAN2(STH3,CTH3)

C

C CALCULATE THE COUPLER POINT POSITION.

TH6=TH3+ALPHA

XP=R2*CTH2+R5*COS(TH6)

YP=R2*STH2+R5*SIN(TH6)

C

C PLOT THIS SEGMENT OF THE COUPLER CURVE.

IF(IPEN.LT.0)THEN

IPEN=1

CALL MOVEA(XP,YP)

ELSE

IPEN=-1

CALL DRAWA(XP,YP)

END IF

TH2=TH2+DTH2

2 CONTINUE

C

C DRAW THE LINKAGE.

CALL MOVEA(0.0,0.0)

CALL DRAWA(R2,0.0)

XC=R2+R3*COS(TH3)

YC=R3*SIN(TH3)

CALL DRAWA(XC,YC)

CALL DRAWA(XP,YP)

CALL DRAWA(R2,0.0)

CALL MOVEA(XC,YC)

CALL DRAWA(R1,0.0)

1 CALL FINITT(0,0)

CALL EXIT

STOP

END



2.24 For each linkage illustrated in Fig. P2.24, find the path of point P: (a) inverted slider-

crank mechanism; (b) second inversion of the slider-crank mechanism; (c) Scott-Russell

straight-line mechanism; and (d) drag-link mechanism.

(a) (b)

(c) (d)

(a) 40 mmCA

R , 70 mmBA

R , 80 mmPC

R ; (b) 100 mmCA

R , 50 mmBA

R ,

162.5 mmPB

R ; (c) 125 mmBA CB PB

R R R ; (d) 10 mmDA

R , 20 mmBA

R ,

30 mmCC DC

R R , 40 mmPB

R .



2.25 Using the offset slider-crank mechanism in Fig. P2.13, find the crank angles

corresponding to the extreme values of the transmission angle.

As shown, 390 .

Also from the figure

2 2 3sin cose r r .

Differentiating with

respect to 2 ,

2 2 3

2

cos sind

r rd

;

2 2

2 3

cos

sin

rd

d r

.

Now, setting 2/ 0d d , we get 2cos 0 .

Therefore, we conclude that 2 2 1 / 2 90 , 270 ,k Ans.

2.26 In Section 1.10 it is pointed out that the transmission angle reaches an extreme value for

the four-bar linkage when the crank lies on the line between the fixed pivots. Referring

to Fig. 2.19, this means that reaches a maximum or minimum when crank 2 is colinear

with the line 2 4O O . Show, analytically, that this statement is true.

From 4 2 :O O A

2 2 2

1 2 1 2 22 coss r r rr .

Also, from 4 :ABO 2 2 2

3 4 3 42 coss r r r r .

Equating these we

differentiate with respect to 2

to obtain

1 2 2 3 4

2

2 sin 2 sind

r r r rd

or

1 2 2

2 3 4

sin

sin

r rd

d r r

.

Now, for 2

0d

d

, we have 2sin 0 . Thus, 2 0, 180 , 360 , Q.E.D.



2.27 Figure P2.27 illustrates a crank-and-rocker four-bar linkage in the first of its two limit

positions. In a limit position, points 2 , , and BO A lie on a straight line; that is, links 2

and 3 form a straight line. The two limit positions of a crank-rocker describe the extreme

positions of the rocking angle. Suppose that such a linkage has 1 100 mmr ,

2 50 mmr , 3 125 mmr , and 4 100 mmr .

(a) Find 2 4 and corresponding to each limit position.

(b) What is the total rocking angle of link 4?

(c) What are the transmission angles at the extremes?

(a) From isosceles triangle 4 2O O B we

can calculate or measure 2 29 ,

4 58 and 2 248 , 4 136 .

Ans.

(b) Then 4 4 4 78 Ans.

(c) Finally, from isosceles triangle

4 2O O B , 29 and 68 . Ans.

2.28 A double-rocker four-bar linkage has a dead-center position and may also have a limit

position (see Prob. 2.27). These positions occur when links 3 and 4 in Fig. P2.28 lie

along a straight line. In the dead-center position the transmission angle is 180 and the

mechanism is locked. The designer must either avoid such positions or provide the

external force, such as a spring, to unlock the linkage. Suppose, for the linkage

illustrated in Fig. P2.28, that 1

140 mmr , 2

55 mmr , 3

50 mmr , and 4 120 mmr .

Find 2 and 4 corresponding to the dead-center position. Is there a limit position?

For the given dimensions, there are two

dead-center positions, and they

correspond to the two extreme travel

positions of crank 2O A . From 4 2O AO

using the law of cosines, we can find

2 114.0 , 4 162.8 and, symmetrically,

2 114.0 , 4 162.8 . There are

also two limit positions; these occur at

2 56.5 , 4 133.1 and, symmetrically,

at 2 56.5 , 4 133.1 . Ans.



2.29 Figure P2.29 illustrates a slider-crank linkage that has an offset e and that is placed in one

of its limiting positions. By changing the offset e, it is possible to cause the angle that

crank 2 makes in traversing between the two limiting positions to vary in such a manner

that the driving or forward stroke of the slider takes place over a larger angle than the

angle used for the return stroke. Such a linkage is then called a quick-return mechanism.

The problem here is to develop a formula for the crank angle traversed during the

forward stroke and also develop a similar formula for the angle traversed during the

return stroke. The ratio of these two angles would then constitute a time ratio of the drive

to return strokes. Also determine which direction the crank should rotate.

From the figure we can see that 3 2 2 3 2 2sin sin 180e r r r r or

1

2

3 2

sine

r r

, 1

2

3 2

180 sine

r r

1 1

2 2

3 2 3 2

180 sin sindrive

e e

r r r r

Ans.

1 1

2 2

3 2 3 2

360 180 sin sinreturn

e e

r r r r

Ans.

Assuming driving is when B is sliding to the right, the crank should rotate clockwise. Ans.



Chapter 3

Velocity

3.1 The position vector of a point is given by the equation 100 j te R , where R is in meters.

Find the velocity of the point at 0.40 s.t

100 mj tt e R

100 m/sj tt j e R

0.400.40s 100 m/s

100 cos0.40 sin 0.40 m/s

100 sin 72 100 cos72 m/s

jj e

j j

j

R

0.40s 298.783 97.080 m/s 314.159 m/s 162j R Ans.

3.3 If automobile A is traveling south at 70.4 km/h and automobile B north 60 east at 51.2

km/h, what is the velocity difference between B and A? What is the apparent velocity of

B to the driver of A?

ˆ70.4 km/h 90 70.4 km/hA V j

ˆ ˆ51.2 km/h 30 44.34 25.6 km/hB V i j

ˆ ˆ44.34 96 km/hBA B A V V V i j

105.74 km/h 65.2 =105.74 km/h N 24.8 EBA V Ans

Naming B as car 3 and A as car 2, we have2B AV V since 2 is

translating. Then 3 3 2/ 2B B B BA V V V V

3 /2 105.74 km/h 65.2 =105.74 km/h N 24.8 EB V Ans.

3.4 In Fig. P3.4, wheel 2 rotates at 450 rev/min and drives wheel 3 without slipping. Find

the velocity difference between points B and A.



2

450 rev/min 2 rad/rev

60 s/min

15 rad/s cw

2 22 1500 4710 mm/sAO AOV R

4200 mm/s 90B V

ˆ ˆ4710 4200

6300 mm/s 138.4

BA B A

V V V

i j Ans.

3.5 Two points A and B, located along the radius of a wheel (see Fig. P3.5), have speeds of

80 and 140 in/s, respectively. The distance between the points is .75 mmBA

R .

(a) What is the diameter of the wheel?

(b) Find ,

,AB BA

V V and the angular velocity of the wheel.

ˆ ˆ ˆ3.5 2 1.5 m/sBA B A V V V j j j Ans.

ˆ ˆ ˆ2 3.5 1.5 m/sAB A B V V V j j j Ans.

2

1.5 m/s20 rad/s cw

0.075 m

BA

BA

V

R Ans.

2

2

2

3.5 m/s175 mm

20 rad/s

BO

BO

VR

2

2 2 17.5 mm 350 mmBOD R Ans.



3.6 A plane leaves point B and flies east at 448 km/h. Simultaneously, at point A, 320 km

southeast (see Fig. P3.6), a plane leaves and flies northeast at 499.2 km/h.

(a) How close will the planes come to each other if they fly at the same altitude?

(b) If they both leave at 6:00 p.m., at what time will this occur?

ˆ ˆ499.2 45 km/h 352.96 352.96 km/hA V i j ;

ˆ448 km/hB V i

ˆ ˆ95 352.96 km/hBA B A V V V i j

At initial time ˆ ˆ0 320 120 160 276.8 kmBA R i j

Later ˆ ˆ( ) 0 160 95 276.8 352.96 kmBA BA BAt t t t R R V i j

To find the minimum of this:

2 22 160 95 276.8 352.96BAR t t

2 2 160 95 95 2 276.8 352.96 352.96 0BAdR dt t t

269211 225 798 0t 0.838 h 51 mint or 6:51p.m. Ans.

320 kmAB

R ˆ ˆ0.845 h 79.68 21.44 82.56 km 165BA R i j Ans.

3.7 To the data of Problem 3.6, add a wind of 48 km/h from the west.

(a) If A flies the same heading, what is its new path?

(b) What change does the wind make in the results of Problem 3.6?

With the added wind ˆ ˆ400.96 352.96 km/h 534.24 km/h 41.4A V i j

Since the velocity is constant, the new path is a straight line at N 48.6º E. Ans.

Since the velocities of both planes change by the same amount, the velocity difference

BAV does not change. Therefore the results of Problem 3.6 do not change. Ans.

3.8 The velocity of point B on the linkage illustrated in Fig. P3.8 is 1 m/s. Find the velocity

of point A and the angular velocity of link 3.

0.1 mAB

R

A B AB V V V

1.24 m/s 165A V Ans.

0.37 m/s 120AB V



3

0.37 m/s3.7 rad/s ccw

0.1 m

AB

AB

V

R Ans.



3.9 The mechanism illustrated in Fig. P3.9 is driven by link 2 at 2

45 rad/s ccw. Find the

angular velocities of links 3 and 4.

100 mm,

2AO

R 250 mm,BA

R 2

250 mm,4

O OR and 300 mm.

4BO

R

2 22 45 rad/s 100 mm 4 500 mm/sAO AOV R

4 4B A BA O BO V V V V V

358.5 mm/sBAV ; 4

4 619 mm/sBOV .

3

358.5 in/s1.43 rad/s ccw

250 in

BA

BA

V

R Ans.

4

4

4

4 619 mm/s15.40 rad/s ccw

300 mm

BO

BO

V

R Ans.



3.10 Crank 2 of the push-link mechanism illustrated in Fig. P3.10 is driven at

260 rad / s cw. Find the velocities of points B and C and the angular velocities of

links 3 and 4.

2

6 in,AO

R 12 in,BA

R 4 2

3 in,O O

R 12 in,4

BOR 6 in,

DAR and 4 in.

CDR

2 22 60 rad/s 6 in 360 in/sAO AOV R


520.8 in/sBAV ; 454.4 in/s 41B V Ans.

C A CA B CB V V V V V

153.2 in/s 60C V Ans.

3

520.8 in/s43.40 rad/s cw

12 in

BA

BA

V

R ; 4

4

4

454.4 in/s37.87 rad/s cw

12 in

BO

BO

V

R Ans.

3.11 Find the velocity of point C on link 4 of the mechanism illustrated in Fig. P3.11 if crank

2 is driven at 2

48 rad / s ccw. What is the angular velocity of link 3?

2 22 48 rad/s 200 mm 9 600.0 mm/sAO AOV R


2

200 mm,AO

R 800 mm,BA

R 4 2

400 mm,O O

R 4

400 mm,BO

R and 4

300 mm.CO

R

268 mm/sBAV ; 3

268 mm/s0.335 rad/s ccw

800 mm

BA

BA

V

R Ans.

4 4C O CO B CB V V V V V ; 7 118 mm/s 75.8C V Ans.



3.12 Figure P3.12 illustrates a parallel-bar linkage, in which opposite links have equal lengths.

For this linkage, demonstrate that 3

is always zero and that 4 2

. How would you

describe the motion of link 4 with respect to link 2?

Referring to Fig. 2.19 and using 1 3r r and 2 4r r , we compare Eqs. (2.26) and (2.27) to

see that . Then Eq. (2.29) gives 3 0 and its derivative is 3 0 . Ans.

Next we substitute Eq. (2.25) into Eq. (2.33) to see that 2 . Then Fig. 2.19 shows

that, since link 3 is parallel to link 1 3 0 , then 4 2 . Finally, the derivative of

this gives 4 2 . Ans.

Since 4/ 2 4 2 0 , link 4 is in curvilinear translation with respect to link 2. Ans.



3.13 Figure P3.13 illustrates the antiparallel or crossed-bar linkage. If link 2 is driven at

21 rad/s ccw, find the velocities of points C and D.

2 4

300 mmAO BO

R R , 4 2

150 mmBA O O

R R , and 75 mmCA DBR R

2 22 1 rad/s 300 mm 300 m/sAO AOV R


Construct the velocity image of link 3.

402.5 mm/s 151C V Ans.

290 mm/s 249D V Ans.

3.14 Find the velocity of point C of the linkage illustrated in Fig. P3.14 assuming that link 2

has an angular velocity of 60 rad/s ccw. Also find the angular velocities of links 3 and 4.

2

150 mm,AO BA

R R 4 2 4

250 mm,O O BO

R R and 200 mm.CA

R



4 235 mm/sBAV ; 4

7 940 mm/sBOV

3

4 235 mm/s28.23 rad/s cw

150 in

BA

BA

V

R Ans.

4

4

4

7 940 mm/s31.76 rad/s cw

250 mm

BO

BO

V

R Ans.

Construct the velocity image of link 3:

12 680 mm/s 156.9C V Ans.



3.15 The inversion of the slider-crank mechanism illustrated in Fig. P3.15 is driven by link 2

at 2

60 rad/s ccw. Find the velocity of point B and the angular velocities of links 3 and

4.

2

75 mm,AO

R 400 mm,BA

R and 2

125 mm.4

O OR

2 22 60 rad/s 75 mm 4500 mm/sAO AOV R

3 3 4 3 / 4P A P A P P V V V V V

3

3

4 3


194 mm

P A

P A

V

R Ans.


4790 mm/s 96.5B V Ans.

3.16 Find the velocity of the coupler point C and the angular velocities of links 3 and 4 of the

mechanism illustrated if crank 2 has an angular velocity of 30 rad/s cw.

2

75 mm,AO

R 125 mm,BA CB

R R 2

250 mm,4

O OR and

4

150 mm.BO

R

2 22 30 rad/s 75 mm 2 250.0 mm/sAO AOV R



2 250.0 mm/s 126.9C V Ans.

3

2 250.0 mm/s18.00 rad/s ccw

125 mm

BA

BA

V

R ; 4

4

4

00

200.0 mm

BO

BO

V

R Ans.



3.17 Link 2 of the linkage illustrated in Fig. P3.17 has an angular velocity of 10 rad/s ccw.

Find the angular velocity of link 6 and the velocities of points B, C, and D.

2

62.5 mm,AO

R 250 mm,BA

R 200 mm,CB

R 100 mm,CA DC

R R

2

200 mm,6

O OR and

6

150 mm.DO

R

2 22 10 rad/s 62.5 mm 625.0 mm/sAO AOV R

B A BA V V V

289.25 mm/s 180B V Ans.

Construct velocity image of link 3:

C A CA B CB V V V V V 605.5 mm/s 207.6C V Ans.

6 6D C DC O DO V V V V V 604.5 mm/s 206.2D V Ans.

6

6

6

604.5 mm/s4.03 rad/s ccw

150 mm

DO

DO

V

R Ans.

3.18 The angular velocity of link 2 of the drag-link mechanism illustrated in Fig. P3.18 is 16

rad/s cw. Plot a polar velocity diagram for the velocity of point B for all crank positions.

Check the positions of maximum and minimum velocities by using Freudenstein’s

theorem. 2

350 mm,AO

R 425 mm,BA

R 2

100 mm,4

O OR and 400 mm.

4BO

R

The graphical construction is shown in the position where

2 135 , where the result is 5760 mm/s 7.2B V .

It is repeated at increments of 2 15 . The maximum

and minimum velocities are

9130 mm/s 146.6B,max V at 2 15 and

4590 mm/s 63.7B,min V at 2 225 , respectively.

Within graphical accuracy these two positions

approximately verify Freudenstein’s theorem.

A numeric solution for the same problem can be found from Eq. (3.22) using Eqs. (2.25)

through (2.33) for position values. The accuracy of the values reported above have been



verified in this way.

3.19 Link 2 of the mechanism illustrated in Fig. P3.19 is driven at 2

36 rad/s cw. Find the

angular velocity of link 3 and the velocity of point B.

2

125 mm,AO

R 200 mm,4

BA BOR R and

2

175 mm.4

O OR



5 070 mm/s 56.3B V Ans.

3


200 mm

BA

BA

V

R Ans.

3.20 Find the velocity of point C and the angular velocity of link 3 of the push-link mechanism

illustrated in Fig. P3.20. Link 2 is the driver and rotates at 8 rad/s ccw.

2

150 mm,AO

R 250 mm,4

BA BOR R

2

75 mm,4

O OR 300 mm,

CAR and 100 mm.

CBR




C A CA B CB V V V V V

3847.5 mm/s 136.8C V Ans.



3


250 mm

BA

BA

V

R Ans.

3.21 Link 2 of the mechanism illustrated in Fig. P3.21 has an angular velocity of 56 rad/s ccw.

Find the velocity of point C.

2150 in,AOR 250 mm,

4BA BO

R R 2

100 mm,4

O OR and 300 mm.

CAR



Construct the velocity image of link3:

C A CA B CB V V V V V 927.5 mm/s 137.8C V Ans.

3.22 Find the velocities of points B, C, and D of the double-slider mechanism illustrated in

Fig. P3.22 if crank 2 rotates at 42 rad/s cw.

2

50 mm,AO

R 250 mm,BA

R 100 mm,CA

R 175 mm,CB

R 200 mm.DC

R

2 22

42 rad/s 50 mm 2100 mm/s

AO AOV R

B A BA V V V 1635 mm/s 180B V Ans.




C A CA B CB V V V V V 1695 mm/s 154.2C V Ans.

530 mm/s 90D C DC V V V Ans.

3.23 Figure P3.23 illustrates the mechanism used in a two-cylinder 60 V engine consisting,

in part, of an articulated connecting rod. Crank 2 rotates at 2000 rev/min cw. Find the

velocities of points B, C, and D.

2

50 mm,AO

R 150 mm,BA CB

R R 50 mm,CA

R 125 mm.DC

R

2

2000 rev/min 2 rad/rev209.4 rad/s

60 s/min

2 22

209.4 rad/s 50 mm 10 470 mm/s

AO AOV R

10 648 mm/s 120B A BA V V V Ans.


12 180 mm/s 93C V Ans.

9 468 mm/s 60D C DC V V V Ans.



3.24 Make a complete velocity analysis of the linkage illustrated in Fig. P3.24 given that

224 rad/s cw. What is the absolute velocity of point B? What is its apparent velocity

to an observer moving with link 4?

2

200 mm,AO

R 2

500 mm.4

O OR


Using the path of P3 on link 4, we write

3 3 4 3 / 4P A P A P P V V V V V

3

3

3

4045 mm/s6.16 rad/s cw

656.75 mm

P A

P A

V

R

From this, or graphically, we complete the velocity image of link 3, from which

33945 mm/s 39.7B V Ans.

Then, since link 4 remains perpendicular to link 3, we have 4 3 and we find the

velocity image of link 4:

3 /4 2582.5 mm/s 12.4B V Ans.

3.25 Find B

V for the linkage illustrated in Fig. P3.25 if 300 mm/s.A

V

300 mm/sAV

Using the path of P3 on link 4, we write

3 3 4 3 / 4P A P A P P V V V V V

Next construct the velocity image of link 3 [or 3 3B P BP A BA V V V V V ]:

312.5 mm/s 23.0B V Ans.



3.26 Figure P3.26 illustrates a variation of the Scotch-yoke mechanism. The mechanism is

driven by crank 2 at 2

36 rad/s ccw. Find the velocity of the crosshead, link 4.

2

250 mm.AO

R


Using the path of A2 on link 4, we write

2 4 2 / 4A A A V V V

(Note that the path is unknown for4 / 2AV !)

44657.5 m/s 180A V Ans.

All other points of link 4 have this same velocity; it is in translation.

3.27 Make a complete velocity analysis of the linkage illustrated in Fig. P3.27 for 2

72

rad/s ccw.

2

37.5 mm,AO DC

R R 262.5 mm,BA

R 2

150 mm,4

O OR 125 mm,

4BO

R 2

175 mm,6

O OR

and 6

200 mm.EO

R

2 22 72 rad/s 37.5 mm 2700 mm/sAO AOV R


Construct velocity image of link 3: 3C A CA B CB V V V V V

31908 mm/s 203.2C V Ans.

Using the path of C3 on link 6, we next write

3 6 3 6 6 6/ 6 and C C C C C O V V V V V , from which 6

1076.75 mm/s 241.4C V Ans.

From this, graphically, we can complete the velocity image of link 6, from which

1934.75 mm/s 98.9E V Ans.

Since link 5 remains perpendicular to link 6, 6 6

6 6

5 6 9.67 rad/s cwC O

C O

V

R Ans.



From these we can get 5 5 5 5

1612.25 mm/s 210.1D C D C V V V Ans.

3.28 The mechanism illustrated in Fig. P3.28 is driven such that VC = 250 mm/s to the right.

Rolling contact is assumed between links 1 and 2, but slip is possible between links 2 and

3. Determine the angular velocity of link 3.

Using the path of C2 on link 3, we write

2 3 2 /3C C C V V V and 3 3 3 3C D C D V V V

3 3

3 3

3

105.65 mm/s1.569 rad/s cw

67.25 mm

C D

C D

V

R Ans.

3.29 The circular cam illustrated in Fig. P3.29 is driven at an angular velocity of 2

15 rad/s

ccw. There is rolling contact between the cam and the roller, link 3. Find the angular

velocity of the oscillating follower, link 4.

2 15 rad/s 31.25 mm 468.75 mm/sBA BAV R

4 2 4 / 2D D D V V V and 4 4D E D E V V V

4

4

4


87.5 mm

D E

D E

V

R Ans.



3.30 The mechanism illustrated in Fig. P3.30 is driven by link 2 at 10 rad/s ccw. There is

rolling contact at point F. Determine the velocity of points E and G and the angular

velocities of links 3, 4, 5, and 6.

2 10 rad/s 25 mm 250 mm/sBA BAV R

C B CB D CD V V V V V

3

333.25 mm/s3.333 rad/s ccw

100 mm

CB

CB

V

R Ans.

4

166.75 mm/s3.333 rad/s ccw

50 mm

CD

CD

V

R Ans.

Construct velocity image of link 3: 3 3 3

251.5 mm/s 220.9E B E B C E C V V V V V Ans.

Using the path of E3 on link 6, we write 3 6 3 / 6E E E V V V and

6 6E H E H V V V

6

6

6

121.45 mm/s3.774 rad/s cw

32.2 mm

E H

E H

V

R Ans.

Construct velocity image of link 6: 6 6

298.25 mm/s 57.1G E GE H GH V V V V V Ans.

5 6F FV V ; 5 3E EV V ; 5

5

5

319.5 mm/s25.56 rad/s cw

12.5 mm

F E

F E

V

R Ans.



3.31 Figure P3.31 is a schematic diagram for a two-piston pump. The pump is driven by a

circular eccentric, link 2, at 2

25 rad/s ccw. Find the velocities of the two pistons,

links 6 and 7.

2 2 2 25 rad/s 25 mm 625 mm/sF F E FEV V R

Using the path of F2 on link 3, we write

2 3 2 /3F F F V V V and 3 3F G F G V V V


3 3C F CF G CG V V V V V and

3 3D F DF G DG V V V V V . Then

32.55 mm/s 180A C AC V V V Ans.

144.42 mm/s 180B D BD V V V Ans.



3.32 The epicyclic gear train illustrated in Fig. P3.32 is driven by the arm, link 2, at 2

10

rad/s cw. Determine the angular velocity of the output shaft, attached to gear 3.

2 10 rad/s 75 mm 750 mm/sB BA BAV V R

Using 0D V construct the velocity image of link 4 from which 1500 mm/s 0C V .

3

1500 mm/s30.00 rad/s cw

50 mm

CA

CA

V

R Ans.



3.33 The diagram in Fig. P3.33 illustrates a planar schematic approximation of an automotive

front suspension. The roll center is the term used by the industry to describe the point

about which the auto body seems to rotate with respect to the ground. The assumption is

made that there is pivoting but no slip between the tires and the road. After making a

sketch, use the concepts of instant centers to find a technique to locate the roll center.

By definition, the “roll center” (of the vehicle body, link 2, with respect to the road, link

1,) is the instant center I12. It can be found by the repeated application of Kennedy’s

theorem as shown.

In the automotive industry it has become common practice to use only half of this

construction, assuming by symmetry that I12 must lie on the vertical centerline of the

vehicle. Notice that this is true only when the right and left suspension arms are

symmetrically positioned. It is not true once the vehicle begins to roll as in a turn.

Having lost sight of the relationship to instant centers and Kennedy’s theorem, and

remembering only the shortened graphical construction on one side of the vehicle, many

in the industry are now confused about the movement of the roll center along the

centerline of the vehicle (called the “jacking coefficient”!). They should be thinking

about the fixed and moving centrodes (Section 3.21), which are more horizontal than

vertical!



3.34 Locate all instant centers for the linkage of Problem 3.22.

Instant centers 12I , 23I ,

34I , 14I (at infinity), 35I ,

56I , and 16I (at infinity)

are found by inspection.

All others are found by

repeated applications of

Kennedy’s theorem except

46I .

One line can be found for

46I ; however, no second

line can be found by

Kennedy’s theorem since

no line can be drawn (in

finite space) between 14I

and 16I . Now it must be

seen that 46I must be

infinitely remote because

the relative motion

between links 4 and 6 is

translation; that is, the

angle between lines on

links 4 and 6 remains

constant.



3.35 Locate all instant centers for the mechanism of Problem 3.25.

Instant centers 12I (at infinity), 23I ,

34I (at infinity), and 14I are found

by inspection. All others are found

by repeated applications of

Kennedy’s theorem.


Instant centers 12I , 23I , 34I (at

infinity), and 14I (at infinity) are

found by inspection. All others

are found by repeated applications

of Kennedy’s theorem except 13I .

One line ( 12 23 I I ) can be found for

13I ; however, no second line can

be found by Kennedy’s theorem

since no line can be drawn (in

finite space) between 14I and 34I .

Now it must be seen that 13I must

be infinitely remote because the

relative motion between links 1

and 3 is translation; the angle

between links 1 and 3 remains

constant.




Instant centers 12I , 23I ,

34I , 14I , 35I , 56I (at

infinity), and 16I are

found by inspection. All

others are found by

repeated applications of

Kennedy’s theorem.


Instant centers 12I and 13I are

found by inspection.

One line for 23I is found by

Kennedy’s theorem. The other is

found by drawing perpendicular

to the relative velocity of slipping

at the point of contact between

links 2 and 3.




Instant centers 12I , 23I , 34I , and 14I are found by inspection. The other two are found by

use of Kennedy’s theorem.



3.40 For the mechanism illustrated in Fig. P3.40, the input link 2 is in the position

4150 mmAOR and is moving to the right at a velocity of 18.75 mm/sAV . Determine

the first-order kinematic coefficients for the mechanism in the given position, and

determine the angular velocities of links 3 and 4.

4150 mmBO BAR R

Let the following vectors be defined: 42

j

AOR e r , 3

3

j

BAR e

r , and 4

44

j

BOR e

r . Then

the loop-closure equation is 2 3 4 r r r 0 . The two scalar position equations are

2 3 3 4 4

3 3 4 4

cos cos 0

sin sin 0

r r r

r r

With the given data, at the position 2 150 mmr , the solution is 3 60 and 4 120 .

Taking the derivative of the position equations with respect to input 2r gives

3 3 3 4 4 4

3 3 3 4 4 4

1 sin sin 0

cos cos 0

r r

r r

or, in matrix format,

3 3 4 4 3

3 3 4 4 4

sin sin 1

cos cos 0

r r

r r

The determinant of the Jacobian is 3 4 4 3sinr r and goes to zero when 3 4 or

when 3 4 180 .

The solutions for the first-order kinematic coefficients are 3

3 4 4cos 3.85 10 rad/mmr and 3

4 3 3cos 3.85 10 rad/mmr Ans.

The input velocity is given as 2 15.0 m/sr .

3 3 2 72.17 rad/s (ccw)r and 4 4 2 72.17 rad/s (cw)r Ans.



3.41 For the mechanism illustrated in Fig. P3.41 pinion 3 is rolling without slipping on rack 4

at point D. Input link 2 is in the position 4

250 mmGOR , and the input velocity is ˆ75 mm/sG V i . Determine the first-order kinematic coefficients of the mechanism.

Find the angular velocities of both the rack 4 and the pinion 3.

Rack and pinion mechanism. 3 125 mmDGR .

Let the following vectors be defined as 4

0

2

j

GOR er , 4

44

j

DOR e

r , and 4

3 3

jj e ρ .

Then the loop-closure equation is 2 3 4 r ρ r 0 . The two scalar position equations are

2 3 4 4 4

3 4 4 4

sin cos 0

cos sin 0

r r

r

At the position 2 250 mmr , the solution is 4 150 and 4 3 4tan 216.5 mmr .


3 4 4 4 4 4 4 4

3 4 4 4 4 4 4 4

1 cos sin cos 0

sin cos sin 0

r r

r r

or, simplifying by use of the position equations and putting into matrix format,

4 4

2 4 4

0 cos 1

sin 0r r

The determinant of the Jacobian is 2 4cosr and goes to zero when 4 90 or 2 0r .

The solutions for the first-order kinematic coefficients are

4 4sin 0.002 309 rad/mm and 4 2 1.154 7 mm/mmr r Ans.

The input velocity is given as 2 1875 mm/sr . From this we can get

4 4 2 0.173 2 rad/s (cw)r Ans.

However, we must notice that vector 3ρ is not attached to link 3. To find 3 we start

with the constraint for rolling with no slip. If we designate rotation of link 3 by the angle

3 then 3 3 4 4r . Dividing this by t and taking the limit, we get the

angular velocity of the pinion, link 3

3 3 4 4 3 0.519 6 rad/s (ccw)r Ans.



3.42 For the mechanism of Example 2.9, see Fig. 2.34, the dimensions are 1 800 mmR ,

9 550 mmR , and 3 500 mm . In the position where 2 750 mmR , the input link 2 has

a velocity of ˆ150 mm/sA V j . Determine the first-order kinematic coefficients for this

mechanism. Find the velocity of rack 4, and the angular velocity of pinion 3.

Using the vectors defined in Example 2.9, the complex algebra loop-closure equation is 34 34

2 9 34 1 4 0j j

jR jR e R e jR R

and the two scalar position equations are

9 34 34 34 4

2 9 34 34 34 1

sin cos 0

cos sin 0

R R R

R R R R

At 2 750 mmR with the given dimensions these give 34 259.8 mmR and 4 606.2 mmR

with the no-slip condition that 34 3 3R .

Taking the derivative of the position equations with respect to input 2R gives

34 34 4

34 34

cos 0

1 sin 0

R R

R

with the condition that 34 3 3R .

From these, the first-order kinematic coefficients are

34 1.154 7 mm/mmR , 3

3 2.3 10 rad/mm , and 4 0.577 35 mm/mmR Ans.

The velocity of the rack is 4 4 2ˆ ˆ86.6 mm/sR R V i i Ans.

The angular velocity of the pinion is 3 3 2 0.3464 rad/s ccw.R Ans.



3.43 For the mechanism illustrated in Fig. P3.43, in the current position 4

250 mmAOR , and

the input velocity is ˆ125 mm/sA V i . Determine the first-order kinematic coefficients

of the mechanism. Find the angular velocity of link 3 and the slipping velocity between

links 3 and 4.

125 mmPAR and

490 .APO

Let the following vectors be defined as 4

0

2

j

AO r eR , 3

3

j

PA r e

R , 3

4 4

j

PO jr e

R . Then

the loop-closure equation is 3 3

2 3 4 0j j

r r e jr e

The two scalar equations are

2 3 3 4 3

3 3 4 3

cos sin 0

sin cos 0

r r r

r r

which, at the input position 2 250 mmr , has the solution 1

3 3 2cos 240r r and

4 3 3tan 216.5 mmr r .


3 3 3 4 3 3 3 4

3 3 3 4 3 3 3 4

1 sin cos sin 0

cos sin cos 0

r r r

r r r


3 3

2 3 4

0 sin 1

cos 0r r

The determinant of the Jacobian is 2 3sinr and goes to zero when 3 0 at 2 125 mmr .

From the solution of these equations, the first-order kinematic coefficients are

3 3cos 0.002 309 rad/mm and 4 2 1.1547 mm/mmr r Ans.

For the given input velocity of 2 125 mm/sr ,

the angular velocity of link 3 is 3 3 3 2 0.289 rad/s (cw)r Ans.

and the slipping velocity is 3/4 4 4 2 144.34 mm/sV r r r . Ans.



3.44 For the mechanism illustrated in Fig. 3.30 there is rolling contact at point F. The input

has an angular velocity of 2 10 rad/s ccw and there is rolling contact between links 5

and 6 at point F. Determine the first-order kinematic coefficients for links 3, 4, 5, and 6.

Find the angular velocities for links 3, 4, 5, and 6 and the velocities of points E and G.

Let the following vectors be defined: 2

2 ,j

BA r e R 3

3 ,j

CB r e

R 0

1 ,j

DA reR

4

4 ,j

CD r e

R 3 3

312.5 37.5 mm,j j

EB r e j e R

25 37.5 mm,HA j R and

6 6

612.5 .j j

EH j e r e R

Then there are two loop-closure equations

BA CB CD DA

BA EB EH HA

R R R R 0

R R R R 0

and four corresponding scalar equations

2 2 3 3 4 4 1

2 2 3 3 4 4

2 2 3 3 3 6 6 6

2 2 3 3 3 6 6 6

cos cos cos 0

sin sin sin 0

cos ½ cos 1.5sin 0.5sin cos 25 0

sin ½ sin 1.5cos 0.5cos sin 37.5 0

r r r r

r r r

r r r

r r r

Numerical solution of these with the dimensions specified at the position 2 180 gives

the current position as 3 28.955 , 4 75.522 , 6 14.478 , 6 29.65 mm.r

Taking derivatives of these four equations with respect to input 2 gives

2 2 3 3 3 4 4 4

2 2 3 3 3 4 4 4

2 2 3 3 3 3 3 6 6 6 6 6 6 6

2 2 3 3 3 3 3 6 6 6 6 6 6 6

sin sin sin 0

cos cos cos 0

sin ½ sin 1.5cos 0.5cos sin cos 0

cos ½ cos 1.5sin 0.5sin cos sin 0

r r r

r r r

r r r r

r r r r

Numerical solution gives the solution as 3 0.33334 rad/rad, 4 0.33334 rad/rad,

6 0.37751 rad/rad, and 6 27.24 mm/rad.r

The no-slip condition gives the displacement constraint 6 5 5 6r from which

we find 6 5 5 6r , which gives 5 6 6 5 2.55663 rad/rad.r



Therefore the first-order kinematic coefficients are

3 0.3333 rad/rad, 4 0.3333 rad/rad,

5 2.5566 rad/rad, 6 0.3775 rad/rad. Ans.

The angular velocities are

3 3 2 3.333 rad/s ccw, 4 4 2 3.333 rad/s ccw, 5 5 2 25.566 rad/s (cw),

and 6 6 2 3.775 rad/s (cw). Ans.

The positions of point E and G are

2 2 3 3 3

2 2 3 3 3

6 6

6 6

cos ½ cos 37.5sin

sin ½ sin 37.5cos

25 25sin 75cos

37.5 25cos 75sin

E

E

G

G

x r r

y r r

x

y

The derivatives of these give the first-order kinematic coefficients

2 2 3 3 3 3 3

2 2 3 3 3 3 3

6 6 6 6

6 6 6 6

sin ½ sin 37.5cos 19 mm/rad

cos ½ cos 37.5sin 16.468 mm/rad

25cos 75sin 16.22 mm/rad

25sin 75cos 25.05 mm/rad

E

E

G

G

x r r

y r r

x

y

And the velocities are

2 2ˆ ˆ+ y 190 164.67 251.47mm / s 139.09

ˆ ˆ2 6.487 10.022 298.45mm / s 57.09

E E E

G G G

x

x y

V î j î j

V î j î j

&

& & Ans.



3.45 For the mechanism illustrated in Fig. P3.45, input link 2 is moving vertically upwards

with a velocity of 187.5 mm/sAV . Pinion 4 has a radius of 25 mm and is rolling without

slipping on rack 3 at point B. The distance from point E to point B is equal to the

distance from point B to pin A. The distance from O4 to A is 50 mm. Determine the first-

order kinematic coefficients for the rack 3 and the pinion 4, and find the angular velocity

of rack 3 and pinion 4 and the velocity of point E. Also find the velocity along rack 3 of

the point of contact between links 3 and 4 (that is, point B).

Let the following vectors be defined: 4 2 ,AO jrR

3

3 ,j

BA r e

R and 34

4 4 4 .jj

BO r e jr e

R

Then the loop-closure equation is 4 4AO BA BO R R R 0

and the scalar equations are

3 3 4 3

2 3 3 4 3

cos sin 0

sin cos 0

r r

r r r

The position solution for the given data at 2 50 mmr is

3 120 , 3 43.3 mmr .

Taking derivatives of these equations with respect to input 2r gives

3 3 3 3 3 4 3 3

3 3 3 3 3 4 3 3

cos sin cos 0

25 sin cos sin 0

r r r

r r r


3 2 3

3 3

cos 0

sin 0 25

r r

The determinant of the Jacobian is 2 3sinr and goes to zero when 3 0 or 180 .

From these, the first-order kinematic coefficients are 3 31 sinr and 3 2 31 tanr

The no-slip condition gives the displacement constraint 3 4 4 3r from

which we find 3 4 4 3r , which gives 4 3 3 4r .

Therefore the first-order kinematic coefficients are

3 15470 mm/mm,r 3 0.0115 rad/mm, 4 0.0346 rad/mm. Ans.

For 2 187.5 mm/sr , the angular velocities of links 3 and 4 are

3 3 2 2.165 rad/s ccwr and 4 4 2 6.495 rad/s (cw)r . Ans.

Given that REA = 2r3 = 86.6 mm, the position of point E is 3

2 86.6j

E E Ex jy jr e R

386.6cos 43.3 mmEx and 2 386.6sin 25 mmEy r

The derivative with respect to input 2r gives

3 386.6sin 0.866 03 mm/mmEx and

3 31 3.464cos 0.500 mm/mmEy

2 162.38 mm/sE Ex x r and 2 93.75 mm/sE Ey y r

The velocity of point E is 187.5 mm/s 150EV Ans.

The velocity along rack 3 of the point of contact between links 3 and 4 is

4 /3 3 3 2 1.154 70 mm/mm 187.5 mm/s 216.5 mm/sBV r r r



4 /3 216.5 mm/s 60BV Ans.

3.46 For the mechanism illustrated in Fig. P3.46, the dimensions are 2

250 mmAOR and

4 500 mmPOR . At the position illustrated, where 4 2 30O O A ,

4PA AOR R , and

PB BAR R , the angular velocity of the input link 2 is ω2 = 5 rad/s cw. Determine the

first-order kinematic coefficients for links 3, 4, and 5. Then find: (i) the angular

velocities of links 3 and 4; (ii) the velocity of link 5; and (iii) the velocity of point P fixed

in link 4.

Using instant centers, the first-order kinematic coefficients for link 3 and link 4 are

23 12

23 13

3

250.00 mm0.500 rad/rad

500.00 mm

I I

I I

R

R

Ans.

24 12

24 14

4

144.35 mm0.500 rad/rad

288.70 mm

I I

I I

R

R

Ans.

For link 5

0,Bx 25 12

216.50 mm/radB B I Ir y R Ans.

From these, with 2 5 rad/s (cw),

(i) 3 3 2 2.50 rad/s ccw, 4 4 2 2.50 rad/s ccw Ans.

(ii) 2ˆ1 082.5 mm/sB Br V j Ans.

(iii) 144 0.500 rad/rad 500.00 mm 250.00 mm/radP PIr R

2 1 250.0 in/s 120P Pr V Ans.



3.47 For the mechanism illustrated in Fig. P3.47, the input link 2 is moving parallel to the X-

axis with a constant velocity 375 mm/sBV to the right. At the instant indicated, the

angle θ4 = 60º. (i) Determine the first-order kinematic coefficients for links 3 and 4, and

find the angular velocities of links 3 and 4. (ii) Determine the conditions for the

determinant of the coefficient matrix of part (i) to be zero; then sketch the mechanism in

the position where the determinant is zero.

4100 mm.BA AOR R

The two scalar loop-closure equations are

4

4

2 3 4

2 3 4

cos cos 0

sin sin 0

BA AO

BA AO

r R R

y R R

The solution at the current geometry, with 2 50 mm,r 2 86.6 mm,y is 4 60 and 3 90 .

(i) Taking the derivative with respect to input 2r gives

4

4

3 3 4 4

3 3 4 4

1 sin sin 0

cos cos 0

BA AO

BA AO

R R

R R

which in matrix format becomes

4

4

3 4 3

43 4

sin sin 1

cos cos 0

BA AO

BA AO

R R

R R

The determinant of the Jacobian matrix is 4

2

4 3sin 5 000 mBA AOR R .

The first-order kinematic coefficients for links 3 and 4 are

43 4cos 0.010 rad/mmAOR and 4 3cos 0BAR Ans.


3 3 2 3.75 rad/s (cw)r and 4 4 2 0r Ans.

(ii) The conditions for which 0 are that 3 4 or 3 4 180 ; e.g., this will

happen when 3 4 68.907 and 2 71.975 mmr as shown below.



3.48 For the mechanism illustrated in Fig. P2.15, the dimensions are 2

50 mmAOR ,

150 mmBAR , and 5

62.5 mmCOR . In the position indicated, the angle 2BAO is 150°

and the distance 80 mmBCR . The input link 2 is vertical and the angular velocity is

2 10 rad/s cw . (i) Show the locations of all instant centers. (ii) Using instant centers,

determine the first-order kinematic coefficients of link 3, rack 4, and pinion 5. (iii)

Determine the angular velocity of link 3, the velocity of rack 4, and the angular velocity

of pinion 5.

(ii) The first-order kinematic coefficients are

23 12

23 13

3

50 mm0.385 rad/rad

130 mm

I I

I I

R

R

Ans.

24 124 29 mm/radI Ir R Ans.

25 12

25 15

5

146 mm0.462 rad/rad

316 mm

I I

I I

R

R Ans.

(iii) The requested velocities are

3 3 2 0.385 rad/rad 10 rad/s 3.85 rad/s (ccw) Ans.

4 4 2 29 mm/rad 10rad/s 290 mm/s 90V r Ans.

4 4 2 0.462 rad/rad 10 rad/s 4.62 rad/s (cw) Ans.



3.49 For the mechanism illustrated in Fig. P2.16, the dimensions are 177 mmBAR and

150 mmBCR . The radius of gear 2 is 2 25 mm and the radius of gear 5 is

5 50 mm . In the position indicated, the angular velocity 2 5 rad/s ccw .

Determine the first-order kinematic coefficients of links 3, 4, and 5. Find the angular

velocities of links 3, 4, and 5.

The two scalar loop closure equations are

5 2 5 5 4 3 2 2

5 5 4 3 2 2

cos cos cos cos 0

sin sin sin sin 0

O O BC BA

BC BA

R R R

R R

with the rolling contact constraint equation 5 5 2 2 .

At the position 2 90 with the given dimensions the solution is 3 45 ,

4 90 , 5 0 .

The derivatives of these equations with respect to input 2 are

5 5 5 4 4 3 3 2 2

5 5 5 4 4 3 3 2 2

sin sin sin sin 0

cos cos cos cos 0

BC BA

BC BA

R R

R R

with the constraint 5 5 2 .

In matrix form, these appear as

3 4 5 5 5 2 2 2 5 23

3 4 5 5 5 2 2 2 5 24

sin sin sin sin sin sin

cos cos cos cos cos cos

BA BC

BA BC

R R

R R

The determinant is 2

3 4sin 18 750 mBA BCR R .

At 2 90 with the given dimensions the first-order kinematic coefficients are

3 2 4 5 4 2sin ( sin 0.200 rad/radBCR Ans.

4 2 3 5 3 2sin ( sin 0BCR Ans.

5 2 5 0.500 rad/rad Ans.



4 4 2 0 Ans.




3.50 For the mechanism illustrated in Fig. P2.17, the radius of wheel 3 is 3 15 mm and the

other dimensions are 2 5

140 mmO OR , 110 mmBAR and 5

52 mmAOR . For the given

position, link 4 is parallel to the X-axis and link 5 is coincident with the Y-axis. Also, the

input link 2 has an angular velocity of 2 15 rad/s cw . Determine the first-order

kinematic coefficients for links 3, 4, and 5. Find the angular velocities of links 3, 4, and

5.

The two scalar loop closure equations are

2 5 2 5

2 5

2 4 5

2 4 5

cos cos cos 0

sin sin sin 0

O O BO BA AO

BO BA AO

R R R R

R R R

with the rolling contact constraint equation 3 3 2 1 1 2 , which,

since 1 0 , reduces to 3 3 1 3 2 .

At the position 2 120 with 2

60 mm,BOR 1 45 mm, and the given dimensions,

the solution is 4 0,

5 90 .

The derivatives of the loop-closure equations with respect to input 2 are

2 5

2 5

2 4 4 5 5

2 4 4 5 5

sin sin sin 0

cos cos cos 0

BO BA AO

BO BA AO

R R R

R R R

and the constraint equation derivative gives 3 3 1 3 .

In matrix form, these appear as

5 2

5 2

4 5 24

54 5 2

sin sin sin

cos cos cos

BA AO BO

BA AO BO

R R R

R R R

The determinant is 5

2

5 4sin 5 720 mBA AOR R .

At 2 120 with the given dimensions the first-order kinematic coefficients are

3 1 3 3 4.000 rad/rad Ans.

2 54 5 2sin 0.273 rad/radBO AOR R Ans.

25 2 4sin 1.000 rad/radBA BOR R Ans.



4 4 2 0.273 rad/rad 15 rad/s 4.09 rad/s (ccw) Ans.




Chapter 4

Acceleration

4.1 The position vector of a point is defined by the equation 3 ˆ ˆ4 3 10t t R i j where R is

in meters and t is in seconds. Find the acceleration of the point at 3 st .

3 ˆ ˆ4 3 10t t t R i j

2 ˆ4t t R i

ˆ2t t R i 2ˆ ˆ3 s 2 3 6 m/s R i i Ans.

4.2 Find the acceleration at 2t s of a point that moves according to the equation

2 3 3ˆ ˆ= 6 3t t t R i j . The units are mm and seconds.

2 3 3ˆ ˆ= 6 3t t t t R i j

2 2ˆ ˆ= 2 2t t t t R i j

ˆ ˆ= 2 2t t t R i j 2ˆ ˆ ˆ2 s = 2 2 2 2 4 mm/s R i j j Ans.



4.3 The path of a point is described by the equation 2 /10= ( 4) j tt e R where R is in

metre and t is in seconds. For t = 15 s, find the unit tangent vector for the path, the

normal and tangential components of the point’s absolute acceleration, and the radius of

curvature of the path.

2 /10= ( 4) j tt t e R

/10 2 /10= 2 ( 4)10

j t j tjt te t e R

2

/10 /10 2 /10= 2 ( 4)5 5 100

j t j t j tj t j tt e e t e

R

Noticing, at t = 15 s, that /10 1.5j t je e j , we find that

215 s = (15 4) 229 m R

220 s = 2 15 (15 4) 71.94 30.00 77.95 m/s 22.610

jj j j R

2

2 215 1520 s = 2 (15 4) 18.850 20.601 m/s

5 5 100

j jj j j j

R

From the direction of the velocity we find the unit tangent and unit normal vectors

ˆ ˆ ˆ ˆˆ 1.0 22.6 cos 22.6 sin 22.6 0.92296 0.38489t u i j i j Ans.

ˆ ˆ ˆ ˆˆˆ ˆ sin 22.6 cos 22.6 0.38489 0.92296n t u k u i j i j

From these, the components of the point’s absolute acceleration are

2 2ˆ ˆ ˆ ˆˆ 0.38489 0.92296 18.850 20.601 ft/s 26.269 m/sn nA u R i j i j Ans.

2 2ˆ ˆ ˆ ˆˆ 0.92296 0.38489 18.850 20.601 ft/s 9.468 m/st tA u R i j i j Ans.

Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is

2 2

2

77.95 m/s231.3 m

26.269 m/snA

R Ans.

where the negative sign indicates that the center of curvature is in the negative ˆ nu

direction from the point.

4.4 The motion of a point is described by the equations 34 cos x t t and 3 6 sin 2y t t

where x and y are in meters and t is in seconds. Find the acceleration of the point at

1.26 st .

3 3ˆ ˆ4 cos 6 sin 2t t t t t R i j

3 3 3 2 3ˆ ˆ4 cos 12 sin 2 sin 2 3 cos2t t t t t t + t t R i j

2 3 2 5 3 2 2 2ˆ ˆ48 sin 36 cos 1 2 3 sin 2 cos2t t t t t t t t +2 t t R i j

2ˆ ˆ1.26 s 1 128.379 10.626 1 128.429 m/s 180.54 R i j Ans.



4.5 Link 2 in Fig. P4.5 has an angular velocity of 2 120 rad/s ccw and an angular

acceleration of 4 800 rad/s2 ccw at the instant indicated. Determine the absolute

acceleration of point A.

2500 mmAOR

2 2 2

2 2

2

2 2

2 2

ˆ

ˆ ˆˆ120 rad/s 500 mm mm 4 800 rad/s 500 mm

n t

A O AO AO

AO AO

A A A A

R k R

i k i

2 2ˆ ˆ7200 m 2400 m m/s 7589.5 m/s 161.6A A i j Ans.

4.6 Link 2 is rotating clockwise as illustrated in Fig. P4.6. Find its angular velocity and

acceleration and the acceleration of its midpoint C.

500 mm BAR

n t

B A BA BA A A A A

Construct the acceleration polygon. 2

2

178.4 m/s18.9 rad/s cw

0.5 m

n

BA

BA

A

R Ans.

Note the ambiguous sign of this square root. The sense of cannot be determined from

the accelerations, but here is found from the problem statement. 2

2

2

51 m/s102 rad/s cw

0.5 m

t

BA

BA

A

R Ans.

292.76m/s 44.0C A Ans.



4.7 For the data given in Fig. P4.7, find the velocity and acceleration of points B and C.

400 mm, 250 mm, 200 mmBA CA CBR R R

B A BA V V V

2 24 rad/s 400 mm 9600 mm/sBA BAV R

Construct the velocity polygon.

3600 mm/s 270B V Ans.

2510 mm/s 12.1C V Ans.

n t

B A BA BA A A A A

22 2

2 24 rad/s 400 mm 230.4m/sn

BA BAA R

2 2

2 160 rad/s 400 mm 63.99 m/st

BA BAA R

Construct the acceleration polygon. 2118.53 m/s 165B A Ans.

263.06 m/s 240.3C A Ans.



4.8 For the straight-line mechanism illustrated in Fig. P4.8, 2 20 rad/s cw and

2

2 140 rad/s cw . Determine the velocity and acceleration of point B and the angular

acceleration of link 3.

2100 mmAO CA BAR R R

2 2A O AO V V V


C A CA V V V


3864mm/s 270B V Ans.

2 2 2

n t

A O AO AO A A A A

2 2

22 2

2 20 rad/s 100 mm 40,000 mm/sn

AO AOA R

2 2

2 2

2 140 rad/s 100 m 14,000 mm/st

AO AOA R

n t

C A CA CA A A A A

22 22000 mm/s 100 mm 40,000.0 mm/sn

CA CA CAA V R

Construct the acceleration image of link 3. 26340mm/s 90B A Ans.

22

3

14000 mm/s140 rad/s cw

100 mm

t

BA

BA

A

R Ans.



4.9 In Fig. P4.8, the slider 4 is moving to the left with a constant velocity of 200 mm/s. Find

the angular velocity and angular acceleration of link 2.

2 2A C AC O AO V V V V V .


2

2

2


100 mm

AO

AO

V

R Ans.

2 2 2AO AO

n t n t

A C AC AC O A A A A A A A

22 2/ 386.4 mm/s / 100 mm 1493 mm/sn

AC AC ACA V R

2 2 2

22 2/ 386.4 mm/s / 100 mm 1493 mm/sn

AO AO AOA V R

2

2

22

2

5572 mm/s55.72 rad/s cw

100 mm

t

AO

AO

A

R Ans.

4.10 Solve Problem 3.8 using constant input velocity, for the acceleration of point A and the

angular acceleration of link 3.

n t

A B AB AB A A A A

22

2366 mm/s

1339.5 mm/s100 mm

n ABAB

AB

VA

R

275.78 in/s 15A A Ans.

22

3

1339.5mm/s13.39 rad/s cw

100 mm

t

AB

AB

A

R Ans.



4.11 For Problem 3.9, using constant input velocity, find the angular accelerations of links 3

and 4.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 45 rad/s 100 mm 202 500 mm/sn

AO AOA R

4 4 4BO BO

n t n t

B A BA BA O A A A A A A A

22 2/ 358.5 mm/s / 250.0 mm 514.09 mm/sn

BA BA BAA V R (Ignore compared to other

parts.)

4 4 4

22 2/ 4 619 mm/s / 300 mm 71 117 mm/sn

BO BO BOA V R

22

3

140 818 mm/s563.3 rad/s ccw

250 mm

t

BA

BA

A

R Ans.

4

4

22

4

37 103 mm/s123.7 rad/s ccw

300 mm

t

BO

BO

A

R Ans.



4.12 For Problem 3.10, using constant input velocity, find the acceleration of point C and the

angular accelerations of links 3 and 4.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 60 rad/s 150 mm 21 600 in/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 13,020 mm/s / 300 mm 540,000 mm/sn

BA BA BAA V R

4 4 4

22 2/ 11,360 mm/s / 300 mm 430,75 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3. 2209,615 mm/s 10.6C A Ans.

22

3

136,500 mm/s455.0 rad/s ccw

300 mm

t

BA

BA

A

R Ans.

4

4

22

4

46,050 mm/s153.5 rad/s cw

300 mm

t

BO

BO

A

R Ans.





2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 48 rad/s 200 mm 460 800 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 268 mm/s / 800 mm 89.78 mm/sn

BA BA BAA V R (Ignore compared to other parts.)

4 4 4

22 2/ 9 365 mm/s / 400 mm 219 258 mm/sn

BO BO BOA V R

Construct the acceleration image of link 4. 2931 350 mm/s 114.4C A Ans.

22

3

1 393 250 mm/s1 741.6 rad/s ccw

800 mm

t

BA

BA

A

R Ans.

4

4

22

4

1 222 300 mm/s3 055.8 rad/s ccw

400 mm

t

BO

BO

A

R Ans.



4.14 Using the data of Problem 3.13 and assuming constant input velocity, solve for the

accelerations of points C and D and the angular acceleration of link 4.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 1 rad/s 300 mm 300 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 263.5 mm/s / 150 mm 462.87 mm/sn

BA BA BAA V R

4 4 4

22 2/ 227 mm/s / 300 mm 171.77 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3. 2515.4 mm/s 119.8C A Ans. 2492.5 mm/s 21.8D A Ans.

4

4

22

4

221.2 mm/s0.737 3 rad/s cw

300 mm

t

BO

BO

A

R Ans.





2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 60.0 rad/s 150 mm 540 000 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 4 235 mm/s / 150 mm 119 570 mm/sn

BA BA BAA V R

4 4 4

22 2/ 7 940 mm/s / 250.0 mm 252 170 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3.

2781 250 mm/s 68.9C A Ans.

4

4

22

4

373 750 mm/s1 495 rad/s ccw

250.0 mm

t

BO

BO

A

R Ans.



4.16 Solve Problem 3.16, using constant input velocity, for the acceleration of point C and the


2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 30.0 rad/s 75 mm 67 500 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 2 250.0 mm/s / 125 mm 40 500 mm/sn

BA BA BAA V R

4 4 4

22 2/ 0.0 mm/s / 125.0 mm 0.0 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3. 2148 500 mm/s 216.9C A Ans.

4

4

22

4

108 000 mm/s720 rad/s ccw

150 mm

t

BO

BO

A

R Ans.



4.17 For Problem 3.17 using constant input velocity, find the acceleration of point B and the


2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 10.0 rad/s 62.5 mm 6250 mm/sn

AO AOA R

n t

B A BA BA A A A A

22 2/ 467.5 mm/s / 250 mm 874.75 mm/sn

BA BA BAA V R

25022.5 mm/s 0B A Ans.

22

3

4372.5 mm/s17.49 rad/s ccw

250 mm

t

BA

BA

A

R Ans.

Construct the acceleration image of link 3, or CB CB

n t n t

C A CA CA B A A A A A A A

6 6 6DO DO

n t n t

D C DC DC O A A A A A A A

22 2/ 15 mm/s / 100 mm 2.25 mm/sn

DC DC DCA V R (Ignore compared to other

parts.)

6 6 6

22 2/ 604.5 mm/s / 150 mm 2436.14 mm/sn

DO DO DOA V R

6

6

22

6

1621.9 mm/s10.81 rad/s cw

150 mm

t

DO

DO

A

R Ans.



4.18 For the data of Problem 3.18, what angular acceleration must be given to link 2 for the

position indicated to make the angular acceleration of link 4 zero?

4 4 4BO BO

n t

B O A A A A

4 4 4

22 2/ 5760 mm/s / 400 mm 3 317.8 in/sn

BO BO BOA V R

2 2 2AO AO

n t n t

A B AB AB O A A A A A A A

22 2/ 5005 mm/s / 425 mm 58940 mm/sn

AB AB ABA V R

2 2 2

22 2/ 5600 mm/s / 350 mm 89600 mm/sn

AO AO AOA V R

2

2

22

2


350 mm

t

AO

AO

A

R Ans.

4.19 For the data of Problem 3.19, what angular acceleration must be given to link 2 for the

angular acceleration of link 4 to be 100 rad/ 2s cw at the instant indicated?

4 4 4BO BO

n t

B O A A A A

4 4 4

22 2/ 5 070 mm/s / 200 mm 128 525 mm/sn

BO BO BOA V R

4 4

2 2

4 100 rad/s 200 in 20 000 mm/st

BO BOA R

2 2 2AO AO

n t n t

A B AB AB O A A A A A A A

22 2/ 647 mm/s / 200 mm 2 093 mm/sn

AB AB ABA V R (Ignore compared to other parts.)

2 2 2

22 2/ 4 500 mm/s / 125 mm 162 000 mm/sn

AO AO AOA V R

2

2

22

2

522 575 mm/s4 180.6 rad/s ccw

125 mm

t

AO

AO

A

R Ans.



4.20 Solve Problem 3.20 using constant input velocity for the acceleration of point C and the


2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 8.0 rad/s 150 mm 9600 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 3785 mm/s / 250 mm 57305 mm/sn

BA BA BAA V R

4 4 4

22 2/ 3483.75 mm/s / 250 mm 48545 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3. 263410 mm/s 22.9C A Ans.

22

3

15,647.5 mm/s62.59 rad/s cw

250 mm

t

BA

BA

A

R Ans.



2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 56.0 rad/s 150 mm 470400 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 6965 mm/s / 250 mm 194045 mm/sn

BA BA BAA V R

4 4 4

22 2/ 11380 mm/s / 250 mm 518017.5 mm/sn

BO BO BOA V R

Construct the acceleration image of link 3. 2450600 mm/s 255.6C A Ans.

22

3

18520 mm/s74.08 rad/s cw

250 mm

t

BA

BA

A

R Ans.



4.22 Find the accelerations of points B and D of Problem 3.22 using constant input velocity.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 42.0 rad/s 50 mm 88200 mm/sn

AO AOA R

22 2/ 1066 mm/s / 250 mm 4547.5 mm/sn

BA BA BAA V R

22117 in/s 0B A Ans.


n t n t


n t

D C DC DC A A A A

22 2/ 1541.5 mm/s / 200 mm 11880 mm/sn

DC DC DCA V R

249400 mm/s 90D A Ans.

n t

B A BA BA A A A A



4.23 Find the accelerations of points B and D of Problem 3.23 using constant input velocity.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 209.4 rad/s 50 mm 2 192 420 mm/sn

AO AOA R

22 2/ 5469 mm/s / 150 mm 199 400 mm/sn

BA BA BAA V R

2732 180 mm/s 240B A Ans.


n t n t


n t

D C DC DC A A A A

22 2/ 6 725 mm/s / 125 mm 361 805 mm/sn

DC DC DCA V R

21 209 300 mm/s 120D A Ans.

4.24 to 4.30 The nomenclature for this group of problems is illustrated in Fig. P4.24, and the

dimensions and data are given in Table P4.24 to P4.30. For each problem,

determine 3 4 3 4 3 4, , , , , and . The angular velocity 2 is constant for each

problem, and a negative sign is used to indicate the clockwise direction. The dimensions

of even-numbered problems are given in inches and odd-numbered problems are given in

millimeters.

This group of problems was solved on a programmable calculator. The position solution

values were found from Eqs. (2.25) through (2.32). The velocity values were found from

Eqs. (3.22). The acceleration values were found from Eqs. (4.31) and (4.32).

Prob. 3, deg 4 , deg 3, rad/s 4 , rad/s 2

3, rad/s 2

4 , rad/s

4.24 105.29 159.60 0.809 3 0.525 0 0.230 28 0.008 09

4.25 171.01 195.54 70.452 7 47.566 8 3 196.657 49 3 330.841 37

4.26 45.57 91.15 -4.000 0 -4.000 0 -1.120 22 54.890 98

4.27 28.32 55.88 -0.632 6 -2.155 7 7.822 40 6.704 18

4.28 24.17 63.73 -1.516 7 1.712 9 41.414 93 74.975 93

4.29 38.42 155.60 -6.855 2 -1.234 5 62.500 44 -96.514 21

4.30 73.16 138.51 -0.505 1 7.275 3 -206.384 28 -94.122 01

n t

B A BA BA A A A A



4.31 Crank 2 of the system illustrated in Fig. P4.31 has a constant speed of 60 rev/min ccw.

Find the velocity and acceleration of point B and the angular velocity and acceleration of

link 4.

4 2 2 4300 mm, 175 mm, 700 mmO O AO BOR R R

2

rev rad 1 min60 2 6.283 rad/s

min rev 60 s

2 2 2 2 4 2 / 4A O A O A A V V V V V

2 2 2 22 6.283 rad/s 175 mm 1 099.5 mm/sA O A OV R

Construct the velocity polygon.

4 4

4 4

4

910.75 mm/s6.571 rad/s cw

138.6 mm

A O

A O

V

R Ans.

4 4B O BO V V V

4 44 6.571 rad/s 700 mm 4 600 mm/sBO BOV R

4 600 mm/s 19.1B V Ans.

Since we know the path of A2 on link 4, we write

2 2 2 2 2 2 4 2 4 2 2/ 4 / 4

n t c n t

A O A O A O A A A A A A A Α Α A A A A ; 4 4 4 4 4 4

n t

A O A O A O A A Α Α

2 2 2 2

22 2

2 6.283 rad/s 175 mm 6 908.3 mm/sn

A O A OA R

2 4 2

2

/42 2 6.571 rad/s 616 mm/s 8 095.5 mm/s 19.1c

A A A 4A V×

2

2

2

22

/4

/4

/4

616 mm/s0

An

A

A

VA

Construct the acceleration polygon.

4 4

4 4

22

4

11 970 mm/s86.36 rad/s ccw

138.6 mm

t

A O

A O

A

R Ans.

Construct the acceleration image of link 4. 267 568 mm/s 187.5B A Ans.



4.32 Determine the acceleration of link 4 of Problem 3.26 assuming constant input velocity.

Since we know the path of A2 on link 4, we write

2 2 2 2 2 2 4 2 4 2 2/ 4 / 4

n t c n t

A O A O A O A A A A A A A Α Α A A A A

2 2 2 2

22 2

2 36.0 rad/s 250 mm 324000 mm/sn

A O A OA R

4 2 24 /42 2 0.0 rad/s 6587.5 mm/s 0c

A A A A V × ;

2

4

2

22

/4

/2

/4

6587.5 mm/s0

An

A

A

VA

Next we construct the acceleration polygon.

4

22905000 mm/s 180.0A A Ans.



4.33 For Problem 3.27 using constant input velocity, find the acceleration of point E.

2 2 2AO AO

n t

A O A A A A

2 2

22 2

2 72.0 rad/s 37.5 mm 194400 mm/sn

AO AOA R

4 4 4BO BO

n t n t


22 2/ 3780 mm/s / 262.5 mm 54432.5 mm/sn

BA BA BAA V R

4 4 4

22 2/ 1800 mm/s / 125 mm 25920 mm/sn

BO BO BOA V R


Since we know the path of C3 on link 6, we write

3 6 3 6 3 3/ 6 /6

c n t

C C C C C C A A A A A and 6 6 6 6 6 6

n t

C O C O C O A A Α Α

3 6 3

2

6 /62 2 9.673 rad/s 1253.75 mm/s 24254 mm/sc

C C C A V ×

3

3

3

22

/6

/6

/6

1253.75 mm/s0

Cn

C

C

VA

6 6

6 6

6 6

22

21046.75 mm/s

10415 mm/s111.325 mm

C On

C O

C O

VA

R

Construct the acceleration image of link 6. 2180822.5 mm/s 252.8E A Ans.



4.34 Find the acceleration of point B and the angular acceleration of link 4 of Problem 3.24

using constant input velocity.

2 2

22 2

2 24.0 rad/s 200 mm 115200 mm/sn

AO AOA R

Since we know the path of P3 on link 4, we write

3 3 3 3 3 3 4 3 4 3 3/ 4 / 4

n t c n t

P A P A P A P P P P P A A Α Α A A A A

3 3 3 3 3 3

22 2/ 4045 mm/s / 656.75 mm 24915 mm/sn

P A P A P AA V R

3 4 3

2

/42 2 6.159 rad/s 2592.5 mm/s 31937.5 in/s 102.4c

P P P 4A V ×

3

3

3

22

/4

/4

/4

2592.5 mm/s0

Pn

P

P

VA

Construct the acceleration image of link 3. 2123765 mm/s 25.7B A Ans.

Since links 3 and 4 remain perpendicular,

3 3

3 3

22

4 3


656.75 mm

t

P A

P A

A

R Ans.



4.35 For Problem 3.25, using constant input velocity, find the acceleration of point B and the


Since we know the path of P3 on link 4, we write

3 3 3 3 3 3 4 3 4 3 3/ 4 / 4

n t c n t

P A P A P A P P P P P A A Α Α A A A A

3 3 3 3 3 3

22 2/ 68.375 mm/s / 225 mm 20.775 mm/sn

P A P A P AA V R

3 4 3

2

/42 2 0.3039 rad/s 307.75 mm/s 187 mm/s 102.8c

P P P 4A V ×

3

3

3

22

/4

/4

/4

307.75 mm/s0

Pn

P

P

VA

Construct the acceleration image of link 3. 2505.75 mm/s 102.1B A Ans.

3 3

3 3

22

3

280.5 mm/s1.247 rad/s cw

225 mm

t

P A

P A

A

R Ans.



4.36 Solve Problem 3.31 for the accelerations of points A and B assuming constant input

velocity.

Since we know the path of F2 on link 3, we write

2 2 2 2 2 2 3 2 3 2 2/3 /3

n t c n t

F E F E F E F F F F F A A Α Α A A A A and 3 3 3 3 3 3

n t

F G F G F G A A A A

2 2 2 2

22 2

2 25.0 rad/s 25 mm 15625 mm/sn

F E F EA R

3 3

3 3

3 3

22

2102.75 mm/s

69.45 mm/s152 mm

F Gn

F G

F G

VA

R

2 3 2

2

3 /32 2 0.676 rad/s 616.5 mm/s 833.5 mm/s 80.5c

F F F A V ×

2

2

2

22

/3

/3

/3

616.5 mm/s0

Fn

F

F

VA

Construct the acceleration image of link 3. n t

A C AC AC A A A A and n t

B D BD BD A A A A

22 2/ 195.175 mm/s / 150 mm 254 mm/sn

AC AC ACA V R

2169.2 in/s 0A A Ans.

22 2/ 190.55 mm/s / 150 mm 242 mm/sn

BD BD BDA V R

220152.5 mm/s 0B A Ans.



4.37 For Problem 3.32, determine the acceleration of point 4C and the angular acceleration of

link 3 if crank 2 is given an angular acceleration of 2 rad/ 2s ccw.

n t

B A BA BA A A A A

22 2

2 10.0 rad/s 75 mm 7500 mm/sn

BA BAA R

2 2

2 2.0 rad/s 75 mm 150 mm/st

BA BAA R

4 4 /1

r n t

D D B DB DB A A A A A

22 2/ 750 mm/s / 25 mm 22500 mm/sn

DB DB DBA V R

Draw the acceleration image of link 4.

4

215002.5 mm/s 91.1C A Ans.

3 4 3 / 4

r n t

C C C A CA CA A A A A A A

22 2/ 1500 mm/s / 50 mm 45000 mm/sn

CA CA CAA V R

Draw the acceleration image of link 3.

3 3

3 3

22

3


50 m

t

C A

C A

A

R Ans.



4.38 Determine the angular accelerations of links 3 and 4 of Problem 3.29 assuming constant

input velocity.

Since we know the path of D4 on link 2, we write

2 2 2 2 2 2

n t

D A D A D A A A A A

2 2 2 2

22 2

2 15.0 rad/s 31.25 mm 7031.25 mm/sn

D A D AA R

4 2 4 2 4 4 4 4 4 4 4/ 2 / 2

c n t n t

D D D D D D E D E D E A A A A A A A A

4 2 4

2

2 /22 2 15.0 rad/s 1235.25 mm/s 37057.5 mm/s 80.8c

D D D A V ×

4 4 4

22 2

/2 /2 /2 1235.25 mm/s 62.5 mm 24414.5 mm/sn

D D DA V

4 4 4 4 4 4

22 2381 mm/s 87.5 mm 1659.75 mm/sn

D E D E D EA V R

Draw the acceleration images of links 2 and 4.

4 4

4 4

22

4


87.5 mm

t

D E

D E

A

R Ans.

3 2 3 3 3 3 3 3/ 2

r n t

C C C D C D C D A A A A A A

3 3 3 3 3 3

22 21047.75 mm/s 12.5 mm 87825 mm/sn

C D C D C DA V R

3 3

3 3

22

3

5792.5 mm/s463.4 rad/s ccw

12.5 mm

t

C D

C D

A

R Ans.



4.39 For Problem 3.30 using constant input velocity, determine the acceleration of point G and

the angular accelerations of links 5 and 6.

BA BA

n t

B A A A A A

22 2

2 10 rad/s 25 mm 2500 mm/sn

BA BAA R

n t n t

C B CB CB D CD CD A A A A A A A

22 2/ 333.25 mm/s / 100 mm 1111 mm/sn

CB CB CBA V R

22 2/ 166.5 mm/s / 50 mm 555.5 mm/sn

CD CD CDA V R


Since we know the path of E3 on link 6, we write

3 6 3 6 3 3/6 /6

c n t

E E E E E E A A A A A and 6 6 6 6 6 6

n t

E H E H E H A A Α Α

3 6 3

2

6 /62 2 3.774 rad/s 272.25 mm/s 2055.75 mm/s 104.5c

E E E A V ×

3 3 3

22

/6 /6 /6/ 292.75 mm/s / 0n

E E EA V

6 6 6 6 6 6

22 2121.5 mm/s 32.25 mm 458.25 mm/sn

E H E H E HA V R

Construct the acceleration image of link 6. 28785 mm/s 64.5G A Ans.

6 6

6 6

22

6

3547.5 mm/s110.2 rad/s cw

32.25 in

t

E H

E H

A

R Ans.

5 6 5 5 5 5 5 5/ 6

r n t

F F F E F E F E A A A A A A

5 5 5 5 5 5

22 2319.5 mm/s 12.5 mm 8167.5 mm/sn

F E F E F EA V R

5 5

5 5

2

5

0 mm/s0

12.5 mm

t

F E

F E

A

R Ans



4.40 Continue with Problem 3.40 and find the second-order kinematic coefficients of links 3

and 4. Assuming an input acceleration of 2

218750 mm/sAA find the angular

accelerations of links 3 and 4.

4150 mmBO BAR R

From the solution of Problem 3.40 we have the derivative of the loop-closure equations

with respect to the input r2. In matrix form this is

3 3 4 4 3

3 3 4 4 4

sin sin 1

cos cos 0

r r

r r

From these we found the determinant of the Jacobian and the first-order kinematic

derivatives. At the position shown these are 2

3 4 4 3sin 19485.625 mmr r ,

3

3 4 4cos 3.849 10 rad/mmr , and 3

4 3 3cos 3.849 10 rad/mmr .

The next derivative of the above equations with respect to input r2, in matrix form, gives

2 23 3 4 4 3 3 3 3 4 4 4

2 23 3 4 4 4 3 3 3 4 4 4

sin sin cos cos

cos cos sin sin

r r r r

r r r r

The solution to this set of equations is

24 4 4 4 3 3 4 43 3

23 3 3 3 3 3 4 44 4

2 2

3 4 3 4 4 3

2 2

3 3 4 3 4 4

cos sin cos cos1

cos sin sin sin

cos1

cos

r r r r

r r r r

r r r

r r r

For the specified position these give values of 6 2

3 8.5536 10 rad/mm Ans.

and 6 2

4 8.5536 10 rad/mm . Ans.

From these we find angular accelerations of

2 2

3 3 2 3 2 2 934.9 rad/s (cw)r r Ans.

and 2 2

4 4 2 4 2 2 934.9 rad/s ccwr r Ans.



4.41 Continue with Problem 3.49 and find the second-order kinematic coefficients of links 3,

4, and 5. Assuming constant angular velocity for link 2 find the angular accelerations of

links 3, 4, and 5.

From the solution of Problem 3.49 we have the

derivative of the loop-closure equations with respect to

the input 2 . In matrix form this is

3 4 2 5 23

3 4 2 5 24

sin sin sin sin

cos cos cos cos

BA BC

BA BC

R R

R R

with the constraint 5 5 2 and the determinant

3 4sinBA BCR R .

At the position shown these give values of 218 750 m

3 2 4 5 4 2sin ( sin 0.200 rad/radBCR

4 2 3 5 3 2sin ( sin 0BCR

5 2 5 0.500 rad/rad

The next derivative of these equations gives

23 4 3 43 3

23 4 3 44 4

2 5 2 2 5

2 5 2 2

sin sin cos cos

cos cos sin sin

cos cos

sin sin 1

BA BC BA BC

BA BC BA BC

R R R R

R R R R

with the derivative of the constraint giving 5 0 . The solution of the above equations

gives

2 23 3 4 3

2 24 3 4 4

4 5 2 4 52

3 5 2 3

cos1

cos

cos cos

cos cos 1

BA BC BC

BA BA BC

BC BC

BA BA

R R R

R R R

R R

R R

At the position shown the values of the second-order kinematic derivatives are 2

3 0.240 rad/rad , 2

4 0.150 rad/rad , and 5 0 . Ans.

With 2 5 rad/s const , the requested angular accelerations are 2 2

3 3 2 6.0 rad/s ccw , 2 2

4 4 2 3.75 rad/s ccw , 2

5 5 2 0 . Ans.



4.42 Continue with Problem 3.50 and find the second-order kinematic coefficients of links 3,

4, and 5. Assuming constant angular velocity for link 2 find the angular accelerations of

links 3, 4, and 5.

From the solution of Problem 3.50 we have the derivative of the loop-closure equations

with respect to the input 2 . In matrix form this is

5 2

5 2

4 5 24

54 5 2

sin sin sin

cos cos cos

BA AO BO

BA AO BO

R R R

R R R

with the constraint 3 3 1 3 and the determinant 5 5 4sinBA AOR R .

At the position shown these give values of 25 720 m

3 1 3 3 4.000 rad/rad

2 54 5 2sin 0.273 rad/radBO AOR R

25 2 4sin ( 1.000 rad/radBA BOR R

The next derivative of these equations gives

5 5 2

5 5 2

24 5 4 5 24 4

254 5 4 5 25

sin sin cos cos cos

cos cos sin sin sin

BA AO BA AO BO

BA AO BA AO BO

R R R R R

R R R R R

with the derivative of the constraint giving 3 0 . The solution of the above equations

gives

5 25 5

25

2 22 54 54 4

225 2 454 5

coscos1 1

coscos

AO BOBA AO AO

BA BOBA BA AO

R RR R R

R RR R R

At the position shown the values of the second-order kinematic derivatives are

3 0 , 2

4 0.000 35 rad/rad , and 2

5 0.420 rad/rad . Ans.

With 2 15 rad/s const , the requested angular accelerations are 2

3 3 2 0 , 2 2

4 4 2 0.078 8 rad/s ccw , 2 2

5 5 2 94.41 rad/s (cw) . Ans.



4.43 Find the inflection circle for motion of the coupler of the double-slider mechanism

illustrated in Fig. P4.43. Select several points on the centrode normal and find their

conjugate points. Plot portions of the paths of these points to demonstrate for yourself

that the conjugates are indeed the centers of curvature.

125 mmBAR



4.44 Find the inflection circle for motion of the coupler relative to the frame of the linkage

illustrated in Fig. P4.44. Find the center of curvature of the coupler curve of point C and

generate a portion of the path of C to verify your findings.

2 4 462.5 mm, 22.5 mm, 87.5 mm, 29.25 mmcA AO BO POR R R R

Since point C is on the inflection circle, its center of curvature is at infinity and its point

path is a straight line in the vicinity of the position shown. Ans.



4.45 For the motion of the coupler relative to the frame, find the inflection circle, the centrode

normal, the centrode tangent, and the centers of curvature of points C and D of the

linkage of Problem 3.13. Choose points on the coupler coincident with the instant center

and inflection pole and plot nearby portions of their paths.



4.46 The planar four-bar linkage illustrated in Fig. P4.46 has link dimensions 4 2

50 mmO OR ,

220 mmAOR , 63 mmBAR , and

430 mmBOR . For the position indicated, link 2 is

30 counterclockwise from the ground link 2 4O O and the angular velocity and angular

acceleration of the coupler link AB are 3 5 rad/s ccw and 2

3 20 rad/s cw ,

respectively. For the instantaneous motion of the coupler link AB show: (a) the velocity

pole I, the pole tangent T, and the pole normal N; (b) the inflection circle and the Bresse

circle; (c) the acceleration center of the coupler link AB. Then determine; (d) the radius

of curvature of the path of coupler point C where 25 mmCBR , (e) the magnitude and

direction of the velocity of coupler point C, (f) the magnitude and direction of the angular

velocity of link 2, (g) the magnitude and direction of the velocity of the pole, (h) the

magnitude and direction of the acceleration of coupler point C, and (i) the magnitude and

direction of the acceleration of the velocity pole.

(a) The pole I is coincident with the instant center 13I shown in the figure below. The

instant center 24I and the collineation axis are as shown in the figure. From Bobillier's

theorem, the angle from the collineation axis to the first ray (say link 2) is measured as

84° cw. This is equal to the angle from the second ray (link 4) to the pole tangent T; that

is, 84° cw. Therefore, the pole tangent T is as shown in the figure and the pole normal N,

which is perpendicular to the pole tangent T, is also shown.

(b) The inflection point AJ for point A on link 3 can be obtained from the Euler-Savary

equation; that is,

22 13.8 mm9.5 mm

20 mmA

A

AIAJ

AO

RR

R

The location of the inflection point AJ is shown in the figure.

Similarly, the inflection point BJ for point B on link 3 can be obtained from the Euler-

Savary equation; that is,

22 56.2 mm105.3 mm

30 mmB

B

BIBJ

BO

RR

R

The location of the inflection point BJ is shown on the figure.

Knowing the pole normal and the two inflection points, the inflection circle can be

drawn. The inflection circle for the motion of link 3 with respect to 1, the inflection pole

J, and the center of the inflection circle (denoted as point O) are shown on the figure.

Note that the pole normal N points from the pole I toward the inflection pole J and the

pole tangent T is 90 clockwise from the pole normal. The diameter of the inflection

circle for the motion 3/1 is measured as

49.2 mmJIR Ans.

The diameter of the Bresse circle is



22

3

2

3

5 rad/s49.2 mm 61.5 mm

20 rad/sJIb R

Ans.

Since the angular acceleration of the coupler link is clockwise (that is, negative), the

Bresse circle must lie on the positive side of the pole tangent, as shown on the figure.

(c) The point of intersection of the inflection circle and the Bresse circle (other than pole I) is the

acceleration center of the coupler link; see the figure.

(d) From the Euler-Savary equation, the radius of curvature of the coupler point C is

22 37.4 mm16.4 mm

85.2 mmC

C

CIC CO

CJ

RR

R Ans.

The location of the center of curvature of the path of point C (that is, CO ) is as shown on

the figure.

(e) The velocity of coupler point C is

3 5 rad/s 37.4 mm 187 mm/sC CIV R Ans.

The direction of the velocity vector of point C is as shown on the figure.

(f) The angular velocity of link 2 can be written as

23 13

23 12

2 3

13.7 mm5 rad/s 3.43 rad/s (cw)

20 mm

I I

I I

R

R

Ans.

(g) The velocity of the pole I is

3 5 rad/s 49.2 mm 246 mm/sJIv R Ans.

Since the angular velocity of the coupler link is positive (counterclockwise) the velocity

of the pole must be negative; that is, in the direction opposite to the pole tangent T (as

shown on the figure).

(h) The acceleration of coupler point C can be written as

4 2

3 3

24 273.6 mm 5 rad/s 20 rad/s

C CA R

22 356 mm/s Ans.

The angle from the line I to the pole normal N is measured as 38.25 ccw, as shown

in the figure. Therefore, the direction of the acceleration vector of point C is 38.25 ccw

from the line connecting to C, as shown in the figure.

(i) The acceleration of the pole I can be written as

22 2

3 3 5 rad/s 49.2 mm 1 230 mm/sI JIA v R

The acceleration of the pole is directed along the pole normal N as shown on the figure.

The angle from the horizontal axis to the pole normal N is measured as 33.0 .



As a check, the acceleration of the pole can be written as

4 2

3 3

24 238 mm 5 rad/s 20 rad/s

I IA R

=1 217 mm/s2

Ans.



4.47 Consider the double-slider mechanism in the position given in Problem 3.8. Point B

moves with a constant velocity 1000 mm/sBV to the left as illustrated in the figure.

The angular velocity and angular acceleration of coupler link AB are 3 3.66 rad/s ccw

and 2

3 13.40 rad/s cw , respectively. For the absolute motion of coupler link AB in

the specified position, draw the inflection circle and the Bresse circle. Then determine:

(a) the radius of curvature of the path of point C, which is a point in link 3 midway

between points A and B; and

(b) the magnitude and direction of the velocity of the velocity pole I. Using the

acceleration pole determine:

(c) the magnitude and direction of the acceleration of the pole I;

(d) the magnitude and direction of the velocity of points A and C; and

(e) the magnitude and direction of the acceleration of points A and C.

The pole I is the point coincident with the instant center 13I at the intersection of the

vertical line through point B and the line perpendicular to the direction of motion of

slider 2 through point A. Since point A moves on a straight line, the center of curvature

for point A is at infinity. Hence, the inflection point AJ is coincident with point A.

Similarly, the inflection point BJ is coincident with point B since point B also moves on

a straight line and the center of curvature of point B is at infinity. Knowing the two

inflection points AJ and BJ and the pole I, the center O of the inflection circle is

obtained as the intersection of the perpendicular bisectors of AIJ and BIJ as shown in the

figure below. The centrode normal passes through I and O and intersects the inflection

circle at inflection point J. The diameter of the inflection circle is measured as

386.25 mmJIR .



We keep in mind that the centrode normal N points from I toward J and the centrode

tangent is 90 clockwise from the centrode normal. The diameter of the Bresse circle is

22

3

2

3

3.66 rad/s (386.25 mm) 1386.25 mm

13.40 rad/sJIb R

Since the angular acceleration of link 3 (that is, 3 ) is clockwise (negative), the Bresse

circle must lie on the positive side of the centrode tangent. The Bresse circle is

positioned as shown in the figure.

As a check, note that point B, fixed in link 3, moves on a straight line with a constant

velocity. Hence point B must be the acceleration center for the absolute motion of link 3.

With the above construction of the inflection circle and the Bresse circle, the acceleration

center (the point of intersection of the inflection circle and the Bresse circle)

coincides with point B.

The angle from the line I to the centrode normal N is measured as 45 ccw . To

check, in Eq. (4.48),

21 13

22

3

13.40 rad/stan tan 45 ccw

3.66 rad/s

(a) The radius of curvature of point C, from the Euler-Savary equation, is

2 2(301.325 mm)

10939.25 mm8.3 mm

C

CIC CC

CJ

RR

R Ans.

Since the center of curvature C for point C does not lie on the paper, the direction is

indicated by an arrow on the figure.

(b) The magnitude of the velocity of the pole I is

3 (386.25 mm) 3.66 rad/s 1413.75 mm/sJIR Ans.

Since the angular velocity of link 3 is positive (counterclockwise) the pole velocity is

in the negative pole tangent direction as shown in the figure.

(c) The magnitude of the acceleration of the pole I is

2

3 (1413.75 mm/s) 3.66 rad/s 51.75 mm/sIA Ans.

The acceleration of the pole points along the positive pole normal as shown in the figure.

(d) The magnitude and direction of the velocity of points A and C are found as follows.

Since point A is fixed in link 3 the velocity of point A is

3 3.66 rad/s (334.5 mm) 1224.275 mm/sA AIV R Ans.

The direction of the velocity of A is perpendicular to the line AIR as shown in the figure.

Since point C is fixed in link 3 the velocity of point C is

3 3.66 rad/s (301.25 mm) 1102.575 mm/sC CIV R Ans.

The direction of the velocity of C is perpendicular to the line CIR as shown in the figure.

(e) The magnitude and direction of the acceleration of points A and C are found as

follows:

From Eq. (4.49), the acceleration of point A is given by



4 2

3 3

4 2 2

2

100 mm (3.66 rad/s) (13.40 rad/s )

1894.75 mm/s

A AA R

.Ans

The angle between the line I and the centrode normal is 45 ccw. Therefore, the

acceleration of point A is directed at an angle of 45 ccw from the line A as shown in

the figure.

The acceleration of point C can be written as

4 2

3 3

4 2 2

2

50 mm (36.6 rad/s) (1 340 rad/s )

947.5 mm/s

C CA R

.Ans

The acceleration of point C is at an angle of 45 ccw from line C as shown in the

figure.



4.48 For the mechanism of Problem 3.17, the input link 2 is rotating with an angular velocity

2 15 rad/s ccw and an angular acceleration 2

2 320.93 rad/s cw . For the

instantaneous motion of the connecting rod 3, find:

(a) the inflection circle and the Bresse circle;

(b) the location of the acceleration pole;

(c) the center of curvature of the path traced by the coupler point C fixed in link 3;

(d) the magnitude and direction of the velocity and acceleration of points A, B, and C;

(e) the magnitude and direction of the velocity and acceleration of the inflection pole J.

(a) The velocity pole I for the connecting rod 3 is coincident with the instant center 13I .

Since point B on link 3 travels on a straight line, it is an inflection point; that is, JB

coincides with point B (the inflection circle for the motion 3/1 has to pass through point

B). The inflection point for point A on link 3 can be obtained from the Euler-Savary

equation; that is,

22 334.25 mm1787.5 mm

62.5 mmA

A

AIAJ

AO

RR

R

The inflection circle is drawn through points I, JA, and JB as shown in the figure below.



The diameter of the inflection circle is measured as

2213.75 mmJIR Ans.

The centrode tangent T and the centrode normal N are also shown in the figure. Note that

the centrode normal passes through the pole I and the center of the inflection circle O and

intersects the inflection circle at the inflection pole J. The centrode normal points from I

to J and the centrode tangent is 90° clockwise from the centrode normal.

In order to draw the Bresse circle, the angular velocity and angular acceleration of link 3

must be known. The method of kinematic coefficients (see Sections 3.12 and 4.12) is

used here to determine 3

and 3

.

The vectors for the slider-crank portion of the mechanism are shown in the figure. The

vector loop equation can be written as

2 3 4 1 R R R R 0

The X and Y components can be written as

2 2 3 3 4cos cos 0R R R

2 2 3 3 1sin sin 0R R R

Differentiating these with respect to the input 2

gives

2 2 3 3 3 4sin sin 0R R R

2 2 3 3 3cos sin 0R R

where 3 3 2

d d and 4 4 2

R dR d are the first-order kinematic coefficients of links

3 and 4, respectively. Writing these equations in matrix form gives

3 3 3 2 2

3 3 2 24

sin 1 sin

cos 0 cos

R R

R RR

The determinant of the coefficient matrix is 3 3cosR . The length of the input link is

262.5 mmR and the length of the coupler link is

3250 mm.R The slider offset is

1 37.5 mmR . For the given input position o

2135 , the coupler angle (found from

trigonometry) is 3

19.07 . Substituting the known data gives

3

4

3.267 1 1.768

9.451 0 1.768R

Therefore, the first-order kinematic coefficients of link 3 and link 4 are

3

0.187 1 rad/rad and 4

28.925 mm/radR

The angular velocity of link 3 is

3 3 2

(0.187 1 rad/rad)(15 rad/s) 2.81 rad/s (ccw)

Differentiating the above matrix equation with respect to the input variable 2

gives



2

3 3 3 2 2 3 3 3

2

3 3 4 2 2 3 3 3

sin 1 cos cos

cos 0 sin sin

R R R

R R R R

Then substituting the numerical data gives

3

4

3.267 1 1.437

9.451 0 1.653R

Using Cramer's rule, the second-order kinematic coefficients of the mechanism are

2

30.175 rad/rad and 2

450.25 mm/rad R

The angular acceleration of link 3 can be written (see Table 4.2) as

2

3 3 2 3 2

2 2 2

2

(0.187 1 rad/rad)( 320.93 rad/s ) (0.175 rad/rad )(15 rad/s)

20.67 rad/s (cw)

The negative sign indicates that the angular acceleration of link 3 is clockwise.

The diameter of the Bresse circle for the motion 3/1 can now be written as

22

3

2

3

2.81 rad/s2213.75 mm 845.75 mm

20.67 rad/sJIb R

Ans.

Since the angular acceleration of link 3 is clockwise (negative) the Bresse circle must lie

on the positive side of the centrode tangent as shown in the figure.

(b) The acceleration center for the absolute motion of link 3 is the point of

intersection of the inflection circle and the Bresse circle. The angle from the line I to

the centrode normal N is measured as 69.09 ccw . As a check, from Eq. (4.48), the

angle is given by

2

1 13

22

3

20.67 rad/stan tan 69.09 ccw

2.81 rad/s

(c) From the Euler-Savary equation, the radius of curvature of point C can be written as

2 2(323.75 mm)

86.5 mm1210.25 mmC

C

CIC CO

CJ

RR

R

The center of curvature OC of point C is as shown in the figure.

(d) The velocity of point A is

3 (2.81 rad/s)(334.25 mm) 937.75 mm/sA AIV R Ans.

As a cross-check, the velocity of point A can also be found as

22 (15 rad/s)(62.5 mm) 937.55 mm/s.A AOV R

The direction of velocity of point A is 135° as shown in the figure.

The velocity of point B is

3 (2.81 rad/s)(154.5 mm) 433.75 mm/sB BIV R Ans.

The direction of velocity of point B is 180° as shown in the figure.

The velocity of point C is

3 (2.81 rad/s)(323.75 mm) 908.5 mm/sC CIV R Ans.



The direction of velocity of point C is 152.39° as shown in the figure.

From Eq. (4.49), the acceleration of point A can be written as

44 2 2 2 2

3 3 1109.75 mm 2.81 rad/s ( 20.67 rad/s ) 24550 mm/sA AA R Ans.

With o69.09 ccw, the direction of the acceleration of point A is 9.94° as shown in the

figure. The acceleration of point B can be written as

44 2 2 2 2

3 3 932.5 mm 2.81 rad/s ( 20.67 rad/s ) 20628.5 mm/sB BA R Ans.

The direction of the acceleration of point B is as shown in the figure. The acceleration of

point C can be written as

44 2 2 2 2

3 3 1113.5 mm 2.81 rad/s ( 20.67 rad/s ) 24632 mm/sC CA R Ans.

The direction of the acceleration of point C is 4.79° as shown in the figure.

(e) The velocity of the inflection pole J is the same as the velocity of the pole I.

Therefore, the velocity of the inflection pole J is

3 2.81 rad/s 2213.95 in 6212.75 mm/sJIv R Ans.

The direction of the velocity of inflection pole J is perpendicular to line IJ as shown in

the figure.

The acceleration of the inflection pole J is

44 2 2 2 2

3 3 2068 mm 2.81 rad/s ( 20.67 rad/s ) 45747.5 mm/sJ JA R Ans.

The acceleration of the inflection pole J makes an angle of 69.09 with the line J .



4.49 Figure P3.32 illustrates an epicyclic gear train driven by the arm, link 2, with an angular

velocity 2 3.33 rad/s cw and an angular acceleration 2

2 15 rad/s ccw . Define

point E as a point on the circumference of the planet gear 4 horizontal to the right of

point B such that the angle 90 . DBE For the absolute motion of the planet gear 4,

draw the inflection circle and the Bresse circle on a scaled drawing of the epicyclic gear

train. Then determine:

(a) The location of the acceleration center of the planet gear.

(b) The radii of curvature of the paths of points B and E.

(c) The locations of the centers of curvature of the paths of points B and E.

(d) The magnitudes and the directions of the velocities of points B and E and the pole I.

(e) The magnitudes and the directions of the accelerations of points B and E and the pole I.

Since the problem is for the motion 4/1 where 4 is the planet gear which is in internal

rolling contact with the fixed ring gear 1 then the pole I is coincident with the instant

center 14I which is the point of contact between the two gears. Note that the fixed

centrode is gear 1 and the moving centrode is gear 4. Recall that the centrode normal N

points from the fixed centrode toward the moving centrode (in the neighborhood of the

pole I). Therefore, the centrode normal N points vertically downward. Also, recall that

the Euler-Savary equation can be written as

1 1 1

F MJI IO IOR R R

The center of curvature of the fixed centrode FO is coincident with the center of the

fixed gear, that is, point A, and the center of curvature of the moving centrode MO is

coincident with the center of the planet gear; that is, point B. Recall that I with respect to

FO and I with respect to MO are both vertically upward; therefore, the radii of curvature

of the fixed and the moving centrodes, respectively, are

1 4100 mm and 25 mmF MIO IOR R

Substituting into the Euler-Savary equation gives

1 1 1 3

100 mm 25 mm 100 mmJIR

Therefore, the diameter of the inflection circle is

100 3 33.33 mmJIR

The inflection circle is shown in the figure below. Recall that the centrode tangent T is

90 clockwise from the centrode normal N and is, therefore, horizontal and is positive to

the left.

As a check: The inflection point for point C is coincident with the inflection pole J;

therefore, the radius of curvature of the path of point C, from the Euler-Savary equation,

is

2 2(50 mm)

150 mm16.66 mmC

C

CIC CO

CJ

RR

R

To determine the angular velocity and the angular acceleration of the planet gear 4, the

rolling contact equation between the planet gear 4 and the fixed ring gear 1 (see Chapter



3, Example 3.9) can be written as

4 2

1 2

1

4

100 mm 4

25

We use the positive sign here since there is internal rolling contact between the planet

gear 4 and the ring gear 1. Differentiating this equation with respect to the input position,

the rolling contact equation, in terms of first-order kinematic coefficients, is written as

4 2

1 2

1

4

4

Since the input is the arm (link 2) then 2

1 , and since the ring gear 1 is fixed then

10 . Therefore, the first- and second-order kinematic coefficients of the planet gear 4

from the above equation are

4

3 rad/rad and 4

0

The angular velocity of the planet gear 4 is

4 4 2 ( 3 rad/rad)( 3.33 rad/s) 10 rad/s (ccw)

The angular acceleration of the planet gear 4 is

2 2

4 4 2 4 2 ( 3 rad/rad) 15 rad/s 0 45 rad/s (cw)

Therefore, the diameter of the Bresse circle for the motion 4/1 is

22

4

2

4

10 rad/s33.33 mm 74 mm

45 rad/sJIb R

Since the angular acceleration of planet gear 4 is clockwise (negative), the Bresse circle

lies on the positive side of the centrode tangent T. The Bresse circle is as shown in the

figure.



(a) The acceleration center for the absolute motion of planet gear is the intersection

of the inflection circle and the Bresse circle. The acceleration center is shown in the

figure. The angle from the line I to the centrode normal N is measured as

24.23 ccw

Check: The angle is given by the relation

21 1 o4

22

4

45 rad/stan tan 24.23 ccw

10 rad/s

(b) and (c) The radius of curvature of the path of point B, from the Euler-Savary

equation, is

22 25 mm75.75 mm

8.25 mmB

B

BIB BO

BJ

RR

R Ans.

Note that the center of curvature BO of point B is coincident with point A and the

inflection point BJ is coincident with the inflection pole J. The radius of curvature of the

path of point E, from the Euler-Savary equation, is

22 35.25 mm105.75 mm

11.75 mmE

E

EIE EO

EJ

RR

R Ans.

The center of curvature EO of point E is shown on the figure.

(d) The magnitude and direction of the velocity of points B, E and the pole I.

The velocity of point B is

4 10 rad/s 25 mm 25 mm/sB BIV R Ans.

The direction of the velocity of point B is horizontal to the right as shown in the figure.

The velocity of point E is

4 10 rad/s 35.25 mm 352.5 mm/sE EIV R Ans.

The direction of the velocity of point E is perpendicular to line IE and is inclined at an

angle of 45° from the horizontal.

The velocity of the pole I can be written as

4 10 rad/s 33.25 mm 332.5 mm/sJIR Ans.

The direction of the velocity of the pole I is along the negative centrode tangent T since

the angular velocity of link 4 is counterclockwise; that is, positive.

(e) The acceleration of point B can be written as

44 2 2 2 2

4 4 12.75 mm 10 rad/s ( 45 rad/s ) 1400 mm/sB BA R Ans.

With 24.23o ccw the direction of the acceleration of point B is as shown in the figure.

The acceleration of point E can be written as

44 2 2 2 2

4 4 37.75 mm 10 rad/s ( 45 rad/s ) 4139.5 mm/sE EA R Ans.

With 24.23 ccw the direction of the acceleration of point E is as shown in the

figure. The acceleration of the pole I is

2

4 (333.25 mm/s) 10 rad/s 3332.5 mm/sIA Ans.



The acceleration of the pole is directed along the positive pole normal as shown in the

figure.

Check: The acceleration of the pole I can also be obtained from the equation

44 2 2 2 2

4 4 30.4 mm 10 rad/s ( 45 rad/s ) 3332.5 mm/sI IA R



4.50 On 450 by 600 mm paper, draw the linkage illustrated in Fig. P4.50 in full size, placing A at 150 mm from the lower edge and 175 mm from the right edge. Better utilization of

the paper is obtained by tilting the frame through about 15 as indicated.

(a) Find the inflection circle.

(b) Draw the cubic of stationary curvature.

(c) Choose a coupler point C coincident with the cubic and plot a portion of its

coupler curve in the vicinity of the cubic.

(d) Find the conjugate point C . Draw a circle through C with center at C and

compare this circle with the actual path of C.

(e) Find Ball’s point. Locate a point D on the coupler at Ball’s point and plot a

portion of its path. Compare the result with a straight line.

25 mm, 125 mm, 43.75 mm, 81.25 mmAA BA B A BBR R R R

Drawn with a precise CAD system above, the circle around center C matches the

coupler curve near C to better than visual comparison can detect for the 30 of crank

rotation shown. Similarly, Ball’s point D follows an almost perfect straight line over the

same range as shown.



Chapter 5

Multi-Degree-of-Freedom Planar Linkages

5.1 The slotted links 2 and 3 are driven independently at constant speeds of 2

30 rad/s cw

and 3

20 rad/s cw, respectively. Find the absolute velocity and acceleration of the

center of the pin 4

P carried in the two slots.

XB = 100 mm; YB = 25 mm.

Identifying the pin as separate body 4 and, noticing the two paths it travels on bodies 2

and 3, we write

2 22 30 rad/s 54.9 mm 1647 mm/sP P AV R

3 33 20 rad/s 102.6 mm 2052 mm/sP P BV R

4 2 4 3 4/2 /3P P P P P V V V V V

Construct the velocity polygon

42355 mm/s 15.6P V Ans.

2 2 2 2 2 2P O P O

n t

P O A A A A ; 3 3 3 3 3 3P O P O

n t

P O A A A A

2 2 2 2

22 2

2 30.0 rad/s 54.9 mm 49410 mm/sn

P O P OA R .

3 3 3 3

22 2

3 20.0 rad/s 102.6 mm 41040 mm/sn

P O P OA R

4 2 4 2 4 4 3 4 3 4 4/ 2 / 2 /3 /3

c n t c n t

P P P P P P P P P P P A A A A A A A A A

4 2 4

2

2 /22 2 30.0 rad/s 1683 mm/s 100980 mm/s 30.0c

P P P A V ×



4 3 4

2

3 /32 2 20.0 rad/s 1155 mm/s 46200 mm/s 225.0c

P P P A V ×

4

4

4

22

/2

/2

/2

1683 mm/s0

Pn

P

P

VA

;

4

4

4

22

/2

/2

/2

1155 mm/s0

Pn

P

P

VA

Construct the acceleration polygon.

4

2125730 mm/s 66.6P A Ans.

For comparison, let us now solve the same problem by use of kinematic coefficients. The

loop-closure constraint equations can be written as

2 2 3 3

2 2 3 3

cos cos 0

sin sin 0

B

B

r r x

r r y

Recognizing that 2 and 3 are the two independent degrees of freedom, and that r2 and

r3 are dependent position unknowns. Since these appear linearly (which is not true in

other problems), the loop-closure equations can be written in matrix form as follows:

2 3 2

2 3 3

cos cos

sin sin

B

B

r X

r Y

For the given position 2 60 and 3 135 , and the dimensions are 100 mmBX and

25 mm.BY The determinant of this set is 3 2sin 0.966 , and the solutions

for the two unknown position values are 2 54.9 mmr and 3 102.6 mmr . The position

coordinates of the center of the pin are

2 2 3 3

2 2 3 3

cos cos 27.45 mm

sin sin 47.55 mm

P B

P B

x r x r

y r y r

Taking the derivatives of the loop-closure equations with respect to both 2 and 3 , in

turn, we find the following two sets of equations for the first-order kinematic coefficients.

2 3 22 23 2 2 3 3

2 3 32 33 2 2 3 3

cos cos sin sin

sin sin cos cos

r r r r

r r r r

and the solutions for these are

22 23 2 3 2 3

32 33 2 3 3 2

cos1

cos

r r r r

r r r r

At the current position, the numeric values of the first-order kinematic coefficients are

22 23

32 33

14.71 mm/rad 106.2 mm/rad

56.84 mm/rad 27.49 mm/rad

r r

r r

Using these, the first-order kinematic coefficients for the center of the pin are

2 22 2 2 2 32 3

2 22 2 2 2 32 3

cos sin cos 40.19 mm/rad

y sin cos sin 40.19 mm/rad

P

P

x r r r

r r r

3 23 2 33 3 3 3

3 23 2 33 3 3 3

cos cos sin 53.11 mm/rad

sin sin cos 91.98 mm/rad

P

P

x r r r

y r r r

With the given independent input velocities of 2

30 rad/s cw and 3

20 rad/s cw,



the velocity of the pin P4 is

2 2 3 3

2 2 3 3

2268 mm/s

634 mm/s

P P P

P P P

x x x

y y y

ˆ ˆ ˆ ˆV 2268 634 mm/s 2355 mm/s 15.62P P Px y i j i j Ans.

Taking the second derivatives of the loop-closure equations with respect to both 2 and

3 , in turn, we find the following three sets of equations for the second-order kinematic

coefficients:

2 3 222 223 233

2 3 322 323 333

22 2 2 2 23 2 32 3 33 3 3 3

22 2 2 2 23 2 32 3 33 3 3 3

cos cos

sin sin

2 sin cos sin sin 2 sin cos

2 cos sin cos cos 2 cos sin

r r r

r r r

r r r r r r

r r r r r r

and the solutions for these are

222 223 233 22 3 2 2 3 2 23 3 2 32 33

322 323 333 22 23 32 3 2 33 3 2 3 3 2

2 cos sin cos 21

2 cos 2 cos sin

r r r r r r r r

r r r r r r r r

At the current position, the numeric values of the second-order kinematic coefficients are 2 2 2

222 223 233

2 2 2322 323 333

62.787 mm/rad 87.307 mm/rad 56.92 mm/rad

30.46 mm/rad 125.195 mm/rad 117.31 mm/rad

r r r

r r r

The second-order kinematic coefficients for the center of the pin are 2

22 222 2 22 2 2 2 322 3

2

22 222 2 22 2 2 2 322 3

2

23 223 2 23 2 323 3 32 3

cos 2 sin cos cos 21.538 mm/rad

sin 2 cos sin sin 21.538 mm/rad

cos sin cos sin 48.334 mm/rad

y

P

P

P

P

x r r r r

y r r r r

x r r r r

2

23 223 2 23 2 323 3 32 3

2

33 233 2 333 3 33 3 3 3

2

33 233 2 333 3 33 3 3 3

sin cos sin cos 128.719 mm/rad

cos cos 2 sin cos 28.461 mm/rad

sin sin 2 cos sin 49.296 mm/rad

P

P

r r r r

x r r r r

y r r r r

With the given independent input velocities and accelerations of 2

30 rad/s cw,

320 rad/s cw, and 2 3 0 , the acceleration of the pin P4 is

2 2 2

2 2 3 3 22 2 23 2 3 33 3

2 2

2 2 3 3 22 2 23 2 3 33 3

2 50 mm/s

2 115360 mm/s

P P P P P P

P P P P P P

x x x x x x

y y y y y y

2 2ˆ ˆ ˆ ˆ50 115360 mm/s 125730 mm/s 66.6P P Px y A i j i j Ans.

It should be noted that the exact match between the graphic and the analytic solutions

achieved for this problem is not at all typical, nor can such matches be expected. Graphic

results are usually far less accurate than analytic ones. The reason for the agreement

achieved here is that a very precise CAD system was used for the graphic constructions.



5.2 For the five-bar linkage in the position illustrated in Fig. P5.2, the angular velocity of

link 2 is 15 rad/s cw and the angular velocity of link 5 is 15 rad/s cw. Determine the

angular velocity of link 3 and the apparent velocity 4 /5BV .

2 5200 mm 23.1O O R ,

2300 mmAOR , and 200 mmBAR

Let us define 2 51 200 mm,O Or R 1 23.1 ,

22 300 mm,AOr R 3 200 mm,BAr R

and 54 .BOr R Then the loop-closure equation can be written as

1 1 2 2 3 3 4 5 0r r r r

with horizontal and vertical components of

1 1 2 2 3 3 4 5

1 1 2 2 3 3 4 5

cos cos cos cos 0

sin sin sin sin 0

r r r r

r r r r

Solution of these position equations give two unknowns, 4 300 mmr and 3 203.1 .

Derivatives of the loop-closure equations with respect to each of the independent degrees

of freedom, 2 and 5 , give the first-order kinematic coefficients

3 3 5 32 35 2 2 4 5

3 3 5 42 45 2 2 4 5

sin cos sin sin

cos sin cos cos

r r r

r r r r r

The determinant of the Jacobian is 3 3 5cosr and this will go to zero whenever

3 5 2 1 2k ; that is, whenever link 3 is perpendicular to link 5. At the current

position, 159.94 mm. The solutions for the first-order kinematic coefficients are

32 35 2 5 2 4

42 45 2 3 3 2 3 4 3 5

cos1

sin sin

r r

r r r r r r

At the current position, with the given data, the values for the first-order kinematic

coefficients are

32 35

42 45

1.875 74 rad/rad 1.875 74 rad/rad

225.25 mm/rad 225.25 mm/radr r

Therefore, with 2 5 15 rad/s, we have

43 32 2 35 5 /5 42 2 45 50 and 0BV r r Ans.



5.3 For the five-bar linkage in the position illustrated in Fig. P5.2, the angular velocity of

link 2 is 2 25 rad/s ccw and the apparent velocity 4 /5BV is 5000 mm/s upward along

link 5. Determine the angular velocities of links 3 and 5.

Here we can continue the solution of Problem 5.2. However, the problem is now

expressed in terms of two different input variables, 2 and 4r , as independent degrees of

freedom. Therefore, we can use the same vectors and the same loop-closure equations.

However, we must now take derivatives with respect to 2 and 4r to find the first-order

kinematic coefficients. The result, in matrix form, is

3 3 4 5 32 34 2 2 5

3 3 4 5 52 54 2 2 5

sin sin sin cos

cos cos cos sin

r r r

r r r

The determinant of the Jacobian is now 3 4 3 5sinr r , which goes to zero whenever

link 3 is aligned with link 5. The solutions for the first-order kinematic coefficients are

332 34 2 4 5 2 4

352 54 2 3 3 2 3 3 5

sin 0 8.3276 10 rad/mm1

sin cos 1.000 00 rad/rad 4.4396 10 rad/mm

r r r

r r r

With the given input velocities, 2 25 rad/s and 4 5000 mm/sr , the requested

velocities are

3 32 2 34 4 41.64 rad/s (cw)r and 5 52 2 54 4 47.20 rad/s ccwr Ans.



5.4 For Problem 5.2, assuming that the two given input velocities are constant, determine the

angular acceleration of link3 at the instant indicated.

Starting with the equations of Problem 5.2 for the first-order kinematic coefficients,

3 3 5 32 35 2 2 4 5

3 3 5 42 45 2 2 4 5

sin cos sin sin

cos sin cos cos

r r r

r r r r r

we can take derivatives with respect to each independent variable to find equations for

the second-order kinematic derivatives

3 3 5 322 325 355

3 3 5 422 425 455

2 2

3 3 32 2 2 3 3 32 35 5 42 3 3 35 5 45 4 5

2

3 3 32 2 2 3 3 32 35

sin cos

cos sin

cos cos cos sin cos 2sin cos

sin sin sin co

r

r r r r

r r r r r r r

r r r

2

5 42 3 3 35 5 45 4 5s sin 2cos sinr r r r

With satisfaction, we notice that the Jacobian is identical with that of Problem 5.2. The

solution to these equations gives

2 2

3 3 5 32 2 5 2 3 3 5 32 35 42 3 3 5 35 45

2 2 2 2 2

3 32 2 3 3 2 3 3 5 32 35 3 3 5 42 3 35 3 3 5 45 3

322 325 355

422 425 455

1 sin sin sin sin 2

cos cos sin 2 sin

r r r r r r

r r r r r r r r r r

r r r

4 3 5cosr

Substituting the numeric data, including results of Problem 5.2, we get

2 2 2

322 325 355

2 2 2422 425 455

2.641 69 rad/rad 4.050 06 rad/rad 5.458 43 rad/rad

579.95 mm/rad 534.56 mm/rad 1518.19 mm/radr r r

Therefore, given that 2 5 15 rad/s (cw) and 2 5 0, the angular acceleration

of link 3 is

2 2

3 32 2 35 5 322 2 325 2 5 355 52 0 Ans.



5.5 Figure P5.5 illustrates link 2 rotating at a constant angular velocity of 10 rad/s ccw while

the sliding block 3 slides toward point A at the constant rate of 125 mm/s. At the instant

indicated RPA = 100 mm. Find the absolute velocity and absolute acceleration of point P

of block 3.

275 mmAOR and 150 mmBAR

With the dimensions given, O2AP is a 3-4-5 right triangle and 2

ˆ125 mmPO R j .

For velocity, we write

3 2 3

2 2 2 2

/2

2

ˆ ˆˆ10 rad/s 125 mm 1 250 mm/s

P P P

P O P O

V V V

V V ω R

k j i

Also, we are given 3 /2 125 mm/sPV . From these we construct the velocity polygon

shown in the figure, and from this we measure

3

1 179.25 mm/s 175.13P V Ans.

For acceleration, we write

3 2 3 2 3 3

2 2 2 2 2 2

2 2

/2 /2

2 2ˆ ˆ10 rad/s 125 mm 12 500 mm/s

c n t

P P P P P P

n t

P O P O P O

n

P OA

A A A A A

A A A A

j j

3 2 3

3 3 3

3

2

2 /2

2 2

/2 /2 /2

/2

2 2 10 rad/s 125 mm/s 2 500 mm/s

0

0

c

P P P

n

P P P

t

P

A V

A V V

A

From these we construct the acceleration polygon shown in the figure, and from this we

measure

3

211 180 in/s 79.70P A Ans.



5.6 For Problem 5.5, determine the value of the sliding velocity 3 /2PV that minimizes the

absolute velocity of point P of block 3. Find the value of 3 /2PV that minimizes the

absolute acceleration of point P of block 3.

By careful inspection of the velocity polygon of Problem 5.5 we can see that the absolute

velocity is minimized when it becomes perpendicular to 3 /2PV . Reconstructing the

velocity polygon in this condition, as shown in the figure, we find

3

1 000.0 mm/s 143.13P V and 3 /2 750.0 mm/s 53.13P V Ans.

Similarly, the absolute acceleration of P3 is minimized when it becomes perpendicular to

3 2

c

P PA . Reconstructing the acceleration polygon in this condition, as shown in the figure,

we find 3

210 000 mm/s 53.13P A and 3 2

27 500 mm/s 143.13c

P P A . Then, from

this, we can calculate

3 2

3

2

/2

2

7 500 mm/s375.0 mm/s

2 2 10.00 rad/s

c

P P

P

AV

Ans.

Notice how the visualization of the inherent geometry has dramatically simplified this

problem, compared to a totally mathematical approach. Notice also that 3 /2PV must

increase in both cases.



5.7 The two-link planar robots illustrated in Fig. P5.7 have the link lengths

2 4300 mmAO BOR R and 400 mmPA PBR R . The two robots are carrying a small

object labeled P. At the instant indicated the angular positions are 2 45 and

3/2 15 . (Notice that the angle 3/2 3 2 is given because that is the angle

controlled by the motor in joint A.) A second robot having identical dimensions is

stationed at position O4, 1000 mm to the right. What angular positions must the two

joints 4 and 5/4 of the second robot have at this moment to allow it to take over

possession of the object P?

Dimensions are:

2300 mmAOR , 400 mmPAR .

The loop-closure constraint equations at this instant allow us to write

4 5

4 5

1000 300cos 400cos 400cos30 300cos 45 0

300sin 400sin 400sin30 300sin 45 0

These can be rearranged to read

5 4

5 4

cos 0.750cos 1.10364

sin 0.750sin 1.03033

Now, by squaring and adding, we eliminate the variable 5.

4 41.0 0.562 5 1.655 47cos 1.545 50sin 2.279 60

or 4 41.655 47cos 1.545 50sin 1.842 10 0

Next, by defining 4tan 2x , and by use of the standard identities, this becomes

2 21.65547 1 1.545 50 2 1.842 10 1 0x x x

or 20.18663 3.09100 3.49757 0x x

The roots of this equation give

15.340 54x and 1.221 64x

and from the definition of x these give two values of 4

4 172.54 and 4 101.39

Now, returning these to the above equations, we can solve for values of 5

5 111.10 and 5 162.84

Of these, the second value of each pair fits our figure. Therefore,



4 101.39 and 5/4 61.45 Ans.

5.8 For the transfer of the object described in Problem 5.7 it is necessary that the velocities of

point P of the two robots match. If the two input velocities of the first robot are

2 10 rad/s cw and 3/2 15 rad/s ccw, what angular velocities must be used for 4

and 5/4 ?

First we find

3 2 3/2 10 rad/s (cw) 15 rad/s (ccw) 5 rad/s (ccw)

Then, the velocity of point P is given by

2 2 4 4

2 22

3

10 rad/s 300 mm 3000 mm/s

5 rad/s 400 mm 2000 mm/s

P O AO PA O BO PB

AO AO

PA PA

V R

V R

V V V V V V V

From these data and equations, we can construct the velocity polygon shown in the

figure. This allows us to find data for the following calculations:

4

4

4

1350.5 mm/s4.502 rad/s cw

300 mm

BO

BO

V

R Ans.

5


400 mm

PB

PB

V

R



5/4 5 4 1.716 rad/s 4.502 rad/s 6.218 rad/s (ccw) Ans.

If an analytical solution is preferred, we start with the robot on the left, where we find

3 2 3/2 10 rad/s (cw) 15 rad/s (ccw) 5 rad/s (ccw)

22

3

ˆ ˆ ˆ ˆˆ10.0 rad/s 300cos 45 300sin 45 mm 2121.32 2121.32 mm/s

ˆ ˆ ˆ ˆˆ5.0 rad/s 400cos30 400sin 30 mm 1000 1732.05 mm/s

ˆ ˆ1121.32 389.27 mm/s

A AO

PA PA

P A PA

V ω ×R k × i j i j


V V V i j

Similarly, for the robot on the right, we have

44 4

5 5

4 5

ˆ ˆˆ rad/s 300cos101.39 300sin101.39 mm


294.09 59.25 mm 118.02 382.18 mm

B BO

PB PB

P B PB

V ω ×R k × i j

V ω ×R k × i j

V V V i j i j

Next, by setting the two equations for PV equal to each other, and then separating the i

and j components, we obtain a set of two equations for 4 and 5

4

5

294.09 mm 118.02 mm 1121.32 mm/s

59.25 mm 382.18 mm 389.27 mm/s

The solutions to these equations give

4

5

4.502 rad/s (cw)

1.716 rad/s (ccw)

Ans.

5/4 5 4 1.716 rad/s 4.502 rad/s 6.218 rad/s (ccw) Ans

We see that these results agree precisely with those obtained above by the graphical

approach. It must be pointed out that this is not usual for graphic solutions, but is the

result of the high-precision CAD system used here.

As yet a third approach to the solution of this problem, we can find the instant centers of

velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.

Since we are given velocities for 2 and 3 , the location of instant center 13I is defined

by

23 13

23 13 23 12

23 12

2

3

10.0 rad/s2.0 or 2.0

5.0 rad/s

I I

I I I I

I I

RR R

R

Therefore 13I takes the position shown in the figure below. When the other instant

centers are found through the Aronhold-Kennedy theorem, this results in the instant

centers shown for 24 34,I I and for 25 35,I I .

Once the remaining instant centers are found, we may find the information requested in

the problem. We find



24 12

24 14

4 2

818.84 mm10 rad/s 4.502 rad/s (cw)

1818.84 mm

I I

I I

R

R Ans.

25 12

25 15

5 2

193.87 mm10 rad/s 1.716 rad/s (ccw)

1129.51 mm

I I

I I

R

R


Again, the very high degree of agreement is a consequence of the precision of the CAD

system used in finding the distances between instant centers.





2 10 rad/s cw and 3/2 10 rad/s ccw, what angular velocities must be used for 4

and 5/4 ?

First we find

3 2 3/2 10 rad/s (cw) 10 rad/s (ccw) 0 rad/s

Notice that this implies that link 3 is in translation relative to the ground.


2 2 4 4

2 22

3

10 rad/s 300 mm 3000 mm/s

0 rad/s 400 mm 0.0 in/s

P O AO PA O BO PB

AO AO

PA PA

V R

V R

V V V V V V V



4

4

4

3020 mm/s10.067 rad/s cw

300 mm

BO

BO

V

R Ans.

5

2844.5 mm/s7.111 rad/s ccw

400 mm

PB

PB

V

R





3 2 3/2 10 rad/s (cw) 10 rad/s (ccw) 0 rad/s

22

3

ˆ ˆ ˆ ˆˆ10.0 rad/s 300cos 45 300sin 45 in 2121.32 2121.32 mm/s

ˆ ˆ ˆ ˆˆ0 rad/s 400cos30 400sin 30 in 0.000 00 0.000 00 mm/s

ˆ ˆ2.121 32 2.121 32 mm/s

A AO

PA PA

P A PA



V V V i j


44 4

5 5

4 5



294.09 59.25 mm 118.02 382.18 mm

B BO

PB PB

P B PB

V ω ×R k × i j

V ω ×R k × i j

V V V i j i j



4

5

294.09 mm 118.02 mm 2121.32 mm/s

59.25 mm 382.18 mm 2121.32 mm/s


4

5

10.067 rad/s (cw)

7.111 rad/s (ccw)

Ans.

5/4 5 4 7.111 rad/s 10.067 rad/s 17.178 rad/s (ccw) Ans





velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.

Since we are given velocities for 2 and 3 , the location of instant center 13I is defined

by

23 13

23 13

23 12

2

3

10.0 rad/s or

0.0 rad/s

I I

I I

I I

RR

R

Therefore 13I goes to infinity in the direction shown in the figure below. This indicates

that link 3 is in translation with respect to the ground, which we could see when we found

that 3 0 . When the other instant centers are found through the Aronhold-Kennedy

theorem, this results in the instant centers shown for 24 34,I I and for 25 35,I I . However,

we find that the lines toward instant center 24I are essentially parallel, and that instant

center 24I appears to also be at infinity. This implies that link 4 is in translation with

respect to link 2, which means that 4 2 10 rad/s (cw) .



Once the remaining instant centers are found, we may find the information requested in

the problem. We find

24 12

24 14

4 2 10.000 rad/s (cw)I I

I I

R

R Ans.

25 12

25 15

5 2

18.503 60 in10 rad/s 7.112 rad/s (ccw)

26.017 20 in

I I

I I

R

R


Notice that the precision is not perfect this time, in spite of the use of a high-precision

CAD system. However, this probably stems from the possibility that the apparent

parallelism and intersections at infinity were likely not perfect. Still, the precision is

amazing for a graphical solution.





2 10 rad/s cw and 3/2 0, what angular velocities must be used for 4 and 5/4 ?

First we find

3 2 3/2 10 rad/s (cw) 0 rad/s 10 rad/s (cw)

Notice that this implies that link 3 and link 2 rotate together as a single unit.


2 2 4 4

2 22

3

10 rad/s 300 mm 3000 mm/s

10 rad/s 400 mm 4000 mm/s

P O AO PA O BO PB

AO AO

PA PA

V R

V R

V V V V V V V



4

4

4

6359.5 m/s21.198 rad/s cw

300 mm

BO

BO

V

R Ans.



5

7160.5 mm/s17.901 rad/s ccw

400 mm

PB

PB

V

R

5/4 5 4 17.901 rad/s 21.198 rad/s 39.099 rad/s ccw Ans.


22

3

ˆ ˆ ˆ ˆˆ10.0 rad/s 300cos 45 300sin 45 in 2121.32 2121.32 mm/s

ˆ ˆ ˆ ˆˆ10.0 rad/s 400cos30 400sin 30 in 80.000 00 138 56 mm/s

ˆ ˆ164.852 80 223.416 80 in/s

A AO

PA PA

P A PA



V V V i j


44 4

5 5

4 5

ˆ ˆˆ rad/s 12cos101.39 12sin101.39 in

ˆ ˆˆ rad/s 16cos162.84 16sin162.84 in

11.763 60 2.370 00 in 4.720 80 15.287 20 in

B BO

PB PB

P B PB

V ω ×R k × i j

V ω ×R k × i j

V V V i j i j



4

5

294.09 mm 118.02 mm 4121.3 mm/s

59.25 mm 382.18 mm 5585.42 mm/s


4

5

21.198 rad/s (cw)

17.901 rad/s ccw

Ans.

5/4 5 4 17.901 rad/s 21.198 rad/s 39.099 rad/s ccw Ans





velocity. In doing this we follow exactly the approach shown in Example 5.5. However,

since we have equal velocities for 2 and 3 , the location of instant center 13I is defined

by

23 13

23 12

2

3

10.0 rad/s1.0

10.0 rad/s

I I

I I

R

R

and therefore 13I becomes coincident with 12I as shown in the figure below. When the

other instant centers are found through the Kennedy-Aronhold theorem, this also results

in coincident instant centers for 24 34,I I and for 25 35,I I as shown in the figure. Under

these input velocity conditions, links 2 and 3 act as a single solid unit.



Once the remaining instant centers are found, however, we may still find the information

requested in the problem. We find

24 12

24 14

4 2

1892.675 mm10 rad/s 21.198 rad/s (cw)

892.85 mm

I I

I I

R

R Ans.

25 12

25 15

5 2

694.075 mm10 rad/s 17.901 rad/s (ccw)

387.725 mm

I I

I I

R

R

Ans.

Again, the perfect agreement results from the precision of the CAD system used in

finding the distances between instant centers.



5.11 To successfully transfer an object between two robots, as described in Problems 5.7 and

5.8, it is helpful if the accelerations are also matched at point P. Assuming that the two

input accelerations are 2 3 0 at this instant for the robot on the left, what angular

accelerations must be given to the two input joints of the robot on the right to achieve

this?

Starting after the solutions of Problem 5.8 (the velocity analysis) is completed, the

condition for the acceleration of point P is written as

2 2 2 4 4 4

2 4

2 4

2 4

2 22 2

2 2

222

3000 mm/s 1350.6 mm/s30000 mm/s 6080.5 mm/s

300 mm 300 mm

2000 mm/s1000 mm/s

400 mm

n t n t n t n t

P O AO AO PA PA O BO BO PB PB

AO BOn n

AO BO

AO BO

n PAPA

PA

V VA A

R R

VA

R

A A A A A A A A A A A

22

2686.5 mm/s

1178.25 mm/s400 mm

n PBPB

PB

VA

R

Notice that a difficulty arises in the graphic solution of the above acceleration equation in

that the two unknowns do not arise consecutively in the equation. Nevertheless, recalling

that vector addition is commutative (independent of order), we can proceed with the

vectors appearing out of order, as is shown in the dotted lines in the figure above. Once



the solution with the dotted lines is completed, we have obtained the correct magnitudes

and directions of the two unknown tangential components. However, we do not have a

valid acceleration polygon unless we now arrange the components in their correct order,

according to the original acceleration equation, as is shown in the dashed lines in the

figure. If this is not done, the acceleration image point B cannot be labeled, and the

absolute acceleration of B and acceleration images of links 4 and 5 cannot be correctly

shown.

Whether or not the vectors are arranged in their correct order, however, we can proceed

with the solution for the two unknown angular accelerations as follows:

4

4

22

4

27583 mm/s91.943 rad/s ccw

300 mm

t

BO

BO

A

R Ans.

22

4

15127 mm/s37.818 rad/s ccw

400 mm

t

PB

PB

A

R Ans.



PART 2

DESIGN OF MECHANISMS



Chapter 6

Cam Design

6.1 The reciprocating radial roller follower of a plate cam is to rise 40 mm with simple

harmonic motion in 180 of cam rotation and return with simple harmonic motion in the

remaining 180 . If the roller radius is 7.5 mm and the prime-circle radius is 40 mm,

construct the displacement diagram, the pitch curve, and the cam profile for clockwise

cam rotation.

6.2 A plate cam with a reciprocating flat-face follower has the same motion as in Problem

6.1. The prime-circle radius is 40 mm, and the cam rotates counterclockwise. Construct

the displacement diagram and the cam profile, offsetting the follower stem by 15 mm in

the direction that reduces the bending stress in the follower during rise.



6.3 Construct the displacement diagram and the cam profile for a plate cam with an

oscillating radial flat-face follower that rises through 30 with cycloidal motion in 150

of counterclockwise cam rotation, then dwells for 30 , returns with cycloidal motion in

120 , and dwells for 60 . Determine the necessary length for the follower face, allowing

6.25 mm clearance at the free end. The prime-circle radius is 37.5 mm, and the follower

pivot is 150 mm to the right.

Notice that, with the prime circle radius given, the cam is undercut and the follower will

not reach positions 7 and 8. The follower face length shown is 250 mm but can be made

as short as 243.75 mm (position 9) from the follower pivot. Ans.



6.4 A plate cam with an oscillating roller follower is to produce the same motion as in

Problem 6.3. The prime-circle radius is 75 mm, the roller radius is 12.5 mm, the length

of the follower is 125 mm, and it is pivoted at 156.25 mm to the right of the cam rotation

axis. The cam rotation is clockwise. Determine the maximum pressure angle.

From a graphical analysis, max 39 Ans.

6.5 For a full-rise simple harmonic motion, write the equations for the velocity and the jerk at

the midpoint of the motion. Also, determine the acceleration at the beginning and the end

of the motion.

Using Eqs. (6.12) and (6.11) we find

1sin

2 2 2 2

L Ly

1

2 2

Ly

Ans.

3 3

3 3

1sin

2 2 2 2

L Ly

3

3

3

1

2 2

Ly

Ans.

2 2

2 20 cos0

2 2

L Ly

2

2

20

2

Ly

Ans.

2 2

2 21 cos

2 2

L Ly

2

2

21

2

Ly

Ans.



6.6 For a full-rise cycloidal motion, determine the values of for which the acceleration is

maximum and minimum. What is the formula for the acceleration at these points? Find

the equations for the velocity and the jerk at the midpoint of the motion.

Using Eqs. (6.13), we know that acceleration is an extremum when jerk is zero. This

occurs when cos2 0 ; that is, when 1 4 or when 3 4 ,

max2 2

1 2 2sin

4 2

L Ly y

Ans.

min2 2

3 2 3 2sin

4 2

L Ly y

Ans.

1 2

1 cos2

L Ly

Ans.

2 2

3 3

1 4 4cos

2

L Ly

Ans.

6.7 A plate cam with a reciprocating follower is to rotate clockwise at 400 rev/min. The

follower is to dwell for 60 of cam rotation, after which it is to rise to a lift of 50 mm.

During 20 mm of its return stroke, it must have a constant velocity of 800 mm/s.

Recommend standard cam motions from Section 6.7 to be used for high-speed operation

and determine the corresponding lifts and cam rotation angles for each segment of the

cam.



The curves shown are initially only sketches and not drawn to scale. They suggest the

standard curve types that might be chosen. The actual choices are shown in the table

below.

400 rev/min 2 rad/rev41.888 rad/s cw

60 s/min

To match the required velocity condition in Seg. DE we must have

4 4

4

4 4 4 4

4

800.000 mm/s 41.888 rad/s

19.098 600 mm/rad 20.000 mm

1.047 198 rad 60.000

y y

y

y L

Matching the first derivatives at D and E we find

5 5 42 19.098 593 mm/radL y 5 59.549 297 mm/radL (1)

3 3 42 19.098 593 mm/radL y 3 312.158 542 mm/radL (2)

Matching the second derivatives at C we find

2 2 2

2 3 35.26830 2.5000 4L 2 2

2 3 3 3106.758 078 8.780 500L (3)

For geometric continuity, we have

1 2 3 4 5L L L L L or 3 5 30.000 000 mmL L (4)

1 2 3 4 5 2 or 2 3 5 4.188 790 rad (5)

Equations (1) through (5) are now solved simultaneously for 2 , 3L , 3 , 5L , and 5 .

The results are summarized in the following table:

Seg. Type Eq. L, mm rad , deg

AB dwell --- 0 1.047 198 60.000

BC 8th

order poly. (6.14) 50.000 1.089 824 62.442

CD half harmonic (6.20) 1.648 0.135 268 7.750

DE constant velocity --- 20.000 1.047 198 60.000

EA half cycloidal (6.25) 28.352 2.963 698 169.808

6.8 Repeat Problem 6.7 except with a dwell for 20 of cam rotation.

The procedure is the same as for Problem 6.7. The results are:

Seg. Type Eq. L, mm rad , deg

AB dwell --- 0 0.349 066 20.000

BC 8th

order poly. (6.14) 50.000 1.870 958 107.198

CD half harmonic (6.20) 4.872 0.398 667 22.842

DE constant velocity --- 20.000 1.047 198 60.000

EA half cycloidal (6.25) 25.128 2.617 296 149.960



6.9 If the cam of Problem 6.7 is driven at constant speed, determine the time of the dwell and

the maximum and minimum velocity and acceleration of the follower for the cam cycle.

The time duration of the dwell is 1 1.047 198 rad 41.888 rad/s 0.025 st Ans.

Working from the equations listed, the maximum and minimum values of the kinematic

coefficients in each segment of the cam are as follows:

Seg. Eq. maxy mm/rad miny mm/rad maxy mm/rad

2 miny mm/rad

2

AB --- 0 0 0 0

BC (6.14) 241.704 0 221.783 221.783

CD (6.20) 0 19.137 0 221.783

DE --- 19.099 19.099 0 0

EA (6.25) 0 19.133 10.141 0

max max 241.704 mm/rad 41.888 rad/s 10 124 mm/sy y Ans.

min min 19.137 mm/rad 41.888 rad/s 800.0 mm/sy y Ans.

22 2

max max 221.783 mm/rad 41.888 rad/s 389 141 mm/sy y Ans.

22 2

min min 221.783 mm/rad 41.888 rad/s 389 141 mm/sy y Ans.



6.10 A plate cam with an oscillating follower is to rise through 20 in 60 of cam rotation,

dwell for 45 , then rise through an additional 20 , return, and dwell for 60 of cam

rotation. Assuming high-speed operation, recommend standard cam motions from

Section 6.7 to be used, and determine the lifts and cam-rotation angles for each segment

of the cam.

From the sketches shown (not drawn to scale), the curve types identified in the table

below were chosen.

Next, equating the second derivatives at D, the remaining entries in the table were found.

3 4

2 2

3 4

5.26830 5.26830L L

2

4 4

2

3 3

2.000L

L

4 32

3 4 31 2 195

3 80.772

4 114.228

Seg. Type Eq. L, deg rad , deg

AB cycloidal (6.13) 20.000 1.047 198 60.000

BC dwell --- 0 0.785 399 45.000

CD 8th

order poly. (6.14) 20.000 1.409 731 80.772

DE 8th

order poly. (6.17) 40.000 1.993 661 114.228

EA dwell --- 0 1.047 198 60.000



6.11 Determine the maximum velocity and acceleration of the follower for Problem 6.10,

assuming that the cam is driven at a constant speed of 600 rev/min.

Working from the equations listed, the maximum and minimum values of the kinematic

coefficients in each segment of the cam are as follows:

Seg. Eq. maxy miny maxy miny

AB (6.13) 0.666 667 0 2.000 000 2.000 000

BC --- 0 0 0 0

CD (6.14) 0.440 004 0 0.925 344 0.924 344

DE (6.17) 0 0.622 264 0.925 402 0.925 402

EA --- 0 0 0 0

600 rev/min 2 rad/rev 60 s/min 62.832 rad/s

max max 0.666 667 rad/rad 62.832 rad/s 41.888 rad/sy y Ans.

22 2

max max 2.000 rad/rad 62.832 rad/s 7 896 rad/sy y Ans.



6.12 The boundary conditions for a polynomial cam motion are as follows: for 0 , 0y ,

and 0 y ; for , , and 0y L y . Determine the appropriate displacement

equation and the first three derivatives of this equation with respect to the cam rotation

angle. Sketch the corresponding diagrams.

Since there are four boundary conditions, we choose a cubic polynomial

2 3

0 1 2 3y C C C C

2

31 232 CC C

y

Then from the boundary conditions:

00 0y C 0 0C

10 0C

y

1 0C

2 31.0y C C L 2 3C L

3232

1.0 0CC

y

3 2C L

Therefore the equation and its three derivatives are:

2 3 2 3

3 2 3 2y L L L

Ans.

2 2

6 6 6L L Ly

Ans.

2 2 26 612 1 2L LLy

Ans.

312Ly

Ans.



6.13 Determine the minimum face width using 2-mm allowances at each end and determine

the minimum radius of curvature for the cam of Problem 6.2.

Referring to Problem 6.2 for the data and figure,

40.000 mmL 180 rad 0 40.0 mmR

From Eqs. (6.12) and (6.15) for simple harmonic motion,

max 2 20.000 mm/rady L min 2 20.000 mm/rady L

From Eq. (6.30)

max minFace width y allowancesy

Face width 20.000 mm 20.000 mm 2 2 mm 44.0 mm Ans.

From Eq. (6.28):

0

0 01 cos cos (constant)2 2 2

R y y

L L LR R

40.0 mm 40.000 mm 2 60.0 mm Ans.

6.14 Determine the maximum pressure angle and the minimum radius of curvature for the cam

of Problem 6.1.

Referring to Problem 6.1 for the figure and data,

40.000 mmL 180 rad 0 40.000 mmR 7.500 mmrR

For simple harmonic motion, Eq. (6.12) can be substituted into Eq. (6.33) to give

sintan

3 cos

. This can be differentiated and d d set to zero to find the angle

70.53 at which max 19.47 . However, it is much simpler to use the nomogram of

Fig. 6.28 to find max 20 directly. For the accuracy needed, the nomogram is

considered sufficient. Ans.

From Fig. 6.30a, using 0 1.0R L , we get 0min1.43rR R . This gives

0min1.43 1.43 40 mm 7.50 mm 49.7 mmrR R Ans.



6.15 A radial reciprocating flat-face follower is to have the motion described in Problem 6.7.

Determine the minimum prime-circle radius if the radius of curvature of the cam is not to

be less than 10 mm. Using this prime-circle radius, what is the minimum length of the

follower face using allowances of 3 mm on each side?

From Problem 6.9, max 241.704 mm/rady , min 19.137 mm/rady ,

2

min 221.783 in/rady

Therefore, from Eq. (6.29),

0 min min 10.000 mm 221.783 mm 50.000 mm 181.783 mmR y y Ans.

Also, from Eq. (6.30),

max minFace width allowances 241.704 mm 19.137 mm 2 3.0 mm 266.84 mmy y Ans.

6.16 Graphically construct the cam profile of Problem 6.15 for clockwise cam rotation.



6.17 A radial reciprocating roller follower is to have the motion described in Problem 6.7.

Using a prime-circle radius of 400 mm, determine the maximum pressure angle and the

maximum roller radius that can be used without producing undercutting.

We will use the nomogram of Fig. 6.28 to find the maximum pressure angle in each

segment of the cam. Calculations are shown in the following table. Asterisks are used to

signify values used with the nomogram to adjust half-return curves to equivalent full-

return curves, and to adjust the prime-circle baseline.

Seg. *

0R , mm *L , mm * *

0R L * , deg max , deg

BC 400.000 50.000 8.0 62.4 12

CD 446.704 3.296 135.5 15.5 1

EA 400.000 56.704 7.0 339.6 3

For the total cam, max 12 . Ans.

Also we use Figs. 6.32 and 6.33 to check for undercutting. Again, asterisks are used to

denote values that are adjusted for use with the charts. Note that doubling as was done

for use of the nomogram is not necessary since we have figures for half-harmonic and

half-cycloidal cam segments. Note also that segment EA need not be checked since

undercutting occurs only in segments with negative acceleration.

Seg. *

0R , mm L, in *

0R L , deg *

0min rR R max

rR mm

BC 400.000 50.000 8.0 62.1 0.725 290

CD 448.352 1.648 272.1 7.7 0.680 304

To avoid undercutting for the entire cam, 290 mmrR . Ans.

6.18 Graphically construct the cam profile of Problem 6.17 using a roller radius of 15 mm.

The cam rotation is to be clockwise.



6.19 A plate cam rotates at 300 rev/min and drives a reciprocating radial roller follower

through a full rise of 75 mm in 180° of cam rotation. Find the minimum radius of the

prime-circle if simple harmonic motion is used and the pressure angle is not to exceed

25 . Find the maximum acceleration of the follower.

Using max 25 and 180 , Fig. 6.28 gives 0 0.75R L . Therefore

0 0.75 0.75 75 mm 56.25 mmR L Ans.

300 rev/min 2 rad/rev31.416 rad/s

60 s/min

222

max 22

75 mm37.5 mm/rad

2 2 rad

Ly

22 2 2

max max 37.5 mm/rad 31.416 rad/s 37000 mm/sy y Ans.

6.20 Repeat Problem 6.19 except that the motion is cycloidal.

Figure 6.28 gives 0 0.95R L . Therefore 0 0.95 0.95 75 mm 71.25 mmR L Ans.

2

max 22

2 75 mm247.7475 mm/rad

rad

Ly

22 2 2


6.21 Repeat Problem 6.19 except that the motion is eighth-order polynomial.

Figure 6.28 gives 0 0.95R L . Therefore 0 0.95 0.95 75 mm 71.25 mmR L Ans.

2

max 22

5.2683 75 mm5.268340.025 mm/rad

rad

Ly

22 2 2


6.22 Using a roller diameter of 0.80 in, determine whether the cam of Problem 6.19 will be

undercut.

Using 0 0.75R L and 180 , Fig. 6.30a gives min 0 1.55rR R .

min 1.55 2.25 in 0.80 in 2.688 in 0 ; thus, this cam is not undercut. Ans.



6.23 Equations (6.30) and (6.31) describe the profile of a plate cam with a reciprocating flat-

face follower. If such a cam is to be cut on a milling machine with cutter radius cR ,

determine similar equations for the center of the cutter.

In complex polar notation, using Eq. (6.27) and using u and v to denote the local

rectangular part coordinates of the cam shape, the loop-closure equation is

0

j j

cue jve jR jy y jR

Dividing this by je

0

j j

cu jv j R R y e y e

Now separating this into real and imaginary parts we find

0 sin coscu R R y y

0 cos sincv R R y y Ans.



6.24 & 6.25 Since programming languages vary so much, particularly with the use of

graphics, no attempt is made to show a “standard” solution for these problems.

6.26 A plate cam with an offset reciprocating roller follower is to have a dwell of 60° and then

rise in 90° to another dwell of 120°, after which it is to return in 90° of cam rotation. The

radius of the base circle is 40 mm, the radius of the roller follower is 15 mm, and the

follower offset is 20 mm. For the rise motion 60 150 , the equation of the

displacement (the lift) is to be

40( sin )y

where y is in millimeters and is the cam rotation angle in radians. (i) Find equations

for the first- and second-order kinematic coefficients of the lift y for this rise motion. (ii)

Sketch the displacement diagram and the first- and second-order kinematic coefficients

for the follower motion described. Comment on the suitability of this rise motion in the

context of the other displacements specified. At the cam angle θ = 120°, determine the

following: (iii) the location of the point of contact between the cam and follower,

expressed in the moving Cartesian coordinate system attached to the cam; (iv) the radius

of the curvature of the pitch curve and the radius of curvature of the cam surface; and (v)

the pressure angle of the cam. Is this pressure angle acceptable?

(i) From the equation given for the lift, the first- and second-order kinematic

coefficients are

1

40 cos mm/rady

and 240sin mm/rady

(ii) Sketches of the first-order and the second-order kinematic coefficients of the

displacement diagram are shown here.

0

1

2

3

4

5

6

60 80 100 120 140

θ (degree)

y' (c

m/r

ad)

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

60 80 100 120 140

θ (degree)

y''

(cm

/ra

ds)

The rise motion specified is not suitable between dwells on either side since the first- and

second-order kinematic coefficients are not zero at the beginning and end of the rise. A

cycloidal rise curve would be preferable, but would have higher values of acceleration

and would lead to higher forces.

At the cam angle θ = 120º, we have 60 60 3 rad .

340( sin ) 40 sin 60 48.0 mmy



1 140 cos 40 cos60 32.7 mm/rady

240sin 40sin60 34.6 mm/rady

The absolute coordinates of the trace point are

0 20 mm,X 2 2 22

0 40 mm 15 mm 20 mm 51.2 mmrY R R

0 20 mm,X X 0 51.2 mm 48.0 mm 99.2 mmY Y y

The cam coordinates of the trace point (the pitch curve) and derivatives are

cos sin 20 mm cos120 99.2 mm sin120 75.9 mmu X Y

sin cos 20 mm sin120 99.2 mm cos120 66.9 mmv X Y

sin cos sin 38.6 mm/radu X Y y

cos sin cos 92.3 mm/radv X Y y

2 2 100.0 mm/radw u v 2cos sin 2 cos sin 13.87 mm/radu X Y y y 2sin cos 2 sin cos 2.76 mm/radv X Y y y

(iii) From Eq. (6.38), the cam coordinates of the point of contact (the cam surface) are

62.1 mmcam r

vu u R

w

61.1 mmcam r

uv v R

w

Ans.

(iv) From Eq. (6.39), the radius of curvature of the pitch curve is 3

72.2 mmw

u v v u

Ans.

(v) From Eq. (6.39) the pressure angle is

cos sin cos 0.9919v u

w w

7.3 Ans.

This pressure angle is less than 30° and is acceptable.



6.27 A plate cam with an offset reciprocating roller follower is to be designed using the input,

the rise and fall, and the output motion shown in Table P6.27. The radius of the base circle

is to be 30 mm, the radius of the roller follower is 12.5 mm, and the follower offset (or

eccentricity) is to be 15 mm.

Table P6.27 Displacement Information for Plate Cam With Reciprocating Roller Follower

Cam angle range Rise or Fall (mm) Follower motion

0° - 20° 0 Dwell

20° - 110° +25 Full-rise simple harmonic motion

110° - 120° 0 Dwell

120° - 200° +5 Full-rise cycloidal motion

200° - 270° 0 Dwell

270° - 360° -30 Full-return cycloidal motion

Comment on the suitability of the motions specified. At the cam angle 50 ,

determine the following: (i) the first-, second-, and third-order kinematic coefficients of

the lift curve, (ii) the coordinates of the point of contact between the roller follower and

the cam surface, expressed in the Cartesian coordinate system rotating with the cam, (iii)

the radius of curvature of the pitch curve, (iv) the unit tangent and the unit normal vectors

to the pitch curve, and (v) the pressure angle of the cam.

The portion of the motions which specifies simple harmonic motion is not suitable for

high-speed operation since there will be discontinuities in the second derivatives at both

ends of that segment where it interfaces with dwells. Cycloidal motion would correct

this problem but would give a higher peak acceleration. Still, simple harmonic motion is

specified.

(i) Eqs. (6.12), with 50 20 30 , 25 mm, 90 2 rad,L gives

25 mm1 cos 1 cos 6.25 mm

2 2 3

Ly

sin 25 mm sin 21.65 mm/rad2 3

Ly

Ans.

2

2

2cos 2 25 mm cos 25.0 mm/rad

2 3

Ly

Ans.

3

3

3sin 4 25 mm sin 86.60 mm/rad

2 3

Ly

Ans.

Therefore the fixed coordinates of the tracepoint are

0 30 mm 12.5 mm 42.5 mm,rR R R 2 22 2

0 0 42.5 mm 15 mm 39.76 mmY R

15 mm,X 0 39.76 mm 6.25 mm 46.01 mmY Y y

The cam coordinates of the trace point (the pitch curve) and derivatives are

cos sin 15 mm cos50 46.01 mm sin50 44.89 mmu X Y

sin cos 15 mm sin50 46.01 mm cos50 18.08 mmv X Y

sin cos sin 34.67 mm/radu X Y y



cos sin cos 30.97 mm/radv X Y y

2 2 46.49 mm/radw u v 2cos sin 2 cos sin 2.10 mm/radu X Y y y

2sin cos 2 sin cos 35.18 mm/radv X Y y y

(ii) From Eq. (6.38), the cam coordinates of the point of contact (the cam surface) are

36.56 mmcam r

vu u R

w

8.76 mmcam r

uv v R

w

Ans.

(iii) From Eq. (6.39), the radius of curvature of the pitch curve is

3 87.02 mmw u v v u Ans.

(iv) The unit tangent and the unit normal vectors to the pitch curve are

ˆ ˆ ˆ ˆˆ 0.746 0.666t u w v w u i j i j Ans.

ˆ ˆ ˆ ˆˆ 0.666 0.746n v w u w u i j i j Ans.

(v) From Eq. (6.39) the pressure angle is

cos sin cos 0.9897v w u w 8.2 Ans.

This pressure angle is less than 30° and is acceptable for the specified cam angle 50 .



6.28 A plate cam with a radial reciprocating roller follower is to be designed using the input,

the rise and fall, and the output motion shown in Table P6.28. The base circle diameter is

3 in and the diameter of the roller is 1 in. Displacements are specified as follows.

Table P6.28 Displacement Information for Plate Cam With Reciprocating Roller Follower Input (deg) Lift L (in) Output y

0 90 3.0 Cycloidal Rise

90 105 0 Dwell

105 195 -3.0 Cycloidal Fall

195 210 0 Dwell

210 270 2.0 Simple Harmonic Rise

270 285 0 Dwell

285 345 -2.0 Simple Harmonic Fall

345 360 0 Dwell

Plot the lift curve (the displacement diagram), and the profile of the cam. (i) Comment

on the lift curves at appropriate positions of the cam, (for example, when the cam angle is

0 , 45 , 180 , 210 , 225 , and 300 ). (ii) Identify on your

cam profile the location(s) and the value of the largest pressure angle. Would this

pressure angle cause difficulties for a practical cam-follower system? (iii) Identify on

your cam profile the location(s) of discontinuities in position, velocity, acceleration,

and/or jerk. Are these discontinuities acceptable (why or why not)? (iv) Identify on your

cam profile any regions of positive radius of curvature of the cam profile. Are these

regions acceptable (why or why not)? (v) For the values given in Table P6.28, what

design changes would you suggest to improve the cam design?

The lift curve (displacement diagram is shown in Fig. 1.

Figure 1. The lift curve or displacement diagram.

The cam profile is plotted in Fig. 2.



Figure 2. The cam profile.

(i) Because of the choice of harmonic motion rise and return curves, there are

discontinuities in acceleration at 210 , 270 , 285 , and 345 . Because

of adjacent dwells, cycloidal motion would be preferable, although it would lead to

slightly higher peak acceleration in these segments.

(ii) The pressure angle, see Sec. 6.10, should be less than 30 . In this design, the

pressure angle is more than the accepted value at the cam angles

16 64 , 131 180 , 216 256 , and 299 341

Therefore, this cam profile is not a good design. The high values of the pressure angle

may be due to the selection of the displacement curves and the diameter of the base circle

and the diameter of the follower.

(iii) Position discontinuities never occur. Discontinuities in the derivatives will occur

only at transitions between dwell segments and lifting/returning segments of motion.

Discontinuities in the derivatives are undesirable. There is an acceleration discontinuity

at the beginning and end of the simple harmonic motions, both rise and return. There is a

jerk discontinuity at the beginning and end of the cycloidal motions, both rise and return.

Whether these discontinuities are acceptable depends on the intended speed of operation,

and on the masses and stiffnesses involved.



(iv) The radius of curvature of a cam profile should always be negative for a good

cam design. Positive curvature means that the cam has a concave surface and there is the

possibility that the follower may lose contact with the cam. If the radius of curvature of

the cam profile is positive then the radius of curvature of the cam must be greater than the

radius of the follower. In the proposed design, the positive values of the radius of

curvature of the cam are always greater than 0.5 in (i.e., the radius of the follower). The

radius of curvature of the cam is positive for the cam angles

3 25 , 170 192 , 215 220 , and 261 270

The radius of curvature is positive, and smaller than the radius of the roller follower, for

the cam angles

9 13 , 181 187 , and 215 216

Note that the radius of curvature of the cam is zero between the cam angles 214 and 215

degrees, meaning that pointing has occurred. Also, it could imply that undercutting has

occurred. Also, with the exception of where the radius of curvature of the cam goes to

zero, there is an inflection point at the boundary of each range of angles for which the

radius of curvature is positive.

(v) Possible design changes to the cam-follower system. (a) Increasing the radius of

the prime circle (with the same lift curve) in general would reduce the pressure angle. (b)

Change the profiles to match acceleration at the transistion (blend) points to eliminate

acceleration discontinuities. (c) Changing both SHM profiles to cycloidal would make

accelerations continuous but would also increase accelerations (and the pressure angle) in

the middle parts of the rise and the return profiles. (d) We may want to increase the

diameter of the roller if the contact stresses are high. (e) We could explore numerically

the effects on forces of changing the offset (or the eccentricity) . These are not obvious

from observation.



6.29 Continue using the same displacement information and the same design parameters as in

Problem 6.28. Use a spreadsheet to determine and plot the following for a complete

rotation of the cam: (i) the first-order kinematic coefficients of the follower center; (ii)

the second-order kinematic coefficients of the follower center; (iii) the third-order

kinematic coefficients of the follower center; (iv) the lift curve (displacement diagram);

(v) the radius of curvature of the cam surface; and (vi) the pressure angle of the cam-

follower system. Is the pressure angle suitable for a practical cam-follower system?

(i) The first-order kinematic coefficients for the cam design are shown in Fig. 3.

0 50 100 150 200 250 300 350 400-6

-4

-2

0

2

4

6First order Kinematics Coeeficients

(Cam angle) - degrees

First

order

KC - i

n

xfc

'

yfc

'

Figure 3. The first-order kinematic coefficients.

(ii) The second-order kinematic coefficients for the cam design are shown in Fig. 4.

0 50 100 150 200 250 300 350 400-15

-10

-5

0

5

10

15Second order Kinematics Coeeficients


Secon

d orde

r KC -

in

xfc

pp"

yfc

"

Figure 4. The second-order kinematic coefficients.



(iii) The third-order kinematic coefficients for the cam design are shown in Fig. 5.

0 50 100 150 200 250 300 350 400-40

-30

-20

-10

0

10

20

30

40Third order Kinematics Coeeficients


Third

order

KC - i

n

xfc

p"'

yfc

"'

Figure 5. The third-order kinematic coefficients.

(iv) The lift curve (displacement diagram) for the cam design is shown in Fig. 6.

0 50 100 150 200 250 300 350 4000

0.5

1

1.5

2

2.5

3Position Profile


Lift - i

n

Figure 6. The lift curve (displacement diagram).



(v) The radius of curvature of the cam surface is shown in Fig. 7.

Figure 7. The radius of curvature of the cam surface.

The radius of curvature of a cam profile should always be negative for a good cam

design. The positive curvature means that the cam has a concave surface and there is the

possibility that the follower may lose contact with the cam. If the radius of curvature of

the cam profile is positive then the radius of curvature of the cam must be greater than the

radius of the follower. In the proposed design, the positive values of the radius of

curvature of the cam are always greater than 0.5 in (i.e., the radius of the follower).

Note that the radius of curvature of the cam surface goes to infinity when the cam angle

is 4°, 26°, 168°, 191°, 225°, and 330°; i.e., there are six inflection points on the cam

surface. Also, note the discontinuity at the start and the end of the simple harmonic

profile at 210°, 270°, 285°, and 345°. This discontinuity is due to the simple harmonic

profiles and for this reason simple harmonic profiles are not generally recommended for

high-speed cam-follower systems.

(vi) The pressure angle for the cam-follower system is shown in Fig. 8.

0 50 100 150 200 250 300 350 4000

5

10

15

20

25

30

35

40

45

50Pressure Angle


Pres

sure a

ngle -

degre

es

Figure 8. The pressure angle of the cam-follower system.

The recommended value of the pressure angle is less than 30 degrees. In this design, the

pressure angle is more than the accepted value at the cam angles 2 18 62 ,

134 178 , 220 256 , and 302 336 . Therefore, this cam profile is not a good

design. The high values of the pressure angle may be due to the selection of the

displacement curves and the dimensions of the cam and the diameter of the roller.



Chapter 7

Spur Gears

7.1 Find the module of a pair of gears having 32 and 84 teeth, respectively, whose center distance is 87 mm.

322 3

mm m 32 84 87 mm2 2 2

NNR R

2 87 mmm 1.5 mm/tooth116 teeth

Ans.

7.2 Find the number of teeth and the circular pitch of a 150-mm pitch diameter gear whose module is 2.5 mm/tooth.

2 m 60 teeth150 mm 2.5 mm/toothN R Ans. m 7.854 mm/tooth2.5 mm/toothp Ans.

7.3 Determine the module pitch of a pair of gears having 18 and 40 teeth, respectively, whose center distance is 90.625 mm.

2 32 3 m

2

5890.625 m

2

2 90.625 m =

583.125 mm/tooth

N NR R

Ans.

7.4 Find the number of teeth and the circular pitch of a gear whose pitch diameter is 10.5 mm in if module is 3.125 mm/tooth.

2125 mm / 3.125 mm per tooth

m

RN Ans.

= 3.14 3.125 = 9.8125 mm/toothp m Ans.



7.5 Find the module and the pitch diameter of a 40-tooth gear whose circular pitch is 37.7 mm/tooth.

m 12 mm/tooth37.7 mm/toothp Ans. 2 480 mm12 mm/tooth 40 teethD R mN Ans.

7.6 The pitch diameters of a pair of mating gears are 42 mm and 102 mm, respectively. If the module is 1.5 mm/tooth how many teeth are there on each gear?

2 2 22 m m 28 teeth1.5 mm/tooth42 mmN R D Ans.

3 3 32 m 68 teeth102 mm 1.5 mm/toothN R D m Ans.

7.7 Find the module and the pitch diameter of a gear whose circular pitch is 9.815 mm/tooth if the gear has 36 teeth.

m is circular pitch

9.815=

3.14= 3.125 mm/tooth

pwhere P

Ans.

36 3.125 = 35.83mm

3.14

NMD

Ans.

7.8 The pitch diameters of a pair of gears are 62 mm and 100 mm, respectively. If their module is 2 mm per/tooth, how many teeth are there on each gear?

2 2 22 / m / m 62 mm / 2 mm per tooth 31 teethN R D Ans.

3 3 32 / m / m 100 mm / 2 mm per tooth 50 teethN R D Ans.

7.9 What is the diameter of a 33-tooth gear if its module is 2 mm/tooth?

2 2 mm/tooth 33 teeth 66 mmD R mN Ans.

7.10 A shaft carries a 30-tooth, 3-mm/tooth module gear that drives another gear at a speed of 480 rev/min. How fast does the 30-tooth gear rotate if the shaft center distance is 105 mm?

2 2 2 30 teeth 2 45 mm3 mm/toothR mN

3 2 3 2 105 mm 45 mm 60 mmR R R R

32 3

2

60 mm480 rev/min 640 rev/min

45 mm

R

R Ans.



7.11 Two gears having an angular velocity ratio of 3:1 are mounted on shafts whose centers are 150 mm apart. If the module of the gears is 3 mm/tooth, how many teeth are there on each gear?

22 3 2 2 2

3

3150 mm 1 1 4

1R R R R R

2 37.5 mm inR

2 22 / m 2 25 teeth37.5 mm / 3 mm / toothN R Ans.

3 3 2 2 112.5 mmR R R R

3 32 / m 2 75 teeth112.5 mm/ 3 mm/ toothN R Ans.

7.12 A gear having a module of 4 mm/tooth and 21 teeth drives another gear at a speed of 240 rev/min. How fast is the 21-tooth gear rotating if the shaft center distance is 170 mm?

2 2m 2 4 21/ 2 42 mmR N

3 2 3 2 170 mm 42 mm 128 mmR R R R

32 3

2

128 mm240 rev/min 731.4 rev/min

42 mm

R

R Ans.

7.13 A 6.35 mm/tooth module, 24-tooth pinion is to drive a 36-tooth gear. The gears are cut on the 20° full-depth involute system. Find and tabulate the addendum, dedendum, clearance, circular pitch, base pitch, tooth thickness, pitch circle radii, base circle radii, length of paths of approach and recess, and contact ratio.

m 6.35 mm / tooth 6.35 mma Ans. 1.25 m 1.25 6.35 mm / tooth 7.9375 mmd Ans.

7.9375 mm 6.35 mm 1.5875 mm c d a Ans.

πm =(π 6.35mm / tooth) 19.94 mm/toothp Ans.

cos 19.94 mm/tooth cos 20 18.737mm/toothbp p Ans.

2 19.94 mm/tooth 2 9.97 mmt p Ans.

2 2

6.35 mm/tooth 24 teethmN 2 76.2 mm

2R

3 3

6.35 mm/tooth 36 teethmN 2 114.3 mm

2R

Ans.



2 2 cos 76.2 mm cos 20 71.6 mmr R ; 3 3 cos 114.3 mm cos 20 107.4 mmr R Ans.

15.875 mmCP [measured or by Eq. (7.10)] Ans. 15 mmPD [measured or by Eq. (7.11)] Ans.

15.875 mm 15 mm1.647 teeth avg.

18.737 mm/toothcb

CP PDm

p

Ans.



7.14 A 5 mm/tooth module, 15-tooth pinion is to mate with a 30-tooth internal gear. The gears are 20° full-depth involute. Make a drawing of the gears showing several teeth on each gear. Can these gears be assembled in a radial direction? If not, what remedy should be used?

Since the addendum circle of internal gear 3 is of lesser radius (71.12 mm) than its base circle (70.475 mm), contact is initiated to the left of point A before proper involute contact is possible. This is similar to undercutting but on an internal gear it is called fouling.

With this condition the involute curves of the internal gear are extended radially to meet the addendum circle and this results in converging radii; therefore the gears cannot be assembled in the radial direction. Ans.

One remedy is to reduce the internal gear addendum to match the base circle radius. However, the internal gear is then non-standard. A better remedy is to increase the module to 6.35 mm/tooth so that the addendum circle of the internal gear is 63.5 mm.



7.15 A 10 mm/tooth module, pitch 17-tooth pinion and a 50-tooth gear are paired. The gears are cut on the 20° full-depth involute system. Find the angles of approach and recess of each gear and the contact ratio.

m 10 mm/tooth = 10 mma πm 10 mm/tooth = 31.4 mmp

cos 31.4 mm/tooth cos 20 29.5 mm/toothbp p

2 2

10 mm/tooth 17 teethm 2 85 mm

2R N

3 3

10 mm/tooth 50 teethm 2 250 mm

2R N

2 2 cos 85 mm cos 20 79.87 mmr R

3 3 cos 250 mm cos 20 234.923 mmr R

26.314 mmCP [Eq. (7.10)] 22.707 mmPD [Eq. (7.11)]

22

26.314 mm0.324 rad 18.58

79.87 mm

CP

r 3

3

26.314 mm0.110 rad 6.32

234.923 mm

CP

r Ans.

22

22.707 mm0.280 rad 16.04

79.87 mm

PD

r 3

3

22.707 mm0.095 rad 5.45

234.973 mm

PD

r Ans.

26.314 mm 22.707 mm1.66 teeth avg.

29.5 mm/toothcb

CP PDm

p

Ans.

7.16 A gearset with a module of 5 mm/tooth has involute teeth with 22½° pressure angle, and has 19 and 31 teeth, respectively. They have 1.0m for the addendum and 1.25m for the dedendum.* Tabulate the addendum, dedendum, clearance, circular pitch, base pitch, tooth thickness, base circle radius, and contact ratio.

1.0 5.0 mma m Ans.

1.35 1.35 5 mm 6.75 mmd m Ans. 1.75 mmc d a Ans.

5 mm/tooth 15.708 mm/toothp m Ans.

cos 15.708 mm/tooth cos 22.5 14.512 mm/toothbp p Ans. 2 7.854 mmt p Ans.

2 2 2 19 teeth 5 mm/tooth 2 47.500 mmR N m


2 2 cos 47.500 mm cos 22.5 43.884 mmr R Ans.

3 3 cos 77.500 mm cos 22.5 71.601 mmr R Ans. 11.325 mmCP [Eq. (7.10)] 10.640 mmPD [Eq. (7.11)]

11.325 mm 10.640 mm1.51 teeth avg.

14.512 mm/toothcb

CP PDm

p

Ans.

* In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.



7.17 A gear with a module of 8 mm/tooth and 22 teeth is in mesh with a rack; the pressure angle is 25°. The addendum and dedendum are 1.0m and 1.25m, respectively.* Find the lengths of the paths of approach and recess and determine the contact ratio.

1.0 8.0 mma m

8 mm/tooth 25.133 mm/toothp m



sin 18.930 mmCP a [Fig. 7.10] Ans. 16.243 mmPD [Eq. (7.11)] Ans.

18.930 mm 16.243 mm1.54 teeth avg.

22.778 mm/toothcb

CDm

p

Ans.

7.18 Repeat Problem 7.15 using the 25° full-depth system.

m 10 mm/tooth = 10 mma m 10 mm/tooth = 31.4 mm/toothp


2 2

10 mm/tooth 17 teethm / 2 85 mm

2R N

3 3

10 mm/tooth 50 teethm / 2 250 mm

2R N

2 2 cos 85 mm cos 25 77 mmr R 3 3 cos 250 mm cos 25 226.576 mmr R

22.225 mmCP [Eq. (7.10)] 19.989 mmPD [Eq. (7.11)]

22

22.225 mm0.288 rad 16.50

77 mm

CP

r 3

3

22.225 mm0.098 rad 5.61

226.576 mm

CP

r Ans.

22

19.989 mm0.259 rad 14.85

77 mm

PD

r 3

3

19.989 mm0.088 rad 5.04

226.576 mm

PD

r Ans.

22.225 mm 19.989 mm1.48 teeth avg.

28.458 mm/toothcb

CDm

p

Ans.



7.19 Draw a 12.7 mm/tooth module, 26-tooth, 20° full-depth involute gear in mesh with a rack. (a) Find the lengths of the paths of approach and recess and the contact ratio. (b) Draw a second rack in mesh with the same gear but offset 3.175 mm in further

away from the gear center. Determine the new contact ratio. Has the pressure angle changed?

12.7 mm

3.175 mm

(a) sin 12.7 mm sin 20 37.132 mmCP a [see figure] Ans. 30.3784 mmPD [Eq. (7.11)] Ans.

37.132 mm 30.3784 mm1.80 teeth avg.

37.49 mm/toothcb

CDm

p

Ans.

(b) Since the pressure angle is a property that has determined the shapes of the teeth on both the rack and the pinion, moving the rack by 3.175 mm does not change the tooth shapes or the pressure angle. The modified contact ratio is:

sin 9.525 mm sin 20 27.849 mmC P a [see figure] Ans. 30.378 mmP D [Eq. (7.11)] Ans.

27.849 mm 30.378 mm1.55 teeth avg.

37.49 mm/toothcb

C Dm

p

Ans.



7.20 through 7.24 Shaper gear cutters have the advantage that they can be used for either

external or internal gears and also that only a small amount of runout is necessary at the end of the stroke. The generating action of a pinion shaper cutter can easily be simulated by employing a sheet of clear plastic. The figure illustrates one tooth of a 16-tooth pinion cutter with 20° pressure angle as it can be cut from a plastic sheet. To construct the cutter, lay out the tooth on a sheet of drawing paper. Be sure to include the clearance at the top of the tooth. Draw radial lines through the pitch circle spaced at distances equal to one fourth of the tooth thickness as shown in the figure. Next, fasten the sheet of plastic to the drawing and scribe the cutout, the pitch circle, and the radial lines onto the sheet. Then remove the sheet and trim the tooth outline with a razor blade. Then use a small piece of fine sandpaper to remove any burrs.

To generate a gear with the cutter, only the pitch circle and the addendum circle need be drawn. Divide the pitch circle into spaces equal to those used on the template and construct radial lines through them. The tooth outlines are then obtained by rolling the template pitch circle upon that of the gear and drawing the cutter tooth lightly for each position. The resulting generated tooth upon the gear will be evident. The following problems all employ a standard 1-tooth/in diametral pitch 20 full-depth template constructed as described above. In each case you should generate a few teeth and estimate the amount of undercutting

Problem No. 7.20 7.21 7.22 7.23 7.24

No. of Teeth 10 12 14 20 36

The diagram used to make the plastic template for Problems 7.20 through 7.24 is shown at the left.

The drawing generated for Problem 7.20 is also shown at left. Note how the tip(s) of the shaper cutter slightly cut away the material at the flank of the tooth so that the tooth is a small amount narrower here than at its thickest radius. This is the meaning of the term undercut.

Problems 7.21 to 7.24 are similar.



7.25 A 10-mm/tooth module gear has 17 teeth, a 20° pressure angle, an addendum of 1.0m, and a dedendum of 1.25m.* Find the thickness of the teeth at the base circle and at the addendum circle. What is the pressure angle corresponding to the addendum circle?

At the pitch circle:

2 10 mm/tooth 17 teeth 2 85.0 mmpr R mN

85.0 mm 17 15.708 mmpt R N

inv inv 20 0.014 904 At the base circle:

cos 85.0 mm cos 20 79.874 mmbr r R

inv inv0 0.0 From Eq. (7.16)

2 inv inv2

ptt r

R

15.708 mm

2 79.874 mm 0.014 904 0.0 17.142 mm2 85.0 mm

t

Ans.

At the addendum circle: 85.0 mm 10.0 mm 95.0 mmar R a

1 1cos cos 79.874 mm 95.0 mm 32.78a b ar r Ans.

15.708 mm

2 95.0 mm 0.014 904 0.071 844 6.737 mm2 85.0 mmat

Ans.

* In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.



7.26 A 15-tooth pinion has 16.9 mm/tooth module, 20° full-depth involute teeth. Calculate the thickness of the teeth at the base circle. What are the tooth thickness and the pressure angle at the addendum circle?

At the pitch circle: MN 2 16.9 mm/tooth 15 teeth 2 126.75 mmpr R

126.75 mm 15 teeth 26.533 mmpt R N




2 inv inv2

ptt r

R

26.533 mm

2 119.1 mm 0.014 904 0.0 28.48 mm2 126.75 mm

t

Ans.

At the addendum circle: 126.75 mm 16.9 mm 143.65 mmar R a

1 1cos cos 119.1 mm 143.65 mm 33.99b ar r Ans.

26.533 mm

2 143.65 mm 0.014 904 0.081 018 11.076 mm2 126.75 mm

t

Ans.

7.27 A tooth is 19.9 mm thick at a pitch circle radius of 200 mm in and a pressure angle of 25°. What is the thickness at the base circle?

At the base circle: cos 200 mm cos 25 181.26 mmbr r R


2 inv inv2

ptt r

R

19.9 mm

2 181.26 mm 0.029 975 0.0 28.9 mm2 200 mm

t

Ans.



7.28 A tooth is 39.9 mm thick at the pitch radius of 400 mm and has a pressure angle of 20°. At what radius does the tooth become pointed?

2 inv inv 02

ptt r

R

inv 2 inv 39.9 mm 2 400 mm 0.014 904 0.064779pt R

cos 400 mm cos 20 cos31.647 441.53 mmbr r Ans.

7.29 A 25˚ full-depth involute, 2.1 mm/tooth module pinion has 18 teeth. Calculate the tooth thickness at the base circle. What are the tooth thickness and pressure angle at the addendum circle?


m / 2 2.1 mm/tooth 18 teeth 2 18.9 mmpr R N

18.9 mm 18 teeth 3.297 mmpt R N



inv inv0 0.0

2 inv inv2

ptt r

R

3.297 mm

2 17.129 mm 0.029 975 0.0 4 mm2 18.9 mm

t

Ans.

At the addendum circle: 18.9 mm 2.1 mm 21 mmar R a

1 1cos cos 17.129 mm 21 mm 35.35b ar r Ans.

3.297 mm

2 21 mm 0.029 975 0.092 339 1.04 mm2 18.9 mm

t

Ans.



7.30 A nonstandard 10-tooth 3.1 mm/tooth module involute pinion is to be cut with a 22½° pressure angle. What maximum addendum can be used before the teeth become pointed?


2 3.1 mm/tooth 10 teeth 2 15.5 mmpr R MN

15.5 mm 10 teeth 4.867 mmpt R N

At the addendum circle:

2 inv inv 02

ptt r

R

inv 2 inv 4.867 mm 2 15.5 mm 0.021 514 0.18214pt R ; 42.772

cos 15.5 mm cos 22.5 cos 42.772 19.5 mmbr r

19.5 mm 15.5 mm 4 mma r R Ans.

7.31 The accuracy of cutting gear teeth can be measured by fitting hardened and ground pins in diametrically opposite tooth spaces and measuring the distance over the pins. For a 2.5 mm/tooth module 20° full-depth involute system 96-tooth gear: (a) Calculate the pin diameter that will contact the teeth at the pitch lines if there is to

be no backlash. (b) What should be the distance measured over the pins if the gears are cut

accurately?

(a) 2 2.5 mm/tooth 96 teeth 2 120 mmR MN

cos 120 mm cos 20 112.76 mmbr R

tan 112.76 mm tan 20 41 mmbr

2 2 96 teeth 0.016362 rad 0.937 5N

tan 112.76 mm tan 20.937 5 41 mm 2.14 mmbs r

2 2 2.14 mm 4.28 mmd s Ans.

(b) cos 112.76 mm cos20.9375 =120.73 mms br r

distance over pins 2 2 120.73 mm 2.14 mm 245.74 mmsr s Ans.



7.32 A set of interchangeable gears with 6.35 mm/tooth module is cut on the 20° full-depth involute system. The gears have tooth numbers of 24, 32, 48, and 96. For each gear, calculate the radius of curvature of the tooth profile at the pitch circle and at the addendum circle.

, teethN 2 mmR MN

cos , mmbr R tan , mmp br

24 76.2 71.6 26.06 32 101.6 95.47 34.74 48 152.4 143.2 52.12 96 304.8 286.4 104.24

, teethN , mmar b ar r 1cos , dega b ar r tan , mma b ar

24 82.55 0.86741 29.84 41.07 32 107.95 0.88442 27.82 50.36 48 158.75 0.90211 25.56 68.50 96 311.15 0.92052 23.00 121.56

7.33 Calculate the contact ratio of a 17-tooth pinion that drives a 73-tooth gear. The gears are 0.26 mm/tooth module and cut on the 20° fine pitch system.

m 0.26 mm/tooth 0.26 mma

m cos 0.26 mm/tooth cos 20 0.767 mm/toothbp

22

0.26 17 teeth2.21 mm

2 2

MNR

3

3

0.26 73 teeth9.49 mm

2 2

MNR

0.708 mmCP [measured or by Eq. (7.10)] 0.591 mmPD [measured or by Eq. (7.11)]

0.708 mm 0.591 mm1.693 teeth avg.

0.767 mm/toothcb

CDm

p

Ans.

7.34 A 25° pressure angle 11-tooth pinion is to drive a 23-tooth gear. The gears have a module of 3.175 mm/tooth and have stub teeth. What is the contact ratio?

0.8 m 0.8 3.175 mm/tooth = 2.54 mma (Notice the stub teeth.)

πm cos 3.175 mm/tooth cos 25 9 mm/toothbp

22

3.175 11 17.4625 mm

2 2

MNR

3

3

3.175 23 teeth36.5125 mm

2 2

MNR


5.3 mm 4.85 mm1.127 teeth avg.

9 mm/toothcb

CDm

p

Ans.



7.35 A 22-tooth pinion mates with a 42-tooth gear. The gears have full-depth involute teeth, have a module of 1.5 mm/teeth, and are cut with a 17½° pressure angle.* Find the contact ratio.

1.5 mm/tooth 1.5 mma M

m cos 1.5 mm/tooth cos17.5 4.49 mm/toothbp

22

1.5 22 teeth16.5 mm

2 2

MNR

3

3

1.5 42 teeth31.5 mm

2 2

MNR


4.43 mm 3.997 mm1.877 teeth avg.

4.49 mm/toothcb

CDm

p

Ans.

7.36 The center distance of two 24-tooth, 20 pressure angle, full-depth involute spur gears with module of 12.7 mm/tooth is increased by 3.17 mm over the standard distance. At what pressure angle do the gears operate?

The original two gears are identical.

2 3

m

212.7 24 teeth

2152.4 mm

NR R

When the gear centers are separated to a non-standard distance, the base circles do not change. The line of contact adjusts to remain tangent to the new locations of the two base circles. The pressure angle changes. Since the two base circles do not change,

2 3 152.4 mm cos 20

143.2 mm

r r

From the figure we can see that the shaft center distance is related to the new pressure angle as follows

3 2 322 3 cos cos cos

r r rrR R

1 12 3

2 3

143.2 mm 143.2 mmcos cos 21.56

307.96 mm

r r

R R

Ans.

* Such gears came from an older standard and are now obsolete.



7.37 The center distance of two 18-tooth, 25 pressure angle, full-depth involute spur gears with module of 8.5 mm/tooth is increased by 1.5875 mm in over the standard distance. At what pressure angle do the gears operate?

Consult the figure and the discussion with the solution of Problem 7.36.

2 3

8.5 mm/tooth 18 teeth76.5 mm

2 2

MNR R

2 3 76.5 mm cos 25 69.3 mmr r

1 12 3

2 3

69.3 mm 69.3 mmcos cos 25.82

153.98 mm

r r

R R

Ans.

7.38 A pair of mating gears have 1 mm/tooth module and are generated on the 20° full-depth involute system. If the tooth numbers are 15 and 50, what maximum addendums may they have if interference is not to occur?

22

1 mm/tooth 15 teeth7.5 mm

2 2

MNR

3

3

1 mm/tooth 50 teeth25 mm

2 2

MNR

2 2 cos 7.5 mm cos 20 7 mmr R

3 3 cos 25 mm cos 20 23.5 mmr R

From Eq. (7.12), using Eqs. (7.10) and (7.11),

22 22 2 2 3 2sin 5.99 mma r R R R Ans.

22 23 3 2 3 3sin 1.1 mma r R R R Ans.

7.39 A set of gears is cut with a 114 mm/tooth circular pitch and a 17½° pressure angle.* The pinion has 20 full-depth teeth. If the gear has 240 teeth, what maximum addendum may it have in order to avoid interference?

114 mm/tooth / 36.3 mm/toothM P

22

36.3 mm/tooth 20 teeth363 mm

2 2

MNR

3

3

36.3 mm/tooth 240 teeth4356 mm

2 2

MNR

3 3 cos 4356 mm cos17.5 4154.39 mmr R

22 23 3 2 3 3sin 34.272 mma r R R R Ans.

* Such gears came from an older standard and are now obsolete.



7.40 Using the method described for Problems 7.20 through 7.24, cut a 25 mm/tooth module 20° pressure angle full-depth involute rack tooth from a sheet of clear plastic. Use a nonstandard clearance of 0.35 m in order to obtain a stronger fillet. This template can be used to simulate the generating action of a hob. Now, using the variable-center-distance system, generate an 11-tooth pinion to mesh with a 25-tooth gear without interference. Record the values found for center distance, pitch radii, pressure angle, gear blank diameters, cutter offset, and contact ratio. Note that more than one satisfactory solution exists.

One solution may be found by the procedure shown in the numeric example in Section 7.11 under the title of Center-Distance Modification. It proceeds as follows:

20 , 25 mm/toothM , πm 78.5 mm/toothp , m 25 mma , 1.35 m 33.75 mmd , 0.35 m 8.75 mmc ,

2 11 teethN , 3 25 teethN ,

22


2 2

MNR

3

3


2 2

MNR

2 2 cos 129.2 mmr R 3 3 cos 293.65 mmr R

2

2 sin 9 mme a R Ans.

2 2 tan 2 46.67 mmt e p 3 2 39.25 mmt p

2 2 3 2

2 2 3

2inv inv 0.022 115 rad

2

N t t R

R N N

, 22.70 Ans.

22

cos142.85 mm

cos

RR

3

3

cos324.67 mm

cos

RR

Ans.

2 3 467.52 mmR R Ans.

Working depth 50.2 mm

2 2 174.28 mmR a 3 3 343.69 mmR a Ans.

43.1 mmCP 58.68 mmPD 1.36 teeth avg.c bm CD p Ans.



7.41 Using the template cut in Problem 7.40 generate an 11-tooth pinion to mesh with a 44-tooth gear with the long-and-short-addendum system. Determine and record suitable values for gear and pinion addendum and dedendum and for the cutter offset and contact ratio. Compare the contact ratio with that of standard gears.

20 , m 25 mm/tooth , m 78.5 mm/toothp , m 25 mma , 1.35 m 33.75 mmd , 0.35 m 8.75 mmc ,


22


2 2

MNR

3

3

25 mm/tooth 44 teeth550 mm

2 2

MNR

2 2 cos 129.2 mmr R 3 3 cos 516.83 mmr R

4159

18.1

26.98

32.7

41.59

Since, with standard gears, point C is to the left of point A, there is interference and undercutting. This problem can be eliminated using the long-and-short-addendum system as shown in the figure at the left. Since the interference is near point C, we reduce the addendum of the gear until point C is coincident with point A. Combining Eqs. (7.10) and (7.12) we can show that

22 23 3 2 3 3sin 18.1 mma r R R R Ans.

Since the working depth of standard gears is retained, the new dedendum of the gear is

3 3m 1.35 m 41.59 mm ind a Ans.

Then, retaining the same clearance and pitch point P,

2 3 0.85 32.7 mma d P , 2 2m 1.35 m 26.98 mmd a Ans.

Assuming a standard rack cutter with m 25 mma , Fig. 7.26 shows the offset is

2 2 1.582 mm e a d , 3 3 16.197 mme a d Ans.

From Eqs. (7.9), (7.10), and (7.11) the contact ratio is 47.779 mmCP 63.982 mmPD

1.49 teeth avg.c bm CD p Ans. It is not easy to find a “contact ratio” for standard gears since these would have undercutting over the range CC . The distance C D is slightly less than the distance CD, but has eliminated the interference.

7.42 A pair of involute spur gears with 9 and 36 teeth are to be cut with a 20 full-depth cutter with module of 8.5 mm/tooth.



(a) Determine the amount that the addendum of the gear must be decreased in order to avoid interference.

(b) If the addendum of the pinion is increased by the same amount, determine the contact ratio.

20 , 8.5 mm/toothM , m 26.69 mm/toothp ,


22

m 8.5 mm/tooth 9 teeth38.25 mm

2 2

NR

3

3

m 8.5 mm/tooth 36 teeth153 mm

2 2

NR

2 2 cos 35.94 mmr R 3 3 cos 143.77 mmr R

(a) From Eqs. (7.10) and (7.12)

22 23 3 2 3 3sin 4.935 mma r R R R

3m - 3.565 mma Ans.

(b) 2 m 12.065 mma 3 m 4.935 mma

From Eqs. (7.9), (7.10), and (7.11) the contact ratio is 13 mmCP 22 mmPD

cos 1.40 teeth avg.cm CD p Ans.

7.43 A standard 20° pressure angle full-depth involute 25 mm/tooth module 20-tooth pinion drives a 48-tooth gear. The speed of the pinion is 500 rev/min. Using the position of the point of contact along the line of action as the abscissa, plot a curve indicating the sliding velocity at all points of contact. Notice that the sliding velocity changes sign when the point of contact passes through the pitch point.

20 , m 25 mm/tooth , m 25 mma , 2 500 rev/min 52.360 rad/s ,


2 2m 2 250mmR N 3 3m 2 600 mmR N

2 2 cos 234.92 mmr R 3 3 cos 563.82 mmr R

Defining X to be the distance from the point of contact to the pitch point along the line of action, then, since this is the distance to the instant center, the sliding velocity at the point of contact is

3 /2 3 2 2 3 21 1854.4 mm/sXV X X R R X

and, using Eqs. (7.9) and (7.10), X varies between 65.5 mminitX CP and

58.37 mmfinalX PD .



mm/s

5000

2500

-2500

-5000

-25-50-75 25 50 75-mm



Chapter 8

Helical Gears, Bevel Gears, Worms, and Worm Gears

8.1 A pair of parallel-axis helical gears has 14½° normal pressure angle, module of

3mm/teeth, and 45° helix angle. The pinion has 15 teeth, and the gear has 24 teeth.

Calculate the transverse and normal circular pitch, the normal module of 3 mm/teeth, the

pitch radii, and the equivalent tooth numbers.


m 3 mm/teeth , Ans.

m 9.43 mm/toothtp , cos 6.668 mm/toothn tp p , Ans.

2 2m 2 22.5 mmR N , 3 3m 2 36 mmR N , Ans.

3

2 2 cos 42.43 teetheN N , 3

3 3 cos 67.88 teetheN N Ans.

8.2 A set of parallel-axis helical gears are cut with a 20° normal pressure angle and a 30°

helix angle. They have module of 1.8 mm/tooth and have 16 and 40 teeth, respectively.

Find the transverse pressure angle, the normal circular pitch, the axial pitch, and the pitch

radii of the equivalent spur gears.


1tan tan cos 22.796t n , m = 1.8 mm/tooth

m 5.657 mm/toothtp , Ans.

cos 4.899 mm/toothn tp p , tan 9.798 mm/toothx tp p , Ans.

2 2m 2 14.4 mmR N , 3 3m 2 36 mmR N ,

2

2 2 cos 19 mmeR R , 2

3 3 cos 47.62 mmeR R Ans.

8.3 A parallel-axis helical gearset is made with a 20° transverse pressure angle and a 35°

helix angle.The gears have module of 2mm/tooth and have 15 and 25 teeth, respectively.

If the face width is 19 mm, calculate the base helix angle and the axial contact ratio.

1tan tan cos 33.34b t , Ans.

m 2 mm/tooth , m 6.2857 mm/toothtp ,

tan 2.006 teeth avgx tm F p Ans.



8.4 A set of helical gears is to be cut for parallel shafts whose center distance is to be about

88.9 mm in to give a velocity ratio of approximately 1.80. The gears are to be cut with a

standard 20° pressure angle hob whose module is 2.1 mm/teeth. Using a helix angle of

30°, determine the transverse values of the diametral and circular pitches and the tooth

numbers, pitch radii, and center distance.

2 3 3 2 1.8R R , 3 21.8R R ,

2 3 2 2 21.8 2.8 88.9 mmR R R R R , 2 331.75 mm, 57.15 mmR R ,

m = 2.1 mm/tooth ,

m 2.1mt = 2.425 mm/tooth

cos cos 20

, m 7.85 mm/toothtp Ans.

22

2 2 31.75 mm26.18 say 26

mt 2.425 mm/tooth

RN

, 3

3

2 2 31.75 mmNow, N 47.15 say 47

mt 2.425 mm/tooth

R ,

Therefore we will use 2 26 teethN and 3 47 teethN Ans.

22

mt 2.425 2631.525 mm

2 2

NR

, 3

3

mt 2.425 4756.987 mm

2 2

NR

, Ans.

2 3 88.512 mmR R Ans.

8.5 A 16-tooth helical pinion is to run at 1800 rev/min and drive a helical gear on a parallel

shaft at 400 rev/min. The centers of the shafts are to be spaced 275 mm apart. Using a

helix angle of 25° and a pressure angle of 20°, determine the values for the tooth

numbers, pitch radii, normal circular and module as well as and face width.

2 3 3 2 1800 400 4.5R R 3 24.5R R

2 3 2 2 24.5 5.5 275 mmR R R R R , 2 350 mm, 225 mm R R , Ans.

2 16 teethN , 3 3 2 2 24.5 72 teethN R R N N Ans.

2 2 2

2 50m 2R / 6.25 mm/tooth

16N

, nmt = m / cos 6.25 / cos25 6.896 mm/tooth , Ans.

π mn = 3.14 6.25 19.625 mm/toothtp , 0.711 8 in/toothn np P , Ans.

tan 19.625 / tan 25 42.08 mm/toothx tp p 2 teeth 2 42.08 84.16 mmxF p

Therefore, we may choose 84 mmF . Ans.



8.6 The catalog description of a pair of helical gears is as follows: 14½° normal pressure

angle, 45° helix angle, module of 3.175 mm/tooth, 25 mm in face width, and normal

module of 2.24 mm/tooth. The pinion has 12 teeth and a 37.5 mm pitch diameter, and the

gear has 32 teeth and a 100 mm pitch diameter. Both gears have full-depth teeth, and

they may be purchased either right- or left-handed. If a right-hand pinion and left-hand

gear are placed in mesh, find the transverse contact ratio, the normal contact ratio, the

axial contact ratio, and the total contact ratio.

2 37.5 mm 2 18.75 mmR , 3 100 mm 2 50 mmR ,

m 3.175 mma , πm = 9.9695 mm/toothtp ,

tan tan cos 0.365 7t n , 1tan 0.365 7 20.09 20.00t ,

cos 9.9695 cos14.5 9.36819 mm/toothb t tp p ,

2 2 cos 7.69 mmtr R , 3 3 cos 46.98 mmtr R

7.69 mmCP [Eq. (7.10)], 6.55 mmPD [Eq. (7.11)],

1.54 teeth avgt bm CD p , Ans.

1tan tan cos 43.22b t , 2cos 2.91 teeth avgn t bm m Ans.

tan 2.55 teeth avgx tm p , 4.09 teeth avgx tm m m Ans.

8.7 In a medium-sized truck transmission a 22-tooth clutch-stem gear meshes continuously

with a 41-tooth countershaft gear. The data indicate normal module of 3.34 mm/tooth,

18½° normal pressure angle, 23½° helix angle, and a 28 mm face width. The clutch-stem

gear is cut with a left-hand helix, and the countershaft gear is cut with a right-hand helix.

Determine the normal and total contact ratio if the teeth are cut full-depth with respect to

the normal diametral pitch.

tan tan cos 0.364 9t n , 1tan 0.364 9 20.04 20.00t

7.600 teeth/innP , mn 3.34 mma ,

mt mncos 3.64 mm/tooth , πm 11.429 mm/tootht tm ,

2 2m 2 40 mmR N , 3 3m 2 74.62 mmR N ,

2 2 cos 37.58 mmtr R , 3 3 cos 70.12 mmtr R

8.55 mmCP [Eq. (7.10)], 6.657 mmPD [Eq. (7.11)],

cos 9.85 mm/toothb t tp p , 1.53 teeth avgt bm CD p ,

1tan tan cos 22.22b t , 2cos 1.79 teeth avgn t bm m Ans.

tan 1.08 teeth avgx tm F p , 2.87 teeth avgx tm m m Ans.



8.8 A helical pinion is right-handed, has 12 teeth, has a 60° helix angle, and is to drive

another gear at a velocity ratio of 3.0. The shafts are at a 90° angle, and the normal

module of the gears is 2.5 mm/tooth. Find the helix angle and the number of teeth on the

mating gear. What is the shaft center distance?

3 2 30 RH , 3 2 3 2 36 teethN N Ans.

2 2 2m 2cos 30 mm R N , 3 3 3m 2cos 51.96 mmR N

2 3 81.96 mmR R Ans.

8.9 A right-hand helical pinion is to drive a gear at a shaft angle of 90°. The pinion has 6

teeth and a 75° helix angle and is to drive the gear at a velocity ratio of 6.5. The normal

module of the gear is 2.5 mm/tooth. Calculate the helix angle and the number of teeth on

the mating gear. Also determine the pitch radius of each gear.

3 2 15 RH , 3 2 3 2 39 teethN N Ans.

2 2 2m 2cos 28.97 mmR N , 3 3 3m 2cos 50.47 mmR N Ans.

8.10 Gear 2 in Fig P8.10 is to rotate clockwise and drive gear 3 counterclockwise at a velocity

ratio of 2. Use a normal module of 4.25mm/tooth, a shaft center distance of about 250

mm, and the same helix angle for both gears. Find the tooth numbers, the helix angles,

and the exact shaft center distance.

2 3 2 25 , Ans.

2 3 3 2 2.0R R , 3 22.0R R ,

2 3 2 2 22.0 3.0 250 mmR R R R R ,

2 383.33 mm, 166.66 mmR R ,

2 2 22cos / mn 35.5 teethN R ,

3 3 32cos / mn 71.1 teethN R

Therefore we choose

2 36 teethN , Ans.

3 72 teethN , Ans.

2 2 2m 2cos 84.4 mmR N ,

3 3 3m 2cos 168.8 mmR N ,

2 3 253.2 mmR R Ans.



8.11 A pair of straight-tooth bevel gears are to be manufactured for a shaft angle of 90°. If the

driver is to have 16 teeth and the velocity ratio is to be 3:1, what are the pitch angles?

2 16 teethN , 3 2 3 2 23 48 teethN N N ,

1

2 2 3tan 18.43N N , 3 290 71.57 Ans.

8.12 A pair of straight-tooth bevel gears has a velocity ratio of 1.5 and a shaft angle of 60°.

What are the pitch angles?

1

2

2 3

sintan 23.41

cos

, 3 260 36.59 Ans.

8.13 A pair of straight-tooth bevel gears is to be mounted at a shaft angle of 120°. The pinion

and gear are to have 16 and 36 teeth, respectively. What are the pitch angles?

1

2

3 2

sintan 26.33

cosN N

, 3 2120 93.67 Ans.

8.14 A pair of straight-tooth bevel gears with module of 12.7mm/tooth have 19 teeth and 28

teeth, respectively. The shaft angle is 90°. Determine the pitch diameters, pitch angles,

addendum, dedendum, face width, and pitch radii of the equivalent spur gears.

2 2m 2 120.65 mmR N , 3 3m 2 177.8 mmR N , Ans.

1

2 2 3tan 34.16N N , 3 290 55.84 , Ans.

Using Table 8.2: 90 3 2 1.474Gm m N N , 3 9.55 mma , Ans.

Whole depth 55.58 2 27.79 mm , 3 3Whole depth 18.24 mm d a Ans.

Working depth 2 m 25.4 mm , 0.122 m 0.05 mm 2.43 mmc

2 3Working depth 15.85 mm a a 2 2Whole depth 64.5 mmd a Ans.

Cone distance, 2 2sin 214.8 mmR Let 0.3 4.465 mmF , say F = 64.5mm Ans.

2 2 2cos 145.8 mmeR R 3 3 3cos 316.7 mmeR R Ans.



8.15 A pair of straight-tooth bevel gears with module of 3.2 mm/tooth have 18 teeth and 30

teeth, respectively, and a shaft angle of 105°. For each gear, calculate the pitch radius,

pitch angle, addendum, dedendum, face width, and equivalent number of teeth. Make a

sketch of the two gears in mesh. Use standard tooth proportions as for a 90° shaft angle.

2 2m 2 28.8 mmR N , 3 3m 2 48 mmR N , Ans.

1

2

3 2

sintan 34.45

cosN N

, 3 2105 70.55 Ans.

Using Table 8.2: 3 2 901.667, 2.032Gm N N m , 3 2.0675 mma , Ans.

Whole depth 6.9469 mm , 3 3Whole depth 4.8793 mmd a Ans.

Working depth 2.188 3.2 6.4 mm , 0.188 m 0.05 mm 0.6516 mmc

2 3Working depth 4.3325 mma a 2 2Whole depth 2.6219 mmd a Ans.

Cone distance, 2 2sin 50.52 mmR Let 0.3 15.16 mmF , say F = 15 mm Ans.

2 2 2cos 34.64 mmeR R 3 3 3cos 143 mmeR R

2 22m 21.65 teethe eN R , 3 32m 89.375 teethe eN R Ans.

8.16 A worm having 4 teeth and a lead of 25 mm drives a worm gear at a velocity ratio of 7.5.

Determine the pitch diameters of the worm and worm gear for a center distance of 44.5

mm.

2 25 mm 4 teeth 6.25 mm/toothxp N

3 2 3 2 7.5 4 teeth 30 teethN N

3 3 2 29.86 mmxR N p 2 344.5 14.64 mmR R Ans.



8.17 Specify a suitable worm and worm gear combination for a velocity ratio of 50 and a

center distance of 150 mm. Use an axial pitch of 12.5 mm/tooth.

Use 2 1 toothN , 3 2 3 2 50 1 teeth 50 teethN N

3 3 2 99.52 mmxR N p 2 3150 mm 50.48 mmR R Ans.

8.18 A double-threaded worm drives a worm gear having 40 teeth. The axial pitch is 31.75

mm and the pitch diameter of the worm is 44.45mm. Calculate the lead and lead angle of

the worm. Find the helix angle and pitch diameter of the worm gear.

2 2 teeth 31.75 mm/tooth 63.5 mmxN p Ans.

1 1

2tan 2 tan 63.5 mm 44.45 mm 24.453R Ans.

24.453 Ans.

3 3 2 40 teeth 31.75 mm/tooth 2 202.23 mmxR N p Ans.

8.19 A double-threaded worm with a lead angle of 20° and an axial pitch of 12.7 mm/tooth

drives a worm gear with a velocity reduction of 16 to 1. Determine the following for the

worm gear: (a) the number of teeth, (b) the pitch radius, and (c) the helix angle. (d)

Determine the pitch radius of the worm. (e) Compute the center distance.

2 2 teeth 12.7 mm/tooth 25.4 mmxN p

2 2 tan 25.4 mm 2 tan 20 11.11 mmR Ans.

3 2 3 2 16 11.11 mm 177.76 mmR R Ans.

3 32 87.9 teethxN R p Ans.

20.0 Ans.

2 3 11.11 mm 177.76 mm 188.87 mmR R Ans.



Chapter 9

Mechanism Trains

9.1 Find the speed and direction of gear 8 in Fig. P9.1. What is the first-order kinematic

coefficient of the train?

5 72 482

3 5 6 8

18 15 33 16 5

44 33 36 48 88

N NN N

N N N N Ans.

8 82 2 5 88 1 200 rev/min ccw 68.18 rev/min ccw Ans.



9.2 Figure P9.2 gives the pitch diameters of a set of spur gears forming a train. Compute the

first-order kinematic coefficient of the train. Determine the speed and direction of

rotation of gears 5 and 7.

175mm

375mm 225mm

750mm

225mm

400mm

2 452

3 5

175 mm 225 mm 7

375 mm 750 mm 50

R R

R R 5 6

72 52

6 7

7 750 mm 225 mm 21

50 225 mm 400 mm 80

R R

R R Ans.

5 52 2 7 50 120 rev/min cw 16.80 rev/min cw Ans.

7 72 2 21 80 120 rev/min cw 31.50 rev/min cw Ans.



9.3 Figure P9.3 illustrates a gear train consisting of bevel gears, spur gears, and a worm and

worm gear. The bevel pinion is mounted on a shaft that is driven by a V-belt on pulleys.

If pulley 2 rotates at 1200 rev/min in the direction indicated, find the speed and direction

of rotation of gear 9.

6 82 492

3 5 7 9

6 in 18 20 3 3

10 in 38 48 36 304

N NR N

R N N N

9 92 2 3 304 1 200 rev/min 11.84 rev/min cw Ans.



9.4 Use the truck transmission of Fig. 9.2 and an input speed of 4 000 rev/min to find the

drive shaft speed for each forward gear and for the reverse gear.

First gear: 6292

3 9

17 17 289

43 43 1 849

NN

N N

9 92 2 289 1 849 4 000 rev/min 625.2 rev/min Ans.

Second gear: 5282

3 8

17 27 153

43 33 473

NN

N N

8 82 2 153 473 4 000 rev/min 1 293.9 rev/min Ans.

Third gear: 2 472

3 7

17 36 51

43 24 86

N N

N N

7 72 2 51 86 4 000 rev/min 2 372.1 rev/min Ans.

Fourth gear: 22 1.0

2 22 2 1.0 4 000 rev/min 4 000 rev/min Ans.

Reverse gear: 62 1192

3 10 9

17 17 22 3 179

43 18 43 16 641

NN N

N N N

9 92 2 3 179 16 641 4 000 rev/min 764.1 rev/min Ans.



9.5 Figure P9.5 illustrates the gears in a speed-change gearbox used in machine tool

applications. By sliding the cluster gears on shafts B and C, nine speed changes can be

obtained. The problem of the machine tool designer is to select tooth numbers for the

various gears so as to produce a reasonable distribution of speeds for the output shaft.

The smallest and largest gears are gears 2 and 9, respectively. Using 20 and 45 teeth for

these gears, determine a set of suitable tooth numbers for the remaining gears. What are

the corresponding speeds of the output shaft? Note that the problem has many solutions.

We also set 6 20 teethN (minimum). Ans.

Since the largest speed reduction will be obtained with gears 2-5-6-9,

62,min

5 9 5

20 20450 rev/min 137 rev/min

45C A

NN

N N N

From this we get 5 29.197N , and we choose 5 30 teeth.N Ans.

Next, using distance units of circular pitch, the distance between shafts B and C is

5 8 6 9 7 10 65 teethBC N N N N N N . 8 5 35 teethN BC N . Ans.

Similarly the distance between shafts A and B is

2 5 3 6 4 7 50 teethAB N N N N N N . 3 6 30 teeth.N AB N Ans.

Since the minimum is 20 teeth and since 4 7 50 teethAB N N we see that

4 720 , 30N N and we choose 4 7 25 teethN N Ans.

and, finally, 10 7 40 teethN BC N . Ans.

With all tooth numbers known, we can now find the output shaft speed for each gear

arrangement. These are:

Arrangement First-order kinematic coefficient, CA Output shaft speed, C , rev/min

2-5-5-8 0.571 257.1

2-5-6-9 0.296 133.3

2-5-7-10 0.416 187.5

3-6-5-8 1.285 578.6

3-6-6-9 0.667 300.0

3-6-7-10 0.938 421.9

4-7-5-8 0.857 385.7

4-7-6-9 0.444 200.0

4-7-7-10 0.625 281.3



9.6 The internal gear (gear 7) in Fig. P9.6 turns at 60 rev/min ccw. What are the speed and

direction of rotation of arm 3?

62 472

4 5 7

20 teeth 40 teeth 36 teeth 20


NN N

N N N

7 3 372/3

2 3 3

60 rev/min 20

0 77

; 3 81.1 rev/min ccw Ans.

9.7 If the arm in Fig. P9.6 rotates at 400 rev/min ccw, find the speed and direction of rotation

of internal gear 7.

62 472

4 5 7



NN N

N N N

7 3 772/3

2 3

400 rev/min 20

0 400 rev/min 77




9.8 In Fig. P9.8a, shaft C is stationary. If gear 2 rotates at 600 rev/min ccw, what are the

speed and direction of rotation of shaft B?

232

3

18 teeth 3

24 teeth 4

N

N 3 32 2

3600 rev/min ccw 450 rev/min cw

4

5 785

6 8

18 teeth 20 teeth 3

42 teeth 40 teeth 14

N N

N N

8 385/3

5 3 5

0 450 rev/min 3

450 rev/min 14

; 5 1 650 rev/min ccw Ans.

9.9 In Fig. P9.8a, consider shaft B as stationary. If shaft C is driven at 450 rev/min ccw,

what are the speed and direction of rotation of shaft A?

5 785

6 8

18 teeth 20 teeth 3


N N

N N

8 3 385/3

5 3 3

450 rev/min 3

0 14

; 3 572.7 rev/min ccw

32 3

2

24 teeth572.7 rev/min ccw 763.6 rev/min cw

18 teethA

N

N Ans.

9.10 In Fig. P9.8a, determine the speed and direction of rotation of shaft C under the

following conditions:

(a) Shafts A and B both rotate at 450 rev/min ccw; and

(b) Shaft A rotates at 450 rev/min cw and shaft B rotates at 450 rev/min ccw.

5 785

6 8

18 teeth 20 teeth 3


N N

N N

(a) 23 2

3

18 teeth450 rev/min ccw 337.5 rev/min cw

24 teeth

N

N

8 3 885/3

5 3

337.5 rev/min 3

450 rev/min 337.5 rev/min 14

; 8 168.8 rev/min cw Ans.

(b) 23 2

3

18 teeth450 rev/min cw 337.5 rev/min ccw

24 teeth

N

N

8 3 885/3

5 3

337.5 rev/min 3

450 rev/min 337.5 rev/min 14




9.11 In Fig. P9.8b, gear 2 is connected to the input shaft. If arm 3 is connected to the output

shaft, what speed reduction can be obtained? What is the sense of rotation of the output

shaft? What changes could be made in the train to produce the opposite sense of rotation

for the output shaft?

52 462

4 5 6



NN N

N N N

6 3 362/3

2 3 2 3

0 5

27

3 2

5

22

The speed reduction is 17 22 77.3 Ans.

The sense of the output rotation is opposite to the input sense. Ans.

The opposite sense of rotation for the output shaft can be produced by replacing gears 4

and 5 by a single 44-tooth gear. Ans.



9.12 The Lévai type-L train illustrated in Fig. 9.10 has N2 = 16T, N4 = 19T, N5 = 17T, N6 =

24T, and N7 = 95T. Internal gear 7 is fixed. Find the speed and direction of rotation of

the arm if gear 2 is driven at 150 rev/min cw.

62 472

4 5 7


19 teeth 17 teeth 95 teeth 1 615

NN N

N N N

7 3 372/3

2 3 3

0 384

150 rev/min 1 615


9.13 The Lévai type-A train of Fig. 9.10 has N2 = 20T and N4 = 32T.

(a) If the module is m = 8 mm/tooth, find the number of teeth on gear 5 and the crank

arm radius.

(b) If gear 2 is fixed and internal gear 5 rotates at 15 rev/min ccw, find the speed and

direction of rotation of the arm.

(a) 5 2 42 20 teeth 2 32 teeth 84 teethN N N Ans.

3 2 4 2 8 mm/tooth 20 teeth 32 teeth 2 208 mmR m N N Ans.

(b) 2 452

4 5

20 teeth 32 teeth 5


N N

N N

5 3 352/3

2 3 3

15 rev/min 5

0 21

3 12.12 rev/min ccw Ans.



9.14 The tooth numbers for the automotive differential illustrated in Fig. 9.22 are N2 = 17T, N3

= 54T, N4 = 11T, and N5 = N6 = 16T. The drive shaft turns at 1200 rev/min. What is the

speed of the right wheel if it is jacked up and the left wheel is resting on the road surface?

23 2

3

17 teeth1 200 rev/min 377.8 rev/min

54 teeth

N

N

5 465

4 6

16 teeth 11 teeth1

11 teeth 16 teeth

N N

N N

6 3 665/3

5 3

377.8 rev/min1

0 377.8 rev/min

6 755.6 rev/min Ans.

9.15 A vehicle using the differential illustrated in Fig. 9.22 turns to the right at a speed of 48

km/h on a curve of 24-m radius. Use the same tooth numbers as in Problem 9.14. The

tire diameter is 375 mm. Use 1500 mm as the distance between treads.

(a) Calculate the speed of each rear wheel.

(b) Find the rotational speed of the ring gear.

(a)

48 km/h 1000 m/km 60 min/h33.33 rad/min

24 m

carcar

v

For the right and left wheels, respectively:

6

33.33 rad/min 24 m 1000 mm/m 750 mm658.108 rev/min

2 rad/rev 375 / 2 mm

Rv

r

Ans.

5

33.33 rad/min 24 m 1000 mm/m 750 mm700.566 rev/min

2 rad/rev 375/2 mm

Lv

r

Ans.

5 465

4 6

16 teeth 11 teeth1

11 teeth 16 teeth

N N

N N

(b) 6 3 365/3

5 3 3

658.108 rev/min1

700.566 rev/min

3 679.34 rev/min Ans.



9.16 Figure P9.16 illustrates a possible arrangement of gears in a lathe headstock. Shaft A is

driven by a motor at a speed of 720 rev/min. The three pinions can slide along shaft A so

as to yield the meshes 2 with 5, 3 with 6, or 4 with 8. The gears on shaft C can also slide

so as to mesh either 7 with 9 or 8 with 10. Shaft C is the mandrel shaft.

(a) Make a table demonstrating all possible gear arrangements, beginning with the

slowest speed for shaft C and ending with the highest, and enter in this table the

speeds of shafts B and C.

(b) If the gears all have a module of 5 mm/tooth, what must be the shaft center

distances?

N2 = 16T, N3 = 36T, N4 = 25T, N5 = 64T, N6 = 66T, N7 = 17T, N8 = 55T, N9 =

79T, N10 = 41T

(a)

Gears , rev/minB , rev/minC

2-5-7-9 180.0 38.7

4-8-7-9 327.3 70.4

3-6-7-9 589.1 126.8

2-5-8-10 180.0 241.5

4-8-8-10 327.3 439.0

3-6-8-10 589.1 790.2

(b) 2 5 2 5 mm/tooth 16 teeth 64 teeth 2 200 mmAB m N N Ans.

7 9 2 5 mm/tooth 17 teeth 79 teeth 2 240 mmBC m N N Ans.



9.17 Shaft A in Fig. P9.17 is the output and is connected to the arm. If shaft B is the input and

drives gear 2, what is the speed ratio? Can you identify the Lévai type for this train?

N2 = 16T, N3 = 18T, N4 = 16T, N5 = 18T, N6 = 50T

3 5262

3 4 6



N NN

N N N

662/

2 2

0 9

25

A AA

A A

; 29 34A Ans.

This train is Lévai type F. Ans.

9.18 In Problem 9.17, shaft B rotates at 150 rev/min cw. Find the speed of shaft A and of

gears 3 and 4 about their own axes.

Step Arm 2 3 4 5 6

Locked +1 +1 +1 +1 +1 +1

Arm fixed 0 +16/9 -128/81 +16/9 +16/9 -1

Total +1 +25/9 -47/81 +25/9 +25/9 0

29 25 9 25 150 rev/min cw 54.0 rev/min cwA Ans.

3 247 81 9 25 47 225 150 rev/min cw 31.3 rev/min ccw Ans.

4 225 9 9 25 1.0 150 rev/min cw 150.0 rev/min cw Ans.



9.19 Bevel gear 2 is driven by the engine in the reduction unit illustrated in Fig. P9.19. Bevel

planets 3 mesh with crown gear 4 and are pivoted on the spider (arm), which is connected

to propeller shaft B. Find the percentage speed reduction.

N2 = 36T, N3 = 21T, N4 = 52T; crown gear 4 is fixed

3242

3 4

36 teeth 21 teeth 9


NN

N N

442/

2 2

0 9

13

B BB

B B

; 29 22B

Speed reduction to 9/22 = 40.9% is speed reduction of 59.1%. Ans.



9.20 In the clock mechanism illustrated in the Fig. P9.20, a pendulum on shaft A drives an

anchor (see Fig. 1.12c). The pendulum period is such that one tooth of the 30T

escapement wheel on shaft B is released every 2 s, causing shaft B to rotate once every

minute. In the figure, note that the second (to the right) 64T gear is pivoted loosely on

shaft D and is connected by a tubular shaft to the hour hand.

(a) Show that the train values are such that the minute hand rotates once every hour

and that the hour hand rotates once every 12 hours.

(b) How many turns does the drum on shaft F make every day?

(a) 1.0 rev/minB

8 teeth 8 teeth 1


CBDB

C D

NN

N N

1 60 rev/minD DB B 60 min/revDt Ans.

1 28 teeth 8 teeth 1

60 42 teeth 64 teeth 720

D EHB DB

E H

N N

N N

1 720 rev/minH HB B 720 min/rev

12 hr/rev60 min/hr

Ht Ans.

(b) 1 8 teeth 1

60 96 teeth 720

DFB DB

F

N

N

1 720 rev/min 60 min/hr 24 hr/day 2 rev/dayF FB B Ans.



Chapter 10

Synthesis of Linkages

10.1 A function varies from 0 to 1. Find the Chebychev spacing for two, three, four, five, and

six precision positions.

With 0 0.0x and 1 10.0Nx , Eq. (10.5) becomes:

(2 1)5.0 5.0cos 1,2,...,

2j

jx j N

N

j \ N 2 3 4 5 6

1 0.146447 0.066987 0.038060 0.024472 0.017037

2 0.853553 0.500000 0.308658 0.206107 0.146447

3 0.933013 0.691342 0.500000 0.370591

4 0.961940 0.793893 0.629409

5 0.975528 0.853553

6 0.982963



10.2 Determine the link lengths of a slider-crank linkage to have a stroke of 600 mm and a

time ratio of 1.20.

After laying out the distance B1B2 = 600 mm, we see that the time ratio is

180 180 1.20Q

and, from this, our design must have 16.40 . Therefore we construct the point C

such that the central angle 1 2 2 32.80 .B CB Using this point C as the center of a

circle ensures that any point O2 on this circle will have the angle

1 2 2 16.40B O B and thus will be a possible solution point. One typical solution

uses the point O2 shown.

Choosing the point O2 shown we measure the two distances

2 1 3 2 1 324.956 mmO B r r and 2 2 3 2 801.946 mmO B r r and from these we find

2 261.50 mmr Ans.

3 1 063.45 mmr Ans.



10.3 Determine a set of link lengths for a slider-crank linkage such that the stroke is 400 mm

and the time ratio is 1.25.

After laying out the distance B1B2 = 400 mm, we see that the time ratio is

180 180 1.25Q and, from this, our design must have 20.00 .

Therefore we construct the point C such that the central angle 1 2 2 40.00 .B CB

Using this point C as the center of a circle ensures that any point O2 on this circle will

have the angle 1 2 2 20.00B O B and thus will be a possible solution point. One

typical solution uses the point O2 shown.

400

392.25

740.5

Choosing the point O2 shown we measure the two distances 2 1 3 2 740.5 mmO B r r

and 2 2 3 2 392.25 mmO B r r and from these we find

2 176.5 mmr Ans.

3 568.75 mmr Ans.



10.4 The rocker of a crank-rocker linkage is to have a length of 500 mm and swing through a

total angle of 45 with a time radio of 1.25. Determine a suitable set of dimensions for

1 2 3, , and r r r .

After laying out the angle 1 4 2 45B O B with BO4 = 500 mm, we see that the time

ratio is 180 180 1.25Q and, from this, we find that 20.00 .


Then, using this point C as the center of a circle ensures that any point O2 on this circle

will have the angle 1 2 2 20.00B O B and thus will be a possible solution point. One


Choosing the point O2 shown we measure the three distances 2 4 556 mmO O ,

2 1 3 2 878 mmO B r r , and 2 2 3 2 588 mmO B r r and from these we find

1 556 mmr Ans.

2 145 mmr Ans.

3 733 mmr Ans.



10.5 A crank-and-rocker mechanism is to have a rocker of 1800 mm length and a rocking

angle of 75 . If the time ratio is to be 1.32, what are a suitable set of link lengths for the

remaining three links?




Then, using this point C as the center of a circle ensures that any point O2 on this circle



1907.25

3807.85

2434.25

Choosing the point O2 shown we measure the three distances 2 4 2434.25 mmO O ,

2 1 3 2 3807.85 mmO B r r , and 2 2 3 2 1907.25 mmO B r r and from these we find

1 2434.25 mmr Ans.

2 950.25 mmr Ans.

3 2857.5 mmr Ans.



10.6 Design a crank and coupler to drive rocker 4 in Fig. P10.6 such that slider 6 will

reciprocate through a distance of 400 mm with a time radio of 1.20. Use

4 400mma r and 5 600mmr with 4r vertical at midstroke. Record the location of

2O and dimensions 2 3 and r r .



Therefore we construct the point D such that the central angle 1 2 2 32.72 .B DB

Then, using this point D as the center of a circle ensures that any point O2 on this circle



549

895.75

895.75

626

Choosing the point O2 shown we measure the three distances 2 4 626 mmO O ,

2 1 3 2 895.75 mmO B r r , and 2 2 3 2 549 mmO B r r and from these we find

1 626 mmr Ans.

2 173.25 mmr Ans.

3 722.5 mmr Ans.



10.7 Design a crank and rocker for a six-link mechanism such that the slider in Fig. P10.6 for

Problem 10.6 reciprocates through a distance of 800 mm with a time ratio of 1.12; use

4 1 200a r mm and 5 1 800r mm. Locate 4O such that rocker 4 is vertical when the

slider is at midstroke. Find suitable coordinates for 2O and lengths for 2 3 and r r .

After laying out the angle 1

1 4 2 2sin 400 1200 38.94B O B with BO4 = 1 200

mm, we see that the time ratio is 180 180 1.12Q and, from this, we find

that 10.19 .

Therefore we construct the point D such that the central angle 1 2 2 20.38 .B DB

Then, using this point D as the center of a circle ensures that any point O2 on this circle



Choosing the point O2 shown we measure the three distances 2 4 1 979 mmO O ,

2 1 3 2 2 634 mmO B r r , and 2 2 3 2 1 942 mmO B r r and from these we find

1 1 979 mmr Ans.

2 346 mmr Ans.

3 2 288 mmr Ans.



10.8 Figure P10.8 illustrates two positions of a folding seat used in the aisles of buses to

accommodate extra passengers. Design a four-bar linkage to support the seat so that it

will lock in the open position and fold to a stable closing position along the side of the

aisle.

The open position is a toggle position with no force tending to open or close the 3-4-5 tri-

angle. Thus a small catch only allowing joint A to rotate very slightly past the 180°

position at A2 will keep the seat open.

10.9 Design a spring-operated four-bar linkage to support a heavy lid like the trunk lid of an

automobile. The lid is to swing through an angle of 80 from the closed to the open

position. The springs are to be mounted so that the lid will be held closed against a stop,

and they should also hold the lid in a stable open position without the use of a stop.

One typical solution has

2 4 2 4O A AB O B O O . A stop

for the closed position may be

provided with point B slightly

below the line 4O A . The open

position is held stable by the

choice of the spring free length.



10.10 For Fig. P10.10, synthesize a linkage to move AB from position 1 to position 2 and

return.

10.11 Synthesize a mechanism to move AB successively through positions 1, 2, and 3 of Fig.

P10.11.



10.12 through 10.21* Figure P10.12 illustrates a function-generator linkage in which the

motion of rocker 2 corresponds to x and the motion of rocker 4 to the function y = f (x).

Use four precision points and Chebychev spacing and synthesize a linkage to generate

the functions illustrated in Table P10.12 to P10.31. Plot a curve of the desired function

and a curve of the actual function that the linkage generates. Compute the maximum

error between them in percent.

Problem

number

Function,

y f x

Range

0 0x ,

deg

0 0y ,

deg

2 1r r 3 1r r 4 1r r Max.

Error, deg

10.12, 10.22 10log x 1 2x 52.628 259.077 -3.352 0.845 3.485 0.0037

10.13, 10.23 sin x 0 2x -62.263 75.606 1.834 2.238 -0.693 0.1900

10.14, 10.24 tan x 0 4x 269.709 124.189 -2.660 7.430 8.685 0.0380

10.15, 10.25 xe 0 1x 241.644 40.422 -3.499 0.878 3.399 0.0258

10.16, 10.26 1 x 1 2x 33.804 120.213 -0.385 1.030 0.384 0.0161

10.17, 10.27 1.5x 0 1x -5.171 211.689 0.625 1.309 -0.401 0.1460

10.18, 10.28 2x 0 1x -29.321 233.836 2.523 3.329 -0.556 0.0673

10.19, 10.29 2.5x 0 1x -88.313 44.492 -1.801 0.908 1.274 0.4120

10.20, 10.30 3x 0 1x -85.921 37.637 -1.606 0.925 1.107 0.5095

10.21, 10.31 2x 1 1x -21.180 -53.670 -0.610 0.565 0.380 2.3400

10.22 through 10.31 Repeat Problems 10.12 through 10.21 using the overlay method.

The overlay method can be used to confirm the above solutions. Other nearby solutions

are also possible but are too numerous to display here.

* Solutions for these problems were among the earliest computer work in kinematic synthesis and results are

shown in F. Freudenstein, “Four-bar Function Generators,” Machine Design, vol. 30, no. 24, pp. 119-123, 1958.



10.32 Figure P10.32 illustrates a coupler curve that can be generated by a four-bar linkage (not

illustrated). Link 5 is to be attached to the coupler point, and link 6 is to be a rotating

member with 6O as the frame connection. In this problem we wish to find a coupler

curve from the Hrones and Nelson atlas or by precision positions, such that, for an

appreciable distance, point C moves through an arc of a circle. Link 5 is then

proportioned so that D lies at the center of curvature of this arc. The result is then called

a hesitation motion because link 6 will hesitate in its rotation for the period during which

point C traverses the approximate circular arc. Make a drawing of the complete linkage

and plot the velocity-displacement diagram for 360 of displacement of the input link.

This coupler curve for point C was found from the Hrones and Nelson atlas, page 150.

The hesitation is shown by the following plot of the first-order kinematic coefficient 6 .



10.33 Synthesize a four-bar linkage to obtain a coupler curve having an approximate straight-

line segment. Then, using the suggestion included in Fig. 10.42b or Fig. 10.44b,

synthesize a dwell motion. Using an input crank angular velocity of unity, plot the

velocity of rocker 6 versus the input crank displacement.

The Hrones and Nelson atlas contains a wide variety of coupler curves similar to the one

shown; this one is from page 93.

The dwell in the rotation of link 6 is shown by the following plot of the first-order

kinematic coefficient 6 .



10.34 Synthesize a dwell mechanism using the idea suggested in Fig. 10.42a and the Hrones

and Nelson atlas. Rocker 6 is to have a total angular displacement of 60 . Using this

displacement as the abscissa, plot a velocity diagram of the motion of the rocker to

illustrate the dwell motion.

The Hrones and Nelson atlas contains a wide variety of coupler curves similar to the one

shown here; this one is from page 93.

The dwell in the rotation of link 6 is shown by the following plot of the first-order

kinematic coefficient 6 .



Chapter 11

Spatial Mechanisms

11.1 Use the Kutzbach criterion to determine the mobility of the SSC linkage illustrated in Fig.

P11.1. Identify any idle freedoms and state how they can be removed. What is the

nature of the path described by point B?

375BA O OR R

mm, 3

150BOR mm, and 2 = 30

3n , 2 3j , 3 1j , 1 4 5 0j j j

6 3 1 4 1 3 2 2m Ans.

There is one idle freedom, the rotation of link 3 about its own axis. This idle freedom may

be eliminated by employing a two-freedom pair, such as a universal joint, in place of one of

the two spheric pairs, either at B or at O3. Ans.

The path described by point B is the curve of intersection of a cylinder of radius BA about

the y axis and a sphere of radius BO3 centered at O3. Ans.



11.2 For the SSC linkage illustrated in Fig. P11.1 express the position of each link in vector

form.

3

ˆ75 mmO O R i , 2 2ˆ ˆˆ ˆ75cos 75sin 64.952 08 37.5 mmBA R i k i k , Ans.

ˆ mmAO AORR j , 3

150 mmBOR .

Substituting these into 3 3O O BO AO BA R R R R gives

2ˆ ˆ450 cos 1 144.899 mmAO R j j , Ans.

3 2 2 2ˆ ˆ ˆ75 cos 1 450 cos 1 75sin

ˆ ˆ ˆ10.048 144.889 37.5 mm

BO

R i j k

i j k

Ans.

11.3 For the linkage of Fig. P11.1 with ˆ50 mm/sA V j , use vector analysis to find the

angular velocities of links 2 and 3 and the velocity of point B at the position specified.

The velocity of point B is given by 3B BO A BA V V V V , or

33 2ˆ2B BO BA V ω ×R j ω ×R

Assuming that the idle freedom is not active we can set 33 0BO ω R , where

3 3 3 3ˆ ˆ ˆx y z ω i j k and 2 2

ˆω j. Expanding these and using the position data from

Problem 11.2 gives four simultaneous equations:

2

3

3

3

0 10.048 144.889 37.5 0

37.5 0 37.5 144.899 0

0 37.5 0 10.048 50

64.952 144.889 10.048 0 0

x

y

z

Solving these gives 2ˆ2.570 rad/s ω j Ans.

3ˆ ˆ ˆ1.158 0.086 0.643 rad/s ω i j k , 3 1.327 rad/s Ans.

ˆ ˆ ˆ96.35 50 166.9 mm/sB V i j k , 199 mm/sBV Ans.



11.4 Solve Problem 11.3 using graphic techniques.

The position and velocity solutions are shown in the figure below. After the top and

front views are drawn to scale, first and second auxiliary views are drawn to view rod

O3B in true length and end views, respectively.

Next a velocity polygon is drawn with origin at point B. The velocity VA is drawn true

length, downward in the front view. The direction of VBA is added horizontal in the front

view and perpendicular to link 2 in the top view. This direction is projected to the first

auxiliary view where it intersects the line of VB, which is perpendicular to rod 3 in this

view. This completes the velocity polygon for the equation B A BA V V V . Projecting

this to all other views, we can measure the true lengths of VB from the second auxiliary

view, and VBA from the top view.

The angular velocities are then found from

2

305 mm/s4.067 rad/s

75 mm

BA

BA

V

R Ans.

3

3

3

222 mm/s1.480 rad/s

150 mm

BO

BO

V

R Ans.

222 mm/sBV Ans.

The results show typical graphical error when compared with the analytical solution in

Problem 11.3.



11.5 For the spherical RRRR illustrated in Fig. P11.5, use vector algebra to make complete

velocity and acceleration analyses at the position given.

2

ˆ175O O R k mm, 4

ˆ50O O R i mm, 2

ˆ75AO R i mm, 4

ˆ225BO R jmm, ˆ ˆ ˆ125 225 175BA R i j k mm,

2ˆ60 ω k rad/s.

2

ˆ75 mmAO R i , 4 2

ˆ ˆ50 175 mmO O R i k , 4

ˆ225 mmBO R j .

Substituting these into 2 4 2 4AO BA O O BO R R R R gives

ˆ ˆ ˆ125 225 175 mmBA R i j k , 311.25 mmBAR

The velocity analysis proceeds as follows:

2 22

ˆ ˆˆ60 rad/s 75 mm 4 500 mm/sA AO AO V V ω ×R k × i j Ans.

4 44 4 4

ˆ ˆ ˆ225 mm 225B BO BO V V ω ×R i × j k

Since the two revolutes at A and B have axes that intersect at O, this is a spherical

linkage; therefore triangle AOB rotates about O as a rigid link with point O stationary.

From this we see that the axis of rotation of link 3 passes through O and is perpendicular

to both AV and BV . Therefore 3 3ˆω i and

3 3 3 3ˆ ˆ ˆ ˆˆ ˆ125 225 175 175 225BA BA V ω ×R i × i j k j k



Substituting these into B A BA V V V or 4 3 3ˆ ˆˆ ˆ225 4 500 175 225 k j j k , and

equating components gives

3 4ˆ4 500 175 25.714 rad/s ω ω i , Ans.

ˆ ˆ4 500 5 785 mm/sBA V j k , and ˆ5 785 mm/sB V k . Ans.

To find accelerations we first calculate

2 2

2 2

2ˆ270 000 mm/sn

AO AO A R i , 2 22

t

AO AO A α ×R 0 Ans.

4 4

2 2

4ˆ148 775 mm/sn

BO BO A R j , 4 44 4

ˆ225t

BO BO A α ×R k

Remembering that link 3 rotates about point O we also find

2

3 3ˆ115 725 mm/sn

AO AO A × ×R k ,

3 3 3 3ˆ ˆ ˆ175 175 75 75t y z y

AO AO x

3A α ×R i j k

2

3 3ˆ148 775 mm/sn

BO BO A × ×R j ,

3 3 3 3ˆ ˆ ˆ225 50 225 50t z z y

BO BO x

3A α ×R i j k

Substituting these into 2

n n t

A AO AO AO A A A A and 4

n n t

B BO BO BO A A A A , and

equating components gives y

3270 000 175 , 3 0 225 z ,

3 3 0 175 75x z , 3148 775 148 775 50 z ,

30 115 725 75 y , and 4 3 3225 225 50 x y .

From these we solve for 2ˆ1 543 rad/s3α j and 2

4ˆ343 rad/s α i . Ans.

ˆ ˆ148 775 77 175B A j k mm/s2 Ans.




To avoid confusion, the position and velocity solutions are shown in a separate figure

above. The acceleration solution is shown below. The results agree with those of the

previous analytic solution of Problem 11.5.



11.7 Solve Problem 11.5 using transformation matrix techniques.

Following the conventions of Sec. 11.6 and Fig. 11.11, the Denavit-Hartenberg parameter

values are:

i,j ,i ja ,i j ,i j ,i js

1,2 0 23.20° 1 0 0

2,3 0 94.90° 2 101.54º 0

3,4 0 77.47° 3 67.30º 0

4,1 0 90.00° 4 90.00º 0

Using Eqs. (11.12) and (11.15) we find

12

1 0 0 0

0 0.91914 0.39394 0

0 0.39394 0.91914 0

0 0 0 1

T

23

0.20005 0.08369 0.97620 0

0.97979 0.01709 0.19932 0

0 0.99635 0.08542 0

0 0 0 1

T

, 13

0.20005 0.08369 0.97620 0

0.90056 0.37679 0.21685 0

0.38598 0.92252 0 0

0 0 0 1

T

,

34

0.38591 0.20015 0.90057 0

0.92254 0.08372 0.37671 0

0 0.97618 0.21695 0

0 0 0 1

T

, 14

0 1 0 0

0 0 1 0

1 0 0 0

0 0 0 1

T

.

Next, from Eqs. (11.22) and (11.25),

1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D

, 2

0 0.91914 0.39394 0

0.91914 0 0 0

0.39394 0 0 0

0 0 0 0

D

,

3

0 0 0.21685 0

0 0 0.97620 0

0.21685 0.97620 0 0

0 0 0 0

D

, 4

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

D

.

Now, from Eq. (11.26), 1 1 2 2 3 3 4 4 0D D D D , we get the following set of

equations:

2

3 1

4

0 0.97620 0 0 0

0.39394 0.21685 1 0 0 rad/s

0.91914 0 0 1 60

From these we find 2 65.275 rad/s , 3 0 , and 4 25.714 rad/s . From these and



Eq. (11.27) we find the velocity matrices

2

0 60 0 0

60 0 0 0

0 0 0 0

0 0 0 0

rad/s, 3

0 0 25.714 0

0 0 0 0

25.714 0 0 0

0 0 0 0

rad/s, 4

0 0 25.714 0

0 0 0 0

25.714 0 0 0

0 0 0 0

rad/s.

These can be used with Eq. (11.28) to find the velocities of all moving points.

Acceleration analysis follows similar steps. From Eq. (11.29) we get the following set of

equations:

2

2

3

4

0 0.97620 0 1 543

0.39394 0.21685 1 0 rad/s

0.91914 0 0 0

From these we find 2 0 , 2

3 1 580 rad/s , and 2

4 343 rad/s . From these and Eq.

(11.30) we find the acceleration matrices

2

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

, 3

0 0 0 0

0 0 1 543 0

0 1 543 0 0

0 0 0 0

rad/s2, 4

0 0 343 0

0 0 0 0

343 0 0 0

0 0 0 0

rad/s2.

These can be used with Eq. (11.31) to find the accelerations of all moving points.



11.8 Solve Problem 11.5 except with 90 .

The position vectors for this new position are:

2

ˆ75 mmAO R j , ˆ ˆ ˆx y z

BA BA BA BAR R R R i j k , 4 2

ˆ ˆ50 175 mmO O R i k , 4 4 4

ˆ ˆ225sin 225cosBO R j k .

Substituting these into 2 4 2 4AO BA O O BO R R R R and separating components gives

50x

BAR , 4225sin 75y

BAR , 4225cos 175z

BAR , and squaring and adding these gives

2 2 2 2 2

4 450 225 78 750cos 175 33 750sin 75 96 875BAR

If we now define 4tan 2 and use the identities 2 2

4cos 1 1 and

2

4sin 2 1 , then the above equation can be reduced to 219 18 23 0 .

The root of interest here is 0.72419 , which corresponds to 4 71.823 . With this

the four position vectors are

2

ˆ75 mmAO R j , ˆ ˆ ˆ50 288.772 104.810 mmBA R i j k , 4 2

ˆ ˆ50 175 mmO O R i k , 4

ˆ ˆ213.772 70.189 mmBO R j k

The velocity analysis proceeds as in Problem 11.5:

2 22

ˆ ˆˆ60 rad/s 75 mm 4 500 mm/sA AO AO V V ω ×R k × j i

4 44 4 4 4

ˆ ˆ ˆˆ ˆ213.772 70.189 mm 70.189 213.772B BO BO V V ω ×R i × j k j k

Since the two revolutes at A and B have axes which intersect at O, this is a spherical

linkage; therefore triangle AOB rotates about O as a rigid link with point O stationary.

From this we see that the axis of rotation of link 3 passes through O and is perpendicular

to both AV and BV . Therefore 3 3 3ˆ ˆ0.95010 0.31195 ω j k and

3 3 3 3ˆ ˆ ˆ189.650 15.600 47.500BA BA V ω ×R i j k

Substituting these into B A BA V V V and equating components gives

3 23.726 rad/s and 4 5.273 rad/s , and from these we get

3ˆ ˆ22.542 7.401 rad/s ω j k , 4

ˆ5.273 rad/sω i , Ans. ˆ ˆ ˆ4 500 370 1 127 mm/sBA V i j k , and ˆ ˆ370 1 127 mm/sB V j k . Ans.

To find accelerations we first calculate

2 2

2 2

2ˆ270 000 mm/sn

AO AO A R j , 2 22

t

AO AO A α ×R 0 ,

4 4

2 2

4ˆ ˆ5 943 1 951 mm/sn

BO BO A R j k , 4 44 4 4

ˆ ˆ70.20 213.78t

BO BO A α ×R j k


2

3 3ˆ ˆ33 300 101 450 mm/sn

AO AO A × R j k ,

3 3 3ˆ ˆ ˆ175 75 175 75t y z

AO AO x x

3 3A α ×R i j k ,

2

3 3ˆ28 150 mm/sn

BO BO A × ×R i ,

3 3 3 3ˆ ˆ ˆ70.20 213.78 70.20 50 213.78 50t y z x z y

BO BO x

3 3 3A α ×R i j k

Substituting these into 2

n n t

A AO AO AO A A A A and4

n n t

B BO BO BO A A A A , and equating

components gives

3 30 175 75y z , 3 30 28 150 70.20 213.78y z ,

3270 000 33 300 175 x

, x

4 3 35 943 70.20 70.20 50 z , 30 101 450 75 x ,

4 3 31 951 213.78 213.78 50x y .

From these we solve for 2

3ˆ ˆ ˆ1 302 49 115 rad/s α i j k and 2

4ˆ1 304 rad/s α i . Ans.

2ˆ270 000 mm/sA A j , 2ˆ ˆ85 750 280 750 mm/st

BO A j k Ans.



11.9 Determine the advance-to-return time ratio for Problem 11.5. What is the total angle of

oscillation of link 4?

Time ratio = 181.1/178.9 = 1.012, 4 46.5 Ans.



11.10 For the spherical RRRR linkage illustrated in Fig. P11.10, determine whether the crank is

free to turn through a complete revolution. If so, find the angle of oscillation of link 4

and the advance-to-return time ratio.

2150O OR mm,

4225O OR mm,

237.5AOR mm,

4262.5BOR mm, 412.5BAR mm,

2 120 , and ˆ30 ω k rad/s.

Time ratio = 187/173 = 1.081, 4 38 Ans.



11.11 Use vector algebra to make complete velocity and acceleration analyses of the linkage of

Fig. P11.10 at the position specified.

The position vectors were found from a graphical analysis done on a CAD software

system (see Problem 11.12). For the position 2 120 they are as follows:

4 2

ˆ ˆ225 150 mmO O R i k , 4

ˆ ˆ236.975 111.75 mmBO R j k ,

2

ˆ ˆ18.75 32.475 mmAO R i j , ˆ ˆ ˆ243.75 204.5 261.75 mmBA R i j k .

The velocity analysis for this position, with 2ˆ30 rad/sω k , proceeds as follows:

22ˆ ˆ974.275 562.5 mm/sA AO V ω ×R i j ,

44 4 4ˆ ˆ ˆ111.75 236.975B BO V i ×R j k .

Since all revolute axes intersect at O, this is a spherical mechanism and triangle AOB

(link 3) rotates about O. Thus the axis of rotation of link 3 passes through O and is

perpendicular to AV and BV . Calling this axis 3 3ˆ ˆ, u u ,

ˆ ˆ ˆˆ 11.5733 20.0432 9.45121B A B A u V ×V V ×V i j k

3 3 3 3ˆ ˆ ˆ132.525 213.325 290.1BA BA V ×R i j k

Substituting these into B A BA V V V , equating components, and solving gives

3 8.820 rad/s and 4 10.797 rad/s . From these

3ˆ ˆ ˆ4.083 7.071 3.334 rad/s i j k , and 4

ˆ10.797 rad/s i . Ans.

For acceleration analysis we first calculate

2 2

2

2 2ˆ ˆ25300 42089 mm/sn

AO AO A × ×R i j , 2 22

t

AO AO A ×R 0 ,

4 4

2

4 4ˆ ˆ25126 13027 mm/sn

BO BO A × ×R j k ,

4 44 4 4ˆ ˆ ˆ111.75 236.975t

BO BO A i ×R j k


2

3 3ˆ ˆ ˆ2251 3898 11023 mm/sn

AO AO A × ×R i j k ,

2

3 3ˆ ˆ ˆ22117 10447 4926 mm/sn

BO BO A × ×R i j k ,

3 3 3 3 3 3 3ˆ ˆ ˆ150 32 19 150 32 19t y z z x x y

AO AO A ×R i j k ,

3 3 3 3 3 3 3ˆ ˆ ˆ112 237 225 112 237 225t y z z x x y

BO BO A ×R i j k

Next, from 2 2

n t n t

A AO AO AO AO A A A A A and 4 4

n t n t

B BO BO BO BO A A A A A , we

separate components and obtain

3 3 24300 2251 150 32y z ; 3 30 22117 112 237y z ;

3 342089 3898 150 19x z ; 4 3 327626 112 10447 112 225x z ;

3 3 0 11023 32 19x y ; 4 3 313027 237 4926 237 225x y ;

From these 2

3ˆ ˆ ˆ273 115 148 rad/s α i j k and 2

4ˆ130 rad/sα i . Ans.




The position and velocity solutions are shown first with the acceleration solution on the

next diagram. The results verify those of the analytical solution in Problem 11.11.




The Denavit-Hartenberg parameters are:

12 12 12 1 12

23 23 23 2 23

34 34 34 3 34

41 41 41 4 41

0, 14.04 , 60.00 , 0,

0, 104.41 , 69.52 , 0,

0, 49.34 , 93.18 , 0,

0, 90.00 64.75 0.

a s

a s

a s

a s

From Eqs. (11.12) and (11.15) the transformation matrices are:

12

0.50000 0.84015 0.21005 0

0.86603 0.48506 0.12127 0

0 0.24260 0.97014 0

0 0 0 1

T

13

0.61206 0.39312 0.68618 0

0.75747 0.04214 0.65151 0

0.22720 0.91852 0.32356 0

0 0 0 1

T

14

0.42650 0.90449 0 0

0 0 1 0

0.90449 0.42650 0 0

0 0 0 1

T

Next, from Eqs. (11.22) and (11.25),

1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D

2

0 0.97015 0.12127 0

0.97015 0 0.21005 0

0.12127 0.21005 0 0

0 0 0 0

D

3

0 0.32357 0.65151 0

0.32357 0 0.68618 0

0.65151 0.68618 0 0

0 0 0 0

D

4

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

D

and from Eq. (11.26) we get the following set of equations

2

3 1

4

0.21005 0.68618 0 0 0

0.12127 0.65151 1 0 0 rad/s

0.97015 0.32357 0 1 36

from which we find 2 33.670 rad/s , 3 10.307 rad/s , and 4 10.798 rad/s .



With these values and Eqs. (11.27) we find the velocity matrices

2

0 36 0 0

36 0 0 0

0 0 0 0

0 0 0 0

, 3

0 3.335 4.083 0

3.335 0 7.072 0

4.083 7.072 0 0

0 0 0 0

, 4

0 0 10.798 0

0 0 0 0

10.798 0 0 0

0 0 0 0

.Ans.

These can be used with Eq. (11.28) to find the velocities of all moving points. Ans.

The acceleration analysis follows parallel steps using Eqs. (11.29) and (11.30)

2

2

3

41

0.21005 0.68618 0 183.0

0.12127 0.65151 1 254.6 m/s

0.97015 0.32357 0 76.4

;

2

2

2

3

2

4

152 rad/s

220 rad/s

130 rad/s

2

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

, 3

0 148 273 0

148 0 115 0

273 115 0 0

0 0 0 0

, 4

0 0 130 0

0 0 0 0

130 0 0 0

0 0 0 0

. Ans.

Although the global axes have changed because of the Denavit-Hartenberg conventions,

these results correlate with and verify those of Problems 11.11 and 11.12. Ans.



11.14 Figure P11.14 illustrates the top, front, and auxiliary views of a spatial slider-crank RSSP

linkage. In the construction of many such mechanisms provision is made to vary the

angle ; thus the stroke of slider 4 becomes adjustable from zero, when = 0, to twice

the crank length, when = 90. With = 30, use vector algebra to make a complete

velocity analysis of the linkage at the given position.

50AOR mm, 150BAR mm, 240 , ˆ24 ω i rad/s



ˆ ˆ ˆ21.651 37.500 25.000 mmAO R i j k , ˆ164.720 mmBO R i ,

ˆ ˆ ˆ143.069 37.500 25.000 mmBA R i j k .

The velocity analysis for this position, with 2 24 rad/s , proceeds as follows:

2ˆ ˆ20.785 12.000 rad/s ω i j , 2

ˆ ˆ ˆ300.000 519.615 1 039.230 mm/sA AO V ω ×R i j k ,

3 3 3 3 3 3 3ˆ ˆ ˆ25.000 37.500 143.069 25.000 37,500 143.069y z z x x y

BA BA V ω ×R i j k , ˆB BVV i .

Substituting these into B A BA V V V and separating into components,

3 3

3 3

3 3

300.000 25.000 37.500

0.0 519.615 25.000 143.069

0.0 1 039.230 37.500 143.069

y z

B

x z

x y

V

However, this is a set of only three equations and there are four unknown variables. This

results from the fact that the linkage has two degrees of freedom and the connecting rod

is free to rotate about the axis AB. If we assume that this second “idle freedom” is

inactive, then we can set 3 0BA R to get a fourth equation:

3 3 30.0 143.069 37.500 25.000x y z

The four equations can now be solved to give

3ˆ ˆ ˆ2.309 6.659 3.228 rad/s ω i j k and ˆ345.400 mm/sB V i Ans.



11.15 Solve Problem 11.14 using graphical techniques.

The graphic solution is shown in the figure above. The results verify those found in

Problem 11.14. They are

3

1 159.6 mm/s7.731 rad/s

150 mm

BA

BA

V

R , and 345.5 mm/sBV Ans.


One choice for the Denavit-Hartenberg parameters gives:

12 0a , 12 90 , 12 1 30 , 12 0s ,

14 0a , 14 30 , 14 0 , 14 4s .

From Eqs. (11.12) and (11.11) we obtain

1 1

1 1

12

cos 0 sin 0

sin 0 cos 0

0 1 0 0

0 0 0 1

T

,

1

1

12

0 2sin

0 2cos

2 0

1 1

AR T

;



4

14

4

1 0 0 0

0 cos sin sin

0 sin cos cos

0 0 0 1

T

, 4

14

4

0 0

0 sin

0 cos

1 1

BR T

.

From these and the length of the connecting rod we write

2 2 22 2

1 1 4 42sin 2cos sin cos 6t

BA B A B AR R R R R

This reduces to 2

4 4 14 sin cos 32 0 , which has a solution

2 2

4 1 12sin cos 4sin cos 32

By differentiating the above equation with respect to time we obtain

4 4 4 1 1 4 12 4 sin cos 4 sin sin 0

which has for a solution

4 1

4 1

1 4

2 sin sin

2sin cos

and, from Eqs. (11.22), (11.23), (11.25), and (11.27),

1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D

, 4

0 0 0 0

0 0 0 sin

0 0 0 cos

0 0 0 0

D

;

1

1

2

0 0 0

0 0 0

0 0 0 0

0 0 0 0

,

4

4

4

0 0 0 sin

0 0 0 cos

0 0 0

0 0 0 0

.

Now, with 30 , 1 30 , and 1 24 rad/s , the above formulae give

4 164.7 mm , 4 345.4 mm/s , and Ans.

25.000 mm

43.300 mm

0

1

AR

,

0

82.350 mm

142.650 mm

1

BR

,

1 039.225 mm/s

600.000 mm/s

0

0

AR

,

0

172.700 mm/s

299.125 mm/s

0

BR

. Ans.

These results agree with those of Problems 11.14 and 11.15 once the Denavit-Hartenberg

coordinate directions are considered. Note, however, that the loop-closure equation was

never used. No coordinate system was fixed to link 3 and no velocity of link 3 was

found. This is because of our lack of information about the degree of freedom

representing the spin of link 3 about the line AB.



11.17 Solve Problem 11.14 with = 60 using vector algebra.



ˆ ˆ ˆ37.500 21.650 25.000 mmAO R i j k , ˆ183.809 mmBO R i ,

ˆ ˆ ˆ146.309 21.651 25.000 mmBA R i j k .

The velocity analysis for this position, with 2 24 rad/s , proceeds as follows:

2ˆ ˆ12.000 20.784 rad/s ω i j ,

2ˆ ˆ ˆ519.615 300.000 1 039.230 mm/sA AO V ω ×R i j k , Ans.

3 3 3 3 3 3 3ˆ ˆ ˆ25.000 21.651 146.309 25.000 21.651 146.309y z z x x y

BA BA V ω ×R i j k , ˆB BVV i .

Substituting these into B A BA V V V and separating into components we get

3 3

3 3

3 3

519.615 25.000 21.650

0.0 300.000 25.000 146.309

0.0 1 039.230 21.651 146.309

y z

B

x z

x y

V

However, this is a set of only three equations and there are four unknown variables. This

results from the fact that the linkage has two degrees of freedom and the connecting rod

is free to rotate about the axis AB. If we assume that this second “idle freedom” is

inactive, then 3 0BA ω R .

3 3 30.0 146.309 21.651 25.000x y z

The four equations can now be solved to give

3ˆ ˆ ˆ1.333 6.905 1.823 rad/s ω i j k and ˆ652.821 mm/sB V i Ans.



11.18 Solve Problem 11.14 with = 60 using graphical techniques.


Problem 11.17. They are

3

1 089.75 mm/s7.265 rad/s

150 mm

BA

BA

V

R , and 652.851 mm/sBV Ans.



11.19 Solve Problem 11.14 with = 60 using transformation matrix techniques.

One choice for the Denavit-Hartenberg parameters gives:

12 0a , 12 90 , 12 1 30 , 12 0s ,

14 0a , 14 60 , 14 0 , 14 4s .

From Eqs. (11.13) and (11.12) we obtain

1 1

1 1

12

cos 0 sin 0

sin 0 cos 0

0 1 0 0

0 0 0 1

T

,

1

1

12

0 2sin

0 2cos

2 0

1 1

AR T

;

4

14

4

1 0 0 0

0 cos sin sin

0 sin cos cos

0 0 0 1

T

, 4

14

4

0 0

0 sin

0 cos

1 1

BR T

.

From these and the length of the connecting rod we write

2 2 22 2

1 1 4 42sin 2cos sin cos 6t

BA B A B AR R R R R

This reduces to 2

4 4 14 sin cos 32 0 , which has a solution

2 2

4 1 12sin cos 4sin cos 32

By differentiating the above equation with respect to time we obtain

4 4 4 1 1 4 12 4 sin cos 4 sin sin 0

which has for a solution

4 1

4 1

1 4

2 sin sin

2sin cos

and, from Eqs. (11.22), (11.23), (11.25), and (11.27),

1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D

, 4

0 0 0 0

0 0 0 sin

0 0 0 cos

0 0 0 0

D

;

1

1

2

0 0 0

0 0 0

0 0 0 0

0 0 0 0

,

4

4

4

0 0 0 sin

0 0 0 cos

0 0 0

0 0 0 0

.

Now, with 60 , 1 30 , and 1 24 rad/s , the above formulae give

4 183.800 mm , 4 652.800 mm/s , and



25.0 mm

43.3 mm

0

1

AR

,

0

159.175 mm

91.900 mm

1

BR

,

1 039.225 mm/s

600.0 mm/s

0

0

AR

,

0

565.35 mm/s

326.40 mm/s

0

BR

.

These results agree with those of Problems 11.17 and 11.18 once the Denavit-Hartenberg

coordinate directions are considered.



11.20 Figure P11.20 illustrates the top, front, and profile views of an RSRC crank and

oscillating-slider linkage. Link 4, the oscillating slider, is rigidly attached to a round rod

that rotates and slides in the two bearings. (a) Use the Kutzbach criterion to find the

mobility of this linkage. (b) With crank 2 as the driver, find the total angular and linear

travel of link 4. (c) Write the loop-closure equation for this mechanism and use vector

algebra to solve it for all unknown position data.



100

100 AOR mm in, 300 BAR mm in, 2 40 , and ˆ48 ω i rad/s.

(a) The RSRC linkage has n = 4, j1 = 2, j2 = 1, j3 = 1. The Kutzbach criterion gives



1 2 36 1 5 4 3 6(4 1) 5(2) 4(1) 3(1) 1m n j j j Ans.

(b) Since vectors do not show the rotation , matrix methods were necessary and are

shown in Problem 11.23. See (c) for the vector solution. Together, they show:

4135 45 ; 4 90 . Ans.

182.85 mm 382.85 mmBy ; 400 mmBy . Ans.

(c) B AB Q AQ R R R R

2 2ˆ ˆ ˆ ˆ ˆˆ ˆ100 100sin 100cosB AB AB ABy x y z j i j k i j k

Separating components,

100ABx , 2100sinB ABy y , 2100cosABz

and, from the length of link 3,

2 2 2 2

ABAB AB ABx y z R

2 2 2 2

2 2100 100sin 100cos (300)By

2

2200sin 2800 0B By y

2

2 2100sin 100 175 sinBy Ans.

2

2 2ˆ ˆ ˆ100 100 175 sin 100cosAB R i j k Ans.

For 2 40 , ˆ ˆ208 mmB By R j j , ˆ ˆ ˆ100 272.275 76.6 mmAB R i j k . Ans.



11.21 Use vector algebra to find VB, 3, and 4 for Problem 11.20.

First we identify, for 2 40 , that ˆ100 mmQ R i , ˆ ˆ64.275 76.6 mmAQ R j k ,

ˆ208 mmB R j , and ˆ ˆ ˆ100 272.275 76.6 mmAB R i j k .

Next we note that the rotation axis of the revolute at B is ˆ ˆ76.6 100AB j×R i k and,

normalizing this, we can express the apparent angular velocity 3/ 4ω axis as

3/ 4ˆ ˆˆ 0.608 0.794 ω i k . Then the angular velocity of link 3 can be written as

3 4 3/ 4 3/ 4 4 3/ 4ˆ ˆ ˆ0.608 0.794 ω ω ω i j k . With this done, we can find

2ˆ ˆ3677 3085.375 mm/sA AQ V ω ×R j k , ˆ

B BVV j , and

3 4 3/4 3/4 4 3/4ˆ ˆ ˆ(76.6 216.15 ) 125.975 100 165.575AB AB V ω ×R i j k . Now,

setting B AB A V V V and equating components, we get the following equations:

4 3/4

3/4

4 3/4

76.6 216.15 0

125.975 147.081

100 165.575 123.415

BV

which can be solved to give 4 19.444 rad/s , 3/ 4 6.891 rad/s , 2808.95 mm/sBV .

3ˆ ˆ ˆ4.190 19.444 5.471 rad/s ω i j k ,

4ˆ19.444 rad/sω j , ˆ2808.95 mm/sB V j . Ans.

Note that if 3ω were written as 3 3 3 3ˆ ˆ ˆx y z ω i j k , then the above set of simultaneous

equations would have four unknowns and could not be solved.



11.22 Solve Problem 11.21 using graphical techniques.


Problem 11.21. 3 and 4 are not apparent in the graphic method, but

2807.5 mm/sBV . Ans.




The Denavit-Hartenberg parameters from the global coordinate system to joint A are:

12 0a , 12 0 , 12 2 40 , 12 100 mms ,

and, proceeding in the other direction around the loop, to joint A, they are:

16 0a , 16 90 , 16 0 , 16 0s ,

65 0a , 65 0 , 65 0 , 65 Bs y ,

54 0a , 54 0 , 54 4 , 54 0s ,

43 0a , 43 90 , 43 0 , 43 0s ,

3 0Ba , 3 0B , 3 3B , 3 0Bs ,

300 mmBAa , 0BA , 90BA , 0BAs .



From Eqs. (11.12) we obtain for a first path to A:

2 2

2 2

12

cos sin 0 0

sin cos 0 0

0 0 1 100

0 0 0 1

T

,

and along the other path to joint A, from Eqs. (11.12) and (11.15), we get:

16

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

T

,

65

1 0 0 0

0 1 0 0

0 0 1

0 0 0 1

B

Ty

, 15

1 0 0 0

0 0 1

0 1 0 0

0 0 0 1

ByT

,

4 4

4 4

54

cos sin 0 0

sin cos 0 0

0 0 1 0

0 0 0 1

T

,

4 4

14

4 4

cos sin 0 0

0 0 1

sin cos 0 0

0 0 0 1

ByT

,

43

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

T

,

4 4

13

4 4

cos 0 sin 0

0 1 0

sin 0 cos 0

0 0 0 1

ByT

,

3 3

3 3

3

cos sin 0 0

sin cos 0 0

0 0 1 0

0 0 0 1

BT

,

3 4 3 4 4

3 3

1

3 4 3 4 4

cos cos sin cos sin 0

sin cos 0

cos sin sin sin cos 0

0 0 0 1

B

B

yT

,

0 1 0 0

1 0 0 300

0 0 1 0

0 0 0 1

BAT

,

3 4 3 4 4 3 4

3 3 3

1

3 4 3 4 4 3 4

sin cos cos cos sin 300sin cos

cos sin 0 300cos

sin sin cos sin cos 300sin sin

0 0 0 1

B

A

yT

.



At the terminations of these two paths, the position of point A must agree. Therefore,

2 2 3 4 3 4 4 3 4

2 2 3 3 3

3 4 3 4 4 3 4

cos sin 0 0 100 sin cos cos cos sin 300sin cos 0

sin cos 0 0 0 cos sin 0 300cos 0

0 0 1 100 0 sin sin cos sin cos 300sin sin 0

0 0 0 1 1 0 0 0 1 1

By

2 3 4

2 3

3 4

100cos 300sin cos

100sin 300cos

100 300sin sin

1 1

By

Noting from the figure that 390 0 and 4180 0 , we can solve these three

equations for the position results

1 2

3 2sin 1 cos 3

1

4 2tan 1 cos

2

2 2100 7 sin 100sinBy

and, at 2 40 , these give 3 24.83 , 4 52.55 , and 208 mmBy .

For velocity analysis we begin by using Eqs. (11.22) and (11.25) to find

1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

Q

, 1 1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D Q

,

3

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

Q

,

4 4

4 41

3 13 3 13

4 4

0 cos 0 cos

cos 0 sin 0

0 sin 0 sin

0 0 0 0

B

B

y

D T Q Ty

,

4

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

Q

, 1

4 14 4 14

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

D T Q T

,

6

0 150 0 0

0 0 0 0

0 0 0 1

0 0 0 0

Q

, 1

6 16 6 16

0 0 0 0

0 0 0 1

0 0 0 0

0 0 0 0

D T Q T

.



Next we write from Eq. (11.27)

2

2

2 2 2

0 0 0

0 0 0

0 0 0 0

0 0 0 0

D

and, along the other path,

4 3 4 4 3

4 3 4 3

3 6 4 4 3 3

4 4 3 4 3

0 cos cos

cos 0 sin

sin 0 sin

0 0 0 0

B

B

B

B

y

yD y D D

y

4

4 6 4 4

4

0 0 0

0 0 0

0 0 0

0 0 0 0

B

B

yD y D

Since the velocity of point A must agree along the two paths

2 2

2 2

2 3

100cos 100cos

100sin 100sin

100 100

1 1

2 2 2 4 3 4 3 4

2 2 2 4 3 4 3

2 4 2 4 3 4 3

100sin 100sin cos cos 4

100cos 100cos cos 100sin

0 100cos 100sin sin sin

0 0

B

B

B

y

y

y

At the position where 2 40 , 3 24.83 , 4 52.55 , 208 mmBy , and

2 48.0 rad/s , these equations can be solved for 3 6.891 rad/s , 4 19.444 rad/s ,

and 112.357 in/sBy . With these values we can evaluate

3

0 4.191 19.444 34.865

4.191 0 5.471 112.357

19.444 5.471 0 45.513

0 0 0 0

and 4

0 0 19.444 0

0 0 0 112.357

19.444 0 0 0

0 0 0 0

from which we write the vector forms of the results: ˆ2808 mm/sB V j , 3

ˆ ˆ ˆ5.471 19.444 4.191 rad/s ω i j k , and 4ˆ19.444 rad/sω j . Ans.

Note that the global x1 and z1 axis orientations in this solution differ from those of

Problem 11.21 because of the conventions of the Denavit-Hartenberg parameters. This is

also the reason that the components of 3 seem switched.



Chapter 12

Robotics

12.1 For the SCARA robot shown in Fig. P12.1, find the transformation matrix T15 relating the

position of the tool coordinate system to the ground coordinate system when the joint

actuators are set to the values 1

30 , 2

60 , 3

50 mm , and 4

0 . Also find

the absolute position of the tool point that has coordinates x5 = y5 = 0, z5 = 35 mm.

12 23250 mma a ,

34 450a a ,

12 34 450 ,

23180 ,

12 1 ,

23 2 ,

340 ,

45 4 ,

12300 mms ,

230s ,

34 3s , and

4550 mms .

See the solution to Problem 12.2 for the formulae before numerical evaluation.

1 15 5

0.866 0.500 0 433 0 433 mm

0.500 0.866 0 0 0 0

0 0 1 200 35 mm 165 mm

0 0 0 1 1 1

R T R

Ans.



12.2 Repeat Problem 12.1 using arbitrary (symbolic) values for the joint variables.

12 23250 mma a ,

34 450a a ,

12 34 450 ,

23180 ,

12 1 ,

23 2 ,

340 ,

45 4 ,

12300 mms ,

230s ,

34 3s ,

4550 mms

1 1 1

1 1 1

12

cos sin 0 250cos mm

sin cos 0 250sin mm

0 0 1 300 mm

0 0 0 1

T

2 2 1

2 2 1

23

cos sin 0 250cos mm

sin cos 0 250sin mm

0 0 1 0

0 0 0 1

T

34

3

1 0 0 0

0 1 0 0

0 0 1

0 0 0 1

T

4 4

4 4

45

cos sin 0 0

sin cos 0 0

0 0 1 50 mm

0 0 0 1

T

1 2 1 2 1 1 2

1 2 1 2 1 1 2

13 12 23

cos sin 0 250cos 250cos mm

sin cos 0 250sin 250sin mm

0 0 1 300 mm

0 0 0 1

T T T

1 2 1 2 1 1 2

1 2 1 2 1 1 2

14 13 34

3



0 0 1 300 mm

0 0 0 1

T T T

1 2 4 1 2 4 1 1 2

1 2 4 1 2 4 1 1 2

15 14 45

3



0 0 1 250 mm

0 0 0 1

T T T

Ans.

5

0

0

35 mm

1

R

1 1 2

1 1 2

1 15 5

3

250cos 250cos mm

250sin 250sin mm

215 mm

1

R T R

Ans.



12.3 For the gantry robot shown in Fig. P12.3, find the transformation matrix T15 relating the

position of the tool coordinate system to the ground coordinate system when the joint

actuators are set to the values 1

450 mm , 2

181.25 mm , 3

50 mm , and 4

0 .

Also find the absolute position of the tool point that has coordinates x5 = y5 = 0, z5 =

43.75 mm.

12 23 34 450a a a a ,

1290 ,

2390 ,

34 450 ,

12 2390 ,

340 ,

45 4 ,

12 1s ,

23 2s ,

34 3s , and 45

50 mms .

See the solution to Problem 12.4 for the formulae before numerical evaluation.

15

0 1 0 181.25 mm

0 0 1 100 mm

1 0 0 450 mm

0 0 0 1

T

1

181.25 mm

143.75 mm

450 mm

1

R

Ans.



12.4 Repeat Problem 12.3 using arbitrary (symbolic) values for the joint variables.

12 23 34 450a a a a

, 1290

, 2390

, 34 450

, 12 2390

, 340

,

45 4

, 12 1s

, 23 2s

, 34 3s

, 45

50 mms

12

1

0 0 1 0

1 0 0 0

0 1 0

0 0 0 1

T

23

2

0 0 1 0

1 0 0 0

0 1 0

0 0 0 1

T

34

3

1 0 0 0

0 1 0 0

0 0 1

0 0 0 1

T

4 4

4 4

45

cos sin 0 0

sin cos 0 0

0 0 1 50 mm

0 0 0 1

T

2

13 12 23

1

0 1 0

0 0 1 0

1 0 0

0 0 0 1

T T T

2

3

14 13 34

1

0 1 0

0 0 1

1 0 0

0 0 0 1

T T T

4 4 2

3

15 14 45

4 4 1

sin cos 0

0 0 1 50 mm

cos sin 0

0 0 0 1

T T T

Ans.

5

0

0

43.75 mm

1

R

2

3

1 15 5

1

93.75 mm

1

R T R

Ans.



12.5 For the SCARA robot of Problem 12.1 in the position described, find the instantaneous

velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 35 mm, if the actuators

have (constant) velocities of 1

0.20 rad/s , 2

0.35 rad/s , and 3 4

0 .

1 1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D Q

1

2 12 2 12

0 1 0 125 mm/rad

1 0 0 216.5 mm/rad

0 0 0 0

0 0 0 0

D T Q T

1

3 13 3 13

0 0 0 0

0 0 0 0

0 0 0 1

0 0 0 0

D T Q T

1

4 14 4 14

0 1 0 0

1 0 0 433 mm/rad

0 0 0 0

0 0 0 0

D T Q T

2 1 1 1

0 0.200 0 0

0.200 0 0 0

0 0 0 0

0 0 0 0

D

3 2 2 2

0 0.150 0 43.75 mm/s

0.150 0 0 75.78 mm/s

0 0 0 0

0 0 0 0

D

4 3 3 3

0 0.150 0 45.75 mm/s

0.150 0 0 75.78 mm/s

0 0 0 0

0 0 0 0

D

5 4 4 4

0 0.150 0 45.75 mm/s

0.150 0 0 75.78 mm/s

0 0 0 0

0 0 0 0

D

2 1 1 1 1 1 1 1 1

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

D D D

3 2 2 2 2 2 2 2 2

2

2

0 0 0 15.16 mm/s

0 0 0 8.75 mm/s

0 0 0 0

0 0 0 0

D D D

4 3 3 3 3 3 3 3 3

2

2

0 0 0 15.16 mm/s

0 0 0 8.75 mm/s

0 0 0 0

0 0 0 0

D D D

5 4 4 4 4 4 4 4 4

2

2

0 0 0 15.16 mm/s

0 0 0 8.75 mm/s

0 0 0 0

0 0 0 0

D D D

5 5 1

43.75 mm/s

10.83 mm/s

0

0

R R

2

2

5 5 5 5 1

13.5 mm/s

2.2 mm/s

0

0

R R

Ans.



12.6 For the gantry robot of Problem 12.3 in the position described, find the instantaneous

velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 43.75 mm, if the

actuators have (constant) velocities of 1 2 0 , 3 40 mm/s , and 4 20 rad/s .

1 1

0 0 0 0

0 0 0 0

0 0 0 1

0 0 0 0

D Q

1

2 12 2 12

0 0 0 1

0 0 0 0

0 0 0 0

0 0 0 0

D T Q T

1

3 13 3 13

0 0 0 0

0 0 0 1

0 0 0 0

0 0 0 0

D T Q T

1

4 14 4 14

0 0 1 450

0 0 0 0

1 0 0 180

0 0 0 0

D T Q T

2 1 1 1 0D 3 2 2 2 0D

4 3 3 3

0 0 0 0

0 0 0 40

0 0 0 0

0 0 0 0

D

5 4 4 4

0 0 500 9000

0 0 0 40

500 0 0 3600

0 0 0 0

D

2 1 1 1 1 1 1 1 1 0D D D 3 2 2 2 2 2 2 2 2 0D D D

4 3 3 3 3 3 3 3 3 0D D D 5 4 4 4 4 4 4 4 4 0D D D

5 5 1

0

40 mm/s

0

0

R R

5 5 5 5 1

0

0

0

0

R R

Ans.



12.7 The SCARA robot of Problem 12.1 is to be guided along a path for which the origin of

the end effector O5 follows the straight line given by

5

1 1 1

ˆ ˆ ˆ( ) 40 100 30 75 50 mmO

t t t R i j k

with t varying from 0.0 to 5.0 s; the orientation of the end effector is to remain constant

with 5 1

ˆ ˆ k k (vertically downward) and 5

i radially outward from the base of the robot.

Find expressions for how each of the actuators must be driven, as functions of time, to

achieve this motion.

From the problem statement we construct the figure shown for the path described.

From this we write the transformation T15.

15

0.800 0.600 0 100 40

0.600 0.800 0 75 30

0 0 1 50

0 0 0 1

t

tT

However, as shown in the solution for Problem 12.2,

1 2 4 1 2 4 1 1 2

1 2 4 1 2 4 1 1 2

15 14 45

3

cos sin 0 250cos 250cos

sin cos 0 250sin 250sin

0 0 1 250

0 0 0 1

T T T

.

Equating these gives

1 1 2

1 1 2

3

1 2 4

250cos 250cos 100 40 100 1.000 0.400

250sin 250sin 75 30 75 1.000 0.400

250 50

36.87

t t

t t

( )

( )

( )

( )

a

b

c

d



Squaring and adding Eqs. (a) and (b) gives

2

2125 000 62 500cos 15 625 1.000 0.800 0.160t t

1 2

2 cos 0.04 0.20 1.75t t 2180 0 Ans.

where the quadrant was found from the figure of the robot.

Expanding the trigonometric functions and recognizing that 2 is now known, Eqs. (a)

and (b) become

2 1 2 1

2 1 2 1

250 1 cos cos 250 sin sin 100 1.000 0.400

250 sin cos 250 1 cos sin 75 1.000 0.400

t

t

which can be solved for 1sin and 1cos

1 2 2sin 3 1 cos 4sin

1 2 2cos 4 1 cos 3sin

where 220 1 cos 1.000 0.400t

From the ratio of these we get

2 21

1

2 2

3 1 cos 4sintan

4 1 cos 3sin

Ans.

where the quadrant of 1 is found by considering the signs of the numerator and

denominator separately.

With both 1 and 2 known, Eqs. (c) and (d) give

3 200 mm Ans.

4 1 2 36.87 Ans.



12.8 The gantry robot of Problem 12.3 is to travel a path for which the origin of the end

effector O5 follows the straight line given by

5

1 1 1

ˆ ˆ ˆ( ) 120 300 150 90 225 mmO

t t t R i j k

with t varying from 0.0 to 4.0 s; the orientation of the end effector is to remain constant

with 5 1

ˆˆ k j (vertically downward) and 5 1

ˆ ˆi i . Find expressions for the positions of

each of the actuators, as functions of time, for this motion.

From Problem 12.4 and the problem statement we can write

4 4 2

3

15

4 4 1

sin cos 0 1 0 0 120 300

0 0 1 50 0 0 1 150

cos sin 0 0 1 0 90 225

0 0 0 1 0 0 0 1

t

Tt

Equating individual elements and solving, we get

1 90 225 mmt Ans.

2 120 300 mmt Ans.

3 100 mm Ans.

4 90 Ans.



12.9 The end effector of the SCARA robot of Problem 12.1 is working against a force loading

of 1 1

ˆ ˆ10 5 Nti k and a constant torque loading of 1

ˆ25 mm Nk as it follows the trajectory

described in Problem 12.7. Find the torques required at the actuators, as functions of

time, to achieve the motion described.

Using the 1 , 2 , 3 , and 4 values from Problem 12.7 and formulae from Problems 12.2

and 12.5,

1 1

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

D Q

1

11

2 12 2 12

0 1 0 250sin

1 0 0 250cos

0 0 0 0

0 0 0 0

D T Q T

1

3 13 3 13

0 0 0 0

0 0 0 0

0 0 0 1

0 0 0 0

D T Q T

1 1 2

1 1 1 2

4 14 4 14

0 1 0 250sin 250sin

1 0 0 250cos 250cos

0 0 0 0

0 0 0 0

D T Q T

From these the Jacobian and loads are

1 1 1 2

1 1 1 2

0 0 0 0

0 0 0 0

1 1 0 1

0 250sin 0 250sin 250sin

0 250cos 0 250cos 250cos

0 0 1 0

J

0

0

25 mm N

10 N

0

5 N

F

Now, from Eq. (12.19), we get

1 25 mm N Ans.

2 125 2500sin mm N Ans.

3 5 N Ans.

4 1 1 225 2500sin 2500sin mm N Ans.



12.10 The end effector of the gantry robot of Problem 12.3 is working against a force loading of

1 1

ˆ ˆ20 10 lbti j and a constant torque loading of 1

ˆ22.5 NMj as it follows the trajectory

described in Problem 12.8. Find the torques required at the actuators, as functions of time,

to achieve the motion described.

Using the 1 , 2 , 3 , and 4 values from Problem 12.8 and formulae from Problems 12.4

and 12.6,

1 1

0 0 0 0

0 0 0 0

0 0 0 1

0 0 0 0

D Q

1

2 12 2 12

0 0 0 1

0 0 0 0

0 0 0 0

0 0 0 0

D T Q T

1

3 13 3 13

0 0 0 0

0 0 0 1

0 0 0 0

0 0 0 0

D T Q T

1

4 14 4 14

0 0 1 1.107 0.406

0 0 0 1.356 0.542

1 0 0 0

0 0 0 0

D T Q T

From these, the Jacobian and loads are

0 0 0 0

0 0 0 1

0 0 0 0

0 1 0 1.107 0.406

0 0 1 1.356 0.542

1 0 0 0

Jt

t

0

22.6 N M

0

89 N

44.5 N

0

F

t

Now, from Eq. (12.19), we get

1 0 Ans.

2 89 N Ans.

3 44.5 Nt Ans. 2

4 2.26 5.424 54.24 Mt t N Ans.



PART 3

DYNAMICS OF MACHINES



Chapter 13

Static Force Analysis

13.1 Figure P13.1 illustrates four mechanisms and the external forces and torques exerted on

or by the mechanisms. Sketch the free-body diagram of each part of each mechanism.

Do not attempt to show the magnitudes of the forces, except roughly, but do sketch them

in their proper locations and orientations.



13.2 What moment 12M must be applied to the crank of the mechanism illustrated in Fig.

P13.2 if 4005 NP ?

275 mm, 350 mm.AO BAR R

Kinematic analysis:

1 1sin sin sin 75 mmsin105 350 mm 11.95r

Force analysis: ˆ4005 N P i

14 34 14 34ˆ ˆ ˆ ˆ ˆ ˆ ˆcos sin 4005 N 0.978 0.207 0F F F F F P j i j i j i j

344005 N 0.978 0F 34 4005 N 0.978 4094 NF

14 340.207 0F F 14 0.207 4094 N 845.5 NF

32 34ˆ ˆ ˆ ˆ4094 N 0.978 0.207 4005 845.5 N F F i j i j

12 2 32 12ˆ ˆ ˆ ˆ75 mm cos105 sin105 4005 845.5 N M M r F M i j × i j 0

12 277.98 N.M M 0 12ˆ277.98 N.M M k Ans.



13.3 If 12 452 N mM for the mechanism illustrated in Fig. P13.2, what force P is required to

maintain static equilibrium?

275 mm, 350 mmAO BAR R

Kinematic analysis:

Recall 11.95 from Problem 13.2.

cos cos 75 mmcos105 350 mmcos11.95 323 mmx r

Force analysis:

14 12 0z

O xF M M

14 12 452 N.M 323 mm 139938 NF M x

Recall the force polygon on link 4 from Problem 13.2.

14 tan 1399.38 N tan11.95 6510.35 NP F Ans.



13.4 Find the frame reactions and torque 12M necessary to maintain equilibrium of the four-

bar linkage shown in Fig. P13.4a.

Kinematic analysis:

2

ˆ ˆ87.5 mm 210 75.775 43.75 mmAO R i j4

ˆ ˆ150 mm 135.53 107.05 105.075 mmBO R i j

ˆ ˆ150 mm 82.83 18.725 148.825 mmBA R i j4

ˆ ˆ100 mm 135.53 71.375 70.05 mmCO R i j

445 N

445 N

Force analysis:

4 4 4 34

34

34

ˆ ˆ ˆ ˆ71.375 70.05 mm 311.732 317.574 N

ˆ ˆ ˆ ˆ 107.05 105.075 in cos82.83 sin82.83

ˆ45.203 N M 119.325 in

O CO BO

F

F

M R × P R ×F 0

i j × i j

i j × i j 0

k 0

34 9.4702 N mF 34ˆ ˆ372.941 N 82.83 46.569 370.022 N F i j

4 34 14i F P F F 0 14ˆ ˆ265.153 52.447 N 270.288 N 168.81 F i j

Ans.

2 32 12i F F F 0 12ˆ ˆ46.569 370.022 N 372.941 N 82.83 F i j

Ans.



2 212 32

12ˆ ˆ ˆ ˆˆ 75.75 43.75 mm 46.569 370.021 N

O AO

M

M M R ×F 0

k i j × i j 0

12 26.408 N MM 12ˆ26.408 N M M k Ans.

13.5 What torque must be applied to link 2 of the linkage illustrated in Fig. P13.4b to maintain

static equilibrium?

22.5 N

2 4 4 2 4 487.5 mm; 150 mm; 100 mm; 50 mm; 175 mmAO BA BO CO O O DOR R R R R R

Kinematic analysis:

2

ˆ ˆ87.5 mm 240 43.75 75.75 mmAO R i j4

ˆ ˆ150 mm 152.64 133.225 68.925 mmBO R i j

ˆ ˆ150 mm 105.26 39.475 144.7 mmBA R i j4

ˆ ˆ175 mm 152.64 155.425 80.425 mmDO R i j

222.5 N

222.5 N

Force analysis:

4 4 4 34

34

34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ155.425 80.425 in 222.5 N 133.225 68.925 mm cos105.26 sin105.26

ˆ18.176 N M 110.375 in

O DO BO

F

F

M R P R F 0

i j × i i j × i j 0

k 0

34 4.116 N MF

34ˆ ˆ162.109 N 105.26 42.66 156.391 N F i j



2 212 32

12ˆ ˆ ˆ ˆˆ 43.75 75.751 mm 42.66 156.39 N

O AO

M

M M R ×F 0

k i j × i j 0

12 10.23 N MM 12ˆ10.23 N M M k Ans.



13.6 Sketch a complete free-body diagram of each link of the linkage illustrated in Fig. P13.6.

What force P is necessary for equilibrium?

2 4 4100 mm, 150 mm, 125 mm, 200 mm, 400 mm, AO BA BO CO CDR R R R R and

2 460 mm.O OR

Kinematic analysis:

2

ˆ100 mm 90 100 mmAO R j 4

ˆ ˆ125 mm 44.3 89 87 mmBO R i j

ˆ ˆ150 mm 4.86 149 13 mmBA R i j 4

ˆ ˆ200 mm 44.3 143 140 mmCO R i j

ˆ ˆ400 mm 20.44 375 140 mmDC R i j

Force analysis:

2 212 32

32

32 32

ˆ ˆ ˆˆ90 N m 100 mm cos 4.86 sin 4.86

ˆ90 N m 99.640 mm 903 N 4.86

O AO

F

F

M M R ×F 0

k j × i j 0

k 0 F

4 4 434 54

54

54 54

ˆ ˆ ˆ ˆ89 87 mm cos175.14 sin175.14 903 N

ˆ ˆ ˆ ˆ 143 140 mm cos 20.44 sin 20.44

ˆ ˆˆ86 N m 181 mm 472 N 20.44 =44 165 N

O BO CO

F

F

M R ×F R ×F 0

i j × i j

i j × i j 0

k 0 F i j

16ˆ ˆ ˆ ˆ443 N 165 N =P F F i i j j 0 ˆ443 NP i Ans.



13.7 Determine the torque 12M required to drive slider 6 of Fig. P13.7 against a load of

112.5 NP at a crank angle of 30 , or as specified by your instructor.

250 mm

150 mm

2 462.5 mm; 400 mm; 200 mm.AO BO BCR R R

Kinematic analysis:

2

ˆ ˆ62.5 mm 30 54.125 31.25 mmAO R i j

4

ˆ ˆ7.466 in 73.37 2.136 7.154 inAO R i j

4

ˆ ˆ400 mm 73.37 114.45 383.275 mmBO R i j ˆ ˆ200 mm 175.20 199.3 16.725 mmCB R i j

Force analysis:

16 56ˆ ˆ ˆ ˆcos175.20 sin175.20P F F F i j i j 0

56 cos175.20 1112.5 N 0.996 1116.416 NF P

56ˆ ˆ1116.416 N 175.20 112.5 93.419 N F i j

4 4 454 34

34

34 34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ114.45 383.275 mm 12.5 93.419 N 53.4 178.85 mm cos163.37 sin163.37

ˆ4439.954 N M 186.65 mm 2341.679 N 163.37 = 2243.

O BO AO

F

F

M R ×F R ×F 0

i j × i j + i j × i j 0

k 0 F ˆ ˆ735 670.165 Ni j

2 212 32

12ˆ ˆ ˆ ˆˆ 54.125 31.25 mm 2243.73 670.165 N

O AO

M

M M R F 0

k i j i j 0

12 106.38 N MM 12ˆ106.38 N M M k Ans.

13.8 Sketch complete free-body diagrams for the four-bar linkage illustrated in Fig. P13.8.

What torque M12 must be applied to link 2 to maintain static equilibrium at the position

shown?



Kinematic analysis:

2

ˆ ˆ200 mm 60 100 173 mmAO R i j 4

ˆ ˆ350 mm 109.05 114 331 mmCO R i j

ˆ ˆ400 mm 46.06 278 288 mmBA R i j , ˆ ˆ700 mm 46.06 486 504 mmCA R i j

Force analysis:

Since the lines of action of all constraint forces cannot be found from two- and three-

force members, the force 34F is resolved into radial and transverse components,

34 34 and r F F . Then

4 414 34 34

ˆ ˆ ˆ ˆˆ45 N m 114 331 mm cos 19.05 sin 19.05O CO F M M + R ×F k + i j × i j 0

34ˆ ˆ45 N m 350 mmF k + k 0 34

ˆ ˆ129 N 19.05 =122 42 N F i j

43 43

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ278 288 mm 350 N 486 504 mm 122 42 N

ˆ ˆ ˆ ˆ + 486 504 mm cos70.95 sin 70.95

r

A BA CA CA

rF

M R P R ×F + R ×F 0

i j × i + i j × i j

i j × i j 0

43ˆ ˆ ˆ101 N m 41 N m+624 mm rF k k k 0 , 43

ˆ ˆ228 N 70.95 =74 215 Nr F i j

43 43 43ˆ ˆ48 257 N 261 N 100.58r F F F i j

Now the lines of action for other forces may be found as shown.

43 43 23

r F F F P F 0

23ˆ ˆ ˆ ˆ ˆ74 215 N 122 42 N 350 N i j i j i F 0 , 23

ˆ ˆ=398 257 N 474 N 32.85 F i j

2 212 32 12

ˆ ˆ ˆ ˆˆ 100 173 mm 398 257 NO AO M M M R ×F k i j × i j 0

12 94.55 N mM 12ˆ94.55 N m M k Ans.

13.9 Sketch free-body diagrams of each link and show in Fig. P13.9 all the forces acting. Find

the magnitude and direction of the moment that must be applied to link 2 to drive the



linkage against the forces shown.

Kinematic analysis:

2

ˆ ˆ200 mm 30 173.2 100 mmAO R i j4

ˆ ˆ500 mm 84.34 1.973 19.902 inCO R i j

ˆ ˆ700 mm 67.81 264.375 648.15 mmBA R i j ˆ ˆ700 mm 34.61 23.045 15.903 inCA R i j

4

ˆ ˆ350 mm 84.34 34.525 348.3 mmDO R i j

445 N

890 N

445 N

890 N

Force analysis:




4 4 4 34

34ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ34.5251 348.3 mm 858.85 231.4 N 49.325 497.55 mm cos 5.66 sin 5.66

O DO D CO

F

M R ×P R ×F 0

i j × i j + i j × i j 0

34ˆ ˆ307 N M 499.95 mmF k k 0

34ˆ ˆ614 lN 5.66 =609.65 62.3 N F i j

43 43

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ264.375 648.15 mm 445 N 576.125 397.5755 mm 609.65 62.3 N

ˆ ˆ ˆ ˆ + 576.125 397.575 mm cos 95.66 sin 95.66

r

A BA B CA CA

rF

M R ×P R ×F R ×F 0

i j × i + i j × i j

i j × i j 0

43ˆ ˆ ˆ288.36 N M 278.347 N M 534.15 mm rF k k k 0 ,

43ˆ ˆ1059.1N 95.66 = 102.35 1054.65 Nr F i j

43 43 43ˆ ˆ712 992.35 N 1219.3 N 125.66r F F F i j


43 43 23

r

B

F F F P F 0

23ˆ ˆ ˆ ˆ ˆ102.35 1054.65 N 609.65 62.3 N 445 N i j i j i F 0 ,



23ˆ ˆ=1157 992.35 N 1526.35 N 40.62 F i j

2 212 32 12

ˆ ˆ ˆ ˆˆ; 173.2 100 mm 1157 992.35 NO AO M M M R ×F 0 k i j × i j 0

12 56 N MM 12ˆ56 N M M k Ans.

13.10 Figure P13.10 illustrates a four-bar linkage with external forces applied at points B and C.

Draw a free-body diagram of each link and show all the forces acting on each. Find the

torque that must be applied to link 2 to maintain equilibrium.

2 4 2150 mm, 600 mm, 600 mm,AO CA O OR R R

4400 mm,CO BAR R and

300 mm.BCR

Kinematic analysis:

2

ˆ ˆ150 mm 30 130 75 mmAO R i j 4

ˆ ˆ400 mm 124.56 227 329 mmCO R i j

ˆ ˆ400 mm 16.00 385 110 mmBA R i j ˆ ˆ600 mm 42.38 443 404 mmCA R i j

Force analysis:

43

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ385 110 mm 354 354 N + 443 404 mm 1 800 N + 443 404 mm cos124.56 sin124.56

A BA B CA C CA

F

i j × i j i j × i i j × i j 0

M R ×P R ×P R ×F 0

43ˆ ˆ ˆ174.876 N m 726.000 N m 594 mm rF k k k 0 ,

43ˆ ˆ927 124.56 N= 526 763 N F i j

43 23B C F F P P F 0



23ˆ ˆ ˆ ˆ ˆ526 763 N 354 354 N+1 800 N i j i j i F 0 ,

23ˆ ˆ= 920 1 117 N 1 447 N 129.48 F i j

2 212 32 12

ˆ ˆ ˆ ˆˆ; 130 75 mm 920 1 117 NO AO M M M R ×F 0 k i j × i j 0

12 214.21 N mM 12ˆ214.21 N m M k Ans.

13.11 Draw a free-body diagram of each of the members of the mechanism illustrated in Fig.

P13.11 and find the magnitudes and the directions of all the forces and moments.

Compute the magnitude and direction of the torque that must be applied to link 2 to

maintain static equilibrium. 534 N

801 N

2 4 2 4 450 mm; 125 mm; 100 mm; 75 mm; 50 mm; 175 mm;AO CA O O CO DO DC BAR R R R R R R

62.5 mm.BCR

Kinematic analysis:

2

ˆ50 mm 180 50 mmAO R i

ˆ ˆ175 mm 55.98 97.9 145 mmBA R i j

4

ˆ ˆ100 mm 124.23 56.25 82.675 mmCO R i j

ˆ ˆ125 mm 41.41 93.75 82.675 mmCA R i j

4

ˆ ˆ75 mm 95.27 6.9 74.675 mmDO R i j

534 N534 N

801 N

Force analysis:






4 4 4 34

34ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ6.9 74.675 mm 694.2 400.5 N 56.25 82.675 mm cos34.23 sin34.23

O DO D CO

F

M R ×P R ×F 0

i j × i j + i j × i j = 0

34ˆ ˆ49 N M 100 mmF k k 0

34ˆ ˆ489.5 N 34.23 =404.95 275.9 N F i j

43 43

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ97.9 145 mm 534 N 93.75 82.675 mm 405 276 N

ˆ ˆ ˆ ˆ + 93.75 82.675 mm cos 55.77 sin 55.77

r

A BA B CA CA

rF

M R ×P + R ×F R ×F 0

i j × i + i j × i j

i j × i j 0

43ˆ ˆ ˆ77.43 N M 7.565 N M 124 mm rF k k k 0 , 43

ˆ ˆ685.3 N 55.77 =387.15 565.15 Nr F i j

43 43 43ˆ ˆ17.8 841 N 841 N 91.21r F F F i j

, 34 43ˆ ˆ17.8 841 N 841 N 88.79 F F i j

Ans.


43 23B F F P F 0 , 23ˆ ˆ ˆ17.8 841 N 534 N i j i F 0 ,

23ˆ ˆ=53.4 841 N 1005.7 N 56.73 F i j , 32 23

ˆ ˆ124 189 lb 226 lb 123.27 F F i j Ans.

34 14D F F P F 0 , 14ˆ ˆ ˆ ˆ4 189 lb 156 90 lb i j i j F 0 ,

14ˆ ˆ=676.4 1241.55 N 1415.1 N 61.42 F i j , 12 32

ˆ ˆ534 841 N 1005.7 N 56.73 F F i j Ans.

2 212 32 12

ˆ ˆ ˆˆ; 50 mm 534 841 NO AO M M M R ×F 0 k i × i j = 0

12 42 N MM 12ˆ42 N M M k Ans.

13.12 Determine the magnitude and direction of the torque that must be applied to link 2 to

maintain static equilibrium.

445 N

222.5 N

275 mm; 350 mm; 175 mm; 200 mm.AO CA BA BCR R R R

Kinematic analysis:

2

ˆ75 mm 90 75 mmAO R j ˆ ˆ350 mm 12.37 341.275 75 mmCA R i j

ˆ ˆ175 mm 34.93 143.475 100.225 mmBA R i j



445 N

222.5 N

Force analysis:

14

14ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ143.475 100.225 mm 222.5 N 341.875 75 mm 445 N 341.875 75 in

A BA B CA C CA

F

M R ×P R ×P R ×F 0

i j × j + i j × i + i j × j = 0

14ˆ ˆ ˆ31.927 N M 33.375 N M 350 mmF k k k 0 14

ˆ4.45 N 90 =4.45 N F j

14 23B C F P P F F 0

23ˆ ˆ ˆ222.5 N 445 N 4.45 N j i j F 0 , 23

ˆ ˆ=445 226.95 N 498.4 N 27.02 F i j

2 212 32O AO M M R F 0 12ˆ ˆ ˆˆ 75 mm 445 226.95 NM k + j × i j 0

12 33.375 N MM 12ˆ33.375 N M M k Ans.



13.13 Figure P13.13 shows a Figee floating crane with leminscate boom configuration. Also

shown is a schematic diagram of the crane. The lifting capacity is 16 T (with 1 T = 1

metric ton =1 000 kg) including the grab, which is about 10 T. The maximum outreach is

30 m, which corresponds to the position 2 49 . Minimum outreach is 10.5 m at

2 132 . Other dimensions are given in the legend to Fig. P13.13. For the maximum

outreach position and a grab load of 10 T (under standard gravity), find the bearing

reactions at A, B, 2 4, and ,O O as well as the moment 12M required. Notice that the

photograph shows a counterweight on link 2; neglect this weight and also the weights of

the members.

214.7 m;AOR 6.5 m;BAR

419.3 m;BOR 22.3 m;CAR 16 m.CBR (Photograph and

dimensions by permission from B.V. Machinefabriek Figee, Haarlem, Holland)



Kinematic analysis:

2

ˆ ˆ14.700 m 49.00 9.644 11.094 mAO R i j , 4

ˆ ˆ19.300 m 59.70 9.739 16.663 mBO R i j

ˆ ˆ6.500 m 2.37 6.494 0.269 mBA R i j , ˆ ˆ22.300 m 14.39 21.600 5.543 mCA R i j

Force analysis:

Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane

would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is

21 000 kg 9.81 m/s 9.810 kNW mg . Therefore, the stated load on the crane is 98.100 kNF .

43

43ˆ ˆ ˆ ˆ ˆ ˆ ˆ6.494 0.269 m cos59.70 sin 59.70 21.600 5.543 m 98.100 kN

A BA CA

F

M R ×F R ×F 0

i j × i j + i j × j 0

43ˆ ˆ5.471 m 2 119 kN mF k k 0 ,

43ˆ ˆ387 59.70 kN=195 334 kN F i j , 14 34

ˆ ˆ387 kN 59.70 =195 334 kN F F i j , Ans.

43 23 F F F F 0

23ˆ ˆ ˆ195 334 kN 98.1 kN i j j F 0 ,

23ˆ ˆ= 195 236 kN 307 kN 129.59 F i j , 12 32

ˆ ˆ= = 195 236 kN 307 kN 129.59 F F i j , Ans.

2 212 32O AO M M R ×F 0 ; 12ˆ ˆ ˆ ˆˆ 9.644 11.094 m 195 236 kNM k i j × i j 0

12 113 kN mM 12ˆ113 kN m M k Ans.



13.14 Repeat Problem 13.13 for the minimum outreach position.

Kinematic analysis:

2

ˆ ˆ14.700 m 132.00 9.836 10.924 mAO R i j , 4

ˆ ˆ19.300 m 120.35 9.751 16.656 mBO R i j

ˆ ˆ6.500 m 3.81 6.486 0.432 mBA R i j , ˆ ˆ22.300 m 15.83 21.454 6.083 mCA R i j

Force analysis:

Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane

would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is

21 000 kg 9.81 m/s 9.810 kNW mg . Therefore, the stated load on the crane is 98.100 kNF .

43

43ˆ ˆ ˆ ˆ ˆ ˆ ˆ6.486 0.432 m cos120.35 sin120.35 + 21.454 6.083 m 98.1 kN

A BA CA

F

M R ×F R ×F 0

i j × i j i j × j = 0

43ˆ ˆ5.815 m 2 105 kN mF k k 0 ,

43ˆ ˆ362 kN 120.35 = 183 312 kN F i j , 14 34

ˆ ˆ362 kN 120.35 = 183 312 kN F F i j Ans.

43 23 F F F F 0 23ˆ ˆ ˆ183 312 kN 98.1 kN i j j F 0

23ˆ ˆ=183 214 kN 282 kN 49.51 F i j , 12 32

ˆ ˆ= =183 214 kN 282 kN 49.51 F F i j , Ans.

2 212 32 12

ˆ ˆ ˆ ˆˆ, 9.836 10.924 m 183 214 kNO AO M M M R ×F 0 k i j × i j 0

12 109 kN mM 12ˆ109 kN m M k Ans.



13.15 Repeat Problem 13.7 assuming coefficients of Coulomb friction 0.20c between links

1 and 6 and 0.10c between links 3 and 4. Determine the torque 12M necessary to

drive the system, including friction, against the load P.

250 mm

150 mm

2 462.5 mm; 400 mm; 200 mm.AO BO BCR R R

See the figure and solution for Problem 13.7 for the kinematic and frictionless solutions.

For friction between links 1 and 6, the friction angle is 1tan 0.20 11.31 . Since

the impending motion 6 /1CV is to the left the friction force 16 16

n

cf F is toward the right.

Also, since the non-friction normal force 16

nF is downward (from the solution for Problem

13.7), the total force 16F acts at the angle 90 11.31 78.69 . Therefore,

16 56ˆ ˆ ˆ ˆ ˆ ˆcos 78.69 sin 78.69 cos175.20 sin175.20P F F F i i j j i j 0

16 561112.5 N cos 78.69 cos175.20 0F F , 16 56sin 78.69 sin175.20 0F F

16 96.89 NF , 56 1135.484 NF , 56ˆ ˆ1135.484 N 175.20 1131.5 95 N F i j . For friction

between links 3 and 4, the friction angle is 1tan 0.10 5.71 . Since the impending

motion 3 / 4AV is upward the friction force 34 34

n

cf F is upward. Also, since the non-

friction normal force 34

nF is toward the left (from the solution for Problem 13.7), the total

force 34F acts at the angle 163.37 5.71 157.66 . Therefore,

4 4 454 34

34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ114.45 383.275 mm 1131.5 95 N 53.4 178.85 mm cos157.66 sin157.66

O BO AO

F

M R ×F R ×F 0

i j × i j i j × i j 0

34ˆ444.55 N M 185.725 mmF k 0 , 34

ˆ ˆ2393.6 N 157.66 = 2213.95 909.815 N F i j

2 212 32O AO M M R ×F 0 , 12ˆ ˆ ˆ ˆˆ 54.125 31.25 mm 2213.95 909.815 NM k i j × i j 0

12 118.22 N MM 12ˆ118.22 N M M k Ans.



13.16 Repeat Problem 13.12 assuming a coefficient of static friction 0.20 between links 1

and 4. Determine the torque 12M necessary to overcome friction.

445 N

222.5 N

275 mm; 350 mm; 175 mm; 200 mm.AO CA BA BCR R R R

See the figure and solution for Problem 13.12 for the kinematic and frictionless solution.

For friction between links 1 and 4, the friction angle is 1tan 0.20 11.31 . Since

the impending motion 4 /1CV is to the right the friction force 14 14

n

cf F is toward the left.

Also, since the non-friction normal force 14

nF is upward (from the solution for Problem

13.12), the total force 14F acts at the angle 90 11.31 101.31 . Therefore,

14

14

ˆ ˆ ˆ ˆ ˆ ˆ143.475 100.225 mm 222.5 N + 341.875 75 mm 445 N

ˆ ˆ ˆ ˆ + 341.875 75 mm cos101.31 sin101.31

A BA B CA C CA

F

M R P R ×P R ×F 0

i j × j i j × i

i j × i j 0

14ˆ ˆ ˆ31.923 N M 33.375 N M 349.95 mmF k k k 0 14

ˆ ˆ4.147 N 101.31 = 0.814 3.964 N F i j

14 23B C F P P F F 0

23ˆ ˆ ˆ ˆ222.5 N 445 N 0.814 3.964 N j i i j F 0 , 23

ˆ ˆ=445.8 226.54 N 500 N 26.94 F i j

2 212 32O AO M M R F 0 , 12ˆ ˆ ˆˆ 75 mm 445.8 226.54 NM k j i j 0

12 33.435 NM 12ˆ33.435 N M M k Ans.



13.17 In each case shown, pinion 2 is the driver, gear 3 is an idler, and the gears have module of

4.20 mm/tooth and 20 pressure angle. For each case, sketch the free-body diagram of

gear 3 and show all forces acting. For (a) pinion 2 rotates at 600 rev/min and transmits

18 hp to the gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits 25 hp

to the gearset.

(a) 22

m 4.20 18 teeth37.8 mm

2 2

NR

3

3

m 4.20 34 teeth71.4 mm

2 2

NR

2

600 rev/min 262.832 rad/s cw

60 s/min

, 2

3 2

3

37.8 mm62.832 rad/s 33.264 rad/s ccw

71.4 mm

R

R

23 3

3 3

18 hp 734.25 N M/s/hp5565 N

71.4 mm 10 m 33.264 rad/s

t PF

R

23 23 cos 5565 N cos20 5922.15 NtF F , 43 23 5922.15 NF F Ans.

23 43 13 F F F F 0 13 23 43 5565 N 5565 N 11130 Nt tF F F Ans.



(b) 22


2 2

NR

3

3

m 4.20 mm/tooth36 teeth75.6 mm

2 2

NR

2

900 rev/min 294.248 rad/s ccw

60 s/min

, 2

3 2

3

37.8 mm94.248 rad/s 47.124 rad/s cw

75.6 mm

R

R

23

3 3

25 hp 734.25 N M/s/hp5152.5 N

75.6 mm 10.3 m 47.124 rad/s

t PF

R

23 23 cos 5152.5 N cos20 5483 NtF F , 43 23 5483 NF F Ans.

23 43 13 F F F F 0

13 23 43

5483 N 20 5483 110

F F F

4634.7 N 225 Ans.

(c) 22

m 4.20 18 37.8 mm

2 2

NR

3

3

m 4.20 36 75.6 mm

2 2

NR

2


60 s/min

, 2

3 2

3

37.8 mm94.248 rad/s 47.124 rad/s cw

75.6 mm

R

R

23

3 3

25 hp 734.25 N M/s/hp5152.5 N

75.6 mm 47.124 rad/s

t PF

R

23 23 cos 5152.5 N cos20 5483 NtF F , 43 23 5483 NF F Ans.

23 43 13 F F F F 0

13 23 43

5483 N 20 5483 N 70

F F F

9931.5 N 135 Ans.



13.18 A 15-tooth spur pinion with module of 5 mm/tooth and 20 pressure angle, rotates at 600

rev/min, and drives a 60-tooth gear. The drive transmits 18 kW. Construct a free-body

diagram of each gear showing upon it the tangential and radial components of the forces

and their proper directions.

22

15 teeth5 mm/tooth37.500 mm

2 2

mNR

33

60 teeth5 mm/tooth150 mm

2 2

mNR

2

600 rev/min 262.832 rad/s

60 s/min

2

3 2

3

1.500 in62.832 rad/s 15.708 rad/s

6.000 in

R

R

32

3 3

18 kW 1000 N m/s/kW 1000 mm/m1 910 N

150 mm 62.832 rad/s

t PF

R

, 32 32 tan 695 Nr tF F

23 32 1 910 Nt tF F 23 32 695 Nr rF F



13.19 A 16-tooth pinion on shaft 2 rotates at 1 720 rev/min and transmits 5 hp to the double-

reduction gear train. All gears have 20 pressure angle. The distances between centers

of the bearings and gears for shaft 3 are shown in Fig. P13.19. Find the magnitude and

direction of the radial force that each bearing exerts against the shaft.

4.20 m,

3.175 mm,

50 mm 200 mm 50 mm

22

m 3.175 mm/teeth 16 teeth25.4 mm

2 2

NR

m 3.175 mm/teeth 64 teeth

101.6 mm2 2

AA

NR

m 4.20mm/teeth 24 teeth50.4 mm

2 2

BB

NR

4


2 2

ANR

2

1 720 rev/min 2180.118 rad/s

60 s/min

2

3 2

25.4 mm180.118 rad/s 45.029 rad/s

101.6 mmA

R

R

4 3

4

50.4 mm45.029 rad/s 30.020 rad/s

75.6 mm

BR

R

23

3

5 hp 734.25 NM/s/hp802.47 N

101.6 mm 45.029 rad/s

t

A

PF

R

, 23 23 cos 855 NtF F

43 23

101.6 mm802.47 N 1617.67 N

50.4 mm

t tA

B

RF F

R 43 43 cos 1721.49 NtF F



Choosing a coordinate system with origin at C as shown we have

23ˆ ˆ855 N 20 802.47 293.79 NA F F i j ˆ ˆ101.6 50.4 mmA R j k

43ˆ ˆ1721.49 N 20 1617.67 527.64 NB F F i j ˆ ˆ50.4 254 mmB R j k

ˆ ˆx y

C C CF F F i j C R 0

ˆ ˆx y

D D DF F F i j ˆ304.8 mmD R k

C A A B B D D M R ×F R ×F R ×F 0

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ101.6 50.4 mm 803.47 293.79 N 50.4 254 mm 1617.67 587.84 N 304.8 mm x y

D DF F j k × i j j k × i j k × i j 0

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ14.79 40.72 81.43 N M 148.4 407.6 81.43 N M 304.8 304.8 mmy x

D DF F i j k i j k i j 0

1495.2 N, 445 Nx y

D DF F ˆ ˆ350 lb 163.42 1495.2 445 ND F i j Ans.

A B C D F F F F F 0

ˆ ˆ ˆ ˆ ˆ ˆ803.47 293.79 N 1617.67 587.84 N 1495.2 445 NC i j i j F i j 0,

ˆ ˆ961.2 N 188.86 952.3 146.85 NC F i j Ans.



13.20 Solve Problem 13.17 if each pinion has right-hand helical teeth with a 30 helix angle

and a 20 pressure angle. All gears in the train are helical, and, of course, the module is

4.20 mm/teeth for each case.

Since the pressure angles and the helix angle are related by cos tan tann t ,

1 1tan tan cos tan tan 20 cos30 22.80t n

(a) 22


2 2

NR

3

3


2 2

NR

2

600 rev/min 262.832 rad/s cw

60 s/min

, 2

3 2

3

37.8 mm62.832 rad/s 33.264 rad/s ccw

71.4 mm

R

R

23

3 3

18 hp 734.25 N M/s/hp5607 N

71.4 mm 33.264 rad/s

t PF

R

23 23 tan 5607 N tan 22.80 2358.5 Nr t

tF F ,

23 23 tan 5607 N tan30 3235.15 Na tF F

23ˆ ˆ ˆ5607 2358.5 3235.15 N F i j k 43

ˆ ˆ ˆ5607 2358.5 3235.15 N F i j k

23 43 13 F F F F 0 13ˆ11218.45 N F i Ans.

3 23 3 43 13ˆ ˆR R M j×F j F M 0

13ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ71.4 mm 5607 2358.5 3235.15 N 71.4 mm 5607 2358.5 3235.15 lb j × i j k j × i j k M 0

13ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ71.4 mm 5607 2358.5 3235.15 N 71.4 mm 5607 2358.5 3235.15 N j × i j k j × i j k M 0

13ˆ458.238 N M M i This moment must be supplied by the shaft bearings. Ans.



(b) 22


2 2

NR

3

3


2 2

NR

2


60 s/min

, 2

3 2

3

37.8 mm94.248 rad/s 47.124 rad/s cw

75.6 mm

R

R

23

3 3

25 hp 734.25 N M/s/hp5152.5 N

75.6 mm 47.124 rad/s

t PF

R

23 23 tan 5152.5 N tan 22.80 2184.95 Nr t

tF F ,

23 23 tan 5152.5 N tan30 2999.3 Na tF F

23ˆ ˆ ˆ5152.5 2184.95 2999.3 N F i j k 43

ˆ ˆ ˆ2184.95 5152.5 2999.3 N F i j k

23 43 13 F F F F 0 13ˆ ˆ3 3 N F i j Ans.

3 23 3 43 13ˆ ˆR R M j×F i ×F M 0

13ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ75.6 mm 5152.5 2184.95 2999.3 N 756 mm 2184.95 5152.5 2999.3 N j × i j k i × i j k M 0

13ˆ ˆ224.95 224.95 N M M i j This moment must be supplied by the shaft bearings. Ans.

(c) 22


2 2

NR

3

3


2 2

NR

2


60 s/min

, 2

3 2

3

37.8 mm94.248 rad/s 47.124 rad/s cw

75.6 mm

R

R



23

3 3

25 hp 734.25 N M/s/hp5152.5 N

75.6 mm 47.124 rad/s

t PF

R

23 23 tan 5152.5 N tan 22.80 2184.95 Nr t

tF F ,

23 23 tan 5607 N tan30 2999.3 Na tF F

23ˆ ˆ ˆ5152.5 2184.95 2999.3 N F i j k 43

ˆ ˆ ˆ2184.95 5152.5 2999.3 N F i j k

23 43 13 F F F F 0 13ˆ ˆ7378.1 7378.1 N F i j Ans.

3 23 3 43 13ˆ ˆR R M j×F i ×F M 0

13ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ75.6 mm 5152.5 2184.95 2999.3 N 75.6 mm 2184.95 5152.5 2999.3 N j × i j k i × i j k M 0

13ˆ ˆ224.95 224.95 N M M i j This moment must be supplied by the shaft bearings. Ans.

13.21 Analyze the gear shaft of Example 13.8 and find the bearing reactions andC DF F .

The solution is shown in Fig. 13.20c.

623 N

1223.75N

207 N623 N

1806.7 N

761 N

316 N

690 N525 N

58.3 N

1117 N 98

.5m

m

94.82 mm

525 623 1171 N

316 6901 N

C

D

F i j k

F i k

ˆ ˆ ˆ

ˆ ˆ Ans.



13.22 In each of the bevel gear drives illustrated in Fig. P13.22, bearing A takes both thrust load

and radial load, whereas bearing B takes only radial load. The teeth are cut with a 20

pressure angle. For (a) 2ˆ20 N M T i and for (b) 2

ˆ26.7 N M T k . Compute the

bearing loads for each case.

17.25 mm

34.5 mm2.5 mm

50 mm 26.25 mm

32 mm

42.5 mm62.5 mm

4.2 mm50 mm

(a) 1tan 32 teeth 16 teeth 63.43

32 2 2 20 N M 17.25 mm 1161.45 NtF T R

32 32 tan cos 189.12 Nr tF F 32 32 tan sin 377.8 Na tF F

23ˆ ˆ ˆ189.1 1161.45 377.8 N F i j k

23 3A BA B PA M R ×F R ×F T 0

3ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ50 mm 34.5 59 mm 189.12.5 1161.45 377.8 Nx y

B BF F T k × i j i k × i j k k 0

3ˆ ˆ ˆ ˆ ˆ ˆ50 mm 50 mm 68.49 1.889 40 N My x

B BF F T i j i j k k 0

3ˆ40 N M T k ˆ ˆ37.825 1370.6 NB F i j Ans.

23A B F F F F 0 ˆ ˆ ˆ226.95 209.15 378.25 NA F i j k Ans.

(b) 1tan 18 teeth 24 teeth 36.87

32 2 2 26.7 N M 32 mm 834.375 NtF T R

32 32 tan cos 242.97 Nr tF F 32 32 tan sin 182 Na tF F

32ˆ ˆ ˆ242.97 834.375 182 N F i j k

32 2A BA B PA M R ×F R ×F T 0

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ50 mm 32 20 mm 242.97 834.375 182 N 26.7 N Mx y

B BF F k × i j i k × i j k k 0

ˆ ˆ ˆ ˆ ˆ ˆ50 mm 50 mm 16.687 0.89 26.7 N M 26.7 N My x

B BF F i j i j k k 0

ˆ ˆ17.8 333.75 lbB F i j Ans.



23A B F F F F 0 ˆ ˆ ˆ222.5 1170.35 182.45 NA F i j k Ans.



13.23 Figure P13.23 illustrates a gear train composed of a pair of helical gears and a pair of

straight-tooth bevel gears. Shaft 4 is the output of the train and delivers 6 hp to the load

at a speed of 370 rev/min. All gears have pressure angles of 20 . If bearing E is to take

both thrust load and radial load, whereas bearing F is to take only radial load, determine

the force that each bearing exerts against shaft 4.

43.7

5 m

m

18.

75 m

m

2m,

12.5 mm,

21.875 mm,

21.875 mm,

3m,

The diameters of the bevel gears at their large faces are

4 4m 2 3 mm/tooth×2 teeth 60 mmR N 3 3m 2 3 mm/tooth 2 teeth 30 mmR N

1

4 3tan 63.43R R 1

3 4tan 26.57R R

The average pitch radii are

4,avg 4 12.5sin 51.3 mmR R 3,avg 3 12.5sin 25.65 mmR R

4


60 s/min

34

4,avg 4

6 hp 734.25 N M/s/hp2216 N

51.3 mm 38.746 rad/s

t PF

R

34 34 tan cos 360.45 Nr tF F 34 34 tan sin 720.9 Na tF F

34ˆ ˆ ˆ720.9 360.45 2216 N F i j k

34 4E FE F PE M R ×F R ×F T 0

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ60 mm 18 51.3 mm 720.9 360.45 2216 N 113.69 N My z

F FF F i × j k i j × i j k i 0

ˆ ˆ ˆ ˆˆ ˆ60 mm 60 mm 113.69 40 30.48 in lb 113.69 N Mz y

F FF F j k i j k i 0

ˆ ˆ489.5 640.8 NF F j k Ans.

34E F F F F F 0 ˆ ˆ ˆ720.9 849.95 1575.3 NE F i j k Ans.



13.24 Using the data of Problem 13.23, find the forces exerted by bearings C and D onto shaft

3. Which of these bearings should take the thrust load if the shaft is to be loaded in

compression?

The pitch radius of the helical gear S is

m 2 2 mm/teeth 35 teeth 2 35 mmS SR N

23 43 3,avg 2216 N 25.65 mm 36.45 mm 1559.4 Nt t

SF F R R

23 23 tan 1559.4 N tan 20 567.575 Nr tF F

23 23 tan 1559.4 N tan30 900.32 Na tF F

23ˆ ˆ ˆ567.575 900.32 1559.4 N F i j k

43 23C DC D PC RC M R ×F R ×F R ×F 0

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ43.75 mm 25.65 67.425 mm 720.9 360.45 2216 N

ˆ ˆ ˆ ˆ ˆ36.45 21.875 in 567.575 900.32 1559.4 N

x z

D DF F

j × i k i j × i j k

i j × i j k 0

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ43.75 mm 43.75 mm 149.4 56.85 39.38 N m 34 56.85 45.278 N mz x

D DF F i k i j k i j k 0

ˆ ˆ133.5 4191.9 ND F i k Ans.

23 43C D F F F F F 0 ˆ ˆ ˆ284.8 538.45 418.3 NC F i j k Ans.

Since the thrust force is in the ˆj direction, C should be a thrust bearing.



13.25 Use the method of virtual work to solve the slider-crank mechanism of Problem 13.2.

1 1sin sin sin 75 mm sin105 350 mm 11.95r

cos cos 75 mmcos105 350 mmcos11.95 342.4 mmx r

The first-order kinematic coefficient is

24 12tan 342.4 mmtan11.95 72.467 mmI Ix dx d R x

12 4005 N 72.467 mm 290.23 N M cwM Px Ans.

13.26 Use the method of virtual work to solve the four-bar linkage of Problem 13.5.

222.5 N

2 4 4 2 4 487.5 mm; 150 mm; 100 mm; 50 mm; 175 mm.AO BA BO CO O O DOR R R R R R

The first-order kinematic coefficient is

24 12 24 144 4 2 64.420 mm 114.425 mm 0.563I I I Id d R R

414 sin152.64 175 mm 222.5 N sin152.64 17.89 N M cwDOM R P

12 14 4 2 14 4 17.89 N M cw 0.563 10.072 N M ccwM M d d M Ans.



13.27 Use the method of virtual work to analyze the crank-shaper linkage of Problem 13.7.

Given that the load remains constant at ˆ100 lb,P i find and plot a graph of the crank

torque 12M for all positions in the cycle using increments of 30 for the input crank.

4 4 4 2cos 2.5cosAO AOx R 4 4 4 2sin 6 2.5sinAO AOy R

1 24

2

6 2.5sintan

2.5cos

4 242.25 30sinAOR

24 14 4 4sinI I AOy R 24 14

4

24 14

44 4

2

61 6sin

I I

AO

I I

ydR

d y

5 48sin 16 16sinBCy 1

5 4sin 2 2sin

46

5 5 4 5

5 4 5

8sin 8cos 16cos 8cos

8 sin 2cos tan

I C BC C CBy y x x

46 144 5 4 516 8sin 16cos tanC I Idx d y

12 4 4 2CM dx d d d P



Values for one cycle are shown in the following table.

2 (deg.) 4 (deg.) 4(mm)AOR 4 2d d 5(deg.) 4 (mm)Cdx d 12 (N m)M

0 67.38 162.5 0.147 93 8.85 393.175 25.88

30 73.37 189.15 0.240 15 4.80 392.875 41.98

60 81.30 206.5 0.281 97 1.32 396.775 49.78

90 90.00 212.5 0.294 12 0.00 400 52.35

120 98.70 206.5 0.281 97 1.32 394 49.43

150 106.63 189.15 0.240 15 4.80 373.65 39.93

180 112.62 162.5 0.147 93 8.85 345.275 22.73

210 114.50 130.5 0.04593 10.37 333.65

240 108.05 100.85 0.41416 5.65 368.05 67.83

270 90.00 87.5 0.71429 0.00 400 127.143

300 71.95 100.85 0.41416 5.65 392.575 72.35

330 65.50 130.5 0.04593 10.37 394.35 8.06

360 67.38 162.5 0.147 93 8.85 393.75 25.88

The values of 12M from this table are graphed as follows:

100

50

-50

-100

-150



13.28 Use the method of virtual work to solve the four-bar linkage of Problem 13.10.

23 12 23 133 3 2 75 mm 688 mm 0.1089I I I Id d R R

132 3

ˆ ˆ0.1089 568 mm 135.33 61.881 mm 45.33B B BId d R R k R k

132 3

ˆ ˆ0.1089 662 mm 124.56 72.153 mm 34.56C C CId d R R k R k

12

500 N 135 61.881 mm 45.33 1 800 N 0 72.153 mm 34.56

30.940cos89.67 N m+129.876cos34.56 N m

B B C CM

P R P R

12 107 N m cwM Ans.



13.29 A car (link 2) that weighs 8900 N is slowly backing a 4450 N trailer (link 3) up a 30

inclined ramp as illustrated in Fig. P13.29. The car wheels are of 325 mm radius, and the

trailer wheels have 250 mm radius; the center of the hitch ball is also 13 in above the

roadway. The centers of mass of the car and trailer are located at 2G and 3,G

respectively, and gravity acts vertically downward in Fig. 13.29. The weights of the

wheels and friction in the bearings are considered negligible. Assume that there are no

brakes applied on the car or on the trailer, and that the car has front-wheel drive.

Determine the loads on each of the wheels and the minimum coefficient of static friction

between the driving wheels and the road to avoid slipping.

1250 m

m

600 mm

800 mm

1200 mm

2000 mm

900 mm

8900 N

4450 N

For the trailer:

3

ˆ ˆ945 863.15 mm 1279.9 mm 42.41G B R i j

313 3ˆ1250 mmB G BF M k R W 0

13ˆ ˆ3364.2 N 120 1682 2914.75 N F i j Ans.

13 23 3 F F F W 0

23ˆ ˆ2278.49 N 42.41 1682 1535.25 N F i j

For the car:

12 32ˆ ˆˆ ˆ2000 mm 800 mm 8900 N 2900 325 mmR

P F M k k i j F 0

12ˆ5513.55 NR F j Ans.

12 12 3212

ˆF Rf F F i F F 0 12ˆ4921.7 NF F j Ans.



12ˆ1682 Nf i Ans.

12 12 1682 N 4921.7 NFf F 0.34 Ans.

13.30 Repeat Problem 13.29 assuming that the car has rear-wheel drive rather than front-wheel

drive.

The entire solution is identical with that of Problem 13.29 except that friction force 12f

acts on the rear wheel of the car instead of on the front wheel. The solution process and

all values are the same until the final step. Then

12 12 1682 N 5513.5 NRf F 0.31 Ans.



13.31 The low-speed disk cam with oscillating flat-faced follower illustrated in Fig. P13.31 is

driven at a constant shaft speed. The displacement curve for the cam has a full-rise

cycloidal motion, defined by Eq. (6.13), with parameters 30L , 30 , and a prime-

circle radius 30 mmoR ; the instant pictured is at 2 112.5 . A force of 8 NCF is

applied at point C and remains at 45 from the face of the follower as demonstrated. Use

the virtual work approach to determine the moment 12M required on the crankshaft at

the instant shown to produce this motion.

2 350 mm; 42 mm; 150 mm.O O B CR R R

The moment on link 3 caused by the output load is

313

ˆ ˆ150 mm 8 N sin 135 0.849 N mCO C M R ×F k k

From Eq. (6.13b),

2 30 112.51 cos 1 cos 2 0.200

150 150

Ly

From virtual work

12 3 2 13 13ˆ ˆ0.200 0.849 N m 0.170 N md d y M M M k k Ans.

13.32 Repeat Problem 13.31 for the entire lift portion of the cycle, finding 12M as a function of

2.

From Problem 13.31

313

ˆ ˆ150 mm 8 N sin 135 0.849 N mCO C M R F k k

22

3602 301 cos 1 cos 0.200 1 cos 2.4

150 150

Ly

12 13 2 2ˆ ˆ0.200 1 cos2.4 0.849 N m 0.170 1 cos2.4 N my M M k k Ans.



13.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an applied

torque T as illustrated in Fig. P13.33. Assume that a coefficient of static friction of

exists between the disk and ground and that all other joints are frictionless. A force F is

acting vertically downward on the bar at a distance d from the pivot 2.O Assume that the

weights of the links are negligible in comparison to F. Find an equation for the torque T

required as a function of the distance 2COx R , and an equation for the final distance x

that is reached when friction no longer allows further movement.

20O M 32

cos

cosB B

d dF F F

x x

0C M 23 sin sinB

RdT F R F

X

But, for geometric compatibility, sinR x . Therefore,

22sinB BT F d F d R x Ans.

Also, 2

2

sec 1 tan 11

sin tan tan tan

Rx R R R

Motion is still possible as long as tan , or as long as

21 1x R Ans.



13.34 Links 2 and 3 are pinned together at B and a constant vertical load 800 kNP is applied

at B as illustrated in Fig. P13.34. Link 2 is fixed in the ground at A and link 3 is pinned to

the ground at C. The length of link 2 is 8 m and it has a 150 mm solid square cross-

section. The length of link 3 is 0.5 m and it has a solid circular cross-section with

diameter D. Both links are made from a steel with a modulus of elasticity 207 GPaE

and a compressive yield strength yc 200 MPa.S Using the theoretical values for the end

condition constants of each link, determine: (i) the slenderness ratio, the critical load, and

the factor of safety guarding against buckling of link 2; and (ii) the minimum diameter

minD of link 3 if the static factor of safety guarding against buckling is to be 2N .

(i) The area moment of inertia and the radius of gyration of link 2, respectively, are

33

44

2

150 mm 150 mm0.421 88 10 m

12 12

bhI

4

22 2

2

/12 0.150 m0.043 30 m

12 12

I b bk

A b

Therefore, the slenderness ratio of link 2 is

2

2r

2

8 m184.75

0.043 30 m

LS

k Ans.

The end-condition constant for link 2 (with fixed-pinned ends) is 2 2.C Therefore, the

slenderness ratio at the point of tangency is

2

9

2r 6

yc

2 2 207 10 Pa2202.14

200 10 PaD

C ES

S

Comparing these gives 2 2

r r DS S . Therefore, the Johnson parabolic equation must be

used to determine the critical load of link 2.

The critical unit load of link 2 is

2

2

cr yc r2yc

2 2

1

2

P S SS

A C E



Substituting the given information into this equation, the critical load of link 2 is

2

6

cr 2.62 10 NP Ans.

The axial compressive load 2F is the component of the vertical load P acting along the

central axis of the link 2 as shown in this figure:

F2

F3

P

30°

60°

Therefore, the axial compressive load in link 2 is

2

ocos30 692.8 kNF P

Similarly, the axial compressive load in link 3 is

3

osin30 400 kNF P

The factor of safety guarding against buckling for link 2 is

2

6cr

2 3

2

2.62 10 N3.782

692.8 10 N

PN

F

Ans.

Therefore, link 2 is safe against this axial compressive load.

(ii) The end condition constant for link 3 (with pinned-pinned ends) is 3 1C . The

slenderness ratio at the point of tangency for link 3 is

3

9

3r 6

yc

2 2 1 207 10 Pa142.93

200 10 PaD

C ES

S

(1)

The cross-sectional area and the area moment of inertia of link 3 are 2

34

DA

and

4

364

DI

Therefore, the radius of gyration for the link 3, in terms of the diameter D, is

33

3 4

I Dk

A

Therefore, the slenderness ratio for link 3, in terms of diameter D, is

3

3

3

0.5 m

4

2 mr

D

LS

k D

To determine the minimum diameter of link 3 to prevent buckling, the critical load must

be derived in terms of the diameter D for both the Euler column formula and the Johnson

parabolic equation. The critical load can be written as

3cr 3 3P N F

Substituting 3 2N and 3F from above, the critical load is



3cr 2 400 kN 800 kNP

CASE (1). The critical load in terms of the diameter D from the Euler column formula is

3

3

211 43

cr 3 2 4

r

N4.011 44 10

m

C EP A D

S

Equating with the load F3 gives

3

11 4 4 cr 4.011 44 10 N m 800 kNP D

Therefore, the minimum diameter from the Euler column formula is

Euler 0.038 mD (2)

CASE (2). The critical load can be written in terms of the diameter D from the Johnson

parabolic equation as

3

3

2

r

cr 3

3

1

2yc

yc

S SP A S

C E

or as

3

8 2 cr 1.570 8 10 Pa 15 377.3 NP D

Equating with the load F3 gives

3

8 2 cr 1.570 8 10 Pa 15 377.3 N 800 kNP D

Therefore, the minimum diameter of the link 3 from the Johnson parabolic equation is

Johnson 0.072 mD (3)

Note that the diameter given by the Euler column formula, Eq.(2), is smaller than the

diameter from the Johnson parabolic equation, Eq. (3), i.e., ( Euler JohnsonD D ).

In certain cases, the claim that the bigger of the two diameters is the correct answer

may not be true. Therefore, to determine the minimum diameter of the column we need

to decide which of the two criteria is valid. The slenderness ratio must be compared with

the slenderness ratio at the point of tangency for both diameters.

Tthe slenderness ratio of link 3 is

3r E

Euler

2 m52.63S

D

Comparing with Eq. (1), the conclusion is

33E Dr rS S

Therefore, the Euler column formula is not appropriate. So min 0.038 m.D From Eq.

(3), the slenderness ratio of link 3 can be written as

3r J

Johnson

2 m27.78S

D

The conclusion is

3 3r J r D

S S

Therefore, the Johnson parabolic equation is the valid equation. The minimum diameter

of the column 3 from the Johnson parabolic equation is

min Johnson 0.072 m 72 mmD D Ans.



13.35 The horizontal link 2 is subjected to the load 150 kNF at C as illustrated in Fig.

P13.35. The link is supported by the solid circular aluminum link 3. The lengths of the

links are 2 5 m,CAL R 3 m,BAR and

3 3 m.BDL R The end D of link 3 is fixed in

the ground and the opposite end B is pinned to link 2 (that is, the effective length of the

link is EFF 30.5L L ). For aluminum, the yield strength is Syc = 370 MPa and the

modulus of elasticity is E = 207 GPa. Determine the diameter d of the solid circular

cross-section of link 3 to ensure that the static factor of safety is N = 2.5.

The cross-sectional area and the area moment of inertia of link 3, respectively, are

A = π d 2

/ 4 and I = π d 4

/ 64

Therefore, the radius of gyration of the link is 4

2

64

4 4

I d dk

A d

Using the effective length EFF 30.5L L , the slenderness ratio of the link is

r EFF 0.5 3 m / 4 6 mS L k d d (1)

The slenderness ratio at the point of tangency is

9 6

r yc2 2 207 10 Pa 370 10 Pa 105.09D

S E S (2)

Taking moments about A gives

3 m 5 m cos60P F

where P is the compressive load acting at B on link 3. Solving this equation, the

compressive load is

5 m 150 000 N 0.5125 000

3 mP N

The factor of safety guarding against buckling of link 3 is defined as

crN P P

Substituting N = 2.5, the critical unit load is

cr 2.5 125 000 N 312 500 NP

Using the Euler column formula, the critical unit load can be written as 2

cr2

r

P EA S



Which, with the available data and Eq. (1), can be written as

2 9

22

207 10 Pa312 500 N

4 6 md d

Rearranging this equation gives 4 8 4701.12 10 md

Therefore, the diameter of link 3 is

0.0515 m 51.5 mmd (3)

Using the Johnson parabolic equation, the critical unit load can be written as 2

y rcry

12

S SPS

A E

which can be written as 2

66

2 9

312 500 N 1 370 10 10 m370 10 Pa

4 207 10 Pa 2

d

d

Rearranging this equation gives 2 3 25.60 10 m d


0.074 8 m 74.8 mmd (4)

To check which answer is valid, that is, Eq. (3) or Eq. (4), recall that the slenderness

ratio, from Eq. (1), is

Sr = 6 m/ d

To check the Euler column formula, the slenderness ratio is

(Sr)EULER = 6 m / (0.0515 m) = 116.5

Comparing this answer with Eq. (2) indicates that

(Sr)EULER > (Sr)D that is 116.5 > 105.09

Therefore, the Euler column formula is a valid equation. The correct diameter of link 3 is

d = 51.5 mm Ans.

Next we check the validity of the Johnson parabolic equation. The slenderness ratio is

(Sr)JOHNSON = 6 m / (0.0748 m) = 80.21

Therefore

(Sr)JOHNSON < (Sr)D; that is, whether 80.21 < 105.09, which is not possible. Therefore, the

Johnson parabolic equation is not valid.



13.36 The horizontal link 2 is subjected to the inclined load 8 000 NF at C as illustrated in

Fig. P13.36. The link is supported by a solid circular cross-section link 3 whose length

35 m L BD . The end D of the vertical link 3 is fixed in the ground link and the end B

supports link 2 (that is, the effective length of the link is 30.5EFFL L ). Link 3 is a steel

with a compressive yield strength yc 370 MPaS and a modulus of elasticity

207 GPaE . Determine the diameter d of link 3 to ensure that the factor of safety

guarding against buckling is 2.5.N Also, answer the following statements true or false

and briefly give your reasons. (i) The slenderness ratio at the point of tangency between

the Euler column formula and the Johnson parabolic equation does not depend on the

geometry of the column. (ii) Under the same loading conditions, a link with pinned-

pinned ends will give a higher factor of safety against buckling than an identical link with

fixed-fixed ends. (iii) If the slenderness ratio ( )r r DS S at the point of tangency then the

critical unit load does not depend on the yield strength of the column material.

The cross-sectional area and the area moment of inertia of link 3, respectively, are

A = π d 2

/ 4 and I = π d 4

/ 64

Therefore, the radius of gyration of the link is

4

2

64

4 4

I d dk

A d




4 m 5 m cos60P F

where P is the compressive load acting at B on link 3. Rearranging this equation, the

compressive load P is

5 m 8 000 N 0.5 4 m 5 000 NP

Using the effective length EFF 30.5L L , the slenderness ratio of the link is

r EFF / 0.5 5 m 4 10 mS L k d d (1)


1 21 2

9 6

r 2 2 207 10 Pa 370 10 Pa 105.087ycDS E S

The factor of safety guarding against buckling can be written as

N = Pcr / P

Therefore, the critical unit load is

cr 2.5 5 000 N 12 500 NP NP

and the critical unit load for this factor of safety is

2

12 500 Ncr

4

P

A d (2)

From the Euler column formula, the critical unit load can be written as 2

cr

2

r

P E

A S

(3)

Equating Eqs. (2) and (3) gives

2 9

22

12 500 N 207 10 Pa

4 10 m

d d

Rearranging and solving, the diameter of the link using the Euler formula is

0.0297 m 29.7 mmd (4)

From the Johnson parabolic equation, the critical unit load can be written as 2

y rcry

1

2

S SPS

A E

Substituting the known data gives 2

66

2 9

12 500 N 1 370 10 Pa 10 m370 10 Pa

4 207 10 Pa 2

d

d

Rearranging and solving, the diameter of the link using the Johnson parabolic equation is

0.0676 m 67.6 mmd (5)

To determine the correct diameter, that is, Eq. (4) or Eq. (5), the slenderness ratio from

Eq. (1) is

r 10 mS d

Using Eq. (4), the slenderness ratio, from the Euler column formula, is

(Sr)EULER = 10 m / 0.0297 m = 336.7

Since 336.7 > 105.087, (Sr)EULER > (Sr)D is valid. Therefore, the diameter of link 3 is

d = 29.7 mm Ans.



To check the Johnson parabolic equation: Using Eq. (5), the slenderness ratio, from the

Johnson parabolic equation, is

(Sr)JOHNSON = 10 m / 0.0676 m = 147.93

Since 147.93 > 105.087, (Sr)JOHNSON > (Sr)D, which is not possible. Therefore, the


Statement (i) is true. Ans.

The reason is that the slenderness ratio at the point of tangency is defined as

1 2

r yc( ) 2 /DS E S

which is a function of only the material properties of the link (and does not depend on the

geometry of the link).

Statement (ii) is false. Ans.

The factor of safety is defined as N = Pcr / P. Therefore, a higher value for the critical

load Pcr will give a higher value for the factor of safety. The critical load Pcr is greater for

fixed-fixed ends (C = 4) than for pinned-pinned ends (C = 1) from both the Euler column

formula and the Johnson parabolic equation.

Statement (iii) is true. Ans.

When Sr > (Sr)D, then we must use the Euler column formula; that is, 2

cr

2

r

P E

A S

, which

does not depend on the yield strength of the material.



13.37 A load AP is acting at A and a load BP is acting at B of the horizontal link 3, as illustrated

in Fig. P13.37. Link 3 is pinned to the vertical link 2 at O and link 2 is fixed in the

ground link 1 at D. The lengths are 1200 mm,AO 600 mm,OB and

1800 mm.DO Both links have solid circular cross-sections with diameter 50 mmD

and are made from a steel alloy with a compressive yield strength yc 585.65 mPa,S a

tensile yield strength yt 516.75 mPa,S and modulus of elasticity

3206.7 10 mPa.E

Assuming that links 2 and 3 are in static equilibrium and using the theoretical value for

the end-condition constant for link 2, determine: (i) the magnitude of the force BP that is

acting as shown at B if 133.5 RN,AP (ii) the critical load, critical unit load, and the

factor of safety to guard against buckling for link 2, and (iii) the diameter of a solid

circular cross-section for link 2 that will ensure the factor of safety guarding against

buckling of the link is N = 4.

(i) The free body diagram of link 3 is as shown in the figure below.

Taking moments about O gives 1200 mm 600 mm 0y y

O A BM P P which can be

written as

1200 mm cos30 600 mm cos30A BP P

Therefore, the reaction force at B is



2B AP P (1)

Substituting the given data into this equation, the reaction force at B is

133 5 RN 267 RNBP Ans.

Taking the sum of the forces in the y-direction on link 3 gives

23 0y y y y

A BF P P F

which can be written as

23 cos30 cos30y

A BF P P

Substituting Eq. (1) gives

23 3 cos30 3 133.5 RN cos30 346.84 RNy

AF P (2)

Taking the sum of the forces in the x-direction gives

23 0x x x x

A BF P P F


12 sin30 sin30x

A BF P P

Substituting Eq. (1), the x-component of the reaction force at O is

23 200.25 RNxF

(ii) The cross-sectional area and the area moment of inertia of link 2, respectively, are

2 2 2(50 mm) 1963.5 mm4 4

A D

and

4 4 4(50 mm) 306796.875 mm64 64

I D

The radius of gyration of link 2 is 4

2

306796.875 mm12.5 mm

1963.5 mm

Ik

A

The slenderness ratio of link 2 can be definded as

r

1800 mm150

12.5 mm

LS

k (3)

and the slenderness ratio of link 2 at the point of tangency can be written as

r

yc

2D

CES

S

where the end condition constant for fixed-pinned end conditions (using the theoretical

value) is C = 2; that is, the effective length for fixed-pinned ends is EFF 0.5L L .

Therefore, the slenderness ratio of link 2 at the point of tangency is

32 2 206.7 10 mPa

118.04585.65 mPa

r DS

(4)

In order to determine the critical load on link 2, we must first determine if this link is an

Euler column or a Johnson column. The criterion for using the Johnson parabolic

equation is

r r DS S



From Eqs. (3) and (4) this implies that 150 118.04 , which is a contradiction. Therefore,

link 2 is an Euler column. The critical load on link 2 from the Euler column equation can

be written as 2

cr 2

r

C EP A

S

(5)

Substituting the known values into this equation, the critical load on link 2 is

3

2 3

cr 2

206.7 10 mPa1963.5 mm 335.695 10 RN or 355.695 RN

150P

Ans.

The critical unit load on link 2 is 3

cr

2

355.695 10 N181.15 mPa

1963.5 mm

P

A

Ans.

Link 2 is in compression as shown in the figure.

The definition of the factor of safety for link 2 is

cr

32

y

PN

F

where, from Eq. (2), 32 77 942 lbyF is the compressive load at point O on link 2.

Therefore, the factor of safety for link 2 is

355.695 N1.025

346840 NN Ans.

(iii) For the circular cross-section of link 2 and a factor of safety N = 4, the critical

load for buckling can be written as

cr 4346840 N

PN

Therefore, the required critical load for the buckling is

cr 346840 N 1387.36 RNP (6)

The cross-section area and the area moment of inertia of the solid circular link 2,

respectively, are 2 4A D and

4 64I D




4

2

64

4 4

I D Dk

A D

The slenderness ratio of link 2 is

r

1800 mm 7200 mm

4

LS

k D D (7)

First consider the Euler column formula. Substituting Eqs. (6) and (7) into Eq. (5), the

critical load is

2 2 2 3

cr 22

r

2 206.7 10 mPa1387.36 RN

4 7200 mm

C E DP A

S D

Rearranging this equation gives 2

4 2

3

4 1387360 N(7200 mm)9556506.65 mm

2 206700 mPaD

Therefore, the diameter of the cross-section of the link 2 is 68.25 mmD ,

Substituting this into Eq. (7), the slenderness ratio of link 2 is

r

7200 mm 7200 mm105.5

68.25 mmS

D (8)

Compaing Eq. (8) with Eq. (4) gives that r r DS S ; that is,105.5 118.04 . Therefore,

the condition for link 2 to be an Euler column is not satisfied; that is, the assumption that

the link is an Euler column is not correct.

Now we assume that link 2 is a Johnson column; the Johnson parabolic equation is 2

y r

cr y

1

2

S SP A S

CE

Substituting Eq. (7) into this equation gives 2

yc2

cr yc

7200 mm1

4 2 206760 mPa 2

S DP D S

Rearranging this equation gives 2

yc2 cr

yc

7200 mm4 1 1

2 206760 mPa 2

SPD

S

Substituting the known values into this equation gives 2

2 24.1387.36 RN 1 585.65 mPa 7200 mm 14781.25 mm

2 206760 mPa 2 585.65 mPaD

Therefore, the diameter of link 2 from the Johnson parabolic equation is 69.146 mmD .

Substituting this into Eq. (7), the slenderness ratio of link 2 is

r

7200 mm 7200 mm104.1

69.146 S

D



Comparing with Eq. (4) now shows that r r DS S ; that is, that 104.1 118.04 .

Therefore, the assertion that the column is a Johnson column is valid. The diameter of

link 2 to guard against buckling is 2.77 inD . Ans.



13.38 The horizontal link 2 is subjected to the load P = 5 000 N as illustrated in Fig. P13.38.

This link is supported by the vertical link 3, which has a constant circular cross-section.

The lengths are 5 m,AC 4 m,AB and 3 5 m.DB L For the vertical link 3, the end

D is fixed in the ground link and the end B supports link 2 (that is, the effective length of

link 3 is 30.5EFFL L ). The yield strength and the modulus of elasticity for the aluminum

link 3 are Sy = 370 MPa and E = 207 GPa, respectively. Determine the diameter d of link

3 to ensure that the static factor of safety guarding against buckling is N = 2.5.

The cross-sectional area and the second moment of area of link 3 can be written as

A = π d 2

/ 4 and I = π d 4

/ 64


4

2

64

4 4

I d dk

A d

Using the effective length 30.5EFFL L , the slenderness ratio of the link is

r EFF 0.5 5 m 4 10 mS L k d d


1 21 2

9 62 2 207 10 Pa 370 10 Pa 105.09r yDS E S

Taking moments about A of all forces on link 2 we find

324 m 5 m sin53.13 0y

AM F P

And therefore the vertical load on link 3 is

32 sin53.13 5 m 4 m 5 000 NyF P

The factor of safety guarding against buckling of link 3 can be written as



2.5crN P P

Therefore, the critical unit load is

2.5 5 000 N=12 500 NcrP NP

(i) Using the Euler column formula, the critical unit load can be written as 2

cr

2

r

P E

A S


2 9

22

207 10 Pa12 500 N

4 10 md d



0.0297 m 29.7 mmd (1)

(ii) Using the Johnson parabolic equation, the critical unit load can be written as 2

y rcry

1

2

S SPS

A E

which can be written as 2

66

2 9

12 500 N 1 370 10 Pa 10 m370 10 Pa

4 207 10 Pa 2

d

d



0.0676 m 67.6 mmd (2)

To check which answer is valid, that is, Eq. (1) or Eq. (2), recall that the slenderness ratio

is defined as

Sr = 10 m/ d

To check the Euler column formula: The slenderness ratio is

(Sr)EULER = 10 m/ (0.0297 m) = 336.7

or (Sr)EULER > (Sr)D since the values are 336.7 > 105.09. Therefore, the Euler column

formula is the valid equation. The correct diameter of the link is

d = 29.7 mm Ans.

Using the Johnson parabolic equation, the slenderness ratio is (Sr)JOHNSON = 10 / 0.0676 =

147.93 or (Sr)JOHNSON > (Sr)D. Since 147.93 < 105.09 is not possible, therefore the




13.39 The horizontal link 2 is pinned to the vertical wall at A and pinned to link 3 at B as

illustrated in Fig. P13.39. The opposite end of link 3 is pinned to the wall at C. A

vertical force 25 kNP is acting on link 2 at B. Link 2 has a 20 × 30 mm solid

rectangular cross-section and link 3 has a 40 × 40 mm solid square cross-section. The

length of link 3 is 1.2 mBC and the angle 30ABC . The two links are made from

a steel alloy with a tensile yield strength yt 190 MPaS , a compressive yield strength

yc 205 MPaS , and a modulus of elasticity GPa207E . Using the theoretical value for

the end-condition constant for link 3, determine: (i) the value of the slenderness ratio at

the point of tangency between the Euler column formula and the Johnson parabolic

formula; (ii) the critical load and the factor of safety guarding against buckling of link 3;

and (iii) the minimum width of the square cross-section of link 3 for the factor of safety to

guard against buckling to be N = 1.

The free-body diagram of link 2 is as shown below.

2

F12Y

F12X

F32Y

F32XA

B

P




32 0y

A BA BAM R F R P

Therefore, the y-component of the reaction force at B is

32 25 kNyF P

The free-body diagram of link 3 is as shown in the next figure.

F13X

F13Y

F23X

F23Y

A

3

B

C Taking moments about C gives

23 23 0y x

C BA ACM R F R F

or

23 23 23tan 60x y yBA

AC

RF F F

R

Therefore, the x- and y-components of the reaction force at B are

23 tan 60 25 kN 43.3 kNxF

and

23 32 25 kNy yF F

The magnitude of the tensile load T on link 2 at B is

32 23 43.3 kNx xT F F (1)

The magnitude of the compressive load appP on link 3 at B is

2 2

app 23 23 23 50 kNx yP F F F (2)

Link 2 is subjected to the tensile load T, which creates a tensile stress in the link. The

factor of safety guarding against yielding for the link is defined as

ytN S (3)

where the tensile stress can be written as

T A (4)



and the cross-sectional area of the link is 4 2(0.02 m)(0.03 m) 6 10 mA (5)

Substituting Eqs. (1) and (5) into Eq. (4), the tensile stress is 3

4 2

43.3 10 N72.17 MPa

6 10 m

(6)

Substituting Eq. (6) and the tensile yield strength into Eq. (3), the factor of safety

guarding against yielding for link 2 is

190 MPa2.63

72.17 MPaN

(i) The slenderness ratio of link 3, at the point of tangency, can be written as

r

yc

2D

CES

S

Using the theoretical value (for pinned-pinned ends), the end-condition constant for the

link is

1C (7)

Substituting 207 GPaE , yc 205 MPaS , and Eq. (7) into Eq. (6), the slenderness

ratio, at the point of tangency, is

9

r 6

2 1 207 10 Pa141.18

205 10 PaDS

Ans. (8)

(ii) The cross-sectional area of link 3 is 2 2 3 2(0.040 m) 1.6 10 mA b (9)

The second moment of area of the link is 4 4

7 4(0.040 m)2.13 10 m

12 12

bI (10)

and the radius of gyration of the link is 7 4

2

3 2

2.13 10 m1.155 10 m

1.6 10 m

Ik

A

Therefore, the slenderness ratio is

r 2

1.2 m103.92

1.155 10 m

LS

k

(11)

In order to determine the critical load for link 3, we must first determine if this column is

an Euler column or a Johnson column. The criterion for using the Johnson parabolic

equation is

r r DS S

From Eqs. (8) and (11) we have r r DS S that is, 18.14192.103 . Therefore, link 3 is

a Johnson column. The critical load for link 3 (that is, the Johnson parabolic equation)

can be written as 2

yc r

cr yc

1

2

S SP A S

CE

(12)



Substituting the known values into this equation, the critical load is 2

63 2 6

cr 9

1 205 10 Pa 103.921.6 10 m 205 10 Pa

1 207 10 Pa 2P

Therefore, the critical load is

cr 239.14 kNP Ans. (13)

The definition of the factor of safety guarding against buckling is

cr

app

PN

P (14)

where appP is the compressive load on column 3 at B and is given by Eq. (2). Substituting

Eqs. (2) and (13) into Eq. (14), the factor of safety guarding against buckling is

239.14 kN4.78

50 kNN Ans.

(iii) The critical load for a factor of safety 1N is cr app 1 50 kN 50 kNP NP .

If we assume that link 3 is an Euler column, then the critical load on link 3 (that is, the

Euler column equation) can be written as 2

cr 2

r

C EAP

S

(15)

From Eqs. (9) and (10), the radius of gyration of link 3 is 4

2

/12

12

I b bk

A b

and from Eq. (11), the slenderness ratio of link 3 is

r 12 /L

S L bk

(16)

Substituting Eqs. (9) and (16) into Eq. (15), the critical load on link 3 can be written as 2 2

cr 2 212 /

C EbP

L b

Rearranging this equation, the minimum width of the link can be written as 1/4

2

cr

2

12L Pb

C E

Substituting the known values into this equation gives

1/4

2 3

9

12 1.2 m 50 10 N0.025 5 m 25.5 mm

21 207 10 Pab

To check whether the assumption of an Euler column is correct, from Eq. (16), the new

slenderness ratio of link 3 is

3

r 12 / 12 1.2 m / (25.5 10 m) 163.02S L b

Comparing this result with Eq. (8) we have r r DS S ; that is, 18.14101.163 . So this

verifies that link 3 is indeed an Euler column.



If we assume that link 3 is a Johnson column, then substituting Eqs. (9) and (16) into

Eq. (12), the critical load can be written as 2

yc2

cr yc

( 12 / )1

2

S L bP b S

CE

or as 2

2yc2

cr yc

12

2

SLP S b

CE

Rearranging this equation, the new width of the link can be written as

22

yc

cr yc

12

2

SLb P S

CE

Substituting the known values into this equation, the width is

2 2

63 6

9

Pa N Pa

Pa

12 1.2 m 205 1050 10 (205 10 ) 26 mm

21 207 10b

Now we must check if the assertion that the link is a Johnson column is correct. From

Eq. (16), the new slenderness ratio of link 3 is

2

r 12 / 12 1.2 m / (2.60 10 m) 159.88S L b

Comparing this result with Eq. (8) we have r r DS S ; that is, 159.88 141.18 . This

means that the assumption that link 3 is a Johnson column is invalid. Link 3 is an Euler

column and the minimum width of the square cross-section (in order for the factor of

safety to guard against buckling to be N = 1) is b = 25.5 mm. Ans.



13.40 The link BC = 1.2 m and 25 mm square cross-section is fixed in the vertical wall at C and

pinned at B to a circular steel cable AB with diameter d = 20 mm as illustrated in Fig.

P13.40. The distance AC = 0.7 m. The mass m of a container, suspended from pin B,

produces a gravitational load at B that results in the moment at point C in the wall

C 8 000 NmM ccw. The yield strength and modulus of elasticity of the steel cable AB

and the steel link BC are y 370 MPaS and 207 GPa,E respectively. Given that m =

2 000 kg, determine: (i) the tension in the cable AB and the factor of safety guarding

against tensile failure; (ii) the compressive load acting in link BC; (iii) the factor of safety

guarding against buckling of link BC. (Use the theoretical value of the end-condition

constant assuming that the link has fixed-pinned ends.) If C 10 000 NmM ccw, then

determine the maximum mass of a container that can be suspended from pin B before

buckling of link BC will begin (that is, the factor of safety guarding against buckling

failure is 1N ).

(i) The angle between the cable and the link is

1 0.7tan 30.256

1.2



The cable AB and the link BC are in static equilibrium. The free body diagram of link BC

is as shown in the following figure.

Summing moments about C at the wall gives

sin30.256 0C CM M L P Lmg (1)

where P is the tension in the cable AB. Rearranging Eq. (1), the tension can be written as

sin 30.256

Cmg M LP

Substituting the given data into this equation, the tension is 22 000 kg 9.81 m/s 8 000 Nm 1.2 m

25 708 Nsin30.256

P

Ans.

The axial stress in the cable can be written as

2 4A

P P

A d

Substituting the given data into this equation, the axial stress in the cable is

2

25 708 N81.83 MPa

0.020 m 4A

The factor of safety guarding against tensile failure in the cable can be written as

y 370 MPa4.52

81.83 MPaA

SN

Ans. (2)

(ii) The compressive load in link BC can be obtained by summing forces in the x-

direction; that is,

cos30.256 0x

cF P P

where cP is the compressive load in the link. Therefore, the compressive load is

cos30.256 22 206 NcP P Ans.

(iii) To determine whether the link is an Euler column or a Johnson column, we first

find the slenderness ratio at the point of tangency between the Euler column formula and

the Johnson parabolic equation

r

y

2D

ECS

S



where the theoretical value of the end-condition constant corresponding to fixed-pinned

end conditions is 2C . Therefore, the slenderness ratio at the point of tangency D is

9

r 6

2 207 10 Pa 2148.62

370 10 PaDS

The radius of gyration of the link can be written as 3 12

12

I bh hk

A bh

Which, for the given data, is

0.025 m0.007 22 m

12k

The slenderness ratio of the link can be written as

r

1.2 m166.2

0.007 22 m

LS

k

Since r r DS S , therefore the link is an Euler column. The critical load using the Euler

column formula can be written as 2 2 2 9

cr 2 2

r

2 (0.025 m) (207 10 Pa)92 452 N

(166.2)

C AEP

S

Therefore, the factor of safety guarding against buckling of the link is

cr

c

92 367 N4.16

22 206 N

PN

P Ans.

Combining Eqs. (1) and (2), the compressive load exerted on the link, as a function of

the mass of the container, is

c cos30.256 cos30.256sin30.256 tan30.256

C Cmg M L mg M LP P

(3)

For a factor of safety guarding against buckling N = 1, the critical load must be equal to

the compressive load; that is,

cr ctan30.256

Cmg M LP P

Rearranging this equation, the mass of the container can be written as

crtan 30.256 CP M Lm

g

Substituting the given data into this equation, the mass of the container is

2

tan30.256 (92 367 N) 10 000 Nm 1.2 m6 342 kg

9.81 m/sm

Ans. (4)

Substituting Eq. (4) into Eq. (3), the compressive load is 2

c

(6 342 kg)(9.81 m/s ) 10 000 Nm 1.2 m92 370 N

tan30.256 tan30.256

Cmg M LP

Note that the compressive load in the link is equal to the critical load. Also, note that it is

important to show that this answer cannot be obtained using the factor of safety found in

part (iii) because the moment has changed; that is, new 2 000 kg 4.16 8 320 kg.m



13.41 A vertically upward force F is applied at C of the horizontal link 4 as illustrated in Fig.

P13.41. The link is pinned to the ground at B and pinned to the vertical link 2 at A. The

lengths 600 mm and 1500 mmAB AC and links 2 and 4 are made from a steel alloy

with a compressive yield strength yc 413.4 mPaS and a modulus of elasticity

3206.7 10 mPa.E Link 2 has a hollow circular cross-section with an outside diameter

50 mm,D wall thickness 6.25 mm,t and length 1500 mm.L Using the theoretical

value for the end-condition constant for link 2, determine: (i) the value of the slenderness

ratio at the point of tangency between Euler’s column formula and Johnson’s parabolic

formula; (ii) the critical load acting on link 2; and (iii) the force F for the factor of safety

of link 2 to guard against buckling to be 1N . (iv) If link 2 has a solid circular cross-

section with diameter 75 mmD and 89 RNF then determine the maximum length

of link 2 for the factor of safety to guard against buckling to be 2N .

(i) The cross-sectional area and the second moment of area for the hollow circular

link 2, respectively, are

2 22 2 24 50 mm 37.5 mm 4 859 mmA D d

and



4 44 4 464 50 mm 37.5 mm 64 209.73 mmI D d

Therefore, the radius of gyration for the link is 4

2

209.73 mm15.62 mm

859 mm

Ik

A

and the slenderness ratio of the link is

r

1500 mm96

15.62 mm

LS

k

The slenderness ratio at the point of tangency can be written as

r

y

2D

CES

S

where the theoretical value for the end-condition constant for link 2 is C = 2 . Therefore,

the slenderness ratio at the point of tangency is

3

r

2 2 206.7 10 mPa140.50

413.4 mPaDS

Ans.

(ii) The critical load is determined by first finding if the link is to be considered an

Euler column or a Johnson column. The criterion for using the Johnson parabolic

equation is r r DS S . In this example, we have 96 < 140.50. Therefore, the link is

regarded as a Johnson column. The critical load from the Johnson parabolic equation is 2

y r

cr y

1

2

S SP A S

CE

Substituting the known values into this equation gives the critical load as

2

2

cr 3

1 413.4 mPa 96859 mm 413.4 mPa 281.297 RN

2 206.7 10 mPa 2P

Ans.

(iii) The definition of the factor of safety of the link is

cr APPN P P

From the given factor of safety for the link, this equation can be written as

cr APP 1N P P

Therefore, the applied force at point A can be written as

APP cr 1 281297 NP P

To determine the force F acting at C in link 4, we take moments about B, which gives

APP900 mm 600 mm 0BM F P

Therefore, the force F acting at point C is

APP600 mm 600 mm 281297 N187531.9 N or 187.53 RN

900 mm 900 mm

PF

Ans.

(iv) Link 2 is a solid circular column with factor of safety of N = 2. To determine the

applied force we take moments about B; that is,



0BM

Therefore, the applied force is

APP

900 mm 900 mm 89 RN133.5 RN

600 mm 600 mm

FP

From the definition of the factor of safety

cr APP 2N P P

Therefore, the critical load is

cr APP2 2 133.5 RN 267 RNP P

The cross-sectional area is

22 24 75 mm 4 4418.75 mmA D

The second moment of area is

44 464 75 mm 64 1554687.5 mmI D

The radius of gyration is 4

2

15546875 mm18.75 mm

4418.75 mm

Ik

A

First, the critical unit load from the Johnson parabolic equation is 2

y rcry

1

2

S SPS

A CE

With known values, this equation can be written as 22

r

2 3

267 RN 413.4 mpa413.4 mPa

4418.75 mm 2.206.7 10 mPa 2

S

Solving this equation gives the slenderness ratio from the Johnson parabolic formula as

r 184.10S

Second, the critical unit load from the Euler column formula is 2

cr

2

r

P C E

A S

Substituting the known values into this equation gives 2 3

2 2

r

267 RN 2 206.7 10 mPa

4418.75 mm S

Therefore, the slenderness ratio from the Euler formula is

264.14rS

A comparison of the two slenderness ratios shows that

264.14 184.10

In other words

(Sr)EULER > (Sr)JOHNSON

The length of link 2 can be written as

rL k S

where the slenderness ratio that is used in this equation is for the Euler column formula.

Therefore, the length of the link is



18.75 mm 264.14 4952.625 mm 4.952 mL Ans.

As an alternative method to determine the critical load, consider the effective length of

the link; that is, LEFF. From the end-condition constant

C = (1/α)2

This defines α = 0.707. Therefore, the effective length of the link can be written as

EFF 0.707 1500 mm 1060.5 mmL L

The slenderness ratio of the link is

EFFr

1060.5 mm67.88

15.625 mm

LS

k


3

r

y

2 206.7 10 mPa99.32

413.4 mPaD

ES

S

The critical load is determined by first finding if the link is an Euler column or a Johnson

column. The criterion for the Johnson column is

r r DS S

In this example, we have

67.88 < 99.32

Therefore, the Johnson parabolic equation must be applied. The equation can be written

as 2

y r

cr y

1

2

S SP A S

E

Substituting the known values into this equation, the critical load is 2

2

cr 3

1 413.4 mPa 67.88859 mm 413.4 mPa 281.3 N

206.7 10 mPa 2P

Note that this answer is in good agreement with Eq. (1).



13.42 A force F is acting at C perpendicular to link 2 as illustrated in Fig. P13.42. The end A is

pinned to the ground and the supporting link 3 is pinned to link 2 at B and pinned to the

ground at D. The lengths are 200 mmAC and 150 mm.AD Link 3 has a circular

cross-section with diameter 5 mm.D Both links are a steel alloy with a compressive

yield strength yc 420 MPaS and modulus of elasticity 206 GPa.E Using the

theoretical value for the end-condition constant for link 3, determine: (i) the slenderness

ratio at the point of tangency between Euler’s column formula and Johnson’s parabolic

formula; (ii) the critical load and the critical unit load acting on the link; and (iii) the

force F for the factor of safety to guard against buckling to be 1N . If the force

3 000 NF then for link 3 determine: (a) the critical load for the factor of safety to

guard against buckling to be 1N ; and (b) the slenderness ratio.

(i) From the pinned-pinned end conditions, the end-condition constant for link 3 is

C = 1. (1)

Therefore, the slenderness ratio of link 3 at the point of tangency is

9

r 6

yc

2 2 1 206 10 Pa98.4

420 10 PaD

CES

S

Ans. (2)

(ii) In order to determine the critical load on link 3, we first determine if this link is an

Euler column or a Johnson column. The Euler and Johnson criterion, respectively, are

r r DS S and r r D

S S (3)



The cross-sectional area of link 3 is 2 2 5 2A 4 (0.005 m) 4 1.963 10 mD (4)

and the second moment of area of link 3 is 4 4 11 464 (0.005 m) 64 3.068 10 mI D

The radius of gyration of link 3 is 11 4

3

5 2

3.068 10 m1.25 10 m

1.963 10 m

Ik

A

The slenderness ratio of link 3 is

r 3

0.15 m120

1.25 10 m

LS

k

(5)

Substituting Eqs. (2) and (5) into Eq. (3) gives

r r DS S ; that is, 120 98.4

Therefore, link 3 is an Euler column. The critical load on link 3 from the Euler column

equation can be written as 2

cr 2

r

C EP A

S

(6)

Substituting the known values into this equation, the critical load on link 3 is 2 9

5 2

cr 2

1 206 10 Pa1.963 10 m 2 772 N

120P

Ans.

Therefore, the critical unit load on link 3 is

6cr

5 2

2 772 N141.2 10 Pa

1.963 10 m

P

A

Ans.

(iii) The free body diagram of link 2 is shown in the figure. Taking moments about pin

A can be written as

32 32( ) ( ) 0x y y x x y y x

A CA CA BA BAM R F R F R F R F

or as



2 2 2 32 3 2 32 3( cos )( sin ) ( sin )( cos ) ( cos )( sin ) ( sin )( cos ) 0CA F CA F BA BAR F R F R F R F

or as

2 32 3 2sin( ) sin( ) 0CA F BAR F R F

Rearranging this equation, the force is

32 2 3

2

sin( )

sin( )

BA

CA F

R FF

R

(7)

The free-body diagram of link 3 is shown in the following figure. Note that link 3 is in

compression.

The reaction force 23 32BF F F ; therefore, the magnitude of the reaction force BF is

equal to the magnitude of the internal force 32F ; that is, Eq. (7) can be written as

2 3

2

sin ( )

sin( )

BA B

CA F

R FF

R

(8)

The factor of safety guarding against buckling for link 3 can be written as

cr

B

PN

F (9)

Rearranging Eq. (9), the compressive load on link 3 can be written as

cr 2 772 N2 772 N

1B

PF

N (10)

Substituting 2 120 , 3 60 , 210 ,F 0.2 m,CAR 0.15 m,BAR and putting Eq.

(10) into Eq. (8), the force is

0.15 m 2 772 N sin(120 60 )1 800.6 N

0.2 m sin(210 120 )F

Ans.

(a) Rearranging Eq. (7) gives



232

2 3

sin( )

sin( )

CA F

BA

R FF

R

Substituting 3 000 NF and the given data into this equation gives

32

0.2 m 3 000 N sin(210 120 )4 618.8 N

0.15 m sin(120 60 )BF F

(11)

Rearranging Eq. (9) and substituting Eq. (11) and the factor of safety N = 1 into the

resulting equation, the critical load applied on link 3 is

cr 1 4 618.8 N 4 618.8 NBP NF Ans. (12)

(b) If we assume that the link is an Euler column: Rearranging Eq. (6), the

slenderness ratio of link 3 can be written as

2

r Eulercr

C EAS

P

(13)

Substituting Eqs. (1), (4), and (12) into Eq. (13), the slenderness ratio of link 3 is

2 9 5 2

r Euler

1 206 10 Pa 1.963 10 m92.97

4 618.8 NS

Next, if we assume that the link is a Johnson column, the Johnson parabolic equation is 2

y rcry

1

2

S SPS

A CE

Rearranging this equation, the slenderness ratio can be written as

crr yJohnson

y

2 PS S CE

S A

or as

6 9

r 6 5 2Johnson

2 4 618.8 N420 10 Pa 1 206 10 Pa 92.3

420 10 Pa 1.963 10 mS

(14)

From Eq. (2), the slenderness ratio of link 3 at the point of tangency is r 98.4D

S .

Note that

r rJohnson DS S ; that is, 92.3 98.4

which is the correct answer. Therefore, the link must be a Johnson column. The correct

value for the slenderness ratio is given by Eq. (14); that is,

r r Johnson92.3S S Ans.

As a check, we note that

r rEuler DS S ; that is, 92.97 98.4 , which is not possible. Therefore, the link must be

a Johnson column. So again the correct value for the slenderness ratio is

r r Johnson92.3S S .



Chapter 14

Dynamic Force Analysis*

14.1 The steel bell crank illustrated in Fig. P14.1 is used as an oscillating cam follower. Using

76.42 RN/ 3m for the density of steel, find the mass moment of inertia of the lever about

an axis through O.

75 mm

150 mm

9.375 mm

18.75 mm

25 mm12.5 mm

For the vertical arm, using Appendix A, Table 5,

318.75 mm 93.75 mm 9.375 mm 76.42 RN/m 1.32 Nm wht

2 22 2 2 2

12 1.32 N 9804 mm/s 18.75 mm 93.75 mm 12 0.10446 N mm sG m a cI

22 2 22

0.10446 N mm s 1.32 N 9804 mm/s 37.5 mm 0.29726 N mm sO GI I md

For the horizontal arm, using Appendix A, Table 5,

318.75 mm 150 mm 9.375 mm 76.42 KN/m 2.12 Nm wht

2 22 2 22

12 2.12 N / 9804 mm/s 18.75 mm 150 mm 12 0.41774 N mm sG m a cI

22 22 2

0.41774 N mm s 2.12 N / 9804 mm / s 93.75 mm 1.98002 N mm sO GI I md

For the roller, using Appendix A, Table 5,

* Unless instructed otherwise, solve all problems without friction and without gravitational loads.



22 312.5 mm 12.5 mm 76.42 RN/m 0.493 Nm r t

22 2 3 22 0.493 N 9804 mm/s 12.5 mm 2 4 10 N mm sGI mr

22 2 22 3

4 N mm s 0.493 N 9804 mm/s 150 mm 1.1529 N mm s10O GI I md

For the composite lever

1.32 N 2.12 N 0.493 N 3.933 Nm

2 2 2 20.29726 N mm s 1.98002 N mm s 1.1529 N mm s 3.4302 N mm sOI

Ans.



14.2 A 5- by 50- by 300-mm steel bar has two round steel disks, each 50 mm in diameter and

20 mm long, welded to one end as shown. A small hole is drilled 25 mm from the other

end. The density of steel is 7.80 Mg/ 3m . Find the mass moment of inertia of this

weldment about an axis through the hole.

Dimensions are in millimeters.

For the rectangular bar, using Appendix A, Table 5,

30.050 m 0.300 m 0.005 m 7.80 Mg/m 0.585 kgm wht

2 22 2 212 0.585 kg 0.050 m 0.300 m 12 0.004 509 kg mGI m a c

22 2 20.004 509 kg m 0.585 kg 0.125 m 0.013 650 kg mO GI I md

For the two circular disks, using Appendix A, Table 5,

22 32 2 0.025 m 0.020 m 7.80 Mg/m 0.613 kgm r t

22 22 0.613 kg 0.025 m 2 0.000 191 kg mGI mr

22 2 20.000 191 kg m 0.613 kg 0.250 m 0.038 480 kg mO GI I md

For the composite lever

0.585 kg 0.613 kg 1.198 kgm

2 2 20.013 650 kg m 0.038 480 kg m 0.052 130 kg mOI Ans.



14.3 Find the external torque that must be applied to link 2 of the four-bar linkage shown to

drive it at the given velocity.

2 4 275 mm, 175 mm,AO A OR R 200 mm, 150 mm,BA BOR R

3100 mm,G AR

3 475 mm,G oR

3 4 43.15 N, 3.47 N,w o w

2 3

2 22.8702 N mm s , 1.7132 N mm s ,G GI I

4

21.246 N mm s ,GI 2ˆ180 rad/s,ω k 2 , 0 2

2ˆ4 950 rad/s , k 2

4ˆ8 900 rad/s , k

3

2ˆ ˆ1896 225 m/s ,G i + j 4

2ˆ ˆ684 225 m/s ,G i + j .

The D’Alembert inertia forces and offsets are:

22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2 2

3 3

ˆ ˆ3.15 N 9.66 m/s 1896 +225 m/s

ˆ ˆ618.5 71.2 N 623 N 186.77

Gm

i j

f A

i j

4

2 2

4 4

ˆ ˆ3.47 N 9.66 m/s 684 +225 m/s

ˆ ˆ244.75 80.1 N 258.1 N 198.21

Gm

i j

f A

i j

33 3

2 2ˆ1.7132 N mm s 4 950 rad/s

ˆ8455 N mm

GI

t α

k

k

44 4

2 2ˆ1.242 N mm s 8 900 rad/s

ˆ11125 N mm

GI

t α

k

k

3 3 3 14240 N mm 623 N 13.625 mmh t f ,

4 4 4 1125 N mm 258.1 N 42.85 mmh t f

Next, the free-body diagrams are drawn with the inertia forces applied. Since the lines of

action for the forces on the free-body diagrams cannot be discovered from two- and three-

force member concepts, the force 34F is divided into radial and transverse components.

(Notice that it is totally coincidental that the reaction components are also exactly aligned



with the radial and transverse axes of link 3. This results from the perpendicularity of

links 3 and 4.)

4 4 4 44 4 34

t

O G O BO M R ×f t R ×F 0

34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ45 60 mm 244.75 80.1 N 1125 N mm 90 120 mm 0.800 0.600t

F i j × i j k i j i j 0×

34ˆ ˆ ˆ182895 N mm 11125 N mm 150 mm tF 0k k k

34 44 lbtF

3 3 3 43

r

A G A BA M R ×f t R ×F 0

43ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ80 60 in 618.55 71.2 lb 8455 N mm 160 120 mm 0.600 0.800

rF i j × i j k i j × i j 0

43ˆ ˆ ˆ31.417 N mm 8455 N mm 200 mm rF 0k k k

43 115.7 NrF 34ˆ ˆ89 209.15 N 226.95 N 67.26 F i j

43 3 23 F F f F 0

23ˆ ˆ707.55 284.8 N 760.95 N 21.82 F i j

2 2 32 12O AO M R ×F M 0 12ˆ854.4 N mm M k Ans.



14.4 Crank 2 of the four-bar linkage illustrated in Fig. P14.4 is balanced. For the given

angular velocity of link 2, find the forces acting at each joint and the external torque that

must be applied to link 2.

2 4 250 mm, 325 mm,AO O OR R

4425 mm, 200 mm,BA BOR R

3212.5 mm,G AR

4 4100 mm,G oR

3 4 411.79 N, 29.9 N,w o w

2 3

2 22.66 N mm s , 6.74 N mm s ,G GI I

4

259.07 N mm s ,GI 2ˆ200 rad/s,ω k 2 , 0 2

3ˆ6 500 rad/s , k 2

4ˆ240 rad/ s , k

3

2ˆ ˆ948 78.3 m/s ,G i + j 4

2ˆ ˆ240 633 m/s ,G i + j


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2 2

3 3

ˆ ˆ11.79 N 9.66 m/s 948 +78.6 m/s

ˆ ˆ1157 97.9 N 1161.45 N 4.74

Gm

i j

f A

i j

4

2 2

4 4

ˆ ˆ29.9 N 9.66 m/s 240 633 m/s

ˆ ˆ743.15 1958 N 2096 N 69.24

Gm

i j

f A

i j

33 3

2 2ˆ6.7417 N mm s 6 500 rad/s

ˆ43832.5 N mm

GI

t α

k

k

44 4

2 2ˆ59.1 N mm s 240 rad/s

ˆ14128.75 N mm

GI

t α

k

k

3 3 3 43832.5 N mm 1161.45 N 37.725 mmh t f ,

4 4 4 141228.75 N mm 2096 N 6.775 mmh t f

Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for

the forces on the free-body diagrams cannot be discovered from two- and three-force

member concepts, the force 34F is divided into radial and transverse components.



4 4 4 44 4 34

t


34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ36.6 93.1 mm 747.15 1958 N 14128.75 N mm 73 186.175 mm 0.931 0.365t


34ˆ ˆ ˆ2407.451 N mm 14128.75 N mm 200 mm tF 0k k k

34 84.55 NtF

3 3 3 43 43

t r

A G A BA BA M R ×f t R ×F R ×F 0

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ181.35 110.77 mm 1157 97.9 N 43832.5 N mm 362.7 221.55 mm 75.65 31.15 N

ˆ ˆ ˆ ˆ 362.7 221.55 mm 0.365 0.931 rF

i j × i j k i j × i j

i j × i j 0

43ˆ ˆ ˆ ˆ145.913 N m 43.83 N m 18.37 N m 0.29717 m rF 0k k k k

43 280.35 NrF 34ˆ ˆ26.7 293.7 N 293.7 N 95.06 F i j Ans.

34 4 14 F F f F 0

14ˆ ˆ720.9 1668.7 N 1815.6 N 113.27 F i j Ans.

43 3 23 F F f F 0

23ˆ ˆ1183.7 195.8 N 1201.5 N 170.57 F i j Ans.

32 12 F F F 0 12ˆ ˆ1183.7 195.8 N 1201.5 N 9.43 F i j Ans.



2 2 32 12 0O AO M R F M 12ˆ54.846 N m M k Ans.



14.5 For the angular velocity of crank 2 given in Fig. P14.5, find the reactions at each joint and

the external torque applied to the crank.

275 mm,AOR 300 mm,BAR

3112 mm,G AR 3 1.54 kg,m 4 1.30 kg,m

2

20.039 8 kg m ,GI 3

20.012 2 kg m ,GI 2ˆ2 1 0 r a d / s ,ω k

2 0,α

2

3ˆ7.670 rad/s , α k

3

2ˆ ˆ2 384 1 486 m/s ,G A i j 4

2ˆ2 393 m/s .G A i


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

2

33 3

ˆ ˆ1.54 kg 2 384 1 486 m/s

ˆ ˆ3 671 2 288 N 4 326 N 31.94

Gm

i j

f A

i j

4

2

4 4

ˆ1.30 kg 2 393 m/s

ˆ3 111 N 3 111 N 0.00

Gm

i

f A

i

33 3

2 2ˆ0.012 2 kg m 7 670 rad/s

ˆ85 904 mm N

GI

t α

k

k

44 4GI t α 0

3 33 85 904 mm N 4 326 lb 19.86 mmt fh , 4 4 4 0h t f

Next, the free-body diagrams with inertia forces are drawn and the solution proceeds.

3 3 3 4 14A G A BA BA M R ×f t R ×f R ×F 0

14

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ110.2 19.8 mm 3 671 2 288 N 85 904 mm N 295.3 53.0 mm 3 111 N

ˆ ˆ ˆ 295.3 53.0 mm F

i j × i j k i j × i

i j × j 0

14ˆ ˆ ˆ ˆ324 902 mm N 85 904 mm N 164 985 mm N 295.3 mm F 0k k k k

14 1 273 NF 14ˆ1 273 N 1 273 N 90 F j Ans.

4 14 34 F f F F 0 34ˆ ˆ3 111 1 273 N 3 361 N 157.75 F i j Ans.

43 3 23 F F f F 0 23ˆ ˆ6 782 1 015 N 6 858 N 171.49 F i j Ans.

32 12 F F F 0 12ˆ ˆ6 782 1 015 N 6 858 N 171.49 F i j Ans.

2 2 32 12O AO M R ×F M 0 12ˆ305.84 N m M k Ans.



14.6 Figure P14.6 illustrates a slider-crank mechanism with an external force BF applied to the

piston. For the given crank velocity, find the reaction forces in the joints and the crank

torque.

275 mm,AOR 300 mm,BAR

2 231.25 mm,G oR

387.5 mm,G AR 2 4.22 N,w

3 15.575 N,w 4 11.125 N,w 2

20.4105 N mm s ,GI 3

212.2375 N mm s ,GI

2ˆ160 rad/s,ω k

2 0,α2

3ˆ3 090 rad/s , α k

2

2 0792 m/s 150 ,G A

3

2 01839 m/s 158.3 ,G A4

2 01.884 m/s 180 ,G A03560 180 ,B N F


2 2

22 2

ˆ ˆ4.2 N 9804 mm/s 685.8 396 m/s

ˆ ˆ298.15 173.55 N 347 N 30.00

Gm

i j

f A

i j

22 2GI t α 0

2 2 2 0h t f

2

33 3

2 ˆ ˆ1708 680 m/s9804 mm/s

ˆ ˆ2754.5 1094.7 N 2963.7 N 21.70

Gm

i j

f A

i j

4

2 2

4 4

ˆ11.125 N 9804 mm/s 1.884 mm/s

ˆ2171.6 N 2171.6 N 0.00

Gm

i

f A

i

33 3

2 2ˆ12.2375 N mm s 3 090 rad/s

ˆ37.83 N m

GI

t α

k

k

44 4GI t α 0

3 33 37825 N mm 2963.7 N 12.9 mmt fh , 4 4 4 0h t f

Next, the free-body diagrams are drawn with inertia forces and the solution proceeds.

3 3 3 4 14A G A BA BA B BA M R ×f t R ×f R ×F R ×F 0



14

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ86.82 11.12 mm 2754.5 1094.7 N 37825 N mm 297.65 375 mm 2171.6 N

ˆ ˆ ˆ ˆ ˆ ˆ 297.65 37.5 mm 3560 N 297.65 375 mm F


i j × i i j × j 0

14ˆ ˆ ˆ ˆ ˆ125.16 N m 37.82 N m 81.43 N m 133.5 N m 297.65 mm F 0k k k k k

14 120.15 NF 14ˆ120.15 N 120.15 N 90 F j Ans.

4 14 34B F f F F F 0

34ˆ ˆ1388.4 120.15 N 1392.85 N 4.95 F i j Ans.

43 3 23 F F f F 0

23ˆ ˆ1366.15 974.5 N 1677.65 N 144.50 i jF Ans.

32 12 F F F 0 12ˆ ˆ1664.3 1148.1 N 2020.3 n 145.40 i jF Ans.

2 2 32 12 0O AO M R F M 12ˆ12015 N mm M k Ans.



14.7 The following data apply to the four-bar linkage illustrated in Fig. P14.7: 2

0.3 mAOR ,

4 20.9 mO OR , 1.5 mBAR ,

40.8 mBOR , 0.85 mCAR , 33C ,

40.4 mDOR ,

53 ,D 2 2

0,G OR 3

0.65 m,G AR 16 , 4 4

0.45 m,G OR 17 , 2 5.2 kg,m

3 65.8 kg,m 4 21.8 kg,m 2

22.3 kg m ,GI 3

24.2 kg m ,GI 4

20.51 kg m .GI A

kinematic analysis at 2 60 , 2ˆ12 rad/s ccw kω , and 2 0α , gives 3 0.7 ,

4 20.4 , 23 85.6 rad/s cw , 2

4 172 rad/s cw , 3

296.4 259 m/sG A , and

4

297.8 270 m/sG A . Find all pin reactions and the torque applied to link 2.


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ65.8 kg 18.394 94.629 m/s

ˆ ˆ1 210 6 227 N 6 343 N 79

Gm

i j

f A

i j

4

2

4 4

ˆ21.8 kg 97.800 m/s

ˆ2 132 N 2 132 N 90

Gm

j

f A

j

33 3

2 2ˆ4.200 kg m 85.6 rad/s

ˆ360 N m

GI

t α

k

k

44 4

2 2ˆ0.51 kg m 172 rad/s

ˆ88 N m

GI

t α

k

k

3 33 360 N m 6 343 N 0.057 mt fh , 4 4 4 88 N m 2 132 N 0.041 mh t f



Next, the free-body diagrams are drawn with inertia forces. Since the lines of action for



4 4 4 44 4 34

t


34

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.357 0.273 m 2 132 N 88 N m 0.750 0.279 m 0.348 0.937t

F i j × j k i j i j 0×

34ˆ ˆ ˆ761 N m 88 N m 0.800 m tF 0k k k

34 1 061 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.622 0.187 m 1 210 6 227 N 360 N m 1.499 0.019 m 370 996 N

ˆ ˆ ˆ ˆ 1.499 0.019 m 0.937 0.348 rF


i j × i j 0

43ˆ ˆ ˆ ˆ3 650 N m 360 N m 1 501 N m 0.505 m rF 0k k k k

43 10 913 NrF 34 ˆ ˆ10 600 2 800 N 10 964 N 14.8 F i j Ans.

34 4 14 F F f F 0 14ˆ ˆ10 600 4 932 N 11 691 N 155.0 F i j Ans.

43 3 23 F F f F 0 23ˆ ˆ9 390 3 427 N 9 996 N 20.0 i jF Ans.

32 12 F F F 0 12ˆ ˆ9 390 3 427 N 9 996 N 20.0 i jF Ans.

2 2 32 12O AO M R ×F M 0 12ˆ2 954 N m M k Ans.



14.8 Solve Problem 14.7 with an additional external force 12 kN 0D F acting at point D.

Here, the method of superposition is used for the solution. The force components of

Problem 14.7 are denoted with primes and the additional force increments with double

primes. The figures here show the incremental forces only.

4 4 4 34O DO D BO M R ×F R ×F 0

34ˆ ˆ ˆ ˆ ˆ ˆ ˆ0.114 0.383 m 12 000 N 0.750 0.279 m 0.999 0.013 F i j × i i j i j 0×

34ˆ ˆ4 600 N m 0.269 m F 0k k 34 17 081 NF

34ˆ ˆ17 079 217 N 17 081 N 179.3 i jF

34ˆ ˆ6 479 2 584 N 6 975 N 158.3 i jF Ans.

14ˆ ˆ5 079 217 N 5 084 N 2.4 F i j

14ˆ ˆ5 521 4 715 N 7 260 N 139.5 i jF Ans.

23ˆ ˆ17 079 217 N 17 081 N 179.3 i jF 23

ˆ ˆ7 689 3 644 N 8 509 N 154.6 i jF Ans.

12ˆ ˆ17 079 217 N 17 087 N 179.3 i jF 12

ˆ ˆ7 689 3 644 N 8 509 N 154.6 i jF Ans.

12ˆ4 405 N m M k 12

ˆ1 451 N m M k Ans.



14.9 Make a complete kinematic and dynamic analysis of the four-bar linkage of Problem 14.7

using the same data, but with 2 2 2170 , 12 rad/s ccw, 0, and an external

force D 8.94 64.3 kN F acting at point D.

Kinematic Analysis:

22

ˆ ˆˆ12 rad/s 0.295 0.052 m

ˆ ˆ0.625 3.545 m/s 3.600 m/s 100

A AO

V ω ×R k × i j

i j

4

3 4

3 3 4 4

3 4

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ0.625 3.545 m/s rad/s 1.304 0.740 m rad/s 0.109 0.793 m

ˆ ˆ ˆ ˆ ˆ ˆ0.625 3.545 m/s 0.740 1.304 m 0.793 0.109 m

B A BA BO

i j k × i j k × i j

i j i j i j

V V ω ×R ω ×R

3ˆ3.020 rad/sω k 4

ˆ3.607 rad/sω k Ans.

ˆ ˆ2.859 0.393 m/s 2.886 m/s 172.16B V i j

2 2 2

22

2

2 2

ˆ ˆ12 rad/s 0.295 0.052 m

ˆ ˆ42.543 7.502 m/s 43.200 m/s 10

A AO AO

A R α ×R × i j

i j



4 4

2 23 3 4 4

2 2 23

2 24

23

ˆ ˆ ˆ ˆ ˆ ˆˆ42.543 7.502 m/s 11.893 6.749 m/s rad/s 1.304 0.740 m

ˆ ˆ ˆ ˆˆ1.419 10.311 m/s rad/s 0.109 0.793 m

ˆ ˆ ˆ32.069 3.940 m/s 0.740 1.

B A BA BA BO BO

A A R α ×R R α ×R

i j i j k × i j

i j k × i j

i j i 3 4 4ˆ ˆ ˆ304 m 0.793 0.109 m j i j

2

3ˆ0.390 rad/s α k 2

4ˆ40.816 rad/s α k Ans.

3 3 3

23 3

2 2 2ˆ ˆ ˆ ˆ ˆ ˆˆ42.543 7.502 m/s 4.149 4.234 m/s 0.390 rad/s 0.455 0.464 m

G A G A G A

A A R α ×R

i j i j k × i j

3

2 2ˆ ˆ38.575 11.913 m/s 40.373 m/s 17.16G A i j Ans.

4 4 4 4 4

24 4

2 2ˆ ˆ ˆ ˆˆ0.932 5.780 m/s 40.816 rad/s 0.072 0.444 m

O OG G G

A R α ×R

i j k × i j

4

2 2ˆ ˆ19.054 2.841 m/s 19.265 m/s 8.48G A i j Ans.

Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ65.8 kg 38.575 11.913 m/s

ˆ ˆ2 538 784 N 2 657 N 162.84

Gm

i j

f A

i j

4

2

4 4

ˆ ˆ21.8 kg 19.054 2.841 m/s

ˆ ˆ415 62 N 420 N 171.52

Gm

i j

f A

i j

33 3

2 2ˆ4.200 kg m 0.390 rad/s

ˆ1.638 N m

GI

t α

k

k

44 4

2 2ˆ0.51 kg m 40.816 rad/s

ˆ21 N m

GI

t α

k

k

3 33 1.638 N m 2 657 N 0.001 mt fh , 4 4 4 21 N m 420 N 0.050 mh t f






4 4 4 4 44 4 34

t

O G O DO D BO M R ×f t R ×F R ×F 0

34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.072 0.444 m 415 62 N 21 N m 0.284 0.282 m 3 877 8 056 N

ˆ ˆ ˆ ˆ0.109 0.793 m 0.991 0.136 tF


i j i j 0×

34ˆ ˆ ˆ ˆ180 N m 21 N m 3 379 N m 0.800 m tF 0k k k k

34 3 972 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.455 0.464 m 2 538 784 N 1.638 N m 1.304 0.740 m 3 935 542 N

ˆ ˆ ˆ ˆ 1.304 0.740 m 0.136 0.991r

F


i j × i j 0

43ˆ ˆ ˆ ˆ1 534 N m 2 N m 3 618 N m 1.192 in rF 0k k k k

43 1 747 NrF 34

ˆ ˆ4 173 1 189 N 4 339 N 164.10 i jF Ans.

34 4 14D F F f F F 0 14ˆ ˆ711 6 929 N 6 966 N 84.14 F i j Ans.

43 3 23 F F f F 0 23

ˆ1 635 1 972 N 2 562 N 129.65 iF Ans.

32 12 F F F 0 12

ˆ1 635 1 972 N 2 562 N 129.65 iF Ans.

2 2 32 12O AO M R ×F M 0 12ˆ668 N m M k Ans.



14.10 Repeat Problem 14.9 using 2 2 2200 , 12 rad/s ccw, 0, and an external force

8.49 45 kNC F acting at point C.

Kinematic Analysis:

22

ˆ ˆˆ12 rad/s 0.282 0.103 m

ˆ ˆ1.231 3.383 m/s 3.600 m/s 70

A AO

V ω ×R k × i j

i j

4

3 4

3 3 4 4

3 4


ˆ ˆ ˆ ˆ ˆ ˆ1.231 3.383 m/s 0.902 1.198 m 0.799 0.016 m

B A BA BO


i j i j i j

V V ω ×R ω ×R

3ˆ2.846 rad/sω k 4

ˆ1.671 rad/sω k Ans.

ˆ ˆ1.336 0.027 m/s 1.337 m/s 178.84B V i j

2 2 2

22

2

2 2

ˆ ˆ12 rad/s 0.282 0.103 m

ˆ ˆ40.595 14.775 m/s 43.200 m/s 20

A AO AO

A R α ×R × i j

i j

4 4

2 23 3 4 4

2 2 23

2 24

23


ˆ ˆ ˆ ˆˆ0.045 2.233 m/s rad/s 0.016 0.799 m

ˆ ˆ ˆ30.937 9.702 m/s 0.902 1.1

B A BA BA BO BO


i j i j k × i j

i j k × i j

i j i 3 4 4ˆ ˆ ˆ98 m 0.799 0.016 m j i j

2

3ˆ8.760 rad/s α k

2




3 3 3

2

3 3


G A G A G A

A A R α ×R

i j i j k × i j

3

2 2ˆ ˆ41.972 7.146 m/s 42.576 m/s 9.66G A i j Ans.

4 4 4 4 4

2

4 4

2 2ˆ ˆ ˆ ˆˆ0.343 1.209 m/s 48.558 rad/s 0.123 0.433 m

G G O G O

A R α ×R

i j k × i j

4

2 2ˆ ˆ21.365 4.754 m/s 21.887 m/s 12.54G A i j Ans.

Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ65.8 kg 41.972 7.146 m/s

ˆ ˆ2 762 470 N 2 802 N 170.34

Gm

i j

f A

i j

4

2

4 4

ˆ ˆ21.8 kg 21.365 4.754 m/s

ˆ ˆ466 104 N 477 N 167.46

Gm

i j

f A

i j

33 3

2 2ˆ4.200 kg m 8.760 rad/s

ˆ37 N m

GI

t α

k

k

44 4

2 2ˆ0.51 kg m 48.558 rad/s

ˆ25 N m

GI

t α

k

k

3 33 37 N m 2 802 N 0.013 mt fh , 4 4 4 25 N m 477 N 0.052 mh t f




4 4 4 44 4 34

t


34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.123 0.433 m 466 104 N 25 N m 0.016 0.799 m 0.999 0.020t

F i j i j k i j i j 0



34ˆ ˆ ˆ214 N m 25 N m 0.800 m tF 0k k k

34 299 NtF

3 3 3 43 43

t r

A G A CA C BA BA M R ×f t R ×F R ×F R ×F 0

43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.391 0.519 m 2 762 470 N 37 N m 0.291 0.799 m 6 003 6 003 N

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1.198 0.902 m 299 6 N 1.198 0.902 m 0.020 0.999r

F


i j × i j i j × i j 0

43ˆ ˆ ˆ ˆ ˆ1 250 N m 37 N m 3 050 N m 277 N m 1.179 in rF 0k k k k k

43 1 260 NrF 34ˆ ˆ274 1 266 N 1 295 N 77.80 F i j Ans.

34 4 14 F F f F 0 14ˆ ˆ192 1 370 N 1 383 N 82.01 F i j Ans.

43 3 23C F F f F F 0 23

ˆ2 968 6 799 N 7 419 N 113.58 iF Ans.

32 12 F F F 0 12

ˆ2 968 6 799 N 7 419 N 113.58 iF Ans.




14.11 At 2 270 , 2 18 rad/s ccw , 2 0 , a kinematic analysis of the linkage whose

geometry is given in Problem 14.7 gives 3 46.6 , 4 80.5 , 23 178 rad/s cw ,

24 256 rad/s cw ,

3

2112 m/s 22.7G A , 4

2119 m/s 352.5G A . An external force

8.94 kN 64.3D F acts at point D. Make a complete dynamic analysis of the linkage.


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ65.8 kg 103.324 43.221 m/s

ˆ ˆ6 799 2 844 N 7 370 N 157.30

Gm

i j

i j

f A

4

2

4 4

ˆ ˆ21.8 kg 117.982 15.533 m/s

ˆ ˆ2 572 339 N 2 594 N 172.50

Gm

i j

f A

i j

33 3

2 2ˆ4.200 kg m 178 rad/s

ˆ748 N m

GI

t α

k

k

44 4

2 2ˆ0.51 kg m 256 rad/s

ˆ131 N m

GI

t α

k

k

3 33 748 N m 7 370 N 0.101 mt fh , 4 4 4 131 N m 2 594 N 0.050 mh t f






4 4 4 4 44 4 34

t

O G O DO D BO M R ×f t R ×F R ×F 0

34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.059 0.446 m 2 572 339 N 131 N m 0.276 0.290 m 3 877 8 056 N

ˆ ˆ ˆ ˆ0.131 0.789 m 0.986 0.164 tF


i j i j 0×

34ˆ ˆ ˆ ˆ1 127 N m 131 N m 3 348 N m 0.800 m tF 0k k k k

34 2 613 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.300 0.577 m 6 799 2 844 N 748 N m 1.031 1.089 m 2 578 429 N

ˆ ˆ ˆ ˆ1.031 1.089 m 0.164 0.986 rF


i j × i j 0

43ˆ ˆ ˆ ˆ3 070 N m 748 N m 3 250 N m 0.839 in rF 0k k k k

43 677 NrF 34ˆ ˆ2 466 1 097 N 2 699 N 156.01 i jF Ans.

34 4 14D F F f F F 0 14ˆ ˆ1 411 9 153 N 9 261 N 98.76 i jF Ans.

43 3 23 F F f F 0 23

ˆ9 265 1 747 lb 9 428 N 10.68 iF Ans.

32 12 F F F 0 12

ˆ9 265 1 747 lb 9 428 N 10.68 iF Ans.




14.12 The following data apply to a four-bar linkage similar to the one illustrated in Fig. P14.7:

2120 mmAOR ,

4 2300 mmO OR , 320 mmBAR ,

4250 mmBOR , 360 mmCAR ,

15C , 4

0DOR , 0D , 2 2

0G OR , 3

200 mmG AR , 8 , 4 4

125 mmG OR ,

0 , 2 0.5 kg,m 3 4 kg,m 4 1.5 kg,m 2

20.005 N m s ,GI 3

20.011 N m s ,GI

4

20.002 3 N m s .GI A kinematic analysis at 2 90 and 2 32 rad/s ccw with 2 0

gave 3 23.9 . 4 91.7 , 2

3 221 rad/s ccw, 2

4 122 rad/s ccw, 3

288.6 m/s 255 ,G A

and 4

232.6 m/s 244 .G A Using an external force 632 N 342C F acting at point C,

make a complete dynamic analysis of the linkage.


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ4.0 kg 22.931 85.581 m/s

ˆ ˆ92 342 N 354 N 75

Gm

i j

i j

f A

4

2

4 4

ˆ ˆ1.5 kg 14.291 29.301 m/s

ˆ ˆ21 44 N 49 N 64

Gm

i j

f A

i j

33 3

2 2ˆ0.011 kg m 221 rad/s

ˆ2.431 N m

GI

t α

k

k

44 4

2 2ˆ0.002 3 kg m 122 rad/s

ˆ0.281 N m

GI

t α

k

k



3 33 2.431 N m 354 N 0.007 mt fh , 4 4 4 0.281 N m 49 N 0.006 mh t f




4 4 4 44 4 34

t


34

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.004 0.125 m 21 44 N 0.281 N m 0.008 0.249 m 0.999 0.030t


34ˆ ˆ ˆ2.844 N m 0.281 N m 0.250 m tF 0k k k

34 13 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.170 0.106 m 92 342 N 2.431 N m 0.280 0.226 m 601 195 N

ˆ ˆ ˆ ˆ ˆ ˆ ˆ0.292 0.130 m 12.5 N 0.292 0.130 m 0.030 0.999 rF


i j × i i j × i j 0

43ˆ ˆ ˆ ˆ ˆ48 N m 2.431 N m 191 N m 1.623 N m 0.296 in rF 0k k k k k

43 496 NrF 34ˆ ˆ2 496 N 496 N 89.71 i jF Ans.

34 4 14 F F f F 0 14ˆ ˆ24 452 N 453 N 93.02 i jF Ans.

43 3 23C F F f F F 0 23

ˆ690 643 lb 944 N 137 iF Ans.

32 12 F F F 0 12

ˆ690 643 lb 944 N 137 iF Ans.




14.13 Repeat Problem 14.12 at 2 260 . Analyze both the kinematics and the dynamics of

the system at this position.

Kinematic Analysis:

22

ˆ ˆˆ32 rad/s 0.021 0.118 m

ˆ ˆ3.782 0.667 m/s 3.840 m/s 10

A AO

V ω ×R k × i j

i j

4

3 4

3 3 4 4

3 4


ˆ ˆ ˆ ˆ ˆ ˆ3.782 0.667 m/s 0.289 0.138 m 0.171 0.183 m

B A BA BO


i j i j i j

V V ω ×R ω ×R

3ˆ10.554 rad/sω k 4

ˆ4.314 rad/s ω k Ans.

ˆ ˆ0.736 0.789 m/s 1.079 m/s 46.99B V i j

2 2 2

22

2

2 2

ˆ ˆ32 rad/s 0.021 0.118 m

ˆ ˆ21.338 121.013 m/s 122.880 m/s 80

A AO AO

A R α ×R × i j

i j

4 4

2 23 3 4 4

2 2 23

2 24


ˆ ˆ ˆ ˆˆ3.402 3.174 m/s rad/s 0.183 0.171 m

B A BA BA BO BO


i j i j k × i j

i j k × i j

23 3 4 4

ˆ ˆ ˆ ˆ ˆ ˆ2.562 92.029 m/s 0.289 0.138 m 0.171 0.183 m i j i j i j

2

3ˆ199.622 rad/s α k 2


3 3 3

2

3 3


G A G A G A

A A R α ×R

i j i j k × i j

3

2 2ˆ ˆ52.748 87.796 m/s 102.423 m/s 59.00G A i j Ans.

4 4 4 4 4

2

4 4

2 2ˆ ˆ ˆ ˆˆ1.701 1.587 m/s 352.356 rad/s 0.091 0.085 m

G G O G O

A R α ×R

i j k × i j

4

2 2ˆ ˆ31.651 30.477 m/s 43.939 m/s 43.92G A i j Ans.



Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ4.0 kg 52.748 87.796 m/s

ˆ ˆ211 351 N 410 N 121.00

Gm

i j

f A

i j

4

2

4 4

ˆ ˆ1.5 kg 31.651 30.477 m/s

ˆ ˆ47 46 N 66 N 136.08

Gm

i j

f A

i j

33 3

2 2ˆ0.011 N m s 199.622 rad/s

ˆ2.196 N m

GI

t α

k

k

44 4

2 2ˆ0.002 3 N m s 352.356 rad/s

ˆ0.810 N m

GI

t α

k

k

3 3 32.196 N m 410 N 0.005 mh t f , 4 4 4

0.810 N m 66 N 0.012 mh t f




4 4 4 44 4 34

t


34



34ˆ ˆ ˆ8.212 N m 0.810 N m 0.250 m tF 0k k k

34 36 NtF

3 3 3 43 43

t r


43


ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ0.138 0.289 m 25 26 N 0.138 0.289 m 0.731 0.682 rF


i j × i j i j × i j 0


43 658 NrF 34ˆ ˆ505 422 N 659 N 39.88 i jF Ans.

34 4 14 F F f F 0 14ˆ ˆ458 468 N 655 N 134.39 i jF Ans.

43 3 23C F F f F F 0 23

ˆ115 124 N 169 N 46.97 iF Ans.

32 12 F F F 0 12

ˆ115 124 N 169 N 46.97 iF Ans.




14.14 Repeat Problem 14.13 at 2 300 .

Kinematic Analysis:

22

ˆ ˆˆ32 rad/s 0.060 0.104 m

ˆ ˆ3.326 1.920 m/s 3.840 m/s 30

A AO

V ω ×R k × i j

i j

4

3 4

3 3 4 4

3 4


ˆ ˆ ˆ ˆ ˆ ˆ3.326 1.920 m/s 0.306 0.093 m 0.202 0.147 m

B A BA BO


i j i j i j

V V ω ×R ω ×R

3ˆ1.597 rad/sω k 4

ˆ14.071 rad/s ω k Ans.

ˆ ˆ2.844 2.066 m/s 3.515 m/s 36.04B V i j

2 2 2

22

2

2 2

ˆ ˆ32 rad/s 0.060 0.104 m

ˆ ˆ61.440 106.417 m/s 122.880 m/s 120

A AO AO

A R α ×R × i j

i j

4 4

2 23 3 4 4

2 2 23

2 24


ˆ ˆ ˆ ˆˆ29.105 39.995 m/s rad/s 0.147 0.202 m

B A BA BA BO BO


i j i j k × i j

i j k × i j

23 3 4 4

ˆ ˆ ˆ ˆ ˆ ˆ90.782 145.632 m/s 0.306 0.093 m 0.202 0.147 m i j i j i j

2

3ˆ671.302 rad/s α k

2


3 3 3

2

3 3


G A G A G A

A A R α ×R

i j i j k × i j

3

2 2ˆ ˆ71.399 85.103 m/s 111.087 m/s 50.00G A i j Ans.

4 4 4 4 4

2

4 4

2 2ˆ ˆ ˆ ˆˆ14.547 20.022 m/s 567.507 rad/s 0.073 0.101 m

G G O G O

A R α ×R

i j k × i j

4

2 2ˆ ˆ71.865 21.406 m/s 74.986 m/s 16.59G A i j Ans.

Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0



2 2 2 0h t f

3

2

3 3

ˆ ˆ4.0 kg 71.399 85.103 m/s

ˆ ˆ286 340 N 444 N 130.00

Gm

i j

f A

i j

4

2

4 4

ˆ ˆ1.5 kg 71.865 21.406 m/s

ˆ ˆ108 32 N 112 N 163.41

Gm

i j

f A

i j

33 3

2 2ˆ0.011 N m s 671.302 rad/s

ˆ7.384 N m

GI

t α

k

k

44 4

2 2ˆ0.002 3 N m s 567.507 rad/s

ˆ1.305 N m

GI

t α

k

k

3 3 37.384 N m 444 N 0.017 mh t f , 4 4 4

1.305 N m 112 N 0.012 mh t f




4 4 4 44 4 34

t


34



34ˆ ˆ ˆ13.273 N m 1.305 N m 0.250 m tF 0k k k

34 58 NtF

3 3 3 43 43

t r


43


ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ0.094 0.306 m 47 34 N 0.094 0.306 m 0.588 0.809 rF


i j × i j i j × i j 0


43 603 NrF 34ˆ ˆ401 453 N 606 N 48.50 i jF Ans.

34 4 14 F F f F 0 14ˆ ˆ293 486 N 567 N 121.13 i jF Ans.

43 3 23C F F f F F 0 23

ˆ86 81 N 119 N 43.26 iF Ans.

32 12 F F F 0 12

ˆ86 81 N 119 N 43.26 iF Ans.




14.15 Analyze the dynamics of the offset slider-crank mechanism illustrated in Fig. P14.15

using the following data: 0.06 ma , 2

0.1 m,AOR 0.38 m,ABR 0.4 m,CAR

32 ,C 3

0.26 m,G AR 22 , 2 2.5 kg,m 3 7.4 kg,m 4 2.5 kg,m

2

20.005 N m s ,GI 3

20.013 6 N m s ,GI 2 120 , and 2 18 rad/s cw, with 2 0,

ˆ2 000 N,B F i ˆ1 000 N.C F i . Assume a balanced crank and no friction forces.

Kinematic Analysis:

22

ˆ ˆˆ18 rad/s 0.050 0.087 m

ˆ ˆ1.559 0.900 m/s 1.800 m/s 30

A AO

V ω ×R k × i j

i j

3

3 3

3

ˆ ˆ ˆ ˆˆ1.559 0.900 m/s rad/s 0.351 0.147 m

ˆ ˆ ˆ ˆ1.559 0.900 m/s 0.147 0.351 m

ˆ

B A BA

BV

i j k × i j

i j i j

V V ω × R

i

3ˆ2.567 rad/s ω k ˆ1.182 m/sB V i Ans.

2 2 2

22

2

2 2

ˆ ˆ18 rad/s 0.050 0.087 m

ˆ ˆ16.200 28.059 m/s 32.400 m/s 60

A AO AO

A R α ×R × i j

i j

23 3

2 2 23

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ16.200 28.059 m/s 2.310 0.966 m/s rad/s 0.351 0.147 mB

B A BA BA

A

i

A A R α ×R

i j i j k × i j

23 3

ˆ ˆ ˆ ˆ ˆ13.890 27.093 m/s 0.147 0.351 mBA i i j i j

2

3ˆ77.188 rad/sα k 2ˆ25.156 m/sB A i Ans.

3 3 3

2

3 3


G A G A G A

A A R α ×R

i j i j k i j

3

2 2ˆ ˆ14.466 7.969 m/s 16.516 m/s 28.85G A i j Ans.



Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ7.4 kg 14.466 7.969 m/s

ˆ ˆ107 59 N 122 N 151.15

Gm

i j

f A

i j

2

4 4

ˆ2.5 kg 25.156 m/s

ˆ63 N 63 N 180

Bm

i

f A

i

33 3

2 2ˆ0.013 6 N m s 77.188 rad/s

ˆ1.050 N m

GI

t α

k

k

44 4GI t α 0

3 3 31.050 N m 122 N 0.009 mh t f ,

4 4 40h t f

Next, the free-body diagrams with inertia forces are drawn. Here the lines of action for the

forces on the free-body diagrams can all be discovered from two- and three-force member

concepts.

3 3 3 4 14A G A CA C BA BA B BA M R ×f t R ×F R ×f R ×F R ×F 0

14

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.260 0.003 m 107 59 N 1.050 N m 0.395 0.065 m 1 000 N

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0.351 0.147 m 63 N 0.351 0.147 m 2 000 N 0.351 0.147 m 1.000 F


i j × i i j × i i j × j 0

14ˆ ˆ ˆ ˆ ˆ ˆ15 N m 1.050 N m 65 N m 9.261 N m 294 N m 0.351 in F k k k k k k 0

14 639 NF 14ˆ639 N 639 N 90.00 jF Ans.

14 4 34B F F f F F 0 34ˆ ˆ2 063 639 N 2 160 N 17.21 i jF Ans.

43 3 23C F F f F F 0 23

ˆ3 170 698 N 3 246 N 12.42ˆ iF j Ans.

32 12 F F F 0 12

ˆ3 170 698 N 3 246 N 12.42ˆ iF j Ans.

2 2 32 12O AO M R ×F M 0 12ˆ241 N m M k Ans.



14.16 Analyze the system of Problem 14.15 for a complete rotation of the crank. Use C F 0

and ˆ1 000 NB F i when BV is toward the right, but use B F 0 when BV is toward the

left. Plot a graph of 12M versus 2 .

Kinematic Analysis: 32

1 2 3 4

jjjR R e R e R

320.060 0.100 0.380jj

Bj e e R

2 30.060 0.100sin 0.380sin 0 1

3 2sin 0.158 0.263sin

2 30.100cos 0.380cosBR

32

3 31 2

jj

G G AjR R e R e

R 32

3

220.060 0.100 0.260

jj

G j e e

R

3 2 3 2 30.100cos 0.260cos 22 0.060 0.100sin 0.260sin 22G j R

The first-order kinematic coefficients are found as follows: 32

3 40.100 0.380jjj e j e R

2 3 30.100cos 0.380cos 0 2 3 3 40.100sin 0.380sin R

3 2 30.263cos cos 2 2 30.100 sin cos tanBR

32

3

22

30.100 0.260jj

G j e j e R

3 2 3 3 2 3 30.100sin 0.260sin 22 0.100cos 0.260cos 22G j R

Similarly, the second-order kinematic coefficients are as follows: 3 32 2

3 30.100 0.380 0.380j jj

Be j e e R

2

2 3 3 3 30.100sin 0.380cos 0.380sin 0 2

2 3 3 3 30.100cos 0.380sin 0.380cos BR

2

3 2 3 3 30.263sin cos tan , 2

3 2 3 30.100cos 0.380 cosBR

3 32

3

22 22 2

3 30.100 0.260 0.260j jj

G e j e e R

3

2

2 3 3 3 3

2

2 3 3 3 3

0.100cos 0.260sin 22 0.260cos 22

0.100sin 0.260cos 22 0.260sin 22

G

j

R

Dynamic Analysis:

By virtual work we can formulate the dynamic input torque requirement as:

312 3 3 3 4ˆ

G B B B M f R t k f R F R

The individual elements of this equation are:

3 3 3

2 2

3 3 3 2 2 397.6 kg/sG G Gm m f A R R

2

3 2 3 3 3 3

2

2 3 3 3 3

240cos 623sin 22 623cos 22

240sin 623cos 22 623sin 22j

f



3

2

3 2 3 3 3 3 2 3 3

2

2 3 3 3 3 2 3 3

240 cos 623sin 22 623cos 22 0.100sin 0.260sin 22

240sin 623cos 22 623sin 22 0.100 cos 0.260 cos 22

G

f R

33 3 2 3 3 3 2 3 3 362sin 22 1 62cos 22 162G f R

3 3

2

3 3 3 2 3ˆ ˆ4.406 N mG GI I t α k k

3 3 3 3ˆ 4.406 N m t k

2 2

4 4 4 2 810 kg/sB B Bm m f A R R

2

4 3 2 3 3ˆ81cos 308 N cos f i

2 2

4 3 2 3 2 3 3sin 8.1cos 30.8 N m cosB f R

2 3 2 3 2 3ˆ ˆ500 1 sgn cos tan sin N=500 1+sgn sin cos NB F i i

3 2 3 3 2 350 1+sgn sin cos sin cos N mB B F R

Finally, putting these pieces together, we obtain:

12 3 2 3 3 3 2 3 3 3

2 2

3 2 3 2 3 3

3 2 3 3 2 3

62sin 22 1 62cos 22 158

sin 8.1cos 30.8 cos

50 1+sgn sin cos sin cos N m

M

The plot of this torque requirement is shown below. The sinusoidal curve in the first half

of the cycle is caused primarily by the mass of the connecting rod; the mass of the piston

is included also. The applied force FB causes the rise in the second half of the cycle.

Note that the mass of link 3 causes dynamic torque, which helps to overcome up to one

third of the applied force effect.



14.17 A slider-crank mechanism similar to that of Problem 14.15 has 0,a 2

100 mm,AOR

450 mm,BAR 0,CBR 0,C 3

200 mm,G AR 0 , 2 14.68 N,w 3 34.26 N,w

4 11.57 N,w 2

29.79 N mm s ,GI 3

258.96 N mm s ,GI and 600.75 N mm.

Corresponding to 2 120 and 2 24 rad/s cw with 2 0, a kinematic analysis

gives 3 10.9 , 392 mm,BR 2

3 89.3 rad/s ccw, 2ˆ40600 mm/s ,B A i and

3

2ˆ ˆ40600 22600 mm/s .G A i j Assume link 2 is balanced and find 14F and 23F .


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

2 ˆ ˆ40600 22600 mm / s34.26 N 9650 mm/s

ˆ ˆ144.18 80.23 N 165 N 150.90

Gm

i j

f A

i j

2

4 4

2 ˆ40600 mm / s11.57 N 9650 mm/s

ˆ48.68 N 48.68 N 180

Bm

i

f A

i

33 3

2 2ˆ58.96 N mm s 89.3 rad/s

ˆ5265.46 N mm

GI

t α

k

k

44 4GI t α 0

3 3 3ˆ5265.5 in lb 165 N 31.9 mmh t f k ,

4 4 40h t f


14

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ220.9 42.5 mm 144.2 802 N 52655 N mm 441.9 85 mm 48.68 N

ˆ ˆ ˆ441.9 85 mm 1.000 F


i j × j 0



14ˆ ˆ ˆ ˆ11596.7 N mm 5265.5 N mm 4139.6 N mm 441.9 mm F k k k k 0

14 4.94 NF 14ˆ4.94 N 4.94 N 90.00 jF Ans.

14 4 3 23 F F f f F 0 23

ˆ192.86 85.2 N 210.8 N 23.83 iF Ans.

14.18 Repeat Problem 14.17 for 2 240 . The results of a kinematic analysis are 3 10.9 ,

392 mm,BR 2

3 112 rad/s ccw, 2ˆ35200 mm/s ,B A i and

3

2ˆ ˆ31600 27700 mm/s .G A i j


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

2 ˆ ˆ3160 27700 mm/s34.26 N 9650 mm/s

ˆ ˆ112.21 98.34 N 158.1 N 138.76

Gm

i j

f A

i j

2

4 4

2 ˆ11.57 N / 9650 mm / s 35200 mm / s

ˆ42.2 N 42.2 N 180

Bm

i

f A

i

33 3

2 2ˆ58.96 N mm s 112 rad/s

ˆ6603.8 N mm

GI

t α

k

k

44 4GI t α 0

3 3 3ˆ6603.8 N mm 149.21 N 44.25 mmh t f k ,

4 4 40h t f




14

ˆ ˆ ˆ ˆ ˆ ˆˆ220.95 42.52 mm 112.2 142.84 N 6603.8 N mm 441.9 85.1 mm 42.2 N

ˆ ˆ ˆ441.9 85.1 mm 1.000 F

i j × i j k i j ×

i j × j 0

14ˆ ˆ ˆ ˆ16958.95 N mm 6603.8 N mm 3587.8 N mm 441.9 mm F k k k k 0

14 15.3 NF 14ˆ15.3 N 15.3 N 90.00 jF Ans.

14 4 3 23 F F f f F 0 23

ˆ154.37 113.65 N 191.7 N 36.36 iF Ans.

14.19 A slider-crank mechanism, as in Problem 14.15, has 0.008 m,a 2

0.25 m,AOR

1.25 m,BAR 1.0 m,CAR , 38 ,C 3

0.75 m,G AR 18 , 2 10 kg,m

3 140 kg,m 4 50 kg,m 2

22.0 N m s ,GI and 3

28.42 N m s ,GI and has a balanced

crank. Make a complete kinematic and dynamic analysis of this system at 2 120 with

2 6 rad/s ccw and 2 0 , using 50 kN 180B F and 80 kN 60 .C F

Kinematic Analysis:

22

ˆ ˆˆ6 rad/s 0.125 0.217 m

ˆ ˆ1.299 0.750 m/s 1.500 m/s 150

A AO

V ω ×R k × i j

i j

3

3 3

3

ˆ ˆ ˆ ˆˆ1.299 0.750 m/s rad/s 1.227 0.225 m

ˆ ˆ ˆ ˆ1.299 0.750 m/s 0.225 1.227 m

ˆ

B A BA

BV

i j k × i j

i j i j

V V ω ×R

i

3ˆ0.611 rad/sω k ˆ1.162 m/sB V i Ans.



2 2 2

22

2

2 2

ˆ ˆ6 rad/s 0.125 0.217 m

ˆ ˆ4.500 7.794 m/s 9.000 m/s 60

A AO AO

A R α ×R × i j

i j

23 3

2 2 23

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ4.500 7.794 m/s 0.459 0.084 m/s rad/s 1.227 0.225 mB

B A BA BA

A

i

A A R α ×R

i j i j k × i j

23 3

ˆ ˆ ˆ ˆ ˆ4.041 7.710 m/s 0.225 1.227 mBA i i j i j

2

3ˆ6.284 rad/sα k 2ˆ5.455 m/sB A i Ans.

3 3 3

2

3 3


G A G A G A

A A R α ×R

i j i j k × i j

3

2 2ˆ ˆ6.496 3.514 m/s 16.516 m/s 28.41G A i j Ans.

Dynamic Analysis:


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2

3 3

ˆ ˆ140 kg 6.496 3.514 m/s

ˆ ˆ909 492 N 1 034 N 151.59

Gm

i j

f A

i j

2

4 4

ˆ50 kg 5.455 m/s

ˆ273 N 273 N 180

Bm

i

f A

i

33 3

2 2ˆ8.42 N m s 6.284 rad/s

ˆ52.911 N m

GI

t α

k

k

44 4GI t α 0

3 3 352.911 N m 1 034 N 0.051 mh t f ,

4 4 40h t f



3 3 3 4 14A G A CA C BA BA B BA M R ×f t R ×F R ×f R ×F R ×F 0

14

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ0.660 0.357 m 909 492 N 52.911 N m 0.263 0.301 m 40 000 69 282 N

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1.227 0.225 m 273 N 1.227 0.225 m 50 000 N 1.227 0.225 m 1.000 F


i j × i i j × i i j × j 0

14

ˆ ˆ ˆ ˆ ˆ ˆ0.207 N m 52.911 N m 6 157 N m 61.425 N m 11 250 N m 1.227 in F k k k k k k 0

14 14 280 NF 14ˆ14 280 N 14 280 N 90.00 jF Ans.

14 4 34B F F f F F 0 34

ˆ ˆ50 273 14280 N 52 262 N 15.86 F i j Ans.

43 3 23C F F f F F 0 23

ˆ ˆ11 182 54 510 N 55 645 N 78.41 i jF Ans.

32 12 F F F 0 12

ˆ ˆ11 182 54 510 N 55 645 N 78.41 i jF Ans.




14.20 Cranks 2 and 4 of the cross-linkage illustrated in Fig. P14.20 are balanced. The

dimensions of the linkage are 2

150 mm,AOR 4 2

450 mm,O OR 450 mm,BAR 4

150 mm,BoR

600 mm,CAR and 3

300 mm.G AR Also, 3 17.8 N,w 2 4

27 N mm s ,G GI I and

3

255.29 N mm s .GI Corresponding to the position shown, and with 2 10 rad/s ccw and

2 0, a kinematic analysis gives results of 3 1.43 rad/s cw, 4 11.43 rad/s cw,

23 4 84.8 rad/s ccw, and

3

2ˆ ˆ7.776 7.374 m/s .G A i j Find the driving torque

and the pin reactions with ˆ133.5 N.C F j


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2 2

3 3

ˆ ˆ17.8 N 7.776 7.374 mm/s 9.65 m/s

ˆ ˆ14.33 13.58 N 19.74 N 136.52

Gm

i j

i j

f A

44 4 Gm f A 0

33 3

2 2ˆ55.29 N mm s 84.8 rad/s

ˆ4688.74 N mm

GI

t α

k

k

44 4

2 2ˆ7 N mm s 84.8 rad/s

ˆ594.3 N mm

GI

t α

k

k

3 3 34688.74 N mm 19.74 N 237.48 mmh t f ,

40h






4 44 34

t

O BO M t R ×F 0

34

ˆ ˆ ˆ ˆˆ594.3 N mm 21.42 148.45 mm 0.990 0.143t

F k i j i j 0×

34 4.99 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ235.72 185.57 m 14.32 13.59 N 4690.6 N mm 471.4 371.15 in 133.5 lb

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ353.57 278.375 mm 4.49 0.716 N 353.57 278.375 mm 0.143 0.990 rF

i j × i j k i j × j

i j × i j i j × i j 0

43ˆ ˆ ˆ ˆ ˆ5861.7 N mm 4690.3 N mm 62935.2 N mm 999 N mm 389.75 mm rF 0k k k k k

43 191.13 NrF 34ˆ ˆ22.38 189.92 N 191.26 N 96.7 F i j Ans.

34 14 F F F 0 14ˆ ˆ22.38 189.92 N 191.26 N 83.3 F i j Ans.

43 3 23C F F f F F 0 23ˆ ˆ8.05 42.8 N 43.56 N 100.67 F i j Ans.

32 12 F F F 0 12ˆ ˆ8.05 42.8 N 43.56 N 100.67 F i j Ans.



14.21 Find the driving torque and the pin reactions for the mechanism of Problem 14.20 under

the same dynamic conditions, but with crank 4 as the driver.

Given the same dynamic conditions, the D’Alembert forces and torques are the same as in

Problem 14.20. However, crank 2 is now a two-force member with no applied moment.

Therefore the free-body diagrams appear as:

The solution now proceeds as follows:

3 3 3 23B G B CB C AB M R ×f t R ×F R ×F 0

ˆ133.5 Nˆ ˆ ˆ ˆ ˆ ˆ117.85 92.8 mm 14.32 13.59 N 4690.6 N.mm 117.85 92.8 mm ji + j i j i j

23ˆ ˆ ˆ ˆ353.575 278.37 mm 0.500 0.866 F 0i j i j

23ˆ ˆ ˆ ˆ2931.2 N mm 4690.6 N mm 15733.97 N mm 445.37 mm F 0k k k k

23 39.275 NF 23ˆ ˆ19.64 34 N 39.27 N 120 F i j Ans.

32 12 F F F 0 12ˆ ˆ19.64 34 N 39.27 N 120 F i j Ans.

23 3 43C F F f F F 0 43ˆ ˆ33.95 181.1 N 184.27 N 79.38 F i j Ans.

34 14 F F F 0 43ˆ ˆ33.951 181.1 N 184.27 N 79.38 F i j Ans.



4 4 34 4 14O BO M R ×F t M 0 14ˆ1754.4 N mm M k Ans.

14.22 A kinematic analysis of the mechanism of Problem 14.20 at 2 210 with

2 10 rad/s ccw and 2 0 gave 3 14.7 , 4 164.7 , 3 4.73 rad/s ccw,

4 5.27 rad/s cw, 3 4 10.39 rad/s cw, and 3

27.8 20.85 m / s .G A Compute the crank

torque and the pin reactions for this posture using the same force CF as in Problem 14.20.


22 2 Gm f A 0 22 2GI t α 0

2 2 2 0h t f

3

2 2

3 3

ˆ ˆ17.8 N 7.29 2.77 m/s 9.65 m/s

ˆ ˆ13.43 5.12 N 14.37 N 159.15

Gm

i j

i j

f A

44 4 Gm f A 0

33 3

2 2ˆ7 N mm s 10.39 rad/s

ˆ72.8 N mm

GI

t α

k

k

44 4

2 2ˆ7 N mm s 10.39 rad/s

ˆ72.8 N mm

GI

t α

k

k

3 3 372.86 N mm 14.37 N 5.1 mmh t f ,

40h

Next, the free-body diagrams are drawn. Since the lines of action for the forces on the

free-body diagrams cannot be discovered from two- and three-force member concepts, the

force 34F is divided into radial and transverse components.

4 44 34

t

O BO M t R ×F 0

34

ˆ ˆ ˆ ˆˆ72.86 N mm 144.7 39.47 mm 0.263 0.965t

F k i j i j 0×

34 0.485 NtF

3 3 3 43 43

t r


43

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ290.12 76.32 mm 13.83 5.12 N 72.86 N mm 580.25 152.65 mm 133.5

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ435.2 114.48 mm 0.129 0.467 N 435.2 114.48 mm 0.965 0.263 rF

i j × i j k i j × j

i j × i j i j × i j 0

43ˆ ˆ ˆ ˆ ˆ459.685 N mm 72.86 N mm 77430 N mm 188.57 N mm 224.92 mm

rF k k k k k 0

43 347 NrF 34ˆ ˆ333.75 89 N 347 N 15.20 F i j Ans.

43 3 23C F F f F F 0 23ˆ ˆ347 48.95 N 351.5 N 8.02 F i j Ans.




14.23 Figure P14.23 illustrates a linkage with an extended coupler having an external force of

CF acting during a portion of the cycle. The dimensions of the linkage are

2400 mm,AOR

4 21000 mm,O O BAR R

41400 mm,BOR

3800 mm,G AR and

4 4500 mm.G OR Also 3 987.9 N,w 4 925.6 N,w

3

225142.5 N mm s ,GI and

4

229370 N mm s ,GI and the crank is balanced. Make a kinematic and dynamic

analysis for a complete rotation of the crank using 2 10 rad/s ccw,

ˆ ˆ222.5 3942.7 NC F i j for 290 300 and C F 0 otherwise. 1

00

mm

400 mm

Kinematic Analysis 32 4

2 3 1 4

jj jR e R e R R e

32 4400 1000 1000 1400jj je e e

2 3 4400cos 1000cos 1000 1400cos 2 3 4400sin 1000sin 1400sin

Eliminating 3 we find 4 from the roots of the quadratic

2

2 4 2 4 2425 200cos tan 2 1400sin tan 2 1200cos 3075 0 Ans.

Then 1

3 4 2 4 2tan 175sin 50sin 175cos 125 50cos Ans.



32

3400 800

jj

G e e R

4

41000 500

j

G e R

Ans.

324.086

400 1403.5jj

C e e

R Ans.

The first-order kinematic coefficients are: 32 4

3 4400 1000 1400jj jj e j e j e

2 3 3 4 4400sin 1000sin 1400sin 2 3 3 4 4400cos 1000cos 1400cos

3 4 2 4 3350sin 875sin 4 2 3 4 3250sin 875sin Ans.

32

3 3 2 3 3 2 3 3400 800 400 sin 50sin 400 cos 50cosjj

G j e j e j R Ans.

4

4 4 4 4 4 4500 500sin 500cosj

G j e j R Ans.

32

2 3 3 2 3 3

4.086

3

400sin 1403.57sin 4.086 400cos 1403.57 cos 4.086

400 1403.57jj

C

j

j e j e

R

Ans.

The second-order kinematic coefficients are: 3 32 4 42 2

3 3 4 4400 1000 1000 1400 1400j jj j je j e e j e e

2 2

2 3 3 3 3 4 4 4 4400cos 1000sin 1000cos 1400sin 1400cos 2 2

2 3 3 3 3 4 4 4 4400cos 1000sin 1000cos 1400sin 1400cos

2 2

3 4 2 4 3 3 4 4 3350cos 875cos 1225 875sin Ans.

2 2

4 4 2 3 4 3 4 4 3250cos 625 875cos 875sin Ans.

3 32

3

2 2

3

2

3 3

2 3 3 3 3 2 3 3 3 3400cos 800sin 800cos 400sin 800cos 800sin

400 800 800j jj

G

j

e j e e

R

Ans.

4 4

4

2 2 2

4 4 4 4 4 4 4 4 4 4500 500 500sin 500cos 500cos 500sin

j j

G j e e j R Ans.

Dynamic Analysis

By virtual work we can formulate the dynamic input torque requirement as:

3 412 3 3 3 4 4 4ˆ ˆ

G G C CM f R t k f R t k F R

The individual elements of this equation are:

2 2

3

3 3 3 3

22 2

3 3 3 2

2 3 3 3 3 2 3 3 3 34094 cos 8188sin 8188cos 4094sin 8188cos 8188sin N

987.9 9804 mm/s 10 rad/s 10 N/mmG G G G

j

m m

f A R R R

3

2

3 2 3 3 3 33 2 3 3

2

2 3 3 3 3 2 3 3

4094cos 8188sin 8188cos 400sin 800sin N mm

4094sin 8188cos 8188sin 400cos 800cos N mm

G

f R

33 3 2 3 3 3 2 3 3 332575.97 N m sin 1 cos 2G f R

3 3

2

3 3 3 2 3ˆ ˆ2514.25 N mG GI I t α k k

3 3 3 3ˆ 2514.25 N m t k

4 4 4 4

22 2

4 4 4 2

2 2

4 4 4 4 4 4 4 4

925.6 N 9804 mm/s 10 rad/s 9.44 N/in

4797sin 4797cos 4797cos 4797sin N

G G G Gm m

j

f A R R R



4

2

4 4 4 4 4 4 4

2

4 4 4 4 4 4

4797sin 4797cos 500sin N mm

4797cos 4797sin 500cos N mm

G

f R

44 4 42298.55 N mG f R

4 4

2

4 4 4 2 4ˆ ˆ2937 N mG GI I t α k k

4 4 4 4ˆ 2937 N m t k

2 3 3

2 3 3

4450 N cos120 400sin 1403.57sin 4.086 mm

4450 N sin120 400cos 1403.57cos 4.086 mm

C C

F R

2 3 31780sin 120 6245.9sin 124.086 N mC C F R

Reassembling the elements we must remember that force FC is nonzero for only a portion

of the cycle. Therefore,

3 2 3 3 3 2 3 3 3 3 3 4 4 4 4

3 3 4 43 2 3 3 2 3 3

12 3275.9 sin 1 cos 2 2514.2 2298.55 2937 N m

N m53353275.9cos 3275.9sin 1 9066=

M

for the entire cycle and, for 290 300 , an additional increment is added:

2 3 312 1780sin 120 6245.9n 124.086 N mM Ans.

This input torque requirement is shown in the following plot. Notice the small

discontinuities in the curve when force FC begins and ends its effect.

11,000

5,500

5,500



14.24 Figure P14.24 illustrates a motor geared to a shaft on which a flywheel is mounted. The

mass moments of inertia of the parts are as follows: flywheel, 2303.7 N mm s ;I

flywheel shaft, 21.724 N mm s ;I gear, 219.135 N mm s ;I pinion, 20.388 N mm s ;I motor, 29.612 N mm s .I If the motor has a starting torque of

8343.75 N mm, what is the angular acceleration of the flywheel shaft at the instant the

motor is turned on ?

225 mm

50 mm

If we identify the motor shaft as 2 and the flywheel shaft as 3 then 2 2 2

2 9.612 N mm s 0.388 N mm s 10 N mm sI 2 2 2 2

3 303.7 N mm s 1.724 N mm s 19.135 N mm s 324.559 N mm sI

3 2 3 2R R 3 2 3 2R R

3 3 23 3 3M R F I 2

23 2 3 3 2F R R I

2 12 2 32 2 2M M R F I 2

12 2 2 2 32 2 2 3 3 2M I R F I R R I

Now, substituting the numeric values,

22 2 2

2 28343.75 N mm 10 N mm s 25 mm 112.5 mm 324.559 N mm s 26.028 N mm s

2 2

2 8343.75 N mm 26.028 N mm s 320.56 rad/s

2 2

3 2 3 2 25 mm 112.5 mm 320.56 rad/s 71.24 rad/sR R Ans.



14.25 The disk cam of Problem 13.31 is driven at a constant input shaft speed of

2 25 rad/s ccw. Both the cam and the follower have been balanced so that the centers

of mass of each are located at their respective fixed pivots. The mass of the cam is 0.075

kg with radius of gyration of 30 mm, and for the follower the mass is 0.030 kg with

radius of gyration of 35 mm. Determine the moment 12M required on the camshaft at the

instant shown in the figure to produce this motion.

3

22

3 3 0.030 0.035 0.000 036 75 2

GI m k kg m kg m

For full-rise cycloidal cam motion, Eq. (5.19),

2 30 112.51 cos 1 cos 2 0.200

150 150

Ly

2 2

360 302 2 112.5sin sin 2 0.480

150150

Ly

22 2

3 2 0.480 25 rad/s 300 rad/sy

3

2 2

3 3 0.000 036 75 kg m 300 rad/s 0.011 025 N mGt I

By virtual work,

312 3ˆ

0.200 0.011 025 N m 0.150 m 8 N sin 45 0.168 N m ccw

CO CM y t

R ×F k

Ans.



14.26 Repeat Problem 14.25 with a shaft speed of 2 50 rad/s ccw.

3

22 2

3 3 0.030 kg 0.035 m 0.000 036 75 kg mGI m k

For full-rise cycloidal cam motion, Eq. (5.19),

2 30 112.51 cos 1 cos 2 0.200

150 150

Ly

2 2

360 302 2 112.5sin sin 2 0.480

150150

Ly

22 2

3 2 0.480 50 rad/s 1 200 rad/sy

3

2 2

3 3 0.000 036 75 kg m 1 200 rad/s 0.044 1 N mGt I

By virtual work,

312 3ˆ

0.200 0.044 1 N m 0.150 m 8 N sin 45 0.161 N m ccw

CO CM y t

R ×F k

Ans.



14.27 A rotating drum is pivoted at 2O and is decelerated by the double-shoe brake mechanism

illustrated in Fig. P14.27. The mass of the drum is 1023.5 N and its radius of gyration is

1414.5 mm. The brake is actuated by force ˆ445 N P j , and it is assumed that the

contacts between the two shoes and the drum act at points C and D, where the

coefficients of Coulomb friction are 0.350. Determine the angular acceleration of

the drum and the reaction force at the fixed pivot 12.F

5 65ˆ ˆ ˆ ˆ300 mm 75 mmP F M i × j j × i 0 65 445 1780 NF P

5 65 35 F P F F 0 35ˆ ˆ1780 445 N 1833.4 N 14.04 F i j

The friction angle is 1tan 0.350 19.29 .

3 53 23ˆ ˆ ˆ ˆ ˆ550 mm 125 225 mm cos sin F M j ×F i j × i j 0

23 3858 NF 32ˆ ˆ3644.5 1277 N 3858 N 19.29F i j

4 64 24ˆ ˆ ˆ ˆ ˆ625 mm 125 225 mm cos sin F M j ×F i j × i j 0

24 6501.45 NF 42ˆ ˆ6136.5 2149.35 N 6501.45b 160.71F i j

2

22 2 2

2 2 1023.5 N 9804.4 mm/s 141.5 mm 2123.1 N mm sGI m k

2 2 2

ˆ ˆ ˆ ˆ ˆ ˆ200 mm 3644.5 1277.2 N 200 mm 6136.5 2149.4 NO GI M i × i j i × i j α

2

2ˆ322 rad/s α k Ans.

2 12 32 42 F F F F 0

12ˆ ˆ2496.45 872.2 N 2643.3 N 19.29F i j Ans.

Note that gravitational effects are not yet included. If gravity acts in the ˆj direction then

the j component is 410 lb. Since the main bearing at O2 supports this weight, it does not

affect the friction forces and can be added by superposition. If weights of the other parts



were known, however, these weights might have some small effect on the friction forces

and the braking forces, and would have to be included simultaneously. Superposition

could not be applied.



14.28 For the mechanism illustrated in Fig. P14.28, the dimensions are 2 4

0.15 m,G OR

20.20 m,EGR and the length of link 4 is 0.20 m, symmetric about O4. The ground

bearing is midway between E and G2. There is a torque 2T acting on the input link 2, and

a torque 4T acting on link 4. Link 2 is in translation with a velocity of

2ˆ0.114 8 m/sV j and an acceleration of 2

2ˆ0.35 m/s A j and the line connecting

mass centers 2G and

3G is horizontal. The kinematic coefficients are 3 11.5 rad/m ,

2

3 380 rad/m , 43 2 m/mR , and 2

43 40 m/mR (where 43R is the vector from

3G

to 4G ). The acceleration of the mass center of link 3 is

3

2ˆ ˆ1.053 0.432 m/sG A i j . The

masses and second moments of mass of the moving links are 2 4m m 0.5 kg,

3m 1 kg, 2 4

22 kg m ,G GI I and 3

2I 5 kg m .G Assume that gravity acts in the

negative Z direction, and that the effects of friction can be neglected. Determine the

unknown internal reaction forces, and the unknown torques 2T and 4T .

The free-body diagram for link 2 is shown in Figure 1. Recall that gravity acts in the

negative Z-direction and the effects of friction can be neglected.

Since the center of mass 2G is translating in the Y-direction, the sum of the external

forces in the X-direction acting on link 2 shows

212 32 2 0X X X X

GF F P m A (1)



Figure 1. Free-body diagram of link 2.

Since friction can be neglected, the sum of the external forces in the Y-direction acting on

link 2 gives

232 2

Y Y Y

GF P m A

Substituting the given information, the Y-component of the reaction force between links 2

and 3 is

2

32 0.5 kg ( 0.35 m/s ) 40 N sin120 34.816 NYF Ans. (2)

Since link 2 is not rotating then the angular acceleration 2 0 . Therefore, the sum of the

external moments on link 2 acting about 2G can be written as

7 12 9 2 0X XR F R P T (3)

Therefore, there are still 4 unknowns for link 2, namely the forces 12 ,XF 32 ,XF and 32

YF and

the torque 2T .

The free-body diagram for link 3 is shown in Figure 2.


The sum of the external forces in the X-direction acting on link 3 can be written as



323 43 3

X X X

GF F m A (4)

The sum of the external forces in the Y-direction acting on link 3 can be written as

323 43 3

Y Y Y

GF F m A (5)

The sum of the external moments acting about the center of mass G3 can be written as

2 3 343 43 23 3

Y

G G GR F R F I (6)

The vector 43R points from the center of mass of link 3 to the location of the reaction

force F43, and the vector 2 3G GR points from the center of mass of link 3 to the center of

mass of link 2.

Equations (4), (5), and (6) contain two new unknown variables, namely the internal

reaction force F43 and the location of this force (i.e., R43). Note that the force F43 is

perpendicular to the slot since friction is neglected. Therefore, 43

XF and 43

YF are not

independent unknowns (the angle is known). Therefore, there are 6 equations and 6

unknown variables, namely the forces 12 ,XF 32 ,XF 32 ,YF 43 ,F the torque T2, and the distance

R43.

These six unknowns can now be solved for by inspection. Substituting Eq. (2) and

the given acceleration of the center of mass 2G into Eq. (5), the Y-component of the

internal reaction force between links 3 and 4 is

2

43 43 sin 60 1 kg ( 0.432 m/s ) (34.816 N) 34.384 NYF F Ans.

Therefore, the force between links 3 and 4 is

43

34.384 N39.70 N

sin 60F

Ans.

Substituting known values into Eq. (4), the internal reaction force between links 2 and

3 is

2

23 ( 39.70 N)cos60 1 kg 1.053 m/s 20.91 NXF Ans.

Substituting known values into Eq. (6) gives

2 2

34( 39.70 N) (0.259 81 m)(34.816 N) 5 kg m ( 0.983 rad/s )R

Rearranging this equation, the unknown distance is

43

13.961 N m0.351 66 m

39.70 NR

Therefore, the distance from the ground pin O4 to the line of action of the internal

reaction force F43 is

43 0.300 m 0.351 66 0.300 m 51.7 mmZ R

Since the distance Z is less than the length of link 4 then link 3 is sliding along link 4 (i.e.,

there is sliding contact and not tipping). The internal reaction force between links 3 and 4

acts within the physical limits of link 4.


12 20.91 N 40 N cos60 0XF Ans.

Rearranging this equation, the unknown force is

12 40.91 NXF Ans.




20.1 m (40.91 N) 0.2 m (40 N)cos120 0T

Rearranging this equation, the unknown torque acting on link 2 is

2 0.091 NmT Ans.

The free-body diagram for link 4 is shown in Figure 3.


Since 4G is coincident with the ground pivot 4O , the sum of the external forces in the X-

direction acting on link 4 can be written as

434 14 4cos60 0X X

GF F m A (7)

The sum of the external forces in the Y-direction acting on link 4 can be written as

34 14sin 60 0YF F (8)

The sum of the external moments acting about the center of mass of link 4 can be written

as

434 4 4GZF T I (9)

Equations (7), (8), and (9) contain three new unknown variables, namely the internal

reaction forces 14 ,XF 14 ,YF and the torque T4. These three unknown variables can now be

solved for as follows. Substituting known values into Eq. (7), the X-component of the

force between links 1 and 4 is

14 19.85 NXF Ans.

Substituting known values into Eq. (8), the Y-component of the internal reaction force

between links 1 and 4 is

14 34.38 NYF Ans.

Substituting known values into Eq. (9), the torque acting on link 4 is

2 2

4 2 kg m ( 0.983 rad/s ) (0.051 7 m)(39.70 N) 4.018 NmT Ans.

The negative sign indicates that the torque acting on link 4 is clockwise.



14.29 For the mechanism illustrated in Fig. P14.29, the kinematic coefficients are

3 2 rad/m, 2

3 6.928 rad/m , 4 1.732 m/m,R and 2

4 8 m/m .R The

velocity and the acceleration of the input link 2 are 2

ˆ5 m/s V j and 2

2ˆ20 m/s A j

and the force acting on link 2 is ˆ200 N. F j The length of link 3 is 1 mBAR and the

distance 3 2

0.5 m.G GR A linear spring is attached between points O and A with a free

length 0.5 mL and a spring constant 2 500 N/m.K A viscous damper with a

damping coefficient 45 N s/mC is connected between the ground and link 4. The

masses and mass moments of inertia of the links are 2 0.75 kg,m

3 2.0 kg,m

4 1.5 kg,m 2

20.25 N m s ,GI 3

21.0 N m sGI and 4

20.35 N m s .GI Assume

that gravity acts in the negative Y-direction (as illustrated in Fig. P14.29) and the effects

of friction in the mechanism can be neglected.

(i) Determine the first-order kinematic coefficients of the linear spring and the viscous

damper.

(ii) Determine the equivalent mass of the mechanism.

(iii) Determine the magnitude and direction of the horizontal force P that is acting on link

4.

A double-slider mechanism.



(i) The vectors for the linear spring are shown in Fig. 2a.

Figure 2a. Vectors for the linear spring.

The vector loop for the linear spring can be written as

?

2S

I

R R 0

From which, the magnitudes can be written as the scalar equation

2SR R

Differentiating this with respect to the input position 2R , the first-order kinematic

coefficient of the spring is

1 m/mSR Ans. (1)

Note that the sign is positive because, for a negative input, the length of the linear spring

is decreasing. Also, note that the first-order kinematic coefficient of the mass center of

input link 2 is

21 m/mG SY R

The vectors for the viscous damper are shown in Fig. 2b.

Fig. 2b. Vectors for the viscous damper.

The vector loop for the damper can be written as

4

?

9 C

R R R 0

Since all components of this equation are horizontal, this gives

9 4 0CR R R

Differentiating with respect to the input position 2R gives

4 0CR R

Rearranging and substituting the given data, the first-order kinematic coefficient of the

viscous damper is

4 1.732 m/mCR R Ans. (2)

The positive sign agrees with our intuition since, for a positive input, the length of the

viscous damper is increasing.

(ii) The equivalent mass of the mechanism can be written as

4

EQ

2

j

j

m A

(3)

For link 2: 2 2 2

2 2 2

2 2 2G G GA m X Y I

(4)

The vector loop for the center of mass of the input link can be written as

2

??

2

I

G

R R 0



The X and Y components of this equation give

20GX

and 2 2GY R

Differentiating these with respect to the input position 2R give

20GX

and

2 2R 1 m/mGY

Substituting these values into Eq. (4) gives

22 2 2

2 0.75 kg 0 1 0.25 kg m 0 0.75 kgA

(5)

For link 3: 2

33

2

3

2

333 GGG IYXmA

(6)

The vector loop for the center of mass of link 3 can be written as

3 3 2

??

2

I

G G G

R R R

The X and Y components are

3 3 2 3cos 0.25 mG G GX R

and 3 3 22 3sin 0.433 mG G GY R R

Differentiating these with respect to the input position 2R give

3 3 2 3 3sin 0.866 m/mG G GX R

and 3 3 2 3 31 cos 0.5 m/mG G GY R

Substituting these and other known data into Eq. (6) gives

22 2 2

3 2.0 kg ( 0.866 m/m) 0.5 m/m 1.0 kg m ( 2 rad/m) 6 kgA

(7)

For link 4: 2

44

2

4

2

444 GGG IYXmA

(8)

Note from given data that 4 4 1.732 m/m;GX R therefore, Eq. (8) can be written as

2 22 2

4 1.5 kg [ 1.732 m/m 0 ] 0.35 kg m 0 4.5 kgA

(9)

Therefore, substituting Eqs. (5), (7), and (9) into Eq. (3), the equivalent mass of the

mechanism is

EQ 0.75 kg 6 kg 4.5 kg 11.25 kgm Ans. (10)

(iii) The power equation for the mechanism can be written as

2 4

fdWdT dU

dt dt dt F V P V

Substituting the time rate of change of energy terms into the right-hand side gives

4 4 4

3 2 2

2 4 2 2 2 2 0 22

2 2 2jj j j G s S S S C

j j j

A R R B R m gY R K R R R R CR R

F V P V

The linear velocity of link 2 and the force acting on link 2 are both in the same direction

(that is, both downward). Assuming that the force P is in the same direction as the

velocity of link 4 (that is, to the right), then the above equation can be written as

4 4 4

3 2 2

2 4 2 2 2 2 0 22


j j j

FV PV A R R B R m gY R K R R R R CR R

The velocity of the input link 2 is 2 2V R and the velocity of link 4 is 4 4;V R therefore,

this can be written as



4 4 43 2 2

2 4 2 2 2 2 0 22


j j j

FR PR A R R B R m gY R K R R R R CR R

Dividing by the input velocity 2R throughout gives the equation of motion for the

mechanism, that is

4 4 42 2

4 2 2 0 2


j j j

F PR A R B R m gY K R R R CR R

(11)

where the first-order kinematic coefficient of link 4 is given as 4 1.732 m/mR . The

sum of the Bj terms can be written as

4 4

2 2 2

1

2

j

j

j j

dAB

d

(12)

For link 2: 2 2 2 2 22 2 2 2( )G G G G GB m X X Y Y I

and this has a value of

2

2 0.75 kg 0 0 0.25 kg m 0 0B

(13)

For link 3: 3 3 3 3 33 3 3 3G G G G GB m X X Y Y I

which has a value of

2 2 2

3 2 kg [( 0.866 m/m)( 4 m/m ) (0.5 m/m)(0)] 1 kg m ( 2 rad/m)( 0.928 rad/m ) 20.78 kg/mB (14)

For link 4: 4 4 4 4 44 4 4 4G G G G GB m X X Y Y I

which has a value of

2 2

4 1.5 kg [( 1.732 m/m)( 8 m/m ) (0)(0)] 0.35 kg m 0 0 20.78kg/mB (15)

Substituting Eqs. (13), (14), and (15) into Eq. (12) gives

4

2 3 4

2

0 20.78 kg/m 20.78 kg/m 41.56 kg/mj

j

B B B B

(16)

The change in potential energy due to gravity is

4

2jj G

j

m gY

For link 2: 2

2

2 (0.75 kg)(9.81 m/s ) 1 m/m 7.36 NGm gY

For link 3: 3

2

3 (2 kg)(9.81 m/s )( 2 m/m) 39.24 NGm gY

For link 4: 4

2

4 (1.5 kg)(9.81 m/s )(0) 0Gm gY

Summing these three values gives

4

2

7.36 N 39.24 N 0 31.80 Njj G

j

m gY

(17)

Then substituting Eqs. (1), (2), (10), (16), and (17) into Eq. (11) gives

2 2

4 2 2 0 211.25 kg 41.56 kg/m 31.80 N (1 m/m) (1.732 m/m)S S SF PR R R K R R C R

Rearranging this equation, the force acting on link 4 can be written as

2

2 2 0 2

4

111.25 kg 41.56 kg/m 31.80 N 3S S SP F R R K R R CR

R



The input velocity is 2 5 m/sR , the input acceleration is 2

2 20 m/s ,R and the force

is 200 N.F Substituting these values and the known data into this equation, the force

acting on link 4 can be written as

2 2200 N 11.25 kg ( 20 m/s ) 41.56 kg/m ( 5 m/s) 31.80 N1

1.732 m/m 2500 N/m 0.866 m 0.5 m 3 45 Ns/m ( 5 m/s)P

or as

1200 N 225 N 1 039 N 31.80 N 915.0 N 675.0 N

1.732P

Therefore, the force acting on link 4 is

705.66 NP Ans.

The negative sign indicates that the force P (acting on link 4) is acting to the left; that is,

in the opposite direction to the velocity of the output link 4. Recall that the force P was

originally assumed to be acting to the right.



14.30 Consider the four-bar linkage of Problem P14.20 modified as illustrated in Fig. P14.30.

The linkage includes a spring and a viscous damper as shown. The spring has a stiffness

2.14 N/mmk and a free length 0 112.5 mm.R The viscous damper has a damping

coefficient 0.04.45 N /mm.C The external force acting at point C of coupler link 3 is

556.25 NCF vertically downward (that is, in the negative Y-direction). The input

crank is rotating with a constant angular velocity ω2 = 10 rad/s ccw and the angular

acceleration of link 3 is 2

3 84.8 rad/s ccw; the acceleration of the mass center of

coupler link 3 is 3

2

Gˆ ˆ7750 7375 mm/s A i j . Use the masses and the second moments of

mass as specified in Problem 14.20 with the exception that the weight of link 3 is w3 =

445 N. Assume that the locations of the centers of mass of links 2 and 4 are coincident

with the ground pivots 2O and

4 ,O respectively, and the center of mass of link 3 is as

indicated by 3G in Fig. P14.30. Also, assume that gravity acts vertically downward (that

is, in the negative Y-direction) and the effects of friction in the mechanism can be

neglected.

1. Write the equation of motion for the mechanism.

2. Determine the equivalent mass moment of inertia of the mechanism.

3. Determine the driving torque 2T on the input crank 2 from the equation of motion.

The mechanism modified from Problem 14.20.



From Problem 14.20, the given data are

2

4 2

4

150 mm

450 mm

450 mm

150 mm

600 mm

AO

O O

AB

BO

AC

R

R

R

R

R

3

3

2 4

3

2

2

300 mm

445 N

55.29 N mm s

7 N mm s

G A

G

G G

R

w

I

I I

2

3

4

2

10 rad/s ccw

1.43 rad/s cw

11.43 rad/s cw

9660 mm/sg

2

2

3

2

4

0

84.8 rad/s ccw

84.8 rad/s ccw

ˆ556.25 NC

F j

0 shows vectors that are used throughout the solution.

The first-order kinematic coefficient of link 3 can be written as

33 33

2

1.43 rad/s0.143 rad/rad

10 rad/s


44

2

11.43 rad/s1.143 rad/rad

10 rad/s

The angular acceleration of link 3 can be written as

2

3 3 2 3 2

Rearranging this, the second-order kinematic coefficient for link 3 can be written as

223 3 2

3 33 22

2

84.8 rad/s ( 0.143 rad/rad)(0)0.84 rad/rad

10 rad/s

Similarly, the second-order kinematic coefficient of link 4 is

224 4 2

4 22

2

84.8 rad/s ( 0.143 rad/rad)(0)0.84 rad/rad

10 rad/s

Since the mass centers G2 and G4 are located at the fixed pivots O2 and O4, respectively,

the first- and second-order kinematic coefficients of these mass centers are

20Gx

20Gx

20Gx

20Gy

20Gy

20Gy

418 inGx

40Gx

40Gx

40Gy

40Gy

40Gy

To find the first- and second-order kinematic coefficients for the center of mass G3, the

vector loop for the center of mass of link 3 can be written as

3 2 33G R R R

where R2 = 150 mm, θ2 = 60º, R33 = 300 mm, and θ33 = θ3 = 321.8º. The X and Y

components of the vector equation for the center of mass of link 3 are



3 2 2 33 3cos cos 150 mm cos60 300 mm cos321.8 310.75 mmGx R R

3 2 2 33 3sin sin 150 mm sin 60 300 mm sin321.8 55.62 mmGy R R

The first-order kinematic coefficients for the center of mass of link 3 are

3 2 2 33 3 3sin sin 150 mm sin60 300 mm sin321.8 ( 0.143 rad/rad) 156.43 mm/radGx R R

3 2 2 33 3 3cos cos 150 mm cos60 300 mm cos321.8 ( 0.143 rad/rad) 41.29 mm/radGy R R

The second-order kinematic coefficients for the center of mass of link 3 can be written as

3

2

2 2 33 3 3 33 3 3cos cos sinGx R R R

3

2

2 2 33 3 3 33 3 3sin sin cosGy R R R

Therefore,

3

2 2 2150 mm cos60 300 in cos321.8 ( 0.143 rad/rad) 300 mm sin321.8 (0.848 rad/rad ) 77.5 mm/radGx

3

2 2150 mm sin60 300 mm sin321.8 ( 0.143 rad/rad) 300 mm cos321.8 (0.848 rad/rad) 73.8 mm/radGy

To determine 4

2

j

j

A

, note that 4

EQ

2

j

j

A I

(that is, the equivalent mass moment of

inertia). Therefore, the units must be 2in lb s .

For link 2:

2 2 2

2 2 2 2 2 2

2 2 2 2( ) (0 0) 7 N mm s (1 rad/rad) 7 N mm sG G GA m x y I m

For link 3:

2 2 2

3 3 3 3 3 3

22 2 2 2

2

( )

445 N[( 156.43 mm/rad) 41.29 mm/rad ] 55.29 N mm s ( 0.143 rad/rad) 121.7 N mm s

9653 mm/s

G G GA m x y I

For link 4:

4 4 4

2 2 2 2 2 2

4 4 4 4( ) (0 0) 7 N mm s ( 1.143 rad/rad) 9.15 N mm sG G GA m x y I m

Therefore, the equivalent mass moment of inertia of the mechanism is 4

2 2 2 2

EQ 2 3 4

2

7 N mm s 121.1 N mm s 9.15 N mm s 137.85 N mm sj

j

I A A A A

Ans.

To determine

4

2j

jB , note that 4 4

2 2 2

1

2

j

j

j j

dAB

d

; therefore, the units must be 2 in lb s .

For link 2:

2

2 2 2 2 2 2 2 2 2 2( ) (0 0) 7 N mm s (1 rad/rad)(0) 0G G G G GB m x x y y I m

For link 3:



3 3 3 3 3 3 3 3 3

2 2

2

2 2

2

( )

445 N[( 156.43 mm/rad)(77.5 mm/rad ) (41.29 mm/rad)(73.8mm/rad )]

9653 mm/s

55.29 N mm s ( 0.143 rad/rad)(0.848 rad/rad )

48.505 N mm s

G G G G GB m x x y y I

For link 4:

4 4 4 4 4 4 4 4 4

2 2 2

4

( )

(0 0) 7 N mm s ( 1.143 rad/rad)(0.848 rad/rad ) 6.786 N mm s


m

Therefore, the sum of these coefficients is 4

2 2 2

2 3 4

2

0 48.505 N mm s 6.786 N mm s 55.291 N mm sj

j

B B B B

The power equation can be written as

fdWdT dU

Pdt dt dt

(1)

The left-hand side of the power equation can be written

2 2 C CP T F V (2)

The unknown torque 2T is taken to be positive in the same direction as the input angular

velocity (that is, counterclockwise). The velocity of point C can be written as

2ˆ ˆ( )C C Cx y V i j (3)

The first-order kinematic coefficients for the path of point C can be obtained from the

vector equation

2 3C R R R

The X and Y components of this vector equation are

2 2 3 3cos cosCx R R

2 2 3 3cos sinCy R R

Therefore, the first-order kinematic coefficients for point C are

2 2 3 3 3sin sin 150 mm sin60 600 mm sin321.8 ( 0.143 rad/rad) 182.963 mm/radCx R R (4a)

2 2 3 3 3cos cos 150 mm cos60 600 mm cos321.8 ( 0.143 rad/rad) 7.5735 mm/radCy R R (4b)

Substituting Eqs. (4) into Eq. (3), the velocity of point C can be written as

2ˆ ˆ182.963 mm/rad 7.5735 mm/radC

V i j

Therefore, the power due to the vertically downward force at point C is

2 2

2 2

ˆ ˆ ˆ556.25 N ( ) 556.25 N

556.25 N (7.5735 mm/rad) 4213 N mm/rad

C C C C Cx y y

F V j i j

(5)

The negative sign indicates that the vertical force and the vertical component of the

velocity of point C are in opposite directions (that is, that the vertical component of the

velocity of point C is upwards).

Substituting Eq. (5) into Eq. (2), the net power is



2 2 24213 N mm/radP T (6)

Now consider the right-hand side of the power equation, see Eq. (1). In general, the time

rate of change of kinetic energy can be written as

4 4

3

2 2

j j

j j

dTA B

dt

However, the generalized inputs for this problem are 2 , 2 2 , and

2 2. Therefore, this equation can be written as

4 4

3

2 2 2

2 2

j j

j j

dTA B

dt

The constant angular velocity of the input link is 2 10 rad/s ccw. Therefore, the time

rate of change of the kinetic energy is

2 2 2

2 2[ 137.838 N mm s (0) 55.291 N mm s (10 rad/s) ] 5529.1 N mmdT

dt

(7)

The time rate of change of the potential energy due to gravity is

3

4

3 2 2 2

2

445 N 41.2875 mm/rad 1837.293 N mmj

g

j G G

j

dUm gy m gy

dt

(8)

The vector loop equation for the spring can be written as

2 10S R R R 0

The X and Y components are

2 2cos cos 0S SR R (9a)

2 2 10sin sin 0S SR R R (9b)

From Eq. (9a) the stretched length of the spring (for this position of the mechanism) is

2 2cos 150 mm cos60 75 mmAsR x R

Differentiating Eqs. (9) with respect to the input position gives

2 2sin cos sin 0S S S S SR R R

2 2cos sin cos 0S S S S SR R R

Substituting the known information gives

150 mm sin 60 0o

SR

Therefore, the first-order kinematic coefficient for the spring is

150 mm sin 60 129.75 mm/radsR

The time rate of change of the potential energy in the spring is

2 2 2( ) 2.136 N/mm (75 mm 112.5 mm)( 129.75 mm/rad) 10.392 N mm/rads so ssdU

k R R Rdt

(10)

The vector loop equation for the viscous damper can be written as

4 11C R R R 0



The X and Y components of this equation are

4 4 11 11cos cos cos 0C CR R R

4 4 11 11sin sin sin 0C CR R R

Differentiating these with respect to the input position gives

4 4 4sin cos sin 0C C C C CR R R

4 4 4cos sin cos 0C C C C CR R R

Substituting the known information (with the angle 4 261.8 , and the first-order

kinematic coefficient 4 1.143 rad/rad ) gives

150 mm sin 261.8 ( 1.143 rad/rad) cos180 0CR

Therefore, the first-order kinematic coefficient for the damper is

150 mm sin 261.8 ( 1.143 rad/rad) 169.75 mm/radCR

The positive sign indicates that the length of the vector CR is increasing for positive

input. The velocity of point B at the end of the damper is

2 169.75 mm/rad (10 rad/s) 1697.5 mm/sB CV R

The time rate of change of the dissipative effect of the damper is

2 2 2

2 2 20.0445 N s/mm (169.75 mm/rad) 10 rad/s 12.784 N mm/radf

c

dWCR

dt

(11)

Therefore, from Eqs. (7), (8), (10), and (11), the right hand side of the power equation,

Eq. (1), can be written as

2 2 2 2

2

5.5291 N m/rad 1.8373 N m/rad 10.392 N m/rad 12.784 N m/rad

30.5424 N m/rad

fdWdT dU

dt dt dt

(12)

Substituting Eqs. (6) and (12) into Eq. (1) gives

2 2 2 24.213 N m/rad 30.5424 N m/radT (13)

The equation of motion is obtained by dividing both sides of Eq. (13) by the input ngular

velocity ω2. Therefore, the equation of motion for this problem can be written as

2 4.213 N m 30.5424 N mT Ans.

The driving torque acting on the input crank is

2 34.755 N mT

The positive sense indicates that the driving torque acting on the input crank is in the

same direction as the input angular velocity. Therefore, the driving torque acting on the

input crank is

2 34.7554 N m ccwT Ans.



14.31 For the Scotch-yoke mechanism in the position illustrated in Fig. P14.31, an external

force ˆ125 NP j is acting on link 4, and an unknown torque 2T is acting on the input

link 2. The length 2 1 mR , the angle 30 , and the angular velocity and acceleration

of link 2 are 2

ˆ15 rad/sω k and 2

2ˆ2 rad/sα k , respectively. The accelerations of the

centers of mass of the links are 2

2ˆ ˆ5.4 11.3 m/sG A i j , 3

2ˆ ˆ10.8 22.6 m/sG A i j , and

4

2ˆ22.6j m/sG A . The centers of mass of links 2 and 3 are at the geometric centers of

links 2 and 3, respectively. The masses and mass moments of inertia of the links are

2 5 kgm , 3 5 kgm , 4 15 kgm , 2

20.02 N m sGI , 3

20.12 N m sGI , and

4

20.08 N m sGI . Gravity is acting vertically downwards (that is, 29.81 m/sg in the

negative Y-direction). Assume that friction in the mechanism can be neglected. (i) Draw

free-body diagrams of all moving links of the mechanism. (ii) Write the governing

equations for all moving links. List all unknown variables. (iii) Determine the

magnitudes and directions of all internal reaction forces. (iv) Determine the magnitude

and the direction of the torque 2T . (v) Indicate the point(s) of contact of link 4 with

ground link 1.

A Scotch-yoke mechanism.



(i) The free-body diagram for link 2 is shown in Fig. 1.

G2

T2

G3

O2

W2

F12X

F12Y

F32X

F32Y

R5

R2

Figure 1. The free-body diagram for link 2. Ans.

The sum of the forces acting on link 2 in the X-direction can be written as

2

12 32 2

X X X X

GF F F m A (1)

The sum of the forces acting on link 2 in the Y-direction can be written as

212 32 2 2 .Y Y Y Y

GF F F W m A (2)

The sum of the moments acting on link 2 about point 2O can be written as

2 2 2 22 32 2 32 5 2 2 2 2 5 5( )X Y Y X X X Y Y X

O G G GM R F R F R W T I m R A R A (3)

Therefore, there are three equations and five unknowns for the free-body diagram of link

2. The unknowns are the four reaction forces 12

XF , 12

YF , 32

XF , 32

YF and the crank torque 2T .

The free-body diagram for link 3 is shown is Fig. 2.

G3F23X

F23Y

W3F43

R7 Figure 2. The free-body diagram for link 3. Ans.


323 3

X X X

GF F m A (4)


323 3

Y Y Y

GF F m A (5)

The sum of the moments acting on link 3 about the center of mass 3G can be written as

3 37 43 3G GM R F I

Since link 3 cannot rotate, the angular acceleration is 03 . Therefore, this equation

can be written as

0437 FR

(6)

Equations (4), (5), and (6) contain two new unknowns, 43F and 7R . Therefore, there are

now a total of six equations and seven unknowns.



If we assume that link 4 is only sliding on ground link 1, then the free-body diagram for

link 4 is as shown in Fig. 3.

F34R8

W4

G4

P

F14

R9

Figure 3. The free-body diagram for link 4 when sliding only.


414 4

X X

GF F m A

Since link 4 can only accelerate in the Y-direction, that is, since 4

0X

GA , this becomes

14 0F Ans. (7)


434 4 4

Y Y

GF F P W m A (8)


4 8 34 9 14 4 4G GM R F R F I

Since link 4 can not rotate, that is, since 04 , this becomes

0149348 FRFR

(9)

From Equations (7) and (9), the distance 9R , which means that link 4 attempts to tip.

For tipping, the free body diagram of link 4 is modified as shown in Figure 4.

F34R8

W4

F14B

R11

G4

P

F14T

R10

Figure 4. The free-body diagram for link 4 with tipping. Ans.




414 14 4 0X X

T B GF F F m A (10)


434 4 4

Y Y

GF F P W m A (11)

Note that Eq. (11) is the same as Eq. (8).

The sum of the moments acting on link 4 about the center of mas 4G can be written as

014111410348 BT FRFRFR

(12)

Equations (10), (11) and (12) contain two new unknowns TF14 and BF14 . Therefore, there

are a total of nine equations and nine unknowns.

(iii) We will solve this problem by the method of inspection.

Substituting kg53 m and3

210.8 m/sX

GA into Eq.(4), we have

2

23 (5 kg)( 10.8 m/s ) 54 NXF Ans. (13)

Substituting kg52 m , 2

25.4 m/sX

GA , and N5423 XF into Eq.(1), we have

2

12 (5kg)( 5.4 m/s ) ( 54 N) 81 NXF Ans. (14)

Equation (6) implies that either 07 R or 043 F . Since there is contact between links 3

and 4, the internal reaction force 43F cannot be zero. Therefore

07 R

(15)


222.6 m/sY

GA , 2

4 4 (15 kg)(9.81 m/s ) 147 NW m g , and

N125P into Eq. (11) we have

2

34 (15 kg)(22.6 m/s ) 147 N 125 N 361 NF Ans. (16)


3 m/s6.22YGA , N49m/skg)8.9)(5( 2

33 gmW , and

N36134 F into Eq. (5), we have

2

23 (5 kg)(22.6 m/s ) 49 N 361 N 523 NYF Ans. (17)


211.3 m/sY

GA , 2

2 2 (5 kg)(9.81 m/s ) 49 NW m g , and

N52323 YF into Eq. (2), we have

2

12 (5 kg)(11.3 m/s ) 49 N 523 N 628.5 NYF Ans. (18)

From Eq. (12), we have

34814111410 FRFRFR BT (19)

Using Eqs. (10) and (19), we have

1011

348

14RR

FRF T

, and

1011

348

14RR

FRF B

(20)

Substituting 8 2 7 2 2 sin30 0.5 mX XR R R R R , m4.010 R , m9.011 R and

N36134 F into Eq. (20), we have

14

(0.5 m)(361 N)361 N

0.9 m 0.4 mTF

Ans. (21a)

and



14

(0.5 m)(361 N)361 N

0.9 m 0.4 mBF

Ans. (21b)

(iv) From Eq. (3), we have

2 2 22 2 2 5 5 5 2 2 23 2 23( )X Y Y X X X Y Y X

G G GT I m R A R A R W R F R F (22a)

Substituting 2 49 N,W 2

20.02 N m s ,GI 2

2 rad/s2 , kg52 m , 5 2

1sin30 0.25 m

2

XR R ,

5 2

1cos30 0.433 m

2

YR R , 2

25.4 m/sX

GA , 2

211.3 m/sY

GA , 2 2 sin30 0.5 mXR R ,

2 2 cos30 0.866 mYR R , 23 54 NXF , and 23 523 NYF into Eq. (22a), we have

2 2 2 2

2 (0.02 N m s )(2 rad/s ) (5 kg) (0.25 m)(11.3 m/s ) ( 0.433 m)( 5.4 m/s )

(0.25 m)(49 N) (0.5 m)(523 N) ( 0.866 m)( 54 N) N m

229.46 N.

T

Ans. (22b)

(v) Since N36114 TF , therefore N36141 TF ; this means that link 4 is pushing to the

right on ground link 1 at the upper-right corner of the slot. Also, since N36114 BF ,

therefore N36141 BF ; this means that link 4 is pushing to the left on ground link 1 at

the lower-left corner of the slot. Ans.



14.32 For the parallelogram four-bar linkage in the position illustrated in Fig. P14.32, the

angular velocity and acceleration of the input link 2 are 2 2 rad / s ccw and 2

2 1 rad / s ccw, respectively. The distances 2 4

0.2 m,BO AOR R 2 4

0.3 m,BA O OR R

and 0.1 m.CBR The force 100 NCF acts at point C on link 3 in the X-direction and a

counterclockwise torque 4 10 N mT acts on link 4. The masses and the second

moments of mass are 2 4 0.5 kg,m m 2 4

22 kg m ,G GI I 3 1 kg,m and

3

25 kg m .GI The mass centers of links 2 and 4 are coincident with pins O2 and O4 and

the mass center of link 3 is coincident with pin A. Gravity acts into the page (in the

negative Z-direction) and friction can be neglected. The first and second–order kinematic

coefficients of the mass center of link 3 are 3

0.141 m/rad,GX 3

0.141 m/rad,GY

3

20.141 m/rad ,GX and 3

20.141 m/rad .GY (i) Determine the acceleration of the

mass center of link 3. (ii) Draw the free-body diagrams for links 2, 3, and 4. List all

unknown variables. (iii) Determine the magnitudes and the directions of the internal reaction

forces 23F and 43F . (iv) Determine the magnitude and the direction of the input torque 2.T

Figure P14.32 A parallelogram four-bar linkage.

(i) The acceleration of the mass center G3 can be written as

3 3 3

22 2 2

2 2 0.141 m 2 rad/s 0.141 m 1 rad/s 0.423 m/sX

G G GA X X

Ans.

3 3 3

22 2 2

2 2 0.141 m 2 rad/s 0.141 m 1 rad/s 0.705 m/sY

G G GA Y Y

Ans.

(ii) The free-body diagram of link 2 is shown in Fig. 1.

Figure 1. Free-body diagram of link 2. Ans.

The sum of the forces in the X-direction acting on link 2 can be written as



12 32 0X X XF F F (1)

The sum of the forces in the Y-direction can be written as

12 32 0Y Y YF F F (2)

The sum of the moments acting about the mass center G2 can be written as

2 232 32 2 2cos sinY X

G B B B B GM R F R F T I (3)

The unknown variables in Eqs. (1), (2), and (3) are 12 12 32 32 2, , , , and .X Y X YF F F F T Therefore,

the total number of unknown variables is five and there are only three equations.

The free body diagram of Link 3 is shown in Fig. 2.



323 43 3

X X X X

C GF F F F m A (4)

The sum of the forces in the Y-direction acting on link 3 can be written as

323 43 3

Y Y Y Y

GF F F m A (5)

The sum of the external moments acting about the mass center 3G can be written as

3 23 23cos sin 0Y X

G BB BB BB BBM R F R F (6a)

The new unknown variables in Eqs. (4), (5), and (6a) are 43 43 and .X YF F Therefore, the

total number of equations is six and the total number of unknown variables is seven; that

is, 12 12, ,X YF F 32 32 2, , ,X YF F T 43

XF and 43.YF Note that 0;BB therefore, Eq. (6a) can be

written as

23 0Y

BBR F (6b)

Therefore, either

0BBR or 23 0YF (6c)

The free body diagram of link 4 is shown in Fig. 3.





14 34 0X X XF F F (7)

The sum of the forces in the Y-direction acting on link 4 can be written as

34 14 0Y Y YF F F (8)

The sum of the moments acting on link 4 about the mass center 4G can be written as

4 434 34 4 4cos sinY X

G A A A A GM R F R F T I (9)

The new unknown variables in Eqs. (7), (8), and (9) are 14 14 and .X YF F Therefore, there are

a total of nine equations and nine unknown variables; that is, 12 12 32 32 2 43 43, , , , , , ,X Y X Y X YF F F F T F F

14 14 and .X YF F Ans.

(iii) The solution procedure will be the method of inspection. From Eq. (6c), the

reaction force

23 0 YF (10)

Substituting Eq. (10) into Eq. (5), the reaction force

3

2

43 3 23 1 kg( 0.705 m/s ) 0 0.705 Y Y Y

GF m A F N (11)

Rearranging Eq. (9), the reaction force 34

XF can be written as

4 4 4 4 34

34

4

2 2

cos

sin

2kg m (1 rad/s ) 10 N m 0.2 m 0.707 0.705 N55.87 N

0.2 m 0.707

Y

G AX

A

I T R FF

R

(12)

Therefore, the total reaction force is

0

43 55.87 N 180.72 F Ans. (13)

Solving Eq. (7), the reaction force 23

XF can be written as

323 3 43

21 kg 0.423 m/s 55.87 N 100 N 43.71 N

X X X

G CF m A F F

(14)

Therefore, the total reaction force is

0

23 43.71 N 180 F Ans. (15)

(iv) Rearranging Eq. (3), the input torque 2T can be written as

22 2 2 32 2 32

2 2

sin cos

2 kg m 1 rad/s 0.2 m 0.707 43.71 N 0.2 m 0.707 0 8.18 N m

X Y

G B BT I R F R F

Ans. (16)

The positive sign indicates that the input torque is counterclockwise. Ans.



14.33 For the mechanism in the position shown, the velocity and acceleration of link 2 are

2ˆ10 m/s V i and 2

2ˆ10 m/s ,A i respectively. The length of link 3 is

33.5 mCGR and

the distance 3

2.5 m.OGR The first- and second-order kinematic coefficients of link 3

are 3 0.200 rad/m and 2

3 0.1386 rad/m . The force 10 NCF acts in the

negative Y direction at point C on link 3 and the line of action of an unknown force P

acting on link 2 is parallel to the X-axis as illustrated in Fig. P14.33. The masses and the

second moments of mass of links 2 and 3 are 2 3 kg,m 3 5 kg,m 2

2

G 1.5 kg m ,I

and 3

2

G 7.5 kg m .I Gravity acts in the negative Y-direction and there is no friction in

the mechanism. (i) Draw the free-body diagrams of links 2 and 3. (ii) Write the

governing equations for links 2 and 3. List all unknown variables. (iii) Determine the

magnitudes and directions of all internal reaction forces. (iv) Determine the magnitude

and direction of the force P acting on link 2. (v) Indicate the point(s) of contact of link 2

with ground link 1.

Figure P14.33 A planar mechanism.

(i) The free-body diagram of link 2 is shown in Fig. 1.


(ii) The sum of the forces acting on link 2 in the X-direction can be written as

232 2

X X X

GF F P m A Ans. (1)

Note that the direction of the external force P is assumed to be in the positive X-direction.




212 32 2 2

Y Y Y Y

GF F F W m A Ans. (2)

Note that 2

210 m/sX

GA and 2

0Y

GA .

The sum of the moments acting on link 2 about the mass center 2G can be written as

2 3 2 3 23 2 3 2 2

12 12 2 2 2( )G G G G G

X Y X Y Y X

G G G G GM R F R W I m R A R A Ans. (3)

Therefore, there are three equations and five unknowns for the free-body diagram of link

2. The unknowns are the three reaction forces 32 32 12, ,X Y YF F F , the distance 12

XR (that is, the

location of 12

YF ), and the external force P.

The free-body diagram of link3 is shown is Fig. 2.



323 13 3

X X X X

GF F F m A Ans. (4)


323 13 3 3

Y Y Y Y

C GF F F F W m A Ans. (5)

Note that the acceleration of the center of gravity of link 3 is the same as the acceleration

of link 2 (that is, 3

2

2 10 m/sX

GA A and 3

0Y

GA ).


3 332 3 13 32 3 13 3 3cos sin cosY X

G C C GM R F R F R F I Ans. (6)

Equations (4), (5) and (6) contain two new unknowns, 13

XF and 13

YF . Therefore, there are

a total of six equations and seven unknowns.

Since links 1 and 3 have contact at a pin in a slot, the direction of the reaction force must

be perpendicular to the slot. This means only the magnitude of the reaction force 13F is

unknown.

The X and Y components of this reaction force can be written as

13 13 3cos( 90 )XF F and 13 13 3sin( 90 )YF F (7a)

This provides a seventh equation allowing the seven unknowns to be solved. Also, since

the internal reaction force 13F is perpendicular to the slot then Eq. (6) can be written as

332 13 3 3cosC C GR F R F I (7b)




2

3 3 2 3 2 R R (8a)

where 2 2 10 m/sR V and 2

2 2 10 m/s .R A This agrees with the observation that

the input velocity 2R must be positive; that is, the vector

2R is increasing in length for

this position. Substituting this information and the known kinematic coefficients for link

3 (that is, 3 0.200 rad/m and 2

3 0.1386 rad/m ) into Eq. (8a), the angular

acceleration of link 3 is

2 2 2 2

3 (0.200 rad/m)( 10 m/s ) ( 0.1386 rad/m )(10 m/s) 15.86 rad/s (8b)

Note that the angular acceleration of link 3 has a negative value; that is, the angular

acceleration of link 3 is clockwise. Also, note that the angular velocity of link 3 is

counterclockwise for this position.

(iii) Substituting the known values into Eq. (7b) gives

2 2

13(2.5 m) (3.5 m)cos( 30 )10 N (7.5 kg m )( 15.86 rad/s )F (9a)

Therefore, the reaction force is

13 35.46 N F Ans. (9b)

The negative sign indicates that the internal reaction force 13F is acting downward; that

is, in the opposite direction to the assumed direction shown in Fig. 2. Therefore, the

point of contact between link 3 and link 1 is on the top side of the ground pin O.

Substituting the known values into Eq. (5) gives

2

23 35.46 N sin(60 ) 10 N (5 kg)(9.81 m/s ) 0YF (10a)


23 89.76 NYF Ans. (10b)


2

23 35.46 N cos(60 ) (5 kg)(10 m/s )XF (11a)


23 67.73 NXF Ans. (11b)

(iv) Substituting the known values into Eq. (1) gives

267.73 N (3 kg)(10 m/s )P (12a)

Therefore, the applied force is

97.73 NP Ans. (12b)


2

12 89.76 N (3 kg)(9.81 m/s ) 0YF (13a)


12 119.19 NYF Ans. (13b)


3 2

2

12 (119.19 N) (3 kg)(9.81 m/s ) 0X

G GR R (14a)

(v) Therefore, the distance is

3 212 0.25X

G GR R (14b)

The negative sign indicates that the location of the internal reaction force 12

YF is to the left

of the mass center of link 3. Given that the distance



3 2

0.5 mG GR (15)

Then the distance from the mass center of link 3 to the point of application of the normal

force is

12 0.25 0.5 m 0.125 mXR Ans. (16)



14.34 Consider the slider-crank mechanism of Problem 14.5. The designer proposes to modify

this mechanism by including a linear spring and a viscous damper as illustrated in Fig.

P14.34. The spring, with a stiffness 3.5 kN/mSk and an unstretched length 0 75 mm,r

is attached from the ground to point E on the input link 2. In the given position, the

spring is parallel to the X-axis. The damper with coefficient 1.25 kN s/mC is attached

between the ground pivot 2O and pin B on link 4. At the input position 2 45 , the

motor driving input link 2 is applying a torque 2 6.7 N mT ccw, causing the link to

rotate with an angular velocity 2 100 rad/sω ccw and an angular acceleration 2

2 10 rad/sα ccw. An external horizontal force BF is acting at pin B on link 4, as

illustrated in Fig. P14.34. The variable positions, velocities, and accelerations of the

mechanism have been determined and are provided in Table P14.34.

Table P14.34

2 3 2AOR

BAR 2EOR

3 4V 3 4A

deg deg mm mm mm rad/s m/s 2rad/s 2m/s

45 10.18 75 300 125 17.96 -6.25 1736.20 -534

Assume: (i) Gravity acts vertically downward (that is, in the negative Y-direction). (ii)

The location of the center of mass of link 2 is coincident with the ground pivot 2O and

the center of mass of link 4 is coincident with pin B. The location of the center of mass

of link 3 is as indicated in Fig. P14.5. (iii) The effects of friction in the mechanism can

be neglected. (iv) The weight and mass moment of inertia of each link are as given in

Fig. P14.5. Determine: (i) the first- and second-order kinematic coefficients of the

mechanism that are necessary for the power equation. (ii) the equivalent mass moment of

inertia of the mechanism. (iii) the equation of motion for the mechanism. (iv) the

magnitude and direction of the external force BF acting on link 4 when the mechanism is

in the given position.

Figure P14.34. The mechanism modified from Problem 14.5.



The vectors for a kinematic analysis of the mechanism are shown in Fig. 1.

Figure 1. The vectors for analysis of the mechanism.

(i) The first-order kinematic coefficient of link 3 can be written as

33

2

17.96 rad/s0.179 6 rad/rad

100 rad/s


44

2

6.25 m/s62.5 mm/rad

100 rad/s

VR

(1)


2

3 3 2 3 2

Rearranging this equation, the second-order kinematic coefficient of link 3 can be written

as 2 2

23 3 23 2 2

2

(1 736.20 rad/s ) ( 0.1796 rad/rad)(10 rad/s )0.173 8 rad/rad

(100 rad/s)

Similarly, the linear acceleration of link 4 can be written as

2

4 4 2 4 2A R R

Therefore, the second-order kinematic coefficient of link 4 is

2 2

24 4 24 2 2

2

534 000 mm/s ( 625 mm/rad)(10 rad/s )54 mm/rad

(100 rad/s)

A RR

Since the mass center of link 2, G2, is located at the fixed pivot O2, their first- and

second-order kinematic coefficients are

2

0Gx , 2

0Gx

2

0Gy , and 2

0Gy .

To determine the first- and second-order kinematic coefficients for the center of mass of

link 3: The vector loop for point G3, see Fig. 1, can be written as

3

??

2 33

I C

G

R R R

where the magnitude of the vector 33R is given as 112.5 mm and the angle 33 3 . This

implies that the first-order kinematic coefficient 33 3 and the second-order kinematic

coefficient 33 3 . The X and Y components of the above equation are

3 2 2 33 3cos cos =163.76 mmGX R R

3 2 2 33 3sin sin 33.145 mmGY R R



The first-order kinematic coefficients of the center of mass of link 3 are

3 2 2 33 3 3sin sin 56.605 mm/radGX R R

3 2 2 33 3 3cos cos 33.145 mm/radGY R R

The second-order kinematic coefficients of the center of mass of link 3 are

3

2 2

2 2 33 3 3 33 3 3cos cos sin 53.148 mm/radGX R R R

3

2 2

2 2 33 3 3 33 3 3sin sin cos 33.145 mm/radGY R R R

The first- and second-order kinematic coefficients of the center of mass of link 4 are

4 4 62.558 mm/radGx R

4

2

4 53.340 mm/radGx R

4

0Gy 4

0Gy

The first-order kinematic coefficient for the damper is

4 62.558 mm/radCR R

The first-order kinematic coefficient for the spring can be obtained from the vector loop

for point E, see Fig. 1; that is,

??

22

C

E

R R

where the magnitude of the vector 22R is given as 5 in and the angle

22 2 180 225 .

This implies that the first-order kinematic coefficient 22 1 rad/rad . The X component

of point E is

22 22cos = 88.388 mmEX R

Therefore, the first-order kinematic coefficient of point E is

22 22 2sin 88.388 mm/radEX R

Note that the first-order kinematic coefficient for the spring is

88.388 mm/radS ER X

(ii) To determine 4

2

j

j

A

: Note that 4

EQ

2

j

j

A I

; (that is, the equivalent mass moment of

inertia) therefore, the units must be 2mm N s .

For link 2:

2 2 2

2 2 2

2 2 2

2 2 2 2

2

( )

(0 0) 0.039 kg m (1 rad/rad) 0.039 kg m 39.0 mm N s

G G GA m x y I

m

For link 3:

3 3 3

2 2 2

3 3 3

2 2 2 2

2 2

( )

[( 0.056 605 m/rad) (0.033 145 m/rad) ] 0.012 kg m ( 0.179 6 rad/rad)1.54 kg

0.007 013 kg m 7.013 mm N s

G G GA m x y I

For link 4:

4 4 4

2 2 2

4 4 4

2 2 2 2

( )

[( 0.062 558 mm/rad) (0) ] 0 0.005 088 kg m 5.088 mm N s1.30 kg

G G GA m x y I

Therefore, the sum of the coefficients is



42 2 2 2

2 3 4

2

39.0 mm N s 7.013 mm N s 5.088 mm N s 51.101 mm N sj

j

A A A A

Ans.

To determine

4

2j

jB : Note that 4 4

2 2 2

1

2

j

j

j j

dAB

d

; therefore, the units must be 2 mm N s .

For link 2:

2 2 2 2 2

2

2 2 2 2 2( ) (0 0) 0.039 kg m (1 rad/rad)(0) 0G G G G GB m x x y y I m

For link 3:

3 3 3 3 33 3 3 3

2 2

2 2

( )

[( 0.056 605 m/rad)( 0.053 148 m/rad) (0.033 145 m/rad)( 0.033 145 m/rad)]1.54 kg

0.012 2 kg m ( 0.179 6 rad/rad)(0.173 8 rad/rad )

0.002 560 kg m 2.560 mm N s


For link 4:

4 4 4 4 44 4 4 4

2

2 2

( )

[( 0.062 558 m/rad)( 0.053 340 in/rad ) 0] 01.30 kg

0.004 338 kg m 4.338 mm N s


Therefore, the sum of the coefficients is

4

2 2 2

2 3 4

2

0 2.560 mm N s 4.338 mm N s 6.898 mm N sj

j

B B B B

Ans.

(iii) The power equation can be written as

fdWdT dU

Pdt dt dt

(2)

The time rate of change of the kinetic energy can be written as

4 4

3

2 2

j j

j j

dTA B

dt

where the generalized inputs for this problem are 2 2 2 2 2, , and .

Therefore, the time rate of change of the kinetic energy is 4 4

3 2 2 2 2

2 2 2 2

2 2

[51.101 mm N s (10 rad/s ) 6.898 mm N s (100 rad/s) ]j j

j j

dTA B

dt

That is,

269.491 N mdT

dt

The time rate of change of the potential energy can be written as

4

3 2 2

23

( ) ( )jj G S S O S G S S O S

j

dUm gy K R R R m gy K R R R

dt

The time rate of change of the potential energy due to gravity is



3 2 3 2

22 2

3 3

1.54 kg (0.033 145 m/rad) 0.500 7 N m9.81 m/s

g

G G

dUm gy w y

dt

The time rate of change of the potential energy due to the linear spring can be written as

0 2

2 2

( )

3.500 N/mm[ 125 mm cos 45 75 mm]( 0.088 388 m/rad) 4.141 8 N m

SS S S S

dUK R R R

dt

Therefore, the time rate of change of the total potential energy is

2 2 20.500 7 N m 4.141 8 N m 3.641 1 N m

g SdU dUdU

dt dt dt

The time rate of change of the dissipative effects due to the viscous damper is

2 2 2

2 21.25 N s/mm( 62.558 mm/rad) (100 rad/s) 489.188 N mf

C

dWCR

dt

Note that the time rate of change of the dissipative effects due to the viscous damper is a

positive value. This must always be true for this term on the right-hand side of the power

equation.

Therefore, the right-hand side of the power equation, see Eq. (2), can be written as

2 2 2

2

69.491 N m 3.641 1 N m 489.188 N m

555.038 N m

fdWdT dU

dt dt dt

(3)

Note that the most influential term is the time rate of change of the dissipative effects due

to the viscous damper. This implies that the damping coefficient 7 lb s/inC is a very

large value.

The left-hand side of the power equation, see Eq. (2), can be written as

2 2 2 26.7 N m ( )X

B B B BP F x T ω F V (4)

Note that the torque 2T is acting in the same direction as the angular velocity of link 2

(that is, counterclockwise) and the external force acting on the piston BF is assumed

positive when acting in the same direction as the velocity of the piston (link 4) (that is, in

the negative X-direction). The first-order kinematic coefficient of point B can be

obtained from the point path vector equation, or by noting that 4Bx R . From Eq. (1),

the first-order kinematic coefficient of link 4 is 4 62.5 mm/radR . Therefore, the first-

order kinematic coefficient of point B is

62.5 mm/radBx (5)

Substituting Eq. (5) into Eq. (4) gives

2 26.7 N m ( 0.062 5 m)X

BP F (6)

Finally, equating the two equations, Eqs. (3) and (6), the power equation can be written as

2 2 26.7 N m (0.062 5 m) 555.038 N mX

BF (7)



The equation of motion is obtained by dividing both sides of the power equation, Eq. (7),

by the input angular velocity 2 . Therefore, the equation of motion can be written as

6.7 N m (0.062 5 m) 555.038 N mX

BF Ans. (8)

(iv) Rearranging Eq. (8), the external force acting on the piston is

8 773.4 NX

BF Ans.

The negative sign indicates that the external force acting on the piston (link 4) is acting to

the left, that is, in the negative X-direction. Therefore, the assumption made in Eq. (4)

was correct; that is, the force is acting in the same direction as the velocity of the piston

(link 4).



14.35 For the mechanism in the position illustrated in Fig. P14.35, the velocity and acceleration

of the input link 2 are 2

ˆ7 m/sV i and 2

2ˆ2 m/s , A i respectively. The first- and

second-order kinematic coefficients of links 3 and 4 are 3 0, 4 1.0 rad/m,

2

3 1.0 rad/m , and 2

4 3.0 rad/m . The radius of massless link 4, which is rolling on

the ground link, is 4 1 m,R the length of link 3 is 2

6 m,G AR and the radius of the

ground link is 1 2 m.R The free length and spring rate of the spring, the damping

constant of the viscous damper, the masses, and the mass moments of inertia of links 2

and 3 (about their mass centers) are as shown in Table P14.35. Assume that gravity acts

in the negative Y direction and the effects of friction can be neglected. Determine: (i) the

kinematic coefficients ,Sr ,Cr 3,Gx

3,Gy

3,Gx and

3.Gy ; (ii) the equivalent mass of the

mechanism; (iii) the equation of motion for the mechanism in symbolic form; (iv) the

magnitude and direction of the horizontal external force P that is acting on link 2.

Table P14.35

Ro K C 2m 3m 2GI

3GI

m N/m Ns/m kg kg 2kg-m 2kg-m

3 25 15 1.20 0.80 0.25 0.10




The vectors for kinematic analysis of the mechanism are shown in Fig. 1.

Figure 1. Vectors for a kinematic analysis of the mechanism.

The vector loop equation for the mechanism can be written as

2 3 7 R R R 0

where link 7 is an arm connecting the ground link to the center of the wheel, link 4. The

X and Y components of this equation are

2 2 3 3 7 7cos cos cos 0R R R

2 2 3 3 7 7sin sin sin 0R R R

(i) Differentiating these equations with respect to the input position gives

2 3 3 3 7 7 7cos sin sin 0R R

2 3 3 3 7 7 7sin cos cos 0R R

Differentiating again with respect to the input position gives

2 2

3 3 3 3 3 3 7 7 7 7 7 7sin cos sin cos 0R R R R

2 2

3 3 3 3 3 3 7 7 7 7 7 7cos sin cos sin 0R R R R

Solving for the first-order kinematic coefficients we get

3 0 and 7 0.333 rad/m

Solving for the second-order kinematic coefficients, they are

2

3 1 rad/m and 2

7 1 rad/m

The rolling contact equation between the wheel, link 4, and the ground link can be written

as

4 71

4 1 7

R

R

The correct sign is negative because there is external contact between link 4 and the

ground link. Differentiating this equation with respect to the input position gives

4 71

4 1 7

R

R



Substituting the known values into this equation, the first-order kinematic coefficient for

link 4 is

4 1 rad/m

Differentiating the above equation with respect to the input position gives

4 71

4 1 7

R

R

Substituting the known values into this equation, the second-order kinematic coefficient

for link 4 is

2

4 3 rad/m

The first-order kinematic coefficient of the spring is

2 1 m/mSR R Ans.

Note that the answer is positive because the length of the spring increases for a positive

change in the input position. The first-order kinematic coefficient of the damper can be

written as

2 1 m/mCR R Ans. (1)

Note that the answer is negative because the change in the length of the vector 1C AOR R

decreases for positive change in the input position.

Check: Since link 4 is rolling on the ground link at point E then point E is the instant

center 14I . Therefore, the velocity of point A (which is directed in the positive X-

direction) can be written as

14 144 4 2( ) ( 1 rad/m 7 m/s)( 1 m) 7 m/sA I A I AV R R R (2)

The first-order kinematic coefficient of the damper is defined as

2

7 m/s1 m/m

7 m/s

AC

VR

R

The correct sign is negative because the change in length of the vector 1C AOR R

decreases as the change in length of the input vector 2R increases.

The vector equation for the center of mass of link 3 can be written as

3 2 33G R R R


3 2 2 33 3cos cosGx R R and

3 2 2 33 3sin sinGy R R

Differentiating these equations with respect to the input position, the first-order kinematic

coefficients for the center of mass of link 3 are

3 2 33 3 3cos sinGx R and

3 2 33 3 3sin cosGy R (3)

Substituting the known data into this equation, the first-order kinematic coefficients for

the center of mass of link 3 are

3

1 m/mGx and 3

0Gy Ans.

Differentiating Eq. (3) with respect to the input position, the second-order kinematic

coefficients of the center of mass of link 3 can be written as

3

2

33 3 3 33 3 33sin cosGx R R and 3

2

33 3 3 33 3 33cos sinGy R R



Substituting the known data into this equation, the second-order kinematic coefficients

for the center of mass of link 3 are

3

21.5 m/mGx and 3

22.598 m/mGy Ans.

The X and Y components of the acceleration of the mass center of link 3 are

3 3 3

2 2 2 2 2

2 2A ( 1.5 m/m )(7m/s) (1 m/m)( 2 m/s ) 75.5 m/sX

G G Gx R x R

3 3 3

2 2 2 2 2

2 2A ( 2.598 m/m )(7 m/s) (0)( 2 m/s ) 127.302 m/sY

G G Gy R y R

Note that the mass center of link 3 has X and Y components, therefore, the path of the

mass center of link 3 is not a horizontal straight line. The magnitude and direction of the

acceleration of the mass center of link 3 are

3

2148.0 m/s 239.33G A

(ii) The equivalent mass of the mechanism can be written as

2 2 2

EQ

2 2

( )j j j

n n

j j G G G j

j j

m A m x y I

(4)

Substituting known data into Eq. (4) gives

2 2 2 2

2 1.2 kg[(1 m/m) 0 ] 0.25 kg m (0) 1.2 kgA


2 2 2 2

3 0.8 kg[(1 m/m) 0 ] 0.10 kg m (0) 0.8 kgA

Since link 4 is massless then

4 0A

Therefore, the equivalent mass of the mechanism is

EQ 1.20 kg 0.80 kg 0 2.00 kgm Ans.

(iii) The power equation for the mechanism can be written as

4 4 4

2 2 2

2 j 2 2 2 2 0 2 2

j=2 2 2

[ ] ( )jj j G S S C

j j

A R B R R m gy R K R R R R CR R

P V

Assume that the force P acting on link 2 is positive in the same direction as the positive

input velocity. Canceling the input velocity 2 2 ,V R the equation of motion for the

mechanism can be written as

4 4

2 2

2 2 0 2

2 2 2

(R )j

n

j j j G S s C

j j j

P A R B R m gy K R R CR R

Ans. (5)

The coefficients jB can be written as

( ) j j j jj j G G G G G j jj

B m x x y y I (6)


2

2 1.20 kg[(1 m/m)(0) (0)(0)] (0.25 kg m )(0)(0) 0B

Substituting known data into Eq. (6) gives 2 2 2

3 0.80 kg[(1 m/m)( 1.5 m/m) (0)( 2.598 m/m )] 0.10 kg m (0)( 1 rad/m ) 1.20 kg/mB

Since link 4 is massless,

4 0B

Therefore, the sum of the coefficients jB is



4

2

0 1.20 kg/m 0 1.20 kg/mj

j

B

(7)

The effects of gravity can be written as

2 3 4

42

2 3 4

2

[ ] 9.81 m/s [(1.20 kg)(0) (0)(0) (4 kg)(0)] 0jj G G G G

j

m gy g m y m y m y

The terms for the spring and the damper in Eq. (5) are

0( )R 25 N/m(5.196 m 3 m)(1 m/m) 54.9 N m/mS sK R R (8a)

2 2

2 (15 N s/m)( 1 m/m) (7 m/s) 105 N m/mCCR R (8b)

(iv) The external force P acting on link 2 can be written from Eq. (5) as

4 4

2 2

EQ 2 2 0 2

2 2

(R )jj j G S s c

j j

P m R B R m gy K R R CR R

Substituting the given data and Eqs. (1), (2), (7), and (8) into this equation, the magnitude

of the external force P acting on link 2 can be written as

2 2(2.0 kg)( 2 m/s ) ( 1.2 kg/m)(7 m/s) 0 54.9 N 105 N 97.1 NP Ans.

The positive sign indicates that the assumption that the external force P is acting to the

right (that is, in the same direction as the input velocity) is correct. Therefore, the external

force P is acting to the right.



14.36 For the mechanism in the position illustrated in Fig. P14.36, link 3 is horizontal. The

constant angular velocity of the input link 2, which rolls without slipping on the inclined

plane, is 2 20 rad/s cw. The first- and second-order kinematic coefficients of links 3

and 4 are 3 0.125 rad/rad, 4 1.299 m/rad,R 3 0, and 2

4 0.094 m/rad .R The

radius of link 2 is 1.5 m,R the length of link 3 is 4 2

6 m,BA G GR R and 2.5 m.SAOR

The free length of the spring is 3 m, the spring rate is 25 N/m,k and the damping

constant of the viscous damper is 15 N s/m.C · The masses and mass moments of

inertia of links 2 and 4 are 2 7 kg,m 4 4 kg,m 2

218 kg m ,GI · and

4

222 kg m .GI ·

Assume that the mass of link 3 is negligible compared with the masses of links 2 and 4,

the effects of friction can be neglected, and gravity acts vertically downward as illustrated

in Fig. P14.36. Determine: (i) the first- and second-order kinematic coefficients of the

mass centers of links 2 and 4, (ii) the equivalent mass moment of inertia of the

mechanism, and (iii) the magnitude and direction of the input torque acting on link 2.


Vectors for the mass centers of links 2 and 4 are shown in Fig. 1.

Figure 1. Vectors for the mass centers of links 2 and 4.



The vector equation for the mass center of link 2, see Fig. 1, can be written as

2

?

9 7G

R R R


2 9 9 7 7cos cosGX R R and

2 9 9 7 7sin sinGY R R

Differentiating with respect to the input position 2 gives

2 9 9cosGX R and

2 9 9sinGY R (1)

Substituting 9 150 and 9 1.5 m/radR gives

2

1.5cos150 1.299 mo

GX and 2 9 9sin 1.5sin150 0.75 mo

GY R Ans.

The rolling contact equation between link 2 and the inclined plane (link 1) can be written

in terms of the first-order kinematic coefficients as

9 2 2 2 1.5 mR

Therefore, the second-order kinematic coefficient is

9 0R

Differentiating Eqs. (1) with respect to the input position 2 gives

2 9 9cos 0GX R and

2 9 9sin 0GY R Ans. (2)

The vectors for the mass center of link 4 are shown in Fig. 1. The first-order kinematic

coefficients of the mass center of link 4 can be written as

4 24 1.299 mG GX R X and

40GY Ans.

The second-order kinematic coefficients of the mass center of link 4 can be written as

4 4 0.094 mGX R and

40GY Ans.

(ii) The power equation for the mechanism can be written as

4 4

2 2 2

2 2 2 2 2 2 0 2 2

2 2

[ ] ( )jEQ j j G s s c

j j

I B m gy K r r r Cr

T ω

The input torque is taken positive in the same direction as the given input angular velocity

(that is, clockwise). Then canceling the input angular velocity, the equation of motion for

the mechanism can be written as

4 4

2 2

2 2 2 0 2

2 2

( )jEQ j j G s s c

j j

T I B m gy K r r r Cr

(3)

The equivalent mass moment of inertia of the mechanism can be written as

2 2 2

EQ ( )j j j jj G G Gj

I A m x y I (4)

Substituting known data for link 2 into this equation gives 2 2 2 2 2 2

2 (7 kg)(( 1.299 m/rad) (0.75 m/rad) ) 18 kg m (1 rad/rad) 33.75 kg m /radA

Substituting known data for link 3 into Eq. (4) gives

3 0A

Substituting known data for link 4 into Equation (4) gives

2 2 2 2 2 2

4 (4 kg)(( 1.299 m/rad) 0 ) 22 kg m (0) 6.75 kg m /radA

Therefore, the equivalent mass moment of inertia of the mechanism is



4

2 2 2

EQ

2

33.75 kg m 0 6.75 kg m 40.5 kg mj

j

I A

Ans.

(iii) The coefficient jB can be written as

( )j j j j jj j G G G G G j jB m x x y y I (5)


2

2 7 kg(( 1.299 m/rad)(0) (0.75 m/rad)(0)) (18 kg m )(1 rad/rad)(0) 0B

Substituting known values for link 3 into Eq. (5) gives

3 0B


2 2 2

4 4 kg(( 1.299 m/rad)( 0.094 m/rad ) (0)(0)) (22 kg m )(0)(0) 0.488 kg mB

Therefore, the coefficient is

4

2 2

2

0 0 0.488 kg m 0.488 kg mj

j

B

The effects of gravity can be written as

2 3 4

3

4

2 3 4

2

2

[ ]

9.81 m/s [(7 kg)(0.75 m/rad) (0)( ) (4 kg)(0)] 51.5 N m/rad

jj G G G G

j

G

m gy m gy m gy m gy

y

The velocity of the mass center of link 2 down the inclined plane is

2 2 20 rad/s 1.5 m 30 m/sGV R

which agrees with the first-order kinematic coefficients in Eq. (2); that is,

2

2 2

2 21.299 m/rad 0.75 m/rad 1.5 m/rad 30 m/sGV

Therefore, the first-order kinematic coefficient of the spring is

Sr R 1 5 m rad. /

and is negative because the length of the spring is decreasing for positive input motion.

Therefore

0( ) 25 N/m(2.5 m 3 m)( 1.5 m/rad) 18.75 N m/radS SK r r r

The first-order kinematic coefficient of the damper can be written as

4 1.299 m/radCr r

Therefore

2 2

2 (15 N s/m)(1.299 m/rad) ( 20 rad/s) 506.22 N m/radcCr

From Eq. (3). the input torque can be written as 4 4

2 2

2 2 2 0 2

2 2

2 2 2

( )

(40.5 kg m )(0) (0.488 kg m )( 20 rad/s) 51.5 N m 18.75 N m 506.22 N m

240.77 N m

jEQ j j G S S C

j j

T I B m gy K r r r Cr

Ans.

The negative sign indicates that the torque is in the opposite direction to the input angular

velocity (which is specified as clockwise). Therefore, the input torque must be acting

counterclockwise.



14.37 Figure P14.37 illustrates a two-throw opposed-crank crankshaft mounted in bearings at A

and G. Each crank has an eccentric weight of 26.7 N, which may be considered as

located at a radius of 50 mm from the axis of rotation, and at the center of each throw

(points C and E). It is proposed to locate weights at B and F to reduce the bearing

reactions, caused by the rotating eccentric cranks, to zero. If these weights are to be

mounted 75 mm from the axis of rotation, how much must they weigh?

150 mm

150 mm

150 mm

50 mm

50 mm

150 mm

45 mm

50 mm

50 mm

75 mm

75 mm

2 2 2 2 2 2 2 250 mm 200 mm 500 mm 650 mm 700 mm 0y

A B B C C E E F F GM m r m r m r m r F

2 2 2 2 2 2 2 2700 mm 650 mm 500 mm 200 mm 50 mm 0y

G A B B C C E E F FM F m r m r m r m r

Dividing by 250 mm and substituting numeric values gives

2 2 25625 mm 400500 mm N 73125 mm 0B Fm m

2 2 273125 mm 400500 mm N 5625 mm 0B Fm m

Solving simultaneously gives

5.932 NB F Bw w m Ans.



14.38 Figure P14.38 illustrates a two-throw crankshaft, mounted in bearings at A and F, with

the cranks spaced 90 apart. Each crank may be considered to have an eccentric weight

of 26.7 N at the center of the throw and 50 mm from the axis of rotation. It is proposed

to eliminate the rotating bearing reactions, which the crank would cause, by mounting

additional correction weights on 75 mm arms at points B and E. Calculate the magnitudes

and angular locations of these weights.

50 mm

150 mm

300 mm

150 mm

50 mm

50 mm

50 mm

2 2 2 2

2 2 2 2

ˆ ˆ ˆ ˆˆ50 mm sin cos 200 mm

ˆ ˆ ˆˆ ˆ 500 mm 650 mm sin cos

y

A B B B B C C

D D E E E E

m r m r

m r m r

M i × j k i × j

i × k i × j k 0

2 2 2 2

2 2 2 2

ˆ ˆ ˆ ˆˆ650 mm sin cos 500 mm


y

F B B B B C C

D D E E E E

m r m r

m r m r

M i × j k i × j

i × k i × j k 0

Dividing by 250 mm , substituting numeric values, and equating vector components

gives

2 2 2 5625 mm cos 667500 N mm 73125 mm cos 0B B E Em m

2 2 2 5625 mm sin 267000 N mm 73125 mm sin 0B B E Em m


2 2 273125 mm sin 667500 N mm 5625 mm sin 0B B E Em m


cos 2.968 NB Bm , sin 8.9 NB Bm , cos 8.9 NE Em , sin 2.968 NE Em

9.38 NBm , 108.43B , 9.38 NEm , 161.57E Ans.



14.39 Solve Problem 14.38 with the angle between the two throws reduced from 90 to 0 .

2 2 2 2

2 2 2 2

ˆ ˆ ˆˆ ˆ50 mm sin cos 200 mm


y

A B B B B C C

D D E E E E

m r m r

m r m r

M i × j k i × k

i × k i × j k 0

2 2 2 2

2 2 2 2

ˆ ˆ ˆˆ ˆ650 mm sin cos 500 mm


y

F B B B B C C

D D E E E E

m r m r

m r m r

M i × j k i × k

i × k i × j k 0

Dividing by 250 mm , substituting numeric values, and equating vector components

gives


2 2 5625 mm sin 73125 mm sin 0B B E Em m

2 2 2 73125 mm cos 934500 N in 5625 mm cos 0B B E Em m

2 273125 mm sin 5625 mm sin 0B B E Em m


cos 11.868 NB Bm , sin 0.000 NB Bm , cos 11.868 NE Em ,

sin 0.000 lbE Em

11.868 NBm , 180.00B , 11.868 NEm , 180.00E Ans.



14.40 The connecting rod illustrated in Fig. P14.40 weighs 35.155 N and is pivoted on a knife

edge and caused to oscillate as a pendulum. The rod is observed to complete 54.5

oscillations in 1 min. Determine the mass moment of inertia of the rod about its own

center of mass.

43.75 mm

56.25 mm

350 mm

100 mmGr , 35.155 Nw , 60 s 54.5 cycles 1.101 s/cycle

From Eq. (14.101),

2 2 22 35.155 N 100 mm 1.101 s/cycle 2 rad/cycle 107.91 N mm sO GI mgr

22 2 2 2107.91 N mm s 35.155 N 9650 mm/s 100 mm 71.53 N mm sG O GI I mr

Ans.

14.41 A gear is suspended on a knife edge at the rim as illustrated in Fig. P14.41 and caused to

oscillate as a pendulum. Its period of oscillation is observed to be 1.08 s. Assume that

the center of mass and the axis of rotation are coincident. If the weight of the gear is 178

N, find the mass moment of inertia and the radius of gyration of the gear.

200 mm



From Eq. (14.101),

2 2 22 178 N 200 mm 1.08 s/cycle 2 rad/cycle 10515.75 N mm sO GI mgr

22 2 2 21051.75 N mm s 178 N 9650 mm/s 200 mm 313.95 N mm sG O GI I mr

Ans.

2 2313.95 N mm s 178 N 9650 mm/s 130.45 mmG Gk I m Ans.



14.42 Figure P14.42 illustrates a wheel whose mass moment of inertia I is to be determined.

The wheel is mounted on a shaft in bearings with very low frictional resistance to

rotation. At one end of the shaft and on the outboard side of the bearings is connected a

rod with a weight bW secured to its end. It is possible to measure the mass moment of

inertia of the wheel by displacing the weight bW from its equilibrium and permitting the

assembly to oscillate. If the weight of the pendulum arm is neglected, show that the mass

moment of inertia of the wheel can be obtained from the equation

2

24b

lI W l

g

Using W for the weight of the wheel, the location of the center of mass of the assembly is

rG where G b GWr W l r or b b GW l W W r and the mass moment of inertia is

2

O bI I W l g . Now, using Eq. (14.101)

22 2

22 4

bb bG

W WW l W lI gr

g g

Rearranging this we get

2

24b

lI W l

g

Q.E.D.



14.43 If the weight of the pendulum arm is not neglected in Problem 14.42, but is assumed to be

uniformly distributed over the length l, show that the mass moment of inertia of the wheel

can be obtained from the equation

2

2 2 34

a ab b

W WlI l W W

g

where aW is the weight of the arm.

Using W for the weight of the wheel and rG for the location of the center of mass of the

assembly,

2G b G a GWr W l r W l r or 2b a G b aW W W r W l W l

The total mass moment of inertia is 22 2 2 2

12 2 3

b a a b aO

W l W l W W l W llI I I

g g g g g

Using Eq. (14.101) 22 2

3 2 2

b a ab

W l W l W lI W l

g g

which can now be rearranged to read 2

24 2 3

a ab b

W WlI l W W

g

Q.E.D.



14.44 Wheel 2 in Fig. P14.44 is a round disk that rotates about a vertical axis z through its

center. The wheel carries a pin B at a distance R from the axis of rotation of the wheel,

about which link 3 is free to rotate. Link 3 has its center of mass G located at a distance r

from the vertical axis through B, and it has a weight 3W and a mass moment of inertia GI

about its own mass center. The wheel rotates at an angular velocity 2 with link 3 fully

extended. Develop an expression for the angular velocity 3 that link 3 would acquire if

the wheel were suddenly stopped.

Consider link 3 alone. The momentum before and after are

3 2ˆm R r L i 3 3

ˆm r L i

The angular momentum before and after about point G are

2

3 2ˆ3G m r H k 2

3 3ˆ3G m r H k

The angular momentum before and after about point B are

2 2

3 2 3 2

2

3 2

ˆ

ˆ ˆ3

ˆ4 3

B G r

m r m Rr r

m Rr r

H H j×L

k k

k

2 2

3 3 3 3

2

3 3

ˆ

ˆ ˆ3

ˆ4 3

B G r

m r m r

m r

H H j×L

k k

k

Since there is no angular impulse on link 3 about point B, B B H H

3 21 3 4R r Ans.



14.45 Repeat Problem 14.44, except assume that the wheel rotates with link 3 radially inward.

Under these conditions, is there a value for the distance r for which the resulting angular

velocity 3 is zero?

Consider link 3 alone. The momentum before and after are

3 2ˆm R r L i 3 3

ˆm r L i

The angular momentum before and after about point G are

2

3 2ˆ3G m r H k 2

3 3ˆ3G m r H k

The angular momentum before and after about point B are

1 1

2 2

3 2 3 2

2

3 2

ˆ

ˆ ˆ3

ˆ4 3

B G r

m r m Rr r

m Rr r

H H j×L

k k

k

2

2 2

3 3 3 3

2

3 3

ˆ

ˆ ˆ3

ˆ4 3

B G r

m r m r

m r

H H j×L

k k

k

Since there is no angular impulse on link 3 about point B, B B H H

3 21 3 4R r

3 0 for 3 4r R Ans.



14.46 Figure P14.46 illustrates a planetary gear-reduction unit that utilizes 3.63mm/tooth spur

gears cut on the 20 full-depth system. All parts are steel with density 38 10 N/mm .

The arm is rectangular and is 100 mm wide by 350 mm long with a 100 mm diameter

central hub and two 75 mm diameter planetary hubs. The segment separating the planet

gears is a 0.5 × 100 mm diameter cylinder. The inertia of the gears can be obtained by

treating them as cylinders equal in diameter to their respective pitch circles. The input to

the reducer is driven with 25 with a torque of 297 N-m at 600 rev/min. The mass

moment of inertia of the resisting load is 2648.5875 N mm s . Calculate the bearing

reactions on the input, output, and planetary shafts. As a designer, what forces would you

use in designing the mounting bolts? Why?

37.5 mm

37.5 mm

40.625 mm

25

mm

50

mm

337.5 mm

37.

5m

m

34.375 mm

12.5 mm

31.25

50

mm

40

.625

mm 268.75 mm

200 mm 200 mm

34.375 mm

25

mm

34.375 mm



400 mm

11


2 2

NR

2

2

m 3.63mm/tooth 35 teeth63.525 mm

2 2

NR

33


2 2

NR

4

4


2 2

NR

1 297 N mout AT T

Assume a symmetric arrangement of (typically = 3) planets, symmetrically arranged on

an pronged planet carrier. Then the tangential component of the force 2 AF for each planet

is

2

297 N m2371.8 N

125.23 mm

t outA

A

TF

R

For equilibrium of each planet

2 3 43 2 12cos20 cos20 0M R F R F 3 43 2 12R F R F

2 12 43 2cos20 cos20 0t t

AF F F F

12 43 2 3 12 21 cos20t

AF F R R F F

3

12 2

2 3 cos 20

t

A

RF F

R R

12 1508.5 / NF

43 2 3 12F R R F 43 1014.6 NF



2 12 43 2sin 20 sin 20 0r r

AF F F F

2 12 43 sin 20r

AF F F 2 169.1 Nr

AF

2 2

2 2 2

t r

A A AF F F 2 2376.3 NAF Ans.

Assuming that the m-pronged planet carrier is arranged symmetrically, there is no net

force on the input or output shafts; 14 1 0AF F Ans.

The input torque is

4 4 34 14cos20 0M R F T 14 28.92 N minT T Ans.

Balancing the moments on the casing

1 2 21 11cos20 400 0M R F mm F 11 222.5 NF Ans.

This force F11 must be absorbed by the mounting bolts.



14.47 It frequently happens in motor-driven machinery that the greatest torque is exerted when

the motor is first turned on, because of the fact that some motors are capable of delivering

more starting torque than running torque. Analyze the bearing reactions of Problem

14.46 again, but this time use a starting torque equal to 250% of the full-load torque.

Assume a normal-load torque and a speed of zero. How does this starting condition

affect the forces on the mounting bolts?

Note that data are given for m = 2 planets. The masses of the moving elements are:

5 3 2 2 28 10 N/mm 35037.5 2 12.5 2 37.5 12.5 2 15.75 87.5 133 NAm

5 3 2 2 2 2

2 8 10 N/mm 62.5 34.5 92.75 34.5 50 12.5 18.75 81.25 109.47 Nm

5 3 2

4 8 10 N/mm 30.25 34.5 8 Nm

The centroidal mass moments of inertia are:

5 3 2 2 4 4

2 2 4 2 2 2

8 10 N/mm [4 350 375 350 100 12 50 12.5 2 2 37.5 12.5 2

2 37.5 12.5 123.25 2 15.75 87.5 2 2 15.75 87.5 123.5 ] 1.479 N m

AI

5 3 4 4 4 4 2

2 8 10 N/mm 62.5 34.5 2 92.75 34.5 2 50 12.5 2 18.75 81.25 2 0.3949 N mI

5 3 4 3 2

4 8 10 N/mm 30.25 34.5 2 3.6434 10 N mI

The angular accelerations are found by the tabular method (see Section 9.7):

Step Number Frame 1 Arm A Planets 2, 3 Sun 4

1. Gears fixed to arm

2. Arm fixed 0 104 35 104 35 52 17

3. Total 0 69 35 6 003 595

The output torque is 292.14 N moutT .

The input torque is 2.5 28.92 N m 72.31 N minT .

Balancing the sun gear and input shaft:

4 4 34 4 42 t

inM R F T I

3 2 2

342 30.35 mm 72.31 N m 3.6434 10 N m 9650 mm/s 6003 595tF

34 1190.8 0.0627tF

Balancing the arm and output shaft:

22 t

A out A A A AM T R F I

2 2

2292.14 N m 2 123.2 mm = 1.479 N m 9650 mm/st

AF

2 1185.48 0.6225t

AF



Balancing a typical planet:

2 3 43 2 12 2 2

t tM R F R F I

1292.85 mm 1190.8 0.0627 62.5 in 0.3949 m 9650 mm / s 69 35t 2 2 F N

12 1769.3 1.3839tF

22 12 43 2 2

t t t t

A GF F F F m A

21769.3 1.3839 1190.8 0.0627 1185.48 0.6225 109.47 N 9650 mm/s 21.93

2512 rad/s

Now, reassembling the above results,

34 1190.8 0.6274 1158.78 NtF 34 34 tan 20 421.86 Nr tF F

12 1769.3 1.3839 1060.88 NtF 12 12 tan 20 386.26 Nr tF F

2 1158.78 0.6225 1504 Nt

AF 2 386.26 421.86 35.6 Nr

AF

The input and output bearing reactions are zero. Ans.

The load on the planet shaft is 2 2

2 2 2 1504.5 Nt r

A A AF F F Ans.

The forces in the mounting bolts to restrain the unbalanced frame moment are:

11 1 212 400 mm 985.23 NtF R F Ans.



14.48 The gear-reduction unit of Problem 14.46 is running at 600 rev/min when the motor is

suddenly turned off, without changing the resisting-load torque. Solve Problem 14.46 for

this condition.

Here we can use the free-body diagrams from Problem 14.46 and the mass data and

angular motion relationships from Problem 14.47. Then, proceeding as in Problem 14.47,

but with Tin = 0, we balance the sun gear and input shaft:

4 4 34 4 42 t

inM R F T I

3 2 2

3460.7 mm 3.6434 10 N m 9650 mm/s 6003 595tF

34 0.6274tF

Balancing the arm and output shaft:

22 t

A out A A A AM T R F I

2 2

20.2921 N m 2 123.2 mm = 1.4796 N m 9650 mm/st

AF

2 1185.48 0.6225t

AF

Balancing a typical planet:

2 3 43 2 12 2 2

t tM R F R F I

2 2

1292.84 mm 0.6274 62.5 mm 0.3949 N m 9650 mm/s 69 35tF

12 1.3839tF

22 12 43 2 2

t t t t t

A GF F F F m A

21.3839 0.6274 1185.48 0.6225 109.47 N 9650 mm/s 123.2

2342 rad/s

600 rev/min 62.8 rad/sA 2

2 2486425 mm/sr

G A AA R

2

2 2

2 12 43 2 2 109.47 N 9650 mm/s 486425 mm/s 5518 Nr r r r r

A GF F F F m A

Reassembling the above results,

34 0.6279 21.36 NtF 34 34 tan 20 8 Nr tF F

12 1.3839 473.48 NtF 12 12 tan 20 172.215 Nr tF F

2 1185.48 0.6225 972.77 Nt

AF 2 172.21 8 5682.65 Nr

AF

The input and output bearing reactions are zero. Ans.

The load on the planet shaft is 2 2

2 2 2 5762.75 Nt r

A A AF F F Ans.

The forces in the mounting bolts to restrain the unbalanced frame moment are:

11 1 212 400 mm 439.66 NtF R F Ans.



14.49 The differential gear train illustrated in Fig. P14.49 has gear 1 fixed and is driven by

rotating shaft 5 at 500 rev/min in the direction shown. Gear 2 has fixed bearings

constraining it to rotate about the positive y axis, which remains vertical; this is the output

shaft. Gears 3 and 4 have bearings connecting them to the ends of the carrier arm, which

is integral with shaft 5. The module of gears 1 and 5 are both 3.175 mm/tooth, while the

module of gears 3 and 4 are both 4.23 mm/tooth. All gears have the 20 pressure angles

and are each 18.75 mm thick, and all are made of steel with density 3 37.9 10 g / mm .

The mass of shaft 5 and all gravitational loads are negligible. The output shaft torque

loading is ˆ133.5 N m T j as shown. Note that the coordinate axes shown rotate with

the input shaft 5. Determine the driving torque required, and the forces and moments in

each of the bearings. (Hint: It is reasonable to assume through symmetry that 13 14t tF F .

It is also necessary to recognize that only compressive loads, not tension, can be

transmitted between gear teeth.)

3 3 2

2 7.98 10 g/mm 100 18.75 47.97 Nm 2

2 247.97 N 4 2 0.2399 N myy

GI

3 3 2

3 7.92 10 g/mm 75 18.75 26.96 Nm 3

2 226.96 N 3 2 0.0759 N mxx

GI

3

2 226.96 N 75 4 0.0379 N myy

GI 3 3

20.0379 N mzz yy

G GI I

4 3 26.96 Nm m 4 3

20.0759 N mxx xx

G GI I

4 3

20.0379 N myy yy

G GI I 4 4

20.0379 N mzz yy

G GI I



5 0m

5ˆ ˆ500 rev/min 52.36 rad/s ω j j 2 5

ˆ2 104.72 rad/s ω ω j

Noting that the primary xyz axes rotate at an angular velocity of 5ω

3ˆ ˆ69.81 52.36 rad/s ω i j 4

ˆ ˆ69.81 52.36 rad/s ω i j

Recognizing that 5ˆ ˆ ˆ52.36 rad/sd dt i ω ×i k , the absolute accelerations are 2 0α ,

2

3ˆ ˆ69.81 52.36 3 655 rad/s α k k , 2

4ˆ ˆ69.81 52.36 3 655 rad/s α k k , 5 0α .

Link 5: Note that we assume the mass of link 5 is negligible. Also, assuming no thrust

bearing at the fixed pivot, 15 35 45 0y y yF F F .

5 15 35 45 0x x x xF F F F

5 15 35 45 0z z z zF F F F

5 15 5 15 0x x zM M d F

5 15 35 454 4 0y y z zM M F F

5 15 35 45 5 15 0z z z z xM M M M d F



7.275

5. 45

7. 275

7.275

5. 45

5. 45

7.275

5.45

Link 4: Note that the forces on bevel gear teeth are related by Eq. (13.20) where, in this

case, 4cos 0.8 , 4sin 0.6 , and 20 . Noting that

4

2 2

5 5 5ˆ ˆ ˆ100 mm 100 mm 274150 mm/sG A ω × ω × i i i

4

2 2

4 14 24 54 45.45 5.45 26.96 N 9650 mm/s 274150 mm/s 765.4 Nx z z x x

GF F F F m A

4 14 247.275 7.275 0y z zF F F 24 14

z zF F

4 14 24 54 0z z z zF F F F

Using Eqs. (14.110) for the moment equations,

4 14 2475 75 0x z zM F F

4 0yM

4 4 4

2

4 14 24 54 4 4 45.45 5.45 ( ) 6460 mm/sz z z z zz z xx yy x y

G G GM F F M I I I

2

54 6460 mm/szM

Link 3: Again 3cos 0.8 , 3sin 0.6 , and 20 and using Eqs. (14.110) we get

similar results

3

2 2

5 5 5ˆ ˆ ˆ100 in 100 mm 274150 mm/sG A ω × ω × i i i

3

2 2

3 13 23 53 35.45 5.45 26.96 N 9650 mm/s 274150 mm/s 765.845 Nx z z x x

GF F F F m A

3 13 237.275 7.275 0y z zF F F 23 13

z zF F

3 13 23 53 0z z z zF F F F

3 13 2375 75 0x z zM F F

3 0yM

3 3 3

2

3 13 23 53 3 3 35.45 5.45 ( ) 6460 mm/sz z z z zz z xx yy x y

G G GM F F M I I I

53 6460 mmz 2M /s



7.275

5.45

7.275

5.45

Link 2: This time 2cos 0.6 , 2sin 0.8 , and 20 .

2 12 32 425.45 5.45 0x x z zF F F F

2 12 32 427.275 7.275 0y y z zF F F F

2 12 32 42 0z z z zF F F F

2 2 12 0x zM d F 12 0zF

2 12 32 42100 100 0y y z zM M F F 32 42 133.5 N m 100 mm 1335 Nz zF F

2 32 42 2 12100 7.275 100 7.275 0z z z xM F F d F

Reviewing these again shows

32 42 667.5 Nz zF F 12 0xF , 12 388.7 NyF

Finally, collecting all results, we have:

12ˆ388.7 N F j 12

ˆ133.5 N m M j Ans.

15 F 0 15ˆ267 N m M j Ans.

35ˆ ˆ1057.3 1335 N F i k 35

ˆ28.74 N m M k Ans.

45ˆ ˆ1057.3 1335 N F i k 45

ˆ28.749 N m M k Ans.



14.50 Figure P14.50 illustrates a flyball governor. Arms 2 and 3 are pivoted to block 6, which

remains at the height shown but is free to rotate around the y axis. Block 7 also rotates

about and is free to slide along the y axis. Links 4 and 5 are pivoted at both ends between

the two arms and block 7. The two balls at the ends of links 2 and 3 weigh 15.575 N

each, and all other masses are negligible in comparison; gravity acts in the ˆj direction.

The spring between links 6 and 7 has a stiffness of 0.178 N/mm and would be unloaded if

block 7 were at a height of ˆ275 mm.D R j All moving links rotate about the y axis with

angular velocities of ˆ j . Make a graph of the height DR versus the rotational speed

rev/min, assuming that changes in speed are slow.

150 m

m

400 m

m

150 mm

400 mm

The free-body diagram below shows only one of the arms, body 2, containing one of the

two flyballs. Force F42 comes from body 4, which is a two-force member, thus defining

its line of action. The force FA comes from the spring. The position of the bottom of the

spring is

400 2 150cos 400 300cos mmDR

The total force in the spring is

0 1 lb/in 400 mm 300cos mm 275 mm 22.25 53.4cos ND Dk R R

Since this total force must balance two flyball arms, force FA of the free-body diagram is

11.17 26.7cos NAF



Taking moments about point B,

2

2 2150cos mm 300sin 150sin mm 150sin mm 11.125 26.7cos N 0m m g

Dividing by 150sin mm

2

2 253.4cos 11.125 N 26.7cos N 0m m g

Substituting values and rearranging we get

2 20.484 N s cos 26.7 26.7cos 6 1 cos N , 2 2 2cos 55.155 1 cos rad /s

1 cos 1 cos7.427 rad/s 70.919 rev/min

cos cos

, 400 300cos mmDR Ans.

400

mm

300

200

100



Chapter 15

Vibration Analysis

15.1 Derive the differential equation of motion for each of the systems illustrated in Fig. P15.1

and write the formula for the natural frequency n for each system.

(a) F k x y mx n

mx kx ky

k m

Ans.

(b) F F cx kx mx n

mx cx kx F

k m

Ans.

(c) 1 2F cx k x k x mx

1 2

1 2

0

n

mx cx k k x

k k m

Ans.



(d) 1 2F k x k x y mx

1 2 2

1 2n

mx k k x k y

k k m

Ans.

(e) F c x y kx mx n

mx cx kx cy

k m

Ans.

(f) Both springs 3 and 4 experience the same spring force F34, and each is deflected

by an amount consistent with its own rate, F34/k3 or F34/k4, respectively. The total

deflection is 34 3 34 4x F k F k or 34 3 4 3 4F k k x k k

3 41 2

3 4

k kF k x k x x mx

k k

3 41 2

3 4

3 41 2

3 4

0

n

k kmx k k x

k k

k kk k

k k

m

Ans.



15.2 Evaluate the constants of integration of the solution to the differential equation for an

undamped free system, using the following sets of starting conditions:

(a) 0 , 0x x x

(b) 00, x x v

(c) 0 0, x x x a

(d) 0 0, x x x b

For each case, transform the solution to a form containing a single trigonometric term.

For each case we use a trial solution of: with initial value of:

sin cosn nx A t B t (0)x B (1)

cos sinn n n nx A t B t 0 nx A (2)

2 2sin cosn n n nx A t B t 20 nx B (3)

3 3cos sinn n n nx A t B t 30 nx A (4)

(a) 0 , 0x x x . Use Eqs. (1) and (2); 00,A B x

0 cos nx x t Ans.

(b) 00, x x v . Use Eqs. (1) and (2); 0 , 0nA v B

0 sinn nx v t Ans.

(c) 0 0, x x x a . Use Eqs. (1) and (3); 2

0 0, nB x B a

These are inconsistent unless 2

0 0na x . Second given condition is not useful.

One more initial condition required, such as 00x v from which

0 nA v .

0 0sin cosn n nx v t x t

22

0 0 sinn nx x v t where 1

0 0tan nx v Ans.

(d) 0 0, x x x b . Use Eqs. (1) and (4); 3

0 0, nB x A b

3

0 0sin cosn n nx b t x t

2

2 3

0 0 sinn nx x b t where 1 3

0 0tan nx b Ans.



15.3 A system like Fig. 15.5 has m = 1 kg and an equation of motion

20cos 8 / 4 mm.x t Determine the following:

(a) The spring constant k

(b) The static deflection st

(c) The period

(d) The frequency in hertz

(e) The velocity and acceleration at the instant t = 0.20 s

(f) The spring force at t = 0.20 s

Plot a phase diagram to scale showing the displacement, velocity, acceleration, and

spring-force phasors at the instant t = 0.20 s.

(a) 8 rad/sn k m

22 8 rad/s 1 kg 631.65 N/mnk m Ans.

(b) 21 kg 9.81 m/s

0.015 53 m 15.53 mm631.65 N/m

st

F mg

k k Ans.

(c) 2 2 rad/rev 8 rad/s 0.250 s/revn Ans.

(d) 1 4 rev/s 4 Hzf Ans.

(e) 8 4 8 0.25 8 0.20 0.25 1.35 rad 243t t

8 rad/s 0.020 m sin 8 4 0.503sin 243 m/s 0.448 m/sx t Ans.

2 2 28 rad/s 0.020 m cos 8 4 12.633cos 243 m/s 5.735 m/sx t Ans.

(f) 631.65 N/m 0.020 m cos243 12.633cos243 N 5.735 NF kx Ans.

0.020cos243 mx , 0.503sin 243 m/sx , 212.633cos243 m/sx

12.633cos243 NF



15.4 The weight 1W in Fig. P15.4 drops through the distance h and collides with 2W with

plastic impact (a coefficient of restitution of zero). Derive the differential equation of

motion of the system, and determine the amplitude of the resulting motion of 2W .

Define 0t at the instant of impact. At the beginning of impact we have 1 2v gh .

By conservation of momentum, 1 1 1 2 2m v m m v . Thus 1 1 2 22W g gh W W v g

or 2 1 1 22v ghW W W . Therefore, at 0t , 0x , 2x v .

Note that, at 0x , the spring force includes a reaction to 2W . And so,

1F kx W W g x where, for convenience, we have defined 1 2W W W . From

the force balance we get the differential equation of motion

1W g x kx W Ans.

with natural frequency of n kg W . Then

1cos sinn nx A t B t W k

sin cosn n n nx A t B t

and with the initial conditions stated above 1A W k and 2 nB v . Therefore

1 2 1cos sinn n nx W k t v t W k

Transforming to a single transient term we get

1sin nx X t W k where

2 2

1 2 nX W k v Ans.

and 1 1

2

tann

W k

v



15.5 The vibrating system illustrated in Fig. P15.5 has 1 3 850 N/mk k , 2 1 850 N/mk ,

and 40 NW . What is the natural frequency in hertz?

Springs 1 and 2 both experience the same spring force F12, and each is deflected by an

amount consistent with its own rate, F12/k1 or F12/k2, respectively. The total deflection is

12 1 12 2x F k F k or 12 1 2 1 2F k k x k k

12 3F F k x mx 1 23

1 2

0k k

mx k xk k

The natural frequency is

1 23

1 2

2

850 N/m 1850 N/m850 N/m

850 N/m 1850 N/m 18.74 rad/s 2.983 Hz40 N 9.81 m/s

n

k kk

k k

W g

Ans.



15.6 Figure P15.6 illustrates a weight W = 80.1 N connected to a pivoted rod which is

assumed to be weightless and very rigid. A spring having a rate of k =13.35 N/M is

connected to the center of the rod and holds the system in static equilibrium at the

position shown. Assuming that the rod can vibrate with a small amplitude, determine the

period of the motion.

150 mm

300 mm

2M a ka W W g 2 2W g ka W

22

2

13.35 N/M 9650 mm/s150 mm20.05 rad/s

300 mm 80.1 Nn

ka a kg

W g W

2 2 rad/rev0.313 s/rev

20.05 rad/sn

Ans.



15.7 Figure P15.7 illustrates an upside-down pendulum of length l retained by two springs

connected a distance a from the pivot. The springs have been positioned such that the

pendulum is in static equilibrium when it is in the vertical position.

(a) For small amplitudes, find the natural frequency of this system.

(b) Find the ratio l/a at which the system becomes unstable.

(a) 2

1 2M W a k a a k a W g

2 2

1 2 0W g k k a W

2

1 21 0

k k ag

W

2

1 21n

k k ag

W

Ans.

(b) The system becomes unstable whenever the natural frequency becomes the square

root of a negative number (imaginary). At such values the system is not oscillatory.

This happens whenever

1 2a k k a W Ans.

15.8 (a) Write the differential equation for the system illustrated in Fig. P15.8 and find

the natural frequency.

(b) Find the response x if y is a step input of height 0y .

(c) Find the relative response z x y to the step input of part (b).



(a) 1 2F k x k x y mx

1 2 2mx k k x k y 1 2n k k m Ans.

(b) The complementary solution is

cos sinn nx A t B t

For a particular solution, arrange the equation to the form

2

2nx x k m y

Since, for a step input the right-hand side is constant, try x C , 0x .

2

2nC k m y 2 1 2C k y k k

The complete solution is x x x

2 1 2cos sinn nx A t B t k y k k

At 0t , 0x and 0y y .

2 0 1 20 1 0A B k y k k 2 0 1 2A k y k k

sin cosn n n nx A t B t

At 0t , 0x

0 0 1n nA B 0B

Therefore, the complete response is

2 0 1 2 2 0 1 2cos nx k y k k t k y k k

2 0

1 2

1 cos n

k yx t

k k

Ans.

(c) 1 2 02 0

1 2 1 2

1 cos n

k k yk yz x y t

k k k k

1 0 2 0

1 2

cos nk y k y tz

k k

Ans.



15.9 An undamped vibrating system consists of a spring whose scale is 35 kN/m and a mass of

1.2 kg. A step force F = 50 N is exerted on the mass for 0.040 s.

(a) Write the equations of motion of the system for the era in which the force acts and

for the era that follows.

(b) What are the amplitudes in each era?

(c) Sketch a time plot of the displacement.

(a) 35 000 N/m 1.2 kg 171 rad/sn

First era: 0 0.040 st : 1.2 kg 35 000 N/m 50 Nx x

From Eq. (15.21)

1 cos 50 N 35 000 N/m 1 cos171nx F k t t

0.001 429 m 1 cos171x t 1.429 mm 1 cos171x t Ans.

Also 0.244 m/s sin171 244 mm/s sin171x t t

At the end of the first era 0.040 st , 6.831 rad 391.4nt

0.000 209 m 0.209 mmx , 0.127 m/s 127 mm/sx

Second era: 0.040 st : 1.2 35 000 0x x

From Eqs. (15.16) and (15.17)

2 2 22

0 0 0 0.209 mm 127 mm/s 171 rad/s 0.773 mmnX x v

1 10 0tan tan 127 mm/s 171 rad/s 0.209 mm 74.3nv x

0 cos 0.773 mm cos 171 74.3nx X t t Ans.

(b) First era; 0 0.040 st : 1.429 mmX Ans.

Second era; 0.040 st : 0.773 mmX Ans.

(c)



15.10 Figure P15.10 illustrates a round shaft whose torsional spring constant is in lb/radtk

connecting two wheels having mass moments of inertia 1 2andI I . Show that the system

is likely to vibrate torsionally with a frequency of

1 2

1 2

t

n

k I I

I I

Designating the angular positions of the two wheels by 1 and 2 , respectively, and

summing moments on each, we get

1 1 2 1 1tM k I 1 1 1 2 0t tI k k

2 1 2 2 2tM k I 2 2 1 2 0t tI k k

Next, we assume a solution of the form

cosj j nC t

for each inertia with j = 1, 2.

Substituting these gives

2

1 1 1 2cos cos cos 0n n t n t nI C t k C t k C t

2

2 2 1 2cos cos cos 0n n t n t nI C t k C t k C t

Dividing each by cos nt and writing these in matrix form they become

211

222

0n t t

t n t

CI k k

Ck I k

For this set of equations to have a non-trivial solution for C1 and C2, the determinant of

the coefficient matrix must vanish. Therefore,

2 2

1 2 0n t n t t tI k I k k k

This expands to a quadratic equation in 2

n

2

2 2

1 2 1 2 0n t nI I k I I

Solving this, we get four roots:

0n 1 2

1 2

t

n

k I I

I I

Q.E.D.

Note that the other frequency of 0n shows the capability for rigid body rotation

since the entire shaft with wheels is free to rotate.



15.11 A motor is connected to a flywheel by a 15.625 mm diameter steel shaft 900 mm long, as

shown. Using the methods of Chapter 15, it can be demonstrated that the torsional spring

constant of the shaft is 531.1 N M/rad. The mass moments of inertia of the motor and

flywheel are 2.712 and 6.328 2N M s , respectively. The motor is turned on for 2 s,

and during this period it exerts a constant torque of 22.6 N M on the shaft.

(a) What speed in revolutions per minute does the shaft attain?

(b) What is the natural circular frequency of vibration o f the system?

(c) Assuming no damping, what is the amplitude of the vibration of the system in

degrees during the first era? During the second era?

This is a difficult problem, but too interesting and challenging not to include.

(a) The angular impulse equation is

00

t

H H Tdt

Since the motor starts from rest, its initial angular momentum is 0 0H . We also

see that 1 2H I I , 22.6 NMT , and 2 st . Substituting,

1 2 00

22.6 NMt

I I H dt

29.04 NM s 22.6 NM t

22.5 rad/s t for 0 2 st At t = 2 s 5.0 rad/s 47.75 rev/min Ans.

(b) From Problem 15.10

2 2

1 2

2 2

1 2

531.1 NM/rad 2.712 NMs 6.328 NMs16.726 rad/s

2.712 NMs 6.328 NMs

t

n

k I I

I I

Ans.

(c) First era; 0 2 st :

The differential equations are:

1 1 1 2t tI k k T

2 2 1 2 0t tI k k

After using the conditions that, at t = 0, 1 2 0 the solutions become

2

2 21

1 1 2

cos2

n

t

I IT tA B t

I I I k

2

2

1 2

cos2

n

TtA B t

I I

Then using the conditions that, at t = 0, 1 2 0 , we find



1 2

2

1 2t

I ITB A

k I I

and the solutions become

2

2 21 1 22

1 21 2

cos2

n

t t

I IT T tI I t

k I I kI I

2

1 22 2

1 21 2

1 cos2

n

t

I IT Ttt

k I II I

But we are interested in the relative motion, the twist in the shaft, which is

2

1 2

1 2

1 cos n

t

I tT

k I I

So the amplitude during the first era is

2

2

21 2

6.328 NMs22.6 NM0.029 rad 1.707

531.1 NM/rad 9.04 NMst

IT

k I I

Ans.

For the completion of the first era we can compute that, at t = 2.0 s,

16.7 rad/s 2.0 s 33.45 rad 1 916.7nt and cos 0.449nt . Thus,

1 5.0302 rad , and 1 5.0302 rad . These are the initial displacements for the

second era.

Second era; 2 st :

The differential equations now become:

1 1 1 2 0t tI k k

2 2 1 2 0t tI k k

Following a similar procedure to that above, we eventually obtain

1 2 0.0555cos 19.6nt

Therefore the amplitude of the second era is

0.0555 rad 3.180 Ans.



15.12 The weight of the mass of a vibrating system is 44.5 N, and it has a natural frequency of

1 Hz. Using the phase-plane method, plot the response of the system to the force

function illustrated in Fig. P15.12. What is the final amplitude of the motion?

22 244.5 n 9650 mm/s 2 rad/s 4.552 n/25 mm 0.182n/mmnk m

1 53.4 n 0.182 n/mm 293.4 mmF k 2 26.7 n 0.182 n/mm 146.7 mmF k

12 rad/s 0.25 s 1.571 rad 90.0

nt , 2

2 rad/s 0.25 s 1.571 rad 90.0n

t

The final amplitude is X = 464 mm. Ans.



15.13 An undamped vibrating system has a spring scale of 35.6 N/mm and a weight of 222.5 N.

Find the response and the final amplitude of vibration of the system if it is acted upon by

the forcing function illustrated in Fig. P15.13. Use the phase-plane method.

235.6 n/mm 50 lb 9650 mm/s 39.298 rad/sn k m

39.298 rad/s 0.040 s 1.572 rad 90n t ,

55.625 n 35.6 n/mm 1.5625 mmF k

The final amplitude is 4.42 mm. Ans.



15.14 A vibrating system has a spring k = 3 936 N/m and a weight of W = 20 N. Plot the

response of this system to the forcing function illustrated in Fig. P15.14:

(a) Using three steps

(b) Using six steps

max 200 NF

23 936 N/m 20 N 9.81 m/s 43.937 rad/sn k m

(a) Three-step solution: 43.937 rad/s 0.1 s 3 1.465 rad 83.91n t

1 0.083 mF k , 2 0.250 mF k , 3 0.417 mF k , 0.500 mF k

(b) Six-step solution: 43.937 rad/s 0.1 s 6 0.732 rad 41.96n t

1 0.042 mF k , 2 0.125 mF k , 3 0.208 mF k , 4 0.292 mF k ,

5 0.375 mF k , 6 0.458 mF k , 0.500 mF k



15.15 (a) What is the value of the coefficient of critical damping for a spring-mass-damper

system in which k = 64 kN/m and m = 36 kg?

(b) If the actual damping is 20% of critical, what is the natural frequency of the

system?

(c) What is the period of the damped system?

(d) What is the value of the logarithmic decrement?

(a) 64 000 N/m 36 kg 42.164 rad/sn k m

2 2 36 kg 42.164 rad/s 3 035.8 N s/mc nc m Ans.

(b) 2 21 42.164 rad/s 1 0.20 41.312 rad/sd n Ans.

(c) 2 2 41.312 rad/s 0.152 s/cycled Ans.

(d) 2 22 1 2 0.2 1 0.20 1.283 Ans.

15.16 A vibrating system has a spring k = 3.6 kN/m and a mass m = 16 kg. When disturbed, it

was observed that the amplitude decayed to one-fourth of its original value in 4.80 s.

Find the damping coefficient and the damping factor.

3 600 N/m 16 kg 15.0 rad/sn k m

Using Eq. (15.32) with ln 1.0 0.25 1.386N and 4.80 sN

1.386 4.80 s 15.0 rad/s 0.019 25N nN Ans.

2 0.01925 2 15 kg 15.0 rad/s 8.663 N s/mnc m Ans.

15.17 A vibrating system has k = 53.44 N/mm, W = 445 N, and damping equal to 20% of

critical.

(a) What is the damped natural frequency d of the system?

(b) What are the period and the logarithmic decrement?

253.4 N/mm 445 N 9650 mm/s 34.03 rad/sn k m

(a) 2 21 34.03 rad/s 1 0.20 33.34 rad/sd n Ans.

(b) 2 2 33.34 rad/s 0.188 s/cycled Ans.

2 22 1 2 0.20 1 0.20 1.283 Ans.



15.18 Solve Problem 15.14 using damping equal to 15% of critical.

23 936 N/m 20 N 9.81 m/s 43.937 rad/sn k m

2 21 43.937 rad/s 1 0.15 43.440 rad/sd n

Six-step solution: 43.440 rad/s 0.10 s 6 0.724 rad 41.48d t

1 0.042 mF k , 2 0.125 mF k , 3 0.208 mF k , 4 0.292 mF k ,

5 0.375 mF k , 6 0.458 mF k , 0.500 mF k

In each step of 0.724 radd t the reduction in amplitude is

20.724 rad 1

41.48 0.896n nX X e

. Therefore,

0 0.0417 mx , 0 90 1 0.896 0.0417 m 0.0376 mx

1 0.1138 mx , 1 77.34 2 0.896 0.1138 m 0.1020 mx

2 0.1653 mx , 2 59.98 3 0.896 0.1653 m 0.1481 mx

3 0.1916 mx , 3 42.86 4 0.896 0.1916 m 0.1716 mx

4 0.1926 mx , 4 27.01 5 0.896 0.1926 m 0.1726 mx

5 0.1719 mx , 5 13.53 6 0.896 0.1719 m 0.1539 mx

6 0.1394 mx , 6 12.64



15.19 A damped vibrating system has an undamped natural frequency of 10 Hz and a weight of

3560 N. The damping ratio is 0.15. Using the phase-plane method, determine the response

of the system to the forcing function illustrated in Fig. P15.19.

10 revs/s 2 rad/rev 62.832 rad/sn

2 21 62.832 rad/s 1 0.15 62.121 rad/sd n

62.121 rad/s 0.01 s 0.621 rad 35.59d t

22 23560 N 9650 mm/s 62.832 rad/s 339.1 N/mmnk m

1 26.25 mmF k , 2 39.375 mmF k , 3 13.125 mmF k , 4 13.125 mmF k ,

In each step of 0.621 radd t the reduction in amplitude is

20.621 rad 1

35.59 0.910n nX X e

. Therefore,

0 26.25 mmx , 0 90 1 23.875 mmx 1 22.75 mmx

1 29.7 mmx , 1 43.53 2 27.025 mmx 2 24.6 mmx

2 43.5 mmx , 2 59.95 3 39.6 mmx 3 36.025 mmx

3 58.5 mmx , 3 113.94 4 53.15 mmx 4 48.375 mmx

4 51.25 mmx , 4 199.90



15.20 A vibrating system has a spring rate of 3 000 N/m, a damping coefficient of 100 N s/m ,

and a weight of 800 N. It is excited by a harmonically varying force 0

50 NF at a

frequency of 60 cycles per minute.

(a) Calculate the amplitude of the forced vibration and the phase angle between the

vibration and the force.

(b) Plot several cycles of the displacement-time and force-time diagrams.

(a) 2 81.549 kg800 N 9.81 m/sm W g

3 000 N/m

6.065 rad/s81.549 kg

n

k

m

100 N s/m0.101

2 2 81.549 kg 6.065 rad/sn

c

m

60 cycles/min 2 rad/cycle 60 s/min1.036

6.065 rad/sn

0

2 22 21 2n n

F kX

2 22

50 N 3 000 N/m0.075 m 75 mm

1 1.036 2 0.101 1.036

X

Ans.

1 1

2 2 2

2 2 0.101 1.036tan tan 109.25

1 1 1.036

n

n

Ans.

15.21 A spring-mounted mass has k = 44.5 N/mm, c = 1.424 / NS/mm, and weighs 1557.5 N.

This system is excited by a force having an amplitude of 890 N at a frequency of 2 Hz.

Find the amplitude and phase angle of the resulting vibration and plot several cycles of

the force-time and displacement-time diagrams.

22 0.161 Ns /mm1557.5 n. 9650 mm/sm w g

244.5 N/mm 16.605 rad/s0.161 ns /mmn k m

22 1.424 Ns/mm 2 0.16 Ns /mm 16.605 rad/s 0.266nc m

2 Hz 2 rad/cycle 16.605 rad/s 0.757n

0

2 22 21 2n n

F kX

2 22

890 N 44.5 N/mm34.075 mm

1 0.757 2 0.266 0.757

X

Ans.

1 1

2 2 2

2 2 0.266 0.757tan tan 43.26

1 1 0.757

n

n

Ans.



15.22 When a 26700-N press is mounted upon structural-steel floor beams, it causes them to

deflect18.75 mm. If the press has a reciprocating unbalance of 2002.5. N and it operates

at a speed of 72 rev/min, how much of the force will be transmitted from the floor beams

to other parts of the building? Assume no damping. Can this mounting be improved?

26700 N 18.75 mm 1.424 N/mmk 2 226700 N 9650 mm/s 2.767 Ns /mmm

21424 N/mm 2.76 Ns /mm 22.714 rad/sn k m

72 rev/min 2 rad/rev 60 s/min 22.689 rad/s 0.332n

Assuming no damping, Eq. (15.62) gives

2 2 22

2 2

1 1 11.124

1 1 0.3321 n

n

T

0 1.124 2002.5 N 2250.8 NtrF TF

Fig. 15.37 shows that, with 0.332n , small changes in either damping or n will do

little to reduce transmissibility. Therefore the mounting cannot be improved. The

primary opportunity for improvement would be to reduce the unbalance.



15.23 Four vibration mounts are used to support a 450-kg machine that has a rotating unbalance

of 0.35 kg m and runs at 250 rev/min. The vibration mounts have damping equal to 36

percent of critical. What must the spring constant of the mounting be if 20 percent of the

exciting force is transmitted to the foundation? What is the resulting amplitude of motion

of the machine?

From Eq. (15.63)

2 2 22 2

2 22 2 22 2

/ 1 2 0.30 / / 1 0.36 /0.20

1 / 2 0.30 / 1 / 0.36 /

n n n n

n n n n

T

2

2 2 2 20.20 1 / 0.36 / / 1 0.36 /n n n n

2

2 2 4 20.04 1 / 0.36 / / 1 0.36 /n n n n

6 4 2

0.36 / 0.96 / 0.0656 / 0.04 0n n n

Numerically searching for the root we find

2

/ 0.168 423n / 0.410 39n

250 rev/min 26.180 rad/s 63.792 83 rad/sn

22 450 kg 63.792 83 rad/s 1 831 286 N/mnk m Ans.

Now, from Eq. (15.56)

2

2 2 22 2

/ 0.168 4230.186 81

1 0.168 423 0.72 0.168 4231 / 0.36 /

n

un n

mX

m e

0.186 81 0.35 kg m 450 kg 0.145 mmX Ans.



15.24 A 600-mm long steel shaft is simply supported by two bearings at A and C as illustrated

in Fig. P15.24. Flywheels 1 and 2 are attached to the shaft at locations B and D,

respectively. Flywheel 1 at location B weighs 50 N, flywheel 2 at location D weighs 20

N, and the weight of the shaft can be neglected. The known stiffness coefficients are

11 20 000 N/m,k 12 50 000 N/m,k and

22 40 000 N/m.k Determine: (i) the first

and second critical speeds of the shaft using the exact solution and the first critical speed

using (ii) the Dunkerley and (iii) Rayleigh-Ritz approximations. (iv) If flywheel 2 is then

placed at location B and flywheel 1 is placed at location D, determine the first critical

speed of the new system using the Dunkerley approximation.

(i) The exact solutions for the first and second critical speeds of the shaft are 2

11 1 22 2 11 1 22 2 1 2 11 22 12 21

2 2

1 2

( ) ( ) 4 ( )1 1,

2

a m a m a m a m m m a a a a

(1)

The influence coefficients are the reciprocals of the stiffness coefficients; that is,

1ii iia k and 1jk kj jka a k (2)

Therefore, the influence coefficients are

4 5

11 1 2 10 N/m 5 10 m/Na , 4 5

22 1 4 10 N/m 2.5 10 m/Na (3a)

and 4 5

12 1 5 10 N/m 2 10 m/Na (3b)

The masses of the two flywheels are

1 2

50 N5.10 kg

9.81 m/sm and 2 2

20 N2.04 kg

9.81 m/sm (4)

Substituting Eqs. (3) and (4) into Eq. (1) gives 5 2 2 4 4 10

5 2

2 2

1 2

30.6 10 s (30.6 s 208.080 s ) 101 1, (15.3 13.49) 10 s

2

Using the positive sign for the first critical speed and the negative sign for the second

critical speed gives

1 86.10 rad/s and 2 235.05 rad/s Ans.



(ii) Using the Dunkerley approximation, the first critical speed of the shaft with the

two flywheels can be written as

2221112

1

1mama

(5)

Substituting Eqs. (3) and the masses into Eq. (5) gives

5 5 4 2

2

1

15 10 m/N 5.10 kg 2.5 10 m/N 2.04 kg 3.06 10 s

Therefore, the first critical speed of the shaft is

157.17 rad / s Ans.

(iii) Using the Rayleigh-Ritz approximation, the first critical speed of the shaft with

the two flywheels can be written as

2 1 1 2 21 2 2

1 1 2 2

( )

( )

g W x W x

W x W x

(6)

The total deflections of the shaft at the mass particles can be written as

1 11 1 12 2x a W a W and 2 12 1 22 2x a W a W (7)

Substituting the known values and Eq. (2) into Eqs. (7) the total deflections are 3

1 2.9 10 mx and 3

2 1.5 10 mx (8)

Substituting Eqs. (8) and the known values into Eq. (6) gives 2 3 3

2

1 3 2 3 2

9.81 m/s [(50 N)(2.9 10 m) (20 N)(1.5 10 m)]

[(50 N)(2.9 10 m) (20 N)(1.5 10 m) ]

or 2

2 2 2

1 6

1.72 m/s3 688 rad /s

466 10 m


1 60.72 rad/s Ans.

(iv) When the two flywheels are interchanged then the Dunkerley approximation, see

Eq. (5), can be written as

newnew mama 2221112

1

1

(9)

Note that the influence coefficients of the shaft do not change (even though the two

flywheels were interchanged). Therefore, substituting these values into Eq. (9) gives

5 5 4 2

2

1

15 10 m/N 2.04 kg 2.5 10 m/N 5.10 kg 2.30 10 s


1 66.01 rad/s Ans.



15.25 The first critical speeds of a rotating shaft with two mass disks, obtained from three

different mathematical techniques, are 110 rad/s, 112 rad/s, and 100 rad/s, respectively.

(i) Which values correspond to the first critical speed of the shaft from the exact solution,

the Dunkerley approximation, and the Rayleigh-Ritz approximation? (ii) If the influence

coefficients are 4

11 22 10 m/Na a and the masses of the two disks are the same, that is,

m1 = m2 = m, then use the Dunkerley approximation to calculate the mass m. (iii) If the

influence coefficients are 4

11 22 10 m/Na a and the masses of the two disks are

specified as m1 = m2 = m = 0.5 kg, use the Rayleigh-Ritz approximation to calculate the

influence coefficient a12.

(i) The first critical speed from the Rayleigh-Ritz approximation is an upper bound;

therefore, the value 1 112 rad/s corresponds to the answer from the Rayleigh-Ritz

approximation. The Dunkerley approximation gives a lower limit to the first critical

speed; therefore, the value 1 100 rad/s corresponds to the answer from the Dunkerley

approximation. The value of the first critical speed from the exact method is

110 rad/s1

. Ans.

(ii) Since the first critical speed and the influence coefficients are given then the

Dunkerley approximation can be used to calculate the two masses; that is, 2

1 11 1 22 21 a m a m (1)

Substituting the given information and the first critical speed from the table into Eq. (1)

gives

2 4 41 100 rad/s 10 m/N 10 m/Nm m

Therefore, the mass is

24

10.5 kg

2 10 m/N 100 rad/sm

Ans.

(iii) The Rayleigh-Ritz equation can be written as

2 1 1 2 21 2 2

1 1 2 2

( )

( )

g W x W x

W x W x

Since the two masses are the same then this equation can be written as

2 2 2

1 1 2 1 2( )g x x x x (2)

The total deflections of the shaft at locations 1 and 2 can be written as

1 11 1 12 2x a W a W and 2 12 1 22 2x a W a W (3)

Since the influence coefficients 11 22a a and 12 21a a and the masses m1 = m2 = m then

the deflections 1 2x x x . Therefore, Eq. (2) can be written as 2

1 g x (4)

Substituting the known data and Eqs. (3) into Eq. (4) gives

2

2

2 4

12

9.81 m/s112 rad/s

(0.5 kg)(9.81 m/s )(10 m/N )a

Solving for the influence coefficient gives 5

12 5.94 10 m/Na Ans.



15.26 A steel shaft is simply supported by two rolling element bearings at A and B as illustrated

in Fig. P15.26. The length of the shaft is 1.45 m and two flywheels with weight 300 N

are attached to the shaft at the locations shown. One flywheel is 0.35 m to the right of

the left bearing at A and the other flywheel is 0.35 m to the left of the right bearing at B.

The weight of the shaft can be neglected. The influence coefficients are specified as 5

11 126 10 mm/Na and 5

21 92.5 10 mm/N.a (i) Determine the first and second

critical speeds of the shaft using the exact solution. Determine the first critical speed of

the shaft using: (ii) the Dunkerley approximation and (iii) the Rayleigh-Ritz equation.

(i) The exact solutions for the first and second critical speeds of the shaft can be written 2

11 1 22 2 11 1 22 2 11 22 12 21 1 2

2 2

1 2

( ) ( ) 4( )1 1,

2

a m a m a m a m a a a a m m

(1)

From the symmetry of the loading, we find the influence coefficients 6

11 22 1.26 10 m/Na a (2)

From Maxwell's reciprocity theorem, we get the influence coefficients 6

21 12 0.925 10 m/Na a (3)

The mass of the flywheels are

2

1 2 2

300 N30.581 N s /m

9.81 m/sm m m (4)

Substituting Eqs. (2), (3), and (4) into Eq. (1), the exact solutions can be written as

11 212 2

1 2

1 1, a a m

(5)

Equation (5) can be written as

2 2

1 2

11 21

1,

a a m

(6)

Substituting the numerical values into Eq. (6), the exact solutions can be written as

62 2

1 2 2

10,

(1.26 m/N 0.925 m/N) 30.581 N s /m

(7)

Using the positive sign in the denominator of Eq. (7), the first critical speed of the shaft is

obtained from the relation

6 62 4 2 2

1 22

10 101.4966 10 rad / s

66.819 s2.185 m/N 30.581 N s /m


1ω 122.3 rad / s Ans. (8)



Similarly, using the negative sign in the denominator of Eq. (7), the second critical speed

of the shaft can be obtained from the relation

6 62 4 2 2

1 22

10 109.761 10 rad / s

10.245 s0.335 m/N 30.581 N s /m

Therefore, the second critical speed of the shaft is

2 312.4 rad/s Ans.

Note that the second critical speed is about three times the first critical speed.

(ii) The Dunkerley approximation to the first critical speed of the shaft can be written as

11 22 112

1

12a a m a m

(9)

Substituting the numerical values into Eq. (9), the Dunkerley approximation to the first

critical speed of the shaft is

6 2 6 2

2

1

12 1.26 10 m/N 30.581 N s /m 77.064 10 s

Therefore, the Dunkerley approximation to the first critical speed of the shaft is

1 113.9 rad/s Ans.

Note that the the Dunkerley approximation to the first critical speed of the shaft is less

than the exact answer, see Eq. (8); that is, the Dunkerley approximation always gives a

lower bound.

(iii) The Rayleigh-Ritz equation can be written as

2 1 1 2 21 2 2

1 1 2 2

W x W xg

W x W x

(10)

where the deflections are 6 6

1 11 1 12 2 300 N(1.260 0.925) 10 m/N 655.5 10 mx a W a W

and 6 6

2 21 1 22 2 300 N(0.925 1.260) 10 m/N 655.5 10 mx a W a W

Substituting these values into Eq. (10), the Rayleigh-Ritz equation can be written as 2

2 4 2 2

1 2 6

2Wx 1 9.81 m/sω g g 1.4966 10 rad /s

2Wx x 655.5 10 m

Therefore, the Rayleigh-Ritz equation to the first critical speed of the shaft is

1 122.3 rad/s Ans.

Note that the Rayleigh-Ritz approximation to the first critical speed of the shaft gives the

same as the exact answer, see Eq. (8). In general, the Rayleigh-Ritz equation will give a

slightly greater value than the exact answer; that is, the Rayleigh-Ritz equation will give

an upper bound.



15.27 A steel shaft is simply supported by two rolling element bearings at A and C as illustrated

in Fig. P15.27. The length of the shaft is 0.6 m and two flywheels are attached to the

shaft at the locations B and D as shown. The flywheel at location B weighs 200 N and the

flywheel at location D weighs 90 N. The weight of the shaft can be neglected. It was

found that with flywheel 1 alone, the first critical speed of the shaft is 800 rad/s and with

flywheel 2 alone, the first critical speed of the shaft is 1 200 rad/s. (i) Determine the first

critical speed for the two mass system. (ii) If the two flywheels are interchanged (that is,

flywheel 2 is placed at location B and flywheel 1 is placed at location D), determine the

first critical speed of the new system using the Dunkerley approximation.

(i) Using the Dunkerley approximation, the first critical speed of the shaft with the two

flywheels can be written as 2

1 11 1 22 21 ω a m a m

(1)

or as

2 2 2

1 11 22

1 1 1

(2)

From the given data, the critical speeds are 11 800 rad/s and 22 1200 rad/s.

Therefore, Eq. (2) can be written as

4 2

2 22

1

1 1 1 1 110 s

64 144800 rad/s 1 200 rad/s

(3a)

or as 2 4 2 2

1 44.308 10 rad /s (3b)


1 665.6 rad/s Ans. (4)

(ii) When the two flywheels are interchanged then Eq. (1) can be written as 2

1 11 1 22 21 new newa m a m (5)

Note that the influence coefficients of the shaft (by definition) do not change (even

though the two flywheels were interchanged). Therefore, the influence coefficient

2

11 11 11a m (6a)


8

11 2 2

17.6641 10 m/N

800 rad/s (200 N / 9.81 m/s )a

(6b)



Similarly, the influence coefficient

2

22 22 21a m (7a)


8

22 2 2

17.5694 10 m/N

1 200 rad/s (90 N / 9.81 m/s )a (7b)

Substituting Eqs. (6b) and (7b) into Eq. (5) gives

8 8

2 2 2

1

1 90 N 200 N(7.6641 10 m/N) (7.5694 10 m/N)

ω 9.81 m/s 9.81 m/s

(8)

From this equation, the first critical speed of the shaft is 1 667.2 rad/s. Ans.



15.28 A steel shaft, which is 1250 mm in length, is simply supported by two bearings at B and

D as illustrated in Fig. P15.28. Flywheels 1 and 2 are attached to the shaft at A and C,

respectively. The flywheel at location A weighs 66.75 N the flywheel at location C

weighs 133.5 N, and the weight of the shaft can be neglected. The stiffness coefficients

are specified as 11 3560 N/mmk and

22 720 m/mmk . (i) Determine the first critical

speed for the two-mass system using the Dunkerley approximation. (ii) If the two

flywheels are interchanged (that is, flywheel 2 is placed at location A and flywheel 1 is

placed at location C), determine the first critical speed of the new system.

(i) Using the Dunkerley approximation, the first critical speed of the shaft with the two

flywheels can be written as 2

1 11 1 22 21 a m a m (1)

The influence coefficients are the inverse of the spring stiffness coefficients.

1ii iia k

Therefore, the influence coefficients are 4

11 2.8 10 mm/Na and 4

22 1.4 10 mm/Na

Substituting these values and the masses into Eq. (1) gives

4 4 6 2

2 2 2

1

1 66.75 N 135.5 N2.8 10 mm/N 1.4 10 mm/N 3.9 10 s

9650 mm/s 9650 mm/s

which gives

1 507.3 rad/s Ans.

(ii) When the two flywheels are interchanged then Eq. (1) becomes 2 new new

1 11 1 22 21 a m a m (2)

Note that the influence coefficients of the shaft do not change (even though the two

flywheels are interchanged). Therefore, substituting values into Eq. (2) gives

4 4 6 2

2 2 2

1

1 133.5 N 66.75 N2.8 10 mm/N 1.4 10 mm/N 4.08 10 s

9650 mm/s 9650 mm/s

which gives

1 453.7 rad/s Ans.



Chapter 16

Dynamics of Reciprocating Engines

16.1 A one-cylinder, four-stroke engine has a compression ratio of 7.6 and develops brake

power of 2.25 kW at 3 000 rev/min. The crank length is 22 mm with a 60-mm bore.

Develop and plot a rounded indicator diagram using a card factor of 0.90, a mechanical

efficiency of 72%, a suction pressure of 100 kPa and a polytropic exponent of 1.30.

22 24 0.060 m 4 0.002 827 mA D

2 32 2 0.022 m 0.002 827 m 1 000 L/m 0.124 4 L 124.4 mLv rA

1 1 124.4 mL 7.6 7.6 1 143.2 mLv v R R

2 1 143.2 mL 124.4 mL 18.8 mLv v v

2 18.8 mL 124.4 mL 0.1511 15.11%C v v

2

2

2.25 kW 60 s/min 1 000 N m/ kW s 0.001 kPa m / N724 kPa

0.044 m 0.002 827 m 3 000 rev/min 2 rev/work strokebp

724 kPa 0.72 1 005 kPai b mp p e

1 100 kPap

4 1 1.3

1 7.6 1 1005 kPa1 1.30 1 100 kPa 447 kPa

7.6 7.6 0.90

i

k

c

pRp k p

R R f

As in Example 16.1, we calculate the values:

X(%) v(mL) pc(kPa) pe(kPa)

0 18.8 1401 6266

5 25.0 966 4321

10 31.2 724 3238

15 37.5 572 2557

20 43.7 468 2094

25 49.9 394 1761

30 56.1 338 1512

35 62.3 295 1319

40 68.6 261 1165

45 74.8 233 1041

50 81.0 210 938

55 87.2 191 852

60 93.4 174 779

65 99.7 160 717



70 105.9 148 662

75 112.1 138 615

80 118.3 128 573

85 124.5 120 536

90 130.8 113 503

95 137.0 106 474

100 143.2 100 447

Then we sketch and round the following diagram:

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 20 40 60 80 100 120

Displacement X , %

Pre

ssu

re p

, kP

a



16.2 Construct a rounded indicator diagram for a four-cylinder, four-stroke gasoline engine

having a 85-mm bore, a 90-mm stroke, and a compression ratio of 6.25. The operating

conditions to be used are 22.4 kW at 1 900 rev/min. Use a mechanical efficiency of 72%,

a card factor of 0.90, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.

22 24 0.085 m 4 0.001 806 mA D

2 30.090 m 0.001 806 m 1 000 L/m 0.162 54 L 162.54 mLv A

1 1 162.54 mL 6.25 6.25 1 193.5 mLv v R R

2 1 193.5 mL 162.54 mL 30.96 mLv v v

2 30.96 mL 162.54 mL 0.1905 19.05%C v v

2

2

22.4 kW 60 s/min 1 000 N m/ kW s 0.001 kPa m / N8 704 kPa


8 704 kPa 0.72 12 090 kPai b mp p e

1 100 kPap

4 1 1.3

1 6.25 1 12 090 kPa1 1.30 1 100 kPa 4 719 kPa

6.25 6.25 0.90

i

k

c

pRp k p

R R f



0 31.0 1 083 51 102

5 39.1 800 37 744

10 47.2 626 29 526

15 55.3 509 24 019

20 63.5 426 20 100

25 71.6 364 17 186

30 79.7 317 14 944

35 87.8 279 13 172

40 96.0 249 11 741

45 104.1 224 10 564

50 112.2 203 9 580

55 120.4 185 8 748

60 128.5 170 8 036

65 136.6 157 7 420

70 144.7 146 6 883

75 152.9 136 6 411

80 161.0 127 5 994

85 169.1 119 5 622

90 177.2 112 5 289

95 185.4 106 4 990

100 193.5 100 4 719




0

5000

10000

15000

20000

25000

30000

35000

40000

0 10 20 30 40 50 60 70 80 90 100

Displacement X , %

Pre

ssu

re p

, kP

a



16.3 Construct an indicator diagram for a V6 four-stroke gasoline engine having a 100-mm

bore, a 90-mm stroke, and a compression ratio of 8.40. The engine develops 150 kW at

4 400 rev/min. Use a mechanical efficiency of 72%, a card factor of 0.88, a suction

pressure of 100 kPa, and a polytropic exponent of 1.30.

22 24 0.100 m 4 0.007 854 mA D

2 30.090 m 0.007 854 m 1 000 L/m 0.707 L 707 mLv A

1 1 707 mL 8.40 8.40 1 803 mLv v R R

2 1 803 mL 707 mL 96 mLv v v

2 96 mL 707 mL 0.1358 13.58%C v v

2

2

150 000 W 6 cyl 60 s/min 0.001 kPa m / N965 kPa


965 kPa 0.72 1 340 kPai b mp p e

1 100 kPap

4 1 1.3

1 8.40 1 1 340 kPa1 1.30 1 100 kPa 550 kPa

8.40 8.40 0.88

i

k

c

pRp k p

R R f



0 96 1 590 8 751

5 131 1 056 5 812

10 167 774 4 259

15 202 603 3 315

20 237 488 2 687

25 272 408 2 243

30 308 348 1 914

35 343 302 1 661

40 378 266 1 462

45 414 237 1 302

50 449 213 1 170

55 484 193 1 061

60 520 176 968

65 555 161 888

70 590 149 820

75 626 138 760

80 661 129 708

85 696 120 661

90 732 113 620

95 767 106 583

100 802 100 550



16.4 A single-cylinder, two-stroke gasoline engine develops 30 kW at 4 500 rev/min. The

engine has an 80-mm bore, a stroke of 70 mm, and a compression ratio of 7.0. Develop a

rounded indicator diagram for this engine using a card factor of 0.990, a mechanical

efficiency of 65%, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.

22 24 0.080 m 4 0.005 027 mA D

2 30.070 m 0.005 027 m 1 000 L/m 0.352 L 352 mLv A

1 1 352 mL 7.0 7.0 1 411 mLv v R R

2 1 411 mL 352 mL 59 mLv v v

2 59 mL 352 mL 0.1662 16.62%C v v

2

2

30 000 W 60 s/min 0.001 kPa m / N1 137 kPa


1 137 kPa 0.65 1 749 kPai b mp p e

1 100 kPap

4 1 1.3

1 7.0 1 1 749 kPa1 1.30 1 100 kPa 673 kPa

7.0 7.0 0.990

i

k

c

pRp k p

R R f


x(%) v(mL) pc(kPa) pe(kPa)

0 59 1 259 8 473

5 76 894 6019

10 94 682 4 592

15 111 546 3 672

20 129 451 3 034

25 147 382 2 569

30 164 329 2 217

35 182 288 1 942

40 199 256 1 722

45 217 229 1 542

50 235 207 1 394

55 252 188 1 268

60 270 173 1 162

65 287 159 1 070

70 305 147 991

75 323 137 921

80 340 128 860

85 358 120 805

90 375 112 756

95 393 106 712

100 411 100 673



16.5 The engine of Problem 16.1 has a connecting rod 80 mm long and a mass of 0.100 kg,

with the mass center 15 mm from the crankpin end. Piston mass is 0.175 kg. Find the

bearing reactions and the crankshaft torque during the expansion stroke corresponding to

a piston displacement of X = 30% ( 60t ). To find ,ep see the answer to Problem

16.1 in Appendix B.

0.080 m , 3 0.100 kgm , 4 0.175 kgm , 0.015 mA , 0.065 mB A ,

3 3 0.100 kg 0.065 m 0.080 m 0.081 25 kgA Bm m ,

3 3 0.100 kg 0.015 m 0.080 m 0.018 75 kgB Am m ,

3 000 rev/min 2 rad/rev 60s/min 314.16 rad/s , 0.022 mr ,

0.022 m 0.080 m 0.275r , 22 20.022 m 314.16 rad/s 2 171 m/sr ,

60t ,

2 2 2

cos 1 sin 0.022 m cos60 0.080 m 1 0.275sin 60 0.088 7 mx r r ,

30%X , 1 512 kPaep (from Prob. 16.1), 2 20.060 m 4 0.002 827 mA ,

21 512 kPa 0.002 827 m 4 274 NeP p A

2 2 2(cos cos 2 171 m/s cos60 0.275cos120 787 m/s2 )r

x r t t

2 22 2

2

0.275tan sin 1 sin 0.275sin 60 1 sin 60 0.244 9

2 2

r rt t

41 3 4

2

ˆtan

ˆ0.018 75 kg 0.175 kg 787 m/s 4 274 N 0.244 9

Bm m x P

F j

j

ˆ 1 013 N j Ans.

34 4 3 4

2

ˆ ˆtan

ˆ ˆ0.175 kg 787 m/s 4 274 N 1 013 N

Bm x P m m x P

F i j

i j

ˆ ˆ 4 136 1 013 N 4 258 N 13.8 i j Ans.

2 232 3 3 4 3 3 4

2

ˆ ˆ[ cos ] sin tan

ˆ ˆ0.081 25 kg 2 171 m/s cos60 4 274 N N153 1 013

A B A Bm r t m m x P m r t m m x P

F i j

i j

ˆ ˆ 4 186 1 166 N 4 345 N 164.4 i j Ans.

21 3 4ˆ ˆ ˆtan 0.088 7 m 1 013 N 89.82 N mBx m m x P T k k k Ans.



16.6 Repeat Problem 16.5, but do the computations for the compression cycle ( 660ωt ).

0.080 m , 3 0.100 kgm , 4 0.180 kgm , 0.010 mA , 0.070 mB A ,

3 3 0.100 kg 0.065 m 0.080 m 0.081 25 kgA Bm m ,

3 3 0.100 kg 0.015 m 0.080 m 0.018 75 kgB Am m ,


0.022 m 0.080 m 0.275r , 22 20.022 m 314.16 rad/s 2 171 m/sr ,

660t ,

2 2 2

cos 1 sin 0.022 m cos660 0.080 m 1 0.275sin 660 0.088 7 mx r r ,

30%X , 338 kPacp (from Prob. 16.1), 2 20.060 m 4 0.002 827 mA ,

2338 kPa 0.002 827 m 956 NcP p A

2 2 2(cos cos 2 171 m/s cos660 0.275cos1320 787 m/s2 )r

x r t t

2 22 2

2

0.275tan sin 1 sin 0.275sin 660 1 sin 660 0.230 36

2 2

r rt t

41 3 4

2

ˆtan

ˆ0.012 5 0.175 kg 787 m/s 956 N 0.230 36

Bm m x P

kg

F j

j

ˆ 186 N j Ans.

34 4 3 4

2

ˆ ˆtan

ˆ ˆ0.175 kg 787 m/s 956 N 186 N

Bm x P m m x P

F i j

i j

ˆ ˆ 818 186 N 839 N 12.8 i j Ans.

2 232 3 3 4 3 3 4

2

ˆ ˆ[ cos ] sin tan

ˆ ˆ0.081 25 kg 2 171 m/s cos660 1 105 N N153 186

A B A Bm r t m m x P m r t m m x P

F i j

i j

ˆ ˆ 1 017 339 N 1 071 N 161.5 i j Ans.

21 3 4ˆ ˆ ˆtan 0.088 7 m 3.02 N m34 NBx m m x P T k k k Ans.



16.7 Make a complete force analysis of the engine of Problem 16.5. Plot a graph of the

crankshaft torque versus crank angle for 720 of crank rotation.

0.080 m , 3 0.100 kgm , 4 0.175 kgm , 0.010 mA , 0.070 mB A ,

3 3 0.100 kg 0.065 m 0.080 m 0.081 25 kgA Bm m ,

3 3 0.100 kg 0.015 m 0.080 m 0.018 75 kgB Am m ,


0.022 m 0.080 m 0.275r , 22 20.022 m 314.16 rad/s 2 171 m/sr ,

2 20.060 m 4 0.002 827 mA , and/or e cp p p as taken from Prob. 16.1,

20.002 827 mP Ap p ,

2

2cos 1 sin 0.022 m cos 0.080 m 1 0.275sin

rx r t t

,

2 2(cos cos 2 171 m/s cos 0.275cos 22 )r

x r t t

2

2 2

2tan sin 1 sin 0.275sin 1 0.037 8sin

2

r rt t

14 3 4 tan

0.193 75 kg tan

BF m m x P

x P

21 3 4 14tanBT m m x P x F x

t, x, m X,% P, N 2, m/sx tan

14F , N 21T , N·m

0 0.102 0 14 276 -2 768 0 0 0

15 0.101 2.27 15 218 -2 614 0.071 36 1 048 106

30 0.098 8.43 10 115 -2 179 0.138 80 1 345 132

45 0.094 18.12 6 412 -1 535 0.198 13 1 211 114

60 0.089 30.23 4 249 -787 0.244 91 1 003 89

75 0.083 43.59 3 042 -45 0.275 00 834 69

90 0.077 57.01 2 326 597 0.285 40 697 54

105 0.071 69.47 1 888 1 078 0.275 00 577 41

120 0.067 80.23 1 615 1 384 0.244 91 461 31

135 0.063 88.83 1 444 1 535 0.198 13 345 22

150 0.060 95.03 1 340 1 582 0.138 80 229 14

165 0.059 98.76 1 283 1 580 0.071 36 113 7

180 0.058 100.00 1 264 1 574 0 0 0

195 0.059 98.76 1 264 1 580 -0.071 36 -112 -7

210 0.060 95.03 1 264 1 582 -0.138 80 -218 -13

225 0.063 88.83 1 264 1 535 -0.198 13 -309 -19

240 0.067 80.23 1 264 1 384 -0.244 91 -375 -25

255 0.071 69.47 1 264 1 078 -0.275 00 -405 -29



270 0.077 57.01 1 264 597 -0.285 40 -394 -30

285 0.083 43.59 1 264 -45 -0.275 00 -345 -29

300 0.089 30.23 1 264 -787 -0.244 91 -272 -24

315 0.094 18.12 1 264 -1 535 -0.198 13 -192 -18

330 0.098 8.43 1 264 -2 179 -0.138 80 -117 -11

345 0.101 2.27 1 264 -2 614 -0.071 36 -54 -5

360 0.102 0 774 -2 768 0 0 0

375 0.101 2.27 283 -2 614 0.071 36 -16 -2

390 0.098 8.43 283 -2 179 0.138 80 -19 -2

405 0.094 18.12 283 -1 535 0.198 13 -2 0

420 0.089 30.23 283 -787 0.244 91 32 3

435 0.083 43.59 283 -45 0.275 00 75 6

450 0.077 57.01 283 597 0.285 40 114 9

465 0.071 69.47 283 1 078 0.275 00 135 10

480 0.067 80.23 283 1 384 0.244 91 135 9

495 0.063 88.83 283 1 535 0.198 13 115 7

510 0.060 95.03 283 1 582 0.138 80 82 5

525 0.059 98.76 283 1 580 0.071 36 42 2

540 0.058 100.00 283 1 574 0 0 0

555 0.059 98.76 287 1 580 -0.071 36 -42 -2

570 0.060 95.03 300 1 582 -0.138 80 -84 -5

585 0.063 88.83 324 1 535 -0.198 13 -123 -8

600 0.067 80.23 361 1 384 -0.244 91 -154 -10

615 0.071 69.47 422 1 078 -0.275 00 -173 -12

630 0.077 57.01 521 597 -0.285 40 -181 -14

645 0.083 43.59 681 -45 -0.275 00 -185 -15

660 0.089 30.23 950 -787 -0.244 91 -196 -17

675 0.094 18.12 1 434 -1 535 -0.198 13 -226 -21

690 0.098 8.43 2 262 -2 179 -0.138 80 -256 -25

705 0.101 2.27 3 402 -2 614 -0.071 36 -207 -21

720 0.102 0 3 961 -2 768 0 0 0

T21 (N∙m) vs. ωt (deg.)



16.8 The engine of Problem 16.3 uses a connecting rod 300 mm long. The masses are

3 0.90Am kg, 3 0.30Bm kg, and 4 1.64m kg. Find all the bearing reactions and the

crankshaft torque for one cylinder of the engine during the expansion stroke at a piston

displacement of X = 30% ( 63.2t ). The pressure should be obtained from the

indicator diagram, Fig. AP16.3 in Appendix B.

0.300 m , 3 0.80 kgAm , 3 0.30 kgBm , 4 1.64 m kg ,


0.045 m 0.300 m 0.150r , 22 20.045 m 460.8 rad/s 9 554 m/sr ,

63.2t ,

2 2 2

cos 1 sin 0.045 m cos63.2 0.300 m 1 0.15sin 63.2 0.317 6 mx r r ,

30.0%X , 1 914 kPaep (from Prob. 16.3), 2 20.100 m 4 0.007 854 mA ,

21 914 kPa 0.007 854 m 15 033 NeP p A

2 2 2(cos cos 9 554 m/s cos63.2 0.150cos126.4 3 457 m/s2 )r

x r t t

2 22 2

2

0.15tan sin 1 sin 0.15sin 63.2 1 sin 63.2 0.135

2 2

r rt t

41 3 4

2

ˆtan

ˆ0.30 kg 1.64 3 457 m/s 15 033 N 0.135

Bm m x P

kg

F j

j

ˆ 1 124 N j Ans.

34 4 3 4

2

ˆ ˆtan

ˆ ˆ1.64 3 457 m/s 15 033 N 1 124 N

Bm x P m m x P

kg

F i j

i j

ˆ ˆ 9 364 1 124 N 9 431 N 6.8 i j Ans.

2 2

2 232 3 3 4 3 3 4

ˆ ˆ0.80 kg 9 554 m/s cos 63.2 8 326 N 0.80 kg 9 554 m/s sin 63.2 1 124 N

ˆ ˆ[ cos ] sin tanA B A Bm r t m m x P m r t m m x P

i j

F i j

ˆ ˆ 4 880 7 946 N 9 325 N 121.6 i j Ans.

21 3 4ˆ ˆ ˆtan 0.317 6 m 1 124 N 357 N mBx m m x P T k k k Ans.



16.9 Repeat Problem 16.8, but do the computations for the same position in the compression

cycle ( 656.8t ).

0.300 m , 3 0.90 Am kg , 3 0.30 Bm kg , 4 1.64 m kg ,


0.045 m 0.300 m 0.150r , 22 20.045 m 460.8 rad/s 9 554 m/sr ,

656.8t ,

2 2 2

cos 1 sin 0.045 m cos656.8 0.300 m 1 0.15sin 656.8 0.317 6 mx r r ,

30.0%X , 348 kPacp (from Prob. 16.3), 2 20.100 m 4 0.007 854 mA ,

2348 kPa 0.007 854 m 2 733 NcP p A

2 2 2(cos cos 9 554 m/s cos 656.8 0.150cos1313.62 ) 3 457 m/s

rrx t t

2 22 2

2

0.15tan sin 1 sin 0.15sin 656.8 1 sin 656.8 0.135

2 2

r rt t

41 3 4

2

ˆtan

ˆ0.30 kg 1.64 kg 3 457 m/s 2 733 N 0.135

Bm m x P

F j

j

ˆ 536 N j Ans.

34 4 3 4

2

ˆ ˆtan

ˆ ˆ1.64 kg 3 457 m/s 2 733 N 536 N

Bm x P m m x P

F i j

i j

ˆ ˆ 2 936 536 N 2 985 N 169.6 i j Ans.

2 2

2 232 3 3 4 3 3 4

ˆ ˆ0.90 kg 9 554 m/s cos 656.8 4 250 N 0.90 kg 9 554 m/s sin 656.8 536 N

ˆ ˆ[ cos ] sin tanA B A Bm r t m m x P m r t m m x P

i j

F i j

ˆ ˆ 7 850 7 139 N 10 611 N 42.3 i j Ans.

21 3 4ˆ ˆ ˆtan 0.317 6 m 536 N 170 N mBx m m x P T k k k Ans.



16.10 Additional data for the engine of Problem 16.4 are 3 110l mm, 3 15G AR mm,

4 0.24m kg, and 0.133m kg. Make a complete force analysis of the engine and plot

a graph of the crankshaft torque versus crank angle for 360 of crank rotation.

0.110 m , 3 0.13 m kg , 4 0.24 m kg , 0.015 mA , 0.095 mB A ,

3 3 0.13 0.095 m 0.110 m 0.112 A Bm m kg kg ,

3 3 0.13 0.015 m 0.110 m 0.018 B Am m kg kg ,

4 500 rev/min 2 rad/rev 60 s/min 471.24 rad/s , 0.035 mr ,

0.035 m 0.110 m 0.318r , 22 20.035 m 471.24 rad/s 7 772 m/sr ,

22 24 0.080 m 4 0.005 027 mA D , or e cp p p as taken from Prob. 16.4,

20.005 027 mP Ap p ,

2 22cos 1 sin 0.035 m cos 0.110 m 1 0.318sinx r rt t ,


x r t t

2

2 2


2

r rt t

14 3 4 tanBF m m x P

21 3 4

2

tan

0.005 027 m 2005 N cos 0.318cos 2 tan

BT m m x P x

p x



t, P, N 2, m/sx x, m tan

14F , N 21T , N·m

0 31 945 -10 245 0.145 00 0 0 0

15 35 849 -9 649 0.143 43 0.082 63 2 757 395.4

30 24 949 -7 968 0.138 91 0.161 10 3 688 512.3

45 16 102 -5 496 0.131 93 0.230 68 3 387 446.9

60 10 845 -2 650 0.123 24 0.286 02 2 906 358.2

75 7 812 130 0.113 74 0.321 86 2 525 287.2

90 6 022 2 473 0.104 28 0.334 29 2 226 232.2

105 4 943 4 153 0.095 62 0.321 86 1 936 185.1

120 4 263 5 123 0.088 24 0.286 02 1 597 140.9

135 3 831 5 496 0.082 43 0.230 68 1 211 99.8

150 3 567 5 495 0.078 29 0.161 10 803 62.9

165 3 429 5 366 0.075 82 0.082 63 398 30.2

180 1 943 5 299 0.075 00 0 0 0

195 510 5 366 0.075 82 -0.082 63 -157 -11.9

210 531 5 495 0.078 29 -0.161 10 -314 -24.6

225 568 5 496 0.082 43 -0.230 68 -458 -37.8

240 635 5 123 0.088 24 -0.286 02 -560 -49.4

255 733 4 153 0.095 62 -0.321 86 -581 -55.5

270 897 2 473 0.104 28 -0.334 29 -513 -53.5

285 1 160 130 0.113 74 -0.321 86 -384 -43.7

300 1 609 -2 650 0.123 24 -0.286 02 -265 -32.6

315 2 394 -5 496 0.131 93 -0.230 68 -225 -29.7

330 3 706 -7 968 0.138 91 -0.161 10 -266 -36.9

345 5 508 -9 649 0.143 43 -0.082 63 -249 -33.8

360 31 945 -10 245 0.145 00 0 0 0

-100

0

100

200

300

400

500

600

0 90 180 270 360 450

wt, deg

T2

1, N

.m

T21 (N∙m) vs. ωt (deg)



16.11 The four-stroke engine of Problem 16.1 has a stroke of 66 mm and a connecting rod

length of 183 mm. The mass of the rod is 0.386 kg, and the center of mass center is 42

mm from the crankpin. The piston assembly has mass of 0.576 kg. Make a complete

force analysis for one cylinder of this engine for 720 of crank rotation. Use 110 kPa for

the exhaust pressure and 70 kPa for the suction pressure. Plot a graph to show the

variation of the crankshaft torque with the crank angle. Use Fig. 16.23 for the pressures.

0.183 m , 3 0.386 kgm , 4 0.576 kgm , 0.042 mA , 0.141 mB A ,

3 3 0.386 kg 0.141 m 0.183 m 0.297 4 kgA Bm m ,

3 3 0.386 kg 0.042 m 0.183 m 0.088 6 kgB Am m ,


0.033 m 0.183 m 0.180r , 22 20.033 m 314.16 rad/s 3 257 m/sr ,

2 20.060 m 4 0.002 83 mA , or e cp p p as taken from Example 16.1,

20.002 83 mP Ap p ,

2

2cos 1 sin 0.033 m cos 0.183 m 1 0.180sin

rx r t t

,


x r t t

2

2 2


2

r rt t

14 3 4 tanBF m m x P

21 3 4 tanBT m m x P x

t, P, N 2, m/sx x, m tan

14F , N 21T ,N·m

0 14 279 -3 843 0.216 0 0 0

15 16 066 -3 654 0.215 0.046 64 636 136.8

30 10 688 -3 114 0.211 0.090 36 778 164.3

45 6 792 -2 303 0.205 0.128 31 675 138.4

60 4 445 -1 335 0.197 0.157 78 561 110.6

75 3 230 -335 0.189 0.176 49 531 100.3

90 2 432 586 0.180 0.182 92 516 92.9

105 1 976 1 351 0.172 0.176 49 507 87.2

120 1 649 1 922 0.164 0.157 78 462 75.7

135 1 461 2 303 0.158 0.128 31 384 60.6

150 1 357 2 528 0.154 0.090 36 274 42.3

165 1 288 2 638 0.151 0.046 64 142 21.4

180 1 263 2 671 0.150 0 0 0

195 1 263 2 638 0.151 -0.046 64 -137 -20.7

210 1 263 2 528 0.154 -0.090 36 -266 -40.9



225 1 263 2 303 0.158 -0.128 31 -358 -56.6

240 1 263 1 922 0.164 -0.157 78 -401 -65.7

255 1 263 1 351 0.172 -0.176 49 -381 -65.6

270 1 263 586 0.180 -0.182 92 -302 -54.4

285 1 263 -335 0.189 -0.176 49 -184 -34. 7

300 1 263 -1 335 0.197 -0.157 78 -59 -11.7

315 1 263 -2 303 0.205 -0.128 31 34 7.0

330 1 263 -3 114 0.211 -0.090 36 73 15.4

345 1 263 -3 654 0.215 -0.046 64 54 11.7

360 773 -3 843 0.216 0 0 0

375 283 -3 654 0.215 0.046 64 -100 -21.5

390 283 -3 114 0.211 0.090 36 -161 -34.1

405 283 -2 303 0.205 0.128 31 -160 -32.8

420 283 -1 335 0.197 0.157 78 -95 -18.8

435 283 -335 0.189 0.176 49 11 2.0

450 283 586 0.180 0.182 92 123 22.1

465 283 1 351 0.172 0.176 49 208 35.8

480 283 1 922 0.164 0.157 78 246 40.4

495 283 2 303 0.158 0.128 31 233 36.8

510 283 2 528 0.154 0.090 36 177 27.3

525 283 2 638 0.151 0.046 64 95 14.3

540 283 2 671 0.150 0 0 0

555 289 2 638 0.151 -0.046 64 -95 -14.4

570 305 2 528 0.154 -0.090 36 -179 -27.6

585 327 2 303 0.158 -0.128 31 -238 -37.6

600 371 1 922 0.164 -0.157 78 -260 -42.6

615 440 1 351 0.172 -0.176 49 -236 -40.6

630 543 586 0.180 -0.182 92 -171 -30.7

645 723 -335 0.189 -0.176 49 -88 -16.7

660 993 -1 335 0.197 -0.157 78 -17 -3.3

675 1 521 -2 303 0.205 -0.128 31 1 0.3

690 2 378 -3 114 0.211 -0.090 36 -28 -5.9

705 3 588 -3 654 0.215 -0.046 64 -54 -11.6

720 3 962 -3 843 0.216 0 0 0

-100

-50

0

50

100

150

200

0 90 180 270 360 450 540 630 720 810

T21, N

.m

wt, deg

T21 (N∙m) vs. ωt (deg)



Chapter 17

Balancing

17.1 Determine the bearing reactions at A and B for the system illustrated in Fig. P17.1 if the

speed is 350 rev/min. Determine the magnitude and the angular orientation of the

balancing mass if it is located at a radius of 50 mm.

1 25 mm,R 2 35 mm,R 3 40 mm,R 1 2 kg,m 2 1.5 kg,m 3 3 kg.m


22

1 1 1 2 kg 0.025 m 36.652 rad/s 67.168 NF m R

22

2 2 2 1.5 kg 0.035 m 31.416 rad/s 70.527 NF m R

22

3 3 3 3 kg 0.040 m 31.416 rad/s 161.204 NF m R

1ˆ67.168 N 90 67.168 N F j

2ˆ ˆ70.527 N 165 68.123 18.254 N F i j

3ˆ ˆ161.204 N 75 41.723 155.711 N F i j

1 2 3ˆ ˆ26.401 106.796 N 110.011 N 103.9 F F F F i j

Since all rotating masses are in a single plane, the correction mass must be in that plane.



ˆ ˆ0.200 m 110.011 N 103.9 1.000 mA B M k × k ×F 0

22.002 76.1 NB F Ans.

ˆ ˆ0.800 m 110.011 N 103.9 1.000 mA A M k × k ×F 0

88.008 76.1 NA F Ans.

110.011 N 76.1C F F

22 0.050 m 31.416 rad/s 110.011 NC C C CF m R m

22 110.011 N 0.050 m 31.416 rad/s 1.638 kgC C Cm F R

Ans.

76.1C Ans.



17.2 Figure P17.2 illustrates three weights connected to a shaft that rotates in bearings at A

and B. Determine the magnitude of the bearing reactions if the shaft speed is 350 rev/min.

A counterweight is to be located at a radius of 250 mm. Find the value of the weight and

its angular orientation.

150 mm 300 mm

1 200 mm,R 2 300 mm,R 3 150 mm,R

1 0.556 N,w 2 0.417 N,w 3 0.834 N.w


2

2

1 1 1 2

0.556 N 200 mm 36.652 rad/s15.486 N

9650 mm /F m R

s

2

2

2 2 2 2

0.417 N 300 mm 36.652 rad/s17.42 N

9650 mm/sF m R

2

2

3 3 3 2

0.834 N 150 mm 36.652 rad/s17.42 N

9650 mmF m R

1ˆ15.486 N 90 15.486 N F j

2ˆ ˆ17.42 N 135 12.317 12.317 N F i j

3ˆ ˆ17.42 N 30 15.0891 8.713 N F i j

1 2 3ˆ ˆ2.767 5.544 N 6.198 N 63.5 F F F F i j

Since all rotating masses are in a single plane, the correction mass must be in that plane.

6.198 lb 116.5C F F

22 250 mm 36.652 rad/s 6.198 NC C C CF m R m

2

22

6.198 N 9650 mm/s0.178 N

250 N 31.416 rad/s

CC

C

Fm

R Ans.

116.5C Ans.

Without the correction mass ˆ ˆ450 mm 300 mm A B M k ×F k × F 0

4.134 N 116.5B F Ans.

ˆ ˆ450 mm 150 mm B A M k F k F 0

2.064 N 116.5A F Ans.



17.3 Figure P17.3 illustrates two weights connected to a rotating shaft and mounted outboard

of bearings A and B. If the shaft rotates at 150 rev/min, what are the magnitudes of the

bearing reactions at A and B? Suppose the system is to be balanced by reducing a weight

at a radius of 125 mm. Determine the amount and the angular orientation of the weight

to be removed.

100 mm 50 mm

1 100 mm,R 2 150 mm,R 1 17.8 N,w 2 13.35 N.w


2

2

1 1 1 2

17.8 mm 100 mm 15.709 rad/s45.514 N

9650 mm/sF m R

2

2

2 2 2 2

17.8 N 150 mm 15.708 rad/s51.201 N

9650 mm/sF m R

1ˆ45.514 N 90 45.514 N F j

2ˆ ˆ51.201 N 135 36.205 36.205 N F i j

1 2ˆ ˆ36.205 9.309 N 37.384 N 165.6 F F F i j

ˆ ˆ150 mm 100 mm

ˆ ˆ ˆ ˆˆ1396.187 mm N 5430.78 mm N 100 mm

B A

x y

A AF F

M k × F k ×F

i j k × i j 0

ˆ ˆ54.307 13.959 lb 56.074 N 14.4A F i j Ans.

ˆ ˆ18.102 4.65 N 18.690 N 165.6B A F F F i j Ans.

The correction should be

37.384 N 165.6C F F

This can be done by mass removal at a radius of 125 mm of

2

22

37.384 N 9650 mm/s11.699 N

125 mm 15.708 rad/s

CC

C

Fm

R

Ans.

14.4C Ans.



17.4 For a speed of 250 rev/min, calculate the magnitudes and relative angular orientations of

the bearing reactions at A and B for the two-mass system illustrated in Fig. P17.4.

1 60 mm,R 2 40 mm,R 1 2 kg,m 2 1.5 kg.m


22

1 1 1 2 kg 0.060 m 26.180 rad/s 82.247 NF m R

22

2 2 2 1.5 kg 0.040 m 26.180 rad/s 41.123 NF m R

1ˆ82.247 N 90 82.247 N F j

2ˆ41.123 N 90 41.123 N F j

ˆ ˆˆ ˆ ˆ0.250 m 82.246 N 0.550 m 41.123 N 0.500 mB A M k × j k × j k ×F 0



17.5 The rotating system illustrated in Fig. P17.5 has 1 2 60 mm,R R 300 mm,a c

600 mm,b 1 21 kg, and 3 kg.m m Find the bearing reactions at A and B and their

angular orientations measured from a rotating reference mark if the shaft speed is 150

rev/min.


22

1 1 1 1 kg 0.060 m 15.708 rad/s 14.804 NF m R

22

2 2 2 3 kg 0.060 m 15.708 44.413 NF m R

1ˆ14.804 N 90 14.804 N F j

2ˆ44.413 N 90 44.413 N F j

ˆ ˆˆ ˆ ˆ0.300 m 14.804 N 0.900 m 44.413 N 1.200 mB A M k × j k × j k ×F 0

ˆ29.609 N 29.609 N 90.0A F j

1 2B A F F F F 0 Ans.



17.6 The rotating shaft illustrated in Fig. P15.5 supports two masses 1 2 and m m whose weights

are 17.8 N and 22.25 N, respectively. The dimensions are 1 100 mm,R 2 75 mm,R

50 mm,a 200 mm,b and 75 .c mm Find the magnitudes of the rotating-bearing

reactions at A and B and their angular orientations measured from a rotating reference

mark if the shaft speed is 350 rev/min.


2

2

1 1 1 2

17.8 N 100 mm 36.683 rad/s247.789 lb

9650 mm/sF m R

2

2

2 2 2 2

22.25 N 75 mm 36.683 rad/s232.303 N

9650 in/sF m R

1ˆ247.789 N 90 247.789 N F j

2ˆ232.303 N 90 232.303 N F j

1 2

ˆ ˆ ˆ50 mm 250 mm 325 mm

ˆ ˆ ˆ ˆˆ12389.57 mm N 58076.06 mm N 325 in

B A

x y

A AF F

M k ×F k ×F k ×F

i i k × i j 0

ˆ140.575 N 140.575 N 90A F j Ans.

1 2ˆ156.061 N 156.061 lb 90B A F F F F j Ans.



17.7 The shaft illustrated in Fig. P17.7 is to be balanced by placing masses in the correction

planes L and R. The weights of the three masses 1 2 3, , and m m m are 1.1125 N, 0.834 N,

and 1.39 N, respectively. The dimensions are 1 125 mm,R 2 100 mm,R 3 125 mm,R

25 mm,a 200 mm,b e 250 mm,c and 225 mm.d Calculate the magnitudes

of the corrections in N mm and their angular orientations.

1 1ˆ ˆ1.1125 N 125 in 139.062 N mmm R j j

2 2ˆ ˆ0.834 N 100 mm 150 83.437 N mm 150 72.256 41.718 N mmm R i j

3 3ˆ ˆ1.39 N 125 mm 60 173.828 N mm 60 86.914 150.542 N mmm R i j

Using Eqs. (17.6) and (17.7),

1 1ˆ450 mm 875 mm 71.519 N mmm R j

2 2ˆ ˆ675 mm 875 mm 55.743 32.186 N mmm R i j

3 3ˆ ˆ200 mm 875 mm 19.865 34.411 N mmm R i j

ˆ ˆ35.878 4.922 N mm 36.211 N mm 7.8L Lm R i j Ans.

ˆ ˆ50.535 58.121 N mm 77.019 N mm 131.0R Rm R i j Ans.



17.8 The shaft of Problem 17.7 is to be balanced by removing weight from the two correction

planes. Determine the corrections to be subtracted in ounce-inches and their angular

orientations.

36.211 N mm 172.2L Lm R Ans.

77.019 N mm 49.0R Rm R Ans.



17.9 The shaft illustrated in Fig. P17.7 is to be balanced by subtracting masses in the two

correction planes L and R. The three masses are 1 2 36 g, 8 g, and 5 gm m m . The

dimensions are 1 125 mm,R 2 150 mm,R 3 100 mm,R 25 mm,a 300 mm,b

600 mm,c 150 mm,d and 75 mm.e Calculate the magnitudes and angular

locations of the corrections.

1 1ˆ ˆ6 g 125 mm 750.000 g mmm R j j

2 2ˆ ˆ8 g 150 mm 150 1 200 g mm 150 1 039.230 600.000 g mmm R i j

3 3ˆ ˆ5 g 100 mm 60 500 g mm 60 250.000 433.013 g mmm R i j

Using Eqs. (17.6) and (17.7),

1 1ˆ900 mm 1 125 mm 600.000 g mmm R j

2 2ˆ ˆ1 050 mm 1 125 mm 969.948 560.000 g mmm R i j

3 3ˆ ˆ300 mm 1 125 mm 66.667 115.470 g mmm R i j

The masses to be removed are: ˆ ˆ903.281 75.470 g mm 906.428 g mm 175.2L Lm R i j Ans.

ˆ ˆ114.051 207.543 g mm 236.816 g mm 61.2R Rm R i j Ans.



17.10 Repeat Problem 17.9 if masses are to be added in the two correction planes.

ˆ ˆ903.281 75.470 g mm 906.428 g mm 4.8L Lm R i j Ans.

ˆ ˆ114.051 207.543 g mm 236.816 g mm 118.8R Rm R i j Ans.



17.11 Solve the two-plane balancing problem as stated in Section 17.8.

This is an experimental procedure and is explained in Section 17.8; no further solution

process is shown here.

17.12 A rotor to be balanced in the field yielded an amplitude of 5 at an angle of 142 at the

left-hand bearing and an amplitude of 3 at an angle of 22 at the right-hand bearing

because of unbalance. To correct this, a trial mass of 12 was added to the left-hand

correction plane at an angle of 210 from the rotating reference. A second run then gave

left-hand and right-hand responses of 8 160 and 4 260 , respectively. The first trial

mass was then removed and a second mass of 6 added to the right-hand correction plane

at an angle of 70 . The responses to this were 2 74 and 4.5 80 for the left- and

right-hand bearings, respectively. Determine the original unbalances.

5 142A X , 3 22B X ,

12 210L m , 8 160AL X , 4 260BL X ,

6 70R m , 2 74AR X , 4.5 80BR X .

These gave the following results from a programmable calculator run:

6.05 234L M , 5.98 65.2R M Ans.



Chapter 18

Cam Dynamics

18.1 In Fig. P18.1a, the mass m is constrained to move only in the vertical direction. The

circular cam has an eccentricity of 50 mm, a speed of 25 rad/s, and the weight of the

mass is 26.7 N. Neglecting friction, find the angle t at the instant the cam jumps.

F W F my

my mg F

0 1 cosy y t 2

0 cosy y t 2

0 cosF m y t mg

Jump begins when F = 0; that is, when 2

0 cos 0m y t mg

2

22

0

9650 mm/scos 0.308 8

25 rad/s 50 mm

mgt

m y

1cos 0.308 8 107.99t Ans.



18.2 In Fig. P18.1a, the mass m is driven up and down by the eccentric cam and it has a

weight of 8 lb. The cam eccentricity is 0.75 in. Assume no friction.

(a) Derive the equation for the contact force.

(b) Find the cam velocity corresponding to the beginning of the cam jump.

(a) From the solution to Problem 18.1, we have for the contact force 2

0 cosF m y t mg Ans.

(b) Jump begins when cos 1t and F = 0: that is, when 2

0 0m y mg

2

0

386 in/s22.69 rad/s

0.75 in

g

y Ans.



18.3 In Fig. P18.1a, the slider has a mass of 2.5 kg. The cam is a simple eccentric and causes

the slider to rise 30 mm with no friction. At what cam speed in revolutions per minute

will the slider first lose contact with the cam? Sketch a graph of the contact force at this

speed for 360 of cam rotation.

From Problem 18.1, we have for the contact force 2

0 cosF m y t mg

Jump begins when cos 1t and F = 0; that is, when 2

0 0m y mg

2

0

9.81 m/s18.08 rad/s 172.7 rev/min

0.030 m

g

y Ans.



18.4 The cam-and-follower system illustrated in Fig. P18.1b has 0.9 kN / mk ,

0.80 kg,m 15 15 cos mm,y t and 60 rad / .s The retaining

spring is assembled with a preload of 2.4 N.

(a) Compute the maximum and minimum values of the contact force.

(b) If the follower is found to jump off the cam, compute the angle t corresponding

to the very beginning of jump.

(a) Let Fc = contact force, and P = preload.

cF F ky P my cmy ky P F

0.015 0.015cos 60 my t , 254cos60 m/sy t

20.80 kg 54cos60 m/s 900 N/m 0.015 0.015cos60 m 2.4 N

15.9 29.7cos60 N

cF t t

t

,max 15.9 29.7 N 45.6 NcF , ,min 0cF Ans.

(b) Jump begins when Fc = 0; that is, when

160 cos 15.9 N 29.7 N 122.37t Ans.



18.5 Figure P18.1b illustrates the mathematical model of a cam-and-follower system.

The motion machined into the cam is to move the mass to the right through a

distance of 50 mm with parabolic motion in 150 of cam rotation, dwell for 30 ,

return to the starting position with simple harmonic motion, and dwell for the

remaining 30 of cam angle. There is no friction or damping. The spring rate is

7.12 N/mm, and the spring preload is 26.7 N corresponding to the 0y position.

The weight of the mass is 160.2 N.

(a) Sketch a displacement diagram showing the follower motion for the entire 360

of cam rotation. Without computing numerical values, superimpose graphs of the

acceleration and cam contact force onto the same axes. Show where jump is most

likely to begin.

(b) At what speed in revolutions per minute would jump begin?

(a) Just as in Problem 18.4, if we let Fc = contact force and P = preload:


Using first-order kinematic coefficients and assuming that the input shaft speed is

constant, then 2y y and

2 2160.2 9650 mm/s 7.12 N/mm 26.7 NcF N y y

Going through the different phases of the motion defined above, we can sketch

the approximate curve shown for the cam contact force

This sketch shows that jump is very possible at point A 75t or point B

180t or point C 330t , the three points where the contact

force drops discontinuously, depending on whether is large enough for the

contact force to indicate a negative value.

(b) For point A 75t , 50 mmL , 150 2.618 rad . From Eq. (6.6a),

25y mm and, from Eq. (6.6c), 21.167 in/rady . Therefore,



2 2

,

2 2

160.2 N 9650 mm/s 7.12 N/mm 26.7 N

0.485 N s 204.7 N

c AF y y

Thus, , 0c AF for 20.559 rad/s

For point B 180t , 50 mmL , 150 2.618 rad . From Eq.

(6.12a), 50 mmy and, from Eq. (6.21c), 236 mm/rady . Therefore,

2 2

,

2 2

160.2 N 9650 mm/s 7.12 N/mm 26.7 N

0.485 N s 382.7 N

c BF y y

Thus, , 0c BF for 25.308 rad/s

For point C 330t , 0y y , and , 26.7 Nc CF for all values of .

Of these cases, jump begins at A when 20.559 rad/s 196.3 rev/min . Ans.



18.6 A cam-and-follower mechanism is illustrated in abstract form in Fig. P18.1b. The cam is

cut so that it causes the mass to move to the right a distance of 25 mm with harmonic

motion in 150 of cam rotation, dwell for 30 , and then return to the starting position in

the remaining 180 of cam rotation, also with harmonic motion. The spring is assembled

with a 20-N preload and it has a rate of 4.25 kN/m. The follower mass is 18 kg.

Compute the cam speed in revolutions per minute at which jump would begin.

Just as in Problem 18.4, if we let Fc = contact force and P = preload:


Using first-order kinematic coefficients and assuming that the input shaft speed is

constant, then 2y y and

218 kg 4 250 N/m 20 NcF y y

Going through the different phases of the motion defined above shows that jump is most

likely at the transition from the dwell to the full-return simple-harmonic motion since, at

that position, y and Fc suddenly drop. For that position 180t , 0.025 mL ,

180 3.1416 rad . From Eq. (6.15c), 0.025 my and 20.0125 m/rady .

Therefore,

2

2 2

18 kg 4 250 N/m 20 N

0.630 N s 132 N

cF y y

Thus, 0cF for 23.688 rad/s 226.2 rev/min. Ans.



18.7 Figure P18.7 illustrates a lever OAB driven by a cam cut to give the roller a rise of 37.5

mm with parabolic motion and a parabolic return with no dwells. The lever and roller are

to be assumed weightless, and there is no friction. Calculate the jump speed if

125 l mm

Taking moments about the fixed pivot

2

2 2O AM F mg m

4 2AF m mg 500 mmAF m mg

Going through the different phases of the motion defined above shows that jump is most

likely at the transition from the concave to the convex parabolic rise motion since, at that

position, y and Fc suddenly drop. For that position 90t , 37.5 mmL ,

180 3.1416 rad . From Eqs. (6.6a) and (6.6c), 18.75 mmy and 215.2 mm/rady . Therefore,

2 2 2 215.2 mm/rad 125 mm 0.121 6y

2 2

2 2

500 mm 0.1216 9650 mm/s

60.8 mm 9650 mm/s

AF m m

m m

Thus, 0AF for 12.6 rad/s 120.3 rev/min. Ans.



18.8 A cam-and-follower system similar to the one in Fig. 18.6 uses a plate cam driven at a

speed of 600 rev/min and employs simple harmonic rise and parabolic return motions.

The events are rise in 150 , dwell for 30 , and return in 180 . The retaining spring has

a rate k = 14 kN/m with a precompression of 12.5 mm. The follower has a mass of 1.6

kg. The external load is related to the follower motion y by the equation

0.325 10.75 ,F y where y is in meters and F is in kilonewtons. Dimensions

corresponding to Fig. 18.6 are R = 20 mm, r = 5 mm, 60 mm,Bl and 90 mm.Cl

Using a rise of L = 20 mm and assuming no friction, plot the displacement, cam-shaft

torque, and radial component of the cam force for one complete revolution of the cam.


For simple harmonic rise motion, we use Eqs. (6.12) with 0.020 mL and 150 .

For the first part of the parabolic return motion, following Example 6.1,

2

0.020 1 2 my

, 0.080 my , 20.080 0.008 106 my

For the second part of the parabolic return motion,

2

0.040 1 my , 0.080 1 my , 20.080 0.008 106 my

Then we can use Eq. (18.11)

2

23 325 10 750 14 000 0.0125 1.6 N

500 3 250 6 317 N

yF y y y

y y

and Eqs. (18.9) and (18.13)

tana y y

12 23 23tan y yT a F y F



t , deg y, m y , m/s y , m/s2

23

yF , N 12T , N·m

0 0 0 0.008 106

0.014 400

551.2

591.0

0

15 0.000 489 0.003 708 0.013 695 588.1 -2.181

30 0.001 910 0.007 053 0.011 650 579.8 -4.089

45 0.004 122 0.009 708 0.008 464 566.9 -5.503

60 0.006 910 0.011 413 0.004 450 550.6 -6.284

75 0.010 000 0.012 000 0 532.5 -6.390

90 0.013 090 0.011 413 -0.004 450 514.4 -5.871

105 0.015 878 0.009 708 -0.008 464 498.1 -4.836

120 0.018 090 0.007 053 -0.011 650 485.2 -3.422

135 0.019 511 0.003 708 -0.013 695 476.9 -1.768

150 0.020 000 0 -0.014 400

0

474.0

565.0

0

165 0.020 000 0 0 565.0 0

180 0.020 000 0 0

-0.008 106

565.0

513.8

0

195 0.019 722 -0.002 122 -0.008 106 512.9 1.088

210 0.018 889 -0.004 244 -0.008 106 510.2 2.165

225 0.017 500 -0.006 366 -0.008 106 505.7 3.219

240 0.015 556 -0.008 488 -0.008 106 499.4 4.239

255 0.013 056 -0.010 610 -0.008 106 491.2 5.212

270 0.010 000 -0.012 732 -0.008 106

0.008 106

481.3

583.7

6.128

7.430

285 0.006 944 -0.010 610 0.008 106 573.8 6.088

300 0.004 444 -0.008 488 0.008 106 565.6 4.801

315 0.002 500 -0.006 366 0.008 106 559.3 3.561

330 0.001 111 -0.004 244 0.008 106 554.8 2.355

345 0.000 278 -0.002 122 0.008 106 552.1 1.172

360 0 0 0.008 106

0.014 400

551.2

591.0

0



18.9 Repeat Problem 18.8 with the speed of 900 rev/min, 0.110 10.75 kN,F y where y is

in meters, and the coefficient of sliding friction is 0.025.


For simple harmonic rise motion, we use Eqs. (6.12) with 0.020 mL and 150 .

For the first part of the parabolic return motion, following Example 6.1,

2

0.020 1 2 my

, 0.080 my , 20.080 0.008 106 my

For the second part of the parabolic return motion,

2

0.040 1 my , 0.080 1 my , 20.080 0.008 106 my


2

23

110 10 750 14 000 0.0125 1.6 N

1 1.666 667 0.033 333 tan sgn

285 24 750 14 212 N

1 1.666 667 0.033 333 tan sgn

yy y y

Fy y

y y

y y

and Eqs. (18.9) and (18.13)

tana y y





23

yF , N 12T , N·m

0 0 0 0.008 106

0.014 400

400

490

0

15 0.000 489 0.003 708 0.013 695 498 -1.85

30 0.001 910 0.007 053 0.011 650 509 -3.59

45 0.004 122 0.009 708 0.008 464 521 -5.05

60 0.006 910 0.011 413 0.004 450 533 -6.08

75 0.010 000 0.012 000 0 545 -6.54

90 0.013 090 0.011 413 -0.004 450 556 -6.35

105 0.015 878 0.009 708 -0.008 464 565 -5.49

120 0.018 090 0.007 053 -0.011 650 572 -4.04

135 0.019 511 0.003 708 -0.013 695 576 -2.13

150 0.020 000 0 -0.014 400

0

575

780

0

165 0.020 000 0 0 780 0

180 0.020 000 0 0

-0.008 106

780

665

0

195 0.019 722 -0.002 122 -0.008 106 660 1.40

210 0.018 889 -0.004 244 -0.008 106 641 2.72

225 0.017 500 -0.006 366 -0.008 106 608 3.87

240 0.015 556 -0.008 488 -0.008 106 562 4.77

255 0.013 056 -0.010 610 -0.008 106 529 5.61

270 0.010 000 -0.012 732 -0.008 106

0.008 106

428

664

5.45

8.45

285 0.006 944 -0.010 610 0.008 106 586 6.22

300 0.004 444 -0.008 488 0.008 106 522 4.43

315 0.002 500 -0.006 366 0.008 106 471 3.00

330 0.001 111 -0.004 244 0.008 106 433 1.84

345 0.000 278 -0.002 122 0.008 106 410 0.87

360 0 0 0.008 106

0.014 400

400

490

0



18.10 A plate cam drives a reciprocating roller follower through the distance L = 31.25 mm

with parabolic motion in 120 of cam rotation, dwells for 30 , and returns with cycloidal

motion in 120 , followed by a dwell for the remaining cam angle. The external load on

the follower is 14 160.2F N during the rise and zero during the dwells and the return.

In the notation of Fig. 18.6, R = 75 mm, r = 25 mm, 150 mm,Bl

200 mm, andCl 26.7 N/mm.k The spring is assembled with a preload of 166.875 N

when the follower is at the bottom of its stroke. The weight of the follower is 8 N, and

the cam velocity is 140 rad/s. Assuming no friction, plot the displacement, the torque

exerted on the cam by the shaft, and the radial component of the contact force exerted by

the roller against the cam surface for one complete cycle of motion.

For 0 60 , we use Eqs. (6.5a) (6.5c) with 31.25 mmL and 120 .

2

62.5 mmy , 59.675 mmy , 28.5 mmy

For 60 120 , we use Eqs. (6.6a) – (6.6c) with 31.25 mmL and 120 .

2

31.25 1 2 1 mmy

, 59.675 1 mmy , 28.5 mmy

For 150 270 , we use Eqs. (6.13) with 31.25 mmL and 120 .


23 14 166.875 687.5 406.63 NyF F y y

and Eqs. (18.9) and (18.13)

tana y y





23

yF , N 12T , N·m

0 0 0 0

1.139 863

37.5

177.7

0

15 0.039 063 0.298 416 1.139 863 183.5 54.8

30 0.156 250 0.596 831 1.139 863 201.1 120.0

45 0.351 563 0.895 247 1.139 863 230.4 206.3

60 0.625 000 1.193 662 1.139 863

-1.139 863

271.4

63.1

324.0

75.3

75 0.898 438 0.895 247 -1.139 863 104.1 93.2

90 1.093 750 0.596 831 -1.139 863 133.4 79.6

105 1.210 938 0.298 416 -1.139 863 151.0 45.1

120 1.250 000 0 -1.139 863

0

156.8

225.0

0

135 1.250 000 0 0 225.0 0

150 1.250 000 0 0 225.0 0

165 1.234 424 -0.174 808 -1.266 070 107.0 -18.7

180 1.136 444 -0.596 831 -1.790 493 44.4 -26.5

195 0.921 924 -1.018 854 -1.266 070 60.1 -61.2

210 0.625 000 -1.193 662 0 131.3 -156.7

225 0.328 076 -1.018 854 1.266 070 202.4 -206.2

240 0.113 556 -0.596 831 1.790 493 218.1 -130.2

255 0.015 576 -0.174 808 1.266 070 155.5 -27.2

270 0 0 0 37.5 0

285 0 0 0 37.5 0

300 0 0 0 37.5 0

315 0 0 0 37.5 0

330 0 0 0 37.5 0

345 0 0 0 37.5 0

360 0 0 0

1.139 863

37.5

177.7

0



18.11 Repeat Problem 18.10 if friction exists with 0.04 and the cycloidal return takes place

in 180 .


2

62.5 mmy , 59.675 mmy , 28.5 mmy


2

31.25 1 2 1 mmy

, 59.675 1 mmy , 28.5 mmy

For 150 330 , we use Eqs. (6.13) with 31.25 mmL and 180 .


14

23

166.875 667.5 406.63 N

1 5.6 16.8 tan sgn

y F y yF

y y

and Eqs. (18.9) and (18.13)

tana y y





23

yF , N 12T , N·m

0 0 0 0

1.139 863

38

178

0

15 0.039 063 0.298 416 1.139 863 185 55

30 0.156 250 0.596 831 1.139 863 204 122

45 0.351 563 0.895 247 1.139 863 235 211

60 0.625 000 1.193 662 1.139 863

-1.139 863

278

65

332

77

75 0.898 438 0.895 247 -1.139 863 135 95

90 1.093 750 0.596 831 -1.139 863 152 80

105 1.210 938 0.298 416 -1.139 863 152 45

120 1.250 000 0 -1.139 863

0

157

225

0

135 1.250 000 0 0 225 0

150 1.250 000 0 0 188 0

165 1.245 305 -0.053 307 -0.397 887 188 -10

180 1.213 957 -0.198 944 -0.689 161 157 -31

195 1.136 444 -0.397 887 -0.795 775 136 -54

210 1.005 624 -0.596 831 -0.689 161 127 -76

225 0.828 639 -0.742 468 -0.397 887 127 -94

240 0.625 000 -0.795 775 0 133 -106

255 0.421 361 -0.742 468 0.397 887 139 -104

270 0.244 376 -0.596 831 0.689 161 139 -83

285 0.113 556 -0.397 887 0.795 775 129 -51

300 0.036 043 -0.198 944 0.689 161 106 -21

315 0.004 695 -0.053 307 0.397 887 75 -4

330 0 0 0 38 0

345 0 0 0 38 0

360 0 0 0

1.139 863

38

178

0



Chapter 19

Flywheels, Governors, and Gyroscopes

19.1 Table P19.1 lists the output torque for a one-cylinder engine running at 4 500 rev/min.

(a) Find the mean output torque.

(b) Determine the mass moment of inertia of an appropriate flywheel using

0.018s

C .

Table P19.1 Torque data for Problem 19.1

degi

N mi

T

degi

N mi

T

degi

N mi

T

degi

N mi

T

0 0 180 0 360 0 540 0

10 17 190 -344 370 -145 550 -344

20 812 200 -540 380 -150 560 -540

30 963 210 -576 390 7 570 -577

40 1 016 220 -570 400 164 580 -572

50 937 230 -638 410 235 590 -643

60 774 240 -785 420 203 600 -793

70 641 250 -879 430 490 610 -893

80 697 260 -814 440 424 620 -836

90 849 270 -571 450 571 630 -605

100 1 031 280 -324 460 814 640 -379

110 1 027 290 -190 470 879 650 -264

120 902 300 -203 480 785 660 -300

130 712 310 -235 490 638 670 -368

140 607 320 -164 500 570 680 -334

150 594 330 -7 510 576 690 -198

160 544 340 150 520 540 700 -56

170 345 350 145 530 344 710 -2

(a) Using n = 72 and h = 4π/72, we enter the data from Table P19.1 into Simpson’s

rule to find 2 1 890.7 N mU U .

2 1 4 890.7 N m 4 rad 70.88 N mmT U U Ans.

(b) 4 500 rev/min 471.24 rad/s

22

2 1

2890.7 N m 0.018 471.24 rad/s 0.223 N m s

sU U CI

Ans.



19.2 Using the data of Table 19.2, determine the moment of inertia for a flywheel for a two-

cylinder 90 V engine having a single crank. Use Cs = 0.010 and a nominal speed of

4 600 rev/min. If a cylindrical or disk-type flywheel is to be used, what should be the

thickness if it is made of steel and has an outside diameter of 250 mm? Use = 7.8

Mg/m3 as the density of steel.

Table 19.2 Torque data for a four-cylinder, four-cycle internal combustion engine

degi

T

m 180

T

m

+360

N

T

m 540

N

T

m

Ntotal

T

m

0 0 0 0 0 0

15 311.5 -11.9 -9.5 -11.9 278.2

30 232.52 -22.9 -13.9 -22.9 172.7

45 270.3 -31.1 -9.9 -32.5 196.8

60 240.3 -35.9 0.9 -39.5 165.8

75 204.7 -34.5 14 -41.3 142.9

90 176.8 -26.9 26.9 -40.3 136.6

105 134.6 -14 34.5 -34.7 120.4

120 118.6 -0.9 35.9 -30.3 123.4

135 89.3 9.9 31.1 -30.5 99.9

150 59.2 13.9 22.9 -60.9 35

165 20.5 9.5 11.9 -84.5 -42.7

Using n = 48 and h = 4π/48, we integrate the data from columns 2-5 of Table 19.2 by

Simpson’s rule to find 2 1 394.32 N mU U .

4 600 rev/min 481.71 rad/s

22

2 1

2394.32 N m 0.0100 481.71 rad/s 0.1699 kg m

sU U CI

22 22 2 0.1699 kg m 0.350 m 2.7744 kgm I R

3 3/ 2.7744 kg 7 800 kg/m 0.000 356 mV m

230.000 356 m 0.350 m 0.000924 m 0.924 mmt V A

Ans.

19.3 Using the data of Table 19.1, find the mean output torque and the flywheel inertia

required for a three-cylinder in-line engine corresponding to a nominal speed of 2 400

rev/min. Use Cs = 0.03.

. Table 19.1 Example 19.1: Torque data for Fig. 19.3

degi

N.i

T

m

degi

N.i

T

m

degi

.i

T

N m

degi

.i

T

N m

degi

.i

T

N m

0 0 150 59.2 300 -0.9 450 26.9 600 -39.5

15 311.5 165 20.5 315 9.9 465 34.5 615 -41.3

30 232.5 180 0 330 13.9 480 35.9 630 -40.3

45 270.3 195 -11.9 345 9.5 495 31.1 645 -34.7

60 240.32 210 -22.9 360 0 510 22.9 660 -30.3

75 204.7 225 -31.1 375 -9.5 525 11.9 675 - 30.5

90 176.8 240 -35.9 390 -13.9 540 0 690 -60.9



105 134.6 255 -34.5 405 -9.9 555 -11.9 705 -84.5

120 118.6 270 -26.9 420 0.9 570 -22.9

135 89.3 285 -14 435 14 585 -32.5

Using n = 48 and h = 4π/48, we integrate the data from Table 19.1 by Simpson’s rule to

find 2 1 388.27 NmU U .

2 1 4 388.27 N 4 rad 30.89 NmT U U m m Ans.

2 400 rev/min 251.3 rad/s

22

2 1

2388.27 N.m 0.03 251.3 rad/s 0.2 m s

sU U CI

Ans.



19.4 The load torque required by a 200-tonne punch press is displayed in Table P19.4 for one

revolution of the flywheel. The flywheel is to have a nominal angular velocity of 2 400

rev/min and to be designed for a coefficient of speed fluctuation of 0.075.

(a) Determine the mean motor torque required at the flywheel shaft and the motor

horsepower needed, assuming a constant torque-speed characteristic for the

motor.

(b) Find the moment of inertia needed for the flywheel.

Table P19.4 Torque data for Problem 19.4

degi

mi

T

degi

mi

T

degi

N mi

T

degi

mi

T

0 95.3 90 877.5 180 200.3 270 95.3

10 95.3 100 925.3 190 181.2 280 95.3

20 95.3 110 944.3 200 162.2 290 95.3

30 95.3 120 953.8 210 152.6 300 95.3

40 95.3 130 934.8 220 124 310 95.3

50 143.2 140 858.5 230 114.5 320 95.3

60 286.1 150 391 240 104.9 330 95.3

70 572.3 160 238.5 250 95.3 340 95.3

80 763 170 219.4 260 95.3 350 95.3

(a) Using n = 36 and h = 2π/36, we integrate the data from Table P19.4 by Simpson’s

rule to find 1857.875 N mU .

2 1857.875 N m 2 rad 295.7mT U Nm Ans.

2 400 rev/min 251.3 rad/s

295.7 N m 2 400 rev/min 2 rad/rev6 073 HP

734.25 N m/min/HPP T

Ans.

(b) The torque data show a constant requirement of 95.34 N·m, probably friction, in

addition to the torque for the punching operation. If this constant torque is

subtracted from the data in the table (for speed fluctuation), and the integration

repeated, then we get 1658.18 N mU

2 2 21658.18 N 0.075 251.3 rad/s 0.349 N ss

U CI m

Ans.



19.5 Find Tm for the four-cylinder engine whose torque displacement is that of Fig. 19.4.

N M

333.75

222.5

111.25

Table 19.2 Torque data for a four-cylinder, four-cycle internal combustion engine

degi

m

T

180

N m

T

+360

N m

T

540

N m

T

mtotal

T

0 0 0 0 0

15 311.5 -11.9 -9.5 -11.9 278.2

30 232.5 -22.9 -13.9 -22.9 172.7

45 270.3 -31.1 -9.9 -32.5 196.8

60 240.3 -35.9 0.9 -39.5 165.7

75 204.7 -34.5 14 -41.3 142.9

90 176.8 -26.9 26.9 -40.3 136.6

105 134.6 -14 34.5 -34.7 120.4

120 118.6 -0.9 35.9 -30.3 123.4

135 89.3 9.9 31.1 -30.5 99.9

150 59.2 13.9 22.9 -60.9 35

165 20.5 9.5 11.9 -84.5 -42.7

Using n = 12 and h = π/12, we integrate the data from column 6 of Table 19.2 by

Simpson’s rule to find 2 1 388.3 N mU U .

2 1 388.3 N m rad 123.6 NmT U U m Ans.



19.6 A pendulum mill is illustrated schematically in Fig. P19.6. In such a mill, grinding is

done by a conical muller that is free to spin about a pendulous axle that, in turn, is

connected to a powered vertical shaft by a Hooke universal joint. The muller presses

against the inner wall of a heavy steel pan, and it rolls around the inside of the pan

without slipping. The weight of the muller is W = 436 N; its principal mass moments of

inertia are 213.46 m ssI N and 29.79 N m sI . The length of the muller axle is

1

1000 mmn

GA i ii

l R X Y

and the radius of the muller at its center of mass is

250 mmGB

R . Assuming that the vertical shaft is to be inclined at 30 and will be

driven at a constant angular velocity of 240 rev/minp

, find the crushing force

between the muller and the pan. Also determine the minimum angular velocity p

required to ensure contact between the muller and the pan.

240 rev/min 25.133 rad/sp 4/3ˆ ˆsin coss s ω ω i j

3ˆ ˆ25.133 rad/sp ω j j 4 3 4/3

ˆ ˆsin coss p s ω ω ω i j

3

ˆ ˆ ˆ1000sin 1000cos

ˆ1000sin

G GA

p

p

V ω R

j i j

k

4

ˆ ˆ ˆ ˆsin cos 10cos 10sin

ˆ250 250cos

s p s

G GB

s p

i j i j

V ω R

k

Equating these with 30 we find 4sin cos 2.866s p p . Then



2 2 2ˆ ˆ ˆ ˆ ˆsin cos sin 4sin sin cos ˆ1.433p s p sp s p p j× i j k kω ×ω k

22 2 22 4361 N 9650 mm/s 250 mm 2 564.7 N m sSI mr

2 22 2 2 24 4361 9650 mm/s 250 4 1000 444.755 sI mr ml N mm mm Nm

Now Eq. (19.36) shows

cosps s

p ss

I I I

M

2 2 2 ˆ4sin sin cos

cos14.12 430.64 N m s

4sin cos

p

pp

kM

2 2 ˆ4sin cos14.12sin 430.64sin cos N s p

m kM

2 2 ˆ ˆ10946.555173.29 N m s N m p

k kM

Now formulating the externally applied moments,

ˆ ˆGA c BAW F M j×R i ×R

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4361 N sin cos 1000 mm sin cos 1000 mm cos sin 250 mmcF M j× i j i × i j i j

ˆ ˆ4361 200sin N m 4cos sin 250 nn cF M k k

1ˆ ˆ2180.5 N m 741 mm cFn

M k k

and setting the two expressions equal

ˆ10946.555ˆ ˆ2180.5 N m 741 mm N m cF kM k k

we can now solve for the crushing force

14769.55 kNcF Ans.

If we start before setting the angular velocity, then we have 2 2 ˆ ˆ ˆ173.29 N m s 2180.5 N m 741.025 mm p cF kM k k

Now by setting Fc to zero we can determine the minimum angular velocity p required to

ensure contact between the muller and the pan: 22180.5 N m 173.29 N m s 3.547 rad/s 33.87 rev/minp Ans.



19.7 Use the gyroscopic formulae of this chapter to solve again the problem presented in

Example 14.9 of Chapter 14.

From the given data we can identify

2ˆ5 rad/sp ω ω k 3

ˆ ˆ350 5 rad/ss ω ω i k

22 24.5 kg 0.050 m 0.011 3 kg msI mk

From Eq. (19.37)

2 ˆ ˆˆ ˆ0.011 3 kg m 5 rad/s 350 5 rad/s 19.8 N ms

p sI M ω ×ω k i k j

This brings us precisely to the formulation of Eq. (2) of Example 14.9. From there on the

solution procedure, and the results, are identical. Notice how much more simply this

approach can be accomplished. Q.E.D.



19.8 The oscillating fan illustrated in Fig. P19.8 precesses sinusoidally according to the

equation sin1.5p

t , where 30 ; the fan blade spins at ˆ1800 rev/mins i . The

weight of the fan and motor armature is 23.36 N, and other masses can be assumed

negligible; gravity acts in the ˆj direction. The principal mass moments of inertia are

27.23 N mm ssI and 22.78 N mm sI ; the center of mass is located at

100 mmGC

R to the front of the precession axis. Determine the maximum moment

zM that must be accounted for in the clamped tilting pivot at C.

ˆ ˆ1.5 sin cos rad/sp ω i j ˆ ˆ1800 rev/min 188.5 rad/ss i i

2 2ˆ ˆ ˆ ˆ1.5 sin cos rad/s 188.5 rad/s 282.7cos rad/s ˆ244.9 rad/sp s i j i kω ω k

cosps s

p ss

I I I

M

2 2 21.5 rad/s ˆ7.23 mm s 4.45 N mm s cos 244.9 rad/s188.5 rad/s

1.77 NN m

kM

On the other side of the equation, the external moments are

ˆ ˆˆ ˆ ˆ ˆ ˆsin cos 100 mm cos 2.02 N z z z

GCM W M W M m M k R i j k k k k

Now, equating the two we can solve for the moment zM .

ˆ ˆ ˆ2.02 N m 1.77 NzM m k k k ˆ0.24 N mz M k Ans.

Here we see that the gyroscopic moment is almost large enough to support the weight of



the fan motor.

19.9 The propeller of an outboard motorboat is spinning at high speed and is caused to precess

by steering to the right or left. Do the gyroscopic effects tend to raise or lower the rear of

the boat? What is the effect and is it of noticeable size?

In this case sω refers to the angular velocity of the propeller and is directed fore or aft

depending on the direction of rotation. pω is vertical and refers to the angular velocity

of the turn. The moment required to maintain the turn is proportional to p sω ω as

shown in Eq. (19.36) or (19.37) and this axis is lateral on the boat. Therefore the moment

(or its reaction) can tend to raise or lower the rear of the boat. The direction depends on

both the direction of rotation of the engine and the direction of the turn. Eq. (19.37)

shows that the effect is likely to be very small since, for any reasonable rate of turn,

p sω ω will be at least an order of magnitude smaller than 2

s , which is the order of

the usual accelerations of the engine. In a very extreme case, a knowledgeable person

might be able to detect this moment, but most would not. It would never be a danger.



19.10 A large and very high-speed turbine is to operate at an angular velocity of

18 000 rev/min and will have a rotor with a principal mass moment of inertia of 225 N m ssI . It has been suggested that because this turbine will be installed at the

North Pole with its axis horizontal, perhaps the rotation of the earth will cause gyroscopic

loads on its bearings. Estimate the size of these additional loads.

ˆ ˆ18 000 rev/min 1885 rad/ss ω i i

ˆ ˆ1.0 rev/day 0.0000115 rad/sp ω j j

2ˆ ˆ ˆ0.0000115 rad/s 1885 rad/s 0.022 rad/sp s ω ω j i k

2 2ˆ ˆ25 N m s 0.022 rad/s 0.5461 N ms

p sI ω ω k k Ans.

Thus, if the bearings were separated by only 125es, they would experience less than one

additional pound of loading. This is totally negligible.

THEORY OF MACHINES AND MECHANISMS

Documents

Transcript of THEORY OF MACHINES AND MECHANISMS