ENGR:2750 Mech. Def. Bodies 11.1The University of Iowa

Period #11: STATICALLY INDETERMINATE SHAFTS

A. REVIEW

B. STATICALLY INDETERMINATE SHAFTS

Too many support conditions and not enough equations of static equilibrium.

Or too many redundant members to carry the applied loads.

Must generally augment conditions of static equilibrium with knowledge of how the indeterminate system deforms. This is called the kinematic constraint.

Can use the principle of linear superposition to solve these systems.

Fig. 11.1

max

max

Torsion Formula: ;

Maximum shear strain:

Incremental angle of twist:

Tc T

J J

Tc dc

GJ dx

Td dx

GJ

The University of IowaENGR:2750 Mech. Def. Bodies 11.2

From statics: TA+TB=T

Kinematic constraint: B/A=0

/

/ /

0

=T

B A

C A B C

A AC B BC

A AC B BC

BCA B

AC

T L T L

JG JG

T L T L

LT

L

Fig. 11.2 (a) Indeterminate shaft fixed at both ends; and (b) the free-body diagram.

AT

BT(a) (b)

1

T =

1

A B

BCB

AC

ACB

BC

AC

BCA

T T T

LT T

L

LT T

L L

L

LT T

L

The University of IowaENGR:2750 Mech. Def. Bodies 11.3

Alternative Solution Using Principle of Superposition:

T

A

B

CCase 1

TB

A

B

Case 2

/ /1 2

B

Combining the cases:

0

= +

B A B A

AC

ACB

TL T L

JG JG

LT T

L

B/ 2

Case 2:

AB BB A

T L T L

JG JG

/ 1

Case 1:

ACB A

TL

JG

BCA

LT T

L

ENGR:2750 Mech. Def. Bodies 11.4The University of Iowa

C. SHAFTS OF SOLID, NON-CIRCULAR CROSS-SECTIONS (TEXTBOOK, SECTION 5.6)

To this point, we’ve been concerned with shafts that have circular cross-sections.

What about shafts with non-circular cross-sections?

Consider the torsion of a rubber shaft with a square cross-section shown below (Fig. 11.3):

Note that the white square drawn on the square shaft “warps”. Some edges that were originally straight, become curved.

(a) undeformed square shaft (b) deformed square shaft when subjected to a torque

Figure 11.3 Twisting of rubber shaft with square cross-section

The University of IowaENGR:2750 Mech. Def. Bodies 11.5

With solid non-circular cross-sections, the shear stress distribution over the cross-section cannot vary linearly with radial distance from the center (Fig. 11.4a). This leads to warping of the shaft cross-section (Fig. 11.4b).

(a) (b)

Fig. 11.4 Noncircular shaft response to torsion

The University of IowaENGR:2750 Mech. Def. Bodies 11.6

Although the torsion formulae do not apply to shafts with non-circular cross-sections, the relevant properties of three sections are as indicated in the table below.

Table 11.1. Properties of non-circular sections.

The University of IowaENGR:2750 Mech. Def. Bodies 11.7

D. THIN-WALLED CLOSED SECTION SHAFTS (TEXTBOOK 5.7)

Though you will not be responsible for this material, be aware that thin-walled, closed cross-sections can be treated using the concept of shear flow.

Fig. 11.5. Examples of thin-walled closed sections

constant = shear flow

h = moment arm of segment ds about section centroid

ds = infinitesimal length of wall segment.

T qhds

q t

E. EXAMPLESExample 11.1: The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T=50lb.ft is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear and maximum shear strain in the brass and steel. Take

ksiGst )10(5.11 3 ksiGbr )10(6.5 3

,

The University of IowaENGR:2750 Mech. Def. Bodies 11.8

Along the length of the shaft the internal torque is T=50∙lbft.

In the composite section of the shaft, T=Tst + Tbr =-50∙lbft.

Kinematic constraint in the composite section CB:

/ /C B C Bst br

st BC br BC

st br

T L T L

JG JG

4 4 3

4 3

2

2

1 .5 11.5 102

.5 5.6 102

16,935 =T 30.8

550

stst br br

br

br br

in in ksiJGT T T

JGin ksi

k inT

k in

30.8 50

1.57

48.43

br br

br

st

T T lb ft

T lb ft

T lb ft

/

4 6 2

600 36.001196

11.5 102

BA BAB A

AB

lbin inT Lrad

JGin lb in

/ 6 2

48.43 12 24

16.935 10

.000824

BC stC B

st

T LBC lb in in

JG lb in

rad

/ / / .00202C A B A C B rad

max

4 4 4

48.43 12 lb in 1In BC segment 395

1 .52

st st

steelst

inT cpsi

Jin

max

4

1.57 12 lb in .5In BC segment 96

.52

br br

brbr

inT cpsi

Jin

The University of IowaENGR:2750 Mech. Def. Bodies 11.9

Example 11.2. The two 3-ft-long shafts are made of 2014-T6 aluminum. Each has a diameter of 1.5in. and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts about their axes. If a torque of 600 lb-ft is applied to the top gear as shown, determine the maximum shear stress in each shaft.

TA

fFE

E

fEF TB

F

Statics:

Gear E:

600

=T

600 =T 2

A E FE

EA B

F

A B

lb ft T r f

rT

r

lb ft T

BFE EF

F

Tf f

r

Gear F: BEF

F

Tf

r

Kinematic constraint:

E E F Fr r

/

36"0

A

E A E A

T

JG

/

36"0

B

F B F B

T

JG

36" 36"

1

2

E E F F

A B

E F

FA B B

E

r r

T Tr r

JG JG

rT T T

r

1600 2

2

240 2.88

120 1.44

B B

B

A

lb ft T T

T lb ft k in

T lb ft k in

max

4

1.44 .752.17

.752

AE

AE

k in inT cksi

Jin

max

4

2.88 .754.34

.752

BF

BF

k in inT cksi

Jin

The University of IowaENGR:2750 Mech. Def. Bodies 11.10

Example 11.3: The A-36 steel shaft is made from two segments: AC has a diameter of 0.5in and CB has a diameter of 1 in. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 60lb.in/in along segment CB, determine the absolute maximum shear stress in the shaft.

Solve by method of linear superposition

Case 1: distributed torque-loading shown, free at support Ato find A1.

Case 2: Concentrated torque TA at A to find A2.

Solve for TA such that A1+A2=0

Case 1:

1 60 20 1200BT lb in in lb in

( ) 1200 60T x x

2 1202

1

0

2

4 6 2

1200 20 60 201200 60

12,000 =- .01063

.5 11.5 102

A

BC BC

xdx

JG JG

lb inrad

in lb in

Case 2:TB

TA

2

4 46 2

2 2

5 1 1

3 20

5 20 =

11.5 10 .25 .5

= 8.857 10

AA

st AC BC

A

A

T in in

G J J

T in in

lb in in in

lb in T

2 1

5 1 1

8.857 10 .01063

120

A A

A

A

lb in T rad

T lb in

1200 120

T =1080 lb in

B

B

T lb in lb in

max 4

2

max 4

2

1080 .55.50

.5

120 .254.89

.25

BC

AC

lb in inksi

in

lb in inksi

in