5/15/2013 1 Mechanical Engineering Design III MEEN :3393 LOAD AND STRESS ANALYSIS Equilibrium and...

5/15/2013

Dr. Sathyan Krishnan

Dept. of Mechanical Engineering MACHINE DESIGN

College of Engineering, PMU

Mechanical Engineering Design IIIMEEN :3393

Dr. Sathyan KrishnanRoom- S044

Ph: 8498532, Email: skrishnan@pmu.edu.auDepartment of Mechanical EngineeringPrince Mohammed Bin Fahd University

LOAD AND STRESS ANALYSIS

Equilibrium and Free-Body DiagramsEquilibriumIf a system to be studied is motionless under the action of forces, it is said to be in equilibrium.For equilibrium, the sum of the forces and moments acting on the system must be equal to zero

Free-Body DiagramSimple line diagrams which represent various sections of a complex system, that is used for the purpose of study and analysis.

Load and stress analysis

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Advantages:1. Easy to study and analyze complicated systems by breaking down to segments2. Easy way of communicating thoughts to others3. It simplifies the thinking process4. Helps to understand the dimensions and various forces and moments acting on

the system

Cartesian Stress Components

τxy(τx)net

σ = normal stressτ = shear stress

Note: double subscripts are necessary to represent shear stress. The first subscriptindicate the direction of the surface normal and second subscript is the direction of theshear stress.

σxσx

Shear and Moment

F1 F3F2

Sign conventions

Force -ve +ve

-ve+ve

+ve -ve

Moment

Assume that a portion of the beam is cut at a distance xfrom left end and removed. To ensure equilibrium of thecut section, an internal shear force V and bendingmoment M should act on the section.• The shear force is obtained by summing the forces on

the left of the cut section.• Bending moment is the sum of the moments caused

by the forces to the left of the section

Shear force and Bending moment are related by:

In case of distributed load

dV d M=

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Normal Stresses in Flexure

The bendingstress is given by :

where,c = y , M = Bending Moment

and I = Moment of inertia of the section

MIt can also be written as =

Iwhere Z = is called the section modulus

compression

tension

Neutral axis

Problem-1

A beam having a T-section is subjected to a bending moment 1600N-m. Locate the neutralaxis and maximum tensile and compressive bending stress. The sign of the momentproduces tension at the top surface. The dimensions are in mm

Solution

The bendingstress

Given Bending Moment M = 1600 N - m

To calculate the compressive and tensile bending

stress, we need I and c(i.e c and c sincetop side

of the section is in tension and bottom sideis in

compression)

To calculate the moment of inertia I of the section, we divide the section into two rectangles

and calculate the moment of inertia of each. Then apply the parallel-axis theorem to find I.

First locate the centre of gravity. Since the section is symmetrical about the Y-axis, the c.g. lie

on the Y-axis.

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Area of the section = 0.075 x 0.012 + 0.088 x 0.012 = 0.001956 m2

Tolocatec , take sumof moments of areas about top edge

0.0880.001956×c = 0.012×0.075 0.006 + 0.012×0.088 ( +0.012)2

Now calculate the second moment of area of each rectangle about its own centroidal axis

bhi.e. I =

bhand I =

0.000005400 +0.000059136c = = 0.033m

0.001956

c = 0.100 -0.033 = 0.067 m

Now apply the parallelaxis theorem toobtain thesecond moment ofarea of the

composite figure

xx cg cg

I = I + Ad I second moment of area about its own centroid axis

I second moment of area about xx - axis , A= area,

d = distance between cg of the section and each rectangle

0.01.0800 10 m

0.06.81472 10 m

0.075 12

0.012 88

0.033 0.006 0.027 m

0.067 0.044 0.023m

-8 2 -7 2

Now Nowapply the parallel axis theorem twice

I = 1.08 10 + 0.075 0.012 0.027 6.81472 10 + 0.088 0.012 0.023

1.0800 10 m

6.81472 10 m

7 6 6 46.669 10 1.24 10 1.9069 10 m

M cNowmaximumtensile stress =

M cNowmaximumcompressive stress =

27688919 N / m or Pa10

27.68MPa

56216896 N / m or Pa10

56.21MPa

1600 0.033

1.9069

1600 0.067

1.9069

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Class Exercise-1For the section given in the figure, find the second moment of inertia, location of theneutral axis and the distance from the neutral axis to the top and bottom surfaces.Suppose a bending moment of 80 kN.m is applied, find the resulting stresses at the top,bottom and every abrupt change in cross section.

Solution

1. Divide the figure into three sections and calculate the area of each

section and total area

2. Then locate the cg by equating the moment of areas

3. Calculate the moment of inertia of each section , I1, I2 and I3

4. Apply the parallel axis theorem to calculate the second moment of area

of the composite figure

5. Then calculate the stress at A, B, C and D using

Area of 1= 2 = 3 = 0.04 0.006 = 0.00024m

total area of the section 0.00024

0.00072m

Tolocatec , take sumof moments of areas about bottomedge

0.00072×c = 0.00024×0.02 + 0.00024×0.015

0.0000132c = = 0.01833m

0.00072

c = 0.040 -0.01833 = 0.02166 m

Now apply the parallelaxis theorem toobtain thesecond moment ofarea of the

composite figure

Now calculate the moment of inertia of each rectangle about its own centroidal axis

bhi.e. I I =

bhand I =

+ Ad I second moment of area about its own centroid axis

I second moment of area about xx - axis , A= area, d = depth

0.02 0.01833 0.00167m

0.01833 0.015 0.00333m

from the figure, d d

3.2 10 m

7.2 10 m

0.006 0.04

0.04 0.006

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8 4 10 42 3.2 10 m 7.2 10 m2 2 2 2

Now Nowapply the parallel axis theorem to the sections

I = + 0.00024m 0.00167 + 0.00024m 0.00333

6.87 10 m

6.87 10

Now calculate the stresses at variuos points

M 80000N.m 0.01933mstress at A, =

80000N.m 0.01933m -0.012mMstress at B, =

80000N.m 0.01933m - 0.012m+0.006mMstress at C, =

6.87 10 m

DM 80000N.m 0.02066mstress at D, =

8 9 8 46.53 10 3.38 10 6.87 10 m

2.28 10 N / m

22.8GPa

8.069 10 N / m

8.06GPa

4.99 10 N / m

4.99GPa

2.382 10 N / m

23.82GPa

Problem-2

Determine the diameter for the solid round shaft, 450mm long. The shaft is supportedby self aligning bearings at the ends. Mounted upon the shaft are a V-belt sheave,which contributes a radial load of 1880N on the shaft and a gear, which contributes aradial load of 670N. The loads are in the same plane and have the same direction. Thebending stress is not to exceed 70 Mpa.

Solution

Assumptions1. The weight of the shaft is neglected2. The shaft is assumed simply supported and the loads and bearings reactions to be

concentrated3. The normal bending stress is assumed govern the design at the location of the

maximum bending moment

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R1=1402N R2=1148N

1880N 670N

150 200

Free-body diagram

To find the reactions R1 and R2, take moments about C

M 450R 300 1880 100 670N 0

R 1402 N

Also, R R 1880 N 670N 2550 N

R 2550 N 1402 N 1148 N

Calculate the shear force and draw the shear force diagram BA C

478NShear forceat O 1402 N

Shear forceat A 1402 N 1880N 478N

Shear forceat B 478N 670N 1148 N

Shear forceat D 1148 N 1148 N 0

Calculate the bending moment and draw the bending moment diagram

Bending Moment at O 0 N.mm

Bending Moment at A 1402.2 N 150mm 210300 N.mm

Bending Moment at B 1402.2 N 350mm 1880N 200mm 114700 N.mm

Bending Moment at C 1402.2 N 450mm 1880N 300mm 670N 100mm 0

210300114700

Shear force diagram

Bending moment diagram

First draw the free body diagram of the system

We know the maximum bending moment occurs where shear force changes its sign. Here, the shear force changes its sign at the point A.

M 1402.2N 150mm

210330 N.mm 210.33N.m

Maximum bending moment

MWe know bending stress σ =

πdThe section modulus I c = = 0.0982 d

Given, max.bending stress σ =70 MPa = 70×10 Pa or N/m

substituting M and I c ,

210300114700

Shear force diagram

Bending moment diagram

210.33 N.m 70×10 N/m =

0.0982 d

210.33d = = 0.03127 m = 31.27mm

70×10 0.0982

diameter of the shaft should be32mm

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This problem can also be solved by vector method.Vector method is useful for three-dimensionalproblems. We define unit vector triad i j k along x, yand z axes respectively. Then the position vectors fromthe origin O to the loading points A, B, and C are:rA = 150i rB = 350i rC = 450iAlso, the force vectors areR1 = R1j FA = -1880j FB = -670j R2 = R2j

1880N 670N

150 200

Free-body diagram

0 1A A B B C 2

x y z x y z

y z z y z x x

Taking moments about O yields the vector equation

The eq.(1) is called thecross products. Thecross product of two vectors

A A A B B B

A B A B A B A

r F r F r R

A i j k B i j k

A B = i +

282000

z x y y x

A A B B

B A B A B

but this relation is more conveniently viewed as determinant

We compute the terms in Eq.(1) as foloows

150 0 0

0 -1880 0

i j k i

r F = k, r F = 234500 C 2 2

350 0 0 450 0 0 450R

0 -670 0 0 R 0

j k i j k

k, r R = k

1142.8N

Hence 1142.8 N

A A B B C 2

Substituting these terms into eq.(1) and solvinf the resulting algebraic equation

-282000 - 234500 450R R

We know R R

r F r F r R

k k + k = 0

1880 670 1142.8 1402.2N

1 A B 2

R F + F R

Next step is to drawthe shear force and bending moment diagram as done previously

We know the maximum bending moment occurs where shear force changes its sign.

Here, the s

M 1402.2N 150mm

210330 N.mm 210.33N.m

hear force changes its sign at the point A.

Maximum bending moment

MWe know bending stress σ =

πdThe section modulus I c = = 0.0982 d

Given, max.bending stress σ =70 MPa = 70×10 Pa

or N/m

210.33 N.msubstituting M and I c , 70×10 N/m =

0.0982 d

210.33d = = 0.03127 m = 31.27mm

70×10 0.0982

diameter of the shaft should be32mm

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Shear Stress in Beams

Most beams have both shear forces and bending moment present. It is very rarely weencounter beams with pure bending. The bending equation is developed by assuming purebending. To avoid complications due to shear force we generally assume pure bending. Forengineering purposes, the bending formula is valid no matter whether a shear force ispresent or not.

• Assume bending moment on the near sideof the cut section as M and the far side asM+dM.

• The moment produces a normal stress σ onthe near side and σ +d σ on the far side asshown.

• These normal stress produced by normalforces on the vertical faces (FN and FF). Thecompressive force on the far side is greaterthan near side.

• The resultant of these forces try to slide thesection towards –X direction. So thisresultant must be balanced by a shear force(FB) in +X direction. This shear force result ina shear stress τ

Consider a small area dA on the near side. The stress acting on this area is σ. The force on the area is σ dA.

The force acting ontheentire near faceis

F = σ dA

M y M y Musing σ = F = dA ydA

The force acting ontheentire far faceis

M +dMF = σ +dσ dA ydA

Shear force F at thebottomof thesection= Shear stress τ ×shear area(b dx)

Summing all these forces in the x - direction gives

M +dM M dMF F F ydA ydA ydA

Now substituting F τ bdx, we get

dM 1 dM V τ= ydA but = V the shear stress τ = ydA

dx Ib dx Ib

in this equation,theintegral is the moment of inertia of thevertical face about the neutral axis and this is usually

designated as Q. Q = y

with this simplification, the shear stress

VQτ =

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Problem-4

A beam 1.6 m long is to support a load of 250 kN at a distance of 0.4 m from the leftsupport. Basing the design only on bending stress, the designer has selected 75mmaluminium channel with cross section shown in the figure. If the direct shear is neglected,the stress in the beam may be actually higher than the designer thinks. Determine theprincipal stresses considering the bending and direct shear and compare them with thatconsidering bending only.

I= 6.63x10-7 m4

I/c= 1.88x10-5 m3

First draw the free-body diagram

Solution

0.4m 1.2m

Find the reactions at the supports O and B, take moments about B.

M 1.6m R 1.2M 250kN 0

R 187.5kN

Also, R R 250kN

R 250kN 187.5k N 62.5kN

Calculate the shear force and draw the shear force diagram

Shear forceat O 187.5k N

Shear forceat A 187.5k N 250kN 62.5kN

Shear forceat B 0

187.5kN

62.5kN

Bending Moment at O 0 N.mm

Bending Moment at A 187.5k N 0.4m 75kN.m

Bending Moment at B 187.5k N 1.6m 250kN 1.2m 0

Calculate the bending moment and draw the bending moment diagram

75kN.m

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0.4m 1.2m

187.5kN

62.5kN

75kN.m

The maximum bending stressMc

44.24 10 N / m 4.24GPa

75000N.m 0.0375σ =

6.63 10

However, the maximum stress due to thecombined bending and direct shear stressesmay be maximum at the point (0.4−, 0.068) thatis just to the left of the applied load, where theweb joins the flange. The normal stress atsection ab, with x = 0.4m, is

My 75000N.m×0.034mσ = - = - = 3.846×10 N/m

I 6.63×10 m

= 3.846 GPa

For the shear stress at section ab, considering the area above ab and using Q = y area above ab

0.007Q = 0.034+ × 0.035×0.007

= 9.18×10 m

The shear stress can be calculated fromVQ

τ= - I b

Where, V = 187.5 kN, I = 7.1x10-7 m4 , Q = 9.18x10-6 m3, and b = 0.005m yields

187.5kN×9.18×10 mτ= -

6.63×10 m × 0.005m

The principal stresses at the point can now be determined. Using the following equation, we find that at x = 0.4− m, y = 0.034m,

000.519

x y x y

1 2 =2

3.8463.846

518853.63-9

0.00172= - = - =

3.3315×10- 0.519GPa

03.967 1.923 1.99

0.067, 3.913

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Shear Stresses in Standard-Section Beams

The shear stress distribution in a beam depends on how Q/b varies as a function of y1

Shear stress

• Consider a portion of a beam with arectangular cross section, subjected to ashear force V and a bending moment M.

• As a result of the bending moment, anormal stress σ is developed on a crosssection A-A, which is in compression abovethe neutral axis and in tension below.

• To investigate the shear stress at a distancey1 above the neutral axis, we select anelement of area dA at a distance y abovethe neutral axis. i.e., d A = b dy

• We know, in the equation for shear stress,

by b= b y dy c - y

y= y dAQ

Substitute for dA

Substituting this value for Q into Eq. for gives 2 2

V= c - y

We know the second moment of area for a rectangular section, = 3bh

substituting h = 2c and A = bh = 2bc gives

3 23 2b 2c 4 2bc c8bc AcI

12 12 12 3

Now if we use this value of I in equation for and rearrange

2 2 11 2

yV V= c - y 1 -

2 2 c2

It shows that when y1 is zero (at the neutral axis), shear stress is maximum.

Therefore, Maximum shear stress for rectangular cross section is given by

3Vτ =

2AAs we move away from the neutral axis, the

shear stress decreases parabolically until it is

zero at the outer surfaces where y1 = ± c

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Torsion

Any moment vector that is collinear with an

axis of a mechanical element is called a

torque vector, because the moment causes the

element to be twisted about that axis.

Shear stress in different cross section beams

The angle of twist, in radians, for a solid round bar is

where T = torque, l = length, G = modulus of rigidity

J = polar second moment of area

T lθ =

The torque (T) will cause shear stresses () develop throughout

the cross section.

For a round bar in torsion, these stresses are proportional to the

radius ρ and are given by

T ρτ =

JThis equation applies only to circular sections. For a solid round section

4 44o iπ d dπd

J = and for hollowcircular sections J = 32 32

The torque T can be obtained from a consideration of the power and speed of a rotating shaft

H = (in F P S system) H is in horse power, T in lbs.in, shaft speed rpm 63025

H = Tω, H = power in Watts, T = torque in N - m, ω= angular velocity in rad/s

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Shear stress in rectangular cross section bars

Maximum shearing stress in a rectangular b × c section bar occurs in the middle of the

longest side b and is of the magnitude

max 2 2

where b is the longer side, c the shorter side, and a factor that is a function of the ratio

b/c as shown in the following

T Tτ =

α b c b c

The angle of twist is given by

3where β is a function of b/c, as shown

T lθ =

β b c in the

G table

A pin in a knuckle joint carrying a tensile load F deflects somewhat on account of thisloading, making the distribution of reaction and load as shown in part b of the figure. Theusual designer’s assumption of loading is shown in part c; others sometimes choose theloading shown in part d. If a = 0.5 in, b = 0.75 in, d = 0.5 in, and F = 1000 lbf, estimate themaximum bending stress and the maximum shear stress due to V for each approximation.

Problem-5

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Solution - Case-1

A B C D

• First draw the free-body diagram of the first case showing all forces and dimensions

• Calculate the shear force and bending moment and draw the SF and BM diagrams

• Calculate the moment of inertia of the section• Calculate the bending stress and shear stress

using standard formulae

Distance AB= (a/2)+(b/4) = (0.5/2)+(0.75/4)= 0.4375 in

Distance BC= (b/2) = 0.75/2 = 0.375 inDistance AD= (a+b) = 0.5 + 0.75 = 1.25 in

Since the loading is symmetrical, the support reactions are equal, i.e., RA = RD = 1000/2 lbf

Bending moment at A = 0

Bending moment at B = 500 0.4375 = 218.75 lbf - in

Bending moment at C = 500 0.4375+0.375 - 500 0.375

= 218.75 lbf - in

Bending moment at D =

500 lbf

218.75 lbf-in

Mcσ =

4Vτ =

a = 0.5 in, b = 0.75 in, d = 0.5 in,

F = 1000 lbf

π dMoment of inertia of the pin, I =

π dArea of the pin, I =

McBending stress σ =

4V Maximum shear stress τ =

Solution - Case-2

• In this case the load is applied as uniformly distributed.

• First draw the free-body diagram of the first case showing all forces and dimensions

• Calculate the shear force and bending moment and draw the SF and BM diagrams

• Calculate the moment of inertia of the section• Calculate the bending stress and shear stress using

standard formulae

π 0.5= = 3.068×10 in

π 0.5=0.1963 in

218.75×(0.25)= = 17 825 psi = 17.8 kpsi

3.068(10.3)

4×500= = 3400 psi

3×0.1963

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500 lbf

Since the loading is symmetrical, the support reactions are equal, i.e., RA = RD = 1000/2 lbf

Bending moment at A = 0

Bending moment at B = 500 0.25 = 125 lbf - in

Maximum Bending moment occurs at the point E,

where shear force chnges its direction

1Bending momentat E = 500 0.25 + 500 0.3

= 218.75 lbf - in

Bending moment at D =

Here, it is seen that the maximum Bending moment and Shear forec are same as that of

previous case. Therefore, the

bending stress and maximum shear stress also will be same.

218.75 lbf-in

Problem-6

The 1.5-in-diameter solid steel shaft shownin Figure is simply supported at the ends.Two pulleys are keyed to the shaft wherepulley B is of diameter 4.0 in and pulley C isof diameter 8.0 in. Considering bending andtorsional stresses only, determine thelocations and magnitudes of the greatesttensile, compressive, and shear stresses inthe shaft.

Solution

First of all, draw the free body diagram of the system showing all the forces and moments

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Although this is a three-dimensional problem and vectors might seem appropriate, wewill look at the components of the moment vector by performing a two-plane analysis(since we have forces in the vertical and horizontal directions).

Now we can generate two dimensional diagrams of the system. First draw the xyplane as viewed from the z direction (1). Then the xz plane as viewed from the ydirection (2).

(1) (2)

Draw the shear force and moment diagram for both the planes as follows

o ox x

20004000

The net moment on a section is the vector sum of the components. That is,

At point B

At point c

Thus the maximum bending moment is 8246lbf · in and the maximum bending stress atpulley B is

The maximum torsional shear stress occurs between B and C and is

The maximum bending and torsional shearstresses occur just to the right of pulley B atpoints E and F as shown in Figure.

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At point E, the maximum tensile stress will be σ1 given by

At point F, the maximum compressive stress will be σ2 given by

The extreme shear stress also occurs at E and F and is

Closed Thin-Walled Tubes (t « r)

In closed thin-walled tubes shear stress τis inversely proportional to the wallthickness t.

where Am is the area enclosed by the section median line and T is the torque transmitted

For constant wall thickness t, the angular twist (radians) perunit of length of the tube θl is given by

where Lm is the perimeter of the section median line and G is the modulus of rigidity

A welded steel tube is 40 in long, has a 1/8 -in wall thickness, and a 2.5-in by 3.6-inrectangular cross section as shown in Figure. Assume an allowable shear stress of 11 500 psiand a shear modulus of 11.5(106) psi.(a) Estimate the allowable torque T.(b) Estimate the angle of twist due to the torque.

Problem-7

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The torque T is given by

T = 2A t τ = 2(8.253)0.125(11 500)

The angle of twist θ is given by

T Lmθ = θ l = l

Within the section median line, the area enclosed is

The length of the median perimeter is

Am = (2.5 − 0.125)(3.6 − 0.125) = 8.253 in2

Lm = 2[(2.5 − 0.125) + (3.6 − 0.125)] = 11.70 in

Solution

Compare the shear stress on a circular cylindrical tube with an outside diameter of 1 in

and an inside diameter of 0.9 in, predicted by , to that estimated by

Problem-8

= 23 730 lbf · in

23 730(11.70) = (40) = 0.0284 rad = 1.62

4(11.5 ×10 )(8.253 )(0.125)

From Equation.

Taking Eq. as correct, the error in the thin-wall estimate is −4.7 percent.

Solution o i

4 4π d - dJ =

Open Thin-Walled Sections

When the median wall line is not closed, it is said to be open Open sections in torsion, wherethe wall is thin, have relations derived from the membrane analogy theory resulting in:

where τ is the shear stress, G is the shearmodulus, θl is the angle of twist per unitlength, T is torque, c is the wall thickness and L is the length of the median line.

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Stress Concentration

In the development of the basic stress equations for tension, compression, bending,and torsion, it was assumed that no geometric irregularities occurred in themember

It is quite often that machine members have additional features made on it such asholes, key ways, collars, threads etc.

Any discontinuity in a machine part alters the stress distribution in theneighborhood of the discontinuity and result in stress concentration

• where Kt is used for normal stresses and Kts for shear stresses.• σmax or τmax: the actual maximum stress at the discontinuity• σ0 or τ0 : The nominal stress calculated by using the elementary stress equations and

the net area, or net cross section.

Stress-concentration factors for a variety of geometries may be found from standard Tables

A theoretical, or geometric, stress-concentration factor Kt or Kts is used to relate the actual maximum stress at the discontinuity to the nominal stress.

Thin plate loaded in tension with a centrally located hole

In this case, if we know the force F applied, the nominal stress can be calculated as

w - d t

Then from the table, for the given d/w ratio, obtain the Kt , then the actual maximum stress can be calculated as σmax = σ x Kt

Problem-9

A rotary lawn-mower blade rotates at 3000 rev/min. The steel blade has a uniform crosssection 3mm thick by 32mm wide, and has a 12mm-diameter hole in the center as shownin the figure. Estimate the maximum tensile stress at the central section due to rotation.Density of steel is 8050 kg/m3

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The nominal stress is given by

Force Fσ = =

actual cross sectional area w - d t

Here,the force is due to the rotation of the bladei.e.,centrifugal force

F = m ω r where m is the mass of the blade,r is the radius, ω is the angular velocity

Solution

mass m =volume×density = 0.610×0.032×0.003 ×8050 = 0.471kg

2π n 2×3.14×3000ω= = = 314.2 rad/s

Nominal area of cs = w - d t = 0.032 - 0.012 ×0.003=

we need toconsider only half of the mass sin0.471F = kg×0.305m× 314.2 rad/s

0.00006m

max nom

cethebladeis

rotating about the hole as centre

= 7090.92 kg m/s =

7090.92 Nσ = =11818214 N/m or Pascal =

0.00006m

d 12From table for = = 0.375, K = 2.25w 32

the actual maximum st

7090.92 N

ress σ

= σ × K

t = 118.1MPa× 2. 26525 = .9 MPa

Stresses in Pressurized Cylinders

Cylindrical pressure vessels, hydraulic cylinders, gun barrels, and pipes carrying fluids at high pressures develop both radial and tangential stresses

The tangential stress is given by

The radial stress is given by

Where ri the inside radius of the cylinder, ro is the outside radius, pi is the internal pressure, and po is the external pressure.

Tangential stress distribution Radial stress distribution

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The longitudinal stresses exist in the vessel when the end reactions to the internalpressure are taken by the pressure vessel itself

Longitudinal stress is obtained by

Thin-Walled Vessels

The wall thickness of a cylindrical pressure vessel is less than one-tenth of its radius The radial stress that results from pressurizing the vessel is quite small compared with

the tangential stress

The average tangential stress is given by

The maximum tangential stress is obtained by

where (di + t) is the average diameter

In case of a closed cylinder, the longitudinal stress σl is obtained by

An AISI 1040 cold-drawn steel tube has an OD = 50 mm and wall thickness 6 mm. Whatmaximum external pressure can this tube withstand if the largest principal normalstress is not to exceed 80 percent of the minimum yield strength of the material? Takeminimum yield strength as Sy=490 MPa

Solution

Problem-10

The tangential stress

Given: ro = 25mm, thickness t = 6mm, therefore ri = 25-6 = 19mmallowable stress = 80% of yield strength = 0.8x -490 = -392 Mpa (compressive)

the internal pressure is not given, assume pi = 0

We know the maximum tangential stress will occur at the inner wall of the tube, where r = ri

2 2 2 2

t max o i

σ r - r -392 0.025 -0.019p = - = -

2 r 2×0.025

0.103488= =

0.00182. a

2579 MP

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Stresses in Rotating Rings

Many rotating elements, such as flywheels and blowers, can be simplified to a rotating ring to determine the stresses

The tangential and radial stresses so found are subject to the following restrictions:• The outside radius of the ring, or disk, is large compared with the thickness ro ≥ 10t.• The thickness of the ring or disk is constant.• The stresses are constant over the thickness.

The tangential and radial stress are obtained by

where r is the radius to the stress element under consideration, ρ is the mass density, ν isthe Poisson ratio and ω is the angular velocity of the ring in radians per second. For arotating disk, use ri = 0 in these equations.

The maximum recommended speed for a 250-mm-diameter abrasive grinding wheel is 2000

rev/min. Assume that the material is isotropic; use a bore of 20 mm, ν = 0.24, and a mass

density of 3320 kg/m3, and find the maximum tensile stress at this speed.

Problem-11

Solution

Given:

Speed n = 2000rpm

Outer diameter D = 250mm ro = 125mm = 0.125 m

Inner diameter d = 20mm ri = 10mm = 0.01m

Poisson’s ratio ν = 0.24

mass density ρ = 3320 kg/m3

2πn 2×3.14×2000the angular velocity ω = = = 209.4rad/s

The grinding wheel can be considered as a rotating ring

The tensile stress is given by the equation

0.01 0.125 1 3 0.243 0.240.01

8 0.01 3 0.24

1850000N / m or Pascal 1850000 10 1.85MPa

t = 3320 209.4

Substitute the values

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After assembly, an interference contact pressure pdevelops between the members at the nominal radiusR, causing radial stresses σr = −p in each member atthe contacting surfaces.

Where δ is the radial interference, the subscripts o and i on the material properties correspond to the outer and inner members, respectively.

If the two members are of the same material with Eo = Ei = E, νo = vi , the relation simplifiesto

The p, can be used to determine the radial and tangential stresses in each member using theequation for pressurized cylinders.For the inner member, po = p and pi = 0, For the outer member, po = 0 and pi = p.

Press and Shrink Fits

When two cylindrical parts are assembled by shrinking or press fitting one part uponanother, a contact pressure is created between the two parts

The table-1 gives data concerning the shrink fit of two cylinders of differing materials anddimensional specification in inches. Elastic constants for different materials may be found inTable 2. Identify the radial interference δ, then find the interference pressure p, and thetangential normal stress on both sides of the fit surface for steel/carbon steel.

Problem-12

Table-1

Table-2

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SolutionGiven:

From Table -1, ri = 0,

Outer diameter D = 3 in ro = 1.5 in = 38.1mm

R = 1 in = 25.4mm

From Table -2, Poisson’s ratio νo = νi = 0.292

Ei = Eo = 207 Gpa

The pressure at the interface is given by

The radial interference can be calculated from the values from the Table -1 as

δ = 2.002 - 2.000 =0.0001in=0.0001×25 0.002.4=2

10 410 0.0254 0.0254 0

1.6 10 3.584 100

207 10 0.00254 0.0381p = =

2 0.0254 0.0381

=5734470 N/m or Pa =

5.73 MPa

The tangential stresses at the interface for the inner and outer members are given by

Equations

it 2 2i

ot 2 2o

R +rσ = - p

r + Rσ = p

Temperature Effects

When the temperature of an unrestrained body is uniformly increased, the body expands,

and the normal strain is

where α is the coefficient of thermal expansion and T is the temperature change, in degrees.

If a straight bar is restrained at the ends so as to prevent lengthwise expansion and then is

subjected to a uniform increase in temperature, a compressive stress will develop because of

the axial constraint. The stress is given by

If a uniform flat plate is restrained at the edges and also subjected to a uniform

temperature rise, the compressive stress developed is given by the equation

0.0254 +0-5734470

0.0254 -0

-5734470 Pa =

0.0381 +0.02545734470

0.0381 -0.0254

= 149096

5.73 MPa

14.9 a2 MP2 Pa =

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The stresses developed by temperature changes are called thermal stresses. They arise

because of a temperature change in a clamped or restrained member.

Example, stresses occur during welding, bracing etc. since parts to be welded must be

clamped before welding.

Contact Stresses

When two bodies having curved surfaces are pressed together, point or line contact changes

to area contact, and the stresses developed in the two bodies are three dimensional.

Example : in the contact of a wheel and a rail, in automotive valve cams and tappets, in

mating gear teeth, and in the action of rolling bearings.

Spherical Contact

When two solid spheres of diameters d1 and d2 are pressed together with a force F, a

circular area of contact of radius a is obtained. The radius a is given by the equation

Where E1, ν1 and E2, ν2 are the respective elastic constants of the two spheres,

The maximum pressure occurs at the center of the

contact area and is

The pressure distribution within the contact area of each sphere

wikipedia.org

5/15/2013

The maximum stresses occur on the z axis, and these are principal stresses. Their

values are given by

These equations are valid for either

sphere, but the value used for Poisson’s

ratio must correspond with the sphere

under consideration.

The maximum shear stress is given by

The chart is based on a

Poisson ratio of 0.30

Cylindrical Contact

The area of contact in this case is a narrow

rectangle of width 2b and length l, and the

pressure distribution is elliptical. The half-

width b is given by the equation

The maximum pressure is

Above equations apply to a cylinder and a plane surface, such as a rail, by making d = ∞ for

the plane surface. The equations also apply to the contact of a cylinder and an internal

cylindrical surface; in this case d is made negative for the internal surface.

The stress state along the z axis is given by the equations

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The largest value of max occurs at z/b = 0.786. Its maximum value is 0.30pmax. The chart is

based on a Poisson ratio of 0.30. Note that all normal stresses are compressive stresses.

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A carbon steel ball with 25-mm diameter is pressed together with an aluminum ball 40-mm

diameter by a force of 10 N. Determine the maximum shear stress, and the depth at which it

will occur for the aluminum ball. Assume figure below, which is based on a typical Poisson’s

ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for

these materials. Assume E1 = 207000Pa , ν1 = 0.292 for steel and E2 = 71700Pa, ν2 = 0.333

for aluminium

Problem-13

Solution

Given:

d1 = 25mm, d2 = 40mm

F= 10N, ν1 = 0.292, ν2 = 0.333

E1 = 207000Pa, E2 = 71700Pa

Radius of contact

The maximum contact pressure

From the figure, the maximum shear stress occurs at a depth of z = 0.48 a.

The principal stresses are obtained at a depth of z/a = 0.48. using the equations

5/15/2013

The maximum shear stress

5/15/2013 1 Mechanical Engineering Design III MEEN :3393 LOAD AND STRESS ANALYSIS Equilibrium and...

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