196634050 Mechanics of Materials

8/16/2019 196634050 Mechanics of Materials

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CCHHAAPPTTEER R 77


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PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it

without permission.

PROBLEM 7.2

For the given state of stress, determine the normal and shearing stresses

exerted on the oblique face of the shaded triangular element shown. Use amethod of analysis based on the equilibrium of that element, as was done in the

derivations of Sec. 7.2.

SOLUTION

Stresses Areas Forces

0:Σ = F

80 cos55 cos55 40 sin 55 sin 55 0σ − ° ° + ° ° = A A A

2 280 cos 55 40sin 55σ = ° − ° 0.521 MPaσ = −

0:Σ = F

80 cos 55 sin 55 40 sin 55 cos 55τ − ° ° − ° ° A A A

120 cos 55 sin 55τ = ° ° 56.4 MPaτ =


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PROBLEM 7.5

For the given state of stress, determine (a) the principal planes, (b) the principal

stresses.

SOLUTION

60 MPa 40 MPa 35 MPa x y xyσ σ τ = − = − =

(a)2 (2)(35)

tan 2 3.5060 40

xy

p

x y

τ θ

σ σ = = = −

− − +

2 74.05 pθ = − ° 37.0 , 53.0 pθ = − ° °

(b)

2

2max, min

2

2

2 2

60 40 60 40(35)

2 2

50 36.4 MPa

x y x y

xy

σ σ σ σ σ τ

+ − = ± +

− − − + = ± +

= − ±

max 13.60 MPaσ = −

min 86.4 MPaσ = −


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PROBLEM 7.12

For the given state of stress, determine (a) the orientation of the planes of maximum

in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) thecorresponding normal stress.

SOLUTION

8 ksi 12 ksi 5 ksi x y xyσ σ τ = − = =

(a) 8 12

tan 2 2.02 2(5)

x y s

xy

σ σ θ

τ

− − −= − = − = +

2 63.435 sθ = ° 31.7 , 121.7 sθ = ° °

(b)

2

2max

2

x y xy

σ σ τ τ

− = +

228 12 (5)

2

− − = +

max 11.18 ksiτ =

(c) ave2

σ σ σ σ

+′ = =

x y

8 12

2

− += 2.00 ksiσ ′ =


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PROBLEM 7.16

For the given state of stress, determine the normal and shearing stresses after the element

shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.

SOLUTION

0 80 MPa 50 MPa

40 MPa 40 MPa2 2

cos 2 sin 22 2

sin 2 + cos 22

cos 2 sin 22 2

x y xy

x y x y

x y x y

x xy

x y x y xy

x y x y

y xy

σ σ τ

σ σ σ σ

σ σ σ σ σ θ τ θ

σ σ τ θ τ θ

σ σ σ σ σ θ τ θ

′

′ ′

′

= = − = −

+ −= − =

+ −= + +

−= −

+ −= − −

(a) 25θ = − ° 2 50θ = − °

40 40 cos ( 50 ) 50 sin ( 50°) xσ ′ = − + − ° − − 24.0 MPa xσ ′ =

40 sin ( 50°) 50 cos ( 50 ) x yτ ′ ′ = − − − − ° 1.5 MPa x yτ ′ ′ = −

40 40 cos ( 50 ) 50 sin ( 50 ) yσ ′ = − − − ° + − ° 104.0 MPa yσ ′ = −

(b) 10 2 20θ θ = ° = °

40 40 cos (20°) 50 sin (20°) xσ ′ = − + − 19.5 MPa xσ ′ = −

40 sin (20°) 50 cos (20°) x yτ ′ ′ = − − 60.7 MPa x yτ ′ ′ = −

40 40 cos (20°) + 50 sin (20°) yσ ′ = − − 60.5 MPa yσ ′ = −


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PROBLEM 7.20

Two members of uniform cross section 50 80 mm× are glued together along plane a-a that forms an angle of 25° with the horizontal. Knowing that the allowablestresses for the glued joint are 800 kPaσ = and 600 kPa,τ = determine the largestcentric load P that can be applied.

SOLUTION

For plane a-a, 65 .θ = °

2 2 2

3 3 33

2 2

2 2

3

0, 0,

cos sin 2 sin cos 0 sin 65 0

(50 10 )(80 10 )(800 10 )3.90 10 N

sin 65 sin 65

( )sin cos (cos sin ) sin 65 cos65 0

(50 10 )(8

sin65 cos65

x xy y

x y xy

x y xy

P

A

P

A

A P

P

A

A P

σ τ σ

σ σ θ σ θ τ θ θ

σ

τ σ σ θ θ τ θ θ

τ

− −

−

= = =

= + + = + ° +

× × ×= = = ×

° °

= − − + − = ° ° +

×= =

° °

3 330 10 )(600 10 ) 6.27 10 N

sin65 cos65

−× ×= ×

° °

Allowable value of P is the smaller one. 3.90 kN P =


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PROBLEM 7.26

The axle of an automobile is acted upon by the forces and couple

shown. Knowing that the diameter of the solid axle is 32 mm,determine (a) the principal planes and principal stresses at point H

located on top of the axle, (b) the maximum shearing stress at thesame point.

SOLUTION

31 1 (32) 16 mm 16 10 m2 2

c d −= = = = ×

Torsion: 63 3 3

2 2(350 N m)54.399 10 Pa 54.399 MPa

(16 10 m)

Tc T

J c

τ

π π −

⋅= = = = × =

×

Bending: 4 3 4 9 4

3

36

9

(16 10 ) 51.472 10 m4 4

(0.15m)(3 10 N) 450 N m

(450)(16 10 )139.882 10 Pa 139.882 MPa

51.472 10

I c

M

My

I

π π

σ

− −

−

−

= = × = ×

= × = ⋅

×= − = − = − × = −

×

Top view: Stresses:

ave

2

2 2 2

139.882 MPa 0 54.399 MPa

1 1( ) ( 139.882 0) 69.941 MPa

2 2

( 69.941) ( 54.399) 88.606 MPa2

x y xy

x y

x y xy R

σ σ τ

σ σ σ

σ σ τ

= − = = −

= + = − + = −

− = + = − + − =

(a) max ave 69.941 88.606 Rσ σ = + = − + max 18.67 MPaσ =

min ave 69.941 88.606 Rσ σ = − = − − min 158.5 MPaσ = −


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PROBLEM 7.26 (Continued)

2 (2)( 54.399)tan 2 0.77778 2 37.88

139.882

xy p p

x y

τ θ θ

σ σ

−= = = = °

− −

18.9 and 108.9° pθ = °

(b) max 88.6 MPa Rτ = = max 88.6 MPaτ =


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PROBLEM 7.32

Solve Probs 7.7 and 7.11, using Mohr’s circle.

PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses.

PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) theorientation of the planes of maximum in-plane shearing stress, (b) the maximum in- plane shearing stress, (c) the corresponding normal stress.

SOLUTION

ave

4 ksi,

12 ksi,

15 ksi

4 ksi2

x

y

xy

x y

σ

σ

τ

σ σ σ

=

= −

= −

+= = −

Plotted points for Mohr’s circle:

ave

: ( , ) (4 ksi, 15 ksi)

: ( , ) ( 12 ksi, 15 ksi)

: ( , 0) ( 4 ksi, 0)

x xy

y xy

X

Y

C

σ τ

σ τ

σ

− =

= − −

= −

(a)15

tan 1.8758

FX

CF α = = =

61.93

130.96

2a

α

θ α

= °

= − = − ° 31.0aθ = − °

180 118.07

159.04

2b

β α

θ β

= ° − = °

= = ° 59.0bθ = °

2 2 2 2( ) ( ) (8) (15) 17 ksi R CF FX = + = + =

(b) max ave 4 17a Rσ σ σ = = + = − + max 13.00 ksiσ =

min min ave 4 17 Rσ σ σ = = − = − − min 21.0 ksiσ = −

(a′) 45 14.04d aθ θ = + ° = ° 14.0d θ = °

45 104.04e bθ θ = + ° = ° 104.0eθ = °

max Rτ = max 17.00 ksiτ =

(b′) aveσ σ ′ = 4.00 ksiσ ′ = −


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PROBLEM 7.41

Solve Prob. 7.19, using Mohr’s circle.

PROBLEM 7.19 A steel pipe of 12-in. outer diameter is fabricated from14

-in.-thick plate by welding along a helix which forms an angle of 22.5° witha plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial forceP and an 80-kip ⋅ in. torque T, each directed as shown, are applied to the pipe,determine σ and τ in directions, respectively, normal and tangential to theweld.

SOLUTION

( )

( )

2 2 2

1 2

2 2 2 2 22 1

4 4 4 4 42 1

112 in. 6 in., 0.25 in.

2

5.75 in.

(6 5.75 ) 9.2284 in

(6 5.75 ) 318.67 in2 2

d c d t

c c t

A c c

J c c

π π

π π

= = = =

= − =

= − = − =

= − = − =

Stresses:

2

404.3344 ksi

9.2284

(80)(6)1.5063 ksi

318.67

0, 4.3344 ksi, 1.5063 ksi x y xy

P

A

Tc

J

σ

τ

σ σ τ

= − = − = −

= − = − =

= = − =

Draw the Mohr’s circle.

: (0, 1.5063 ksi)

: ( 4.3344 ksi,1.5063 ksi)

: ( 2.1672 ksi, 0)

X

Y

C

−

−

−

2 2

1.5063tan 0.69504 34.8

2.1672

(2)(22.5 ) 10.8

(2.1672) (1.5063) 2.6393 ksi R

α α

β α

= = = °

= ° − = °

= + =

2.1672 2.6393 cos 10.8w

σ = − − ° 4.76 ksiw

σ = −

2.6393sin10.2wτ = − ° 0.467 ksiwτ = −


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PROBLEM 7.44


PROBLEM 7.22 Two steel plates of uniform cross section 10 80 mm× are welded together as shown. Knowing that centric 100-kN forces are

applied to the welded plates and that the in-plane shearing stress parallel

to the weld is 30 MPa, determine (a) the angle β, (b) the correspondingnormal stress perpendicular to the weld.

SOLUTION

36

3 3

100 10125 10 Pa 125 MPa

(10 10 )(80 10 ) x

P

Aσ

− −

×= = = × =

× ×

0 0 y xyσ τ = =

From Mohr’s circle:

(a)30

sin 2 0.4862.5

β = = 14.3 β = °

(b) 62.5 62.5cos2σ β = +

117.3 MPaσ =


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PROBLEM 7.45


PROBLEM 7.23 A 400-lb vertical force is applied at D to a gearattached to the solid 1-in.-diameter shaft AB. Determine the principal

stresses and the maximum shearing stress at point H located as shown

on top of the shaft.

SOLUTION

Equivalent force-couple system at center of shaft in section at point H :

400 lb (400)(6) 2400 lb inV M = = = ⋅

(400)(2) 800 lb inT = = ⋅

Shaft cross section

1

1 in. 0.5 in.2d c d = = =

4 4 410.098175 in 0.049087 in2 2

J c I J π

= = = =

Torsion: 3(800)(0.5)

4.074 10 psi 4.074 ksi0.098175

Tc

J τ = = = × =

Bending: 3(2400)(0.5)

24.446 10 psi 24.446 ksi0.049087

Mc

I σ = = = × =

Transverse shear: Stress at point H is zero.

Resultant stresses:

ave

2

2

2 2

24.446 ksi, 0, 4.074 ksi

1( ) 12.223 ksi

2

2

(12.223) (4.074) 12.884 ksi

x y xy

x y

x y xy R

σ σ τ

σ σ σ

σ σ τ

= = =

= + =

− = +

= + =

avea Rσ σ = + 25.107 ksiaσ =

aveb Rσ σ = − 0.661ksibσ = −

max Rτ = max 12.88 ksiτ =


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PROBLEM 7.47


PROBLEM 7.25 The steel pipe AB has a 102-mm outer diameterand a 6-mm wall thickness. Knowing that arm CD is rigidly

attached to the pipe, determine the principal stresses and the

maximum shearing stress at point K.

SOLUTION

102 51 mm 45 mm2 2o

o i od r r r t = = = = − =

( )4 4 6 4 6 44.1855 10 mm 4.1855 10 m2

o i J r r π −= − = × = ×

6 41 2.0927 10 m2

I J −= = ×

Force-couple system at center of tube in the plane containing points H and K :

3

3 3

3 3

10 10 N

(10 10 )(200 10 ) 2000 N m

(10 10 )(150 10 ) 1500 N m

x

y

z

F

M

M

−

−

= ×

= × × = ⋅

= − × × = − ⋅

Torsion:

3

3

6

2000 N m

51 10 m

(2000)(51 10 )24.37 MPa

4.1855 10

y

o

xy

T M

c r

Tc

J τ

−

−

−

= = ⋅

= = ×

×= = =

×

Note that the local x-axis is taken along a negative global z -direction.

Transverse shear: Stress due to xV F = is zero at point K.

Bending:

3

6

(1500)(51 10 )36.56 MPa2.0927 10

z y

M c

I σ

−

−

×= = =×

Point K lies on compression side of neutral axis. 36.56 MPa yσ = −


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Total stresses at point K : 0, 36.56 MPa, 24.37 MPa x y xyσ σ τ = = − =

ave

1( ) 18.28 MPa

2 x yσ σ σ = + = −

2

2 30.46 MPa2

x y xy R

σ σ τ

− = + =

max ave 18.28 30.46 Rσ σ = + = − +

max 12.18 MPaσ =

min ave 18.28 30.46 Rσ σ = − = − −

min 48.74 MPaσ = −

max Rτ = max 30.46 MPaτ =


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PROBLEM 7.61

For the element shown, determine the range of values of xyτ for which themaximum tensile stress is equal to or less than 60 MPa.

SOLUTION

ave

20 MPa 120 MPa

1( ) 70 MPa

2

x y

x y

σ σ

σ σ σ

= − = −

= + = −

Set max ave

max ave

60 MPa

130 MPa

R

R

σ σ

σ σ

= = +

= − =

But

2

2

2

2

2 2

2

2

130 50

120 MPa

x x xy

x x xy

R

R

σ σ τ

σ σ τ

− = +

− = −

= −

=

Range of : xyτ 120 MPa 120 MPa xyτ − ≤ ≤


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PROBLEM 7.66

For the state of plane stress shown, determine the maximum shearing

stress when (a) 6 ksi xσ = and 18 ksi yσ = , (b) 14 ksi xσ = and2 ksi yσ = . ( Hint: Consider both in-plane and out-of-plane shearing

stresses.)

SOLUTION

(a)

ave

6ksi, 18 ksi, 8 ksi

1( ) 12 ksi

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

2

2

2 2

2

6 8 10 ksi

x y xy R

σ σ τ

− = +

= + =

ave

ave

12 10 22 ksi (max)

12 10 2 ksi

0 (min)

a

b

c

R

R

σ σ

σ σ

σ

= + = + =

= − = − =

=

max(in-plane) 10 ksi Rτ = =

max max min

1( )

2τ σ σ = −

max

11ksiτ =

(b)

2

2

2 2

14 ksi, 2 ksi, 8 ksi

1( ) 8 ksi

2

2

6 8 10 ksi

x y xy

a x y

x y xy R

σ σ τ

σ σ σ

σ σ τ

= = =

= + =

− = +

= + =

ave

ave

18 ksi (max)

2 ksi (min)

0

a

b

c

R

R

σ σ

σ σ

σ

= + =

= − = −

=

max

min

max max min

18 ksi

2 ksi

1( )

2

σ

σ

τ σ σ

=

= −

= − max 10 ksiτ =


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PROBLEM 7.72

For the state of stress shown, determine the maximum shearing stress

when (a) 0, z σ = (b) 45 z σ = + MPa, (c) 45 MPa. z σ = −

SOLUTION

100 MPa, 20 MPa, 75 MPa x y xyσ σ τ = = =

ave

2

2

2 2

ave

ave

1( ) 60 MPa

2

2

40 75 85 MPa

145 MPa

25 MPa

x y

x y

xy

a

b

R

R

R

σ σ σ

σ σ τ

σ σ

σ σ

= + =

− = +

= + =

= + =

= − = −

(a) 0, 145 MPa, 25 MPa z a bσ σ σ = = = −

max min max max min

1145 MPa, 25 MPa, ( )

2

σ σ τ σ σ = = − = − max 85 MPaτ =

(b) 45 MPa, 145 MPa, 25 MPa z a bσ σ σ = + = = −

max min145 MPa, 25 MPa,σ σ = = − max max min1

( )2

τ σ σ = − max 85 MPaτ =

(c) 45 MPa, 145 MPa, 25 MPa z a bσ σ σ = − = = −

max 145 MPa,σ = min 45 MPa,σ = − max max min1

( )2

τ σ σ = − max 95 MPaτ =


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PROBLEM 7.83

The state of plane stress shown occurs in a machine component made of a steel with

45 ksi.Y σ = Using the maximum-distortion-energy criterion, determine whether

yield will occur when (a) 9 ksi, xyτ = (b) 18 ksi, xyτ = (c) 20 ksi. xyτ = If yield

does not occur, determine the corresponding factor of safety.

SOLUTION

36 ksi, 21 ksi, 0 x y z σ σ σ = = =

For stresses in xy-plane, ave1

( ) 28.5 ksi2

x yσ σ σ = + =

7.5 ksi2

x yσ σ − =

(a)

2

2 2 2

ave ave

9 ksi (7.5) (9) 11.715 ksi2

40.215 ksi, 16.875 ksi

x y xy xy

a b

R

R R

σ σ τ τ

σ σ σ σ

− = = + = + =

= + = = − =

2 2 34.977 ksi 45 ksia b a bσ σ σ σ + − = < (No yielding)

45. .

39.977 F S = . . 1.287 F S =

(b)

2

2 2 2

ave ave

18 ksi (7.5) (18) 19.5 ksi2

48 ksi, 9 ksi

x y xy xy

a b

R

R R

σ σ τ τ

σ σ σ σ

− = = + = + =

= + = = − =

2 2 44.193 ksi 45 ksia b a bσ σ σ σ + − = < (No yielding)

45. .

44.193 F S = . . 1.018 F S =

(c)

2

2 2 2

ave ave

20 ksi (7.5) (20) 21.36 ksi

249.86 ksi, 7.14 ksi

x y xy xy

a b

R

R R

σ σ τ τ

σ σ σ σ

− = = + = + =

= + = = − =

2 2 46.732 ksi 45 ksia b a bσ σ σ σ + − = > (Yielding occurs)


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PROBLEM 7.84

Solve Prob. 7.83, using the maximum-shearing-stress criterion.

PROBLEM 7.83 The state of plane stress shown occurs in a machine componentmade of a steel with 45 ksi.Y σ = Using the maximum-distortion-energy criterion,determine whether yield will occur when (a) 9 ksi,

xyτ = (b) 18 ksi,

xyτ = (c)

20 ksi. xyτ = If yield does not occur, determine the corresponding factor of safety.

SOLUTION

36 ksi, 21 ksi, 0 x y z σ σ σ = = =

For stress in xy-plane, ave1

( ) 28.5 ksi 7.5 ksi2 2

x y x y

σ σ σ σ σ

−= + = =

(a)

2

2

ave ave

max min34.977 ksi,

9 ksi 11.715 ksi2

40.215 ksi, 16.875 ksi

0

x y xy xy

a b

R

R R

τ σ

σ σ τ τ

σ σ σ σ

− = = + =

= + = = − =

= =

max max min2 40.215 ksi 45 ksiτ σ σ = − = < (No yielding)

45. .

40.215 F S = . . 1.119 F S =

(b)

max

2

2

ave ave

min48 ksi

18 ksi 19.5 ksi2

48 ksi, 9 ksi

0

x y xy xy

a b

R

R R

σ

σ σ τ τ

σ σ σ σ

σ

− = = + =

= + = = − =

= =

max max min2 48 ksi 45 ksiτ σ σ = − = > (Yielding occurs)

(c)

max

2

2

ave ave

min49.86 ksi

20 ksi 21.36 ksi2

49.86 ksi 7.14 ksi

0

x y xy xy

a b

R

R R

τ σ

σ σ τ τ

σ σ σ σ

− = = + =

= + = = − =

= =

max max min2 49.86 ksi 45 ksiτ σ σ = − = > (Yielding occurs)


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PROBLEM 7.85

The 36-mm-diameter shaft is made of a grade of steel with a 250-MPa tensile yield

stress. Using the maximum-shearing-stress criterion, determine the magnitude of thetorque T for which yield occurs when 200 kN. P =

SOLUTION

3 3

2 3 2 3 2

36

3

ave

1200 kN 200 10 N 18 mm 18 10 m

2

(18 10 ) 1.01788 10 m

200 10

196.488 10 Pa1.01788 10

196.488 MPa

1 10 ( ) 98.244 MPa

2 2

y

x x y y

P c d

A c

P

A

π π

σ

σ σ σ σ σ

−

− −

−

= = × = = = ×

= = × = ×

×

= − = − = ××

=

= = + = =

2

2 2 2

ave

ave

(98.244)2

(positive)

(negative)

2

x y xy xy

a

b

a b a b a a b b

R

R

R

R

σ σ τ τ

σ σ

σ σ

σ σ σ σ σ σ σ σ

− = + = +

= +

= −

− = − > − >

Maximum shear stress criterion under the above conditions:

2 250 MPa 125 MPaa b Y R Rσ σ σ − = = = =

Equating expressions for R,

2 2

2 2 6

125 (98.244)

(125) (98.244) 77.286 MPa 77.286 10 Pa

xy

xy

τ

τ

= +

= − = = ×

Torsion: 4 3 4 9 4(18 10 ) 164.896 10 m2 2

xy

J c

Tc

J

π π

τ

− −= = × = ×

=

9 6

3

(164.846 10 )(77.286 10 )

18 10

708 N m

xy J T

c

τ −

−

× ×= =

×

= ⋅ 708 N mT = ⋅


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PROBLEM 7.86

Solve Prob. 7.85, using the maximum-distortion-energy criterion.

PROBLEM 7.85 The 36-mm-diameter shaft is made of a grade of steel with a250-MPa tensile yield stress. Using the maximum-shearing-stress criterion,

determine the magnitude of the torque T for which yield occurs when 200 kN. P =

SOLUTION

3 3

2 3 2 3 2

36

3

ave

1200 kN 200 10 N 18 mm 18 10 m

2

(18 10 ) 1.01788 10 m

200 10

196.488 10 Pa1.01788 10

196.448 MPa

1 10 ( ) 98.244 MPa

2 2

y

x x y y

P c d

A c

P

A

π π

σ

σ σ σ σ σ

−

− −

−

= = × = = = ×

= = × = ×

×

= − = − = ××=

= = + = =

2

2 2 2

ave ave

(98.244)2

x y

xy xy

a b

R

R R

σ σ τ τ

σ σ σ σ

− = + = +

= + = −

Distortion energy criterion:

2 2 2

2 2 2ave ave ave ave

2 2 2ave

2 2 2 2

( ) ( ) ( )( )

3

(98.244) (3)[(98.244) ] (250)

89.242 MPa

a b a b Y

Y

Y

xy

xy

R R R R

R

σ σ σ σ σ σ σ σ σ σ

σ σ

τ

τ

−

+ − =+ + − − + − =

+ =

+ + =

= ±

Torsion: 4 3 4 9 4(18 10 ) 164.846 10 m2 2

xy

J c

Tc

J

π π

τ

− −= = × = ×

=

9 6

3

(164.846 10 )(89.242 10 )

18 10

818 N m

xy J T

c

τ −

−

× ×= =

×

= ⋅ 818 N mT = ⋅


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without permission.

PROBLEM 7.98

A spherical gas container made of steel has a 5-m outer diameter and a wall thickness of 6 mm. Knowing that

the internal pressure is 350 kPa, determine the maximum normal stress and the maximum shearing stress inthe container.

SOLUTION

36

5 m 6 mm 0.006 m, 2.494 m2

(350 10 Pa)(2.494 m)72.742 10 Pa

2 2(0.006 m)

d d t r t

pr

t σ

= = = = − =

×= = = ×

72.7 MPaσ =

max

min

72.742 MPa

0 (Neglectingsmallradialstress)

σ

σ

=

≈

max max min

1( )

2τ σ σ = − max 36.4 MPaτ =


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without permission.

PROBLEM 7.104

A steel penstock has a 750-mm outer diameter, a 12-mm wall

thickness, and connects a reservoir at A with a generating stationat B. Knowing that the density of water is 31000 kg/m , determine

the maximum normal stress and the maximum shearing stress inthe penstock under static conditions.

SOLUTION

3

3

3 2

6

6 36

1 3

1 1(750) 12 363 mm 363 10 m

2 2

12 mm 12 10 m

(1000 kg/m )(9.81 m/s )(300 m)

2.943 10 Pa

(2.943 10 )(363 10 )89.0 10 Pa

12 10

r d t

t

p gh

pr

t

ρ

σ

−

−

−

−

= − = − = = ×

= = ×

= =

= ×

× ×= = = ×

×

max 1σ σ = max 89.0 MPaσ =

min 0 pσ = − ≈

max max min

1( )

2τ σ σ = − max 44.5 MPaτ =


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without permission.

PROBLEM 7.114

For the tank of Prob. 7.112, determine the range of values of β that can beused if the shearing stress parallel to the weld is not to exceed 12 MPawhen the gage pressure is 600 kPa.

PROBLEM 7.112 The pressure tank shown has a 8-mm wall thicknessand butt-welded seams forming an angle 20 β = ° with a transverse plane.For a gage pressure of 600 kPa, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

SOLUTION

3 11.6 m 8 10 mm 0.792 m2

d t r d t −= = × = − =

3

13

6

2

1 2

(600 10 )(0.792)

8 10

59.4 10 Pa = 59.4 MPa

29.7 MPa2

14.85 MPa2

|sin 2 |w

pr

t

pr

t

R

R

σ

σ

σ σ

τ β

−

×= =

×= ×

= =

−= =

=

12|sin 2 | 0.80808

14.85

N a

R

τ β = = =

2 53.91a β = − ° 27.0a β = + °

2 53.91b β = + ° 27.0b β = °

2 180 53.91 126.09c β = ° − ° = + ° 63.0 β = °c

2 180 53.91 233.91d β = ° + ° = + ° 117.0d β = °

Let the total range of values for β be 180 < 180 β − ° ≤ °

Safe ranges for β : 22.0 27.0° β − °≤ ≤

and 63.0 117.0° β °≤ ≤


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without permission.

PROBLEM 7.116

The pressurized tank shown was fabricated by welding strips of plate along ahelix forming an angle β with a transverse plane. Determine the largestvalue of β that can be used if the normal stress perpendicular to the weld isnot to be larger than 85 percent of the maximum stress in the tank.

SOLUTION

1 22

pr pr

t t σ σ = =

ave 1 2

1 2

ave

1 3( )

2 4

1

2 4

cos 2w

r

t

pr R

t

R

σ σ σ

σ σ

σ σ β

= + =

−= =

= −

3 10.85 cos 2

4 4

pr pr

t t β

= −

3cos 2 4 0.85 0.4

4

2 113.6

β

β

= − − = −

= ° 56.8 β = °


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without permission.

PROBLEM 7.124

A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness is

fabricated from a 4-ft section of spirally-welded pipe AB and is equippedwith two rigid end plates. The gage pressure inside the vessel is 300 psi

and 10-kip centric axial forces P and P′ are applied to the end plates.Determine (a) the normal stress perpendicular to the weld, (b) theshearing stress parallel to the weld.

SOLUTION

( )

1

2

0

2 2 2 2 20

3

1 1(10) 5 in. 0.25 in.

2 2

(300)(5)6000 psi 6 ksi

0.25

(300)(5)3000 psi 3 ksi

2 (2)(0.25)5 0.25 5.25 in.

(5.25 5.00 ) 8.0503 in

10 101242 psi 1.242 ksi

8.0803

σ

σ

π π

σ

= = = =

= = = =

= = = =

= + = + =

= − = − =

×= − = − = − = −

r d t

pr

t

pr

t r r t

A r r

P

A

Total stresses. Longitudinal: 3 1.242 1.758 ksi xσ = − =

Circumferential: 6 ksi yσ =

Shear: 0 xyτ =

Plotted points for Mohr’s circle:

: (1.758, 0)

: (6, 0)

: (3.879)

X

Y

C

ave

2

2

2

1( ) 3.879 ksi

2

2

(1.758 6) 0 2.121 ksi2

x y

x y

xy R

σ σ σ

σ σ τ

= + =

− = +

− = + =

(a) ave cos 70 3.879 2.121 cos 70 x Rσ σ ′ = + ° = − ° 3.15 ksi xσ ′ =

(b) | | sin 70 2.121 sin 70 xy Rτ = ° = ° | | 1.993 ksi x yτ ′ ′ =


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PROBLEM 7.128

For the given state of plane strain, use the method of Sec. 7.10 todetermine the state of plane strain associated with axes x′ and y′ rotated through the given angle θ .

500 , 250 , 0, 15 x y xyε µ ε µ γ θ = − = + = = °

SOLUTION

15θ = + °

125 375 02 2 2

cos 2 sin 22 2 2

125 ( 375 )cos30 0

x y x y xy

x y x y xy

x

ε ε ε ε γ µ µ

ε ε ε ε γ ε θ θ

µ µ

′

+ −= − = − =

+ −= + +

= − + − ° + 450 xε µ ′ = −

cos 2 sin 22 2 2

125 ( 375 )cos 30 0

x y x y xy

y

ε ε ε ε γ ε θ θ

µ µ

′

+ −= − −

= − − − ° − 200 yε µ ′ =

( )sin 2 cos 2

( 500 250 )sin 30 0

x y x y xyγ ε ε θ γ θ

µ µ

′ ′ = − − +

= − − − ° + 375 x yγ µ ′ ′ =


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without permission.

PROBLEM 7.136

The following state of strain has been measured on the surface of a thin plate. Knowing that the surface of the

plate is unstressed, determine (a) the direction and magnitude of the principal strains, (b) the maximum in- plane shearing strain, (c) the maximum shearing strain. (Use 1

3.v = )

260 , 60 , 480 x y xy

ε µ ε µ γ µ = − = − = +

SOLUTION

For Mohr’s circle of strain, plot points:

: ( 260 , 240 )

: ( 60 , 240 )

: ( 160 , 0)

X

Y

C

µ µ

µ µ

µ

− −

−

−

480tan 2 2.4

260 60

xy

p

x y

γ θ

ε ε = = = −

− − +

2 67.38 pθ = − ° 33.67bθ = − °

56.31aθ = °

2 2(100 ) (240 )

260

R

R

µ µ

µ

= +

=

(a) ave 160 260a Rε ε µ µ = + = − + 100aε µ =

ave 160 260b Rε ε µ µ = − = − − 420bε µ = −

(b) max (in-plane) max (in-plane)1

22

R Rγ γ = = max (in-plane) 520γ µ =

1/3( ) ( ) ( 260 60)

1 1 2/3160

c a b x y

v v

v vε ε ε ε ε

µ

= − + = − + = − − −− −

=

max min160 420ε µ ε µ = = −

(c) max max min 160 420γ ε ε µ µ = − = + max 580γ µ =


29/34




without permission.

PROBLEM 7.140

For the given state of plane strain, use Mohr’s circle to determine (a) the orientation and magnitude of the

principal strains, (b) the maximum in-plane shearing strain, (c) the maximum shearing strain.

60 , 240 , 50 x y xyε µ ε µ γ µ = + = + = −

SOLUTION

Plotted points:

: (60 , 25 )

: (240 , 25 )

: (150 , 0)

X

Y

C

µ µ

µ µ

µ

−

50tan 2 0.277778

60 240

xy

p

x y

γ θ

ε ε

−= = =

− −

2 15.52 pθ = ° 97.76aθ = °

7.76bθ = °

2 2(90 ) (25 ) 93.4 R µ µ µ = + =

(a) ave 150 93.4a Rε ε µ µ = + = + 243.4aε µ =

ave 150 93.4b Rε ε µ µ = − = − 56.6bε µ =

(b) max (in-plane) 2 Rγ = max (in-plane) 186.8γ µ =

(c) max min0, 243.4 , 0cε ε µ ε = = =

max max minγ ε ε = − max 243.4γ =


30/34




without permission.

PROBLEM 7.147

The strains determined by the use of the rosette attached as shown during the test of a

machine element are

61

62

63

93.1 10 in./in.

385 10 in./in.

210 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

Determine (a) the orientation and magnitude of the principal strains in the plane of therosette, (b) the maximum in-plane shearing strain.

SOLUTION

Use1 1

( ) ( )cos 2 sin 2

2 2 2

xy

x x y x y

γ ε ε ε ε ε θ θ ′ = + + − +

where 75θ = − ° for gage 1,

0θ = for gage 2,

and 75θ = + ° for gage 3.

1

1 1( ) ( )cos( 150 ) sin ( 150 )

2 2 2

xy

x y x y

γ ε ε ε ε ε = + + − − ° + − ° (1)

2

1 1( ) ( )cos 0 sin 0

2 2 2

xy

x y x y

γ ε ε ε ε ε = + + − + (2)

3 1 1( ) ( )cos (150 ) sin (150 )2 2 2

xy x y x y

γ ε ε ε ε ε = + + − ° + ° (3)

From Eq. (2), 6385 10 in/in x z ε ε −= = ×

Adding Eqs. (1) and (3),

1 3

1 3

6 6 6

6

( ) ( )cos 150

(1 cos 150 ) (1 cos 150 )

(1 cos 150 )

(1 cos 150 )

93.1 10 210 10 385 10 (1 cos 150 )

1 cos 15035.0 10 in/in

x y x y

x y

x y

ε ε ε ε ε ε

ε ε

ε ε ε ε

− − −

−

+ = + + − °

= + ° + − °

+ − + °=

− °

− × + × − × + °=

− °= ×


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without permission.


Subtracting Eq. (1) from Eq. (3),

3 1

6 63 1

6

sin 150

210 10 ( 93.1 10 )

sin 150 sin 150

606.2 10 in/in

xy

xy

ε ε γ

ε ε γ

− −

−

− = °

− × − − ×= =° °

= ×

6

6 6

606.2 10tan 2 1.732

385 10 35.0 10

xy

p

x y

γ θ

ε ε

−

− −

×= = =

− × − × (a) 30.0 , 120.0a bθ θ = ° = °

6 6ave

6

2 2

2 26 66

6 6ave

1 1( ) (385 10 35.0 10 )

2 2

210 10 in/in

2 2

385 10 35.0 10 606.2350.0 10

2 2

210 10 350.0 10

x y

x y xy

a

R

R

ε ε ε

ε ε γ

ε ε

− −

−

− −−

− −

= + = × + ×

= ×

− = +

× − × = + = ×

= + = × + × 6560 10 in/inaε −= ×

6 6ave 210 10 350.0 10b Rε ε

− −= − = × − × 6140.0 10 in/inbε −= − ×

(b)max (in-plane) 6350.0 10 in/in

2 R

γ −= = × 6max (in-plane) 700 10 in/inγ

−= ×


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without permission.

PROBLEM 7.150

A single strain gage is cemented to a solid 4-in.-diameter steel shaft at anangle 25 β = ° with a line parallel to the axis of the shaft. Knowing that

611.5 10 psi,G = × determine the torque T indicated by a gage reading of6300 10 in./in.−×

SOLUTION

For torsion, 0

0 0

0,

1( ) 0

1( ) 0

1

2 2

x y

x x y

y y x

xy xy

v E

v E

G G

σ σ τ τ

ε σ σ

ε σ σ

τ τ γ γ

= = =

= − =

= − =

= =

Draw the Mohr’s circle for strain.

0

0

2

sin 2 sin 22

x

RG

RG

τ

τ ε β β ′

=

= =

But 0 322

sin 2

xGTc T

J c

ε τ

β π

′= = =

3

3 6 6

sin 2

(2) (11.5 10 )(300 10 )

sin 50

xc GT π ε

β

π

′

−

=

× ×=

°

3113.2 10 lb in= × ⋅ 113.2 kip inT = ⋅


33/34




without permission.

PROBLEM 7.157

Solve Prob. 7.156, assuming that the rosette at point A indicates the

following strains:

61

62

63

30 10 in./in.

250 10 in./in.

100 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

PROBLEM 7.156 A centric axial force P and a horizontal force Q x are both

applied at point C of the rectangular bar shown. A 45° strain rosette on thesurface of the bar at point A indicates the following strains:

61

62

63

60 10 in./in.

240 10 in./in.

200 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

Knowing that 629 10 psi E = × and 0.30,v = determine the magnitudes of

P and Q x .

SOLUTION

61

63

62 1 3

2 2

2 2

3

3

30 10

100 10

2 430 10

29( ) [ 30 (0.3)(100)]

1 1 (0.3)

0

29( ) [100 (0.3)( 30)]

1 1 (0.3)

2.9 10 psi

(2)(6)(2.9 10 )

x

y

xy

x x y

y y x

y y

E v

v

E v

v

P P A

A

ε ε

ε ε

γ ε ε ε

σ ε ε

σ ε ε

σ σ

−

−

−

= = − ×

= = + ×

= − − = ×

= + = − +− −

=

= + = + −− −

= ×

= = = ×

334.8 10 lb= × 34.8 kips P =


34/34



66

3

3 3 4

3

29 1011.1538 10 psi

2(1 ) (2)(1.30)

(11.1538)(430) 4.7962 10 psi

1 1(2)(6) 36 in

12 12ˆ (2)(3)(1.5) 9 in

2 in.

xy xy

E G

v

G

I bh

Q Ay

t

τ γ

×= = = ×

+

= = = ×

= = =

= = =

=

33

ˆ

(36)(2)(4.7962 10 )38.37 10 lb

ˆ 9

xy

xy

VQ

It

It V

Q

Q V

τ

τ

=

×= = = ×

= 38.4 kipsQ =

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196634050 Mechanics of Materials

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