PRO OBLEM 7.2 2 - Prexams.com

PRONot on a

SO

OPRIETARY MAauthorized for sal

a website, in whole

OLUTION

F

F

60 MPa

90 M

608

TERIAL. Copyrile or distribution ie or part.

PRO

For ththe obased

0: 9A

180sin 30 c

0: 90 A290(cos 30

MPa

ight © 2015 McGin any manner. Th

OBLEM 7.2

he given stateblique face of

d on the equilib

90 sin 30 cosA

cos30 60co

0 sin 30 sin 3A2sin 30 )

Graw-Hill Educathis document may

1028

2

e of stress, detf the shaded tbrium of that e

s30 90 coA

2os 30

30 90 cos A

60cos30 sin

tion. This is proprnot be copied, sca

termine the notriangular elemelement, as wa

s30 sin 30

30 cos 30

30

rietary material soanned, duplicated,

ormal and shement shown. Uas done in the

60 cos30 cA

60 cos 30 sinA

olely for authorized, forwarded, distri

earing stressesUse a methodderivations of

cos30 0

3

n 30 0

7

d instructor use. buted, or posted

s exerted on d of analysis f Sec. 7.1A.

32.9 M Pa

71.0 M Pa

PROPRIETARY MATERIAL. © 2013 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.

For the given state of stress, determine the normal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element, as was done in the derivations of Sec. 7.2.

SOLUTION

0: 18 cos15 sin 15 45 cos15 cos15 27 sin 15 sin 15 +18 sin 15° cos15 0F A A A A AσΣ = + ° ° + ° ° − ° ° ° =

2 218 cos 15 sin 15 45 cos 15 27 sin 15 18 sin 15 cos 15σ = − ° ° − ° + ° − ° °

49.2 MPaσ = −

0: 18 cos 15 cos 15 45 cos 15 sin 15 27 sin 15 cos 15 18 sin 15° sin 15 0F A A A A AτΣ = + ° ° − ° ° − ° ° − ° =

2 218(cos 15 sin 15 ) (45 27)cos 15 sin 15 τ = − ° − ° + + ° °

2.41 MPaτ =

3

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

1026

BeerMOM_ISM_C07.indd 1026 1/8/2015 10:04:31 AM




without permission.

PROBLEM 7.6

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

SOLUTION

28 MPa 140 MPa 42 MPax y xyσ σ τ= = = −

(a) 2 (2)( 42)

tan 2 = 0.75028 140

xyp

x y

τθ

σ σ−= =

− −

2 36.87pθ = ° 18.43 ,108.43pθ = ° °

(b) 2

2max, min

22

+2 2

28 140 28 140( 42)

2 2

84 70

x y x yxy

σ σ σ σσ τ

+ − = ±

+ − = ± + −

= ±

max 154 MPaσ =

min 14 MPaσ =


1033

PROBLEM 7.7

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

SOLUTION

150 MPa, x 30 MPa, y 80 MPa xy

(a)

2 2( 80 MPa)tan 2 1.33333 MPa( 150 MPa 30 MPa)

xyp

x y

2 53.130 and 126.870 p

26.6 and 63.4 p ◄

(b) 2max,min 2 2

x y x yxy

2

2150 MPa 30 MPa 150 MPa 30 MPa ( 80 MPa)2 2

90 MPa 100 MPa

max 190.0 MPa ◄

min 10.00 MPa ◄

30 MPa

80 MPa

150 MPa


1030





without permission.

PROBLEM 7.1

For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the corresponding normal stress.

SOLUTION

63 MPa 42 MPa 28 MPax y xyσ σ τ= = − =

(a) 63 42

tan 2 1.8752 (2)(28)x y

sxy

σ σθ

τ− += − = − = −

2 61.93sθ = − ° 30.96 , 59.04sθ = − ° °

(b) 2

2max 2

x yxy

σ στ τ

− = +

2

263 42(28) 59.5 MPa

2

+ = + =

(c) ave63 42

10.5 MPa2 2

x yσ σσ σ

+ −′ = = = =

2




without permission.

PROBLEM 7.1

For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated (a) 25° clockwise, (b) 10° counterclockwise.

SOLUTION

0 56 MPa 35 MPa

28 MPa 28 MPa2 2

cos 2 sin 22 2

sin 2 cos 22

cos2 sin 22 2

x y xy

x y x y

x y x yx xy

x yx y xy

x y x yy xy

σ σ τσ σ σ σ

σ σ σ σσ θ τ θ

σ στ θ τ θ

σ σ σ σσ θ τ θ

′

′ ′

′

= = =

+ −= = −

+ −= + +

−= − +

+ −= − −

(a) 25θ = − ° 2 50θ = − °

28 28 cos( 50 ) 35 sin( 50 ) 16.8 MPaxσ ′ = − − ° + − ° = −

28 sin( 50 ) 35 cos( 50 ) 1.05 MPax yτ ′ ′ = − ° + − ° =

28 28 cos( 50 ) 35 sin( 50 ) 72.8 MPayσ ′ = + − ° − − ° =

(b) 10 2 20θ θ= ° = °

28 28 cos(20 ) 35 sin(20 ) 13.7 MPaxσ ′ = − ° + ° =

28 sin(20 ) 35 cos(20 ) 42.5 MPax yτ ′ ′ = ° + ° =

28 28 cos(20 ) 35 cos(20 ) 42.4 MPayσ ′ = + ° − ° =

3


1036


PRONot on a

SO

For

All

50 m


a website, in whole

OLUTION

r plane a-a,

lowable value

P

mm

a


PR

Twoa-a strescent

65 .

0

s

s

x

P

P

of P is the sm

a 25�


OBLEM 7.2

o members of that forms an

sses for the glutric load P that

2

2

0, 0,

cos

(50sin 65

( )sin

sin 65 cos65

xy

x y

x y

A

A

maller one.


1048

22

uniform crossn angle of 25ued joint are t can be applie

2

3

2

3

sin 2

10 )(80 1sin 6

cos (

(50 10 )(

y

y xy

xy

PA

sin


s section 50 5 with the h

800 kPa aed.

3 3

2 2

sin cos

0 )(800 1065

(cos sin

(8

30 10 )(60n 65 cos65


80 mm are ghorizontal. Knand 600 k

2

33

0 sin 65

) 3.90 10

) sin 65

PA

PA

30 10 ) 6.27


glued togethernowing that thkPa, determin

5 0

N

cos65 0

37 10 N

P


r along plane he allowable ne the largest

3.90 kN


1049

PROBLEM 7.23

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 the same 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 TJ c

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 MPa51.472 10

I c

M

MyI

Top view: Stresses:

ave

22 2 2

139.882 MPa 0 54.399 MPa

1 1( ) ( 139.882 0) 69.941 MPa2 2

( 69.941) ( 54.399) 88.606 MPa2

x y xy

x y

x yxyR

(a) max ave 69.941 88.606R max 18.67 MPa

min ave 69.941 88.606R min 158.5 MPa

3 kN

3 kN

350 N · m

0.15 mH

0.2 m


1046





without permission.

PROBLEM 7.2

A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 100 N force at A, determine the principal stresses and the maximum shearing stress at point H located as shown on top of the 18 mm diameter shaft.

SOLUTION

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

100 N (100)(150) 15000 N mm

(100)(250) 25000 N mm

V M

T

= = = ⋅= = ⋅

Shaft cross section:

4 4 4

118 mm, 9 mm

21

10306 mm 5153 mm2 2

d c d

J c I Jπ

= = =

= = = =

Torsion: 2(25000)(9)21.8 N/mm 21.8 MPa

10306

Tc

Jτ = = = =

Bending: 2(15000)(9)26.2 N/mm 26.2 MPa

5153

Mc

Iσ = = = =

Transverse shear: At point H stress due to transverse shear is zero.

Resultant stresses:

ave

22 2 2

26.2 MPa, 0, 21.8 MPa

1( ) 13.1 MPa

2

13.1 21.8 25.4 MPa2

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = + = + =

ave 38.5 MPaa Rσ σ= + =

ave 12.3 MPab Rσ σ= − = −

max 25.4 MPaRτ = =

5

PRONot aon a w

SO

Forc

Tor

10 kN

OPRIETARY MATauthorized for salewebsite, in whole

LUTION

ce-couple syst

sion: At po

200 mm

6 m

z

N

C

TERIAL. Copyrige or distribution inor part.

tem at center o

oint K, place l

mm

1

T

A

y

KH

A

B

ght © 2015 McGrn any manner. Thi

P

Ththdeat

4

12

24.185512

oo

o

dr

J r

I J

of tube in the p

10 kN

10 10

(10 1

2000 N

(101500

x

y

z

F

M

M

local x-axis in

51

(

24.3724.37

y

o

xy

T M

c r

TcJ

150 mm

51 mm

x

D

raw-Hill Educatiois document may n

1053

PROBLEM 7

he steel pipe Ahickness. Knoetermine the pt point K.

4

6 4

6

102 51 mm2

4.1855

5 10 m

2.0927 10 m

ir

plane containi

3

3 3

3

0 N

0 )(200 10

N m

10 )(150 100 N m

negative glob

3

6

6

2000 N m

1 10 m

(2000)(51 104.1855 1010 Pa

MPa

on. This is proprinot be copied, scan

7.26

AB has a 102wing that armprincipal stres

6 4

4

5 10 mm

m

i or r t

ing points H a

3

3

)

)

bal z-direction

3

60 )

etary material solenned, duplicated,

2-mm outer dim CD is rigidsses and the m

45 mm

and K:

n.

ely for authorizedforwarded, distrib

ameter and a dly attached tmaximum she


6-mm wall to the pipe, aring stress


1050





without permission.

PROBLEM 7.2

A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 100 N force at A, determine the principal stresses and the maximum shearing stress at point H located as shown on top of the 18 mm diameter shaft.

SOLUTION

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

100 N (100)(150) 15000 N mm

(100)(250) 25000 N mm

V M

T

= = = ⋅= = ⋅

Shaft cross section:

4 4 4

118 mm, 9 mm

21

10306 mm 5153 mm2 2

d c d

J c I Jπ

= = =

= = = =

Torsion: 2(25000)(9)21.8 N/mm 21.8 MPa

10306

Tc

Jτ = = = =

Bending: 2(15000)(9)26.2 N/mm 26.2 MPa

5153

Mc

Iσ = = = =

Transverse shear: At point H stress due to transverse shear is zero.

Resultant stresses:

ave

22 2 2

26.2 MPa, 0, 21.8 MPa

1( ) 13.1 MPa

2

13.1 21.8 25.4 MPa2

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = + = + =

ave 38.5 MPaa Rσ σ= + =

ave 12.3 MPab Rσ σ= − = −

max 25.4 MPaRτ = =

5

PRONot aon a w

SO

Forc

Tor

10 kN


LUTION

ce-couple syst

sion: At po

200 mm

6 m

z

N

C


tem at center o

oint K, place l

mm

1

T

A

y

KH

A

B


P

Ththdeat

4

12

24.185512

oo

o

dr

J r

I J

of tube in the p

10 kN

10 10

(10 1

2000 N

(101500

x

y

z

F

M

M

local x-axis in

51

(

24.3724.37

y

o

xy

T M

c r

TcJ

150 mm

51 mm

x

D


1053

PROBLEM 7

he steel pipe Ahickness. Knoetermine the pt point K.

4

6 4

6

102 51 mm2

4.1855

5 10 m

2.0927 10 m

ir

plane containi

3

3 3

3

0 N

0 )(200 10

N m

10 )(150 100 N m

negative glob

3

6

6

2000 N m

1 10 m

(2000)(51 104.1855 1010 Pa

MPa


7.26

AB has a 102wing that armprincipal stres

6 4

4

5 10 mm

m

i or r t

ing points H a

3

3

)

)

bal z-direction

3

60 )


2-mm outer dim CD is rigidsses and the m

45 mm

and K:

n.


ameter and a dly attached tmaximum she


6-mm wall to the pipe, aring stress


1051



1054

PROBLEM 7.26 (Continued)

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

Bending:

3

66

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

zy

M cI

Point K lies on compression side of neutral axis.

36.56 MPay

Total stresses at point K:

ave

22

0, 36.56 MPa, 24.37 MPa

1 ( ) 18.28 MPa2

30.46 MPa2

x y xy

x y

x yxyR

max ave 18.28 30.46R max 12.18 MPa

min ave 18.28 30.46R min 48.7 MPa

max R max 30.5 MPa


1055

PROBLEM 7.27

For the state of plane stress shown, determine the largest value of y for which the maximum in-plane shearing stress is equal to or less than 75 MPa.

SOLUTION

60 MPa, ?, 20 MPax y xy

Let

.2

x yu

Then

2 2

2 2 2 2

2

75 MPa

75 20 72.284 MPa

2 60 (2)(72.284) 84.6 MPa or 205 MPa

y x

xy

xy

y x

u

R u

u R

u

Largest value of y is required. 205 MPay

60 MPa

20 MPa

�y


1052


PRONot on a

SO

Plo

(a)


a website, in whole

OLUTION

otted points fo

25

90 MP

3


r Mohr’s circl

:::

XYC

tan 2 p

2 p

R

. 2 5

2

x

x y

y

Pa

30 MPa

60 MPa


PROBLEM

Solve Prob.

PROBLEMnormal and s(a) 25 clock

ave

60 MP90 MPa,

30 MPa

2

x

y

xy

x y

le:

: ( 60 MPa,: (90 MPa, 30: (15 MPa, 0)

30 075

FXFC

21.80 P

2FC FX

50

2 2 50P

ave cos R

sinR

ave cos R


1064

M 7.36

7.14, using M

M 7.13 througshearing streskwise, (b) 10

Pa,,

15 MPa

30 MPa)0 MPa)

0.4

10.90

2 275 30X

21.80 28


Mohr’s circle.

gh 7.16 For thses after the e counterclock

20 80.78 MP

8.20


he given stateelement shownkwise.

Pa


e of stress, den has been rota

5 x

3 x y

8y


etermine the ated through

56.2 MPa

38.2 MPa

86.2 MPa

PRONot aon a w

(b)


10


. 2 20

2

x

x y R

y


PROB

0

2 21.8 p

ave cos R

sinR

ave cos R


1065

BLEM 7.36 (

80 20 41


(Continued)

.80


d)


45 x

53 x y

75 y


5.2 MPa

3.8 MPa

5.2 MPa


1062


PRONot on a

SO

Plo

(a)


a website, in whole

OLUTION

otted points fo

25

90 MP

3


r Mohr’s circl

:::

XYC

tan 2 p

2 p

R

. 2 5

2

x

x y

y

Pa

30 MPa

60 MPa


PROBLEM

Solve Prob.

PROBLEMnormal and s(a) 25 clock

ave

60 MP90 MPa,

30 MPa

2

x

y

xy

x y

le:

: ( 60 MPa,: (90 MPa, 30: (15 MPa, 0)

30 075

FXFC

21.80 P

2FC FX

50

2 2 50P

ave cos R

sinR

ave cos R


1064

M 7.36

7.14, using M

M 7.13 througshearing streskwise, (b) 10

Pa,,

15 MPa

30 MPa)0 MPa)

0.4

10.90

2 275 30X

21.80 28


Mohr’s circle.

gh 7.16 For thses after the e counterclock

20 80.78 MP

8.20


he given stateelement shownkwise.

Pa


e of stress, den has been rota

5 x

3 x y

8y


etermine the ated through

56.2 MPa

38.2 MPa

86.2 MPa

PRONot aon a w

(b)


10


. 2 20

2

x

x y R

y


PROB

0

2 21.8 p

ave cos R

sinR

ave cos R


1065

BLEM 7.36 (

80 20 41


(Continued)

.80


d)


45 x

53 x y

75 y


5.2 MPa

3.8 MPa

5.2 MPa


1063



1072

PROBLEM 7.43

7.43 Solve Prob. 7. , using Mohr’s circle.7.21 The centric force P is applied to a short post as shown. Knowing

that the stresses on plane a-a are s = – 100 MPa and t = 35 MPa,determine (a) the angle b that plane a-a forms with the horizontal,(b) the maximum compressive stress in the post.

SOLUTION

sx = 0

txy = 0

sy = – P/A

From the Mohr’s circle

tan b = 35

100 = 0.35 b = 19.29º 180° – 2b

2bb

Y

35

b X

X¢

Y¢100

P A/

s

– s = P

A2 +

P

A2cos 2b

21

P

A=

2

1 2

( )

cos

-+

sb

= 2 100

1 2

( )

cos+ b= 112.25 MPa


1073

PROBLEM 7.44

Solve Prob. 7.22, using Mohr’s circle.

PROBLEM 7.22 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 allowable stresses for the glued joint are 800 kPa and 600 kPa, determine the largest centric load P that can be applied.

SOLUTION

3 3

3 2

00

/

(50 10 )(80 10 )

4 10 m

(1 cos50 )2

21 cos50

x

xy

y P A

A

PA

AP

3 3

3

(2)(4 10 )(800 10 )1 cos50

3.90 10 N

P

P

3 3

32 (2)(4 10 )(600 10 )sin 50 6.27 10 N2 sin 50 sin 50P APA

Choosing the smaller value, 3.90 kNP

P

a 25�

50 mm

a


1071



1087

PROBLEM 7.55

Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown.

SOLUTION

Consider the state of stress on the left. We shall express it in terms of horizontal and vertical components.

Principal axes and principal stress:

ave1 (118.3 56.7) 87.52

1 (118.3 56.7) 30.82 2

yx

2 2(30.8) (75) 81.08 R

75tan 2 2 67.6730.8

p p 33.8 p , and 123.8

avemax 87.5 81.08 R max 168.6 MPa

in avem 87.5 81.08 R min 6.42 MPa

100 MPa

50 MPa

50 MPa

75 MPa+

308

50cos3043.30

43.30

50sin 30

25.0

x

y

xy

PRONot on a

SO

Exp

Ad


a website, in whole

OLUTION

press each sta

dding the two s

�0


te of stress in

states of stress

��0

30�


terms of horiz

s,

�0


1088

zontal and ver

�0

30�


PROBLEM

Determine thstresses for tthe superposi

rtical compone


M 7.56

he principal pthe state of plition of the two

ents.


planes and thlane stress reso states of stre

0 ap

m


he principal sulting from ess shown.

and 90°

max 0

min 0


1084


PROBLEM 7.67

7.6 For the state of plane stress shown, determine the maximum shearing stress when (a) sx = 0 andsy = 82 MPa (b) sx = 145 MPa and sy = 62 MPa. (Hint: Consider both in-plane and out-of-planeshearing stresses.)

y

x

�x

56 MPa

�y

z

SOLUTION

(a) sx = 0, sy = 82 MPa, txy = 56 MPa

save = 1

2(sx + sy) = 41 MPa

Y

AB s (MPa)

t (MPa)

41 69.4

X

R = s s

tx yxy

-��

�� +

2

22

= ( )- +41 562 2

= 69.4 MPasa = save + R = 110.4 MPa (max)sb = save – R = – 28.4 MPa (min)sc = 0

smax = 110.4 MPa smin = – 28.4 MPa

tmax = 1

2(smax – smin) = 69.4 MPa

(b) sx = 145 MPa sy = 62 MPa txy = 56 MPasave = 103.5 MPa

R = s s

tx yxy

-��

�� +

2

22

t (MPa)

O B C A s (MPa)

103.5 69.7

Y

X

= ( . )- +41 5 562 2 = 69.7 MPa

sa = save + R = 173.2 MPa (max)sb = save – R = 33.8 MPasc = 0 (min)

smax = 173.2 MPa, smin = 0

tmax = 1

2(smax – smin) = 86.6 MPa

7


1102

PROBLEM 7.68

For the state of stress shown, determine the maximum shearing stress when (a) 40y MPa, (b) 120y MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION

(a)

ave

140 MPa, 40 MPa, 80 MPa

1 ( ) 90 MPa2

x y xy

x y

2

2 2 250 80 94.34 MPa2

x yxyR

ave

ave

184.34 MPa (max)4.34 MPa (min)

0

a

b

c

RR

max(in-plane)1 ( ) 94.34 MPa2 a b R

max max min1 1( ) ( ) 94.3 MPa2 2 a b max 94.3 MPa

(b)

ave

140 MPa, 120 MPa, 80 MPa

1 ( ) 130 MPa2

x y xy

x y

2

2 2 210 80 80.62 MPa2

x yxyR

ave

ave

210.62 MPa (max)49.38 MPa

0 (min)

a

b

c

RR

max min210.62 MPa 0a c

max(in-plane) 86.62 MPaR

max max min1 ( ) 105.3 MPa2

max 105.3 MPa

80 MPa

y

zx

140 MPa

σy


1097



1105

PROBLEM 7.71

For the state of stress shown, determine the maximum shearing stress when (a) z 0, (b) z 60 MPa, (c) z 60 MPa.

SOLUTION

ave

1 (170 100) 1352

1 1( ) (170 100) 352 2 x y

2 2(35) (84) 91 R

135 91 226 MPa A

135 91 44 MPa B

(a) 0. z Point Z corresponding to the z axis is located at O, outside the circle drawn through A and B. The largest of the 3 Mohr’s circles is the circle through O and A. We have

max1 1 1( ) (226)2 2 2

AOA max 113.0 MPa

(b) 60 MPa. z Point Z is located between B and A. The largest of the 3 circles is the one drawn through A and B.

max 91.0 MPa R

(c) 60 MPa. z Point Z is located outside the circle drawn through A and B. The largest of the 8 Mohr’s circles is the circle through Z and A. We have

max1 1( ) (60 226)2 2

ZA max 143.0 MPa

zx

84 MPa

y

170 MPa

100 MPa

z

6464

PROBLEM 7.727.72


1100





without permission.

PROBLEM 7.7

For the state of stress shown, determine the value of xyτ for which the maximum shearing stress is (a) 63 MPa, (b) 84 MPa.

SOLUTION

ave

105 MPa 42 MPa

1( ) 73.5 MPa

2

31.5 MPa2

x y

x y

x yU

σ σ

σ σ σ

σ σ

= =

= + =

−= =

τ (MPa)

(a) maxFor 63 MPaτ =

Center of Mohr’s circle lies at point C. Lines marked (a) show the limits on max .τ Limit on maxσ is max max2 126 MPa.σ τ= = For the Mohr’s circle maxaσ σ= corresponds to point Aa.

ave

2 2

2 2

2 2

126 73.552.5 MPa

52.5 31.542 MPa

a

xy

xy

R

R U

R U

σ σ

τ

τ

= −= −=

= +

= ± −

= ± −= ±

(b) maxFor 84 MPaτ =

Center of Mohr’s circle lies at point C.

84 MPaR =

2 2 78.7 MPaxy R Uτ = ± − = ±

Checking 73.5 84 157.5 MPa

73.5 84 10.5 MPa

0

a

b

c

σσσ

= + == − = −=

max max min1

( ) 84 MPa O.K.2

τ σ σ= − =

5


1111

PROBLEM 7.76

For the state of stress shown, determine two values of y for which the maximum shearing stress is 73 MPa.

SOLUTION

50 MPa, 48 MPax xy

Let 22

y xy xu u

ave1 ( )2 x y x u

2 2 2 2xy xyR u u R

Case (1) 2 2max 73 MPa, 73 48 55 MPaR u

(1a) 55 MPa 2 60 MPay xu u

ave

ave ave

1 ( ) 5 MPa2

78 MPa, 68 MPa

x y

a bR R

max min max0 78 MPa, 68 MPa, 73 MPaa

(1b) 55 MPa 2 160 MPa (reject)y xu u

ave ave

ave max

1 ( ) 105 MPa, 32 MPa2

178 MPa, 0, 0

x y a

b c

R

R

min max max min1178 MPa, ( ) 89 MPa 73 MPa2

60.0 MPay

48 MPa

50 MPa

y

zx

σy


1104



1116

PROBLEM 7.79

For the state of stress shown, determine two values of y for which the maximum shearing stress is 80 MPa.

SOLUTION

90 MPa, 0, 60 MPax z xz

Mohr’s circle of stresses in zx plane:

ave

22 2 2

1 ( ) 45 MPa2

45 60 75 MPa2

x z

x yzxR

ave ave120 MPa, 30 MPaa bR R

Assume max 120 MPa.a

min max max2

120 (2)(80)y

40.0 MPa y

Assume min 30 MPa.b

max min max2

30 (2)(80)y

130.0 MPa y

z

y

x

σy

90 MPa

60 MPa


1117

PROBLEM 7.80*

For the state of stress of Prob. 7.69, determine (a) the value of y for which the maximum shearing stress is as small as possible, (b) the corresponding value of the shearing stress.

SOLUTION

Let 22

x yy xu u

ave

2 2

2 2ave

2 2ave

1 ( )2 x y x

xy

a x xy

b x xy

u

R u

R u u

R u u

Assume max is the in-plane shearing stress. max R

Then max (in-plane) is minimum if 0.u

ave

ave

ave

max min max max min

2 140 MPa, 140 MPa

80 MPa

140 80 220 MPa140 80 60 MPa

1220 MPa, 0, ( ) 110 MPa2

y x x x

xy

a

b

u u

R

RR

Assumption is incorrect.

Assume 2 2max ave

min max max min

2 2

1 10 ( )2 2

1 0 (no minimum)

a x xy

a

a

xy

R u u

d udu u

80 MPa

y

zx

140 MPa

σy


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PRO OBLEM 7.2 2 - Prexams.com