345021434-Mechanics-of-Materials-7th-Edition-Beer-Johnson-Chapter-7.pdf

CHAPTER 7

PR ROBLEM 7.1 7

4 ksi 3 ksi 708 8 ksi

Foor the given sttate of stress, determine thee normal and shearing stressses exerted onn the oblique face of the shhaded trianguular element shown. s Use a method of anaalysis based on o the equilibrrium of that ellement, as waas done in the derivations of Sec. 7.1A.

SOLUTION

F

0:

A

8 A cos 20 2 cos 20

8cos 2 20

3cos 20 sin 20

3 A cos 20 sin 20 2 3 sin 20 cos 20

3 A sin 200 cos 20 4sin 2 20

4 A sin 20 sin 20 2

0

0

9.46 ksi F

0:

A

8 A cos 200 sin 20

8coos 20 sin 20

3(cos2 20

3A A cos 20 cos 20

sin 2 20 )

3 A sin 200 sin 20

4A A sin 20 cos 200

0

4 20 cos 200 4sin

1.013 ksi

PRO OPRIETARY MAT TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use. Not authorized a for salee or distribution inn any manner. Thiis document may not n be copied, scannned, duplicated, forwarded, distribbuted, or posted on a website, w in whole or part. 1027

PRO OBLEM 7.2 2

60 MPa

For th he given statee of stress, dettermine the noormal and sheearing stressess exerted on the oblique face off the shaded triangular t elem ment shown. Use U a methodd of analysis based d on the equilibbrium of that element, e as waas done in the derivations off Sec. 7.1A.

608 90 MPa M

SO OLUTION

F

0:

90 9 A sin 30 coss 30

A

180sin 30 cos c 30

90 A cos 30 sin 30

60 A cos 30 ccos 30

0

60 coos 2 30 3 32.9 M Pa

F

0:

A

90 0 A sin 30 sin 30 3

90(cos 2 30

sin 2 30 )

90 A cos 30 cos 30

60 A cos 30 sinn 30

0

60 cos 30 sin 30 7 71.0 M Pa

PRO OPRIETARY MATERIAL. Copyriight © 2015 McG Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use. Not authorized for salle or distribution in i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted on a website, in wholee or part. 1028

PROBLEM M 7.3

10 ksi

For the giveen state of sttress, determiine the normaal and sheariing stresses exerted on thhe oblique faace of the shaaded triangulaar element shoown. Use a method of annalysis based on the equilibbrium of that element, as was w done in the derivationns of Sec. 7.1A A.

6 ksi

758

4 ksi

SOLUTION

F

0:

A

4 A cos15 sin15

4 co os15 sin15

10 cos 2 15

100 A cos15 cos115

6sin 2 15

6 A sin15 sin15

4 A sin15 cos155

0

4 4sin15 cos155 1 10.93 ksi

F

0:

A

4(ccos2 15

4 A cos15 cos15

10 A cos15 sin 15

sin 2 15 )

6) cos15 sin15

(10

6 A sin15 cos15

4 A sin15 sin155

0

0 0.536 ksi


80 MPa M

PROBLEM P 7.4 For F the given state s of stress, determine thhe normal andd shearing streesses exerted on o the obliquee face of the shaded trianggular element shown. Use a method of analysis a based on the equilibbrium of that element, e as was w done in thee derivations of o Sec. 7.1A.

40 MPa 558

SO OLUTION

Streesses

F

0 0:

A

80 A cos 55 cos555

80 cos 2 55

F

0 0:

A

Areas

Forces

40 A sin 55 sin 55

40sin 2 55

80 A cos 55 sin 55 5

0 0.5521 MPa

40 A sin 55 cos 55 5 MPa 56.4

1 cos 55 sin 55 120


PROBLEM 7.5

40 MPa 35 MPa 60 MPa

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

SOLUTION 60 MPa

x

(a)

tan 2

2

xy

p x

2

p

y

(2)(35) 60 40

y

40 MPa

xy

35 MPa

3.50

74.05

37.0 , 53.0

p 2

(b)

x max, min

y

x

2 60 40 2

y

2 xy

2 60 40 2

2

(35)2

50 36.4 MPa max min

13.60 MPa 86.4 MPa

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. 1031

PROBLEM 7.6

10 ksi


2 ksi

3 ksi

SOLUTION x

(a)

tan 2

2 ksi

2 p

p

y

3 ksi

xy

(2)( 3) 2 10

xy

x

2

10 ksi

y

0.750

36.87

p

18.4 , 108.4 ◄

2

(b)

x

max,min

x

y

2 2

10 2

6

y

2 2

10 2

2 xy

2

( 3)2

5 ksi

max

11.00 ksi ◄

min

1.000 ksi ◄


PROBLEM 7.7

30 MPa


150 MPa

80 MPa

SOLUTION x

(a)

tan 2

150 MPa,

2 p

2

p

y

53.130

80 MPa

xy

2( 80 MPa) ( 150 MPa 30 MPa)

xy

x

30 MPa,

y

1.33333 MPa

and 126.870 p

(b)

max,min

x

y

x

2

y

2

150 MPa

30 MPa 2

90 MPa

26.6 and 63.4 ◄

2 xy

150 MPa

30 MPa 2

2

( 80 MPa)2

100 MPa

max min

190.0 MPa ◄ 10.00 MPa ◄


PROBLEM 7.8

12 ksi 8 ksi


18 ksi

SOLUTION x

(a)

tan 2

18 ksi

2 p

2

p

(2)(8) 18 12

xy

x

12 ksi

y

y

xy

8 ksi

0.5333

28.07

14.0 , 104.0 ◄

p

2

(b)

max,min

x

y

x

2 18

12 2

y

2 18

12 2

2 xy

2

(8)2

3 17 ksi max min

20.0 ksi ◄ 14.00 ksi ◄


PROBLEM 7.9

40 MPa 35 MPa

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) the corresponding normal stress.

60 MPa

SOLUTION x

(a)

tan 2 2

x s

s

60 40 (2)(35)

y

2

xy

60 MPa

y

40 MPa

xy

35 MPa

0.2857

15.95

s

8.0 , 98.0

2

(b)

(c)

x max

y

2 xy

2 60 40 2

2

x

y

ave

2

(35) 2

max

60 40 2

36.4 MPa

50.0 MPa


PROBLEM 7.10

10 ksi


2 ksi

3 ksi

SOLUTION 2 ksi

x

(a)

tan 2 2

y

x

s

2 10 (2)( 3)

y

2

10 ksi

xy

xy

3 ksi

1.33333

53.13

s

s

26.6 , 63.4

2

(b)

x

max

y

2 xy

2 2

10

2

( 3)2

2

max

(c)

ave

x

y

2

2

5.00 ksi

10 2

6.00 ksi


PROBLEM 7.11

30 MPa


150 MPa

80 MPa

SOLUTION 150 MPa,

x

(a)

tan 2 2

x

s

30 MPa,

150 30 2( 80)

y

2

y

xy

xy

80 MPa

0.750

36.87 and 216.87

s

s

18.4 and 108.4

2

(b)

x

max

y

2 xy

2 150

30

2

( 80)2

2

max

(c)

ave

x

100.0 MPa

y

2 150 30 2

90.0 MPa


PROBLEM 7.12

12 ksi 8 ksi


18 ksi

SOLUTION 18 ksi

x

(a)

tan 2 2

x

s

18 12 (2)(8)

y

2

12 ksi

y

xy

xy

8 ksi

1.875

61.93

s

31.0 , 59.0

s

2

(b)

x

max

y

2 xy

2 18

12

2

(8)2

2

max

(c)

ave

x

y

2

17.00 ksi

18 12 2

3.00 ksi


PROBLEM 7.13

8 ksi 5 ksi

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 x x

y

2

0

8 ksi

y

x

4 ksi y

25

2

x

xy y

y

2

x

y

2

4 sin ( 50 ) 5 cos ( 50 )

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

2.40 ksi

x

0.1498 ksi

xy

4 4 cos ( 50 ) 5 sin ( 50)

y

2

cos 2 +

sin 2 +

4 4 cos ( 50°) + 5 sin ( 50°)

xy

10

y

50

x

(b)

y

2 x

4 ksi

2

x

(a)

x

2

xy

y

y

2

x x

5 ksi

xy

10.40 ksi

y

20 4 4 cos (20°) + 5 sin (20°)

4 sin (20°) + 5 cos (20°) 4 4 cos (20°)

5 cos (20°)

x

1.951 ksi

xy

6.07 ksi

y

6.05 ksi


PROBLEM 7.14

90 MPa 30 MPa


60 MPa

SOLUTION x x

y

2

60 MPa

x

y

(a)

25

2

x

2

sin 2 +

y

2

x

y

2

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

x

56.2 MPa

xy

38.2 MPa

15 75 cos ( 50 ) 30 sin ( 50 )

y

86.2 MPa

20 15 75 cos (20°) + 30 sin (20°)

75 sin (20°) + 30 cos (20°)

xy y

y

75 sin ( 50 ) 30 cos ( 50 )

y

2

cos 2 +

15 75 cos ( 50 ) 30 sin ( 50 )

xy

10

y

50

x

(b)

x

30 MPa

xy

75 MPa

2

x

x

y

2

2

xy

y

x

15 MPa

x

90 MPa

y

15 75 cos (20°) 30 sin (20°)

x

45.2 MPa

xy

53.8 MPa

y

75.2 MPa


PROBLEM 7.15

12 ksi


8 ksi

6 ksi

SOLUTION x x

y

2

8 ksi

x

2 ksi x

x

y

2

x

2

y

2

xy

xy

sin 2

cos 2

cos 2

xy

sin 2

2 10 cos ( 50 ) 6 sin ( 50 )

9.02 ksi

x

10 sin ( 50 ) 6 cos ( 50 )

y

2

y

sin 2 +

cos 2 +

2 10 cos ( 50 ) 6 sin ( 50 )

xy

10

y

6 ksi

50

x

(b)

y

2 2

x

25

x

xy

10 ksi

2

x

(a)

y

2

xy

y

12 ksi

y

xy

3.80 ksi

13.02 ksi

y

20

x xy

y

2 10 cos (20°) 6 sin (20°)

x

10 sin (20°) 6 cos (20°)

2 10 cos (20°) + 6 sin (20°)

5.34 ksi

xy

9.06 ksi

y

9.34 ksi


PROBLEM 7.16

80 MPa

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. 50 MPa

SOLUTION x x

y

2

0

80 MPa

y

x x

y

(a)

25

2

2

sin 2 +

y

2

x

y

2

xy

xy

sin 2

xy

sin 2

cos 2

cos 2

40 sin ( 50°) 50 cos ( 50 ) 40 40 cos ( 50 ) 50 sin ( 50 )

y

2

y

cos 2

40 40 cos ( 50 ) 50 sin ( 50°)

xy

10

y

40 MPa

50 x

(b)

x

2

x

x

y

2

2

xy

y

x

40 MPa

50 MPa

xy

x

24.0 MPa 1.498 MPa

xy

y

104.0 MPa

x

19.51 MPa

xy

60.7 MPa

y

60.5 MPa

20 x

xy y

40 40 cos (20°) 50 sin (20°)

40 sin (20°) 50 cos (20°) 40 40 cos (20°) + 50 sin (20°)


PROBLEM 7.17

250 psi

The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain. 158

SOLUTION x

(a)

0

y

x

xy

0

250 psi

xy

y

sin 2 2 250cos( 30 )

15

xy cos 2

xy

(b)

x

x

0

y

2 0

x

y

cos 2 2 250sin( 30 )

217 psi

xy sin 2

x

125.0 psi


PROBLEM 7.18

1.8 MPa

The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

3 MPa

158

SOLUTION x

3 MPa

15 (a)

2

x

xy

1.8 MPa

y

y

2 3

1.8 2

xy

0

30

sin 2

xy sin 2

sin( 30 )

0

0.300 MPa

xy

(b)

x

x

y

x

2 3

y

2 1.8

2

3

1.8 2

cos 2 cos( 30 )

xy sin 2

0 x

2.92 MPa


P'

80 mm m

P PROBLEM 7 7.19

1200 mm

Tw wo wooden members m of 800 120-mm uniform u rectanngular cross seection are joined by the simpple glued scarrf splice shownn. Knowing 22 and thhat a that the maximum m allow wable stresses in the joint arre, respectivelly, 400 kPa inn tension (perrpendicular to the splice) annd 600 kPa inn shear (parallel to the splicce), determinee the largest ceentric load P thhat can be appplied.

b P

SOLUTION

Forces

Areeas

A (80) (120) 9.6 103 mm 2 N all

a all

Fy

0: N

Sall

Fx

aall

A/sin

P sin

A/sin

0:: S

P cos

9.6 10 3 m 2

(4400 103 )(9.6 10 3 ) 10.2251 103 N sin 22 0

P

N sinn

10.251 103 sin 222

27.4 1103 N

(6600 103 )(9.6 10 3 ) 15.3376 103 N sin 22

0

P

S coos

Thee smaller valuee for P governns.

15.376 103 cos 22 2

16.58 103 N P 16.58 kN


P'

PROBLEM 7.20 7

1220 mm

80 mm

Tw wo wooden members m of 800 120-mm uniform u rectanngular cross section are joineed by the simp mple glued scarrf splice show wn. Knowing 25 and a that centriic loads of magnitude m P 10 kN are that t in-plane appplied to the members ass shown, dettermine (a) the shhearing stresss parallel to the splice, (b) the noormal stress peerpendicular too the splice.

b P

SO OLUTION

Forcess

A Areas

A (80)(1220) 9.6 103 mm m 2 (a)

Fx

0: S N A/sin

(b)

Fy

0: N N A/sin

P cos

0

S

P cos

(9.063 103 )sinn 25 9.6 10 3

P sin n

0

N

P sin

(4.226 103 )sin 25 9.6 103

9.6 100 3 m 2

(10 103 ) cos 25

9.063 103 N

399 1003 Pa

(10 103 )sin 25 186.0 103 Pa

399 kPa

4.226 103 N 1 186.0 kPa


P

PROBL LEM 7.21 The centrric force P is applied to a short post as shown. Know wing that the stresses on 5 ksi, determinne (a) the anggle that planne a-a forms 15 ksi and plane a-a are with the horizontal, h (b) the maximum m compressivee stress in the post. p

a

!

a

SOLUTION

x

0

xy

0

y

(a)

P/ A

From the Mohr’s M circle,

tan

5 15

P 2A (b)

P A

2( ) 1 co os 2

0.33333

18.4

P cos 2 2A

(2)(115) 1 coss 2

P 16.67 1 ksi A


PROBLEM 7.2 22 a a

25"

50 mm m

Two o members of uniform crosss section 50 80 mm are glued g togetherr along plane a-a that forms ann angle of 255 with the horizontal. h Knnowing that thhe allowable 800 kPa and a 600 kPa, k determinne the largest stressses for the gluued joint are centtric load P thatt can be applieed.

P

SO OLUTION Forr plane a-a,

65 .

x

0, 0 x

P

0,

cos 2

x

y

y

P A

sin 2

2

xy

sin

cos

0

(50 10 3 )(80 10 3 )(800 103 ) sin 2 65 6

A sin s 2 65 (

P

xy

y )sin

A sin s 65 cos 65

P 2 sin 655 A

0

3.90 103 N

P sin 65 cos 65 0 A (50 10 3 )((80 10 3 )(600 103 ) 6.277 103 N sinn 65 cos 65

cos

( xy (cos

Alllowable value of P is the sm maller one.

2

sin 2 )

P

3.90 kN


PROBLEM 7.23

0.2 m 0.15 m 3 kN

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.

H

350 N · m 3 kN

SOLUTION 1 d 2

c Tc J

Torsion:

I

Bending:

4

2T c3

c4

4

1 (32) 2

2(350 N m) (16 10 3 m)3

(16 10 3 )4

(0.15m)(3 103 N)

M

16 mm 16 10 3 m 54.399 MPa

51.472 10 9 m 4 450 N m

(450)(16 10 3 ) 51.472 10 9

My I

54.399 106 Pa

Top view:

139.882 106 Pa

139.882 MPa

Stresses:

x

139.882 MPa 1 ( 2

x

y)

R

x

y

ave

y

0

1 ( 139.882 2

xy

0)

54.399 MPa 69.941 MPa

2

(a)

2

max

ave

R

69.941

88.606

min

ave

R

69.941

88.606

2 xy

( 69.941)2

( 54.399) 2

88.606 MPa

max min

18.67 MPa 158.5 MPa


PROBLEM 7.23 (Continued)

tan 2

2 p x

xy y

(2)( 54.399) 139.882

0.77778

2

p

37.88

p

(b)

max

R

88.6 MPa

18.9

max

and 108.9°

88.6 MPa


6 in.

PROBLEM 7.24

C H

A 400-lb vertical force is applied at D to a gear attached to the solid l-in. diameter shaft AB. Determine the principal stresses and the maximum shearing stress at point H located as shown on top of the shaft.

B

A D 2 in. 400 lb

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

Shaft cross section:

V

400 lb

T

(400)(2)

d

1 in. c

J

c4

2

1 d 2

(400)(6)

2400 lb in.

800 lb in.

0.5 in.

0.098175 in 4

Torsion:

Tc J

(800)(0.5) 0.098175

Bending:

Mc I

(2400)(0.5) 0.049087

Transverse shear:

M

1 J 2

I

0.049087 in 4

4.074 103 psi

4.074 ksi

24.446 103 psi

24.446 ksi

Stress at point H is zero. x ave

24.446 ksi, 1 ( 2

x

y)

x

y

y

0,

xy

4.074 ksi

12.223 ksi 2

R

2

2 xy

(12.223) 2

(4.074) 2

12.884 ksi a

ave

R

b

ave

R

max

R

a

25.1 ksi

b

0.661 ksi

max

12.88 ksi


PROBLEM 7.25

H

E

A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 34 -in. diameter shaft.

6 in.

B 24 lb A

10 in.



d

V

24 lb M

T

(24)(10)

1 d 2

0.75 in., c

J

2

c4

Tc J

(240)(0.375) 0.031063

Bending:

Mc I

(144)(0.375) 0.015532

Transverse shear: Resultant stresses:

144 lb in.

240 lb in.

0.375 in. 1 J 2

0.031063 in 4 I

Torsion:

(24)(6)

0.015532 in 4

2.897 103 psi

2.897 ksi

3.477 103 psi 3.477 ksi

At point H, the stress due to transverse shear is zero. x ave

3.477 ksi, 1 ( 2

y

x

y)

x

y

0,

2.897 ksi

xy

1.738 ksi 2

R

2

a

ave

R

b

ave

R

max

2 xy

1.7382

2.897 2

3.378 ksi

a

1.640 ksi

b

R

5.12 ksi

max

3.38 ksi


P PROBLEM 7 7.26

y

m 6 mm A 200 mm

Thhe steel pipe AB A has a 1022-mm outer diameter and a 6-mm wall thhickness. Knowing that arm m CD is rigiddly attached to t the pipe, deetermine the principal p stressses and the maximum m shearing stress att point K.

51 mm

A T

D

10 kN N C

1 mm 150 H

K

B x

z

SOLUTION ro J

I

do 2

1102 2

51 mm

ri

ro

t

45 mm

ro4 ri4 4.18555 106 mm 4 2 4.18555 10 6 m 4 1 J 2

2.0927 10 6 m 4

Forcce-couple systtem at center of o tube in the plane p containiing points H and a K:

Fx

10 kN 10 1003 N

My

(10 103 )(200 10 3 ) 2000 N m

Mz

(10 103 )(150 10 3 ) 15000 N m

Torsion:

At po oint K, place local l x-axis in negative globbal z-directionn. T

My

c

ro

xy

2000 N m 511 10 3 m

((2000)(51 100 3 ) 4.1855 106 24.37 106 Pa 24.37 MPa Tc J



Transverse shear:

Stress due to transverse shear V

Fx is zero at point K.

Bending: |

y|

(1500)(51 10 3 ) 2.0927 10 6

|M z |c I

36.56 106 Pa

36.56 MPa

Point K lies on compression side of neutral axis.

36.56 MPa

y

Total stresses at point K: x ave

0,

36.56 MPa,

y

1 ( 2

x

y)

x

y

xy

24.37 MPa

18.28 MPa 2

R

2

2 xy

30.46 MPa

max

ave

R

18.28 30.46

max

12.18 MPa

min

ave

R

18.28 30.46

min

48.7 MPa

max

R

max

30.5 MPa


PROBLEM 7.27

#y 20 MPa 60 MPa

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 x

60 MPa,

y

?,

xy

20 MPa

Let

u

x

y

y

x

R

u2

2

.

Then

Largest value of

y

2 xy

R2

u y

2u

x

2u

75 MPa 2 xy

752

202

60 (2)(72.284)

is required.

72.284 MPa 84.6 MPa or 205 MPa y

205 MPa


PROBLEM 7.28

8 ksi

$xy 10 ksi

For the state of plane stress shown, determine (a) the largest value of xy for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the corresponding principal stresses.

SOLUTION x

10 ksi,

8 ksi,

y

xy

2 max

x

R

z 92

122

(a)

xy

(b)

ave

1 ( 2

y

2 xy

2 xy

? 10 ( 8) z

2 2 xy

12 ksi

92

xy y)

x

7.94 ksi

1 ksi

a

ave

R 1 12 13 ksi

b

ave

R 1 12

a

11 ksi

b

13.00 ksi 11.00 ksi


P PROBLEM 7.29

2 MPaa

$xy

12 MPa

75"

For the state of plane stress shown, determ F mine (a) the vaalue of xy foor which the inn-plane shearring stress paarallel to the weld is zeroo, (b) the corrresponding p principal stressses.

SOLUTION x

Sincce

xy

12 MPa,

y

tan 2

?

15 2

1 ( 2

x

y)

tan 2

x

y

2 xyy

y)

7 MPa M

p

xy

p x

xy

xy

0, x -direction is a principal direection. p

(a)

2 MPaa,

y

1 (12 2)) tan( 30 ) 2

xy

2.89 MPa

2

R

ave

(b)

2 1 ( 2

x

52

2.8992

5.7735 MPa M

a

ave

R

7 5.77735

a

12..77 MPa

b

ave

R

7 5.77735

b

1.2 226 MPa


PROBLEM 7.30

15 ksi 8 ksi

Determine the range of values of is equal to or less than 10 ksi.

x

for which the maximum in-plane shearing stress

#x

SOLUTION x

Let u

R

x

x

2 u2

2 xy

R2

u x

y

y

max 2 xy

y

?,

y

15 ksi,

xy

8 ksi

2u

10 ksi 102

2u 15 (2)(6)

82 z

6 ksi

27 ksi or 3 ksi 3 ksi

Allowable range:

x

27 ksi


PR ROBLEM 7.31 7 40 MPaa

Soolve Probs. 7.55 and 7.9, usinng Mohr’s circcle.

355 MPa

PR ROBLEM 7.55 through 7..8 For the givven state of stress, s determ mine (a) the priincipal planes, (b) the principal stresses.

60 MPa

PR ROBLEM 7.99 through 7.12 For the giiven state of stress, determ mine (a) the oriientation of thhe planes of maximum m in-pplane shearing stress, (b) thee maximum in--plane shearinng stress, (c) thhe correspondiing normal strress.

SOLUTION x

6 MPa, 60

y

4 MPa, 40 355 MPa

xy

x ave

y

2

50 MPa

Plottted points forr Mohr’s circlee:

(a)

a

R

(a )

(b ) (c )

x,

xy )

Y:(

y,

xy )

( 40 MPa, 35 MPa)

C:(

ave ,

0)

( 50 MPa, 0)

X 35 GX CG G 10 74 4.05

tan

b

(b)

X :(

1 2 180 1 2

( 60 MPa, 35 MPa)

3.5000

37.03 105.995 52.97

CG C

2

GX

2

10 2

ave a

R

50 36.4

max

a ave

R

50 36.4

d

B

45

7.97

e

A

45

97.977

352

36.4 MPa min max

86 6.4 MPa 13 3.60 MPa d e

R 36.4 MPa ave a

53.0

a

min

max

37.0

b

max

50 MPaa

8.0 98.0

36 6.4 MPa 50 0.0 MPa


PROBLEM 7.32 30 MPa

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

150 MPa

80 MPa

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) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

SOLUTION x

150 MPa

y

30 MPa 80 MPa

xy

x

ave

y

2

90 MPa

Plotted points for Mohr’s circle:

x

tan 2

p

2

p

y , xy )

C:(

ave ,

30)

(60)2

xy )

0)

(80)2

100

p

90

100

min

ave

R

90

100

max

(90 MPa, 0)

53.130 R

(b′)

(30 MPa, 80 MPa)

80 60

ave

s

(150 MPa, 80 MPa)

60

max

(a′)

(c )

Y:(

2

R

(b)

x,

(150

y

2

(a)

X :(

p

26.6 max min

45

s

R

18.4 max

and 63.4 190.0 MPa 10.00 MPa and 108.4 100.0 MPa

90.0 MPa

ave


PROBLEM 7.33

10 ksi

Solve Prob. 7.10, using Mohr’s circle.

2 ksi

PROBLEM 7.9 through 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) the corresponding normal stress.

3 ksi

SOLUTION x

2 ksi y

x

ave

10 ksi

y

2

2

10 2

xy

3 ksi

6 ksi


FX FC

tan

X: (

x,

Y: (

y , xy )

(10 ksi, 3 ksi)

C: (

ave ,

(6 ksi, 0)

3 4

xy )

0)

(2 ksi, 3 ksi)

0.75

36.87 B

(a)

(b) (c)

1 2

18.43

D

B

45

E

B

45

R

CF

max

R

5.00 ksi

ave

6.00 ksi

2

26.6 63.4

FX

2

26.6

D E

42

32

63.4

5 ksi max

5.00 ksi

6.00 ksi


PROBLEM 7.34

12 ksi 8 ksi

Solve Prob. 7.12, using Mohr’s circle. 18 ksi

PROBLEM 7.9 through 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) the corresponding normal stress.

SOLUTION 18 ksi

x

x

ave

12 ksi

y

y

xy

8 ksi

3 ksi

2


FX CF

tan

X: (

x,

Y: (

y,

xy )

C: (

ave ,

0)

8 15

xy )

(18 ksi, 8 ksi) ( 12 ksi, 8 ksi) (3 ksi, 0)

0.5333

28.07 A

(a)

(b) (c)

1 2

14.04

D

A

45

E

A

45

R

CF

max

R

17.00 ksi

ave

3.00 ksi

2

59.0

D

30.1

FX

2

E

152

82

59.0 30.1

17 ksi max

17.00 ksi

3.00 ksi


PROBLEM 7.35

8 ksi 5 ksi

Solve Prrob. 7.13, usinng Mohr’s circcle.

PROBL LEM 7.13 through 7.16 For the given staate of stress, determine d the normal and shearingg stresses afterr the element shown has beeen rotated thrrough (a) 25 clockwise, (b) 10 counterclockw c wise.

SOLUTION x

0 0,

y

8 ksi,

xy

5 ksi x

ave

y

4 kssi

2


X : (0, 5 ksi) k 5 ksi) Y : (8 ksi, k 0) C : (4 ksi, tan 2

p

2

p

FX 5 1 1.25 FC 4 51.34

R (a)

25

FC .

2

X FX

2

xy

y

10

.

42

52

6.4031 ksi

50

x

(b)

2

2

51.34

50

ave

R cos

1.34

R sin ave

2.40 ksi

x

0.1497 ksi

xy

R cos

y

10.40 ksi

20

51.34 x xy y

ave

20 R cos

R sin ave

R cos

71.34 x

1.951 1 ksi

xy

6.07 ksi

y

6 6.05 ksi


PROBLEM M 7.36

90 MP Pa 3 MPa 30

Solve Prob. 7.14, using Mohr’s M circle.

60 MPa

PROBLEM M 7.13 througgh 7.16 For thhe given statee of stress, deetermine the normal and shearing s stresses after the element e shownn has been rotaated through (a) 25 clockkwise, (b) 10 counterclockkwise.

SO OLUTION 60 MP Pa,

x y

90 MPa,,

xy

30 MPa x

ave

y

15 MPa

2

Plootted points for Mohr’s circlle:

X : ( 60 MPa, 30 MPa) Y : (90 MPa, 300 MPa) C : (15 MPa, 0) tan 2

p

2

p

FX FC

21.80

R

(a)

25

.

30 75

FC

2

0 0.4

10.90

P

FX X

2

752

300 2

80.78 MP Pa

50 5

2

2 x

xy y

2 ave

P

50

21.80

288.20

R cos

R sin ave

R cos

x

5 56.2 MPa

xy

3 38.2 MPa y

8 MPa 86.2


PROB BLEM 7.36 (Continued) ( d)

(b)

10

.

2

200

2 x

p ave

xy

R sin

y

ave

2

21.880

20

41.80

R cos

R cos

x

45 5.2 MPa

xy

53 3.8 MPa

y

755.2 MPa


PR ROBLEM 7..37

12 ksi

8 ksi

6 ksi

Solv ve Prob. 7.15, using Mohr’ss circle.

PRO OBLEM 7.133 through 7.16 For the giveen state of stress, determinee the normal and shearing streesses after thee element shoown has beenn rotated through (a) 25 clocckwise, (b) 10 counterclockkwise.

SO OLUTION 8 ksi,

x

12 ksi,

y

6 ksi

xy

x

y

ave

2 ksi k

2


X : (8 ksi, 6 ksi) Y : ( 12 ksi, 6 ksi)) C : ( 2 ksi, 0)

(a)

tan 2

p

2

p

25

FX 6 CF 100 30.96

R

CF

.

2

2

0.6

FX

2

x

xy y

10

.

62

11.66 ksi k

5 50

5 50

(b)

102

2

ave

30.96

19.04

R cos

R sin ave

R cos

x

9.02 ksi

xy

3.80 ksi 13.02 ksi

y

2 20

3 30.96 x

xy y

20 R cos

ave

R sin ave

50.96

R cos

x

5.34 ksi

xy

9.06 ksi

y

9.34 ksi


80 MPa

PROB BLEM 7.38 Solve Prob. P 7.16, usiing Mohr’s cirrcle.

50 MPa

PROBL LEM 7.13 thrrough 7.16 Foor the given sttate of stress, determine d the normal and shearinng stresses afteer the elementt shown has been rotated thhrough (a) 25 clockwise, (b) 10 counterclockkwise.

SOLUTION x

0,

y

M 80 MPa,

xy

50 MPa M x

ave

y

40 MPa

2

Plotted points for Moohr’s circle:

X : (0, 50 MPa) M MPa, 50 MPa)) Y : ( 80 M M 0) C : ( 40 MPa,

tann 2

p

2

p

R (a)

25

.

2

FX 50 1.25 CF 40 51.34 2

FX CF 64.031 MPa

x

xy y

10

.

402

502

50

51.34

(b)

2

2

x

xy y

50

1.34

R cos

ave

x

R sinn

1.4497 MPa

xy

R cos

ave

244.0 MPa

y

1044.0 MPa

x

199.51 MPa

xy

600.7 MPa

y

0.5 MPa 60

20

51.34

20

ave

R cos

R sinn ave

R cos

71.34


250 psi

PROBLEM 7.39 Solve Prob. 7.17, using Mohr’s circle.

158

PROBLEM 7.17 The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

SOLUTION x xy

y

0

250 psi


(a)

xy

X

(0, 250 psi)

Y

(0, 250 psi)

C

(0, 0)

R cos 2 (250 psi)cos30 217 psi xy

(b)

x

217 psi

R sin 2 (250 psi) sin 30 125.0 psi x

125.0 psi


PROBLEM 7.40

1.8 MPa

Solve Prob. 7.18, using Mohr’s circle. 3 MPa

PROBLEM 7.18 The grain of a wooden member forms an angle of 15° with the vertical. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

158

SOLUTION x

ave

3 MPa x

1.8 MPa

y

y

xy

0

2.4 MPa

2

Points. X: (

x,

xy )

( 3 MPa, 0)

Y: (

y,

xy )

( 1.8 MPa, 0)

C: (

ave ,

0)

( 2.4 MPa, 0)

15 CX

(a)

xy

CX sin 30

(b)

x

ave

CX cos 30

2

0.6 MPa

R sin 30 2.4

30

R

0.6 MPa

0.6sin 30 0.6 cos 30

0.300 MPa 2.92 MPa

0.300 MPa

xy x

2.92 MPa


PROBLEM 7.41 P'

80 mm


120 mm

b P

PROBLEM 7.19 Two wooden members of 80 120-mm uniform rectangular cross section are joined by the simple glued scarf 22 and that the maximum splice shown. Knowing that allowable stresses in the joint are, respectively, 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine the largest centric load P that can be applied.

SOLUTION P , A

x

0

y

xy

0


X:

P ,0 , A

C:

P ,0 2

R

CX

Y : (0, 0)

P 2A

Coordinates of point Y : P (1 cos 2 ) 2A P sin 2 2A

Data:

A

If

(80)(120) 400 kPa

P

9.6 103 mm 2 400 103 Pa,

(2)(9.6 10 3 )(400 103 ) (1 cos 44 )

2A 1 cos 2 27.4 103 N

If

600 kPa

P

2A sin 2

9.6 10 3 m 2

27.4 kN

600 103 Pa,

(2)(9.6 10 3 )(600 103 ) (sin 44 )

16.58 103 N

16.58 kN

The smaller value of P governs.

P

16.58 kN


PROBLEM 7.42 P'

80 mm


120 mm

b P

PROBLEM 7.20 Two wooden members of 80 120-mm uniform rectangular cross section are joined by the simple glued scarf 25 and that centric loads of splice shown. Knowing that magnitude P 10 kN are applied to the members as shown, determine (a) the in-plane shearing stress parallel to the splice, (b) the normal stress perpendicular to the splice.

SOLUTION

x

P A

0

y

xy

0


X:

P ,0 A

Y : (0, 0)

C:

P ,0 2A

R

CX

P 2A

Coordinates of point Y: P (1 cos 2 ) 2A P sin 2 2A

Data:

A

(80)(120)

9.6 103 mm 2

(a)

(10 103 )sin 50 (2)(9.6 10 3 )

(b)

(10 103 )(1 cos 50 ) (2)(9.6 10 3 )

9.6 10 3 m 2

399 103 Pa

399 kPa

186.0 103 Pa

186.0 kPa


PROBLEM 7.43

P

Solve Prob. 7.21, using Mohr’s circle. PROBLEM 7.21 The centric force P is applied to a short post as shown. Knowing that 5 ksi, determine (a) the angle that 15 ksi and the stresses on plane a-a are plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.

a

!

a

SOLUTION

x

0

xy

0

y

P A

(a)

From the Mohr’s circle, tan

(b)

P A

5 0.3333 15 P P cos 2 2A 2A

1

2( ) cos 2

1

18.4

(2)(15) cos 2

16.67 ksi 16.67 ksi


PROBLEM 7.44 Solve Prob. 7.22, using Mohr’s circle.

a a

25"

50 mm

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.

P

SOLUTION

x

0

xy

0

y

P/A

A (50 10 3 )(80 10 3 ) 4 10 3 m 2

P

P (1 cos50 ) 2A 2A 1 cos 50

(2)(4 10 3 )(800 103 ) 1 cos 50 P 3.90 103 N P

P sin 50 2A

P

2A sin 50

Choosing the smaller value,

(2)(4 10 3 )(600 103 ) sin 50

6.27 103 N P

3.90 kN


PROBLEM 7.45

0.2 m 0.15 m 3 kN


H

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.

350 N · m 3 kN

SOLUTION c

Torsion:

1 d 2

1 (32) 2

Tc J

2T c3

16 mm 16 10 3 m

2(350 N m) (16 10 3 m)3 Bending:

I

M

4

c4

4

54.399 106 Pa

(16 10 3 )4

(0.15m)(3 103 N) My I

54.399 MPa

51.472 10 9 m 4

450 N m

(450)(16 10 3 ) 51.472 10 9

139.882 106 Pa

Top view

Stresses

x

Plotted points:

139.882 MPa

139.882 MPa,

y

X : ( 139.882, 54.399); ave

1 ( 2

0,

xy

54.399 MPa

Y: (0, 54.399); C: ( 69.941, 0)

x

y)

x

y

69.941 MPa 2

R

2 xy

2 139.882 2

2

(54.399)2

88.606 MPa



tan 2

2 p

xy

x

y

(2)( 54.399) 139.882

0.77778 (a)

(b)

a a

ave

R

69.941

88.606

b

ave

R

69.941

88.606

max

R

18.9

,

b

108.9 158.5 MPa

a b

max

18.67 MPa 88.6 MPa


6 in.

PROBLEM 7.46

C H


B

PROBLEM 7.24 A 400-lb vertical force is applied at D to a gear attached 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.

A D 2 in. 400 lb


V

400 lb

M

T

(400)(2)

800 lb in.


(400)(6)

2400 lb in.

d

1 in.

J

c4

Resultant stresses:

(800)(0.5) 0.098175

Mc I

Bending:

0.5 in.

0.098175 in 4

Tc J

Torsion:

Transverse shear:

2

1 d 2

c

I

1 J 2

0.049087 in 4

4.074 103 psi

(2400)(0.5) 0.049087

4.074 ksi

24.446 103 psi

24.446 ksi

Stress at point H is zero. x ave

24.446 ksi, 1 ( 2

x

y)

x

y

y

0,

xy

4.074 ksi

12.223 ksi 2

R

2 (12.223) 2

a

ave

R

b

ave

R

max

R

2 xy

(4.074) 2

12.884 ksi a

25.1 ksi

b

0.661 ksi

max

12.88 ksi


PROBLEM 7.47

H


E

PROBLEM 7.25 A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A, determine the principal stresses and the maximum shearing stress at point H located as shown as on top of the 34 -in.-diameter shaft.

6 in.

B 24 lb 10 in.

A


V

24 lb

M

T

(24)(10)

240 lb in.


(24)(6)

144 lb in.

d

0.75 in.

J

c4

Resultant stresses:

1 J 2

I

(240)(0.375) 0.031063

Mc I

Bending:

0.375 in.

0.031063 in 4

Tc J

Torsion:

Transverse shear:

2

1 d 2

c

0.015532 in 4

2.897 103 psi

(144)(0.375) 0.015532

3.477 103 psi

2.897 ksi

3.477 ksi

At point H, stress due to transverse shear is zero. x ave

3.477 ksi, 1 ( 2

0,

y

x

y)

x

y

xy

2.897 ksi

1.738 ksi 2

R

2 1.7382

a

ave

R

b

ave

R

max

R

2.8972

xy

2

3.378 ksi a

5.12 ksi

1.640 ksi

b max

3.38 ksi


PROBLEM 7.48

y

6 mm A 200 mm


51 mm

A T

PROBLEM 7.26 The steel pipe AB has a 102-mm outer diameter and 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.

D

10 kN C

150 mm H

K

B z

x

SOLUTION ro

J I

do 2

102 2

ro4

2

51 mm

ri4

1 J 2

ri

ro

t

4.1855 106 mm 4

45 mm

4.1855 10

6

m4

2.0927 10 6 m 4

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

10 103 N (10 103 )(200 10 3 ) (10 103 )(150 10 3 )

T

My

c

ro

xy

2000 N m 1500 N m

2000 N m 51 10 3 m (2000)(51 10 3 ) 4.1855 10 6

Tc J

24.37 MPa

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

Stress due to V y

Fx is zero at point K. (1500)(51 10 3 ) 2.0927 10 6

Mz c I

Point K lies on compression side of neutral axis.

y

36.56 MPa

36.56 MPa



Total stresses at point K:

x

ave

0, 1 ( 2

36.56 MPa,

y

x

y)

x

y

xy

24.37 MPa

18.28 MPa 2

R max

min

max

2 ave

ave

R

R

R

2 xy

18.28

18.28

30.46 MPa

30.46 max

12.18 MPa

min

48.7 MPa

30.46

max

30.5 MPa


PROBLEM 7.49

#y 20 MPa 60 MPa

Solve Prob. 7.27, using Mohr’s circle. 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 x

60 MPa,

y

?,

xy

20 MPa

Given: max

R

XY

2 R 150 MPa

DY

(2)(

XD y

75 MPa

XY x

xy ) 2

40 MPa DY

2

1502

XD 60 144.6

402

144.6 MPa y

205 MPa


PROBLEM 7.50

8 ksi

$xy 10 ksi

Solve Prob. 7.28, using Mohr’s circle. PROBLEM 7.28 For the state of plane stress shown, determine (a) the largest value of xy for which the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the corresponding principal stresses.

SOLUTION The center of the Mohr’s circle lies at point C with coordinates x

y

2 The radius of the circle is

max (in-plane)

10 8 , 0 2

,0

(1 ksi, 0).

12 ksi.

The stress point ( x , xy ) lies along the line X1 X 2 of the Mohr circle diagram. The extreme points with R 12 ksi are X 1 and X 2 . (a)

The largest allowable value of

xy 2

(b)

The principal stresses are

is obtained from triangle CDX. 2

DX 1

DX 2

xy

122

a

1 12

b

1 12

2

CX 1

CD

2

92

xy a b

7.94 ksi 13.00 ksi 11.00 ksi


2 MPa

PROBLEM 7.51 $xy

Solve Prob. 7.29, using Mohr’s circle. 12 MPa

75"

PROBLEM 7.29 For the state of plane stress shown, determine (a) the value of xy for which the in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal stresses.

SOLUTION Point X of Mohr’s circle must lie on X X that y 2 MPa. The coordinates of C are

so that

2 12 , 0 2

x

12 MPa. Likewise, point Y lies on line Y Y

(7 MPa, 0).

Counterclockwise rotation through 150° brings line CX to CB, where

R (a)

x xy

y

2

x

y

2

sec 30

12 2 sec 30 2

0.

5.7735 MPa

tan 30

12 2 tan 30 2

(b)

so

xy

2.89 MPa

a

ave

R

7 5.7735

a

12.77 MPa

b

ave

R

7 5.7735

b

1.226 MPa


PROBLEM 7.52

15 ksi 8 ksi

Solve Prob. 7.30, using Mohr’s circle. #x

PROBLEM 7.30 Determine the range of values of in-plane shearing stress is equal to or less than 10 ksi.

x

for which the maximum

SOLUTION For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is R 10 ksi. Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.

C1Y

10 ksi

C2Y

10 ksi

Noting right triangles C1 DY and C2 DY ,

C1D

2

DY

2

C1Y

2

C1D

2

82

102

C1D

6 ksi

Coordinates of point C1 are (0, 15 6) (0, 9 ksi). Likewise, coordinates of point C2 are (0, 15 6) (0, 21 ksi). Coordinates of point X1: (9 6, 8) (3 ksi, 8 ksi) Coordinates of point X2: (21 6, 8) (27 ksi, 8 ksi) The point (

x,

xy )

must lie on the line X1 X2. 3 ksi

Thus,

x

27 ksi


2 MPa

PROBLEM 7.53 $xy

75"

Solve Problem 7.29, using Mohr’s circle and assuming that the weld forms an angle of 60 with the horizontal. 12 MPa

PROBLEM 7.29 For the state of plane stress shown, determine (a) the value of xy for which the in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal stresses.

SOLUTION 12

Locate point C at Angle XCB x

y

2

2

7 MPa with

2

0.

120

12

2

2 5 MPa

R

5sec 60 10 MPa

5 tan 60

xy

8.66 MPa

xy ave

a

7

10 ave

b

7

R

10

a

17.00 MPa

b

3.00 MPa

R


3 ksi

6 ksi

5 ksi

+

458

PROBLEM 7.54 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.

2 ksi 4 ksi

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

We now can add the two stress elements by superposition.

Principal planes and principal stresses:



ave

x

y

2

1 (6 2

2)

1 (6 2

2)

(4)2

R tan 2

p

2

p

2

4

(3)2

5

3 4

36.87 p

max

ave

2

R

18.4 , 108.4

5 max

min

ave

R

2

7.00 ksi

5 min

3.00 ksi


PROBLEM 7.55


+

308

75 MPa

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.

x

50 cos 30 43.30

y

43.30

xy

50sin 30 25.0

Principal axes and principal stress:

ave

y

x

2 R tan 2

p

1 (118.3 2

56.7)

1 (118.3 2

56.7)

(30.8)2

(75)2

75 30.8

2

p

87.5

30.8 81.08

67.67

max

ave

R

87.5

81.08

min

ave

R

87.5

81.08

p

33.8

, and 123.8 max min

168.6 MPa 6.42 MPa


#0 #0

PROBLEM M 7.56

#0 #0

Determine thhe principal planes p and thhe principal stresses for the t state of pllane stress ressulting from the superposiition of the twoo states of streess shown.

30" 30"

SO OLUTION Exppress each state of stress in terms of horizzontal and verrtical componeents.

s of stresss, Addding the two states

p

0 and a 90° max m min

0 0


PROBLEM 7.57

$0

$0

30"

+


SOLUTION Mohr’s circle for 2nd state of stress: x

0

y

0 0

xy

xy

3 2

0 sin 60

x

1 2

0 cos 60

0

y

0

y

3 2

0 sin 60

0

0

Resultant stresses: x

3 2 1 2

0

xy

0

ave

1 ( 2

3 2

0

3 2

0

x

y)

x

y

tan 2

2 x

2

2 xy

(2)

xy y

p

60

a

ave

R

b

ave

R

0

0

2

p

3 2

0

0

2

R

3 2

0

3 2 3

3 2

2 0

3 2

2 0

3

0

3 b

30

a

a b

60

3

0

3

0


PROBLEM 7.58

120 MPa

$xy

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

xy

for which the

20 MPa

SOLUTION x ave

Set

max

R

20 MPa 1 ( 2

y)

x

60 MPa max

120 MPa

y

70 MPa

R

ave

130 MPa

ave

But 2 x

R

2 xy

x

2

2 xy

R

2

x

x

2

1302 502 120.0 MPa Range of

xy :

120.0 MPa

xy

120.0 MPa


PROBLEM 7.59

120 MPa

$xy

For the element shown, determine the range of values of xy for which the maximum in-plane shearing stress is equal to or less than 150 MPa. 20 MPa

SOLUTION 20 MPa

x

1( 2

y)

x

Set

max (in-plane)

But

R

120 MPa

y

50 MPa

R 150 MPa 2 x

y

2 xy

2

2 xy

R

2

1502

x

y

2 502

141.4 MPa

Range of

xy :

141.4 MPa

xy

141.4 MPa


!y' 6 ksi

PROBLEM 7.60

"x'y'

!x' #

16 ksi

For the state of stress shown, determine the range of values of for which the magnitude of the shearing stress x y is equal to or less than 8 ksi.

SOLUTION x xy ave

16 ksi,

0

y

6 ksi 1 ( 2

x

y)

x

y

8 ksi 2

R

( 8)2 tan 2

2 p x

2

2 xy

2 (6) 2

xy y

p

36.870

b

18.435

10 ksi

(2)(6) 16

0.75

8 ksi for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle

xy

is

calculated from R sin 2

2

8

8 10

sin 2

53.130

0.8

26.565

k

b

18.435

26.565

45

k

b

18.435

26.565

8.13

u

h

90

45

v

k

90

98.13

Permissible range of : Also,

h

k

u

v

135

45 45

188.13

and 225

8.13 98.13

278.13


PROBLEM 7.61

#y' #x'

For the state of stress shown, determine the range of values of for which the normal stress x is equal to or less than 50 MPa.

%

90 MPa

$x'y' 60 MPa

SOLUTION x

90 MPa,

0

60 MPa

xy ave

y

1 ( 2

x

y)

x

y

45 MPa 2

R

452 2

tan 2

p

602

xy

x

2

x

2 xy

2

y

p

53.13

a

26.565

75 MPa (2)( 60) 90

4 3

50 MPa for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle,

R cos 2

50

cos 2

5 75

2

5 MPa

0.066667

86.177 h

2

45

43.089 26.565

a

k

2

k

110.085

h

360

Permissible range of

:

4

43.089

16.524

32.524

360

h

172.355

220.169

k

16.5

Also,

196.5

110.1

290.1


PROBLEM 7.62

#y' #x'

For the state of stress shown, determine the range of values of for which the normal stress x is equal to or less than 100 MPa.

%

90 MPa

$x'y' 60 MPa

SOLUTION x xy ave

90 MPa,

y

0

60 MPa 1 ( 2

x

y)

x

y

45 MPa 2

R

2 452 2

tan 2

p

xy

x

2

x

602

y

p

53.13

a

26.565

2 xy

75 MPa (2)( 60) 90

4 3

100 MPa for states of stress corresponding to arc HBK of Mohr’s circle. From the circle, R cos 2

100

45

cos 2

55 75

0.73333

2

42.833 h

2

55 MPa

21.417 26.565

a

k

2

k

132.02

h

Permissible range of

Also,

360

4

21.417 10.297

is

h

5.15 360

85.666

264.037

k

5.1

132.0

174.8

312.0


PROBLEM 7.63

#y $xy

For the state of stress shown, it is known that the normal and shearing stresses are directed as shown and that x 14 ksi, y 9 ksi, and min 5 ksi. Determine (a) the orientation of the principal planes, (b) the principal stress max, (c) the maximum in-plane shearing stress.

#x

SOLUTION 14 ksi,

x min

9 ksi,

y

R

ave

1 ( 2

ave

R

ave

x

y)

11.5 ksi

min

11.5 5 6.5 ksi 2 x

R

y

2 xy

2

2 xy

But it is given that (a)

tan 2

x

R2

2 xy

2

is positive, thus

6.52

xy

2.52

6 ksi

6 ksi.

xy

p x

2

y

p

y

(2)(6) 5 67.38

2.4

a b

(b)

max

ave

max (in-plane)

123.7

R max

(c)

33.7

18.00 ksi

R max (in-plane)

6.50 ksi


!

PROBLEM 7.64 "y "y'

The Mohr’s circle shown corresponds to the state of stress given in Fig. 7.5a and b. Noting that x OC (CX )cos (2 p 2 ) and that x y (CX )sin (2 p 2 ), derive the expressions for x and given in Eqs. (7.5) and (7.6), respectively. [Hint: Use xy sin( A B) sin A cos B cos A sin B and cos ( A B) cos A cos B sin A sin B.]

Y Y' C

O

2#p 2#

!x'y' X'

" !xy

X

"x "x'

SOLUTION OC

1 ( 2

y)

x

CX

CX

x

y

CX cos 2

p

CX cos 2

p

CX sin 2

p

CX sin 2

p

x

OC CX cos (2

p

2 )

OC CX (cos 2

p

cos 2

2 xy

OC CX cos 2 x

y

x

2 xy

p

CX sin 2

p

cos 2

cos 2

y

2

CX sin (2

xy

p

cos 2

sin 2

p

sin 2 )

CX sin 2 xy

p

sin 2

2 )

CX (sin 2

p

cos 2

cos 2

CX cos 2

p

sin 2

x

y

2

sin 2

cos 2

p

sin 2 )

sin 2


PROBLEM 7.65 2 (a) Prove that the expression x y x y , where x , y , and x y are components of the stress along the rectangular axes x and y , is independent of the orientation of these axes. Also, show that the given expression represents the square of the tangent drawn from the origin of the coordinates to Mohr’s circle. (b) Using the invariance property established in part a, express the shearing stress xy in terms of x , y , and the principal stresses max and min .

SOLUTION (a)

From Mohr’s circle, R sin 2

xy x y

x

p

ave

R cos 2

p

ave

R cos 2

p

2 xy

y 2 ave

R 2 cos2 2

R 2 sin 2 2

2 ave

R 2 ; independent of

p

p

p.

Draw line OK from origin tangent to the circle at K. Triangle OCK is a right triangle. OC OK

2 2

OK OC

2 2

CK

2 ave x

(b)

2

CK

2

R2 2 xy

y

x,

Applying above to x

y

But x

2 xy ab

y

a

max

2 xy

x

xy

xy

2 ave

R2

max ,

a

2 xy

and

2 ab

b

0,

y,

b

, and to

a,

b,

min

min max

y x

y

min max

min

The sign cannot be determined from above equation.


PROBLEM 7.66

y

For the state of plane stress shown, determine the maximum shearing stress when (a) x 14 ksi and y 4 ksi, (b) x 21 ksi and y 14 ksi. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

σy

12 ksi

σx

z

x

SOLUTION (a)

ave

1 ( 2

x

y)

1 (14 2 1 (14 2

4)

9

4)

5

(5)2

R

(12)2

13

max

ave

R

9

13

22

min

ave

R

9

13

4

Since max and min have opposite signs, the maximum shearing stress is equal to the maximum inplane shearing stress. max

R

13.00 ksi

max

13.00 ksi



(b)

ave

1 ( 2

x

y)

1 (21 14) 2 1 (21 14) 2

(3.5)2

R

17.5 3.5

(12)2

12.5

max

ave

R

17.5

12.5

30

min

ave

R

17.5

12.5

5

Since max and min have the same sign, O and A, we have max

1 2

max

max

is out of the plane of stress. Using Mohr’s circle through

1 (30 ksi) 2

max

15.00 ksi


PROBLEM 7.67

y

For the state of plane stress shown, determine the maximum shearing stress when (a) x 20 ksi and y 10 ksi, (b) x 12 ksi and y 5 ksi. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

σy

12 ksi

σx

z

x

SOLUTION

(a)

ave

1 (20 2 1 ( x 2

10)

15

y)

1 (20 2

R

(5)2

10)

5

(12)2

13

max

ave

R

15

13

28

min

ave

R

15

13

2

Since max and min have the same sign, O and A, we have

max

max

1 2

max

is out of the plane of stress. Using Mohr’s circle through 1 (28 ksi) 2

max

14.00 ksi



(b)

ave

1 (12 2 1 ( 2

x

5)

8.5

y)

1 (12 2

R

5)

(3.5)2

3.5

(12)2

12.5

max

ave

R

8.5

12.5

21

min

ave

R

8.5

12.5

4

Since max and min have opposite signs, the maximum shearing stress is equal to the maximum in-plane shearing stress. max

R

12.50 ksi


y

PROBLEM 7.68 σy

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

80 MPa 140 MPa z x

SOLUTION (a)

x ave

140 MPa, 1 ( 2

y

x

y)

x

y

40 MPa,

80 MPa

xy

90 MPa 2

R

502 802

a

ave

R 184.34 MPa (max)

b

ave

R

c

4.34 MPa

x ave

94.34 MPa

(min)

0 1 ( 2 1 ( max 2 140 MPa,

max (in-plane)

(b)

2 xy

2

1 ( 2

b)

a

y y)

x

y

1 ( 2 120 MPa, min )

max

x

94.34 MPa

R

b)

a xy

94.3 MPa

max

94.3 MPa

80 MPa

130 MPa 2

R

2 xy

2

102 802

a

ave

R

210.62 MPa (max)

b

ave

R

49.38 MPa

c max

80.62 MPa

0 (min) a

max (in-plane)

max

210.62 MPa

min

c

0

R 86.62 MPa 1 ( 2

max

min )

105.3 MPa

max

105.3 MPa


y

PROBLEM 7.69 σy

For the state of stress shown, determine the maximum shearing stress when (a) y 20 MPa, (b) y 140 MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses.)

80 MPa 140 MPa z x

SOLUTION (a)

x ave

140 MPa, 1 ( 2

y

x

y)

x

y

20 MPa,

xy

80 MPa

80 MPa 2

R

602

802

100 MPa

a

ave

R 80 100 180 MPa (max)

b

ave

R 80 100

c

x ave

20 MPa (min)

0 1 ( 2 1 ( max 2 140 MPa,

max (in-plane)

(b)

2 xy

2

1 ( 2

b)

a

min )

max y

x

y)

x

y

100 MPa 100 MPa

140 MPa,

xy

max

100.0 MPa

max

110.0 MPa

80 MPa

140 MPa 2

R

2 xy

2

0 802

a

ave

R

220 MPa (max)

b

ave

R

60 MPa

c

80 MPa

0 (min)

max (in-plane)

max

1 ( 2 1 ( 2

a

max

b)

80 MPa min )

110 MPa


PROBLEM 7.70

y

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

100 MPa 84 MPa

σz

30 MPa x

z

SOLUTION The z axis is a principal axis. We determine the other two principal axes by drawing Mohr’s circle for a rotation in the x y plane.

ave

1 ( 2

x

y)

1 (30 2 1 (30 2

R

(35)2

A

ave

B

ave

100) 100)

(84)2 R R

65 65

65 35

91 91 91

156 MPa 26 MPa

(a)

0. Point Z corresponding to the z axis is located at O between A and B. Therefore, the largest of the 3 Mohr’s circles is the circle we drew through A and B. We have R 91.0 MPa max

(b)

60 MPa. Point Z is located between A and B. The largest of the 3 circles is still the circle z through A and B, and we still have R 91.0 MPa max

(c)

60 MPa. Point Z is now outside the circle through A and B. The largest circle is the circle through Z and A. 1 1 108.0 MPa ( ZH ) (60 156) max max 2 2

z

z


PROBLEM 7.71

y

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


z

x

z

SOLUTION

ave

1 ( 2

x

y)

1 (170 2 1 (170 2

(35)2

R A B

(a)

135

100)

35

(84)2

91 91

91

226 MPa 44 MPa

0. 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 z

max

(b)

135 135

100)

1 (OA) 2

1 2

A

1 (226) 2

max

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

(c)

113.0 MPa

R

91.0 MPa

60 MPa. 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 z

max

1 ( ZA) 2

1 (60 2

226)

max

143.0 MPa


PRO OBLEM 7.72

y

For thhe state of strress shown, determine the maximum sheearing stress when (a) yz 17.5 ksi, (b) yz 8 ksi, (c) yz 0.

τyz 12 ksi

3 ksi x

z

SO OLUTION (a)

(b)

17.55 ksi

yz

R

(6) 2

A

6 18.5

B

6 18.5 A

min

B

max

1 ( 2

24.5 12.5

min )

max

(6))2

(8) 2

6 10 1 16

B

6 10 1

18.50 ksi

max

10.00 ksi

10

4

max

A

16 ksi

min

B

4 ksi

max

1 ( 2

0

max

3 kssi

x

A

yz

18.5

12.5 ksi

8 kssi

R

(c)

(17.5) 2

24.5 ksi

max

yz

3 ksi

x

min )

max

3 ksi

x

max

z

12 ksi

min

x

3 ksi

max

1 ( 2

max

min )

max

7.50 ksi


PROBLEM 7.73

y

For the state of stress shown, determine the maximum shearing stress when (a) yz 17.5 ksi, (b) yz 8 ksi, (c) yz 0.

τyz 12 ksi

10 ksi x

z

SOLUTION (a)

17.5 ksi

yz

(6)2

R A B

max

6 6 A

min

B

max

1 ( 2

(17.5)2

18.5

18.5 24.5 18.5 12.5 24.5 ksi 12.5 ksi max

min ) max

(b)

yz

18.50 ksi

8 ksi



(6)2

R

(c)

yz

(8)2

A

6

10

16

B

6

10

4

max

A

min

x

max

1 ( 2

10

16 ksi

10 ksi min )

max

max

13.00 ksi

max

11.00 ksi

0

max

z

min

x

max

1 ( 2

12 ksi 10 ksi max

min )


PROBLEM 7.74

y

For the state of stress shown, determine the value of maximum shearing stress is (a) 9 ksi, (b) 12 ksi.

6 ksi

xy

for which the

τ xy 15 ksi z x

SOLUTION 15 ksi 1 ( x 2

x

(ksi)

ave

y)

x

u (a)

y

y

R xy

a

u

18

ave

2

2 xy

R2

u2

R2

c max

xy

6.00 ksi

u2

11.24 ksi

b

7.5 ksi

For max 12 ksi, center of Mohr’s circle lies at point C. R 12 ksi xy

10.5 12 10.5 12 0 1 ( max 2

10.5

4.52

6.00 ksi

a

4.5 ksi

2

7.52

Checking,

10.5 ksi

For max 9 ksi, center of Mohr’s circle lies at point C. Lines marked (a) show the limits on max . Limit on max is max 2 max 18 ksi . The Mohr’s circle a max corresponds to point Aa. R

(b)

6 ksi

xy

11.24 ksi

22.5 ksi 1.5 ksi min )

12 ksi

(o.k.)


PROBLEM 7.75

y

For the state of stress shown, determine the value of maximum shearing stress is 80 MPa.

70 MPa

xy

for which the

τ xy 120 MPa z x

SOLUTION 120 MPa

x

1 ( 2

ave

x

2 Assume

min

0

max

2

y)

x

120

y

y

70 MPa

95 MPa

70

25 MPa

2

160 MPa

max

a

max

ave

R

R

max

ave

160

95

65 MPa

652

252

2 x

R2

y

2 xy

2

2 2 xy

R

x

2

y

2

602 xy

b

a

2R

160

130

30 MPa

0

60.0 MPa

(o.k.)


y

PROBLEM 7.76 σy

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

y

for which the

48 MPa 50 MPa z x

SOLUTION 50 MPa,

x

y

u

Let

1 ( 2

(1a)

u ave

55 MPa 1 ( 2

a

a

(1b)

u ave b

min

R

ave

0

max

ave

78 MPa,

min

2u

y

R

178 MPa,

2 xy

u R2

u

732

u

x

482

2 xy

55 MPa

5 MPa b

y)

x

y)

x

60 MPa

x

78 MPa,

55 MPa 1 ( 2

2u

y y)

x

73 MPa,

R

max

2u

y

x

u2

R

Case (1)

x

2

ave

48 MPa

xy

ave

1 ( 2

max

73 MPa

160 MPa (reject)

105 MPa,

max

68 MPa

68 MPa,

x

178 MPa,

R

a c

max

R

ave

0,

0

max min )

32 MPa

89 MPa

73 MPa y

60.0 MPa



Assume

Case (2)

0.

x

u

u2

2 xy

2(

x

b )u

u2

146 MPa

min b

ave

u2

2 xy

x

u2

2 xy

2u

max

(

2 xy

min )

73 MPa

b

b)

x

max

b

R u

1 ( 2

max

(

2

b)

x

x

2

(48)2

b

u

36 MPa

R

u2

a

b

2 xy

2R

y

2u

( 50 146)2 50 146 x

72 MPa

122 MPa

60 MPa

146

120

26 MPa

(o.k.) y

122.0 MPa


y

PROBLEM 7.77 σy

For the state of stress shown, determine two values of maximum shearing stress is 10 ksi.

y

for which the

y

2.00 ksi

8 ksi 14 ksi z x

SOLUTION 14 ksi,

x

y

u

Let

1 ( 2

u ave

(1b)

6 ksi 1 ( 2

x

max

30 ksi,

u

6 ksi

ave

max

1 ( 2

x

18 ksi,

y y)

min

y y)

min

2u

20 ksi,

2u

a

max

2 ksi,

u

x

R2

u

2 xy

6 ksi

u

1 ( 2

30 ksi,

R

ave

min )

max

b

15 ksi

R

ave

10 ksi

7.5 ksi

2 ksi

x

8 ksi,

x

26 ksi (reject)

x

0,

2 xy

10 ksi

max

2u

y)

10 ksi,

R

max

8 ksi, y

x

u2

R

(1a)

x

2

ave

Case (1)

xy

a

ave

max

R

1 ( 2

18 ksi,

max

min )

b

ave

R

2 ksi

10 ksi (o.k.)



Assume

Case (2) a

ave

a

x

x

(

a

x)

2u

u2

u)2

a

2

x)

u ave

a max

y)

x

ave

R

20 ksi,

2 xy

u2

u2

(20

x

2.3333 ksi 1 ( 2

x )u

2 xy

a

u2

y

20 ksi =

a

2 xy

14)2 82 20 14

2u

x

0,

b max

4.6667 ksi

9.3333 ksi u2

11.6667 ksi R

20 ksi min

max

2 xy

a

2

2

max

2 xy

u2

2(

a

u

x

u

(

(

R

0.

min

ave

2 xy

R

8.3333 ksi

3.3334 ksi

10 ksi y

9.33 ksi


PROBLEM 7.78

y

σ y & 100 MPa

For the state of stress shown, determine the range of values of xz for which the maximum shearing stress is equal to or less than 60 MPa.

60 MPa z

x

τ xz

SOLUTION

60 MPa,

x

0,

z

y

100 MPa

For Mohr’s circle of stresses in zx plane,

Assume

max

y

min

b

30

(2)(60) ( 20)

30

xz

u

x

z)

x z

2

30 MPa

30 MPa

max

20 MPa

b

50 MPa

R

ave

R

2

max

ave

a

1 ( 2

100 MPa

100 R

ave

50

u2 R2 502

80 MPa <

y

2 xz

u2 302

40 MPa 40.0 MPa

xz

40.0 MPa


PROBLEM 7.79

y

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

σy

y

for which the

90 MPa z

x 60 MPa

SOLUTION x

90 MPa,

0,

z

xz

60 MPa

Mohr’s circle of stresses in zx plane: ave

1 ( 2

x

z)

x

y

45 MPa 2

R

a

Assume

ave

a

max y

min

120 Assume

min y

b max

30

R

120 MPa,

b

ave

2

R

2 zx

452

602

75 MPa

30 MPa

120 MPa. max

2

max

(2)(80)

40.0 MPa

y

30 MPa. min

2

max

(2)(80)

y

130.0 MPa


y

PROBLEM 7.80* σy

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.

80 MPa 140 MPa z x

SOLUTION x

u

Let

1 ( 2

ave

Assume Then

max

y y

2 y)

x

u

x

u2

a

ave

R

x

u

u2

2 xy

b

ave

R

x

u

u2

2 xy

is minimum if u

2 xy

R

max

0.

2u

140 MPa,

y

x

R

xy

80 MPa

a

ave

R 140 80

b

ave

R 140 80 60 MPa

max

2u

R

is the in-plane shearing stress.

max (in-plane)

x

x

220 MPa,

x

u 140 MPa

220 MPa

0,

min

ave

max

1 ( 2

max

min )

110 MPa

Assumption is incorrect. Assume

max min

d a du

a

R

ave

0

max

1

u u

2

2 xy

x

1 ( 2

u max

0

u2 min )

2 xy

1 2

a

(no minimum)


PROBLEM 7.80* (Continued)

Optimum value for u occurs when

1 ( 2 (

a

x

R) u )2 2u

max (out-of-plane)

R or 2 x

2u

2 x

2 xy x

(a) (b)

y

R

x

u2

a

x

max (in-plane)

R or u

2

u

u

x 2

140 2 80 2 140

u2

2 xy

2 xy

94.286 MPa

u

2u 140 94.286 2 xy

max

47.143 MPa y

92.857 MPa

max

45.7 MPa 92.857 MPa


σ0

PROBLEM 7.81 100 MPa

σ0

The state of plane stress shown occurs in a machine component made of a steel with 325 MPa. Using the maximum-distortion-energy criterion, determine whether Y yield will occur when (a) 0 200 MPa, (b) 0 240 MPa, (c) 0 280 MPa. If yield does not occur, determine the corresponding factor of safety.

SOLUTION 2 ave

(a)

0

200 MPa

ave a

2 a

F . S.

(b)

0

240 MPa

ave a

2 a

F . S.

(c)

0

280 MPa

ave a

2 a

2 b

a

b

R

0

x

y

2 xy

2

100 MPa

200 MPa ave

2 b

R a

100 MPa, b

b

ave

R

300 MPa

264.56 MPa < 325 MPa

(No yielding)

325 264.56

F . S . 1.228

240 MPa ave

2 b

R a

140 MPa, b

b

ave

R

340 MPa

295.97 MPa < 325 MPa

(No yielding)

325 295.97

F . S . 1.098

280 MPa ave

R

180 MPa,

329.24 MPa > 325 MPa

b

ave

R

380 MPa (Yielding occurs)


PROBLEM 7.82

σ0 100 MPa

σ0

Solve Prob. 7.81, using the maximum-shearing-stress criterion. PROBLEM 7.81 The state of plane stress shown occurs in a machine component made of a steel with Y 325 MPa. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) 0 200 MPa, (b) 0 240 MPa, (c) 0 280 MPa. If yield does not occur, determine the corresponding factor of safety.

SOLUTION 2 ave

(a)

0

200 MPa:

ave a max

2

max

F . S.

(b)

0

240 MPa:

ave a max

2 (c)

0

280 MPa:

max ave a max

2

max

R

0

x

y

2 xy

2

100 MPa

200 MPa ave

0,

R

100 MPa

ave

R

300 MPa

300 MPa

min max

b

min

300 MPa

325 MPa

(No yielding)

325 300

F . S . 1.083

240 MPa ave

0,

R

140 MPa,

min

max

min

b

ave

R

340 MPa

340 MPa 340 MPa > 325 MPa

(Yielding occurs)

280 MPa ave

0, max

R

180 MPa,

min min

b

ave

R

380 MPa

380 MPa 380 MPa

325 MPa

(Yielding occurs)


PROBLEM 7.83

21 ksi

τ xy

The state of plane stress shown occurs in a machine component made of a steel with 45 ksi. Using the maximum-distortion-energy criterion, determine whether Y yield will occur when (a) xy 9 ksi, (b) xy 18 ksi, (c) xy 20 ksi. If yield does not occur, determine the corresponding factor of safety.

36 ksi

SOLUTION

36 ksi,

x

For stresses in xy plane,

1 ( 2

ave

x

y

21 ksi,

y y)

x

z

0

28.5 ksi

7.5 ksi

2 2

(a)

xy

9 ksi

a

ave

2 a

F .S .

2 b

x

R R a

y

2 xy

2 40.215 ksi,

ave

b

34.977 ksi

b

(7.5) 2 R

(9) 2

11.715 ksi

16.875 ksi

45 ksi

(No yielding)

45 39.977

F .S .

1.287

2

(b)

x

18 ksi R

xy

ave

a

2 a

F .S .

2 b

R a

y

2 xy

2 48 ksi,

ave

b

44.193 ksi

b

(7.5)2 R

(18)2

19.5 ksi

9 ksi

45 ksi

(No yielding)

45 44.193

F .S .

1.018

2

(c)

xy

20 ksi

a

ave

2 a

2 b

R R a

x

2 xy

2

49.86 ksi, b

y

b

46.732 ksi

ave

(7.5) 2 R

(20) 2

21.36 ksi

7.14 ksi

45 ksi

(Yielding occurs)


PROBLEM 7.84

21 ksi

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

τ xy

PROBLEM 7.83 The state of plane stress shown occurs in a machine component made of a steel with Y 45 ksi. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) xy 9 ksi, (b) xy 18 ksi, (c) xy 20 ksi. If yield does not occur, determine the corresponding factor of safety.

36 ksi

SOLUTION

36 ksi,

x

21 ksi,

y

0

z

For stress in xy plane,

1 ( 2

ave

y)

x

28.5 ksi

x

y

2

7.5 ksi

2

(a)

xy

9 ksi

a

ave

max

2

max

F .S.

x

R R

2 40.215 ksi,

34.977 ksi, max

y

b

11.715 ksi

ave

R

16.875 ksi

0

min

40.215 ksi

min

2 xy

(No yielding)

45 ksi

45 40.215

F .S .

1.119

2

(b)

xy

a

2

x

18 ksi R ave

max

48 ksi

max

max

y

2 xy

2

R

48 ksi,

9 ksi

0

min min

R

ave

b

19.5 ksi

48 ksi

(Yielding occurs)

45 ksi 2

(c)

xy

a max

2

max

x

20 ksi R ave

R

min min

2 xy

2 49.86 ksi

49.86 ksi max

y

b

ave

21.36 ksi R

7.14 ksi

0

49.86 ksi

(Yielding occurs)

45 ksi


PROBLEM 7.85 The 38-mm-diameter shaft AB is made of a grade of steel for which the yield strength is Y 250 MPa. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when P 240 kN.

B

T P

A

d = 38 mm

SOLUTION P A

x y ave

240 103 N d2

4

4

1.13411 103 mm 2

240 103 1.13411 10

P A 0 1 ( 2

(38) 2

x

y)

x

y

3

1 2

211.62 106 Pa

2

2R

2 xy

2 Y

4

xy

From torsion:

xy

J c

T

1 2

1 4

2 xy

2

max

2 x

2 xy

4

2 x

2 x

1 250 2 2

2 x

2 Y

211.62 2

66.553 106 Pa

Tc J

J

c4

1 d 2

2 xy

y

66.553 MPa

2

211.62 MPa

x

2

R

1.13411 10 3 m 2

T

2

38 2

xy

c 4

204.71 103 mm 4

204.71 10 9 m 4

19 10 3 m

(204.71 10 9 )(66.553 106 ) 19 10 3 717 N m


PROBLEM 7.86 Solve Prob. 7.85, using the maximum-distortion-energy criterion.

B

PROBLEM 7.85 The 38-mm-diameter shaft AB is made of a grade of steel for which the yield strength is Y 250 MPa. Using the maximum-shearingstress criterion, determine the magnitude of the torque T for which yield occurs when P 240 kN.

T P

A

d = 38 mm

SOLUTION P A

x y ave

240 103 N 4

d2

(38) 2

4

1.13411 103 mm 2

240 103 1.13411 10

P A 0 1 ( 2

x

y)

x

y

1 2

211.62 106 Pa

3

2 a

2 b

1 4

2 xy

2

2 xy

2 x

a

ave

R

1 2

x

1 4

2 x

2 xy

b

ave

R

1 2

x

1 4

2 x

2 xy

a b

1 4

2 x

1 4 2 x 2 xy

xy

1 4

x

2 x

3

x

2 xy

1 4

2 xy

2 x

1 4

211.62 MPa

x

2

R

1.13411 10 3 m 2

2 x

2 xy

1 4

2 x

2 xy 2 x

2 xy

1 4

2 x

1 4

2 x

2 xy

2 Y

1 2 2 Y x 3 1 2502 211.622 3

76.848 MPa

76.848 106 Pa



From torsion,

xy

J c

T

Tc J

2

c4

1 d 2

T 38 2 2

J

xy

c

4

204.71 103 mm 4

204.71 10 9 m 4

19 10 3 m

(204.71 10 9 )(76.848 106 ) 19 10 3 828 N m


PROBLEM 7.87

P T

A

1.5 in.

The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when P 60 kips.

B

SOLUTION

P

60 kips

A

d2

4

y ave

4

1.76715 in 2

60 1.76715

P A

x

(1.5) 2

33.953 ksi

0 1 ( 2

x

y)

x

y

1 2

x

2

R 2

2

max

2R

2 xy

2 Y

4

xy

1 2

2 x

1 4

2 xy

4

2 xy

2 x

2 xy

Y

2 x 2 Y

2 x

1 422 2

33.9532

12.3612 ksi From torsion,

xy

c

J

T

Tc J 1 d 2

2

c4

T

J

xy

c

0.75 in.

0.49701 in 4

(0.49701)(12.3612) 0.75 8.19 kip in.


PROBLEM 7.88

P T

A

Solve Prob. 7.87, using the maximum-distortion-energy criterion. PROBLEM 7.87 The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when P 60 kips.

1.5 in. B

SOLUTION

P

60 kips

A

d2

4

ave

1.76715 in 2

60 1.76715

p A

x

y

4

(1.5) 2

33.953 ksi

0 1 ( 2

x

y)

x

y

1 2

x

2

R

2

a

ave

b

2 a

2 b

a b

ave

(

ave 2 ave 2 ave

1 4

3

2 xy

2 Y

xy

1 3

1 4

2 xy

2 x

2 xy

R R

R) 2 2 3R

2 x

3

(

ave R 2

1 4

R) 2

ave 2

R

2 x

(

2 ave

2 xy

2 2 x

R)(

ave

ave 2

ave R

R

2 xy

2 Y

3

R) 2 ave

R2

2 x 2 Y

2 x

1 422 3

33.9532

14.2734 ksi From torsion,

xy

c J

T

Tc J 1 d 2

T

J

xy

c

0.75 in.

c4 (0.75) 4 0.49701 in 4 2 2 (0.49701)(14.2734) 9.46 kip in. 0.75


PRO OBLEM 7.89

100 MPa

The state s of plane stress shownn is expected to occur in an a aluminum 8 MPa and castinng. Knowing that for the aluuminum alloy used UT 80 nd using Mohrr’s criterion, determine d wheether rupture 200 MPa an UC of thee casting will occur. o

60 MPa

M 10 MPa

SO OLUTION x y xy

10 MPa, M 10 00 MPa, 60 MPa x

ave

10 1000 2

y

2

45 MPaa

2 x

R

y

2 xy

2

(55) 2

(60)2

a

avee

R

45 81.39 36.39 MPa

b

avee

R

45 81.39 8

81.399 MPa

126.39 MPa

Equuation of 4th quadrant q of bo oundary:

36.39 80

a

b

UT

UC

1

( 1226.39) 1.087 1 2000 Rupture will w occur.


PROBLEM M 7.90

75 MPa

The state of plane stress shown s is expeected to occur in an aluminuum casting. Knowing thatt for the alumiinum alloy useed UT 80 MPa M and UC 200 MPa and using Mohr’s M criterioon, determinee whether ruppture of the casting c will occur.

32 MPa

SOLUTION x y xy ave

32 MPa, M 0, M 75 MPa 1 ( 2

x

y)

x

y

M 16 MPa 2

R

(16) 2

2 xy

2

(775) 2

a

ave

R

16 766.69 60.69 MPa M

b

ave

R

16 766.69

76.69 MPa M

92.69 MPa

Equuation of 4th quadrant q of bouundary:

60.69 80

a

b

UT

UC

1

( 92..69) 1.222 1 200 Rupture will w occur.


PROBLEM M 7.91

7 ksi

The state off plane stress shown s is expeected to occurr in an aluminnum casting. 1 ksi and UC 30 ksi Knowing thaat for the alum minum alloy used UT 10 and using Mohr’s M criterionn, determine whether w rupturee of the castingg will occur.

8 ksi

SO OLUTION x y xy ave

8 ksi, 0, 7 ksi 1 ( 2

x

y)

x

y

4 ksi 2

R

2 xy

2

42

a

ave a

R

4 8.062

b

ave a

R

4 8.062

72

8.062 ksi

4.0622 ksi 12.0062 ksi

Equuation of 4th quadrant q of bo oundary:

4.062 10

a

b

UT

UC

( 122.062) 330

1 0.8088 1 No rupture.


PROBLEM M 7.92

15 ksi k

The state of plane p stress shhown is expeccted to occur in an aluminuum casting. 30 ksi Knowing thatt for the alum minum alloy used u 10 ksi and UC U UT and using Mohr’s criterion, determine whhether rupture of the casting will occur.

9 ksi

2 ksi

SOLUTION x

2 ksi, 15 ksi,

y xy ave a

9 ksi 1 ( 2

x

y)

x

y

6.5 ksii 2

R

2 xy

2

a

ave

R 5.879 ksi

b

ave

R

8.52

92

1 12.379 ksi

18.879 ksii

q of bouundary: Equuation of 4th quadrant

5.879 10

a

b

UT

UC

1

( 18.879) 1.217 1 30 Rupture will occur.


PROBLEM M 7.93 The state of plane stress shown s will occcur at a critical point in an a aluminum 25 ksi. casting that is made of ann alloy for whhich UT 10 ksi and UC U Using Mohrr’s criterion, determine d thee shearing stress 0 for which w failure should be exppected.

8 ksi

t0

SO OLUTION x

8 ksi,

y

0,

xyy

0

avee

1 ( 2

x

y)

x

y

4 ksi 2

R

R2

0

Sinnce

ave

42

2 xy

2

2 0

42

a

ave

R

(4 R) ksi

b

ave

R

(4 R) ksi

< R, stress point lies in 4th quaddrant. Equatioon of 4th quaddrant boundaryy is a

b

UT

UC

4 R 10 1 10

4 R 25

1 1

1 4 R 1 25 10 R

4 25

5.429 ksi

0

5.42992

42

0

3.67 ksi


PROBLEM M 7.94 80 MPa

!0

The state off plane stress shown s will occcur at a criticaal point in a piipe made of an aluminum m alloy for which w Using 75 MPa and UC 150 MPa. M UT Mohr’s criteerion, determinne the shearinng stress 0 foor which failurre should be expected.

SOLUTION x y

80 MPa, 0,

xy

0

ave

1 ( 2

x

y)

x

y

40 MPa 2

R

a

ave

R

b

ave

R

0

Sincce

ave

2 xy

2

R2

402

2 0

MPa

40 2

< R, stress point lies in 4th quaddrant. Equationn of 4th quadrrant boundary is a

b

UT

UC

40 R 75 R 75

R 150

1

R 63.33 MP Pa,

40 R 150 40 75

40 150 0

1 1 1.2667

63.332

402

0

8.49 MPa


PR ROBLEM 7.95 7 T'

Thee cast-aluminnum rod shoown is made of an alloyy for which a wing that the magnitude m T 70 MPa and 1775 MPa. Know UT T UC of the t applied torrques is slowlly increased annd using Mohr’s criterion, dettermine the shearing stress 0 that shouldd be expected at a rupture.

t0 T

SO OLUTION x

0

y

0

xy

0

ave

1 ( 2

x

y)

x

y

0 2

R

Sinnce

ave

2 xy

2

a

ave a

R

b

ave a

R

2 xyy

0

xy

R R

< R, stress point lies in 4th quaddrant. Equatioon of boundaryy of 4th quadrrant is a

b

UT

UC

R 700 1 70

1

R 1 175

1 R 1 1175 R 50 M MPa 0

R

0

5 50.0 MPa


P PROBLEM 7.96 T cast-alum The minum rod shhown is madee of an alloyy for which U Mohr’s criterion, 60 MPa and UC 120 MPa. Using UT d determine the magnitude off the torque T for which faiilure should b expected. be

32 mm B T A

26 kN

SOLUTION P

26 103 N

y

2

Sincce

(32) 2

804.25 mm 2

32.3288 106 Pa

6

804.25 10 6 m 2

322.328 MPa

1 1 ( x (32.328 0) 0 16.164 MP Pa y) 2 2 1 (32.328 0)) 16.164 MP Pa 2

ave x

4

26 1003 804.25 10

P A

x

A

a

ave

R 16..164 R MPa

b

ave

R 16.164 R MPa

< R, stress point lies in the 4th quadrant. q Equaation of the 4thh quadrant is

ave a

b

UT

UC C

1 60

1

16.1644 600

R

1 16.1664 R 1 60 120

16.1644 R 1200

1

16.164 120

R

34 4.612 MPa

2

y

x

R

2 xy

2

R

xy

2

x

y

34.6122 116.1642

2

30.6606 MPa

30.606 6 106 Pa For torsion,

xy

T

Tc J 2

c3

2T c3 x xy

wherre c

2

1 d 2

166 mm 16 100 3 m

(16 100 3 )3 (30.606 106 )

T

196 6.9 N m


1! 2 0

1! 2 0

!0

!0

(a)

PROBLEM M 7.97

1! 2 0

(b)

!0

A machine component c is made of a grade g of cast k and UC 20 ksi. For iron for whicch UT 8 ksi each of the states s of stress shown, and using u Mohr’s criterion, dettermine the normal stress 0 at which rupture of thee component should s be expeected.

0

UT

(c)

SO OLUTION (a)

a b

0

1 2

0

Stress poinnt lies in 1st quadrant. q a

(b)

a

0

b

1 2

0

8.00 ksi

0

Stress poinnt lies in 4th quadrant. q Equaation of 4th quuadrant bounddary is a

b

UT

C UC 1 2

0

8

(c)

a

1 2

0,

b

0,

1

0

1

0

6.67 ksi

0

1

0

8.89 ksi

20

4th quadrannt 1 2

0

8

20


PROBLEM 7.98 A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for the steel used all 80 MPa, E 200 GPa, and 0.29, determine (a) the allowable gage pressure, (b) the corresponding increase in the diameter of the vessel.

SOLUTION

(a)

(b)

r

1 d 2

1

2

all

1

2

pr 2t

p

2t r

p

1.290 106 Pa

1

d

1 ( E 1 E

d

1

1 (3) 2

t

12 10

3

1.488 m

80 106 Pa

(2)(12 10 3 )(80 106 ) 1.488

1

p

1.290 MPa

d

0.852 mm

2)

1

1

1 0.29 (80 106 ) 9 200 10

(3)(284 10 6 )

284 10

6

852 10 6 m


PROBLEM 7.99 A spherical gas container having an inner diameter of 5 m and a wall thickness of 24 mm is made of steel 0.29. Knowing that the gage pressure in the container is increased from zero for which E 200 GPa and to 1.8 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the diameter of the container.

SOLUTION

(a)

p

1.8 MPa

r

1 d 2

1

2

1 (5) 2

t

pr 2t

24 10

3

(1.8)(2.476) (2)(24 10 3 )

2.476 m

92.850 MPa 92.9 MPa

1

(b)

d

1 ( E

d

2)

1

1

1 E

1

(5)(329.6 10 6 )

1 0.29 (92.85 106 ) 200 109

1.648 10 3 m

329.6 d

1.648 mm


PROBLEM 7.100 The maximum gage pressure is known to be 1150 psi in a spherical steel pressure vessel having a 10-in. outer diameter and a 0.25-in. wall thickness. Knowing that the ultimate stress in the steel used is U 60 ksi, determine the factor of safety with respect to tensile failure.

SOLUTION r

1

d t 2 10 in. 0.25 in. 2 4.75 in. pr 2t (1150 psi)(4.75 in.) 2(0.25 in.) 2

10.925 ksi F.S.

U max

60 ksi 10.925 ksi F.S.

5.49


PROBLEM 7.101 A spherical pressure vessel of 750-mm outer diameter is to be fabricated from a steel having an ultimate stress 400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa, U determine the smallest wall thickness that should be used.

SOLUTION r

We have and

max

F.S.

1 d t 2 1 (0.750 m) 2 0.375t (m) 1

t

pr 2t

2 U

max

Combining these two equations gives

F.S. or

2

Ut

2t U pr (F.S.) pr

Substituting for r gives 2(400 106 Pa)t 6

816.80 10 t t

(4)(4.2 106 Pa)(0.375 6.30 10

t)

6

7.71 10 3 m

t

7.71 mm


PROBLEM 7.102 A spherical gas container made of steel has a 20-ft outer diameter and a wall thickness of 167 in. Knowing that the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the container.

SOLUTION d t r

20 ft

240 in.

7 in. 0.4375 in. 16 1 d t 119.56 in. 2 (75)(119.56) pr 10.25 103 psi 2t (2)(0.4375)

max

10.25 ksi

min

0

(Neglecting small radial stress)

1 ( 2

max

max

min )

10.25 ksi

max

5.12 ksi


PROBLEM 7.103 A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball is inflated to a 120-kPa gage pressure.

SOLUTION r

1 d t 2 1 (300 mm) 2 147 mm

1

2

3 or

147 10 3 m

pr 2t

(120 103 Pa)(147 10 3 m) 2(3 10 3 m) 2.9400 106 Pa

2.94 MPa


PROBLEM 7.104

8m

14.5 m

h

The unpressurized cylindrical storage tank shown has a 5-mm wall thickness and is made of steel having a 400-MPa ultimate strength in tension. Determine the maximum height h to which it can be filled with water if a factor of safety of 4.0 is desired. (Density of water 1000 kg/m3.)

SOLUTION d0

t

5 mm

r

1 d 2

all

all

p

but

8m

U

F.S. pr t t

all

r

p

gh,

h

p g

0.005 m t

4

0.005

400 MPa 4.0

3.995 m 100 MPa

(0.005 m)(100 106 Pa) 3.995 m

125.156 103 Pa (1000 kg/m3 )(9.81 m/s 2 )

100 106 Pa

125.156 103 Pa

12.7580 m h

12.76 m


PROBLEM 7.105

8m

14.5 m

For the storage tank of Prob. 7.104, determine the maximum normal stress and the maximum shearing stress in the cylindrical wall when the tank is filled to capacity (h 14.5 m). h

PROBLEM 7.104 The unpressurized cylindrical storage tank shown has a 5-mm wall thickness and is made of steel having a 400-MPa ultimate strength in tension. Determine the maximum height h to which it can be filled with water if a factor of safety of 4.0 is desired. (Density of water 1000 kg/m3.)

SOLUTION d0 t r

p

8m 5 mm 0.005 m 1 d t 4 0.005 2

gh

3.995 m

(1000 kg/m3 )(9.81 m/s2 )(14.5 m) 142.245 103 Pa

1

pr t

(142.245 103 Pa)(3.995 m) 0.005 m

113.654 106 Pa max min max

1

max

113.7 MPa

0 1 ( 2

max

min )

max

56.8 MPa


PROBLEM 7.106 The bulk storage tank shown in Photo 7.3 has an outer diameter of 3.3 m and a wall thickness of 18 mm. At a time when the internal pressure of the tank is 1.5 MPa, determine the maximum normal stress and the maximum shearing stress in the tank.

SOLUTION r

d 2

1

pr t

max

1

min

p

0

1 ( 2

max

max

3.3 2

t

18 10

3

1.632 m,

(1.5 106 Pa)(1.632 m) 18 10 3 m 136 106 Pa

min )

t

18 10 3 m

136 106 Pa

max

68 106 Pa

max

136.0 MPa

68.0 MPa


PROBLEM 7.107 A standard-weight steel pipe of 12-in. nominal diameter carries water under a pressure of 400 psi. (a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum tensile stress in the pipe. (b) Solve part a, assuming an extra-strong pipe is used of 12.75-in. outside diameter and 0.5-in. wall thickness.

SOLUTION (a)

d0

12.75 in. t pr t

(b)

d0

(400)(6.00) 0.375

12.75 in. t pr t

0.375 in. r

t

6.00 in.

6400 psi

0.500 in. r

(400)(5.875) 0.500

1 d0 2

6.40 ksi

1 d0 2

t

5.875 in.

4700 psi

4.70 ksi


PROBLEM 7.108 A cylindrical storage tank contains liquefied propane under a pressure of 1.5 MPa at a temperature of 38 C. Knowing that the tank has an outer diameter of 320 mm and a wall thickness of 3 mm, determine the maximum normal stress and the maximum shearing stress in the tank.

SOLUTION r

d 2

t

3 10 3 m

320 2

t

157 mm

157 10 3 m

(1.5 106 Pa)(157 10 3 m) 3 10 3 m

1

pr t

max

1

min

p

0

1 ( 2

max

max

3

78.5 106 Pa

min )

78.5 106 Pa

max

78.5 MPa

max

39.3 MPa

39.25 106 Pa


PROBLEM 7.109 Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5-ft outer diameter and 5 -in. wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0 8 is desired.

SOLUTION 1

r

1

65 ksi 13 ksi 13 103 psi F .S. 5.0 d (5.5)(12) t 0.625 32.375 in. 2 2 U

pr t

p

t

1

r

(0.625)(13 103 ) 32.375

p

251 psi


PROBLEM 7.110

A

A steel penstock has a 36-in. outer diameter, a 0.5-in. wall thickness, and connects a reservoir at A with a generating station at B. Knowing that the specific weight of water is 62.4 lb/ft3, determine the maximum normal stress and the maximum shearing stress in the penstock under static conditions.

500 ft

B 36 in.

SOLUTION r

1 d 2

t

1 (36) 2

p

rh

(62.4 lb/ft 3 )(500 ft)

0.5

17.5 in.

31.2 103 lb/ft 2

216.67 psi 1

pr t

max

1

min

p

max

1 ( 2

(216.67)(17.5) 0.5

7583 psi

7583 psi

max

7.58 ksi

max

3.90 ksi

217 psi max

min )

3900 psi


PROBLEM 7.111

A

A steel penstock has a 36-in. outer diameter and connects a reservoir at A with a generating station at B. Knowing that the specific weight of water is 62.4 lb/ft3 and that the allowable normal stress in the steel is 12.5 ksi, determine the smallest thickness that can be used for the penstock.

500 ft

B 36 in.

SOLUTION p

h

(62.4 lb/ft 3 )(500 ft)

31.2 103 lb/ft 2

216.67 psi 1

r 1

18

t t 18 t

12.5 ksi

12.5 103 psi

1 d t 18 t 2 pr r 1 , t t p 12.5 103 216.67

57.692

t

58.692

0.307 in.


PROBLEM 7.112

600 mm

b

The cylindrical portion of the compressed-air tank shown is fabricated of 8-mm-thick plate welded along a helix forming an angle 30° with the horizontal. Knowing that the allowable stress normal to the weld is 75 MPa, determine the largest gage pressure that can be used in the tank.

1.8 m

SOLUTION r 1

2

ave

R w

p

1 d t 2 pr t 1 pr 2 t 1 ( 1 2

1 (600) 2

6

292 mm

3 pr 4 t 1 pr 1 2 2 4 t R cos 60 ave 2)

5 pr 8 t 8 wt 5 r p

8 (75)(8) 5 292

3.29 MPa


PROBLEM 7.113

600 mm

For the compressed-air tank of Prob. 7.112, determine the gage pressure that will cause a shearing stress parallel to the weld of 30 MPa. b 1.8 m

PROBLEM 7.112 The cylindrical portion of the compressed-air tank shown is fabricated of 8-mm-thick plate welded along a helix forming an angle 30° with the horizontal. Knowing that the allowable stress normal to the weld is 75 MPa, determine the largest gage pressure that can be used in the tank.

SOLUTION r 1

2

R w

p

1 d t 2 pr t 1 pr 2 t 1

2

2 R sin 60

1 (600) 2

8

292 mm

1 pr 4 t

3 pr 8 t 8 wt 3 R p

8 (30)(8) 3 292

3.80 MPa


PROBLEM 7.114 The steel pressure tank shown has a 750-mm inner diameter and a 9-mm wall thickness. Knowing that the butt-welded seams form an angle 50 with the longitudinal axis of the tank and that the gage pressure in the tank is 1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

!

SOLUTION r

d 2

1

pr t

2

1 2

375 mm

0.375 m

(1.5 106 Pa 0.375 m) 0.009 m 1

31.25 MPa

62.5 106 Pa

2

100

ave

1 ( 2

R

1

(a)

2)

1 2

2 w

62.5 MPa

46.875 MPa

15.625 MPa ave

R cos100 w

(b)

w

44.2 MPa

R sin100 w

15.39 MPa


PROBLEM 7.115 The pressurized tank shown was fabricated by welding strips of plate along a helix forming an angle with a transverse plane. Determine the largest value of that can be used if the normal stress perpendicular to the weld is not to be larger than 85 percent of the maximum stress in the tank.

!

SOLUTION 1

pr t

ave

1 ( 2

R w

0.85

pr t

cos 2 2

2

pr 2t

3 pr 4 t 1 pr 1 2 2 4 t R cos 2 ave 2)

1

3 4

1 cos 2 4

4 0.85

3 4

pr t

0.4

113.6

56.8


12 ft

PROBLEM 7.116

12 ft

Square plates, each of 0.5-in. thickness, can be bent and welded together in either of the two ways shown to form the cylindrical portion of a compressed-air tank. Knowing that the allowable normal stress perpendicular to the weld is 12 ksi, determine the largest allowable gage pressure in each case.

45" 20 ft

(a)

(b)

SOLUTION d

12ft

1

pr t

(a)

1

144 in. r 2

1 d 2 pr 2t

t

71.5 in.

12 ksi 1t

p

(12)(0.5) 71.5

r

0.0839 ksi p

(b)

ave

1 ( 2

R

1

w

ave

2)

1

2 45

2

83.9 psi

3 pr 4 t 1 pr 4 t

R cos

3 pr 4 t p

4 wt 3 r

4 (12)(0.5) 3 71.5

0.1119 ksi

p

111.9 psi


PROBLEM 7.117

3m

The pressure tank shown has a 0.375-in. wall thickness and butt-welded 20° with a transverse plane. For a gage seams forming an angle pressure of 85 psi, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

1.6 m

!

SOLUTION d r 1

2

ave

R

(a)

w

(b)

w

ave

R cos 40

R sin 40

5 ft

60 in.

1 d t 30 0.375 29.625 in. 2 (85)(29.625) pr 6715 psi 0.375 t 1 3357.5 psi 1 2 1 ( 1 5036.2 psi 2) 2 1

2

2

1678.75 psi

3750 psi

1079 psi


PROBLEM 7.118 3m 1.6 m

For the tank of Prob. 7.117, determine the largest allowable gage pressure, knowing that the allowable normal stress perpendicular to the weld is 18 ksi and the allowable shearing stress parallel to the weld is 10 ksi. PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness 20° with a transverse plane. and butt-welded seams forming an angle For a gage pressure of 85 psi, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

!

SOLUTION d

5 ft

60 in.

1 d 2 pr t pr 2t 1 ( 2

r 1

2

ave

t

30

1

w

ave

2

R cos 50 1 cos 50 4

0.58930

p w

p

wt

0.5893r

(18)(0.375) (0.58930)(29.625)

R sin 50

0.191511

wt

0.191511r

0.38664 ksi

pr t (10)(0.375) (0.191511)(29.625)

3 pr 4 t 1 pr 4 t

2 3 4

29.625 in.

2)

1

R

0.375

pr t

pr t

387 psi

0.66097 ksi

661 psi

p

Allowable gage pressure is the smaller value.

387 psi


PROBLEM 7.119 3m 1.6 m

For the tank of Prob. 7.117, determine the range of values of that can be used if the shearing stress parallel to the weld is not to exceed 1350 psi when the gage pressure is 85 psi. PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness 20° with a transverse plane. and butt-welded seams forming an angle For a gage pressure of 85 psi, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

!

SOLUTION d r 1

2

R w

sin 2 2

a

53.53

a

2

b

53.53

b

26.8

2

c

53.53

c

63.2

2

d

d

116.8

53.53

180 180

126.47 233.53

26.8

a

5 ft

60 in.

1 3 29.625 in. d t 30 2 8 (85)(29.625) pr 6715 psi 0.375 t 1 3357.5 psi 1 2 1

2

2 R sin 2 w

R

1678.75 all

1350 1678.75

0.80417

26.8 63.2


PRO OBLEM 7.120 4 ft

P'

A

A preessure vessel of o 10-in. inner diameter annd 0.25-in. waall thickness is fabbricated from a 4-ft sectioon of spirally--welded pipe AB and is equippped with two rigid end plattes. The gage pressure p insidde the vessel is 3000 psi and 10-kkip centric axxial forces P and a P are appplied to the end plates. p Determ mine (a) the noormal stress peerpendicular to t the weld, (b) the shearing streess parallel to the weld.

P 35"

B

SOLUTION

r0

1 d 2 pr t pr 2t r t

A

r02

r 1

2

P A Totaal stresses.

1 t 0.25 in. (10) 5 in. 2 (3000)(5) 6000 psi p 6 ksi 0.225 (3000)(5) 3000 psi p 3 ksi (2)(00.25) 5 0.25 5.25 inn. (5.252

r2

100 103 8.0803

5.002 ) 8.05003 in 2

12442 psi

Longitudinal:

x

3 1.242 1.7588 ksi

Circumferential:

y

6 ksi k

Shear:

xy

1.242 ksi

0


X : (1.758, 0) Y : (6, 0) C : (3.879) ave

1 ( 2

x

y)

x

y

3.8879 ksi 2

R

2 xy

2 ((1.758 6) 2

(a) (b)

x

|

xy |

avee

R cos 70

R siin 70

2

0

2.121 kssi

3.879 2.1221 cos 70

2.1211 sin 70

x

|

xy

3.15 ksi

| 1.993 1 ksi


PROB BLEM 7.12 21 Solve Prob. 7.120, assuming thatt the magnituude P of the tw wo forces is increassed to 30 kips.

4 ft

P'

A

PROB BLEM 7.120 A pressure vesssel of 10-in. inner diameterr and 0.25-in. wall thhickness is faabricated from m a 4-ft sectioon of spirally--welded pipe AB andd is equipped with two rigiid end plates. The gage preessure inside the vesssel is 300 psii and 10-kip centric c axial foorces P and P are applied to the end plates. Determine D (a) the normal stress perpenddicular to the weld, (b) ( the shearinng stress parallel to the weldd.

P 35""

B

SO OLUTION

r0

1 d 2 pr t pr 2t r t

A

r02

r 1

2

Tottal stresses.

1 ( (10) 5 in. t 0.25 in. 2 (3000)(5) 6000 psi 6 ksi 0 0.25 (300)(5) 30000 psi 3 ksi (2))(0.25) 5 0.25 5.25 in.

r2

(5.252

52 ) 8.05033 in 2

P A

30 103 8.0503

37727 psi

Longitudinall:

x

3 3.727

0.7727 ksi

Circumferen ntial:

y

6 ksi

Shear:

3.7727 ksi

0

xy


X : ( 0.727, 0) 0 Y : (6, 0) C : (2.66365, 0) av ve

1 ( 2

x

y)

x

y

2.6365 kssi 2

R

0.7277 6 2 (a) (b)

x

|

xy

ave

|

R cos 70

R sin 70

2 xy

2 2

0

3 3.3635 ksi

2.6365 3.3635 cos 70

3.36 635 sin 70

x

|

xy

1.486 ksi

| 3.16 ksi


T

PROBLEM 7.122 2 A torquue of magnituude T 12 kN nd of a tank containing N m is applied to the en compresssed air under a pressure off 8 MPa. Know wing that the tank t has a 1800-mm inner diameterr and a 12-mm m wall thicknness, determinne the maximuum normal strress and the maximuum shearing strress in the tank.

SOLUTION d

1 d 2

180 mm m r

90 mm t

12 mm

Torssion: c1

90 mm m c2

J

c24

2

90

c14

12

102 mm m

66.9668 106 mm 4

(12 103 )(1022 10 3 ) 66.968 10 6

Tc J

66.968 10 6 m 4

188.277 MPa

Presssure: pr t

1

(8)(90) 12

600 MPa

2

pr 2t

30 MP Pa

Sum mmary of stresses: x

ave

60 MPa, M 1 ( 2

30 MPa,

y

x

y)

x

y

xy

18.277 MPa

45 MPa 2

R

2 xxy

2

a

avee

R

M 68.64 MPa

b

avee

R

21.36 MPa M

c

min

max

Pa 23.64 MP

0 max

688.6 MPa

max

344.3 MPa

0

1 ( 2

max

min )


T

PROBLEM 7.123 The tank shown has a 180-mm inner diameter and a 12-mm wall thickness. Knowing that the tank contains compressed air under a pressure of 8 MPa, determine the magnitude T of the applied torque for which the maximum normal stress is 75 MPa.

SOLUTION r 1

2

ave max

R

1 1 d (180) 90 mm 2 2 pr (8)(90) 60 MPa t 12 pr 30 MPa 2t 1 ( 1 45 MPa y) 2 75 MPa max

t 12 mm

30 MPa

ave 2

R xy

1

2

2 R 2 152

2 xy

152

302 152

2 xy

25.98 MPa

6

25.98 10 Pa

Torsion:

c1

90 mm

c2

90 12 102 mm

J xy

T 4 c2 c14 66.968 106 mm 4 66.968 10 6 m 4 2 J xy (66.968 10 6 )(25.98 106 ) Tc T 17.06 103 N m 3 J c 102 10 T

17.06 kN m


PROBLEM 7.124

y 150 mm

The compressed-air tank AB has a 250-mm outside diameter and an 8-mm wall thickness. It is fitted with a collar by which a 40-kN force P is applied at B in the horizontal direction. Knowing that the gage pressure inside the tank is 5 MPa, determine the maximum normal stress and the maximum shearing stress at point K.

B

P

600 mm K

L

A z

150 mm x

SOLUTION Consider element at point K. Stresses due to internal pressure: p r x

y

Stress due to bending moment:

5 MPa 5 106 Pa 1 250 d t 8 117 mm 2 2 pr (5 106 )(117 10 3 ) 73.125 MPa t (8 10 3 ) (5 106 )(117 10 3 ) (2)(8 10 3 )

pr 2t

Point K is on the neutral axis. 0

y

Stress due to transverse shear:

36.563 MPa

V c2 c1 Q

I

xy

P 40 103 N 1 d 125 mm 2 c2 t 117 mm 2 3 3 2 c2 c1 (1253 1173 ) 3 3 234.34 103 mm3 234.34 10 6 m3 (1254 117 4 ) c24 c14 4 4 44.573 106 mm 4 44.573 10 6 m 4 VQ It

PQ I (2t )

(40 103 )(234.34 10 6 ) (44.573 10 6 )(16 10 3 )

13.1436 106 Pa 13.1436 MPa 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. 1163


Total stresses:

x

Mohr’s circle:

ave

73.125 MPa, 1 ( 2

x

y)

36.563 MPa,

y

xy

13.1436 MPa

54.844 MPa 2 x

R

y

2 xy

2 (18.281) 2

(13.1436) 2

a

ave

R

77.360 MPa

b

ave

R

32.328 MPa

22.516 MPa

Principal stresses:

a

77.4 MPa,

b

The 3rd principal stress is the radial stress.

z max

Maximum shearing stress:

32.3 MPa

max

77.4 MPa,

1 ( 2

max

min

min )

0

max

max

0

77.4 MPa 38.7 MPa


PROB BLEM 7.12 25

y 150 0 mm

In Probb. 7.124, deterrmine the maxximum normall stress and thee maximum shearinng stress at poiint L.

PROB BLEM 7.124 The T compresssed-air tank AB B has a 250-m mm outside diametter and an 8-m mm wall thicknness. It is fitteed with a collaar by which a 40-kN N force P is appplied at B in the horizontal direction. Knnowing that the gagge pressure innside the tankk is 5 MPa, determine d thee maximum normall stress and thee maximum shhearing stress at point K.

B

P

600 mm K

L

A z

m 150 mm x

SOLUTION Connsider elementt at point L. Streesses due to in nternal pressurre:

p r x

y

ding moment: Streess due to bend

5 MPa 5 106 Pa 1 250 d t 8 1177 mm 2 2 pr (5 106 )(117 100 3 ) 73.125 MPa t 8 10 3 pr (5 103 )(117 100 3 ) 36.563 MPa 2t (2)(8 10 3 )

M c2 c1 I

y

(40 kN)(600 k mm) 24,000 N m 1 d 125 mm 2 c2 t 125 8 117 mm c24 c14 (1254 117 4 ) 4 4 44.573 106 mm 4 44.573 10 6 m 4 Mc I

(24, 000)(125 10 3 ) 44.573 10 6

Pa 67.305 MP



Stress due to transverse shear:

Point L lies in a plane of symmetry. xy

Total stresses:

x

Principal stresses:

max

73.125 MPa,

0 30.742 MPa,

y

xy

0

Since xy 0, x and y are principal stresses. The 3rd principal stress is in the radial direction, z 0. 73.125 MPa,

min

0,

a

73.1 MPa,

b

Maximum stress: Maximum shearing stress:

max

1 ( 2

max

min )

30.7 MPa,

z

0 max

73.1 MPa

max

51.9 MPa


PROBLEM 7.126

1.5 in.

A brass ring of 5-in. outer diameter and 0.25-in. thickness fits exactly inside a steel ring of 5-in. inner diameter and 0.125-in. thickness when the temperature of both rings is 50 F. Knowing that the temperature of both rings is then raised to 125 F, determine (a) the tensile stress in the steel ring, (b) the corresponding pressure exerted by the brass ring on the steel ring.

STEEL ts # 81 in. Es # 29 $ 106 psi %ss # 6.5 $ 10–6/"F

5 in.

BRASS tb # 14 in. Eb # 15 $ 106 psi %bs # 11.6 $ 10–6/"F

SOLUTION Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the brass ring. Steel ring.

Internal pressure p:

s

pr ts pr Es t s

s

Corresponding strain:

sp

Es

Strain due to temperature change:

sT

s

Total strain:

(1)

T

s

pr Es t s

Ls

2 r

b

pr tb

s

T

Change in length of circumference:

Brass ring.

External pressure p:

Corresponding strains:

s

2 r

pr Es ts

bT

b

pr , Eb tb

bp

s

T

T


Lb Equating

Ls to Lb ,

pr Es t s

r Es t s

s

2 r

(

b

pr Eb tb

2 r

pr Eb tb

T

r p Eb tb

b

b

s)

T

b

T

T

(2)



T 125 F 50 F 75 F

Data:

r From Equation (2),

2.5 (29 106 )(0.125)

1 d 2

1 (5) 2

2.5 in.

2.5 p (15 106 )(0.25)

(11.6 6.5)(10 6 )(75)

1.35632 10 6 p 382.5 10 p From Equation (1),

s

pr ts

(282.0)(2.5) 0.125

6

282.0 psi

5.64 103 psi

(a)

s

(b) p

5.64 ksi

282 psi


PROBLEM 7.127

1.5 in. STEEL ts # 81 in. Es # 29 $ 106 psi %ss # 6.5 $ 10–6/"F

5 in.

BRASS tb # 14 in. Eb # 15 $ 106 psi %bs # 11.6 $ 10–6/"F

Solve Prob. 7.126, assuming that the brass ring is 0.125 in. thick and the steel ring is 0.25 in. thick. PROBLEM 7.126 A brass ring of 5-in. outer diameter and 0.25-in. thickness fits exactly inside a steel ring of 5-in. inner diameter and 0.125-in. thickness when the temperature of both rings is 50 F. Knowing that the temperature of both rings is then raised to 125 F, determine (a) the tensile stress in the steel ring, (b) the corresponding pressure exerted by the brass ring on the steel ring.

SOLUTION Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the brass ring. Steel ring.

Internal pressure p:

s

pr ts pr Es t s

s

Corresponding strain:

sp

Es

Strain due to temperature change:

sT

s

Total strain:

(1)

T

s

pr Es t s

Ls

2 r

s

T


Brass ring.

External pressure p:

2 r

pr Es t s

bT

b

s

T

pr tb

b

Corresponding strains:

s

pr , Eb tb

bp

T


Lb Equating

Ls to Lb ,

pr Es t s

r Es ts

s

2 r

b

pr Eb tb

T

r p ( Eb tb

b

pr Eb tb

2 r

b

s)

T

b

T

T (2)



Data:

T

125 F 50 F 75 F 1 1 (5) 2.5 in. r d 2 2

From Equation (2),

2.5 (29 106 )(0.25)

2.5 p (15 106 )(0.125)

(11.6 6.5)(10 6 )(75)

1.67816 10 6 p 382.5 10 p From Equation (1),

s

pr ts

(227.93)(2.5) 0.25

6

227.93 psi

2279 psi

(a)

s

(b) p

2.28 ksi

228 psi


PROBLEM 7.128 y

y' x'

&

x

For the given state of plane strain, use the method of Sec. 7.7A to determine the state of plane strain associated with axes x and y rotated through the given angle .

800 ,

x

450 ,

y

xy

200 ,

25

SOLUTION 25 x

y

2

x x

xy

y

y

2 175 (

y

x

( 800

x

xy

625

2

2 175 x

y

x

175

y

2

100

xy

cos 2 sin 2 2 2 ( 625 ) cos ( 50 ) (100 )sin ( 50 ) x

y

xy

cos 2 sin 2 2 2 ( 625 ) cos ( 50 ) (100 )sin ( 50 ) y )sin

2

653

x

xy

y

303

cos 2

450 )sin ( 50 ) ( 200 ) cos ( 50 )

xy

829


PROBLEM 7.129 y

y'


x'

&

x

240 ,

x

160 ,

y

xy

150 ,

60

SOLUTION 60 x

y

2

x

x

xy

y

2 y

x

y

xy

40

y

x

y

cos 2

2

75

xy

sin 2 2 2 2 200 40 cos ( 120 ) 75 sin ( 120 ) x

y

x

200

xy

sin 2 2 2 2 200 40cos ( 120 ) 75sin ( 120 ) (

x

y )sin

cos 2

2

xy

115.0

x

y

285

cos 2

(240 160)sin ( 120 ) 150 cos ( 120 )

xy

5.72


PROBLEM 7.130 y

y'


x'

!

x

500 ,

x

250 ,

y

xy

0,

15

SOLUTION 15 x

y

2

x x

xy

2 125

x

y

y

x

( 500

x

y

cos 2

cos 2

2 ( 375 ) cos 30 y )sin

2

xy

375

2 ( 375 ) cos 30

2 125 (

y

2 y

x y

x

125

xy

2 xy

2 0 xy

2 0

0

sin 2 x

450

y

199.8

sin 2

cos 2

250 )sin 30

0

xy

375


PROBLEM 7.131 y

y' x'

!

x

For the given state of plane strain, use the method of Sec 7.7A to determine the state of plane strain associated with axes x and y rotated through the given angle . 0,

x

y

320 ,

xy

100 ,

30

SOLUTION 30 x

y

2

x

160

2

x x

y

y

x

2

y

2

x

y

2

x

y

2

160 160cos 60 xy

(

x

y )sin

cos 2

xy

2

sin 2

100 sin 60 2

160 160 cos 60

y

160

2

(0 320)sin 60

cos 2

xy

2

100 sin 60 2 xy

x

36.7

sin 2 y

283

xy

227

cos 2

100 cos 60


PROBLEM 7.132 y

y'

For the given state of plane strain, use Mohr’s circle to determine the state of plane strain associated with axes x and y rotated through the given angle .

x' x

!

800 ,

x

450 ,

y

xy

200 ,

25

SOLUTION Plotted points:

X : ( 800 , 100 ) Y : ( 450 , 100 ) C : ( 175 , 0) 100 625

tan

(625 ) 2

R

x

ave

9.09 (100 ) 2

2

50

R cos

175

9.09

632.95

40.91

632.95 cos 40.91 653

x y

ave

R cos

175

632.95 cos 40.91 y

1 2

xy

R sin

632.95 sin 40.91

xy

303 829


PR ROBLEM 7.133 7 y

y' x'

!

x

For the given staate of plane sttrain, use Mohhr’s circle to determine d the staate of plane sttrain associateed with axes x and y rotaated through thee given angle . 240 ,

x

y

160 ,

xy

150 ,

60

SO OLUTION Plootted points for Mohr’s circlle:

X : ( 240 2 , 75 ) Y : ( 1160 , 75 ) C : ( 200 2 , 0) 75 1.875 40

tan n

(40 )2

R 2

1 2

61.933 (75 )2

8 85

x

ave

181.93 120 61.93 R cos c 200 (85 ) cos ( 181.93 )

y

ave

R cos c

xy

R sin

200

(85 ) cos ( 181.93 )

85 sin ( 181.93 )

2 2.86

115.0

x y xy

285 5.72


PROBLEM 7.134 y

y' x' x

!

For the given state of plane strain, use Mohr’s circle to determine the state of plane strain associated with axes x and y rotated through the given angle . 500 ,

x

y

250 ,

xy

0,

15

SOLUTION

Plotted points:

X : ( 500 ,0) Y : ( 250 , 0) C : ( 125 , 0)

1 2

R

375

x

ave

R cos 2

125 375cos 30

y

ave

R cos 2

125 375cos 30

xy

R sin 2

375sin 30

x

y

xy

450 199.8 375


PR ROBLEM 7.135 7 y

y' x'

!

x

For the given staate of plane sttrain, use Mohhr’s circle to determine d the staate of plane strrain associated with axes x and y rotaated through thee given angle . 0,

x

y

3200 ,

xy

100 ,

30


X : (0, 50 ) Y : (320 , 50 ) C : (160 , 0) 50 160

tan n

17.35

(160 ) 2

R

2

1 2

(50 ) 2

60

167.63

17.35

42.65

x

ave

R cos c

160

(167.63 ) coos 42.65

y

ave

R cos c

160

(167.63 ) coos 42.65

xy

R sin

(167.63 )ssin 42.65

x

36.7

y

283

xy

227


PR ROBLEM 7.136 Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the platte is unstresseed, determine (a) the direction and magnnitude of the principal p strains, (b) the maaximum inplanne shearing strrain, (c) the maximum m shearring strain. (U Use v 13 . ) 2600 ,

x

600 ,

y

xy

4880

SOLUTION For Mohr’s circlee of strain, plot points:

X : ( 2660 , 240 ) Y : ( 600 , 240 ) C : ( 1660 , 0) tan 2

x

2

ave

R

160

260

b

ave

R

160

260

max (in-plaane)

R

max m

(c)

max

max x

1 v

(

160 min

a

min

160

2

(240 )

a

56.3

a

100

2

420

b

2R

max ((in-plane)

v c

33.7

260

a

1 2

2.4 b

(100 )

R

(b)

y

67.38

p

R

(a)

480 260 60

xy p

max (in-plane))

v

b)

1 v

(

x

y)

1/3 ( 260 60) 2/3 160

420

c

420

520

maxx

160 580


PR ROBLEM 7.137 7 Thee following sttate of strain has h been measuured on the suurface of a thinn plate. Know wing that the suurface of the plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum m inplaane shearing sttrain, (c) the maximum m sheaaring strain. (U Use v 13 . ) 6000 ,

x

y

4000 ,

xy

3 350


X : ( 600 , 175 ) Y : ( 400 , 175 ) C : ( 500 , 0) 0 tan 2

p

2

p

175 100

60.26 30.1

b a

(100 ) 2

R

59.9

(175 ) 2

201.6

(a)

(b)

a

avee

R

500

201.6

a

298

b

avee

R

500

201.6

b

702

max (in-pllane)

2R v ( 1 v

c

(c)

max

max

500

max

min

a

min

500 5

v ( 1 v

b)

x

y)

1/3 ( 600 2/3

max (in-planee)

403

c

500

400 )

702 702

max

1202


PR ROBLEM 7.138 Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the platte is unstresseed, determine (a) the direction and magnnitude of the principal p strains, (b) the maaximum inplanne shearing strrain, (c) the maximum m shearring strain. (U Use v 13 . ) 160 ,

x

4800 ,

y

xy

6 600

SOLUTION (a)

For Mohr’ss circle of straain, plot pointss:

X : (160 , 300 ) Y : ( 480 , 300 ) C : ( 160 , 0) 0 (a)

tan 2

x

2

3000 3200

xy p

p

y

43.15

0.9375

21.58

p

andd

21.58 900

68.42

21.6

a b

R

(b) (c)

1 2 c

(320 ) 2

(3000 ) 2

68.4

438.66

a

ave a

R

1600

438.6

a

279

b

ave a

R

1600

438.6

b

599

R

(max, in-plaane)

v ( 1 v

a

2R

(maxx, in-plane)

b)

v ( 1 v

max

2778.6

max

m max

1/3 (160 2/3

y)

x

min min

(max, in-plane))

480 )

c

877 160.0

598.6

2778.6

598.6

max

877


PR ROBLEM 7.139 7 Thee following sttate of strain has h been measuured on the suurface of a thinn plate. Know wing that the suurface of the plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum m inplaane shearing sttrain, (c) the maximum m sheaaring strain. (U Use v 13 .) x

30 ,

y

5700 ,

xy

7 720


X : (30 , 360 ) Y : (570 , 360 ) C : (300 , 0) tan 2

p

2

p

360 270 53.13

1.3333

(a)

26.6

b a

(b)

R

(2770 )2

a

ave

R

300

450

b

ave

R

300

450

max (in-planee)

(360 ) 2

(c)

max

450 a

max (in-planee)

1 v

maax

a

maax

min

(

a

750 , 750 0

b)

min

1/3 (750 2/33 c

150 )

750

150.0

b

2R v

c

64.4

900

c

300

max

1050

300

( 300 )


PR ROBLEM 7.140 For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the prinncipal strains, (b) the maxim mum in-plane strain, s (c) the maximum m sheearing strain. 60 ,

x

2400 ,

y

xy

5 50

SOLUTION Plottted points:

X : (60 , 25 2 ) Y : (240 , 25 ) C : (150 , 0) tan 2

xy p x

2

y

50 60 240

0 0.277778

15.52

p

97.8

a

7.8

b

(90 ) 2

R (a)

(b)

(c)

a

ave

R 150

933.4

b

ave

R 150

933.4

max (in-plane))

c

0,

max m

(25 ) 2

933.4 a b

2R

max (in-plane)

243.4 ,

max m

max

min

0

243 56.6

186.8

c max x

m min

0

243


PR ROBLEM 7.141 7 Forr the given staate of plane strain, s use Moohr’s circle to determine (aa) the orientation and magnnitude of the prinncipal strains,, (b) the maxim mum in-plane strain, (c) the maximum shearing strain. 4000 ,

x

200 ,

y

xy

3 375


X : ( 400 , 187.5 ) Y : ( 200 , 187.5 ) C : ( 300 , 0) 0 tan 2

xy p x

2

y

375 400 200

1.875

61.93

p

a

121.0

b

(100 )2

R (a)

a

ave

R

300

212.5 2

b

ave

R

300

212.5 2

(b)

max (in-planee)

(c)

c

0

(187.5 ) 2

212.5 a b

2R

max (in-planee)

max

512.5

max

max

min

31.0

0

513 87.5

425

c m max

min

0

513


PR ROBLEM 7.142 For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the prinncipal strains, (b) the maxim mum in-plane strain, s (c) the maximum m sheearing strain. 3000 ,

x

600 ,

y

xy

1000

SOLUTION X : (300 , 500 ) Y : (60 , 50 ) C : (180 , 0) tan 2

xy p x

2

p

y

100 300 60

22.62 a b

R (a)

(120 )2

(50 ) 2

11.3 101.3

1 130

a

ave

R 180

1330

a

310

b

ave

R 180

1330

b

50.0

max (in-plane))

260

(b)

max (in-plane))

(c)

c

0,

max m

2R

310 ,

max m

max

min

0

c

m min

maxx

0

310


PR ROBLEM 7.143 7 Forr the given staate of plane strain, s use Moohr’s circle to determine (aa) the orientation and magnnitude of the prinncipal strains,, (b) the maxim mum in-plane strain, (c) the maximum shearing strain. 1800 ,

x

y

2660 ,

xy

3 315

SO OLUTION


X : ( 180 , 157.5 ) Y : ( 260 , 157.5 ) C : ( 220 , 0) 0 (a)

tan 2

x

2

315 5 80

xy p

p

y

3.9375

7 75.75

a b

(40 )2

R

(15 57.5 )2

ave

R

22 20

162.5

b

ave

R

22 20

162.5

max (in-planee)

(c)

c

0,

max

2R

57.5

a

b

383

325

0,

max

127.9

162..5

a

(b)

37.9

max

min n min

382.5

c

0 382.5

m max

0

383


3

PRO OBLEM 7.1 144

45"

2

Deterrmine the strain x , knowingg that the folloowing strains have been dettermined by use of o the rosette shown: s

30" 15"

x

1

1

4800

1

15

2

1220

3

800

SOLUTION

c x cos

2

n y sin

1

0.9330

os x co

2

0.75

os x co

2

3

2

x

n y sin

0.06699

30

3

75

1

n 1 cos 1 xy sin

0.06699

x

n y sin

2

2

2

x

2

0.25

y

xy

2

xy sin 2 cos 2

0.25

y

3

0..4330

xy

xy sin 3 cos 3

0.9330

y

0.25

xy

1

4880

(1)

2

(2)

1 120 3

(3)

800

Solvving (1), (2), and a (3) simultaaneously, x

253 ,

y

307 ,

xy

8893 x

253


PROB BLEM 7.145

y

The strrains determinned by the usee of the rosettte shown durring the test of o a machine elemen nt are

30" 3

2 1 30"

600

1

x

4500

2

755

3

Determ mine (a) the in--plane principal strains, (b) the in-plane maximum m sheaaring strain.

SO OLUTION

os x co

2

n y sin

1

0.75 x

co os2

2

y

0.75 x

cos2

y

3

2

sin 2 x

2

2

1500

3

90 xy sin 1 cos 1

1

0.433301

xy

6000

cos

2

2

0.433301

xy

4500

y

xy

0.25

sinn 2

30

1

0.25

x

1

3

sin

y

xy

2

sin

0

3 cos 3 y

(1)

(2)

3

0

755

(3)

Sollving (1), (2), and (3) simulttaneously, x ave

725 7 , 1 ( 2

75 ,

y

x

y)

x

y

(a)

(b)

a

ave

R

b

ave

R

max (in-plane))

2

173.21

325 2

R

xy

2

725 75 2

xy

2

2

173.21 2

2

4 409.3

734

a

84.3

b

2 R 819

max (in-planee)

734 84.3

819


PROBLEM 7.146 4

3

45"

45"

The rosette shown has been used to determine the following strains at a point on the surface of a crane hook:

2 45" 1

420 10 6 in./in.

1

x

45 10 6 in./in.

2

165 10 6 in./in.

4

(a) What should be the reading of gage 3? (b) Determine the principal strains and the maximum in-plane shearing strain.

SOLUTION (a) Gages 2 and 4 are 90 apart. 1 ( 2 4) 2 1 ( 45 10 2

ave

ave

6

165 10 6 )

60 10 6 in./in.

Gages 1 and 3 are also 90 apart. 1 ( 1 2 2 ave

ave 3

3) 1

(2)(60 10 6 ) 420 10

6

300 10 6 in./in.

3

(b)

x

xy

420 10 6 in./in.

1

2

2

1

3

y

300 10 6 in./in.

3

(2)( 45 10 6 ) 420 10

6

( 300 10 6 )

210 10 6 in./in. 2 x

R

y

2 xy

2

420 10

2

6

( 300 10 6 ) 2

2

210 10 2

6

2

375 10 6 in./in. a

ave

R

60 10

6

375 10

6

b

ave

R

60 10

6

375 10

6

max (in-plane)

a b

2R

max (in-plane)

435 10 6 in./in. 315 10 6 in./in. 750 10 6 in./in.


" 2

PROBLEM 7.147 !2

3 45#

Using a 45 rosette, the strains 1, 2 , and 3 have been determined at a given point. Using Mohr’s circle, show that the principal strains are:

!3

2

B

O

45#

A

!

C

1 1 2 ( 1 [( 1 ( 3) 2) 2 2 (Hint: The shaded triangles are congruent.) max,min

! min

1

!1 ! max

2

3)

SOLUTION Since gage directions 1 and 3 are 90 apart,

1 ( 2

ave

Let

u

1

ave

1 ( 2

v

2

ave

2

R2

1 ( 2 u2

1 4

R max, min

ave

3)

1

1 ( 4

3)

1

3)

1

v2 3)

1

1 2

2 1

1 2 1 2 1 ( 1 2 1 [( 2

2

2 2

1

1 4

1 3

2 3

2 2

2 1 2)

2( 1

2)

2

2 2

(

2

2

2 1

1 2

2 3

1 ( 2

2

1 ( 4

3)

3)

1

2 3

3)

2

1 4

2 1

1 2

1 3

1 4

2 3

2 3

2

3)

2 1/2

]

R gives the required formula.


1

2 2 ]

PROBLEM 7.148 2 3

Show that the sum of the three strain measurements made with a 60 rosette is independent of the orientation of the rosette and equal to

60" 60"

1

1

&

where circle.

x

2

3

3

avg

is the abscissa of the center of the corresponding Mohr’s

avg

SOLUTION x 1

ave

2

ave

y

2 x

y

2 x

ave

y

2 xy

2

x

cos (2

2

y

2 x

ave

y

2 xy

cos (2

2 xy

2

120 )

sin 120 sin 2 )

3 sin 2 2 (2) xy

240 )

2

(cos 240 cos 2

y

sin (2

3 cos 2 2

(cos 240 sin 2 x

ave

xy

2

sin 120 cos 2 )

1 cos 2 2

y

x ave

(1)

120 )

1 sin 2 2

xy

sin 2

(cos 120 cos 2

2

2

2

(cos 120 sin 2

ave

3

xy

cos 2

240 )

sin 240 sin 2 )

sin 240 cos 2 )

1 cos 2 2

1 sin 2 2

sin (2

3 sin 2 2

3 cos 2 2

(3)

Adding (1), (2), and (3), 1

2

3

3 ave

3

0 0

ave

1

2

3


PROBLEM 7.149 The strains determined by the use of the rosette attached as shown during the test of a machine element are

3 75"

2

x

75" 1

1

93.1 10 6 in./in.

2

385 10 6 in./in.

3

210 10 6 in./in.

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

SOLUTION Use

x

1 ( 2

x

y)

1 ( 2

where

sin 2

for gage 2,

0

and

2

for gage 1,

75

for gage 3.

75

From Eq. (2),

xy

y ) cos 2

x

1

1 ( 2

x

y)

1 ( 2

x

y ) cos (

2

1 ( 2

x

y)

1 ( 2

x

y ) cos

0

3

1 ( 2

x

y)

1 ( 2

x

y ) cos

(150 )

x

z

xy

150 ) xy

2

2

sin ( 150 )

sin 0 xy

2

sin (150 )

(1) (2) (3)

385 10 6 in./in.

Adding Eqs. (1) and (3), 1

3

(

x (1 y

y)

x

1

(

y ) cos 150

x

cos 150 )

y (1

cos 150 )

x (1 cos 150 ) (1 cos 150 )

3

93.1 10

6

210 10 6 385 10 6 (1 cos 150 ) 1 cos 150

35.0 10 6 in./in.



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

1

sin 150

xy 3

xy

210 10

1

sin 150

6

( 93.1 10 6 ) sin 150

606.2 10 6 in./in. tan 2

606.2 10 6 385 10 6 35.0 10

xy p x

y

1 1 ( x (385 10 y) 2 2 210 10 6 in./in.

ave

2 x

R

6

2

30.0 ,

b

120.0

xy

2

2 6

35.0 10

6

2

606.2 2

a

ave

R

210 10

6

350.0 10

6

b

ave

R

210 10

6

350.0 10

6

R

a

35.0 10 6 )

2

max (in-plane)

(a)

2

y

385 10

(b)

1.732

6

350.0 10 6 in./in.

2

350.0 10

6

a b

max (in-plane)

560 10 6 in./in. 140.0 10 6 in./in. 700 10 6 in./in.


PROBLE EM 7.150

y 1 in n.

P

A centric axial a force P and a a horizonttal force Qx arre both applieed at point C of the rectaangular bar shown. A 45 sttrain rosette on the surface of the bar at point A indicates the folloowing strains::

Qx

C

x 12 in. i

3 A 3 in.

45!

60 10 6 in./in.

2

240 10 6 in./in.

3

200 10 6 in./in.

29 106 psi and v

Knowing thhat E and Qx.

2

1

0.30, determ mine the magnitudes of P

1 3 in..

SO OLUTION x

1

60 10

6

y

3

200 10

6

xy x

y

P A

2

2

1

E ( 1 v2 E ( 1 v2 P

y

3400 10

3

x

v y)

y

v x) A

y

6

29 [ 60 6 (0.3)(200)] 0 1 (0.3)2 29 [2000 (0.3)( 60)] 5.8 103 psi p 2 1 (0.3) (22)(6)(5.8 103 ) 69.6 103 lb

G xy

I Qˆ xy

V

E 2(1 v) G

xy

69.6 kips

Q

30.3 kips

29 1006 11.1538 106 psi (2)(1.300)

(11.1538)((340) 3.79233 103 psi

1 3 1 (2)(6))3 36 in 4 bbh 12 12 A y (2)(3)(1.5) 9 in 3 t ˆ VQ It It xy Qˆ

P

(36)(2)(3..7923 103 ) 9

Q V

2 in.

30.338 103 lb l


PROBLEM 7.151 y 1 in.

Solve Prob. 7.150, assuming that the rosette at point A indicates the following strains:

P Qx

C

x 12 in.

1

30 10 6 in./in.

2

250 10 6 in./in.

3

100 10 6 in./in.

PROBLEM 7.150 A centric axial force P and a horizontal force Qx are both applied at point C of the rectangular bar shown. A 45 strain rosette on the surface of the bar at point A indicates the following strains: 3 A 3 in.

2 45" 1 3 in.

Knowing that E P and Qx.

1

60 10 6 in./in.

2

240 10 6 in./in.

3

200 10 6 in./in.

29 106 psi and v

0.30, determine the magnitudes of

SOLUTION 6

x

1

30 10

y

3

100 10

xy x

y

2

2

1

E ( 1 v2 0 E ( 1 v2

6

430 10

3

6

x

v y)

29 [ 30 (0.3)(100)] 1 (0.3)2

y

v x)

29 [100 (0.3)( 30)] 1 (0.3)2

2.9 103 psi P A

y

P

A

y

(2)(6)(2.9 103 )

34.8 103 lb

P

34.8 kips


PROB BLEM 7.151 (Continue ed)

G xyy

I Qˆ t xyy

V

E 2(1 v) G

xy

29 106 (2)(1.30)

111.1538 106 pssi

(11.1538)(430)

4.7962 103 psi

1 3 1 bh (2)(6)3 36 in 4 12 12 A y (2)(33)(1.5) 9 in 3 2 in.

VQˆ It It xy Qˆ

(366)(2)(4.7962 103 ) 9

Q V

38.37 103 lb Q

3 38.4 kips


PROBL LEM 7.152 T'

A single strain gage iss cemented to a solid 4-in.-diameter steell shaft at an 25 with a line parallel to the axis off the shaft. Knnowing that angle G 11.5 106 psi, deetermine the toorque T indicaated by a gagee reading of 300 100 6 in./in.

! T

2 in.

SOLUTION For torsion,

x

0,

y

1 ( E 1 ( E

x

y

x

v

y)

0

y

v

x)

0

1 2

0

xy

0

G

xy

0

2G

Draaw the Mohr’s circle for straain. R x

But

0

T

0

2G R sin 2 Tc J

c 3G sin 2

2T c3

0

2 2G

sin 2

2G x ssin 2

x

(2)3(11.5 106 )(300 10 6 ) sinn 50

113.2 103 lbb in.

T 113.2 2 kip in.


PROB BLEM 7.153 3 T'

Solve Prob. P 7.152, asssuming that thhe gage formss an angle line parrallel to the axxis of the shaftt.

!

35 with a

PROBL LEM 7.152 A single gagee is cementedd to a solid 4--in.-diameter steel shhaft at an anglee 25 with a line paralllel to the axis of the shaft. Knowinng that G 11.5 106 psi, determine thee torque T inddicated by a gage reaading of 300 10 6 in./in.

T

2 in.

SO OLUTION Forr torsion,

0 0,

x

1 ( E 1 ( E

x

y

G

xy

0

x

v

y)

0

y

v

x)

0

1 2

0

xy

0,

y

xy

0

2G

Draaw Mohr’s cirrcle for strain. R x

0

2G R sin 2

0

2 2G

sin 2

But 0

T

Tc J c 3G sin 2

2T c3 x

2G x ssin 2 (2)3(11.5 106 )(300 10 6 ) 7 sin 70

92.3 103 lb in.

T

92.3 kip in.


PROBLEM 7.15 54 A singgle strain gage forming an angle 18 with a hoorizontal planee is used to determ mine the gage pressure in thhe cylindrical steel tank shoown. The cylinndrical wall of the tank is 6 mm thick, has a 600-mm 6 insidee diameter, annd is made of a steel with E 200 2 GPa and v 0.30. Dettermine the prressure in the tank t indicatedd by a strain gage reading of 280 .

!

SOLUTION x

pr t

1

1 x, 2 1 ( x E

y

x

0.85

v

0.20 xy xy

z)

v

y

v 2

1

x

E

x

E

1 ( v E

y

0

z

x

v

y

z)

1 2

v

x

E

x

E 0

G

Draaw Mohr’s circcle for strain. ave a

R

1 ( 2 1 ( 2

x

p Data:

ave

t

x

r

x

y)

0.525

x

y)

0.325

R cos 2

x

E

(0.5225 0.325cos 2 )

x

E

tE x r (00.525 0.325ccos 2 )

r

1 d 2

t

6 10 3 m mm E

1 (6600) 2

3

p

x

E

300 mm m

0.300 m

200 109 Pa, 9

x

280 10

6

18

6

(6 10 )(200 ) 10 )(2880 10 ) 1..421 106 Pa (0.300)((0.525 0.325 cos 36 )

p 1.4 421 MPa


PRO OBLEM 7.155 Solvee Prob. 7.154, assuming thaat the gage form ms an angle planee. !

35 with a horizontal

PROBLEM 7.1544 A single straain gage formiing an angle 18 with a horizontal planee is used to deetermine the gaage pressure in i the cylindriical steel tank shown. The cylind drical wall off the tank is 6 mm thick, has a 600-mm m inside diam meter, and is madee of a steel witth E 200 GP Pa and v 0.30. Determinne the pressurre in the tank 0 indicaated by a straiin gage reading of 280 .

SO OLUTION x

y

x

y

pr t

1

1 x, 2 1 ( x E

v

1 ( v E xy x

xy

0

z

x

v

y

y

z)

v

v 2

1 1 2

z)

x

0.85

E x

v

E

x

E

0.20

x

E

0

G

Draaw Mohr’s cirrcle for strain. ave

R

1 ( 2 1 ( 2

x

ave a

x

y)

0.525

x

y)

0.325

x

E x

E

R cos 2

0.525 0.325 cos c 2 ) (0 p Data:

t

x

r

tE x r (0.525 0.325 cos 2 )

r

1 d 2

t

6 10 3 m E

1 (600) 2

3

p

x

E

300 mm

0.300 m

200 109 Pa,

x

280 10 1

6

35

6

9

(6 6 10 )(200 10 )(280 100 ) 1.761 106 Pa (0.300)(0.525 ( 0.325 cos 70 )

p 1.761 MPa


PROBL LEM 7.156

150 MPa

The given state of planne stress is knoown to exist on the surface of o a machine component. c G , determ mine the direection and Knowingg that E 200 GPa andd G 77.2 GPa magnitudde of the threee principal strrains (a) by determining d thhe correspondiing state of strain [usse Eq. (2.43) and Eq. (2.38)] and then using Mohr’ss circle for strrain, (b) by using Moohr’s circle foor stress to deetermine the principal p plannes and princippal stresses and then determining the t correspondding strains.

75 MP Pa

SOLUTION (a)

x

E G x

y

xy xy

2 ave x

y

taan 2

0,

4877.0 1 ( x 2 974

y)

2633 974 974

xyy y

1.000

45.0 x

R

y

a

ave

R

b

ave

R x

2 xy

2

v ( E

22.5

a

2

c

75 106 Pa

E E v 1 0.2987 2(1 v) 2G 1 1 ( x v y) [0 (00.2987)(150 1106 )] E 2000 109 224 1 1 ( y v x) [( 1500 106 ) 0] E 2000 109 7500 75 106 xy 974 G 77 109

a

a

xy

77 109 Pa

200 109 Pa G

x

2

150 1006 Pa,

y

2

y)

689

(0.2987)(0 150 1 106 ) 200 10 1 9

b

67.5

a

426

b

952

c

224


PROB BLEM 7.156 6 (Continued d)

(b)

ave

1 ( 2

x

y)

x

y

75 MPa 2

R

0 1150 2

2 xy

2

2

752

1006.07 MPa a

ave

R

b

ave

R

a

1 ( E

31.07 MPa 181 1.07 MPa v

a

1 200 109 4226 10

tan 2

2

xy

[31.07 106

(0.29987)( 181.07 106 )]

6

000 1.0

a x

b)

2

a

426

a

45

y a

22.5


PR ROBLEM 7.157 Thee following staate of strain haas been determ mined on the suurface of a casst-iron machinne part: 720

x

y

400

xy

6 660

Knoowing that E 69 GPa annd G 28 GP Pa, determinee the principaal planes and principal streesses (a) by deteermining the corresponding c g state of planee stress [use Eq. E (2.36), Eqq. (2.43), and the t first two equations e of Probb. 2.73] and then t using Mohr’s M circle for fo stress, (b) by using Moohr’s circle forr strain to dettermine the orieentation and magnitude m of thhe principal strrains and thenn determining the corresponding stresses.

SOLUTION Thee 3rd principall stress is

z

0.

E 69 E 1 1 0.2321 v 2(1 v) 2G 56 6 69 72.933 GPa 1 (0..232) 2

G E 1 v2

(a)

x

y

E ( x v y) 1 v2 (72..93 109 )[ 720 10 59 9.28 MPa E ( y v x) 1 v2 (72..93 109 )[ 4000 10

6

(0.2232)( 400 100 6 )]

6

(0.22321)( 720 10 1 6 )]

41.36 MPa xy

G

(28 109 )(660 10 6 )

xy x

18.4 48 MPa ave

tan 2

1 ( 2 2

xy

2.06225

b y

x

2

b

500.32 MPa

y)

x

64 4.1 ,

b

32.1 ,

a

57.9

2 x

R

y

2

2 xy

20.54 MP Pa

a

avee

R

a

29.8 MPa

b

avee

R

b

700.9 MPa


PROB BLEM 7.157 (Continue ed)

(b)

ave

tan 2

1 ( 2

xy

b

560 5 2.0625

b x

2

y)

x

y

6 64.1 ,

32.1 ,

b

2 x

R

y

a

5 57.9

2 xy

2

2

a

avve

R

193..26

b

avve

R

926..74

366..74

a

E ( 1 v2

a

v b)

a

2 29.8 MPa

b

E ( 1 v2

b

v a)

b

7 70.9 MPa


P

PROBLEM M 7.158

T

1 4

A steel pipe of 12-in. outter diameter iss fabricated frrom 14 -in. -thiick plate by w a plane peerpendicular welding alonng a helix thatt forms an anggle of 22.5 with to the axis off the pipe. Knoowing that a 40-kip 4 axial foorce P and an 80-kip in. torque T, eaach directed as shown, arre applied to the pipe, dettermine the normal and in-plane i shearring stresses in i directions, respectively, normal and tangential to the weld.

in.

Weld 22.5

SOLUTION 1 d2 2 5.75 in.

d2

12 in., c2

c1

c2

t

A

c22

c12

J

c24

c14

2

(62 2

6 in., t

0.25 in.

5.752 )

9 9.2284 in 2

(664

5.754 ) 318.67 3 in 4

Streesses: P A

40 4.33444 ksi 9.22284 Tc2 J (80))(6) 1.5063 ksi k 318.67 0, 4.33444 ksi, y

x

xy

1..5063 ksi

Chooose the x an nd y axes, resspectively, tanngential and noormal to the weld. w Theen

w

y

y

and

w

xy

x

y

x

y

2 ( 4.3344) 2 4 4.76 ksi x

xy

22.5

cos 2 x sin 2 xy 2 [ ( 4.3344)] cos 45 1.5063 sin s 45° 2 w

4.76 ksi

y

sin 2 xy cos 2 2 [ ( 4.3344)] s 45 1.50663 cos 45 sin 2 0 0.467 ksi

w

0.467 0 ksi


100 kN

!

80 mm

100 kN

PROBLEM 7.159 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 25 , determine (a) the in-plane shearing stress parallel to the weld, (b) the normal stress perpendicular to the weld.

SOLUTION Area of weld: Aw

(10 10 3 )(80 10 3 ) cos 25 882.7 10 6 m 2

(a)

Fs w

(b)

Fn w

0: Fs

Fs Aw

100sin 25

42.26 103 882.7 10 6

0: Fn

Fn Aw

100 cos 25

90.63 103 882.7 10 6

0

Fs

42.26 kN

47.9 106 Pa 0

Fn

w

47.9 MPa

90.63 kN

102.7 106 Pa

w

102.7 MPa


100 kN

PROBLEM 7.160

!

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 , (b) the corresponding normal stress perpendicular to the weld.

80 mm

100 kN

SOLUTION Area of weld:

Aw

(10 10 3 )(80 10 3 ) cos 800 10 cos

(a)

Fs w

0: Fs Fs Aw

sin cos (b)

100sin

Aw

800 10 6 cos14.34 Fn Aw

100sin

100 10 sin 800 10 6 / cos 30 106 125 106

1 sin 2 2

0: Fn

Fs 3

30 106

Fn

0

100 cos

0

kN

100 103 sin

m2

N

125 106 sin cos

0.240

Fn

6

14.34

100cos14.34

96.88 kN

825.74 10 6 m 2

96.88 103 825.74 10 6

117.3 106 Pa

117.3 MPa


PROBLEM 7.161

'0 &

+

'0


SOLUTION Mohr’s circle for 2nd stress state: x

y

xy

1 2 1 2 1 2

1 2 1 2

0

0

0

cos 2

0

cos 2

sin 2

0

Resultant stresses: x

y

0

xy

0

1 2 1 2

1 ( 2 2

ave

tan 2

1 2

0

1 2

0

1 2

0

y)

x xy

1 2

0 y

1 2 0

1 2

0

1 2

0

0

0

cos 2

cos 2

sin 2

0

p 0

sin 2 1 cos 2

3 2

cos 2

cos 2

0

sin 2

0

x

0

sin 2 0 cos 2

tan

p

2 x

R

y

2 1 2

0

2 xy

1 2

1 2 cos 2 + cos 2 2

0

1 2

2 0

sin 2 2

cos 2 2 2

0

1 2

2

(

0 sin 2

1 cos 2

1 2

p

0

) 2

|cos |

a

ave

R

a

0

0

cos

b

ave

R

b

0

0

cos


y

PROBLEM 7.162 2 ksi

For the state of stress shown, determine the maximum shearing stress when (a) z 4 ksi, (b) z 4 ksi, (c) z 0. 6 ksi

σz

7 ksi

z x

SOLUTION 7 ksi,

x

ave

1 ( 2

x

y)

x

y

y

2 ksi,

xy

6 ksi

4.5 ksi 2

R

2.52

(a)

(b)

( 6) 2

6.5 ksi

a

ave

R 11 ksi

b

ave

R

z

4 ksi,

max

11 ksi,

z

max

2 ksi

a

a

11 ksi,

min

11 ksi,

11 ksi,

min

2 ksi

b

2 ksi,

min

11 ksi, 0,

11 ksi,

a

4 ksi,

z

max

(c)

2 xy

2

4 ksi, b

2 ksi,

1 ( 2

max

min )

max

max

6.50 ksi

2 ksi

b

max

1 ( 2

max

min )

max

7.50 ksi

1 ( 2

max

min )

max

6.50 ksi

2 ksi max


y

PROBLEM 7.163 40 MPa

For the state of stress shown, determine the value of

xy

for which the

maximum shearing stress is (a) 60 MPa, (b) 78 MPa.

τ xy 100 MPa z x

SOLUTION x

100 MPa, 1 ( 2

ave

(a)

y)

x

40 MPa,

0

z

70 MPa

60 MPa.

max

If

y

z

is

min ,

then

max max max

2

min

max .

0 (2)(60) 120 MPa R

ave

R

max

b

max

ave

2R

120 70 50 MPa 20 MPa > 0 2

x

R xy

(b)

2 xy

2 502

302

2 xy

50 MPa

302

xy

40.0 MPa

xy

72.0 MPa

78 MPa.

max

If

y

z

is

min ,

then

max

min

2

max

ave

R

R

Set

max

0 (2)(78) 156 MPa.

max

156 70 86 MPa >

ave

R

max

78 MPa.

min

ave

R

max

78 MPa

8 MPa < 0 2

R xy

x

y

2 782

2 xy

302

302

2 xy


PROBLEM 7.164

14 ksi

(xy

24 ksi

The state of plane stress shown occurs in a machine component made of a steel with Y 30 ksi. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) xy 6 ksi, (b) xy 12 ksi, (c) xy 14 ksi. If yield does not occur, determine the corresponding factor of safety.

SOLUTION 24 ksi

x

For stresses in xy-plane, (a)

xy

ave

1 ( 2

x

y

14 ksi

y y)

x

0

z x

19 ksi

y

2

5 ksi

6 ksi 2

R a 2 a

2 b

a

b

F .S .

(b)

xy

2 xy

2 R

ave

(5)2

26.810 ksi,

(6)2 ave

b

7.810 ksi R 11.190 ksi

23.324 ksi < 30 ksi

(No yielding)

30 23.324

F .S . 1.286

12 ksi 2 x

R a 2 a

2 b

a

b

F .S .

(c)

xy

y

2 xy

2 R

ave

32 ksi,

(5)2 b

ave

(12)2 R

13 ksi 6 ksi

29.462 ksi < 30 ksi

(No yielding)

30 29.462

F .S . 1.018

14 ksi 2 x

R a 2 a

2 b

a

b

y

2 xy

2 ave

R

33.866,

32.00 ksi > 30 ksi

(5)2 b

ave

(14)2 R

14.866 ksi 4.134 ksi (Yielding occurs)


PROBLEM 7.165

750 mm 750 mm

The compressed-air tank AB has an inner diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing that the gage pressure inside the tank is 1.2 MPa, determine the maximum normal stress and the maximum in-plane shearing stress at point a on the top of the tank.

b a

B

D A 5 kN 500 mm

SOLUTION Internal pressure:

r 1

2

Torsion:

c1

J T

1 d 2 pr t pr 2t

225 mm t

6 mm

(1.2)(225) 6

45 MPa

22.5 MPa

225 mm, c2

2

c24

c14

225

6

446.9 106 mm 4

(5 103 )(500 10 3 ) Tc J

231 mm

2500 N m 3

(2500)(231 10 ) 446.9 10 6 1.29224 106 Pa

At point a,

1.29224 MPa

0 at point a.

Transverse shear: Bending:

446.9 10 6 m 4

I M

1 J 2

223.45 10 6 m 4 , c

(5 103 )(750 10 3 )

231 10 3 m

3750 N m

Mc I

(3750)(231 10 3 ) 223.45 10 6

3.8767

3.8767 MPa

Total stresses (MPa). Longitudinal:

x

22.5

Circumferential:

y

45 MPa

Shear:

xy

26.377 MPa

1.29224 MPa



ave

1 ( 2

x

y)

x

y

35.688 MPa 2

R max

max(in-plane)

2 xy

2 ave

R

R

45.1 MPa

9.40 MPa

9.4007 MPa max

45.1 MPa

max (in-plane)

9.40 MPa


PROBLEM 7.166

750 mm 750 mm

b a

D A 5 kN 500 mm

B

For the compressed-air tank and loading of Prob. 7.165, determine the maximum normal stress and the maximum in-plane shearing stress at point b on the top of the tank. PROBLEM 7.165 The compressed-air tank AB has an inner diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing that the gage pressure inside the tank is 1.2 MPa, determine the maximum normal stress and the maximum in-plane shearing stress at point a on the top of the tank.

SOLUTION Internal pressure:

r 1

2

Torsion:

c1

J T

1 d 2 pr t pr 2t

Bending: At point b,

(1.2)(225) 6

2

6 mm 45 MPa

22.5 MPa

225 mm, c2

c24

225

c14

6

231 mm

446.9 106 mm 4

(5 103 )(500 10 3 ) Tc J

Transverse shear:

225 mm t

446.9 10 6 m 4

2500 N m 3

(2500)(231 10 ) 446.9 10 6

1.29224 106 Pa

1.29224 MPa

0 at point b.

I

M

1 J 2

223.45 10

(5 103 )(2

6

m4 , c

750 10 3 )

Mc I

(7500)(231 10 3 ) 223.45 10 6 7.7534

231 10 3 m

7500 N m 7.7534 MPa

Total stresses (MPa). Longitudinal:

x

22.5

Circumferential:

y

45 MPa

Shear:

xy

30.253 MPa

1.29224 MPa



ave

1 ( x 2

y)

37.626 MPa 2

x

R max

max (in-plane)

y

2 xy

2 ave

R

R

45.1 MPa

7.49 MPa

7.4859 MPa max

45.1 MPa

max (in-plane)

7.49 MPa


0.12 in.

A

PROBLEM 7.167 The brass pipe AD is fitted with a jacket used to apply a hydrostatic pressure of 500 psi to portion BC of the pipe. Knowing that the pressure inside the pipe is 100 psi, determine the maximum normal stress in the pipe.

B

0.15 in.

C D 2 in. 4 in.

SOLUTION The only stress to be considered is the hoop stress. This stress can be obtained by applying 1

pr t

Using successively the inside and outside pressures (the latter of which causes a compressive stress), pi

100 psi, ri

( po

max )i

1

pi ri t

500 psi, ro

(

max )o max

0.12

0.88 in., t

(100)(0.88) 0.12 1 in.,

po ro t 733.33

t

0.12 in.

733.33 psi

0.12 in.

(500)(1) 0.12 4166.7

4166.7 psi 3433.4 psi max

3.43 ksi (compression)


0.12 in.

A

PROB BLEM 7.168 8 For the assembly of Prob. 7.167, determine the normal streess in the jackket (a) in a o the jacket, (b) in a directtion parallel directionn perpendicular to the longitudinal axis of to that axis. a

B

PROBL LEM 7.167 Thhe brass pipe AD A is fitted with w a jacket ussed to apply a hydrostatic pressuree of 500 psi too portion BC of the pipe. Knnowing that thee pressure inside the pipe is 100 psi, determine the t maximum m normal stresss in the pipe.

0.15 in.

C D 2 in. 4 in.

SOLUTION (a)

Hoop stress.

p ( 1)

500 psii, t pr t

0.15 inn., r

((500)(1.85) 0.15

2

0 0.15 1.85 in.

6166.7 psi 1

(b)

6.17 ksi

Longitudin nal stress. Free body of portion of jacket j above a horizontal seection, consideering vertical forces f only:

Fy Af

0: Af p dA

Aj

pA f

Areas :

Af

r22

r12

[(1.85) 2

Aj

r32

r22

[(2) 2

(11)2 ] (1.855)2 ]

2 dA j

0

2 Aj

0

2

p

Af

(1)

Aj

7.6105 in i 2 1.814277 in 2

Recalling Eq. E (1), 2

p

Af Aj

(500)

7.6105 1 1.81427

20097.4 psi 2

2 2.10 ksi


PROBLEM P 7.169 1

2

Determine D the largest in-planne normal straain, knowing that the follow wing strains haave been obtaiined by the usse of the rosettte shown:

3

1

50 100 6 in./in.

2

360 10 6 in./in.

3

1 6 in./in. 315 10

45! x

45!

SO OLUTION 455 ,

1

x

cos 2

1

sin 2

y

1

0.55 cos 2

x

y

2

sin 2

x

cos 2

3

y

sin 2

sin

xy

0.5

x

2

0.55

45 ,

2

xy

y

sin

0.5

x

3

xy

2 y

sin

cos

1

3

Eq. (1)

Eq. (2):

Eq. (1)

Eq. (2):

x

x

315 3 10

6

xy

50 10

6

y

1

2

y

1

2

ave

1 ( 2

xy

cos

2

0.5

xy

cos

3

360 10

y)

50 10

6

(1)

6

(2)

6

(3)

2

360 10 3

0 0

315 10

4 410 10 6 in./iin.

6

3 360 10

6

3115 10

6

5 10 6 in./in.

1 155 10 6 in./inn. 2

y

xy

2 315 10

6

50 10

2 x

R

1

in.//in.

x

x

1

0 0.5

x

Froom (3),

0

3

2 6

5 10

6

2

410 10 2

2

6

2

260 10 6 in../in. max

ave

R 1555 10

6

260 10

6

max

415 110 6 in./in.


y'

y

"y

PROBLEM 7.C1 1

y

#

"y'

!x'y'

!xy x Q

z

#

"x'

Q

x

"x

x' x

z (a)

( (b)

A statee of plane streess is defined by b the stress compoonents x , y , and xy assoociated with the eleement shown inn Fig. P7.C1a. (a) Write a compuuter program that can be b used to calculaate the stress components x y , and ssociated withh the element after it has x y as rotatedd through an angle a abouut the z axis (Fig. P.7C1 P b). (b) Use U this prograam to solve Probs. 7.13 through 7.16. 7

SOLUTION Proggram followin ng equations: x

Equuation (7.5), Paage 427:

x

Equuation (7.7), Paage 427:

y

x,

y,

xyy

x

2 y

x

2

y

2

x xy

y

2

x

Equuation (7.6), Paage. 427: Enteer

y

y

2

sin 2

c 2 cos

xy

sinn 2

c 2 cos

xy

sinn 2

x xy

cos 2

and

Prinnt values obtaiined for

x

,

y

and

xy

Prooblem Outputts Probblem 7.13 x x xy

0 ksi 8 ksi 5 ksi

Rootation of elem ment (+ counterclockw c wise) 25

Rotation off element (+ countercllockwise) 10

x

2.40 ksi

x

1.995 ksi

y

10.40 ksi

y

6.05 ksi

xy

6..07 ksi

xy

0.15 ksi


PROB BLEM 7.C1 (Continued d) Prooblem 7.14 y

60 MPa M 90 MPa

xy

Pa 30 MP

x

Ro otation of Elem ment ( counterclockw wise) 25

Rotation off Element ( countercclockwise) 1 10

x

56.19 MP Pa

x

455.22 MPa

y

86.19 MP Pa

y

755.22 MPa

xy

38.17 MP Pa

xy

53.84 MPa

Prooblem 7.15 x

8 ksii

y

12 ksi

xy

6 ksi Rotation off Element ( countercclockwise) 1 10

Ro otation of Elem ment ( counterclockw wise) 25 x

9.02 kssi

x

5.344 ksi MPa

y

13.02 kssi

y

9.344 ksi MPa

xy

9.066 ksi MPa

xy

3.80 kssi

Prooblem 7.16 x

0 MPa M

y

80 MPa M

xy

50 MPa M

Ro otation of Elem ment ( counterclockw wise) 25

Rotation off Element ( countercllockwise) 10

x

24.01 MPa M

x

19.51 MPa

y

104.01 MPa M

y

60..49 MPa

xy

60.67 MPa

xy

1.50 MPa M


PROBLE EM 7.C2

y

"y

A state of plane p stress iss defined by thhe stress compponents x , y , and xy associated with the elem ment shown inn Fig. P7.C1aa. (a) Write a computer program thhat can be used to calcullate the princcipal axes, thhe principal stresses, thee maximum inn-plane sheariing stress, andd the maximuum shearing stress. (b) Use U this prograam to solve Prrobs. 7.5, 7.9, 7.68, and 7.699.

!xy x Q

x

"x z

SOLUTION Proggram followin ng equations: 2 x

y

Equuation (7.10)

ave

Equuation (7.14)

max

ave

R

J min

ave

R

Equuation (7.12)

p

2

taan

: R

2

1

s

Sheearing stress: Theen

If

m max

2 xy

y x

2

y xy

0 and

m min

0 and

m min

max(out-of-planne)

m max

R;

max(in-plane) m

1

2

max(out-of-planne)

R;

max(in-plane) m

If

Theen

m max

R;

maax(in-plane)

If

Theen

taan

y

xy

x

Equuation (7.15)

x

0 and

m min

max(out-of-plaane)

0: R

0:

1 2 0: 1 | 2

max

min |

Proogram Outputts Probblems 7.5 and d 7.9 x

60.00 MPa

y

40.00 MPa

xy

35.00 MPa


PROB BLEM 7.C2 (Continued d) Proogram Outpu uts (Continued d) Anngle between xy x axes and priincipal axes (+ + counterclockkwise): p

37.003

and 522.97°

max

13.660 MPa

min

86.440 MPa

x axis and plaanes of maxim mum in-plane shearing s stresss (+ countercloockwise): Anngle between xy s

7.97

and 97.977°

max (in-plane)

36.400 MPa

max

43.200 MPa

Prooblem 7.68 x

140.000 MPa

y

40.000 MPa

xy

80.000 MPa

Anngle between xy x axes and priincipal axes ( counterclockkwise): p

29.000

and 1199°

max

1844.34 MPa

min

4.344

MPa

Anngle between xy x axis and plaanes of maxim mum in-plane in-plane sheariing stress ( counterclockw c wise): s

74.000

and 1644.00°

max (in-plane) (

94.334 MPa

max (out--of-plane)

94.334 MPa

x

140.000 MPa

y

120.000 MPa

xy

80.000 MPa

x axes and priincipal axes (+ + counterclockkwise): Anngle between xy p

41.444

and 1311.44°

max

210.62 MPa

min

49.338 MPa

x axis and plaanes of maxim mum in-plane in-plane sheariing stress (+ counterclockw c wise): Anngle between xy s

86.444

and 1766.44°

max (in-plane) (

80.662 MPa

max (out--of-plane)

105.331 MPa


PROBLEM 7.C2 (Continued) Program Outputs (Continued)

Problem 7.69 x

140.00 MPa

y

20.00 MPa

xy

80.00 MPa

Angle between xy axes and principal axes (+ counterclockwise): p

26.57

and 116.57

max

180.00 MPa

min

20.00 MPa

Angle between xy axis and planes of maximum in-plane in-plane shearing stress (+ counterclockwise): s

71.57

and 161.57

max (in-plane)

100.00 MPa

max (out-of-plane)

100.00 MPa

x

140.00 MPa

y

140.00 MPa

xy

80.00 MPa

Angle between xy axes and principal axes (+ counterclockwise): p

45

and 135.00

max

220.00 MPa

min

60.00 MPa

Angle between xy axis and planes of maximum in-plane in-plane shearing stress ( counterclockwise): s

90.00

and 180.00°

max (in-plane)

80.00 MPa

max (out-of-plane)

110.00 MPa


PR ROBLEM 7.C3 7 (a) Write a com mputer program m that, for a given g state off plane stress and a given yield y strength of a ductile mine whether the t material will w yield. The program shouuld use both thhe maximum maaterial, can be used to determ sheearing-strengthh criterion and d the maximum m-distortion-ennergy criterionn. It should allso print the values v of the prinncipal stressess and, if the material m does noot yield, calculate the factor of safety. (b) Use this proggram to solve Proobs. 7.81, 7.82, and 7.164.

SO OLUTION Principal stressess. 2 x avve

Maaximum-shearring-stress criterion. If

a

and

b

havve same sign,

y

2

a

ave

R

b

ave

R

y

1 2

y

maax

1 2

a

If

max

y,

yieelding occurs.

If

max

y,

no yielding occu urs, and factor of safety

x

; R

y

2

2 xy

y m max

Maaximum-distorrtion-energy criterion. 2 a

Compute radicall

a

b

2 b

r If radical

y,

urs. yielding occu

If radical r

y,

no yielding occcurs, and facttor of safety

y

Radical

Proogram Outpu uts Prooblems 7.81a and a 7.82a

Yield strengthh

325 MPa

x

200.00 MPa M

y

200.00 MPa M

xy

100.00 MPa M

maxx

100.00 MPa M

minn

300.00 MPa M


PROB BLEM 7.C3 (Continued d) Proogram Outputts (Continuedd) Usinng the maximu um-shearing-sstress criterionn, matterial will not yield. y F .S . 1.0083

Usinng the maximu um-distortion-energy criteriion, matterial will not yield. y F .S . 1.2228

a 7.82b Probblems 7.81b and

Yield strenngth

325 MP Pa

y

240.000 MPa 240.000 MPa

xy

100.00 MPa

x

max min

140.000 MPa 340.000 MPa

um-shearing-sstress criterionn, Usinng the maximu matterial will yield d. Usinng the maximu um-distortion-energy criteriion, matterial will not yield. y F .S . 1.098

Probblems 7.81c and 7.82c

Yield strenngth

325 MP Pa

y

280.000 MPa 280.000 MPa

xy

100.000 MPa

x

max min

180.000 MPa 380.000 MPa

um-shearing-sstress criterionn, Usinng the maximu matterial will yield d. Usinng the maximu um-distortion-energy criteriion, matterial will yield d. Probblem 7.164a

Yield strenngth

30 ksi

y

24.00 ksi k 14.00 ksi k

xy

6.00 ksi k

x

max min

26.81 ksi k 11.19 ksi k


PROB BLEM 7.C3 (Continued d) Proogram Outpu uts (Continued d) (a)

Using the maximum-sheearing-stress criterion, c material will w not yield. F .S . 1.119

(b)

Using the maximum-disstortion-energgy criterion, material will w not yield. F .S . 1.286

Prooblem 7.164b

(a)

Yield strenngth

30 ksi

x

24.00 ksi k

y

14.00 kssi

xy

12.00 ksi k

max

32.00 ksi k

min

6.00 kssi

Using the maximum-sheearing-stress criterion, c material will w yield.

(b)

Using the maximum-disstortion-energyy criterion, material will w not yield. F .S . 1.018

Prooblem 7.164c

(a)

Yield strength

30 ksi

x

24.00 kssi

y

14.00 kssi

xy

14.00 kssi

max

33.87 kssi

min

4.13 kssi

Using the maximum-sheearing-stress criterion, c material will w yield.

(b)

Using the maximum-disstortion-energyy criterion, material will w yield.


PR ROBLEM 7.C4 (a) Write W a computer program based on Mohhr’s fracture criterion c for brrittle materialss that, for a givven state of planne stress and given g values of o the ultimatee strength of thhe material inn tension and compression, c c be used can to determine d wheether rupture will w occur. Thhe program shhould also print the values of the princippal stresses. (b) Use U this progrram to solve Probs. P 7.89 andd 7.90 and to check c the answ wers to Probs. 7.93 and 7.944.

SOLUTION Prinncipal stresses. 2 x avee

y

2

a

ave

R

b

ave

R

R

x

y

2

2 xy

c Mohhr’s fracture criterion. If

If

and

b

a

UT

and

a

UT

or

a

a

0 and

hav ve same sign, and a UC ,

b

UC ,

b b

no faailure;

failuree.

0:

Connsider fourth quadrant q of Figgure 7.47. For no rupture to occur, point ( If

b

a,

b)

mustt lie within Moohr’s envelope (Figure 7.477).

Criterion n,

thenn rupture occu urs. If

b

Criterion n,

thenn no rupture occcurs.


PROB BLEM 7.C4 (Continued d)

Proogram Outpu uts Prooblem 7.89

10.00 MPa

x

100.00 MPa

y

60 MPa

xy

Ulttimate strengthh in tension

80 MPaa

Ulttimate strengthh in compressiion

200 MP Pa

max

a

336.39 MPa

min

b

1226.39 MPa

Rupture will occcur Prooblem 7.90

32.000 MPa

x y xy

0.00 MPa M 75.00 MPa M


80 MP P


200 M MP

max

a

60.69 MPa

min

b

92.69 MPa

Rupture will not occur. To check answerrs to the follow wing problemss, we check foor rupture usinng given answeers and an adjacent value. Annswer: Ruppture occurs for f

0

3.67 ksi. k

Prooblem 7.93

x

8.00 ksii

y

0.00 ksii

xy

3.67 ksii


10 ksi


25 ksi

max

a

9.443 ksi

min

b

1.443 ksi


PROB BLEM 7.C4 (Continued d) Proogram Outputts (Continuedd) Ruppture will not occur. o x

8.00 ksi

y

0.00 ksi

xy

3.68 ksi

Ultiimate strength h in tension

10 ksi

Ultiimate strength h in compressioon

25 ksi

m max

a

9.444 ksi

m min

b

1.444 ksi

Ruppture will occu ur. Ansswer: Ruppture occurs fo or

0

49.1 MPa. M

Probblem 7.94

x y xy

80.00 MPa Pa 0.00 MP 49.10 MPa M


75 MPaa


150 MP Pa

max

a

23.33 MPa

min

b

1103.33 MPa

Ruppture will not occur. o x y xy

80.00 MPa Pa 0.00 MP 49.20 MPa M


75 MPaa


150 MP Pa

max

a

min

b

233.41 MPa 1 103.41 MPa

Ruppture will occu ur.


PROBLEM 7.C5 y

A state of plane strain is defined by the strain components x , y , and xy associated with the x and y axes. (a) Write a computer program that can be used to calculate the strain components x , y , and x y associated with the frame of reference x y obtained by rotating the x and y axes through an angle . (b) Use this program to solve Probs 7.129 and 7.131.

y' x'

&

x

SOLUTION Program following equations: x

Equation (7.44):

x

Equation (7.45):

y

Equation (7.46):

xy

Enter

x,

y,

y

2 y

x

2 (

y

2

x

xy ,

x

y

2

cos 2

1 2

xy

sin 2

sin 2

1 2

xy

cos 2

y )sin 2

x

xy

cos 2

and .

Print values obtained for

x

,

y

, and

xy

.

Program Outputs

Problem 7.129

x

240 micro meters

y

160 micro meters

xy

150 micro radians

Rotation of element, in degrees (+ counterclockwise):

Problem 7.131

x

60 115.05 micro meters

y

284.95 micro meters

xy

5.72 micro radians x

0 micro meters

y

320 micro meters

xy

100 micro radians

Rotation of element, in degrees (+ counterclockwise): x

30 36.70 micro meters

y

283.30 micro meters

xy

227.13 micro radians


PROBLEM 7.C6 A state of strain is defined by the strain components x , y , and xy associated with the x and y axes. (a) Write a computer program that can be used to determine the orientation and magnitude of the principal strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this program to solve Probs 7.136 through 7.139.

SOLUTION Program following equations: 2 x

y

Equation (7.50):

ave

Equation (7.51):

max

ave

Equation (7.52):

p

tan

R

2 R

min

y

2 ave

2 xy

2 R

xy

1

x

Shearing strains:

x

y

Maximum in-plane shearing strain 2R

max (in-plane)

Calculate out-of-plane shearing strain and check whether it is the maximum shearing strain.

Let

a

max

b

min

v

Calculate

c

1 v

If

a

b

c,

out-of-plane

a

c

If

a

c

b,

out-of-plane

a

b

If

c

a

c

b

b,

out-of-plane

(

a

b)

2R

Program Printout

Problem 7.136

x

260 micro meters

y

60 micro meters

xy

480 micro radians 0.333


PROBLEM 7.C6 (Continued) Program Printout (Continued)

Angle between xy axes and principal axes ( p a b c

33.69 100.00 micro meters 420.00 micro meters 159.98 micro meters

max (in-plane)

520.00 microradians

max

579.98 microradians

Problem 7.137

counterclockwise):

x

600 micrometers

y

400 micrometers

xy

350 microradians 0.333

Angle between xy axes and principal axes (+ = counterclockwise): p

30.13

a

298.44 micrometers

b

701.56 micrometers

c

500.00 micrometers

max(in-plane)

403.11 microradians

max

Problem 7.138

x y xy

1201.56 microradians 160 micrometers 480 micrometers 600.00 microradians 0.333

Angle between xy axes and principal axes ( p a b c

counterclockwise):

21.58 278.63 micrometers 598.63 micrometers 159.98 micrometers

max(in-plane)

877.27 microradians

max

877.27 microradians

Problem 7.139

x

30 micrometers

y

570 micrometers

xy



PROBLEM 7.C6 (Continued) Angle between xy axes and principal axes ( p a

counterclockwise):

26.57 750.00 micrometers

b

150.00 micrometers

c

300.00 micrometers

max(in-plane) max

900.00 microradians 1050.00 microradians


PROBLEM 7.C7 A state of plane strain is defined by the strain components x , y , and xy measured at a point. (a) Write a computer program that can be used to determine the orientation and magnitude of the principal strains, the maximum in-plane shearing strain, and the magnitude of the shearing strain. (b) Use this program to solve Probs 7.140 through 7.143.

SOLUTION Program following equations: 2 x

y

Equation (7.50)

ave

Equation (7.51)

max

ave

Equation (7.52)

p

tan

2 R

y

2

min

ave

2 xy

2 R

xy

1 x

Shearing strains:

x

R

y

Maximum in-plane shearing strain 2R

xy (in-plane)

Calculate out-of-plane-shearing strain and check whether it is the maximum shearing strain. Let

a b c

If

a

max min

(Plain strain)

0

b

c,

out-of-plane

a

c

out-of-plane

a

b

out-of-plane

c

b

If

a

c

b,

If

c

a

b,

2R

Program Printout

Problem 7.140

x

60 micrometers

y

240 micrometers

xy


Angle between xy axes and principal axes (+ = counterclockwise): 7.76 and 82.24 p a

243.41 micrometers

b

56.59 micrometers

max(in-plane)

0.00 micrometers 186.82 microradians

max

243.41 microradians

c


PROBLEM 7.C7 (Continued) Program Printout (Continued) Problem 7.141

y

400 micrometers 200 micrometers

xy

375 microradians

x

0.000 Angle between xy axes and principal axes (

counterclockwise):

30.96 and 59.04

p

max(in-plane)

512.50 micrometers 87.50 micrometers 0.00 micrometers 425.00 microradians

max

512.50 microradians

a b c

Problem 7.142

x

300 micrometers

y

60 micrometers 100 microradians

xy

0.000


11.31 and

78.69

a

310.00 micrometers

b

50.00 micrometers

max(in-plane)


max

310.00 microradians

c

Problem 7.143

x

180 micrometers

y

260 micrometers

xy



37.87 and

52.13

a

57.50 micrometers

b

382.50 micrometers

c

0.00 micrometers

max(in-plane)

325.00 microradians

max

382.50 microradians


PROBLEM 7.C8 A rosette consisting of three gages forming, respectively, angles of 1 , 2 , and 3 with the x axis is attached to the free surface of a machine component made of a material with a given Poisson’s ratio v. (a) Write a computer program that, for given readings 1 , 2 , and 3 of the gages, can be used to calculate the strain components associated with the x and y axes and to determine the orientation and magnitude of the three principal strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this program to solve Probs 7.144, 7.145, 7.146, and 7.169.

SOLUTION For n 1 to 3, enter Enter: NU

n

and

n.

V

Solve Equation (7.60) for

x,

y,

and

xy

using method of determinates or any other method. 2

x

Enter

y

ave

2

a

max

ave

R

b

max

avg

R

V c

p

Shearing strains:

x

; R

1 V 1 tan 2

(

y

2 xy

2

b)

a xy

1 x

y

Maximum in-plane shearing strain max (in plane)

2R

Calculate out-of-plane shearing strain, and check whether it is the maximum shearing strain. If

c

b,

out-of-plane

a

c

If

c

a,

out-of-plane

c

b

Otherwise,

out-of-plane

2R

Problem Outputs

Problem 7.144 Gage

Theta Degrees

Epsilon Micro Meters

1

–15

480

2

30

–120

3

75

80


PROBLEM 7.C8 (Continued) Program Outputs (Continued) x

253.21 micrometers

y


xy

727.21 micrometers

a

167.21 micrometers

b

894.43 microradians

max (in-plane)

Problem 7.145

Gage

Theta Degrees

Epsilon Micro Meters

1 2 3

30 –30 90

600 450 –75

725.00 micrometers

x y

75.000 micrometers

xy

173.205 microradians

a

734.268 micrometers

b

84.268 micrometers

max (in-plane)

818.535 microradians

Problem 7.146 Observe that Gage 3 is orientated along the y axis. Therefore, enter

4

and

4

as

3

and

3,

the value of Gage

y

that is obtained is also the expected reading of Gage 3.

Theta Degrees

Epsilon in./in.

1

0

420

2

45

–45

4

135

165

x y xy a b max (in-plane)

420.00 in./in. 300.00 in./in. 210.00 microradians 435.00 in./in. 315.00 in./in. 750.00 microradians


PROBLEM 7.C8 (Continued) Program Outputs (Continued)

Problem 7.169 Gage

Theta Degrees

Epsilon in./in.

1 2 3

45 –45 0

–50 360 315

x

315.000 in./in.

y

5.000 in./in.

xy a b max (in-plane)

410.000 microradians 415.048 in./in. 105.048 in./in. 520.096 microradians


345021434-Mechanics-of-Materials-7th-Edition-Beer-Johnson-Chapter-7.pdf

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