PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.1

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

SOLUTION

Stresses Areas Forces

0: 15 sin 30 cos 30 15 cos 30 sin 30 + 10 cos 30 cos 30 0!= " ° ° " ° ° ° =F A A A A

230 sin 30 cos 30 10 cos 30! = ° ° " °

5.49 ksi! = W

0: 15 sin 30 sin 30 15 cos 30 cos 30 10 cos 30 sin 30 0#$ = + ° ° " ° ° " ° ° =F A A A A

2 215(cos 30 sin 30 ) + 10 cos 30 sin 30 # = ° " ° ° °

11.83 ksi# = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.4

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

SOLUTION

Stresses Areas Forces

0: 18 cos 15 sin 1545 cos 15 cos 15 27 sin 15 sin 15

+ 18 sin 15 cos 15 0

!$ = + ° °+ ° ° " ° °

° ° =

F A A

A A

A

2

2

18 cos 15 sin 15 45 cos 15

27sin 15 18 sin 15 cos 15

! = " ° ° " °+ ° " ° °

49.2 MPa! = " W

0: 18 cos 15 cos 1545 cos 15 sin 1527 sin 15 cos 1518 sin 15 sin 15 0

#$ = + ° °" ° °" ° °" ° ° =

F A A

A

A

A

2 218(cos 15 sin 15 ) (45 27)cos 15 sin 15# = " ° " ° + + ° °

2.41 MPa# = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.5

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

SOLUTION

60 MPa 40 MPa 35 MPax y xy! ! #= " = " =

(a) 2 (2) (35)tan 2 3.50

60 40xy

p

x y

#%

! != = = "

" " +

2 74.05p% = " ° 37.0 , 53.0p% = " ° ° W

(b) 2

2max, min

22

2 2

60 40 60 40 (35)2 2

50 36.4 MPa

x y x y

xy

! ! ! !! #

+ "§ ·= ± +¨ ¸

© ¹

" " " +§ ·= ± +¨ ¸© ¹

= " ±

max 13.60 MPa! = " W

min 86.4 MPa! = " W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.8

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

SOLUTION

8 ksi 12 ksi 5 ksix y xy! ! #= " = =

(a) 2 2(5)tan 2 = 0.5

8 12xy

px y

#%

! != = "

" " "

2 26.5651p% = " ° 13.3 , 76.7p% = " ° ° W�

(b) 2

2max, min

22

+2 2

8 12 8 12 (5)2 2

x y x yxy

! ! ! !! #

+ "§ ·= ± ¨ ¸

© ¹

" + " "§ ·= ± +¨ ¸© ¹

�

� 2 11.1803= ± � max 13.18 ksi! = W�

� min 9.18 ksi! = " W�

�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.19

A steel pipe of 12-in. outer diameter is fabricated from 14 -in. -thick plate by

welding along a helix that forms an angle of 22.5° with a plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip in.& torque T, each directed as shown, are applied to the pipe, determine ! and # in directions, respectively, normal and tangential to the weld.

SOLUTION

( )( )

2 2 2

1 2

2 2 2 2 22 1

4 4 4 4 42 1

112 in., 6 in., 0.25 in.2

5.75 in.

(6 5.75 ) 9.2284 in

(6 5.75 ) 318.67 in2 2

d c d t

c c t

A c c

J c c

' '

' '

= = = =

= " =

= " = " =

= " = " =

Stresses:

2

40 4.3344 ksi9.2284

(80)(6) 1.5063 ksi318.670, 4.3344 ksi, 1.5063 ksix y xy

P

A

Tc

J

!

#

! ! #

= "

= " = "

=

= =

= = " =

Choose the x( and y( axes, respectively, tangential and normal to the weld.

Then and 22.5w y w x y! ! # # %( ( (= = = °

cos 2 sin 22 2

( 4.3344) [ ( 4.3344)] cos 45 1.5063 sin 45°2 2

4.76 ksi

x y x y

y xy

! ! ! !! % # %(

+ "= " "

" " "= " ° "

= " 4.76 ksiw! = " W

sin 2 cos 22

[ ( 4.3344)] sin 45 1.5063 cos 452

0.467 ksi

x y

x y xy

! !# % # %( (

"= " +

" "= " ° + °

= " 0.467 ksiw# = " W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.22

Two steel plates of uniform cross section 10 80 mm! are welded together as shown. Knowing that centric 100-kN forces are applied to the welded plates and that the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle ", (b) the corresponding normal stress perpendicular to the weld.

SOLUTION

Area of weld:

3 3

62

(10 10 )(80 10 )cos

800 10 mcos

wA "

"

# #

#

! !=

!=

(a) 3

36 6

6

0: 100sin 0 100sin kN 100 10 sin N

100 10 sin30 10 125 10 sin cos800 10 / cos

" " "

"$ " ""#

¦ = # = = = !

!= ! = = !!

s s s

sw

w

F F F

FA

6

61 30 10sin cos sin 2 0.2402 125 10

" " " != = =!

14.34" = ° W

(b) 6

6 2

0: 100cos 0 100cos14.34 96.88 kN

800 10 825.74 10 mcos14.34

"#

#

¦ = # = = ° =

!= = !

n n n

w

F F F

A

�

�3

66

96.88 10 117.3 10 Pa825.74 10

n

w

FA

% #!= = = !!

� 117.3 MPa% = W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.31

Solve Probs. 7.5 and 7.9, using Mohr’s circle.

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

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

SOLUTION

ave

60 MPa,40 MPa,

35 MPa

50 MPa2

x

y

xy

x y

!!#

! !!

= "= "=

+= = "

Plotted points for Mohr’s circle:

ave

: ( , ) ( 60 MPa, 35 MPa)

: ( , ) ( 40 MPa, 35 MPa)

: ( , 0) ( 50 MPa, 0)

x xy

y xy

X

Y

C

! #! #!

" = " "= "= "

(a) 35tan 3.50010

GX

CG) = = =

74.051 37.032b

)

% )

= °

= " = " ° 37.0b% = " ° W

180 105.951 52.972a

* )

% *

= ° " = °

= = ° 53.0a% = ° W

2 2 2 210 35 36.4 MPaR CG GX= + = + =

(b) min ave 50 36.4R! != " = " " min 86.4 MPa! = " W

max ave 50 36.4R! != + = " + max 13.6 MPa! = " W

(a() 45 7.97d B% %= + ° = ° 8.0d% = ° W

45 97.97e A% %= + ° = ° 98.0e% = ° W

max 36.4 MPaR# = = max 36.4 MPa# = W

(b() ave 50 MPa! !( = = " 50.0 MPa! ( = " W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.35

Solve Prob. 7.13, using Mohr’s circle.

PROBLEM 7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.

SOLUTION

ave

0,8 ksi,

5 ksi

4 ksi2

x

y

xy

x y

!!#

! !!

===

+= =

Plotted points for Mohr’s circle:

: (0, 5 ksi): (8 ksi, 5 ksi): (4 ksi, 0)

X

Y

C

"

2 2 2 2

5tan 2 1.254

2 51.34

4 5 6.40 ksi

p

p

FX

FC

R FC FX

%

%

= = =

= °

= + = + =

(a) 25% = ° . 2 50% = °

51.34 50 1.34+ = ° " ° = °

ave cosx R! ! +( = " 2.40 ksix! ( = " W�

sinx y R# +( ( = 0.15 ksix y# ( ( = W�

ave cosy R! ! +( = + 10.40 ksiy! ( = W

(b) 10% = ° . 2 20% = °

51.34 20 71.34+ = ° + ° = °

ave cosx R! ! +( = " 1.95 ksix! ( = W

sinx y R# +( ( = 6.07 ksix y# ( ( = W

� ave cosy R! ! +( = + � 6.05 ksiy! ( = W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.39

Solve Prob. 7.17, using Mohr’s circle.

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

ave

4 MPa 1.6 MPa 0

2.8 MPa2

x y xy

x y

! ! #! !

!

= " = " =

+= = "

Plotted points for Mohr’s circle:

ave

: ( , ) ( 4 MPa, 0)

: ( , ) ( 1.6 MPa, 0)

: ( , 0) ( 2.8 MPa, 0)

x xy

y xy

X

Y

C

! #! #!

" = "

= "

= "

15 . 2 30

1.2MPa 1.2MPaCX R

% %= " ° = " °

= =

(a) sin 30 sin 30 1.2sin 30x y CX R# ( ( (= " ° = " ° = " ° 0.600 MPax y# ( ( = " W

(b) ave cos30 2.8 1.2cos30x CX! !( (= " ° = " " ° 3.84 MPax! ( = " W

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.44

Solve Prob. 7.22, using Mohr’s circle.

PROBLEM 7.22 Two steel plates of uniform cross section 10 80 mm, are welded together as shown. Knowing that centric 100-kN forces are applied to the welded plates and that the in-plane shearing stress parallel to the weld is 30 MPa, determine (a) the angle ), (b) the corresponding normal stress perpendicular to the weld.

SOLUTION

3

63 3

100 10 125 10 Pa 125 MPa(10 10 )(80 10 )x

P

A! " "

,= = = , =, ,

0 0y xy! #= =

From Mohr’s circle:

(a) 30sin 2 0.4862.5

) = = 14.3) = ° W

(b) 62.5 62.5cos 2! )= +

117.3 MPa! = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.98

A spherical gas container made of steel has a 5-m outer diameter and a wall thickness of 6 mm. Knowing that the internal pressure is 350 kPa, determine the maximum normal stress and the maximum shearing stress in the container.

SOLUTION

36

5 m 6 mm 0.006 m, 2.494 m2

(350 10 Pa)(2.494 m) 72.742 10 Pa2 2(0.006 m)

dd t r t

pr

t!

= = = = " =

#= = = #

72.7 MPa! = W

max

min

72.742 MPa0 (Neglectingsmall radialstress)

!!

=$

max max min1 ( )2

% ! != " max 36.4 MPa% = W�

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 7.109

The unpressurized cylindrical storage tank shown has a 316 -in. wall thickness

and is made of steel having a 60-ksi 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. 3(Specific weight of water 62.4 lb/ft .)=

SOLUTION

( )

0

3all

all

3316 2all

(25)(12) 300 in.1 3150 149.81 in.2 16

60 ksi 15 ksi 15 10 psi. . 4.0

(15 10 )18.77 psi 2703 lb/ft

149.81

U

d

r d t

F S

pr

t

tp

r

!!

!

!

= =

= " = " =

= = = = ,

=

,= = = =

But ,p h-= 2

32703 lb/ft62.4 lb/ft

ph

-= = 43.3 fth = W�