chapter 7 · problem 7.1 for the given state of stress, determine the normal and shearing stresses...

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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 0σ= − ° ° − ° ° ° =F A A A A

230 sin 30 cos 30 10 cos 30σ = ° ° − °

5.49 ksiσ =

0: 15 sin 30 sin 30 15 cos 30 cos 30 10 cos 30 sin 30 0τΣ = + ° ° − ° ° − ° ° =F A A A A

2 215(cos 30 sin 30 ) + 10 cos 30 sin 30 τ = ° − ° ° °

11.83 ksiτ =




without permission.

PROBLEM 7.2


SOLUTION


0:Σ =F

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

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

0:Σ =F

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

120 cos 55 sin 55τ = ° ° 56.4 MPaτ =




without permission.

PROBLEM 7.3


SOLUTION


0:Σ =F

5 cos60 sin 60 6 sin 60 sin 60 5 sin 60 cos60 0σ + ° ° − ° ° + ° ° =A A A A

26sin 60 10cos60 sin 60σ = ° − ° ° 0.1699 ksiσ =

0:Σ =F

5 cos 60 cos60 6 sin 60 cos60 5 sin 60 sin 60 0τ + ° ° − ° ° − ° ° =A A A A

2 25(sin 60 cos 60 ) 6sin 60 cos60τ = ° − ° + ° ° 5.10 ksiτ =




without permission.

PROBLEM 7.4


SOLUTION


0: 18 cos 15 sin 15

45 cos 15 cos 15 27 sin 15 sin 15

+ 18 sin 15 cos 15 0

σΣ = + ° °+ ° ° − ° °

° ° =

F A A

A A

A

2

2

18 cos 15 sin 15 45 cos 15

27sin 15 18 sin 15 cos 15

σ = − ° ° − °

+ ° − ° °

49.2 MPaσ = −

0: 18 cos 15 cos 15

45 cos 15 sin 15

27 sin 15 cos 15

18 sin 15 sin 15 0

τΣ = + ° °− ° °− ° °− ° ° =

F A A

A

A

A

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

2.41 MPaτ =




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 xyσ σ τ= − = − =

(a) 2 (2) (35)

tan 2 3.5060 40

xyp

x y

τθ

σ σ= = = −

− − +

2 74.05pθ = − ° 37.0 , 53.0pθ = − ° °

(b)

2

2max, min

22

2 2

60 40 60 40(35)

2 2

50 36.4 MPa

x y x yxy

σ σ σ σσ τ

+ − = ± +

− − − + = ± +

= − ±

max 13.60 MPaσ = −

min 86.4 MPaσ = −




without permission.

PROBLEM 7.6


SOLUTION

10MPa 50 MPa 15MPax y xyσ σ τ= = = −

(a) 2 2( 15)

tan 2 = 0.75010 50

xyp

x y

τθ

σ σ−= − =

− −

2 36.8699pθ = ° 18.4 , 108.4pθ = ° °

(b) 2

2max, min

22

+2 2

10 50 10 50( 15)

2 2

σ σ σ σσ τ

+ − = ±

+ − = ± + −

x y x yxy

30 25= ± max 55.0 ksiσ =

min 5.00 ksiσ =




without permission.

PROBLEM 7.7


SOLUTION

4 ksi 12 ksi 15 ksix y xyσ σ τ= = − = −

(a) 2 (2)( 15)

tan 2 1.8754 12

xyp

x y

τθ

σ σ−= = = −

− +

2 61.93pθ = − ° 31.0 , 59.0pθ = − ° °

(b)

2

2max, min

22

2 2

4 12 4 1215

2 2

4 17

x y x yxy

σ σ σ σσ τ

+ − = ± +

− + = ± +

= − ±

max 13.00 ksiσ =

min 21.0 ksiσ = −




without permission.

PROBLEM 7.8


SOLUTION

8 ksi 12 ksi 5 ksix y xyσ σ τ= − = =

(a) 2 2(5)

tan 2 = 0.58 12

xyp

x y

τθ

σ σ= = −

− − −

2 26.5651pθ = − ° 13.3 , 76.7pθ = − ° °

(b) 2

2max, min

22

+2 2

8 12 8 12(5)

2 2

x y x yxy

σ σ σ σσ τ

+ − = ±

− + − − = ± +

2 11.1803= ± max 13.18 ksiσ =

min 9.18 ksiσ = −




without permission.

PROBLEM 7.9

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

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

(a) 60 40

tan 2 0.28572 (2)(35)x y

sxy

σ σθ

τ− − += − = − =

2 15.95sθ = ° 8.0 , 98.0sθ = ° °

(b)

2

2max 2

x yxy

σ στ τ

− = +

2

260 40(35)

2

− + = +

max 36.4 MPaτ =

(c) ave60 40

2 2x yσ σ

σ σ+ − −′ = = = 50.0 MPaσ ′ = −




without permission.

PROBLEM 7.10


SOLUTION

10 MPa 50 MPa 15 MPax y xyσ σ τ= = = −

(a) 10 50

tan 2 1.333332 2( 15)x y

sxy

σ σθ

τ− −= − = − = −

−

2 53.130sθ = − ° 26.6 , 63.4sθ = − ° °

(b) 2

2max +

2x y

xy

σ στ τ

− =

2

210 50( 15)

2

− = + −

max 25.0 MPaτ =

(c) ave 2

σ σσ σ

+′ = = x y

10 50

2

+= 30.0 MPaσ′ =




without permission.

PROBLEM 7.11


SOLUTION

4 ksi 12 ksi 15 ksix y xyσ σ τ= = − = −

(a) 4 12

tan 2 0.533332 (2)( 15)x y

sxy

σ σθ

τ− += − = − =

−

2 28.07sθ = ° 14.0 , 104.0sθ = ° °

(b)

2

2max 2

x yxy

σ στ τ

− = +

2

24 12( 15)

2

+ = + −

max 17.00 ksiτ =

(c) ave4 12

2σ σ −′ = = 4.00 ksiσ ′ = −




without permission.

PROBLEM 7.12


SOLUTION


(a) 8 12

tan 2 2.02 2(5)x y

sxy

σ σθ

τ− − −= − = − = +

2 63.435sθ = ° 31.7 , 121.7sθ = ° °

(b) 2

2max 2

x yxy

σ στ τ

− = +

2

28 12(5)

2

− − = +

max 11.18 ksiτ =

(c) ave 2

σ σσ σ

+′ = = x y

8 12

2

− += 2.00 ksiσ′ =




without permission.

PROBLEM 7.13

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

SOLUTION

0 8 ksi 5 ksi

4 ksi 4 ksi2 2

cos 2 + sin 22 2

sin 2 + cos 22

cos 2 sin 22 2

x y xy

x y x y

x y x yx xy

x yx y xy

x y x yy xy

σ σ τσ σ σ σ

σ σ σ σσ θ τ θ

σ στ θ τ θ


′

′ ′

′

= = =

+ −= = −

+ −= +

−= −

+ −= − −

(a) 25 2 50θ θ= − ° = − °

4 4 cos ( 50°) + 5 sin ( 50°)xσ ′ = − − − 2.40 ksixσ ′ = −

4 sin ( 50 ) 5 cos ( 50 )x yτ ′ ′ = − ° + − ° 0.15 ksix yτ ′ ′ =

4 4 cos ( 50 ) 5 sin ( 50)yσ ′ = + − ° − − 10.40 ksiyσ ′ =

(b) 10 2 20θ θ= ° = °

4 4 cos (20°) + 5 sin (20°)xσ ′ = − 1.95 ksixσ ′ =

4 sin (20°) + 5 cos (20°)x yτ ′ ′ = 6.07 ksix yτ ′ ′ =

4 4 cos (20°) 5 cos (20°)yσ ′ = + − 6.05 ksiyσ ′ =




without permission.

PROBLEM 7.14


SOLUTION

60 MPa 90 MPa 30 MPa

15 MPa 75 MPa2 2

cos 2 + sin 22 2

sin 2 + cos 22

cos 2 sin 22 2

x y xy

x y x y

x y x yx xy

x yxy xy

x y x yy xy

σ σ τσ σ σ σ


σ στ θ τ θ


′

′

= − = =

+ −= = −

+ −= +

−= −

+ −= − −

(a) 25 2 50θ θ= − ° = − °

15 75 cos ( 50 ) 30 sin ( 50 )xσ ′ = − − ° + − ° 56.2 MPaxσ ′ = −

75 sin ( 50 ) 30 cos ( 50 )x yτ ′ ′ = + − ° + − ° 38.2 MPax yτ ′ ′ = −

15 75 cos ( 50 ) 30 sin ( 50 )yσ ′ = + − ° − − ° 86.2 MPayσ ′ =

(b) 10 2 20θ θ= ° = °

15 75 cos (20°) + 30 sin (20°)xσ ′ = − 45.2 MPaxσ ′ = −

75 sin (20°) + 30 cos (20°)x yτ ′ ′ = + 53.8 MPax yτ ′ ′ =

15 75 cos (20°) 30 sin (20°)yσ ′ = + − 75.2 MPayσ ′ =




without permission.

PROBLEM 7.15


SOLUTION

8 ksi 12 ksi 6 ksi

2 ksi 10 ksi2 2

cos 2 + sin 22 2

sin 2 + cos 22

cos 2 sin 22 2

x y xy

x y x y

x y x yx xy

x yx y xy

x y x yy xy

σ σ τσ σ σ σ


σ στ θ τ θ


′

′ ′

′

= = − = −

+ −= − =

+ −= +

−= −

+ −= − −

(a) 25 2 50θ θ= − ° = − °

2 10 cos ( 50 ) 6 sin ( 50 )xσ ′ = − + − ° − − ° 9.02 ksi xσ ′ =

10 sin ( 50 ) 6 cos ( 50 )x yτ ′ ′ = − − ° − − ° 3.80 ksix yτ ′ ′ =

2 10 cos ( 50 ) 6 sin ( 50 )yσ ′ = − − − ° + − ° 13.02 ksiyσ ′ = −

(b) 10 2 20θ θ= ° = °

2 10 cos (20°) 6 sin (20°)xσ ′ = − + − 5.34 ksixσ ′ =

10 sin (20°) 6 cos (20°)x yτ ′ ′ = − − 9.06 ksix yτ ′ ′ = −

2 10 cos (20°) + 6 sin (20°)yσ ′ = − − 9.34 ksiyσ ′ = −




without permission.

PROBLEM 7.16


SOLUTION

0 80 MPa 50 MPa

40 MPa 40 MPa2 2

cos 2 sin 22 2

sin 2 + cos 22

cos 2 sin 22 2

x y xy

x y x y

x y x yx xy

x yx y xy

x y x yy xy

σ σ τσ σ σ σ


σ στ θ τ θ


′

′ ′

′

= = − = −

+ −= − =

+ −= + +

−= −

+ −= − −

(a) 25θ = − ° 2 50θ = − °

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

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

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

(b) 10 2 20θ θ= ° = °

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

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

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




without permission.

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

4 MPa 1.6 MPa 0

15 2 30

x y xyσ σ τ

θ θ

= − = − =

= − ° = − °

(a) sin 2 + cos 22

x yx y xy

σ στ θ τ θ′ ′

−= −

4 ( 1.6)

sin ( 30 ) 02

− − −= − − ° + 0.600 MPax yτ ′ ′ = −

(b) cos 2 + sin 22 2

x y x yx xy

σ σ σ σσ θ τ θ′

+ −= +

4 ( 1.6) 4 ( 1.6)

cos ( 30 ) 02 2

− + − − − −= + − ° + 3.84 MPaxσ ′ = −




without permission.

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.

SOLUTION

0 0 400 psi

15 2 30

x y xyσ σ τ

θ θ

= = =

= − ° = − °

(a) sin 2 + cos 22

x yx y xy


−= −

0 400cos( 30 )= + − ° 346 psix yτ ′ ′ =

(b) cos 2 + sin 22 2

x y x yx xy


+ −= +

0 0 400sin( 30 )= + + − ° 200 psixσ ′ = −




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 σ and τ 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.

25.75 in.

(6 5.75 ) 9.2284 in

(6 5.75 ) 318.67 in2 2

d c d t

c c t

A c c

J c c

π π

π π

= = = =

= − =

= − = − =

= − = − =

Stresses:

2

404.3344 ksi

9.2284

(80)(6)1.5063 ksi

318.670, 4.3344 ksi, 1.5063 ksix y xy

P

A

Tc

J

σ

τ

σ σ τ

= −

= − = −

=

= =

= = − =

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

Then and 22.5w y w x yσ σ τ τ θ′ ′ ′= = = °

cos 2 sin 22 2

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

2 24.76 ksi

x y x yy xy


+ −= − −

− − −= − ° −

= − 4.76 ksiwσ = −

sin 2 cos 22

[ ( 4.3344)]sin 45 1.5063 cos 45

20.467 ksi

x yx y xy


−= − +

− −= − ° + °

= − 0.467 ksiwτ = −




without permission.

PROBLEM 7.20

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

SOLUTION

For plane a-a, 65 .θ = °

2 2 2

3 3 33

2 2

2 2

3

0, 0,

cos sin 2 sin cos 0 sin 65 0

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

sin 65 sin 65

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

(50 10 )(8

sin 65 cos 65

x xy y

x y xy

x y xy

P

AP

A

AP

P

A

AP

σ τ σ

σ σ θ σ θ τ θ θ

σ

τ σ σ θ θ τ θ θ

τ

− −

−

= = =

= + + = + ° +

× × ×= = = ×° °

= − − + − = ° ° +

×= =° °

3 330 10 )(600 10 )

6.27 10 Nsin 65 cos65

−× × = ×° °

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




without permission.

PROBLEM 7.21

Two steel plates of uniform cross section 10 80mm× are welded together as shown. Knowing that centric 100-kN forces are applied to the welded plates and that 25β = ° , determine (a) the in-plane shearing stress parallel to the weld, (b) the normal stress perpendicular to the weld.

SOLUTION

Area of weld:

3 3

6 2

(10 10 )(80 10 )

cos 25

882.7 10 m

wA− −

−

× ×=°

= ×

(a) 0: 100sin 25 0 42.26 kN = − ° = =s s sF F F

3

66

42.26 1047.9 10 Pa

882.7 10s

ww

F

Aτ −

×= = = ××

47.9 MPawτ =

(b) 0: 100cos 25 0 90.63 kN = − ° = =n n nF F F

3

66

90.63 10102.7 10 Pa

882.7 10n

ww

F

Aσ −

×= = = ××

102.7 MPawσ =




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

SOLUTION

Area of weld:

3 3

62

(10 10 )(80 10 )

cos

800 10m

cos

wAβ

β

− −

−

× ×=

×=

(a) 3

36 6

6

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

100 10 sin30 10 125 10 sin cos

800 10 / cos

β β β

βτ β ββ−

= − = = = ×

×= × = = ××

s s s

sw

w

F F F

F

A

6

6

1 30 10sin cos sin 2 0.240

2 125 10β β β ×= = =

× 14.34β = °

(b) 6

6 2

0: 100cos 0 100cos14.34 96.88 kN

800 10825.74 10 m

cos14.34

β−

−

= − = = ° =

×= = ×

n n n

w

F F F

A

3

66

96.88 10117.3 10 Pa

825.74 10n

w

F

Aσ −

×= = = ××

117.3 MPaσ =




without permission.

PROBLEM 7.23

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.

SOLUTION

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

400 lb (400)(6) 2400 lb in

(400)(2) 800 lb in

V M

T

= = = ⋅= = ⋅

Shaft cross section:

4 4 4

11 in. 0.5 in.

21

0.098175 in 0.049087 in2 2

d c d

J c I Jπ

= = =

= = = =

Torsion: 3(800)(0.5)4.074 10 psi 4.074 ksi

0.098175

Tc

Jτ = = = × =

Bending: 3(2400)(0.5)24.446 10 psi 24.446 ksi

0.049087

Mc

Iσ = = = × =

Transverse shear: Stress at point H is zero.

ave

22 2 2

24.446 ksi, 0, 4.074 ksi

1( ) 12.223 ksi

2

(12.223) (4.074)2

12.884 ksi

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = + = +

=

avea Rσ σ= + 25.1 ksiaσ =

aveb Rσ σ= − 0.661 ksibσ = −

max Rτ = max 12.88 ksiτ =




without permission.

PROBLEM 7.24

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 3

4 -in. diameter shaft.

SOLUTION


24 lb (24)(6) 144 lb in

(24)(10) 240 lb in

V M

T

= = = ⋅= = ⋅

Shaft cross section:

4 4 4

10.75 in., 0.375 in.

21

0.031063 in 0.015532 in2 2

d c d

J c I Jπ

= = =

= = = =

Torsion: 3(240)(0.375)2.897 10 psi 2.897 ksi

0.031063

Tc

Jτ = = = × =

Bending: 3(144)(0.375)3.477 10 psi 3.477 ksi

0.015532

Mc

Iσ = = = × =

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

Resultant stresses:

ave

22 2 2

3.477 ksi, 0, 2.897 ksi

1( ) 1.738 ksi

2

1.738 2.897 3.378 ksi2

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = + = + =







without permission.

PROBLEM 7.25

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.

SOLUTION

( )4 4 6 4

6 4

6 4

10251 mm 45 mm

2 2

4.1855 10 mm2

4.1855 10 m

12.0927 10 m

2

oo i o

o i

dr r r t

J r r

I J

π

−

−

= = = = − =

= − = ×

= ×

= = ×

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

3

3 3

3 3

10 kN

10 10 N

(10 10 )(200 10 )

2000 N m

(10 10 )(150 10 )

1500 N m

x

y

z

F

M

M

−

−

=

= ×

= × ×

= ⋅

= − × ×= − ⋅

Torsion: At point K, place local x-axis in negative global z-direction.

3

3

6

6

2000 N m

51 10 m

(2000)(51 10 )

4.1855 10

24.37 10 Pa

24.37 MPa

y

o

xy

T M

c r

Tc

Jτ

−

−

= = ⋅

= = ×

×= =×

= ×=




without permission.

PROBLEM 7.25 (Continued)

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

Bending:

3

66

| | (1500)(51 10 )| | 36.56 10 Pa 36.56 MPa

2.0927 10z

yM c

Iσ

−

−×= = = × =

×

Point K lies on compression side of neutral axis:

36.56 MPayσ = −

Total stresses at point K:

ave

2

2

0, 36.56 MPa, 24.37 MPa

1( ) 18.28 MPa

2

30.46 MPa2

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = − =

= + = −

− = + =

max ave 18.28 30.46Rσ σ= + = − + max 12.18 MPaσ =

min ave 18.28 30.46Rσ σ= − = − − min 48.7 MPaσ = −

max Rτ = max 30.5 MPaτ =




without permission.

PROBLEM 7.26

The axle of an automobile is acted upon by the forces and couple shown. Knowing that the diameter of the solid axle is 32 mm, determine (a) the principal planes and principal stresses at point H located on top of the axle, (b) the maximum shearing stress at the same point.

SOLUTION

31 1(32) 16 mm 16 10 m

2 2c d −= = = = ×

Torsion: 63 3 3

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

(16 10 m)

Tc T

J cτ

π π −⋅= = = = × =

×

Bending: 4 3 4 9 4

3

36

9

(16 10 ) 51.472 10 m4 4

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

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

51.472 10

I c

M

My

I

π π

σ

− −

−

−

= = × = ×

= × = ⋅

×= − = − = − × = −×

Top view: Stresses:

ave

22 2 2

139.882 MPa 0 54.399 MPa

1 1( ) ( 139.882 0) 69.941 MPa

2 2

( 69.941) ( 54.399) 88.606 MPa2

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= − = = −

= + = − + = −

− = + = − + − =

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

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




without permission.


2 (2)( 54.399)

tan 2 0.77778 2 37.88139.882

xyp p

x y

τθ θ

σ σ−= = = = °

− −

18.9 and 108.9°pθ = °

(b) max 88.6 MPaRτ = = max 88.6 MPaτ =




without permission.

PROBLEM 7.27

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

SOLUTION

2 22 2

max

2 2

10 ksi, 8 ksi, ?

10 ( 8)

9 12 ksi

x y xy

x yxy xy

xy

Rz z

σ σ τ

σ στ τ τ

τ

= = − =

− − − = = + = +

= + =

(a) 2 212 9xyτ = − 7.94 ksixyτ =

(b) ave1

( ) 1 ksi2 x yσ σ σ= + =

ave 1 12 13 ksia Rσ σ= + = + = 13.00 ksiaσ =

ave 1 12 11 ksib Rσ σ= − = − = − 11.00 ksibσ = −




without permission.

PROBLEM 7.28

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

SOLUTION

60 MPa, ?, 20 MPax y xyσ σ τ= = =

Let

.2

x yuσ σ−

=

Then

2 2

2 2 2 2

2

75 MPa

75 20 72.284 MPa

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

y x

xy

xy

y x

u

R u

u R

u

σ σ

τ

τ

σ σ

= −

= + =

= ± − = ± − =

= − = = −

Largest value of yσ is required. 205 MPayσ =




without permission.

PROBLEM 7.29

Determine the range of values of xσ for which the maximum in-plane shearing stress is equal to or less than 10 ksi.

SOLUTION

?, 15 ksi, 8 ksix y xyσ σ τ= = =

Let

2 2max

2 2 2 2

22

10 ksi

10 8 6 ksi

2 15 (2)(6) 27 ksi or 3 ksi

x yx y

xy

zxy

x y

u u

R u

u R

u

σ σσ σ

τ τ

τ

σ σ

−= = +

= + = =

= ± − = ± − = ±

= + = ± =

Allowable range: 3 ksi 27 ksixσ≤ ≤




without permission.

PROBLEM 7.30

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

SOLUTION

12 MPa, 2 MPa, ?x y xyσ σ τ= = =

Since 0,x yτ ′ ′ = x ′-direction is a principal direction.

15

2tan 2

p

xyp

x y

θτ

θσ σ

= − °

=−

(a) 1 1

( ) tan 2 (12 2) tan( 30 )2 2xy x y pτ σ σ θ= − = − − ° 2.89 MPaxyτ = −

2

2 2 2

ave

5 2.89 5.7735 MPa2

1( ) 7 MPa

2

x yxy

x y

Rσ σ

τ

σ σ σ

− = + = + =

= + =

(b) ave 7 5.7735a Rσ σ= + = + 12.77 MPaaσ =

ave 7 5.7735b Rσ σ= − = − 1.226 MPabσ =




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

σστ

σ σσ

= −= −

=

+= = −

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

σ τσ τσ

− = − −

= −

= −

(a) 35

tan 3.50010

GX

CGβ = = =

74.05

137.03

2b

β

θ β

= °

= − = − ° 37.0bθ = − °

180 105.95

152.97

2a

α β

θ α

= ° − = °

= = ° 53.0aθ = °

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

(b) min ave 50 36.4Rσ σ= − = − − min 86.4 MPaσ = −

max ave 50 36.4Rσ σ= + = − + max 13.6 MPaσ = −

(a′) 45 7.97d Bθ θ= + ° = ° 8.0dθ = °

45 97.97e Aθ θ= + ° = ° 98.0eθ = °

max 36.4 MPaRτ = = max 36.4 MPaτ =

(b′) ave 50 MPaσ σ′ = = − 50.0 MPaσ ′ = −




without permission.

PROBLEM 7.32

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

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


SOLUTION

ave

4 ksi,

12 ksi,

15 ksi

4 ksi2

x

y

xy

x y

σστ

σ σσ

== −

= −

+= = −


ave

: ( , ) (4 ksi, 15 ksi)

: ( , ) ( 12 ksi, 15 ksi)

: ( , 0) ( 4 ksi, 0)

x xy

y xy

X

Y

C

σ τσ τσ

− =

= − −

= −

(a) 15

tan 1.8758

FX

CFα = = =

61.93

130.96

2a

α

θ α

= °

= − = − ° 31.0aθ = − °

180 118.07

159.04

2b

β α

θ β

= ° − = °

= = ° 59.0bθ = °

2 2 2 2( ) ( ) (8) (15) 17 ksiR CF FX= + = + =

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

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

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

45 104.04e bθ θ= + ° = ° 104.0eθ = °


(b′) aveσ σ′ = 4.00 ksiσ ′ = −




without permission.

PROBLEM 7.33

Solve Problem 7.10, using Mohr’s circle.


SOLUTION

ave

10 MPa 50 MPa 15 MPa

1 1( ) (10 50) 30 MPa

2 2

x y xy

x y

σ σ τ

σ σ σ

= = = −

= + = + =


,

ave

: ( ) (10 MPa, 15 MPa)

: ( , ) (50 MPa, 15 MPa)

: ( , 0) (30 MPa, 0)

x xy

y xy

X

Y

C

σ τ

σ τ

σ

− =

= −

=

15

tan 0.7520

36.87

118.43

2b

FX

FCα

α

θ α

= = =

= °

= = °

(a) 45d bθ θ= − ° 26.6dθ = − °

45e bθ θ= + ° 63.4eθ = °

(b) 2 2 2 220 15 25 MPaR CF FX= + = + =

max (in-plane)τ = R max (in-plane) 25.0 MPaτ =

(c) aveσ σ′ = 30.0 MPaσ′ =




without permission.

PROBLEM 7.34

Solve Prob. 7.12, using Mohr’s circle.


SOLUTION



ave

: ( , ) ( 8 ksi, 5 ksi)

: ( , ) (12 ksi, 5 ksi)

: ( , 0) (2 ksi, 0)

x xy

y xy

X

Y

C

σ τ

σ τ

σ

− = − −

=

=

5tan 0.5

1026.565

180 153.435

176.718

2a

FX

FCα

αβ α

θ β

= = =

= °= − = °

= = °

(a) 45d aθ θ= + ° 121.7dθ = °

45e aθ θ= − ° 31.7eθ = °

(b) 2 2 2 210 5 11.1803 ksiR CF FX= + = + =

max (in-plane)τ = R max (in-plane) 11.18 ksiτ =

(c) aveσ σ′ = 2.00 ksiσ′ =




without permission.

PROBLEM 7.35


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

σστ

σ σσ

==

=

+= =


: (0, 5 ksi)

: (8 ksi, 5 ksi)

: (4 ksi, 0)

X

Y

C

−

2 2 2 2

5tan 2 1.25

42 51.34

4 5 6.40 ksi

p

p

FX

FC

R FC FX

θ

θ

= = =

= °

= + = + =

(a) 25θ = ° . 2 50θ = °

51.34 50 1.34ϕ = ° − ° = °

ave cosx Rσ σ ϕ′ = − 2.40 ksixσ ′ = −

sinx y Rτ ϕ′ ′ = 0.15 ksix yτ ′ ′ =

ave cosy Rσ σ ϕ′ = + 10.40 ksiyσ ′ =

(b) 10θ = ° . 2 20θ = °

51.34 20 71.34ϕ = ° + ° = °

ave cosx Rσ σ ϕ′ = − 1.95 ksixσ ′ =


ave cosy Rσ σ ϕ′ = + 6.05 ksiyσ ′ =




without permission.

PROBLEM 7.36

Solve Prob 7.14, using Mohr’s circle.


SOLUTION

ave

60 MPa,

90 MPa,

30 MPa

15 MPa2

x

y

xy

x y

σστ

σ σσ

= −=

=

+= =


: ( 60 MPa, 30 MPa)

: (90 MPa, 30 MPa)

: (15 MPa, 0)

X

Y

C

− −

30

tan 2 0.475

θ = = =pFX

FC

2 21.80 10.90θ θ= ° = °p P

2 2 2 275 30 80.78 MPaR FC FX= + = + =

(a) 25θ = ° . 2 50θ = °

2 2 50 21.80 28.20Pϕ θ θ= − = ° − ° = °

ave cosσ σ ϕ′ = −x R 56.2 MPaσ ′ = −x

sinτ ϕ′ ′ = −x y R 38.2 MPaτ ′ ′ = −x y

ave cosσ σ ϕ′ = +y R 86.2 MPayσ ′ =




without permission.


(b) 10θ = ° . 2 20θ = °

2 2 21.80 20 41.80ϕ θ θ= + = ° + ° = °p

ave cosσ σ ϕ′ = −x R 45.2 MPaσ ′ = −x

sinτ ϕ′ ′ =x y R 53.8 MPaτ ′ ′ =x y

ave cosσ σ ϕ′ = +y R 75.2 MPaσ ′ =y




without permission.

PROBLEM 7.37



SOLUTION

ave

8 ksi,

12 ksi,

6 ksi

2 ksi2

x

y

xy

x y

σστ

σ σσ

== −

= −

+= = −


: (8 ksi, 6 ksi)

: ( 12 ksi, 6 ksi)

: ( 2 ksi, 0)

X

Y

C

− −−

2 2 2 2

6tan 2 0.6

102 30.96

10 6 11.66 ksi

θ

θ

= = =

= °

= + = + =

p

p

FX

CF

R CF FX

(a) 25θ = ° . 2 50θ = °

50 30.96 19.04ϕ = ° − ° = °

ave cosx Rσ σ ϕ′ = + 9.02 ksixσ ′ =


ave cosy Rσ σ ϕ′ = − 13.02 ksiyσ ′ = −

(b) 10θ = ° . 2 20θ = °

30.96 20 50.96ϕ = ° + ° = °

ave cosx Rσ σ ϕ′ = + 5.34 ksixσ ′ =

sinx y Rτ ϕ′ ′ = − 9.06 ksix yτ ′ ′ = −

ave cosy Rσ σ ϕ′ = − 9.34 ksiyσ ′ = −




without permission.

PROBLEM 7.38



SOLUTION

ave

0,

80 MPa,

50 MPa

40 MPa2

x

y

xy

x y

σστ

σ σσ

== −

= −

+= = −


: (0, 50 MPa)

: ( 80 MPa, 50 MPa)

: ( 40 MPa, 0)

X

Y

C

− −−

2 2 2 2

50tan 2 1.25

402 51.34

40 50

64.03 MPa

p

p

FX

CF

R CF FX

θ

θ

= = =

= °

= + = +=

(a) 25θ = ° . 2 50θ = °

51.34 50 1.34ϕ = ° − ° = °

ave cosx Rσ σ ϕ′ = + 24.0 MPaxσ ′ =

sinx y Rτ ϕ′ ′ = − 1.5 MPax yτ ′ ′ = −

ave cosy Rσ σ ϕ′ = − 104.0 MPayσ ′ = −

(b) 10θ = ° . 2 20θ = °

51.34 20 71.34ϕ = ° + ° = °

ave cosx Rσ σ ϕ′ = + 19.5 MPaxσ ′ = −

sinx y Rτ ϕ′ ′ = − 60.7 MPax yτ ′ ′ = −

ave cosy Rσ σ ϕ′ = − 60.5 MPayσ ′ = −




without permission.

PROBLEM 7.39


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

σ σ τ

σ σσ

= − = − =

+= = −


ave

: ( , ) ( 4 MPa, 0)

: ( , ) ( 1.6 MPa, 0)

: ( , 0) ( 2.8 MPa, 0)

x xy

y xy

X

Y

C

σ τ

σ τ

σ

− = −

= −

= −

15 . 2 30

1.2MPa 1.2MPaCX R

θ θ= − ° = − °

= =

(a) sin 30 sin 30 1.2sin 30x y CX Rτ ′ ′ ′= − ° = − ° = − ° 0.600 MPax yτ ′ ′ = −

(b) ave cos30 2.8 1.2cos30x CXσ σ′ ′= − ° = − − ° 3.84 MPaxσ ′ = −




without permission.

PROBLEM 7.40


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.

SOLUTION

ave

0, 400psi

02

x y xy

x y

σ σ τ

σ σσ

= = =

+= =

Points:

ave

: ( , ) (0, 400 psi)

: ( , ) (0, 400 psi)

: ( , 0) (0, 0)

x xy

y xy

X

Y

C

σ τ

σ τ

σ

− = −

=

=

15 ,θ = − ° 2 30

400 psiCX R

θ = − °

= =

(a) cos 30 400 cos30x y Rτ ′ ′ = ° = ° 346 psix yτ ′ ′ =

(b) ave sin 30 400 sin 30x Rσ σ′ = − ° = − ° 200 psixσ ′ = −




without permission.

PROBLEM 7.41


PROBLEM 7.19 A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by welding along a helix which 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 σ and τ 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.

25.75 in.

(6 5.75 ) 9.2284 in

(6 5.75 ) 318.67 in2 2

d c d t

c c t

A c c

J c c

π π

π π

= = = =

= − =

= − = − =

= − = − =

Stresses:

2

404.3344 ksi

9.2284(80)(6)

1.5063 ksi318.67

0, 4.3344 ksi, 1.5063 ksix y xy

P

ATc

J

σ

τ

σ σ τ

= − = − = −

= − = − =

= = − =

Draw the Mohr’s circle.

: (0, 1.5063 ksi)

: ( 4.3344 ksi,1.5063 ksi)

: ( 2.1672 ksi, 0)

X

Y

C

−−−

2 2

1.5063tan 0.69504 34.8

2.1672(2)(22.5 ) 10.8

(2.1672) (1.5063) 2.6393 ksiR

α α

β α

= = = °

= ° − = °

= + =

2.1672 2.6393 cos 10.8wσ = − − ° 4.76 ksiwσ = −

2.6393sin10.2wτ = − ° 0.467 ksiwτ = −




without permission.

PROBLEM 7.42


PROBLEM 7.20 Two members of uniform cross section 50 80× mm are glued together along plane a-a that forms an angle of 25° with the horizontal. Knowing that the allowable stresses for the glued joint are 800σ = kPa and 600τ = kPa, determine the largest centric load P that can be applied.

SOLUTION

3 3

3 2

0

0

/

(50 10 )(80 10 )

4 10 m

(1 cos50 )2

2

1 cos50

x

xy

y P A

A

P

AA

P

στσ

σ

σ

− −

−

==

=

= × ×

= ×

= + °

=+ °

3 3

3

(2)(4 10 )(800 10 )

1 cos50

3.90 10 N

P

P

−× ×≤+ °

≤ ×

3 3

32 (2)(4 10 )(600 10 )sin 50 6.27 10 N

2 sin 50 sin 50

P AP

A

ττ−× ×= ° = ≤ = ×

° °

Choosing the smaller value, 3.90 kNP =




without permission.

PROBLEM 7.43


PROBLEM 7.21 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 25β = ° , determine (a) the in-plane shearing stress parallel to the weld, (b) the normal stress perpendicular to the weld.

SOLUTION

3

63) 3)

100 10125 10 Pa 125 MPa

(10 10 (80 10x

P

Aσ − −

×= = = × =× ×

0 0y xyσ τ= =

From Mohr’s circle:

(a) 62.5sin 50wτ = ° 47.9 MPawτ =

(b) 62.5 62.5cos50wσ = + °

102.7 MPawσ =




without permission.

PROBLEM 7.44


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

SOLUTION

3

63 3

100 10125 10 Pa 125 MPa

(10 10 )(80 10 )x

P

Aσ − −

×= = = × =× ×

0 0y xyσ τ= =

From Mohr’s circle:

(a) 30

sin 2 0.4862.5

β = = 14.3β = °

(b) 62.5 62.5cos 2σ β= +

117.3 MPaσ =




without permission.

PROBLEM 7.45


PROBLEM 7.23 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.

SOLUTION


400 lb (400)(6) 2400 lb inV M= = = ⋅

(400)(2) 800 lb inT = = ⋅

Shaft cross section 1

1 in. 0.5 in.2

d c d= = =

4 4 410.098175 in 0.049087 in

2 2J c I J

π= = = =

Torsion: 3(800)(0.5)4.074 10 psi 4.074 ksi

0.098175

Tc

Jτ = = = × =

Bending: 3(2400)(0.5)24.446 10 psi 24.446 ksi

0.049087

Mc

Iσ = = = × =

Transverse shear: Stress at point H is zero.

Resultant stresses:

ave

22

2 2

24.446 ksi, 0, 4.074 ksi

1( ) 12.223 ksi

2

2

(12.223) (4.074) 12.884 ksi

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = +

= + =







without permission.

PROBLEM 7.46

Solve Prob. 7.24 using Mohr’s circle.

PROBLEM 7.24 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 3

4 -in.-diameter shaft.

SOLUTION


24 lb (24)(6) 144 lb inV M= = = ⋅

(24)(10) 240 lb inT = = ⋅

Shaft cross section: 1

0.75 in. 0.375 in.2

d c d= = =

4 4 410.031063 in 0.015532 in

2 2J c I J

π= = = =

Torsion: 3(240)(0.375)2.897 10 psi 2.897 ksi

0.031063

Tc

Jτ = = = × =

Bending: 3(144)(0.375)3.477 10 psi 3.477 ksi

0.015532

Mc

Iσ = = = × =

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

Resultant stresses:

ave

22

2 2

3.477 ksi, 0, 2.897 ksi

1( ) 1.738 ksi

2

2

1.738 2.897 3.378 ksi

x y xy

x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = +

= + =







without permission.

PROBLEM 7.47


PROBLEM 7.25 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.

SOLUTION

102

51 mm 45 mm2 2o

o i od

r r r t= = = = − =

( )4 4 6 4 6 44.1855 10 mm 4.1855 10 m2 o iJ r rπ −= − = × = ×

6 412.0927 10 m

2I J −= = ×

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

3

3 3

3 3

10 10 N

(10 10 )(200 10 ) 2000 N m

(10 10 )(150 10 ) 1500 N m

x

y

z

F

M

M

−

−

= ×

= × × = ⋅

= − × × = − ⋅

Torsion:

3

3

6

2000 N m

51 10 m

(2000)(51 10 )24.37 MPa

4.1855 10

y

o

xy

T M

c r

Tc

Jτ

−

−

−

= = ⋅

= = ×

×= = =×

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

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

Bending: 3

6

(1500)(51 10 )36.56 MPa

2.0927 10z

y

M c

Iσ

−

−×= = =

×

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




without permission.


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

ave1

( ) 18.28 MPa2 x yσ σ σ= + = −

2

2 30.46 MPa2

x yxyR

σ στ

− = + =

max ave 18.28 30.46Rσ σ= + = − +

max 12.18 MPaσ =

min ave 18.28 30.46Rσ σ= − = − −

min 48.74 MPaσ = −

max Rτ = max 30.46 MPaτ =




without permission.

PROBLEM 7.48


PROBLEM 7.26 The axle of an automobile is acted upon by the forces and couple shown. Knowing that the diameter of the solid axle is 32 mm, determine (a) the principal planes and principal stresses at point H located on top of the axle, (b) the maximum shearing stress at the same point.

SOLUTION

31 1(32) 16 mm 16 10 m

2 2c d −= = = = ×

Torsion: 3

2Tc T

J cτ

π= =

63 3

2(350 N m)54.399 10 Pa 54.399 MPa

(16 10 m)τ

π −⋅= = × =

×

Bending: 4 3 4 9 4(16 10 ) 51.472 10 m4 4

I cπ π − −= = × = ×

3(0.15m)(3 10 N) 450 N mM = × = ⋅

3

69

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

51.472 10

My

Iσ

−

−×= − = − = − × = −×

Top view Stresses

139.882 MPa, 0, 54.399 MPax y xyσ σ τ= − = = −

Plotted points: : ( 139.882, 54.399); : (0, 54.399); : ( 69.941, 0)X Y C− − −

ave

22

22

1( ) 69.941 MPa

2

2

139.882(54.399) 88.606 MPa

2

x y

x yxyR

σ σ σ

σ στ

= + = −

− = +

− = + =




without permission.


2 (2)( 54.399)tan 2

139.882

0.77778

xyp

x y

τθ

σ σ−= =

− −

=

(a) 18.9aθ = ° , 108.9bθ = °

ave 69.941 88.606a Rσ σ= − = − − 158.5 MPaaσ = −

ave 69.941 88.606b Rσ σ= + = − + 18.67 MPabσ =

(b) max Rτ = max 88.6 MPaτ =




without permission.

PROBLEM 7.49


PROBLEM 7.27 For the state of plane stress shown, determine (a) the largest value of xyτ 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

( )10 8, 0 , 0 (1 ksi, 0).2 2

x yσ σ+ −= =

The radius of the circle is max(in-plane) 12 ksi.τ =

The stress point ( , )x xyσ τ− lies along the line 1 2X X of the Mohr circle diagram. The extreme points with 12 ksiR ≤ are 1X and 2.X

(a) The largest allowable value of xyτ is obtained from triangle CDX :

2 2 2 21 2 1DX DX CX CD= = −

2 212 9xyτ = − 7.94 ksixyτ =

(b) The principal stresses are 1 12aσ = + 13.00 ksiaσ =

1 12bσ = − 11.00 ksibσ = −




without permission.

PROBLEM 7.50


PROBLEM 7.28 For the state of plane stress shown, determine the largest value of yσ for which the maximum in-plane shearing stress is equal to or less than 75 MPa.

SOLUTION

60 MPa, ?, 20 MPax y xyσ σ τ= = =

Given:

max

2 2 2 2

75 MPa

2 150 MPa

(2)( ) 40 MPa

150 40 144.6 MPa

xy

R

XY R

DY

XD XY DY

τ

τ

= =

= =

= =

= − = − =

60 144.6y x XDσ σ= + = + 205 MPayσ =




without permission.

PROBLEM 7.51


PROBLEM 7.29 Determine the range of values of xσ for which the maximum in-plane shearing stress is equal to or less than 10 ksi.

SOLUTION

For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is 10 ksi.R =

Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.

1

2

10 ksi

10 ksi

C Y

C Y

=

=

Noting right triangles 1C DY and 2 ,C DY

2 2 2 2 2 2

1 1 1 18 10 6 ksiC D DY C Y C D C D+ = + = =

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 xyσ τ− must lie on the line X1 X2.

Thus, 3 ksi 27 ksixσ≤ ≤




without permission.

PROBLEM 7.52


PROBLEM 7.30 For the state of plane stress shown, determine (a) the value of

xyτ 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′ ′′ so that 12 MPa.xσ = Likewise, point Y lies on line Y Y′ ′′ so that 2 MPa.yσ = The coordinates of C are

2 12 , 0 (7 MPa, 0).2+ =

Counterclockwise rotation through 150° brings line CX to CB, where 0.τ =

12 2

sec 30 sec 30 5.77 MPa2 2

x yRσ σ− −= ° = ° =

(a) tan 302

x yxy

σ στ

−= − °

12 2

tan 302

−= − ° 2.89 MPaxyτ = −

(b) ave 7 5.77a Rσ σ= + = + 12.77 MPaaσ =

ave 7 5.77b Rσ σ= − = − 1.23 MPabσ =




without permission.

PROBLEM 7.53

Solve Problem 7.30, using Mohr’s circle and assuming that the weld forms an angle of 60° with the horizontal.

PROBLEM 7.30 For the state of plane stress shown, determine (a) the value of

xyτ for which the in-plane shearing stress parallel to the weld is zero, (b) the corresponding principal stresses.

SOLUTION

Locate point C at 12 2

7 MPa2

σ += = with 0τ = .

Angle 120XCB = °

12 2

2 25 MPa

x yσ σ− −=

=

5sec60

10 MPa

R = °=

5 tan 60xyτ = − °

8.66 MPaxyτ = −

avea Rσ σ= +

7 10= + 17 MPaaσ =

aveb Rσ σ= −

7 10= − 3 MPabσ = −




without permission.

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.

SOLUTION

Mohr’s circle for 1st stress state.

4 ksixσ =

4 ksiyσ = −

0xyτ =

Resultant stresses:

4 4 8 ksixσ = + =

4 7 3 ksiyσ = − + =

6 0 6 ksixyτ = + =

ave1

( ) 5.5 ksi2 x yσ σ σ= + =

2 (2)(6)

tan 2 2.45

xyp

x y

τθ

σ σ= = =

−

2 67.38pθ = ° 33.69aθ = °

123.69bθ = °

2

2

2 2

2

2.5 6 6.5 ksi

x yxyR

σ στ

− = +

= + =

avea Rσ σ= + 12 ksiaσ =

aveb Rσ σ= − 1 ksibσ = −




without permission.

PROBLEM 7.55


SOLUTION

Mohr’s circle for 2nd stress state:

20 20 cos 60°

30 MPa

20 20 cos 60°

10 MPa

20 sin 60°

17.32 MPa

x

y

xy

σ

σ

τ

= +== −

==

=

Resultant stresses:

35 30 65 MPaxσ = + =

25 10 35 MPayσ = + =

0 17.32 17.32 MPaxyτ = + =

ave1 1

( ) (65 35) 50 MPa2 2x yσ σ σ= + = + =

2 (2)(17.32)

tan 2 1.154765 35

xyp

x y

τθ

σ σ= = =

− −

2 49.11 ,pθ = ° 24.6 , 114.6a bθ θ= ° = °

2

2 22.91 MPa2

x yxyR

σ στ

− = + =

avea Rσ σ= + 72.91 MPaaσ =

aveb Rσ σ= − 27.09 MPabσ =




without permission.

PROBLEM 7.56


SOLUTION

Mohr’s circle for 2nd stress state:

0 0

0 0

0

1 1 cos 2

2 21 1

cos 22 21

sin 22

x

y

xy

σ σ σ θ

σ σ σ θ

τ σ θ

= +

= −

=

Resultant stresses:

0 0 0 0 0

0 0 0 0

0 0

ave 0

0

0 0

1 1 3 1 cos 2 cos 2

2 2 2 21 1 1 1

0 cos 2 cos 22 2 2 21 1

0 sin 2 sin 22 2

1( )

22 sin 2

tan 2cos 2

sin 2tan

1 cos 2

x

y

xy

x y

xyp

x y

σ σ σ σ θ σ σ θ

σ σ σ θ σ σ θ

τ σ θ σ θ

σ σ σ σ

τ σ θθσ σ σ σ θ

θ θθ

= + + = +

= + − = −

= + =

= + =

= =− +

= =+

1

2pθ θ=

2 2 22

0 0 0

2 20 0 0

1 1 1 cos 2 sin 2

2 2 2 2

1 21 2 cos 2 + cos 2 sin 2 1 cos 2 |cos |

2 2

x yxyR

σ στ σ σ θ σ θ

σ θ θ θ σ θ σ θ

− = + = + +

= + + = + =

avea Rσ σ= + 0 0 cos aσ σ σ θ= +

aveb Rσ σ= − 0 0 cos bσ σ σ θ= −




without permission.

PROBLEM 7.57


SOLUTION

Mohr’s circle for 2nd state of stress:

0

0

0x

y

x y

σσ

τ τ

′

′

′ ′

==

=

0 0 0 0

0 0

3 3sin 60 sin 60

2 21

cos602

σ τ τ σ τ τ

τ τ τ

= − ° = − = ° =

= ° =

x y

xy

Resultant stresses:

0 0 0 0

0 0 0

3 3 3 30 0

2 2 2 21 3

2 2

σ τ τ σ τ τ

τ τ τ τ

= − = − = + =

= + =

x y

xy

ave

22 22

0 0 0

1( ) 0

2

3 33

2 2 2

x y

x yxyR

σ σ σ

σ στ τ τ τ

= + =

− = + = + =

3(2)

2 2tan 2 3

3xy

px y

τθ

σ σ

= = = −

− −

2 60pθ = − ° 30 60b aθ θ= − ° = °

avea Rσ σ= + 03aσ τ=

aveb Rσ σ= − 03bσ τ= −




without permission.

PROBLEM 7.58

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

SOLUTION

ave

22

2 2

16 ksi, 0

6 ksi

1( ) 8 ksi

2

2

( 8) (6) 10 ksi

2 (2) (6)tan 2 0.75

16

2 36.870

18.435

x y

xy

x y

x yxy

xyp

x y

p

b

R

σ σ

τ

σ σ σ

σ στ

τθ

σ σ

θ

θ

= − =

=

= + = −

− = +

= − + =

= = = −− −

= − °

= − °

8 ksix yτ ′ ′ ≤ for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle ϕ is

calculated from

8

sin 2 8 sin 2 0.810

R ϕ ϕ= = =

2 53.130 26.565

18.435 26.565 45

18.435 26.565 8.13

90 45

90 98.13

k b

k b

u h

v k

ϕ ϕθ θ ϕθ θ ϕθ θθ θ

= ° = °= − = − ° − ° = − °= + = − + ° = °= + ° = °= + ° = °

Permissible range of θ : h kθ θ θ≤ ≤ 45 8.13θ− ° ≤ ≤ °

θ θ θ≤ ≤u v 45 98.13θ° ≤ ≤ °

Also, 135 188.13 and 225 278.13θ θ° ≤ ≤ ° ° ≤ ≤ °




without permission.

PROBLEM 7.59

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

SOLUTION

ave

22

2 2

90 MPa, 0

60 MPa

1( ) 45 MPa

2

2

45 60 75 MPa

2 (2) ( 60) 4tan 2

90 3

2 53.13

26.565

x y

xy

x y

x yxy

xyp

x y

p

a

R

σ σ

τ

σ σ σ

σ στ

τθ

σ σ

θ

θ

= =

= −

= + =

− = +

= + =

−= = = −−

= − °

= − °

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

cos 2 50 45 5 MPaR ϕ = − =

5

cos 2 0.06666775

ϕ = =

2 86.177 43.089

26.565 43.089 16.524

2 2 360 4 32.524 360 172.355 220.169

110.085

h a

k h

k

ϕ ϕθ θ ϕθ θ ϕθ

= ° = °= + = − ° + ° = °= + ° − = ° + ° − ° = °= °

Permissible range of θ : h kθ θ θ≤ ≤

16.524 110.085θ° ≤ ≤ °

Also, 196.524 290.085θ° ≤ ≤ °




without permission.

PROBLEM 7.60

For the state of stress shown, determine the range of values of θ for which the normal stress

xσ ′ is equal to or less than 100 MPa.

SOLUTION

ave

22

2 2

90 MPa, 0

60 MPa

1( ) 45 MPa

2

2

45 60 75 MPa

2 (2) ( 60) 4tan 2

90 3

2 53.13

26.565

x y

xy

x y

x yxy

xyp

x y

p

a

R

σ σ

τ

σ σ σ

σ στ

τθ

σ σ

θ

θ

= =

= −

= + =

− = +

= + =

−= = = −−

= − °

= − °

100 MPaxσ ′ ≤ for states of stress corresponding to arc HBK of Mohr’s circle. From the circle,

cos2 100 45 55 MPaR ϕ = − =

55

cos 2 0.7333375

ϕ = =

2 42.833 21.417

26.565 21.417 5.15

2 2 360 4 10.297 360 85.666 264.037

132.02

h a

k h

k

ϕ ϕθ θ ϕθ θ ϕθ

= ° = °= + = − ° + ° = − °= + ° − = − ° + ° − ° = °= °

Permissible range of θ is h kθ θ θ≤ ≤

5.15 132.02θ− ° ≤ ≤ °

Also, 174.85 312.02θ° ≤ ≤ °




without permission.

PROBLEM 7.61

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

SOLUTION

ave

20 MPa 120 MPa

1( ) 70 MPa

2

x y

x y

σ σ

σ σ σ

= − = −

= + = −

Set max ave

max ave

60 MPa

130 MPa

R

R

σ σσ σ

= = += − =

But

2

2

22

2 2

2

2

130 50

120 MPa

x xxy

x xxy

R

R

σ σ τ

σ στ

− = +

− = −

= −=

Range of :xyτ 120 MPa 120 MPaxyτ− ≤ ≤




without permission.

PROBLEM 7.62

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

SOLUTION

20MPa 120 MPa

1 ( ) 50 MPa2

x y

x y

σ σ

σ σ

= − = −

− =

Set max (in-plane) 150 MPaRτ = =

But

2

2

2

2

2 2

2

2

150 50

141.4 MPa

x yxy

x yxy

R

R

σ στ

σ στ

− = +

− = −

= −=

Range of :xyτ 141.4 MPa 141.4 MPaxyτ− ≤ ≤




without permission.

PROBLEM 7.63

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

SOLUTION

ave

min ave ave min

2

2

2

2 2 2

114 ksi, 9 ksi, ( ) 11.5 ksi

2

11.5 5 6.5 ksi

2

6.5 2.5 6 ksi2

x y x y

x yxy

x yxy

R R

R

R

σ σ σ σ σ

σ σ σ σ

σ στ

σ στ

= = = + =

= − ∴ = −= − =

− = +

− = ± − = ± − = ±

But it is given that xyτ is positive, thus 6 ksi.xyτ = +

(a) 2

tan 2

(2)(6)2.4

52 67.38

xyp

x y

p

τθ

σ σ

θ

=−

= =

= °

33.7aθ = °

123.7bθ = °

(b) max ave Rσ σ= +

max 18.00 ksiσ =

(c) max(in-plane) Rτ =

max(in-plane) 6.50 ksiτ =




without permission.

PROBLEM 7.64

The Mohr’s circle shown corresponds to the state of stress given in Fig. 7.5a and b. Noting that ( )cos (2x pOC CXσ θ′ ′= + − 2θ ) and that ( )sin (2 2 ),x y pCXτ θ θ′ ′ ′= − derive the expressions for xσ ′ and

x yτ ′ ′ given in Eqs. (7.5) and (7.6), respectively. [Hint: Use sin ( ) sin cos cos sinA B A B A B+ = + and cos ( ) cos cosA B A B+ = −sin sin .]A B

SOLUTION

1( )

2

cos 2 cos 22

sin 2 sin 2

x y

x yp p

p p xy

OC CX CX

CX CX

CX CX

σ σ

σ σθ θ

θ θ τ

′= + =

−′ = =

′ = =

cos (2 2 )

(cos 2 cos 2 sin 2 sin 2 )

cos 2 cos 2 sin 2 sin 2

cos 2 sin 22 2

x p

p p

p p

x y x yxy

OC CX

OC CX

OC CX CX

σ θ θ

θ θ θ θ

θ θ θ θσ σ σ σ

θ τ θ

′ ′= + −

′= + +

′ ′= + +

+ −= + +

sin (2 2 ) (sin 2 cos 2 cos 2 sin 2 )

sin 2 cos 2 cos 2 sin 2

cos 2 sin 22

x y p p p

p p

x yxy

CX CX

CX CX

τ θ θ θ θ θ θ

θ θ θ θσ σ

τ θ θ

′ ′ ′ ′= − = −

′ ′= −

−= −




without permission.

PROBLEM 7.65

(a) Prove that the expression 2 ,x y x yσ σ τ′ ′ ′ ′− where , ,x yσ σ′ ′ and x yτ ′ ′ 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 xyτ in terms of , ,x yσ σ and the principal stresses maxσ and min .σ

SOLUTION

(a) From Mohr’s circle,

ave

ave

sin 2

cos 2

cos 2

x y p

x p

y p

R

R

R

τ θσ σ θσ σ θ

′ ′

′

′

=

= +

= −

2x y x yσ σ τ′ ′ ′ ′−

2 2 2 2 2ave cos 2 sin 2p pR Rσ θ θ= − −

2 2ave ;Rσ= − independent of .pθ

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

2 2 2

2 2 2

2 2ave

2x y x y

OC OK CK

OK OC CK

Rσ

σ σ τ′ ′ ′ ′

= +

= −

= −

= −

(b) Applying above to , ,x yσ σ and xyτ , and to , ,a bσ σ

2 2 2 2avex y xy a b ab Rσ σ τ σ σ τ σ− = − = −

But max min0, ,ab a bτ σ σ σ σ= = =

2max minx y xyσ σ τ σ σ− =

2max min

max min

xy x y

xy x y

τ σ σ σ σ

τ σ σ σ σ

= −

= ± −

The sign cannot be determined from above equation.




without permission.

PROBLEM 7.66

For the state of plane stress shown, determine the maximum shearing stress when (a) 6 ksixσ = and 18 ksiyσ = , (b) 14 ksixσ = and

2 ksiyσ = . (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION

(a)

ave

6ksi, 18 ksi, 8 ksi

1( ) 12 ksi

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

22

2 2

2

6 8 10 ksi

x yxyR

σ στ

− = +

= + =

ave

ave

12 10 22 ksi (max)

12 10 2 ksi

0 (min)

a

b

c

R

R

σ σσ σσ

= + = + == − = − ==

max(in-plane) 10 ksiRτ = =

max max min1

( )2

τ σ σ= − max 11 ksiτ =

(b)

22

2 2

14 ksi, 2 ksi, 8 ksi

1( ) 8 ksi

2

2

6 8 10 ksi

x y xy

a x y

x yxyR

σ σ τ

σ σ σ

σ στ

= = =

= + =

− = +

= + =

ave

ave

18 ksi (max)

2 ksi (min)

0

a

b

c

R

R

σ σσ σσ

= + == − = −=

max

min

max max min

18 ksi

2 ksi

1( )

2

σσ

τ σ σ

== −

= − max 10 ksiτ =




without permission.

PROBLEM 7.67

For the state of plane stress shown, determine the maximum shearing stress when (a) 0xσ = and 12 ksiyσ = , (b) 21 ksixσ = and 9 ksiyσ = . (Hint: Consider both in-plane and out-of-plane shearing stresses.)

SOLUTION

(a)

ave

0, 12 ksi, 8 ksi

1( )

26 ksi

x y xy

x y

σ σ τ

σ σ σ

= = =

= +

=

22

2 2

2

( 6) 8

10 ksi

x yxyR

σ στ

− = +

= − +

=

ave

ave

16 ksi (max)

4 ksi (min)

0

a

b

c

R

R

σ σσ σσ

= + == − = −=

max min16 ksi 4 ksiσ σ= = −

max max min1

( )2


(b)

ave

22

2 2

21 ksi 9 ksi 8 ksi

15 ksi

2

( 6) 8 10 ksi

x y xy

x yxyR

σ σ τ

σ

σ στ

= = =

=

− = +

= − + =

ave

ave

25 ksi (max)

5 ksi

0 (min)

a

b

c

R

R

σ σσ σσ

= + == − ==

max min

max max min

25 ksi, 0

1( )

2

σ σ

τ σ σ

= =

= − max 12.5 ksiτ =




without permission.

PROBLEM 7.68

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

SOLUTION

(a)

ave

140 MPa, 40 MPa, 80 MPa

1( ) 90 MPa

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

2

2 2 250 80 94.34 MPa2

x yxyR

σ στ

− = + = + =

ave

ave

184.34 MPa (max)

4.34 MPa (min)

0

a

b

c

R

R

σ σσ σσ

= + == − = −=

max(in-plane)1

( ) 94.34 MPa2 a b Rτ σ σ= − = =

max max min1 1

( ) ( ) 94.3 MPa2 2 a bτ σ σ σ σ= − = − = max 94.3 MPaτ =

(b)

ave

140 MPa, 120 MPa, 80 MPa

1( ) 130 MPa

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

2

2 2 210 80 80.62 MPa2

x yxyR

σ στ

− = + = + =

ave

ave

210.62 MPa (max)

49.38 MPa

0 (min)

a

b

c

R

R

σ σσ σσ

= + == − ==

max min210.62 MPa 0a cσ σ σ σ= = = =

max(in-plane) 86.62 MPaRτ = =

max max min1

( ) 105.3 MPa2

τ σ σ= − = max 105.3 MPaτ =




without permission.

PROBLEM 7.69

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

SOLUTION

(a)

ave

140 MPa, 20 MPa, 80 MPa

1( ) 80 MPa

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

2

2 2 260 80 100 MPa2

x yxyR

σ στ

− = + = + =

ave

ave

80 100 180 MPa (max)

80 100 20 MPa (min)

0

a

b

c

R

R

σ σσ σσ

= + = + == − = − = −=

max(in-plane)1

( ) 100 MPa2 a bτ σ σ= − =

max max min1

( ) 100 MPa2

τ σ σ= − = max 100 MPaτ =

(b)

ave

140 MPa, 140 MPa, 80 MPa

1( ) 140 MPa

2

x y xy

x y

σ σ τ

σ σ σ

= = =

= + =

2

2 20 80 80 MPa2

x yxyR

σ στ

− = + = + =

ave

ave

220 MPa (max)

60 MPa

0 (min)

a

b

c

R

R

σ σσ σσ

= + == − ==

max (in-plane)1

( ) 80 MPa2 a bτ σ σ= − =

max max min1

( ) 110 MPa2

τ σ σ= − = max 110 MPaτ =




without permission.

PROBLEM 7.70

For the state of stress shown, determine the maximum shearing stress when (a) 4zσ = + ksi, (b) 4zσ = − ksi, (c) zσ = 0.

SOLUTION

7 ksi, 2 ksi, 6ksix y xyσ σ τ= = = −

ave

2

2

2 2

ave

ave

1( ) 4.5 ksi

2

2

2.5 ( 6) 6.5 ksi

11 ksi

2 ksi

x y

x yxy

a

b

R

R

R

σ σ σ

σ στ

σ σσ σ

= + =

− = +

= + − == + == − = −

(a) 4 ksi, 11 ksi, 2 ksiz a bσ σ σ= = = −

max min max max min1

11 ksi, 2 ksi, ( )2

σ σ τ σ σ= = − = − max 6.5 ksiτ =

(b) 4 ksi, 11 ksi, 2 ksiz a bσ σ σ= − = = −

max min11 ksi, 4 ksi,σ σ= = − max max min1

( )2

τ σ σ= − max 7.5 ksiτ =

(c) 0, 11 ksi, 2 ksiz a bσ σ σ= = = −

max 11 ksi,σ = min 2 ksi,σ = − max max min1

( )2

τ σ σ= − max 6.5 ksiτ =




without permission.

PROBLEM 7.71

For the state of stress shown, determine the maximum shearing stress when (a) 4zσ = + ksi, (b) 4zσ = − ksi, (c) 0zσ = .

SOLUTION

5 ksi, 10 ksi, 6 ksix y xyσ σ τ= = = −

ave

2

2

2 2

ave

ave

1( ) 7.5 ksi

2

2

( 2.5) ( 6) 6.5 ksi

14 ksi

1 ksi

x y

x yxy

a

b

R

R

R

σ σ σ

σ στ

σ σσ σ

= + =

− = +

= − + − == + == − =

(a) 4 ksi, 14 ksi, 1 ksiz a bσ σ σ= + = =


14 ksi, 1 ksi, ( )2

σ σ τ σ σ= = = − max 6.5 ksiτ =

(b) 4 ksi, 14 ksi, 1 ksiz a bσ σ σ= − = =

max min14 ksi, 4 ksi,σ σ= = − max max min1

( )2


(c) 0, 14 ksi, 1 ksiz a bσ σ σ= = =

max 14 ksi,σ = min 0,σ = max max min1

( )2





without permission.

PROBLEM 7.72

For the state of stress shown, determine the maximum shearing stress when (a) 0,zσ = (b) 45zσ = + MPa, (c) 45 MPa.zσ = −

SOLUTION

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

ave

2

2

2 2

ave

ave

1( ) 60 MPa

2

2

40 75 85 MPa

145 MPa

25 MPa

x y

x yxy

a

b

R

R

R

σ σ σ

σ στ

σ σσ σ

= + =

− = +

= + == + == − = −

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


145 MPa, 25 MPa, ( )2

σ σ τ σ σ= = − = − max 85 MPaτ =

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

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

( )2

τ σ σ= − max 85 MPaτ =

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

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

( )2





without permission.

PROBLEM 7.73

For the state of stress shown, determine the maximum shearing stress when (a) 0,zσ = (b) 45zσ = + MPa, (c) 45 MPa.zσ = −

SOLUTION

150 MPa, 70 MPa, 75 MPax y xyσ σ τ= = =

ave

2

2

2 2

ave

ave

1( ) 110 MPa

2

2

40 75 85 MPa

195 MPa

25 MPa

x y

x yxy

a

b

R

R

R

σ σ σ

σ στ

σ σσ σ

= + =

− = +

= + == + == − =

(a) 0, 195 MPa, 25 MPaz a bσ σ σ= = =


195 MPa, 0, ( )2

σ σ τ σ σ= = = − max 97.5 MPaτ =

(b) 45 MPa, 195 MPa, 25 MPaz a bσ σ σ= + = =

max min195 MPa, 25 MPa,σ σ= = max max min1

( )2


(c) 45 MPa, 195 MPa, 25 MPaz a bσ σ σ= − = =

max 195 MPa,σ = min 45 MPa,σ = − max max min1

( )2





without permission.

PROBLEM 7.74

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

SOLUTION

max14ksi, 8 ksi, 10 ksix xyσ τ τ= = =

Let 22

y xy xu u

σ σσ σ

−= = +

ave1

( )2 x y x uσ σ σ σ= + = +

2 2 2 2xy xyR u u Rτ τ= + = ± −

Case (1) max 10 ksi, 6 ksiR uτ = = = ±

(1a) 6ksi 2 26 ksi (reject)y xu uσ σ= + = + =

ave ave ave1

( ) 20 ksi, 30 ksi, 10 ksi2 x y a bR Rσ σ σ σ σ σ σ= + = = + = = − =


30 ksi, 0, ( ) 15 ksi 7.5 ksi2

σ σ τ σ σ= = = − = ≠

(1b) 6 ksi 2 2 ksiy xu uσ σ= − = + =

ave ave ave1

( ) 8 ksi, 18 ksi, 2 ksi2 x y a bR Rσ σ σ σ σ σ σ= + = = + = = − = −


18 ksi, 2 ksi, ( ) 10 ksi (o.k.)2

σ σ τ σ σ= = − = − =

2.00 ksiyσ =




without permission.


Case (2) minAssume 0.σ = max max2 20 ksi = aσ τ σ= =

2 2avea x xyR u uσ σ σ τ= + = + + +

2 2a x xyu uσ σ τ− − = +

2 2 2( )a x xyu uσ σ τ− − = +

2 2 2 2( ) 2( )a x a x xyu u uσ σ σ σ τ− − − + = +

2 2 2 2( ) (20 14) 8

2 4.6667 ksi20 14

a x xy

a x

uσ σ τ

σ σ− − − −= = = −

− −

2.3333 ksi 2 9.3333 ksiy xu uσ σ= − = + =

2 2ave

1( ) 11.6667 ksi 8.3333 ksi

2 x y xyR uσ σ σ τ= + = = + =

ave 20 ksia Rσ σ= + = ave 3.3334 ksib Rσ σ= − =

max min max20 ksi, 0, 10 ksi σ σ τ= = =

9.33 ksiyσ =




without permission.

PROBLEM 7.75

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

SOLUTION

50 MPa, 48 MPax xyσ τ= − =

Let 22

y xy xu u

σ σσ σ

−= = +

ave1

( )2 x y x uσ σ σ σ= + = +

2 2 2 2xy xyR u u Rτ τ= + = ± −

Case (1) 2 2max 73 MPa, 73 48 55 MPaR uτ = = = ± − = ±

(1a) 55 MPa 2 60 MPay xu uσ σ= + = + =

ave

ave ave

1( ) 5 MPa

278 MPa, 68 MPa

x y

a bR R

σ σ σ

σ σ σ σ

= + =

= + = = − = −

max min max0 78 MPa, 68 MPa, 73 MPaaσ σ σ τ= = = − =

(1b) 55 MPa 2 160 MPa (reject)y xu uσ σ= − = + = −

ave ave

ave max

1( ) 105 MPa, 32 MPa

2178 MPa, 0, 0

x y a

b c

R

R

σ σ σ σ σ

σ σ σ σ

= + = − = + = −

= − = − = =

min max max min1

178 MPa, ( ) 89 MPa 73 MPa2

σ τ σ σ= − = − = ≠

60.0 MPayσ =




without permission.


Case (2) maxAssume 0.σ = max max min1

( ) 73 MPa2

τ σ σ= − =

min 146 MPa bσ σ= − =

2 2aveb x xyR u uσ σ σ τ= − = + − +

2 2xy x bu uτ σ σ+ = − + −

2 2 2 2( ) 2( )τ σ σ σ σ+ = − + − +xy x b x bu u u

2 2 2 2( ) (48) ( 50 146)

2 72 MPa50 146

xy x b

x b

uτ σ σ

σ σ− − − − += = = −

− − +

36 MPa 2 122 MPaσ σ= − = + = −y xu u

2 2 60 MPaxyR u τ= + =

2 146 120 26 MPaa b Rσ σ= + = − + = − (o.k.)

122.0 MPayσ = −




without permission.

PROBLEM 7.76

For the state of plane stress shown, determine the value of xyτ for which the maximum shearing stress is (a) 10 ksi, (b) 8.25 ksi.

SOLUTION

z

ave

15 ksi 3 ksi 0

1( ) 9 ksi

2

x y

x y

σ σ σ

σ σ σ

= − = − =

= + = −

(a) max 10 ksiτ = :

If maxτ is the in-plane maximum shearing stress, then

max

max ave

min ave

10 ksi1 ksi

19 ksi

RRR

τσ σσ σ

= == + == − = −

zσ is the intermediate principal stress; the assumption maxR τ= is correct.

2

2 2 2

2 2

6 102

10 6

x yxy xy

xy

Rσ σ

τ τ

τ

− = + = + =

= − 8 ksixyτ =

(b) max 8.25 ksi.τ =

If maxτ is the in-plane maximum shearing stress, then

max 8.25 ksi.R τ= =

max ave

min ave

9 8.25 0.75 ksi

9 8.25 17.25 ksi

R

R

σ σσ σ

= + = − + = −= − = − − = −

0zσ = is not the intermediate principal stress.

Let max0 .zσ σ= =

min max max

ave min

2 0 (2)(8.25) 16.5 ksi

9 ( 16.5) 7.5 ksiR

σ σ τσ σ

= − = − = −= − = − − − =

2

2 2 2

2 2

6 7.5 ksi2

7.5 6 4.5 ksi

x yxy xy

xy

Rσ σ

τ τ

τ

− = + = + =

= − = 4.5 ksixyτ =




without permission.

PROBLEM 7.77

For the state of stress shown, determine the value of xyτ for which the

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

SOLUTION

ave

100 MPa, 40 MPa, 0

1( ) 70 MPa

2

x y z

x y

σ σ σ

σ σ σ

= = =

= + =

(a) max 60 MPa.τ =

If zσ is min ,σ then max min max2 .σ σ τ= +

max

max ave

max ave

max

0 (2)(60) 120 MPa

120 70 50 MPa

2 20 MPa > 0b

R

R

R

σσ σ

σ σσ σ

= + == += − = − == − =

2

2 2 2

2 2

30 50 MPa2

50 30

x yxy xy

xy

Rσ σ

τ τ

τ

− = + = + =

= − 40 MPaxyτ =

(b) max 78 MPa.τ =

If zσ is min ,σ then max min max2 0 (2)(78) 156 MPa.σ σ τ= + = + =

max ave

max ave max156 70 86 MPa > 78 MPa

R

R

σ σσ σ τ

= += − = − = =

Set

max min ave78 MPa. 8 MPa < 0R Rτ σ σ= = = − = −

2

2 2 2

2 2

302

78 30

x yxy xy

xy

Rσ σ

τ τ

τ

− = + = +

= − 72 MPaxyτ =




without permission.

PROBLEM 7.78

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

SOLUTION

90 MPa, 0, 60 MPax z xzσ σ τ= = =

Mohr’s circle of stresses in zx-plane:

ave

22 2 2

1( ) 45 MPa

2

45 60 75 MPa2

x z

x yzxR

σ σ σ

σ στ

= + =

− = + = + =

ave ave120 MPa, 30 MPaa bR Rσ σ σ σ= + = = − = −

Assume max 120 MPa.aσ σ= =

min max max2

120 (2) (80)

yσ σ σ τ= = −

= − 40 MPayσ = −

Assume min 30 MPa.bσ σ= = −

max min max2

30 (2)(8)

yσ σ σ τ= = +

= − + 130 MPayσ =




without permission.

PROBLEM 7.79

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

SOLUTION

60 MPa, 0, 100 MPax z yσ σ σ= = =

For Mohr’s circle of stresses in zx-plane

ave

1( ) 30 MPa

2

30 MPa2

σ σ σ

σ σ

= + =

−= =

x z

x zu

Assume max 100 MPayσ σ= =

min max max

ave

ave

2 2

2 2

2 2

2

100 (2) (60) 20 MPa

30 ( 20) 50 MPa

30 50 80 MPa <

50 30 40 MPa

b

b

a

y

xz

xz

R

R

R u

R u

σ σ σ τ

σ σ

σ σσ

τ

τ

= = −= − = −= −= − − == += + =

= +

= ± −

= ± − = ±

40 MPa 40 MPaxzτ− ≤ ≤




without permission.

PROBLEM 7.80*

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

SOLUTION

Let 22

x yy xu u

σ σσ σ

−= = −

ave

2 2

2 2ave

2 2ave

1( )

2 x y x

xy

a x xy

b x xy

u

R u

R u u

R u u

σ σ σ σ

τ

σ σ σ τ

σ σ σ τ

= + = −

= +

= + = − + +

= − = − − +

Assume maxτ is the in-plane shearing stress. max Rτ =

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

ave

ave

ave

max min max max min

2 140 MPa, 140 MPa

80 MPa

140 80 220 MPa

140 80 60 MPa

1220 MPa, 0, ( ) 110 MPa

2

y x x x

xy

a

b

u u

R

R

R

σ σ σ σ σ

τ

σ σσ σ

σ σ τ σ σ

= − = = = − =

= =

= + = + == − = − =

= = = − =

Assumption is incorrect.

Assume 2 2max ave

min max max min

2 2

1 10 ( )

2 2

1 0 (no minimum)

a x xy

a

a

xy

R u u

d u

du u

σ σ σ σ τ

σ τ σ σ σ

σ

τ

= = + = − + +

= = − =

= − + ≠+




without permission.

PROBLEM 7.80* (Continued)

Optimum value for u occurs when max (out-of-plane) max (in-plane)τ τ=

2 2

2 2

1( ) or or

2

( ) 2

a a x xy

x x x

R R R u u

u u u

σ σ σ τ

σ σ σ

+ = = − = +

− = − +2

u=2 2

xyτ+

2 2 2 2140 80

2 94.3 MPa140

x xy

x

uσ τ

σ− −= = = 47.14 MPau =

(a) 2 140 94.3y x uσ σ= − = − 45.7 MPayσ =

(b) 2 2max 92.9 MPaxyR u τ τ= + = = max 92.9 MPaτ =




without permission.

PROBLEM 7.81

The state of plane stress shown occurs in a machine component made of a steel with 325 MPa.Yσ = Using the maximum-distortion-energy criterion, determine

whether yield will occur when (a) 0 200 MPa,σ = (b) 0 240 MPa,σ =

(c) 0 280 MPa.σ = If yield does not occur, determine the corresponding factor of

safety.

SOLUTION

2

2ave 0 100 MPa

2x y

xyRσ σ

σ σ τ−

= − = + =

(a) 0 ave200 MPa 200 MPaσ σ= = −

ave ave100 MPa, 300 MPaa bR Rσ σ σ σ= + = − = − = −

2 2 264.56 MPa < 325 MPaa b a bσ σ σ σ+ − = (No yielding)

325

. .264.56

F S = . . 1.228F S =

(b) 0 ave240 MPa 240 MPaσ σ= = −


2 2 295.97 MPa < 325 MPaa b a bσ σ σ σ+ − = (No yielding)

325

. .295.97

F S = . . 1.098F S =

(c) 0 ave280 MPa 280 MPaσ σ= = −


2 2 329.24 MPa > 325 MPaa b a bσ σ σ σ+ − = (Yielding occurs)




without permission.

PROBLEM 7.82

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 325 MPa.Yσ = Using the maximum-distortion-energy

criterion, determine whether yield will occur when (a) 0 200 MPa,σ =

(b) 0 240 MPa,σ = (c) 0 280 MPa.σ = If yield does occur, determine the corresponding factor of safety.

SOLUTION

2

2ave 0 100 MPa

2x y

xyRσ σ

σ σ τ−

= − = + =

(a) 0 ave200 MPa, 200 MPaσ σ= = −

ave ave100 MPa 300 MPaa bR Rσ σ σ σ= + = − = − = −

max min0, 300 MPaσ σ= = −

max max min2 300 MPa 325 MPaτ σ σ= − = < (No yielding)

325

. .300

F S = . . 1.083F S =

(b) 0 ave240 MPa, 240 MPaσ σ= = −



max max min2 340 MPa > 325 MPaτ σ σ= − = (Yielding occurs)

(c) 0 ave280 MPa, 280 MPaσ σ= = −



max max min2 380 MPa 325 MPaτ σ σ= − = > (Yielding occurs)




without permission.

PROBLEM 7.83

The state of plane stress shown occurs in a machine component made of a steel with 45 ksi.Yσ = Using the maximum-distortion-energy criterion, determine whether

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

does not occur, determine the corresponding factor of safety.

SOLUTION

36 ksi, 21 ksi, 0x y zσ σ σ= = =

For stresses in xy-plane, ave1

( ) 28.5 ksi2 x yσ σ σ= + =

7.5 ksi2

x yσ σ−=

(a) 2

2 2 2

ave ave

9 ksi (7.5) (9) 11.715 ksi2

40.215 ksi, 16.875 ksi

x yxy xy

a b

R

R R

σ στ τ

σ σ σ σ

− = = + = + =

= + = = − =

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

45

. .39.977

F S = . . 1.287F S =

(b) 2

2 2 2

ave ave

18 ksi (7.5) (18) 19.5 ksi2

48 ksi, 9 ksi

x yxy xy

a b

R

R R

σ στ τ

σ σ σ σ

− = = + = + =

= + = = − =

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

45

. .44.193

F S = . . 1.018F S =

(c) 2

2 2 2

ave ave

20 ksi (7.5) (20) 21.36 ksi2

49.86 ksi, 7.14 ksi

x yxy xy

a b

R

R R

σ στ τ

σ σ σ σ

− = = + = + =

= + = = − =

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




without permission.

PROBLEM 7.84

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

PROBLEM 7.83 The state of plane stress shown occurs in a machine component made of a steel with 45 ksi.Yσ = Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) 9 ksi,xyτ = (b) 18 ksi,xyτ = (c)

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

SOLUTION

36 ksi, 21 ksi, 0x y zσ σ σ= = =

For stress in xy-plane, ave1

( ) 28.5 ksi 7.5 ksi2 2

x yx y

σ σσ σ σ

−= + = =

(a) 2

2

ave ave

max min34.977 ksi,

9 ksi 11.715 ksi2

40.215 ksi, 16.875 ksi

0

x yxy xy

a b

R

R R

τ σ

σ στ τ

σ σ σ σ

− = = + =

= + = = − == =

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

45

. .40.215

F S = . . 1.119F S =

(b)

max

22

ave ave

min48 ksi

18 ksi 19.5 ksi2

48 ksi, 9 ksi

0

x yxy xy

a b

R

R R

σ

σ στ τ

σ σ σ σσ

− = = + =

= + = = − == =

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

(c)

max

22

ave ave

min49.86 ksi

20 ksi 21.36 ksi2

49.86 ksi 7.14 ksi

0

x yxy xy

a b

R

R R

τ σ

σ στ τ

σ σ σ σ

− = = + =

= + = = − == =

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




without permission.

PROBLEM 7.85

The 36-mm-diameter shaft is made of a grade of steel with a 250-MPa tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when 200 kN.P =

SOLUTION

3 3

2 3 2 3 2

36

3

ave

1200 kN 200 10 N 18 mm 18 10 m

2

(18 10 ) 1.01788 10 m

200 10196.488 10 Pa

1.01788 10196.488 MPa

1 10 ( ) 98.244 MPa

2 2

y

x x y y

P c d

A c

P

A

π π

σ

σ σ σ σ σ

−

− −

−

= = × = = = ×

= = × = ×

×= − = − = ××

=

= = + = =

22 2 2

ave

ave

(98.244)2

(positive)

(negative)

2

x yxy xy

a

b

a b a b a a b b

R

R

R

R

σ στ τ

σ σσ σσ σ σ σ σ σ σ σ

− = + = +

= += −

− = − > − >

Maximum shear stress criterion under the above conditions:

2 250 MPa 125 MPaa b YR Rσ σ σ− = = = =

Equating expressions for R,

2 2

2 2 6

125 (98.244)

(125) (98.244) 77.286 MPa 77.286 10 Pa

xy

xy

τ

τ

= +

= − = = ×

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

xy

J c

Tc

J

π π

τ

− −= = × = ×

=

9 6

3

(164.846 10 )(77.286 10 )

18 10708 N m

xyJT

c

τ −

−× ×= =

×= ⋅ 708 N mT = ⋅




without permission.

PROBLEM 7.86

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

PROBLEM 7.85 The 36-mm-diameter shaft is made of a grade of steel with a 250-MPa tensile yield stress. Using the maximum-shearing-stress criterion, determine the magnitude of the torque T for which yield occurs when 200 kN.P =

SOLUTION

3 3

2 3 2 3 2

36

3

ave

1200 kN 200 10 N 18 mm 18 10 m

2

(18 10 ) 1.01788 10 m

200 10196.488 10 Pa

1.01788 10196.448 MPa

1 10 ( ) 98.244 MPa

2 2

y

x x y y

P c d

A c

P

A

π π

σ

σ σ σ σ σ

−

− −

−

= = × = = = ×

= = × = ×

×= − = − = ××

=

= = + = =

2

2 2 2

ave ave

(98.244)2

x yxy xy

a b

R

R R

σ στ τ

σ σ σ σ

− = + = +

= + = −

Distortion energy criterion:

2 2 2

2 2 2ave ave ave ave

2 2 2ave

2 2 2 2

( ) ( ) ( )( )

3

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

89.242 MPa

a b a b Y

Y

Y

xy

xy

R R R R

R

σ σ σ σ σ

σ σ σ σ σ

σ σ

ττ

−

+ − =

+ + − − + − =

+ =

+ + =

= ±

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

xy

J c

Tc

J

π π

τ

− −= = × = ×

=

9 6

3

(164.846 10 )(89.242 10 )

18 10818 N m

xyJT

c

τ −

−× ×= =

×= ⋅ 818 N mT = ⋅




without permission.

PROBLEM 7.87

The 1.75-in.-diameter shaft AB is made of a grade of steel for which the yield strength is 36 ksi.Yσ = Using the maximum-shearing-stress criterion, determine the magnitude of the force P for which yield occurs when

15 kip in.T = ⋅

SOLUTION

Let the x-axis lie along the shaft axis.

0, 0

,

y z

x xyP Tc

A J

σ σ

σ τ

= =

= =

Section properties: 1

0.875 in.2

c d= =

2 2 2 4 4 4(0.875) 2.4053 in , (0.875) 0.92077 in2 2

A c J cπ ππ π= = = = = =

From torsion, (15)(0.875)

14.254 ksi0.92077xy

Tc

Jτ = = =

From Mohr’s circle, ave1 1

( )2 2x y xσ σ σ σ= + =

2 22 2

22

ave

22

ave

2 4

1

2 4

1

2 4

x y xxy xy

xa x xy

xb x xy

R

R

R

σ σ στ τ

σσ σ σ τ

σσ σ σ τ

− = + = +

= + = + +

= − = − +

Maximum shearing stress criterion:

max min

22

max max min

2 2 2 2

2 2

2 24

4 4

(36) 4(14.254) 21.984 ksi

a b

xa b xy Y

x xy Y x Y xy

x

σ σ σ σ

στ σ σ σ σ τ σ

σ τ σ σ σ τ

σ

= =

= − = − = + =

+ = = −

= − =

(21.984) (2.4053)xP Aσ= = 52.9 kipsP =




without permission.

PROBLEM 7.88

Solve Prob. 7.87, using the maximum-distortion-energy criterion.

PROBLEM 7.87 The 1.75-in.-diameter shaft AB is made of a grade of steel for which the yield strength is 36 ksi.Yσ = Using the maximum-shearing-stress criterion, determine the magnitude of the force P for which yield occurs when 15 kip in.T = ⋅

SOLUTION

Let the x-axis lie along the shaft axis.

0, 0

,

y z

x xyP Tc

A J

σ σ

σ τ

= =

= =

Section properties: 1

0.875 in.2

c d= =

2 2 2 4 4 4(0.875) 2.4053 in , (0.875) 0.92077 in2 2

A c J cπ ππ π= = = = = =

From torsion, (15)(0.875)

14.254 ksi0.92077xy

Tc

Jτ = = =

From Mohr’s circle, ave1 1

( )2 2x y xσ σ σ σ= + =

2 22 2

ave

ave

2 4x y x

xy xy

a

b

R

R

R

σ σ στ τ

σ σσ σ

− = + = +

= += −

Distortion energy criterion:

2 2 2

2 2 2ave ave ave ave

2 2 2ave

( ) ( ) ( )( )

3

a b a b Y

Y

Y

R R R R

R

σ σ σ σ σ

σ σ σ σ σ

σ σ

+ − =

+ + − − + − =

+ =

2 2

2 2 2 23 32 4

x xxy x xy Y

σ σ τ σ τ σ + + = + =

2 2

2 2

3

(36) (3)(14.254)

26.200 ksi

(26.200)(2.4053)

x Y xy

x

xP A

σ σ τ

σ

σ

= −

= −== = 63.0 kipsP =




without permission.

PROBLEM 7.89

The state of plane stress shown is expected to occur in an aluminum casting. Knowing that for the aluminum alloy used 80 MPaUTσ = and

200 MPaUCσ = and using Mohr’s criterion, determine whether rupture of the casting will occur.

SOLUTION

ave

2

2 2 2

10 MPa,

100 MPa,

60 MPa

10 10045 MPa

2 2

(55) (60) 81.39 MPa2

x

y

xy

x y

x yxyR

σστ

σ σσ

σ στ

== −

=

+ −= = = −

− = + = + =

ave

ave

45 81.39 36.39 MPa

45 81.39 126.39 MPaa

b

R

R

σ σσ σ

= + = − + == − = − − = −

Equation of 4th quadrant of boundary:

1

36.39 ( 126.39)1.087 1

80 200

a b

UT UC

σ σσ σ

− =

−− = >

Rupture will occur.




without permission.

PROBLEM 7.90

The state of plane stress shown is expected to occur in an aluminum casting. Knowing that for the aluminum alloy used 80 MPaUTσ = and

200 MPaUCσ = and using Mohr’s criterion, determine whether rupture of the casting will occur.

SOLUTION

ave

2

2 2 2

32 MPa,

0,

75 MPa

1( ) 16 MPa

2

(16) (75) 76.69 MPa2

x

y

xy

x y

x yxyR

σστ

σ σ σ

σ στ

= −=

=

= + = −

− = + = + =

ave

ave

16 76.69 60.69 MPa

16 76.69 92.69 MPaa

b

R

R

σ σσ σ

= + = − + == − = − − = −


1

60.69 ( 92.69)1.222 1

80 200

a b

UT UC

σ σσ σ

− =

−− = >

Rupture will occur.




without permission.

PROBLEM 7.91

The state of plane stress shown is expected to occur in an aluminum casting. Knowing that for the aluminum alloy used 10 ksiUTσ = and 30 ksiUCσ = and using Mohr’s criterion, determine whether rupture of the casting will occur.

SOLUTION

ave

2

2 2 2

8 ksi,

0,

7 ksi

1( ) 4 ksi

2

4 7 8.062 ksi2

x

y

xy

x y

x yxyR

σστ

σ σ σ

σ στ

= −=

=

= + = −

− = + = + =

ave

ave

4 8.062 4.062 ksi

4 8.062 12.062 ksia

b

R

R

σ σσ σ

= + = − + == − = − − = −


1

4.062 ( 12.062)0.808 1

10 30

a b

UT UC

σ σσ σ

− =

−− = <

No rupture.




without permission.

PROBLEM 7.92

The state of plane stress shown is expected to occur in an aluminum casting. Knowing that for the aluminum alloy used 10 ksiUTσ = and 30 ksiUCσ = and using Mohr’s criterion, determine whether rupture of the casting will occur.

SOLUTION

ave

2

2 2 2

2 ksi,

15 ksi,

9 ksi

1( ) 6.5 ksi

2

8.5 9 12.379 ksi2

x

y

xy

x y

x yxyR

σστ

σ σ σ

σ στ

== −

=

= + = −

− = + = + =

ave

ave

5.879 ksi

18.879 ksia

b

R

R

σ σσ σ

= + == − = −


1

5.879 ( 18.879)1.217 1

10 30

a b

UT UC

σ σσ σ

− =

−− = >

Rupture will occur.




without permission.

PROBLEM 7.93

The state of plane stress shown will occur at a critical point in an aluminum casting that is made of an alloy for which 10 ksiUTσ = and 25 ksi.UCσ = Using Mohr’s criterion, determine the shearing stress 0τ for which failure should be expected.

SOLUTION

0

ave

8 ksi,

0,

1( ) 4 ksi

2

x

y

xy

x y

σστ τ

σ σ σ

==

=

= + =

2

2 2 20

2 20

ave

ave

42

4

(4 ) ksi

(4 ) ksi

x yxy

a

b

R

R

R R

R R

σ στ τ

τσ σσ σ

− = + = +

= ± −= + = += − = −

Since ave < ,Rσ stress point lies in 4th quadrant. Equation of 4th quadrant boundary is

1

4 41

10 25

a b

UT UC

R R

σ σσ σ

− =

+ −− =

2 20

1 1 4 41

10 25 10 25

5.429 ksi 5.429 4

R

R τ

+ = − +

= = ± − 0 3.67 ksiτ = ±




without permission.

PROBLEM 7.94

The state of plane stress shown will occur at a critical point in a pipe made of an aluminum alloy for which 75 MPaUTσ = and 150 MPa.UCσ = Using Mohr’s criterion, determine the shearing stress 0τ for which failure should be expected.

SOLUTION

0

ave

2

2 2 20

80 MPa,

0,

1( ) 40 MPa

2

40 MPa2

x

y

xy

x y

x yxyR

σστ τ

σ σ σ

σ στ τ

= −=

= −

= + = −

− = + = +

ave

ave

2 20 40

a

b

R

R

R

σ σσ σ

τ

= += −

= ± −

Since ave < ,Rσ stress point lies in 4th quadrant. Equation of 4th quadrant boundary is

1

40 401

75 150

a b

UT UC

R R

σ σσ σ

− =

− + − −− =

40 40

1 1.266775 150 75 150

R R+ = + − =

2 2063.33 MPa, 63.33 40R τ= = ± − 0 8.49 MPaτ = ±




without permission.

PROBLEM 7.95

The cast-aluminum rod shown is made of an alloy for which 60 MPaUTσ = and 120 MPa.UCσ = Using Mohr’s criterion,

determine the magnitude of the torque T for which failure should be expected.

SOLUTION

3 2 2 6 2

36

6

26 10 N (32) 804.25 mm 804.25 10 m4

26 1032.328 10 Pa 32.328 MPa

804.25 10x

P A

P

A

π

σ

−

−

= × = = = ×

×= = = × =×

ave

1 1( ) (32.328 0) 16.164 MPa

2 2

1(32.328 0) 16.164 MPa

2 2

x y

x y

σ σ σ

σ σ

= + = + =

−= − =

ave

ave

16.164 MPa

16.164 MPaa

b

R R

R R

σ σσ σ

= + = += − = −

Since ave < ,Rσ stress point lies in the 4th quadrant. Equation of the 4th quadrant is

16.164 16.164

1 160 120

a b

UT UC

R Rσ σσ σ

+ −− = − =

1 1 16.164 16.164

160 120 60 120

R + = − +

34.612 MPaR =

2

2 2 2 2

6

34.612 16.164 30.606 MPa2 2

30.606 10 Pa

x y x yxy xyR R

σ σ σ στ τ

− − = + = − = − =

= ×

For torsion, 3

2xy

Tc T

J cτ

π= = where 31

16 mm 16 10 m2

c d −= = = ×

3 3 3 6(16 10 ) (30.606 10 )2 2xyT cπ πτ −= = × × 196.9 N mT = ⋅




without permission.

PROBLEM 7.96

The cast-aluminum rod shown is made of an alloy for which 70 MPaUTσ = and 175 MPa.UCσ = Knowing that the magnitude T

of the applied torques is slowly increased and using Mohr’s criterion, determine the shearing stress 0τ that should be expected at rupture.

SOLUTION

0

ave

2

2 2

ave

ave

0

0

1( ) 0

2

02

x

y

xy

x y

x yxy xy xy

a

b

R

R R

R R

σστ τ

σ σ σ

σ στ τ τ

σ σσ σ

==

= −

= + =

− = + = + =

= + == − = −

Since ave < ,Rσ stress point lies in 4th quadrant. Equation of boundary of 4th quadrant is

1

170 175

1 11

70 175

50 MPa

a b

UT UC

R R

R

R

σ σσ σ

− =

−− =

+ =

=

0 Rτ = 0 50.0 MPaτ =




without permission.

PROBLEM 7.97

A machine component is made of a grade of cast iron for which 8 ksiUTσ = and 20 ksi.UCσ = For each of the states of stress shown, and using Mohr’s criterion, determine the normal stress 0σ at which

rupture of the component should be expected.

SOLUTION

(a) 0

01

2

a

b

σ σ

σ σ

=

=

Stress point lies in 1st quadrant.

0a UTσ σ σ= = 0 8 ksiσ =

(b) 0

01

2

a

b

σ σ

σ σ

=

= −

Stress point lies in 4th quadrant. Equation of 4th quadrant boundary is

1

00 2

1

18 20

a b

UT UC

σ σσ σ

σσ

− =

−− = 0 6.67 ksiσ =

(c) 0 01

, ,2a bσ σ σ σ= = − 4th quadrant

1

0 02 18 20

σ σ−− = 0 8.89 ksiσ =




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 Pa

2 2(0.006 m)

dd t r t

pr

tσ

= = = = − =

×= = = ×

72.7 MPaσ =

max

min

72.742 MPa

0 (Neglectingsmall radialstress)

σσ

=≈

max max min1

( )2

τ σ σ= − max 36.4 MPaτ =




without permission.

PROBLEM 7.99

The maximum gage pressure is known to be 8 MPa in a spherical steel pressure vessel having a 250-mm outer diameter and a 6-mm wall thickness. Knowing that the ultimate stress in the steel used is 400 MPa,Uσ = determine the factor of safety with respect to tensile failure.

SOLUTION

3 3

6 36

1 2 3

2506 119 mm 119 10 m, 6 10 m

2 2

(8 10 Pa)(119 10 m)79.333 10 Pa

2 2(6 10 m)

dr t t

pr

tσ σ

− −

−

−

= − = − = = × = ×

× ×= = = = ××

6

6max

400 10. .

79.333 10UF S

σσ

×= =×

. . 5.04F S =




without permission.

PROBLEM 7.100

A basketball has a 9.5-in. outer diameter and a 0.125-in. wall thickness. Determine the normal stress in the wall when the basketball is inflated to a 9-psi gage pressure.

SOLUTION

1 2

9.50.125 4.625 in. 9 psi

2 2(9)(4.625)

166.5 psi2 2(0.125)

dr t p

pr

tσ σ

= − = − = =

= = = = 166.5 psiσ =




without permission.

PROBLEM 7.101

A spherical pressure vessel of 900-mm outer diameter is to be fabricated from a steel having an ultimate stress 400 MPa.Uσ = Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 3.5 MPa, determine the smallest wall thickness that should be used.

SOLUTION

all

all all

400100 MPa (0.45 ) m

. . 4.0 2

22

2(100) 3.5(0.45 )

203.5 1.575

U dr t t

F Spr

t prt

t t

t

σσ

σ σ

= = = = − = −

= =

= −=

37.74 10 mt −= × min 7.74 mmt =




without permission.

PROBLEM 7.102

A spherical pressure vessel has an outer diameter of 10 ft and a wall thickness of 0.5 in. Knowing that for the steel used all 12 ksi,σ = 629 10 psiE = × , and 0.29,v = determine (a) the allowable gage pressure, (b) the corresponding increase in the diameter of the vessel.

SOLUTION

3

1 2 all

12010 ft 120 in. 0.5 59.5 in.

2 2

12 ksi 12 10 psi

dd r t

σ σ σ

= = = − = − =

= = = = ×

(a) 3

11 2

2 2(0.5)(12 10 )

2 59.5

pr tp

t r

σσ σ ×= = = =

202 psip =

(b) 31 1 2 1 6

6

1 1 1 0.29( ) (12 10 )

29 10

293.79 10

vv

E Eε σ σ σ −

−

− −= − = = ××

= ×

61 (120)(293.79 10 )d dε −Δ = = × 0.0353 in.dΔ =




without permission.

PROBLEM 7.103

A spherical gas container having an outer diameter of 5 m and a wall thickness of 22 mm is made of steel for which 200 GPaE = and 0.29.v = Knowing that the gage pressure in the container is increased from zero to 1.7 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the diameter of the container.

SOLUTION

3 3522 10 2.478 m, 22 10 m

2 2

dr t t− −= − = − × = = ×

(a) 6

61 2 3

(1.7 10 Pa)(2.478m)95.741 10 Pa

2 2(22 10 m)

pr

tσ σ −

×= = = = ××

max 95.7 MPaσ =

(b) 1 1 2 1

66

9

1 1( )

(1 0.29)(95.741 10 Pa)339.88 10

200 10 Pa

vv

E Eε σ σ σ

−

−= − =

− ×= = ××

3 61 (5 10 mm)(339.88 10 )d dε −Δ = = × ×

1.699 mmdΔ =




without permission.

PROBLEM 7.104

A steel penstock has a 750-mm outer diameter, a 12-mm wall thickness, and connects a reservoir at A with a generating station at B. Knowing that the density of water is 31000 kg/m , determine the maximum normal stress and the maximum shearing stress in the penstock under static conditions.

SOLUTION

3

3

3 2

6

6 36

1 3

1 1(750) 12 363 mm 363 10 m

2 2

12 mm 12 10 m

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

2.943 10 Pa

(2.943 10 )(363 10 )89.0 10 Pa

12 10

r d t

t

p gh

pr

t

ρ

σ

−

−

−

−

= − = − = = ×

= = ×

= =

= ×

× ×= = = ××

max 1σ σ= max 89.0 MPaσ =

min 0pσ = − ≈

max max min1

( )2





without permission.

PROBLEM 7.105

A steel penstock has a 750-mm outer diameter and connects a reservoir at A with a generating station at B. Knowing that the density of water is 31000 kg/m and that the allowable normal stress in the steel is 85 MPa, determine the smallest thickness that can be used for the penstock.

SOLUTION

3 2

6

61

3

1

66

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

2.943 10 Pa

85 MPa 85 10 Pa

1 1(750 10 ) 0.375

2 2

(2.943 10 )(0.375 )85 10

p gh

r d t t t

pr

t

t

t

ρ

σ

σ

−

= =

= ×

= = ×

= − = × − = −

=

× −× =

6 6 3(87.943 10 ) 1.103625 10 12.549 10 mt t −× = × = ×

12.55 mmt =




without permission.

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

3 3

66

1 3

3.318 10 1.632 m, 18 10 m

2 2

(1.5 10 Pa)(1.632 m)136 10 Pa

18 10 m

dr t t

pr

tσ

− −

−

= − = − × = = ×

×= = = ××

6max 1 136 10 Paσ σ= = × max 136.0 MPaσ =

min 0pσ = − ≈

6max max min

1( ) 68 10 Pa

2τ σ σ= − = × max 68.0 MPaτ =




without permission.

PROBLEM 7.107

Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5-ft outer diameter and 58

-in. wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0 is

desired.

SOLUTION

31

65 ksi13 ksi 13 10 psi

. . 5.0(5.5)(12)

0.625 32.375 in.2 2

U

F Sd

r t

σσ = = = = ×

= − = − =

3

11

(0.625)(13 10 )

32.375

pr tp

t r

σσ ×= = = 251 psip =




without permission.

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

3

3

6 36

1 3

3203 157 mm 157 10 m

2 2

3 10 m

(1.5 10 Pa)(157 10 m)78.5 10 Pa

3 10 m

dr t

t

pr

tσ

−

−

−

−

= − = − = = ×

= ×

× ×= = = ××

6max 1 78.5 10 Paσ σ= = × max 78.5 MPaσ =

min

6max max min

0

1( ) 39.25 10 Pa

2

pσ

τ σ σ

= − ≈

= − = ×

max 39.3 MPaτ =




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 1660 ksi

15 ksi 15 10 psi. . 4.0

(15 10 )18.77 psi 2703 lb/ft

149.81

U

d

r d t

F Spr

t

tp

r

σσ

σ

σ

= =

= − = − =

= = = = ×

=

×= = = =

But ,p hγ= 2

3

2703 lb/ft

62.4 lb/ft

ph

γ= = 43.3 fth =




without permission.

PROBLEM 7.110

For the storage tank of Prob. 7.109, determine the maximum normal stress and the maximum shearing stress in the cylindrical wall when the tank is filled to capacity ( 48 ft).h =

PROBLEM 7.109 The unpressurized cylindrical storage tank shown has a 3

16-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

3 2

2

31

3(25)(12) 300 in. in. 0.1875 in.

161

149.81 in.2

(62.4 lb/ft )(48 ft) = 2995.2 lb/ft

20.8 lb/in 20.8 psi

(20.8)(149.81)16.62 10 psi

0.1875

d t

r d t

p h

pr

t

γ

σ

= = = =

= − =

= =

= =

= = = ×

max 1σ σ= max 16.62 ksiσ =

min max max min1

0 ( )2

σ τ σ σ≈ = − max 8.31 ksiτ =




without permission.

PROBLEM 7.111

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) 0 01

12.75 in. 0.375 in. 6.00 in.2

d t r d t= = = − =

(400)(6.00)

6400 psi0.375

pr

tσ = = = 6.40 ksiσ =

(b) 0 01

12.75 in. 0.500 in. 5.875 in.2

d t r d t= = = − =

(400)(5.875)

4700 psi0.500

pr

tσ = = = 4.70 ksiσ =




without permission.

PROBLEM 7.112

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

SOLUTION

3

36

1 3

36

2 3

6ave 1 2

61 2

11.6 m 8 10 m 0.792 m

2

(600 10 )(0.792)59.4 10 Pa

8 10

(600 10 )(0.792)29.7 10 Pa

2 (2)(8 10 )

1( ) 44.56 10 Pa

21

( ) 14.85 10 Pa2

d t r d t

pr

t

pr

t

R

σ

σ

σ σ σ

σ σ

−

−

−

= = × = − =

×= = = ××

×= = = ××

= + = ×

= − = ×

(a) 6ave cos 40° 33.17 10 Paw Rσ σ= − = × 33.2 MPawσ =

(b) 6 sin 40° 9.55 10 Paw Rτ = = × 9.55 MPawτ =




without permission.

PROBLEM 7.113

For the tank of Prob. 7.112, determine the largest allowable gage pressure, knowing that the allowable normal stress perpendicular to the weld is 120 MPa and the allowable shearing stress parallel to the weld is 80 MPa.

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

SOLUTION

3

1 2

ave 1 2

1 2

ave

11.6 m 8 10 m 0.792 m

2

21 3

( )2 41 1

( )2 4

cos 40°

3 1cos 40 0.5585

4 4

w

d t r d t

pr pr

t tpr

tpr

Rt

R

pr pr

t t

σ σ

σ σ σ

σ σ

σ σ

−= = × = − =

= =

= + =

= − =

= −

= − ° =

6 36

6 36

(120 10 )(8 10 )2.17 10 Pa 2.17 MPa

0.5585 (0.5585)(0.792)

1sin 40 sin 40 0.1607

4

(80 10 )(8 10 )5.03 10 Pa 5.03 MPa

0.1607 (0.1607)(0.792)

w

w

w

tp

r

pr prR

t t

tp

r

σ

τ

τ

−

−

× ×= = = × =

= ° = ° =

× ×= = = × =

The largest allowable pressure is the smaller value. 2.17 MPap =




without permission.

PROBLEM 7.114

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

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

SOLUTION

3 11.6 m 8 10 mm 0.792 m

2d t r d t−= = × = − =

3

1 3

6

2

1 2

(600 10 )(0.792)

8 10

59.4 10 Pa = 59.4 MPa

29.7 MPa2

14.85 MPa2

|sin 2 |w

pr

t

pr

t

R

R

σ

σ

σ σ

τ β

−×= =

×= ×

= =

−= =

=

12

|sin 2 | 0.8080814.85

Na R

τβ = = =

2 53.91aβ = − ° 27.0aβ = + °

2 53.91bβ = + ° 27.0bβ = °

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

2 180 53.91 233.91dβ = ° + ° = + ° 117.0dβ = °

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

Safe ranges for β: 22.0 27.0°β− °≤ ≤

and 63.0 117.0°β°≤ ≤




without permission.

PROBLEM 7.115

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 50β = ° 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

6

61

2 1

375 mm 0.375 m2

(1.5 10 Pa 0.375 m)62.5 10 Pa 62.5 MPa

0.009 m

131.25 MPa 2 100

2

dr

pr

tσ

σ σ β

= = =

× ×= = = × =

= = = °

ave 1 2

1 2

1( ) 46.875 MPa

2

15.625 MPa2

R

σ σ σ

σ σ

= + =

−= =

(a) ave cos100w Rσ σ= + °

44.2 MPawσ =

(b) sin100w Rτ = °

15.39 MPawτ =




without permission.

PROBLEM 7.116

The pressurized tank shown was fabricated by welding strips of plate along a helix forming an angle β with a transverse plane. Determine the largest value of β 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 2 2

pr pr

t tσ σ= =

ave 1 2

1 2

ave

1 3( )

2 41

2 4cos 2w

pr

tpr

Rt

R

σ σ σ

σ σ

σ σ β

= + =

−= =

= −

3 1

0.85 cos 24 4

pr pr

t tβ = −

3cos 2 4 0.85 0.4

4

2 113.6

β

β

= − − = −

= ° 56.8β = °




without permission.

PROBLEM 7.117

The cylindrical portion of the compressed-air tank shown is fabricated of 0.25-in.-thick plate welded along a helix forming an angle 30β = ° with the horizontal. Knowing that the allowable stress normal to the weld is 10.5 ksi, determine the largest gage pressure that can be used in the tank.

SOLUTION

1 2

110 0.25 9.75 in.

2

,2

r d t

pr pr

t tσ σ

= − = − =

= =

ave 1 2

1 2

ave

1 3( )

2 41

2 4 cos 60°

5

8

w

pr

tpr

Rt

R

pr

t

σ σ σ

σ σ

σ σ

= + =

−= =

= +

=

8 (8)(10.5)(0.25)

0.43077 ksi5 (5)(9.75)

wtp

r

σ= = = 431 psip =




without permission.

PROBLEM 7.118

For the compressed-air tank of Prob. 7.117, determine the gage pressure that will cause a shearing stress parallel to the weld of 4 ksi.

PROBLEM 7.117 The cylindrical portion of the compressed-air tank shown is fabricated of 0.25-in.-thick plate welded along a helix forming an angle 30β = ° with the horizontal. Knowing that the allowable stress normal to the weld is 10.5 ksi, determine the largest gage pressure that can be used in the tank.

SOLUTION

1

10 0.25 9.75 in.2

r d t= − = − =

1 2,2

pr pr

t tσ σ= =

ave 1 2

1 2

1 3( )

2 41

2 4sin 60

3

8

w

pr

tpr

Rt

R

pr

t

σ σ σ

σ σ

τ

= + =

−= =

= °

=

8 8 (4)(0.25)

0.47372 ksi9.753 3

wtp

r

τ= = ⋅ = 474 psip =




without permission.

PROBLEM 7.119

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.

SOLUTION

1 2

112ft 144 in. 71.5 in.

2

2

d r d t

pr pr

t tσ σ

= = = − =

= =

(a) 1

1

12 ksi

(12)(0.5)0.0839 ksi

71.5

tp

r

σσ

=

= = =

83.9 psip =

(b) ave 1 2

1 2

ave

1 3( )

2 41

2 445

cos

3

4

w

pr

tpr

Rt

R

pr

t

σ σ σ

σ σ

βσ σ β

= + =

+= =

= ± °= +

=

4 4 (12) (0.5)

0.1119 ksi3 3 71.5

σ= = ⋅ =wtp

r 111.9 psi=p




without permission.

PROBLEM 7.120

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:

1

2

1225mm 6mm

2(1.2) (225)

45 MPa6

22.5 MPa2

r d t

pr

tpr

t

σ

σ

= = =

= = =

= =

Torsion: 1 2225 mm, 225 6 231 mmc c= = + =

( )4 4 6 4 6 42 1

3 3

36

6

446.9 10 mm 446.9 10 m2

(5 10 )(500 10 ) 2500 N m

(2500)(231 10 )1.292 10 Pa 1.292 MPa

446.9 10

J c c

T

Tc

J

π

τ

−

−

−

−

= − = × = ×

= × × = ⋅

×= = = × =×

Transverse shear: 0τ = at point a.

Bending: 6 4 31223.45 10 m , 231 10 m

2I J c− −= = × = ×

At point a, 3 3(5 10 ) (750 10 ) 3750 N mM −= × × = ⋅

3

6

(3750) (231 10 )3.88 MPa

223.45 10

Mc

Iσ

−

−×= = =

×

Total stresses (MPa).

Longitudinal: 22.5 3.88 26.38 MPaxσ = + =

Circumferential: 45 MPayσ =

Shear: 1.292 MPaxyτ =




without permission.


ave1

( ) 35.69 MPa2 x yσ σ σ= + =

2

2 9.40 MPa2

x yxyR

σ στ

− = + =

max ave 45.1 MPaRσ σ= + = max 45.1 MPaσ =

max(in-plane) 9.40 MPaRτ = = max (in-plane) 9.40 MPaτ =




without permission.

PROBLEM 7.121

For the compressed-air tank and loading of Prob. 7.120, determine the maximum normal stress and the maximum in-plane shearing stress at point b on the top of the tank.

PROBLEM 7.120 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:

1

2

1225 mm 6 mm

2(1.2)(225)

45 MPa6

22.5 MPa2

r d t

pr

tpr

t

σ

σ

= = =

= = =

= =

Torsion: 1 2225 mm, 225 6 231 mmc c= = + =

( )4 4 6 4 6 42 1

3 3

36

6

446.9 10 mm 446.9 10 m2

(5 10 ) (500 10 ) 2500 N m

(2500) (231 10 )1.292 10 Pa 1.292 MPa

446.9 10

J c c

T

Tc

J

π

τ

−

−

−

−

= − = × = ×

= × × = ⋅

×= = = × =×

Transverse shear: 0τ = at point b.

Bending: 6 4 31223.45 10 m , 231 10 m

2I J c− −= = × = ×

At point b, 3 3(5 10 ) (2 750 10 ) 7500 N mM −= × × × = ⋅

3

6

(7500)(231 10 )7.75 MPa

223.45 10

Mc

Iσ

−

−×= = =

×

Total stresses (MPa).

Longitudinal: 22.5 7.75 30.25 MPaxσ = + =

Circumferential: 45 MPayσ =

Shear: 1.292 MPaxyτ =




without permission.


ave1

( ) 37.625 MPa2 x yσ σ σ= + =

2

2 7.487 MPa2

x yxyR

σ στ

− = + =

max ave 45.1 MPaRσ σ= + = max 45.1 MPaσ =

max (in-plane) 7.49 MPaRτ = = max (in-plane) 7.49 MPaτ =




without permission.

PROBLEM 7.122

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.

SOLUTION

Consider element at point K.

Stresses due to internal pressure:

6

6 3

3

6 3

3

5 MP 5 10 Pa

1 2508 117 mm

2 2(5 10 )(117 10 )

73.125 MPa(8 10 )

(5 10 )(117 10 )36.563 MPa

2 (2)(8 10 )

x

y

p

r d t

pr

t

pr

t

σ

σ

−

−

−

−

= = ×

= − = − =

× ×= = =×

× ×= = =×

Stress due to bending moment: Point K is on the neutral axis.

0yσ =

Stress due to transverse shear: 340 10 NV P= = ×

( )

( )

2

1 2

3 3 3 32 1

3 3 6 3

4 4 4 42 1

6 4 6 4

3 6

6 3

1125 mm

2117 mm

2 2(125 117 )

3 3234.34 10 mm 234.34 10 m

(125 117 )4 444.573 10 mm 44.573 10 m

(40 10 )(234.34 10 )

(2 ) (44.573 10 )(16 10 )

13.144

xy

c d

c c t

Q c c

I c c

VQ PQ

It I t

π π

τ

−

−

−

− −

= =

= − =

= − = −

= × = ×

= − = −

= × = ×× ×= = =

× ×= 610 Pa 13.144 MPa× =




without permission.


Total stresses: 73.125 MPa, 36.563 MPa, 13.144 MPax y xyσ σ τ= = =

Mohr’s circle: ave1

( ) 54.844 MPa2 x yσ σ σ= + =

2

2

2 2

ave

ave

2

(18.281) (13.144) 22.516 MPa

77.360 MPa

32.328 MPa

x yxy

a

b

R

R

R

σ στ

σ σσ σ

− = +

= + == + == − =

Principal stresses: 77.4 MPa, 32.3 MPaa bσ σ= =

The 3rd principal stress is the radial stress. 0zσ ≈

max min77.4 MPa, 0σ σ= =

Maximum shearing stress: max max min1

( )2





without permission.

PROBLEM 7.123

In Prob. 7.122, determine the maximum normal stress and the maximum shearing stress at point L.

PROBLEM 7.122 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.

SOLUTION

Consider element at point L.

Stresses due to internal pressure:

6

6 3

3

3 3

3

5 MPa 5 10 Pa

1 2508 117 mm

2 2

(5 10 )(117 10 )73.125 MPa

8 10

(5 10 )(117 10 )36.563 MPa

2 (2)(8 10 )

x

y

p

r d t

pr

t

pr

t

σ

σ

−

−

−

−

= = ×

= − = − =

× ×= = =×

× ×= = =×

Stress due to bending moment: (40 kN)(600 mm) 24000 N mM = = ⋅

( )

2

1 2

4 4 4 42 1

6 4 6 4

3

6

1125 mm

2125 8 117 mm

(125 117 )4 4

44.573 10 mm 44.573 10 m

(24000)(125 10 )67.305 MPa

44.573 10y

c d

c c t

I c c

Mc

I

π π

σ

−

−

−

= =

= − = − =

= − = −

= × = ×

×= − = − = −×




without permission.


Stress due to transverse shear: Point L lies in a plane of symmetry.

0xyτ =

Total stresses: 73.125 MPa, 30.742 MPa, 0x y xyσ σ τ= = − =

Principal stresses: Since 0,xyτ = and x yσ σ are principal stresses. The 3rd principal stress is in the

radial direction, 0.zσ ≈

max min73.125 MPa, 0, 73.1 MPa, 30.7 MPa, 0a b zσ σ σ σ σ= = = = − =

Maximum stress: max 73.1 MPaσ =

Maximum shearing stress: max max min1

( )2





without permission.

PROBLEM 7.124

A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness is fabricated from a 4-ft section of spirally-welded pipe AB and is equipped with two rigid end plates. The gage pressure inside the vessel is 300 psi and 10-kip centric axial forces P and P′ are applied to the end plates. Determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

SOLUTION

( )

1

2

0

2 2 2 2 20

3

1 1(10) 5 in. 0.25 in.

2 2(300) (5)

6000 psi 6 ksi0.25

(300)(5)3000 psi 3 ksi

2 (2)(0.25)

5 0.25 5.25 in.

(5.25 5.00 ) 8.0503 in

10 101242 psi 1.242 ksi

8.0803

σ

σ

π π

σ

= = = =

= = = =

= = = =

= + = + =

= − = − =

×= − = − = − = −

r d t

pr

tpr

t

r r t

A r r

P

A

Total stresses. Longitudinal: 3 1.242 1.758 ksixσ = − =

Circumferential: 6 ksiyσ =

Shear: 0xyτ =


: (1.758, 0)

: (6, 0)

: (3.879)

X

Y

C

ave

2

2

2

1( ) 3.879 ksi

2

2

(1.758 6)0 2.121 ksi

2

x y

x yxyR

σ σ σ

σ στ

= + =

− = +

− = + =

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

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




without permission.

PROBLEM 7.125

Solve Prob. 7.124, assuming that the magnitude P of the two forces is increased to 30 kips.

PROBLEM 7.124 A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness is fabricated from a 4-ft section of spirally-welded pipe AB and is equipped with two rigid end plates. The gage pressure inside the vessel is 300 psi and 10-kip centric axial forces P and P′ are applied to the end plates. Determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.

SOLUTION

( )

1

2

0

2 2 2 2 20

3

1 1(10) 5 in. 0.25 in.

2 2(300)(5)

6000 psi 6 ksi0.25

(300)(5)3000 psi 3 ksi

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

(5.25 5 ) 8.0503 in

30 103727 psi 3.727 ksi

8.0503

r d t

pr

tpr

tr r t

A r r

P

A

σ

σ

π π

σ

= = = =

= = = =

= = = =

= + = + =

= − = − =

×= − = − = − = −

Total stresses: Longitudinal: 3 3.727 0.727 ksixσ = − = −

Circumferential: 6 ksiyσ =

Shear: 0xyτ =


: ( 0.727, 0)

: (6, 0)

: (2.6365, 0)

X

Y

C

−

ave

2

2

2

1( ) 2.6365 ksi

2

2

0.727 60 3.3635 ksi

2

x y

x yxyR

σ σ σ

σ στ

= + =

− = +

− − = + =

(a) ave cos 70 2.6365 3.3635 cos 70x Rσ σ′ = − ° = − ° 1.486 ksixσ ′ =

(b) | | sin 70 3.3635 sin 70x y Rτ ′ ′ = ° = ° | | 3.16 ksix yτ ′ ′ =




without permission.

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: ss

pr

tσ = (1)

Corresponding strain: ssp

s s s

pr

E E t

σε = =

Strain due to temperature change: sT s Tε α= Δ

Total strain: s ss s

prT

E tε α= + Δ

Change in length of circumference:

2 2s s ss s

prL r r T

E tπ ε π α

Δ = = + Δ

Brass ring. External pressure p: bb

pr

tσ = −

Corresponding strains: ,bp bT bb b

prT

E tε ε α= − = Δ


2 2b b bb b

prL r r T

E tπ ε π α

Δ = = − + Δ

Equating sLΔ to ,bLΔ s bs s b b

pr prT T

E t E tα α+ Δ = − + Δ

( )b ss s b b

r rp T

E t E tα α

+ = − Δ

(2)




without permission.


Data: 125 F 50 F 75 FTΔ = ° − ° = °

1 1

(5) 2.5 in.2 2

r d= = =

From Equation (2), 66 6

6 6

2.5 2.5(11.6 6.5)(10 )(75)

(29 10 )(0.125) (15 10 )(0.25)

1.35632 10 382.5 10

282.0 psi

p

p

p

−

− −

+ = −

× ×

× = ×=

From Equation (1), 3(282.0)(2.5)5.64 10 psi

0.125ss

pr

tσ = = = ×

(a) 5.64 ksisσ =

(b) 282 psip =




without permission.

PROBLEM 7.127

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: ss

pr

tσ = (1)

Corresponding strain: ssp

s s s

pr

E E t

σε = =

Strain due to temperature change: sT s Tε α= Δ

Total strain: s ss s

prT

E tε α= + Δ


2 2s s ss s

prL r r T

E tπ ε π α

Δ = = + Δ

Brass ring. External pressure p: bb

pr

tσ = −

Corresponding strains: ,bp bT bb b

prT

E tε ε α= − = Δ


2 2b b bb b

prL r r T

E tπ ε π α

Δ = = − + Δ

Equating to ,s bL LΔ Δ s bs s b b

pr prT T

E t E tα α+ Δ = − + Δ

( )b ss s b b

r rp T

E t E tα α

+ = − Δ

(2)




without permission.


Data: 125 F 50 F 75 F

1 1(5) 2.5 in.

2 2

T

r d

Δ = ° − ° = °

= = =

From Equation (2), 66 6

6 6

2.5 2.5(11.6 6.5)(10 )(75)

(29 10 )(0.25) (15 10 )(0.125)

1.67816 10 382.5 10

227.93 psi

p

p

p

−

− −

+ = −

× ×

× = ×=

From Equation (1), (227.93)(2.5)

2279 psi0.25s

s

pr

tσ = = =

(a) 2.28 ksisσ =

(b) 228 psip =




without permission.

PROBLEM 7.128

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

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

SOLUTION

15θ = + °

125 375 02 2 2

cos 2 sin 22 2 2

125 ( 375 )cos30 0

x y x y xy

x y x y xyx

ε ε ε ε γμ μ

ε ε ε ε γε θ θ

μ μ

′

+ −= − = − =

+ −= + +

= − + − ° + 450xε μ′ = −

cos 2 sin 22 2 2

125 ( 375 )cos 30 0

x y x y xyy


μ μ

′+ −

= − −

= − − − ° − 200yε μ′ =

( )sin 2 cos 2

( 500 250 )sin 30 0

x y x y xyγ ε ε θ γ θμ μ

′ ′ = − − +

= − − − ° + 375x yγ μ′ ′ =




without permission.

PROBLEM 7.129


240 , 160 , 150 , 60x y xyε μ ε μ γ μ θ= + = + = + = °

SOLUTION

60θ = − °

200 40 752 2 2

cos 2 sin 22 2 2

200 40cos ( 120 ) 75 sin ( 120 )

x y x y xy

x y x y xyx

ε ε ε ε γμ μ μ

ε ε ε ε γε θ θ′

+ −= + = =

+ −= + +

= + − ° + − ° 115.0xε μ′ =

cos 2 sin 22 2 2

200 40cos ( 120 ) 75sin ( 120 )

x y x y xyy


+ −= − −

= − − ° − − ° 285yε μ′ =

( )sin 2 cos 2

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

x y x y xyγ ε ε θ γ θ′ ′ = − − +

= − − − ° + − ° 5.72x yγ μ′ ′ = −




without permission.

PROBLEM 7.130


800 , 450 , 200 , 25x y xyε μ ε μ γ μ θ= − = + = + = °

SOLUTION

25θ = − °

175 625 1002 2 2

cos 2 sin 22 2 2

175 ( 625 )cos ( 50 ) (100 )sin ( 50 )

x y x y xy

x y x y xyx

ε ε ε ε γμ μ μ


μ μ μ

′

+ −= − = − = +

+ −= + +

= + − − ° + − ° 653xε μ′ = −

cos 2 sin 22 2 2

175 ( 625 )cos ( 50 ) (100 )sin ( 50 )

x y x y xyy


μ μ μ

′+ −

= − −

= − − − − ° − − ° 303yε μ′ =

( )sin 2 cos 2

( 800 450 )sin ( 50 ) ( 200 )cos ( 50 )

x y x y xyγ ε ε θ γ θμ μ μ

′ ′ = − − +

= − − − − ° + + − ° 829x yγ μ′ ′ = −




without permission.

PROBLEM 7.131

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

0, 320 , 100 , 30x y xyε ε μ γ μ θ= = + = − = °

SOLUTION

30θ = + °

160 1602 2

cos 2 sin 22 2 2

100160 160cos 60 sin 60

2

x y x y

x y x y xyx

ε ε ε εμ μ


+ −= = −

+ −= + +

= − ° − ° 36.7xε μ′ = +

cos 2 sin 22 2 2

100160 160cos 60 sin 60

2

x y x y xyy


+ −= − −

= + ° + ° 283yε μ′ = +

( )sin 2 cos 2

(0 320)sin 60 100cos 60

x y x y xyγ ε ε θ γ θ′ ′ = − − +

= − − ° − ° 227x yγ μ′ ′ = +




without permission.

PROBLEM 7.132

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 θ .

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

SOLUTION

Plotted points:

: ( 500 ,0)

: ( 250 , 0)

: ( 125 , 0)

X

Y

C

μμμ

−+−

375R μ=

ave cos 2 125 375cos30x Rε ε θ′ = + = − − ° 450xε μ′ = −

ave cos 2 125 375cos30y Rε ε θ′ = + = − + ° 200yε μ′ =

1

sin 2 375sin 302

γ θ′ ′ = = °x y R 375x yγ μ′ ′ =




without permission.

PROBLEM 7.133


240 , 160 , 150 , 60x y xyε μ ε μ γ μ θ= + = + = + = °

SOLUTION


: ( 240 , 75 )

: ( 160 , 75 )

: ( 200 , 0)

X

Y

C

μ μμ μμ

+ −++

2 2

ave

75tan 1.875 61.93

40

(40 ) (75 ) 85

2 120 61.93 181.93

cos 200 (85 )cos ( 181.93 )x

R

R

α α

μ μ μβ θ α

ε ε β μ μ′

= = = °

= + == − = − ° − ° = − °= + = + − ° 115.0xε μ′ =

ave cos 200 (85 )cos( 181.93 )y Rε ε β μ μ′ = − = − − ° 285yε μ′ =

1

sin 85 sin ( 181.93 ) 2.862

γ β μ′ ′ = − = − − ° = −x y R 5.72x yγ μ′ ′ = −




without permission.

PROBLEM 7.134


800 , 450 , 200 , 25ε μ ε μ γ μ θ= − + = = + = °x y xy

SOLUTION

Plotted points:

: ( 800 , 100 )

: ( 450 , 100 )

: ( 175 , 0)

X

Y

C

μ μμ μμ

− −+ +−

100

tan 9.09625

α α= = °

2 2(625 ) (100 ) 632.95R μ μ μ= + =

2 50 9.09 40.91β θ α= − = ° − ° = °

ave cos 175 632.95 cos 40.91x Rε ε β μ μ′ = − = − − °

653xε μ′ = −

ave cos 175 632.95 cos 40.91y Rε ε β μ μ′ = + = − + °

303yε μ′ =

1

sin 632.95 sin 40.912 x y Rγ β μ′ ′ = − = − ° 829x yγ μ′ ′ = −




without permission.

PROBLEM 7.135


0, 320 , 100 , 30x y xyε ε μ γ μ θ= = + = − = °

SOLUTION


: (0, 50 )

: (320 , 50 )

: (160 , 0)

X

Y

C

μμ μμ

−

50

tan 17.35160

α α= = °

2 2(160 ) (50 ) 167.63R μ μ μ= + =

2 60 17.35 42.65β θ α= − = ° − ° = °

ave cos 160 (167.63 )cos 42.65x Rε ε β μ μ′ = − = − ° 36.7xε μ′ =

ave cos 160 (167.63 )cos 42.65y Rε ε β μ μ′ = + = + ° 283yε μ′ =

1

sin (167.63 )sin 42.652 x y Rγ β μ′ ′ = − = ° 227x yγ μ′ ′ =




without permission.

PROBLEM 7.136

The following state of strain has been measured on the surface of a thin plate. Knowing that the surface of the plate is unstressed, determine (a) the direction and magnitude of the principal strains, (b) the maximum in-plane shearing strain, (c) the maximum shearing strain. (Use 1

3 .v = )

260 , 60 , 480x y xyε μ ε μ γ μ= − = − = +

SOLUTION

For Mohr’s circle of strain, plot points:

: ( 260 , 240 )

: ( 60 , 240 )

: ( 160 , 0)

X

Y

C

μ μμ μ

μ

− −−−

480

tan 2 2.4260 60

xyp

x y

γθ

ε ε= = = −

− − +

2 67.38pθ = − ° 33.67bθ = − °

56.31aθ = °

2 2(100 ) (240 )

260

R

R

μ μμ

= +=

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

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

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

22

R Rγ γ= = max (in-plane) 520γ μ=

1/3

( ) ( ) ( 260 60)1 1 2/3

160

c a b x yv v

v vε ε ε ε ε

μ

= − + = − + = − − −− −

=

max min160 420ε μ ε μ= = −

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




without permission.

PROBLEM 7.137


3 .v = )

600 , 400 , 350x y xyε μ ε μ γ μ= − = − = +

SOLUTION


: ( 600 , 175 )

: ( 400 , 175 )

: ( 500 , 0)

X

Y

C

μ μμ μμ

− −− +−

175

tan 2100pθ = −

2 60.26θ = − °p

30.13bθ = − °

59.87aθ = °

2 2(100 ) (175 )

201.6

R μ μμ

= +=

(a) ave 500 201.6a Rε ε μ μ= + = − + 298aε μ= −

ave 500 201.6b Rε ε μ μ= − = − − 702bε μ= −

(b) max (in-plane) 2Rγ = max (in-plane) 403γ μ=

1/3

( ) ( ) ( 600 400 )1 1 2/3c a b x y

v v

v vε ε ε ε ε μ μ= − + = − + = − − −

− −

500cε μ=

max min500 702ε μ ε μ= = −

(c) max max min 500 702γ ε ε μ μ= − = + max 1202γ μ=




without permission.

PROBLEM 7.138


3 .v = )

160 , 480 , 600x y xyε μ ε μ γ μ= + = − = −

SOLUTION

(a) For Mohr’s circle of strain, plot points:

: (160 , 300 )

: ( 480 , 300 )

: ( 160 , 0)

X

Y

C

μ μμ μμ

− −−

(a) 300

tan 2 0.9375320

xyp

x y

γθ

ε ε−= = = −

−

2 43.15 21.58p pθ θ= − ° = − ° and 21.58 90 68.42− + = °

21.58aθ = − °

68.42bθ = °

2 2(320 ) (300 ) 438.6R μ μ μ= + =

ave 160 438.6a Rε ε μ μ= + = − + 278.6aε μ= +

ave 160 438.6b Rε ε μ μ= − = − − 598.6bε μ= −

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

22

R Rγ γ= = (max, in-plane) 877γ μ=

(c) 1/3

( ) ( ) (160 480 )1 1 2/3c a b x y

v v

v vε ε ε ε ε μ μ= − + = − + = − −

− − 160.0cε μ=

max min278.6 598.6ε μ ε μ= = −

max max min 278.6 598.6γ ε ε μ μ= − = + max 877γ μ=




without permission.

PROBLEM 7.139


3v = )

30 , 570 , 720x y xyε μ ε μ γ μ= + = + = +

SOLUTION


: (30 , 360 )

: (570 , 360 )

: (300 , 0)

X

Y

C

μ μμ μμ

−+

360tan 2 1.3333

2702 53.13

p

p

θ

θ

−= = −

= − °

(a) 26.565bθ = − °

64.435aθ = °

2 2(270 ) (360 ) 450R μ μ μ= + =

ave 300 450a Rε ε μ μ= + = + 750aε μ=

ave 300 450b Rε ε μ μ= − = − 150bε μ= −


1/3

( ) (750 150 )1 2/3c a b

v

vε ε ε μ μ= − + = − −

− 300cε μ= −

max min750 , 300a cε ε μ ε ε μ= = = = −

(c) max max min 750 ( 300 )γ ε ε μ μ= − = − − max 1050γ μ=




without permission.

PROBLEM 7.140

For the given state of plane strain, use Mohr’s circle to determine (a) the orientation and magnitude of the principal strains, (b) the maximum in-plane shearing strain, (c) the maximum shearing strain.

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

SOLUTION

Plotted points:

: (60 , 25 )

: (240 , 25 )

: (150 , 0)

X

Y

C

μ μμ μμ

−

50

tan 2 0.27777860 240

xyp

x y

γθ

ε ε−= = =

− −

2 15.52pθ = ° 97.76aθ = °

7.76bθ = °

2 2(90 ) (25 ) 93.4R μ μ μ= + =

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

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

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

(c) max min0, 243.4 , 0cε ε μ ε= = =

max max minγ ε ε= − max 243.4γ =




without permission.

PROBLEM 7.141


400 , 200 , 375x y xyε μ ε μ γ μ= + = + =

SOLUTION


: ( 400 , 187.5 )

: ( 200 , 187.5 )

: ( 300 , 0)

X

Y

C

μ μμ μμ

+ −+ ++

375

tan 2 1.875400 200

xyp

x y

γθ

ε ε= = =

− −

2 61.93pθ = ° 30.96aθ = °

120.96bθ = °

2 2(100 ) (187.5 ) 212.5R μ μ μ= + =

(a) ave 300 212.5a Rε ε μ μ= + = + 512.5aε μ=

ave 300 212.5b Rε ε μ μ= − = − 87.5bε μ=


(c) max min0 512.5 0cε ε μ ε= = =

max max minγ ε ε= − max 512.5γ μ=




without permission.

PROBLEM 7.142


300 , 60 , 100x y xyε μ ε μ γ μ= + = + = +

SOLUTION

: (300 , 50 )

: (60 , 50 )

: (180 , 0)

X

Y

C

μ μμ μμ

−

100tan 2

300 60

2 22.62

γθ

ε εθ

= =− −

= °

xyp

x y

p

11.31aθ = °

101.31bθ = °

2 2(120 ) (50 ) 130R μ μ μ= + =

(a) ave 180 130a Rε ε μ μ= + = + 310aε μ=

ave 180 130b Rε ε μ μ= − = − 50bε μ=


(c) max min0, 310 , 0cε ε μ ε= = =

max max minγ ε ε= − max 310γ μ=




without permission.

PROBLEM 7.143


180 , 260 , 315x y xyε μ ε μ γ μ= − = − = +

SOLUTION


: ( 180 , 157.5 )

: ( 260 , 157.5 )

: ( 220 , 0)

X

Y

C

μ μμ μμ

− −− +−

(a) 315

tan 2 3.937580

xyp

x y

γθ

ε ε= = =

−

2 75.75pθ = ° 37.87aθ = °

127.87bθ = °

2 2(40 ) (157.5 ) 162.5R μ μ μ= + =

ave 220 162.5a Rε ε μ μ= + = − + 57.5aε μ= −

ave 220 162.5b Rε ε μ= − = − − 382.5bε μ= −

(b) max (in-plane) 2 325Rγ μ= =

(c) max min0, 0, 382.5cε ε ε μ= = = −

max max min 0 382.5γ ε ε μ= − = + max 382.5γ μ=




without permission.

PROBLEM 7.144

Determine the strain xε knowing that the following strains have been determined by use of the rosette shown:

1 2 3480 120 80ε μ ε μ ε μ= + = − = +

SOLUTION

1

2

3

15

30

75

θθθ

= − °= °= °

2 21 1 1 1 1cos sin sin cosx y xyε θ ε θ γ θ θ ε+ + =

0.9330 0.06699 0.25 480x y xyε ε γ μ+ − = (1)


0.75 0.25 0.4330 120x y xyε ε γ μ+ + = − (2)


0.06699 0.9330 0.25 80x y xyε ε γ μ+ + = (3)

Solving (1), (2), and (3) simultaneously,

253 , 307 , 893x y xyε μ ε μ γ μ= = = −

253xε μ=




without permission.

PROBLEM 7.145

The strains determined by the use of the rosette shown during the test of a machine element are

1 2 3600 450 75ε μ ε μ ε μ= + = + = −

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

SOLUTION

1

2

3

30

150

90

θθθ

= °= °= °


0.75 0.25 0.43301 600x y xyε ε γ μ+ + = (1)


0.75 0.25 0.43301 450x y xyε ε γ μ+ − = (2)


0 0 75yε μ+ + = − (3)

Solving (1), (2), and (3) simultaneously,

ave

2 2 2 2

725 , 75 , 173.21

1( ) 325

2

725 75 173.21409.3

2 2 2 2

x y xy

x y

x y xyR

ε μ ε μ γ μ

ε ε ε μ

ε ε γμ

= = − =

= + =

− + = + = + =

(a) ave 734a Rε ε μ= + = 734aε μ=

ave 84.3b Rε ε μ= − = − 84.3bε μ= −

(b) max (in-plane) 2 819Rγ μ= = max (in-plane) 819γ μ=




without permission.

PROBLEM 7.146

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

6 6 61 2 4420 10 in./in. 45 10 in./in. 165 10 in./in.ε ε ε− − −= + × = − × = + ×

(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.

ave 2 4

6 6 6ave

1( )

21

( 45 10 165 10 ) 60 10 in/in2

ε ε ε

ε − − −

= +

= − × + × = ×

Gages 1 and 3 are also 90° apart.

ave 1 3

6 63 ave 1

1( )

2

2 (2)(60 10 ) 420 10

ε ε ε

ε ε ε − −

= +

= − = × − ×

63 300 10 in/inε −= − ×

(b) 6 61 3420 10 in/in 300 10 in/inx yε ε ε ε− −= = × = = − ×

6 6 6

2 1 3

6

2 (2)( 45 10 ) 420 10 ( 300 10 )

210 10 in/in

xyγ ε ε ε − − −

−

= − − = − × − × − − ×

= − ×

2 22 2 6 6 6

6

420 10 ( 300 10 ) 210 10

2 2 2 2

375 10 in/in

x y xyRε ε γ − − −

−

− × − − × − ×= + = +

= ×

6 6ave 60 10 375 10a Rε ε − −= + = × + × 6435 10 in/inaε −= ×

6 6ave 60 10 375 10b Rε ε − −= − = × − × 6315 10 in/inbε −= − ×

max (in-plane) 2Rγ = 6max (in-plane) 750 10 in/inγ −= ×




without permission.

PROBLEM 7.147

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

61

62

63

93.1 10 in./in.

385 10 in./in.

210 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

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

SOLUTION

Use 1 1

( ) ( )cos 2 sin 22 2 2

xyx x y x y

γε ε ε ε ε θ θ′ = + + − +

where 75θ = − ° for gage 1,

0θ = for gage 2,

and 75θ = + ° for gage 3.

11 1

( ) ( )cos ( 150 ) sin ( 150 )2 2 2

xyx y x y

γε ε ε ε ε= + + − − ° + − ° (1)

21 1

( ) ( )cos 0 sin 02 2 2

xyx y x y

γε ε ε ε ε= + + − + (2)

31 1

( ) ( )cos (150 ) sin (150 )2 2 2

xyx y x y

γε ε ε ε ε= + + − ° + ° (3)

From Eq. (2), 6385 10 in/inx zε ε −= = ×

Adding Eqs. (1) and (3),

1 3

1 3

6 6 6

6

( ) ( )cos 150

(1 cos 150 ) (1 cos 150 )

(1 cos 150 )

(1 cos 150 )

93.1 10 210 10 385 10 (1 cos 150 )

1 cos 150

35.0 10 in/in

x y x y

x y

xy

ε ε ε ε ε εε εε ε εε

− − −

−

+ = + + − °= + ° + − °

+ − + °=

− °− × + × − × + °=

− °= ×




without permission.


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

3 1

6 63 1

6

sin 150

210 10 ( 93.1 10 )

sin 150 sin 150

606.2 10 in/in

xy

xy

ε ε γ

ε εγ− −

−

− = °

− × − − ×= =° °

= ×

6

6 6

606.2 10tan 2 1.732

385 10 35.0 10

xyp

x y

γθ

ε ε

−

− −×= = =

− × − × (a) 30.0 , 120.0a bθ θ= ° = °

6 6ave

6

2 2

2 26 66

6 6ave

1 1( ) (385 10 35.0 10 )

2 2

210 10 in/in

2 2

385 10 35.0 10 606.2350.0 10

2 2

210 10 350.0 10

x y

x y xy

a

R

R

ε ε ε

ε ε γ

ε ε

− −

−

− −−

− −

= + = × + ×

= ×

− = +

× − × = + = ×

= + = × + × 6560 10 in/inaε −= ×

6 6ave 210 10 350.0 10b Rε ε − −= − = × − × 6140.0 10 in/inbε −= − ×

(b) max (in-plane) 6350.0 10 in/in2

Rγ −= = × 6

max (in-plane) 700 10 in/inγ −= ×




without permission.

PROBLEM 7.148

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

122 2

max,min 1 3 1 2 2 31 1

( ) [( ) ( ) ]2 2

ε ε ε ε ε ε ε= + ± − + −

(Hint: The shaded triangles are congruent.)

SOLUTION

Since gage directions 1 and 3 are 90° apart,

ave 1 31

( )2

ε ε ε= +

Let 1 ave 1 3

2 ave 2 1 3

1u ( ).

21

v ( )2

ε ε ε ε

ε ε ε ε ε

= − = −

= − = − +

2 2 2

2 2 21 3 2 2 1 3 1 3

21 1 3

u v

1 1( ) ( ) ( )

4 4

1 1

4 2

ε ε ε ε ε ε ε ε

ε ε ε

= +

= − + − + + +

= −

R

2 2 23 2 2 1 2 3 1 1 3

1 1 1

4 4 2ε ε ε ε ε ε ε ε ε+ + − − + + 2

3

2 2 21 2 1 2 2 3 3

2 21 2 2 3

2 2 1/21 2 2 3

1

4

1 1

2 21 1

( ) ( )2 21

[( ) ( ) ]2

ε

ε ε ε ε ε ε ε

ε ε ε ε

ε ε ε ε

+

= − + − +

= − + −

= − + −R

max, min ave Rε ε= ±

gives the required formula.




without permission.

PROBLEM 7.149

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

1 2 3 avg3ε ε ε ε+ + =

where avgε is the abscissa of the center of the corresponding Mohr’s circle.

SOLUTION

1 ave cos 2 sin 22 2

x y xyε ε γε ε θ θ

−= + + (1)

2 ave

ave

ave

cos (2 120 ) sin (2 120 )2 2

(cos 120 cos 2 sin 120 sin 2 )2

(cos 120 sin 2 sin 120 cos 2 )2

1 3cos 2 sin 2

2 2 2

1 3sin 2 cos 2

2 2 2

x y xy

x y

xy

x y

xy

ε ε γε ε θ θ

ε εε θ θ

γθ θ

ε εε θ θ

γθ θ

−= + + ° + + °

−= + ° − °

+ ° + °

− = + − −

+ − +

(2)

3 ave

ave

ave

cos (2 240 ) sin (2 240 )2 2

(cos 240 cos 2 sin 240 sin 2 )2

(cos 240 sin 2 sin 240 cos 2 )2

1 3cos 2 sin 2

2 2 2

1 3sin 2 cos 2

2 2 2

x y xy

x y

xy

x y

xy

ε ε γε ε θ θ

ε εε θ θ

γθ θ

ε εε θ θ

γθ θ

−= + + ° + + °

−= + ° − °

+ ° + °

− = + − +

+ − −

(3)

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

1 2 3 ave

ave 1 2 3

3 0 0

3

ε ε ε εε ε ε ε

+ + = + += + +




without permission.

PROBLEM 7.150

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

611.5 10 psi,G = × determine the torque T indicated by a gage reading of 6300 10 in./in.−×

SOLUTION

For torsion, 0

0 0

0,

1( ) 0

1( ) 0

1

2 2

x y

x x y

y y x

xy xy

vE

vE

G G

σ σ τ τ

ε σ σ

ε σ σ

τ τγ γ

= = =

= − =

= − =

= =

Draw the Mohr’s circle for strain.

0

0

2

sin 2 sin 22x

RG

RG

τ

τε β β′

=

= =

But 0 3

22

sin 2xGTc T

J c

ετβπ′= = =

3

3 6 6

sin 2

(2) (11.5 10 )(300 10 )

sin 50

xc GT

π εβ

π

′

−

=

× ×=°

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




without permission.

PROBLEM 7.151

Solve Prob. 7.150, assuming that the gage forms an angle 35β = ° with a line parallel to the axis of the shaft.

PROBLEM 7.150 A single gage is cemented to a solid 4-in.-diameter steel shaft at an angle 25β = ° with a line parallel to the axis of the shaft. Knowing that 611.5 10 psi,G = × determine the torque T indicated by a gage reading of 6300 10 in./in.−×

SOLUTION

For torsion, 0

0 0

0, 0,

1( ) 0

1( ) 0

1

2 2

x y xy

x x y

y y x

xy xy

vE

vE

G G

σ σ τ τ

ε σ σ

ε σ σ

τ τγ γ

= = =

= − =

= − =

= =

Draw Mohr’s circle for strain.

0

0

2

sin 2 sin 22x

RG

RG

τ

τε β β′

=

= =

But

0 3

3 3 6 6

22

sin 2

(2) (11.5 10 )(300 10 )

sin 2 sin 70

x

x

GTc T

J c

c GT

ετβπ

π ε πβ

′

−′

= = =

× ×= =°

392.3 10 lb in= × ⋅ 92.3 kip inT = ⋅




without permission.

PROBLEM 7.152

A single strain gage forming an angle 18β = ° with a horizontal plane is used to determine the gage pressure in the cylindrical steel tank shown. The cylindrical wall of the tank is 6 mm thick, has a 600-mm inside diameter, and is made of a steel with

200 GPaE = and 0.30.v = Determine the pressure in the tank indicated by a strain gage reading of 280μ.

SOLUTION

1

1, 0

21

( ) 12

0.85

1 1( )

2

0.20

0

x

y x z

xx x y z

x

xy x y z

x

xyxy

pr

t

vv v

E E

E

v v vE E

E

G

σ σ

σ σ σ

σε σ σ σ

σ

σε σ σ σ

σ

τγ

= =

= ≈

= − − = −

=

= − + − = −

=

= =


ave

ave

1( ) 0.525

21

( ) 0.3252

cos 2 (0.525 0.325cos 2 )

(0.525 0.325cos 2 )

xx y

xx y

xx

x x

E

RE

RE

t tEp

r r

σε ε ε

σε ε

σε ε β β

σ εβ

′

′

= + =

= − =

= + = +

= =+

Data: 1 1

(600) 300 mm 0.300 m2 2

r d= = = =

3 9 6

3 9 66

6 10 mm 200 10 Pa, 280 10 18

(6 10 )(200 10 )(280 10 )1.421 10 Pa

(0.300)(0.525 0.325 cos 36 )

xt E

p

ε β− −′

− −

= × = × = × = °

× × ×= = ×+ °

1.421 MPap =




without permission.

PROBLEM 7.153

Solve Prob. 7.152, assuming that the gage forms an angle 35β = ° with a horizontal plane.

PROBLEM 7.152 A single strain gage forming an angle 18β = ° with a horizontal plane is used to determine the gage pressure in the cylindrical steel tank shown. The cylindrical wall of the tank is 6 mm thick, has a 600-mm inside diameter, and is made of a steel with 200 GPaE = and 0.30.v = Determine the pressure in the tank indicated by a strain gage reading of 280μ.

SOLUTION

1

1, 0

21

( ) 1 0.852

1 1( ) 0.20

2

0

x

y x z

x xx x y z

x xy x y z

xyxy

pr

t

vv v

E E E

v v vE E E

G

σ σ

σ σ σ

σ σε σ σ σ

σ σε σ σ σ

τγ

= =

= ≈

= − − = − = = − + − = − =

= =


ave

ave

1( ) 0.525

21

( ) 0.3252

cos 2

(0.525 0.325 cos 2 )

(0.525 0.325 cos 2 )

xx y

xx y

x

x

x x

E

RE

R

Et tE

pr r

σε ε ε

σε ε

ε ε βσβ

σ εβ

′

′

= + =

= − =

= +

= +

= =+

Data: 1 1

(600) 300 mm 0.300 m2 2

r d= = = =

3 9 6

3 9 66

6 10 m 200 10 Pa, 280 10 35

(6 10 )(200 10 )(280 10 )1.761 10 Pa

(0.300)(0.525 0.325 cos 70 )

xt E

p

ε β− −′

− −

= × = × = × = °

× × ×= = ×+ °

1.761 MPap =




without permission.

PROBLEM 7.154

The given state of plane stress is known to exist on the surface of a machine component. Knowing that 200 GPaE = and 77.2 GPaG = , determine the direction and magnitude of the three principal strains (a) by determining the corresponding state of strain [use Eq. (2.43) and Eq. (2.38)] and then using Mohr’s circle for strain, (b) by using Mohr’s circle for stress to determine the principal planes and principal stresses and then determining the corresponding strains.

SOLUTION

(a) 6 6

9 9

0, 150 10 Pa, 75 10 Pa

200 10 Pa 77 10 Pa

x y xy

E G

σ σ τ= = − × = − ×

= × = ×

69

69

6

9

ave

1 0.29872(1 ) 21 1

( ) [0 (0.2987)(150 10 )]200 10

2241 1

( ) [( 150 10 ) 0]200 10

750

75 10974

77 10

487.02

1( ) 263

2974

tan 2

x x y

y y x

xyxy

xy

x y

x y

xya

x y

E EG v

v G

vE

vE

G

ε σ σ

μ

ε σ σ

μτ

γ μ

γμ

ε ε ε μ

ε ε μγ

θε ε

= = − =+

= − = + ××

=

= − = − × −×

= −− ×= = = −

×

= −

= + = −

− =

−= =−

9741.000

974= −

2 45.0aθ = − ° 22.5aθ = − °

2 2

ave

6892 2

x y xy

a

R

R

ε ε γμ

ε ε

− = + =

= + 426aε μ=

aveb Rε ε= − 952bε μ= −

6

9

(0.2987)(0 150 10 )( )

200 10c x y

v

Eε σ σ − ×= − + = −

× 224cε μ= −




without permission.


(b) ave1

( ) 75 MPa2 x yσ σ σ= + = −

2 22 2

ave

ave

6 69

0 15075

2 2

106.07 MPa

31.07 MPa

181.07 MPa

1( )

1[31.07 10 (0.2987)( 181.07 10 )]

200 10

x yxy

a

b

a a b

R

R

R

vE

σ στ

σ σσ σ

ε σ σ

− + = + = +

== + == − = −

= −

= × − − ××

6426 10−= × 426aε μ=

2

tan 2 1.000xya

x y

τθ

σ σ= = −

− 2 45aθ = − °

22.5aθ = − °




without permission.

PROBLEM 7.155

The following state of strain has been determined on the surface of a cast-iron machine part:

720 400 660x y xyε μ ε μ γ μ= − = − = +

Knowing that 69 GPaE = and 28 GPa,G = determine the principal planes and principal stresses (a) by determining the corresponding state of plane stress [use Eq. (2.36), Eq. (2.43), and the first two equations of Prob. 2.72] and then using Mohr’s circle for stress, (b) by using Mohr’s circle for strain to determine the orientation and magnitude of the principal strains and then determining the corresponding stresses.

SOLUTION

The 3rd principal stress is 0.zσ =

2 2

691 1 0.2321

2(1 ) 2 56

6972.93 GPa

1 1 (0.232)

E EG v

v G

E

v

= = − = − =+

= =− −

(a) 2

9 6 6

( )1

(72.93 10 )[ 720 10 (0.232)( 400 10 )]

59.28 MPa

x x yE

vv

σ ε ε

− −

= +−

= × − × + − ×= −

2

9 6 6

( )1

(72.93 10 )[ 400 10 (0.2321)( 720 10 )]

41.36 MPa

y y xE

vv

σ ε ε

− −

= +−

= × − × + − ×= −

9 6

ave

2

2

(28 10 )(660 10 )

18.48 MPa

1( ) 50.32 MPa

22

tan 2 2.0625

2 64.1 , 32.1 , 57.9

20.54 MPa2

xy xy

x y

xyb

x y

b b a

x yxy

G

R

τ γ

σ σ σ

τθ

σ σθ θ θ

σ στ

−= = × ×

=

= + = −

= = −−

= − ° = − ° = °

− = + =

avea Rσ σ= + 29.8 MPaaσ = −

aveb Rσ σ= − 70.9 MPabσ = −




without permission.


(b) ave1

( ) 5602 x yε ε ε μ= + = −

2 2

ave

ave

tan 2 2.0625

2 64.1 , 32.1 , 57.9

366.742 2

193.26

926.74

xyb

x y

b b a

x y xy

a

b

R

R

R

γθ

ε εθ θ θ

ε ε γμ

ε ε με ε μ

= = −−

= − ° = − ° = °

− = + =

= + = −= − = −

2

( )1

a a bE

vv

σ ε ε= +−

29.8 MPaaσ = −

2

( )1

b b aE

vv

σ ε ε= +−

70.9 MPabσ = −




without permission.

PROBLEM 7.156

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:

61

62

63

60 10 in./in.

240 10 in./in.

200 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

Knowing that 629 10 psiE = × and 0.30,v = determine the magnitudes of P and Qx.

SOLUTION

61

63

62 1 3

2 2

32 2

3

60 10

200 10

2 340 10

29( ) [ 60 (0.3)(200)] 0

1 1 (0.3)29

( ) [200 (0.3)( 60)] 5.8 10 psi1 1 (0.3)

(2)(6)(5.8 10 )

x

y

xy

x x y

y y x

y y

Ev

vE

vv

PP A

A

ε εε ε

γ ε ε ε

σ ε ε

σ ε ε

σ σ

−

−

−

= = − ×= = ×

= − − = ×

= + = − + =− −

= + = + − = ×− −

= = = ×

369.6 10 lb= × 69.6 kipsP =

66

3

3 3 4

3

33

29 1011.1538 10 psi

2(1 ) (2)(1.30)

(11.1538)(340) 3.7923 10 psi

1 1(2)(6) 36 in

12 12ˆ (2)(3)(1.5) 9 in 2 in.

ˆ

(36)(2)(3.7923 10 )30.338 10 lb

ˆ 9

xy xy

xy

xy

EG

v

G

I bh

Q Ay t

VQ

ItIt

VQ

τ γ

τ

τ

×= = = ×+

= = = ×

= = =

= = = =

=

×= = = ×

Q V= 30.3 kipsQ =




without permission.

PROBLEM 7.157

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

61

62

63

30 10 in./in.

250 10 in./in.

100 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

PROBLEM 7.156 A centric axial force P and a horizontal force 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:

61

62

63

60 10 in./in.

240 10 in./in.

200 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

Knowing that 629 10 psiE = × and 0.30,v = determine the magnitudes of P and Qx.

SOLUTION

61

63

62 1 3

2 2

2 2

3

3

30 10

100 10

2 430 10

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

1 1 (0.3)0

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

1 1 (0.3)

2.9 10 psi

(2)(6)(2.9 10 )

x

y

xy

x x y

y y x

y y

Ev

v

Ev

v

PP A

A

ε εε ε

γ ε ε ε

σ ε ε

σ ε ε

σ σ

−

−

−

= = − ×= = + ×

= − − = ×

= + = − +− −

=

= + = + −− −

= ×

= = = ×

334.8 10 lb= × 34.8 kipsP =




without permission.


66

3

3 3 4

3

29 1011.1538 10 psi

2(1 ) (2)(1.30)

(11.1538)(430) 4.7962 10 psi

1 1(2)(6) 36 in

12 12ˆ (2)(3)(1.5) 9 in

2 in.

xy xy

EG

v

G

I bh

Q Ay

t

τ γ

×= = = ×+

= = = ×

= = =

= = ==

33

ˆ

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

ˆ 9

xy

xy

VQ

ItIt

VQ

Q V

τ

τ

=

×= = = ×

= 38.4 kipsQ =




without permission.

PROBLEM 7.158

Two wooden members of 80 × 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that 22β = ° and that the maximum 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

Forces Areas

3 2 3 2

3 33

all all

(80) (120) 9.6 10 mm 9.6 10 m

(400 10 )(9.6 10 ) /sin 10.251 10 N

sin 22σ β

−

−

= = × = ×

× ×= = = ×°

A

N A

3310.251 10

0: sin 0 27.4 10 Nsin sin 22

ββ′

×Σ = − = = = = ×°y

NF N P P

3 3

3all all

(600 10 )(9.6 10 ) /sin 15.376 10 N

sin 22S Aτ β

−× ×= = = ×°

3315.376 10

0: cos 0 16.58 10 Ncos cos 22

ββ′

×Σ = − = = = = ×°x

SF S P P

The smaller value for P governs. 16.58 kNP =




without permission.

PROBLEM 7.159

Two wooden members of 80 × 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that 25β = ° and that centric loads of magnitude 10 kNP = 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

Forces Areas

3 2 3 2(80)(120) 9.6 10 mm 9.6 10 mA −= = × = ×

(a) 3 30: sin 0 sin (10 10 )sin 25 4.226 10 Nβ β′Σ = − = = = × ° = ×yF N P N P

3

33

(4.226 10 )sin 25186.0 10 Pa

/sin 9.6 10

N

Aσ

β× °= = = ×

× 186.0 kPaσ =

(b) 3 30: cos 0 cos (10 10 )cos 25 9.063 10 Nβ β′Σ = − = = = × ° = ×xF S P S P

3

33

(9.063 10 )sin 25399 10 Pa

/sin 9.6 10

N

Aτ

β −× °= = = ×

× 399 kPaτ =




without permission.

PROBLEM 7.160

The centric force P is applied to a short post as shown. Knowing that the stresses on plane a-a are 15 ksiσ = − and 5 ksiτ = , determine (a) the angle β that plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.

SOLUTION

0

0

/

x

xy

y P A

στσ

==

= −

(a) From the Mohr’s circle,

5

tan 0.333315

β = = 18.4β = °

cos 22 2

P P

A Aσ β− = +

(b)2( ) (2)(15)

1 cos 2 1 cos 2

P

A

σβ β

−= =+ +

16.67 ksiP

A=




without permission.

PROBLEM 7.161


SOLUTION

Express each state of stress in terms of components acting on the element shown above.

Add like components of the two states of stress.

0 and 90°pθ =

max 0σ σ=

min 0σ σ= −




without permission.

SOLUTION

42 MPa, 12 MPa, 36 MPax y xyσ σ τ= = = −

ave1

( ) 27 MPa2 x yσ σ σ= + =

2

2

2 2

2

(15) ( 36) 39 MPa

x yxyR

σ στ

− = +

= + − =

ave

ave

66 MPa

12 MPaa

b

R

R

σ σσ σ

= + == − = −

(a) 24 MPa 66 MPaz aσ σ= + = 12 MPabσ = −

max min66 MPa 12 MPaσ σ= = − max max min1

( ) 39 MPa2

τ σ σ= − =

(b) 24 MPa 66 MPaz aσ σ= − = 12 MPabσ = −


( ) 45 MPa2

τ σ σ= − =

(c) 0zσ = 66 MPaaσ = 12 MPabσ = −


( ) 39 MPa2

τ σ σ= − =

PROBLEM 7.162

For the state of stress shown, determine the maximum shearing stress when (a) 24 MPa,zσ = + (b) 24 MPa,zσ = − (c) 0.zσ =




without permission.

PROBLEM 7.163

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

SOLUTION

(a) 17.5 ksi 3 ksiyz xτ σ= = −

2 2(6) (17.5) 18.5

6 18.5 24.5

6 18.5 12.5A

B

R

σσ

= + == + == − = −

max

min

max max min

24.5 ksi

12.5 ksi

1( )

2

A

B

σ σσ σ

τ σ σ

= == = −

= − max 18.5 ksiτ =

(b) 8 ksi 3 ksiyz xτ σ= = −

2 2(6) (8) 10

6 10 16

6 10 4A

B

R

σσ

= + == + == − = −

max

min

max max min

16 ksi

4 ksi

1( )

2

A

B

σ σσ σ

τ σ σ

= == = −

= − max 10.0 ksiτ =

(c) 0 3 ksiyz xτ σ= = −

max

min

max max min

12 ksi

3 ksi

1( )

2

z

x

σ σσ σ

τ σ σ

= == = −

= − max 7.5 ksiτ =




without permission.

PROBLEM 7.164

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

SOLUTION

24 ksi 14 ksi 0x y zσ σ σ= = =

For stresses in xy-plane, ave1

( ) 19 ksi 5 ksi2 2

x yx y

σ σσ σ σ

−= + = =

(a) 6 ksixyτ =

2

2 2 2

ave ave

2 2

(5) (6) 7.810 ksi2

26.810 ksi, 11.190 ksi

23.324 ksi < 30 ksi

x yxy

a b

a b a b

R

R R

σ στ

σ σ σ σ

σ σ σ σ

− = + = + =

= + = = − =

+ − = (No yielding)

30

. .23.324

F S = . . 1.286F S =

(b) 12 ksixyτ =

2

2 2 2

ave ave

2 2

(5) (12) 13 ksi2

32 ksi, 6 ksi

29.462 ksi < 30 ksi

x yxy

a b

a b a b

R

R R

σ στ

σ σ σ σ

σ σ σ σ

− = + = + =

= + = = − =

+ − = (No yielding)

30

. .29.462

F S = . . 1.018F S =

(c) 14 ksixyτ =

2

2 2 2

ave ave

2 2

(5) (14) 14.866 ksi2

33.866, 4.134 ksi

32.00 ksi > 30 ksi

x yxy

a b

a b a b

R

R R

σ στ

σ σ σ σ

σ σ σ σ

− = + = + =

= + = = − =

+ − = (Yielding occurs)




without permission.

PROBLEM 7.165

A torque of magnitude 12 kN mT = ⋅ is applied to the end of a tank containing compressed air under a pressure of 8 MPa. Knowing that the tank has a 180-mm inner diameter and a 12-mm wall thickness, determine the maximum normal stress and the maximum shearing stress in the tank.

SOLUTION

1

180 mm 90 mm 12 mm2

d r d t= = = =

Torsion:

( )1 2

4 4 6 4 6 42 1

3 3

6

90 mm 90 12 102 mm

66.968 10 mm 66.968 10 m2

(12 10 )(102 10 )18.277 MPa

66.968 10

c c

J c c

Tc

J

π

τ

−

−

−

= = + =

= − = × = ×

× ×= = =×

Pressure:

1 2(8)(90)

60 MPa 30 MPa12 2

pr pr

t tσ σ= = = = =

Summary of stresses:

ave

2

2

ave

ave

60 MPa, 30 MPa, 18.277 MPa

1( ) 45 MPa

2

23.64 MPa2

68.64 MPa

21.36 MPa

0

x y xy

x y

x yxy

a

b

c

R

R

R

σ σ τ

σ σ σ

σ στ

σ σσ σσ

= = =

= + =

− = + =

= + == − =≈

max 68.6 MPaσ =

min 0σ =

max max min1

( )2





without permission.

PROBLEM 7.166

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

1

2

ave 1

max

max ave

22 2 21 2

2 2 2 2

6

1 1(180) 90 mm 12 mm

2 2

(8)(90)60 MPa

12

30 MPa21

( ) 45 MPa275 MPa

30 MPa

152

15 30 15 25.98 MPa

25.98 10 Pa

y

xy xy

xy

r d t

pr

tpr

t

R

R

R

σ

σ

σ σ σ

σσ σ

σ σ τ τ

τ

= = = =

= = =

= =

= + =

== − =

− = + = +

= − = − =

= ×

Torsion:

( )

1

2

4 4 6 4 6 42 1

6 63

3

90 mm

90 12 102 mm

66.968 10 mm 66.968 10 m2

(66.968 10 )(25.98 10 )17.06 10 N m

102 10xy

xy

c

c

TJ c c

JTcT

J c

ττ

−

−

−

== + =

= − = × = ×

× ×= = = = × ⋅×

17.06 kN mT = ⋅




without permission.

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.

SOLUTION

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

1pr

tσ =

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

100 psi, 1 0.12 0.88 in., 0.12 in.i ip r t= = − = =

max(100)(0.88)

( ) 733.33 psi0.12

i ii

Pr

tσ = = = +

500 psi, 1 in., 0.12 in.o op r t= = =

max

max

(500)(1)( ) 4166.7 psi

0.12733.33 4166.7 3433.4 psi

o oo

P r

tσ

σ

= − = − = −

= + − = −

max 3.43 ksi (compression)σ =




without permission.

PROBLEM 7.168

For the assembly of Prob. 7.167, determine the normal stress in the jacket (a) in a direction perpendicular to the longitudinal axis of the jacket, (b) in a direction parallel to that axis.

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.

SOLUTION

(a) Hoop stress.

1

500 psi, 0.15 in., 2 0.15 1.85 in.

(500)(1.85)( ) 6166.7 psi

0.15

p t r

pr

tσ

= = = − =

= = =

1 6.17 ksiσ =

(b) Longitudinal stress.

Free body of portion of jacket above a horizontal section, considering vertical forces only:

2

2

0:

0

0

f j

y

f jA A

f j

F

pdA dA

pA A

σ

σ

=

− =

− =

2f

j

Ap

Aσ = (1)

Areas : ( )( )

2 2 2 2 22 1

2 2 2 2 23 2

[(1.85) (1) ] 7.6105 in

[(2) (1.85) ] 1.81427 in

f

j

A r r

A r r

π π

π π

= − = − =

= − = − =

Recalling Eq (1):

27.6105

(500) 2097.4 psi1.81427

f

j

Ap

Aσ = = =

2 2.10 ksiσ =




without permission.

PROBLEM 7.169

Determine the largest in-plane normal strain, knowing that the following strains have been obtained by the use of the rosette shown:

61

62

63

50 10 in./in.

360 10 in./in.

315 10 in./in.

ε

ε

ε

−

−

−

= − ×

= + ×

= + ×

SOLUTION

1 2 345 , 45 , 0θ θ θ= ° = − ° =

2 21 1 1 1 1

6

cos sin sin cos

0.5 0.5 0.5 50 10

x y xy

x y xy

ε θ ε θ γ θ θ ε

ε ε γ −

+ + =

+ + = − × (1)

2 22 2 2 2 2

6

cos sin sin cos

0.5 0.5 0.5 360 10

x y xy

x y xy


ε ε γ −

+ + =

+ − = × (2)

2 23 3 3 3 3

6

cos sin sin cos

0 0 315 10

x y xy

x


ε −

+ + =

+ + = × (3)

From (3), 6315 10 in/inxε −= ×

Eq. (1) − Eq. (2): 6 6 650 10 360 10 410 10 in/inxyγ − − −= − × − × = − ×

Eq. (1) + Eq. (2): 1 2

6 6 6 61 2 50 10 360 10 315 10 5 10 in/in

x y

y x

ε ε ε ε

ε ε ε ε − − − −

+ = +

= + − = − × + × − × = − ×

6ave

1( ) 155 10 in/in

2 x yε ε ε −= + = ×

2 2

2 26 6 6

6

2 2

315 10 5 10 410 10

2 2

260 10 in/in

x y xyRε ε γ

− − −

−

− = +

× + × − ×= +

= ×

6 6max ave 155 10 260 10Rε ε − −= + = × + × 6

max 415 10 in/inε −= ×

chapter 7 · problem 7.1 for the given state of stress, determine the normal and shearing stresses...

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