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12.9 The stresses shown in the figure act at a point in a stressed body.Determine the normal and shear stresses at this point on the inclinedplane shown.

Fig. P12.9

Solution The given stress values are: 4,200 psi, 1,800 psi, 0 psi, 50x y xyσ σ τ θ= = = = + ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(4,200 psi)cos (50 ) (1,800 psi)sin (50 ) 2(0 psi)sin(50 )cos(50 )

2,791.6222 psi 2,790 psi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(4,200 psi) (1,800 psi)]sin(50 )cos(50 ) (0 psi)[cos (50 ) sin (50 )]

1,181.7693 psi 1,182 psi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − ° ° + ° − °

= − = − Ans.

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12.10 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

Fig. P12.10

Solution The given stress values are: 90 MPa, 140 MPa, 0 MPa, 65x y xyσ σ τ θ= − = − = = + ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 90 MPa)cos (65 ) ( 140 MPa)sin (65 ) 2(0 MPa)sin(65 )cos(65 )

131.0697 MPa 131.1 MPa (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − ° + ° °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 90 MPa) ( 140 MPa)]sin(65 )cos(65 ) (0 MPa)[cos (65 ) sin (65 )]

19.1511 MPa 19.15 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − ° ° + ° − °

= − = − Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.11 The stresses shown in the figure act at a point in a stressedbody. Determine the normal and shear stresses at this point on theinclined plane shown.

Fig. P12.11

Solution The given stress values are: 5.5 ksi, 18.7 ksi, 0 ksi, 20x y xyσ σ τ θ= − = = = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 5.5 ksi)cos ( 20 ) (18.7 ksi)sin ( 20 ) 2(0 ksi)sin( 20 )cos( 20 )

2.6691 ksi 2.67 ksi (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= − − ° + − ° + − ° − °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 5.5 ksi) (18.7 ksi)]sin( 20 )cos( 20 ) (0 ksi)[cos ( 20 ) sin ( 20 )]

7.7777 ksi 7.78 ksi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − ° − ° + − ° − − °

= − = − Ans.

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12.12 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at thispoint on the inclined plane shown.

Fig. P12.12

Solution The given stress values are: 17,700 psi, 12,500 psi, 0 psi, 60x y xyσ σ τ θ= = − = = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(17,700 psi)cos ( 60 ) ( 12,500 psi)sin ( 60 ) 2(0 psi)sin( 60 )cos( 60 )

4,950.0000 psi 4,950 psi (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − − ° + − ° − °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(17,700 psi) ( 12,500 psi)]sin( 60 )cos( 60 ) (0 psi)[cos ( 60 ) sin ( 60 )]

13,076.9836 psi 13,080 psi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − ° − ° + − ° − − °

= = Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.13 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

Fig. P12.13

Solution The given stress values are: 8 ksi, 6 ksi, 10 ksi, 75x y xyσ σ τ θ= − = = = ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 8)cos (75 ) (6 ksi)sin (75 ) 2(10 ksi)sin(75 )cos(75 )

10.0622 ksi 10.06 ksi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + ° + ° °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 8 ksi) (6 ksi)]sin(75 )cos(75 ) (10 ksi)[cos (75 ) sin (75 )]

5.1603 ksi 5.16 ksi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° ° + ° − °

= − = − Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.14 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at thispoint on the inclined plane shown.

Fig. P12.14

Solution The given stress values are: 82 MPa, 48 MPa, 26 MPa, 25x y xyσ σ τ θ= = = − = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(82 MPa)cos ( 25 ) (48 MPa)sin ( 25 ) 2( 26 MPa)sin( 25 )cos( 25 )

95.8445 MPa 95.8 MPa (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − ° + − − ° − °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(82 MPa) (48 MPa)]sin( 25 )cos( 25 ) ( 26 MPa)[cos ( 25 ) sin ( 25 )]

3.6897 MPa 3.69 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° − ° + − − ° − − °

= − = − Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.15 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at thispoint on the inclined plane shown.

Fig. P12.15

Solution The given stress values are: 108 MPa, 14 MPa, 72 MPa, 50x y xyσ σ τ θ= = − = − = ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(108 MPa)cos (50 ) ( 14 MPa)sin (50 ) 2( 72 MPa)sin(50 )cos(50 )

34.4987 MPa 34.5 MPa (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + − ° + − ° °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(108 MPa) ( 14 MPa)]sin(50 )cos(50 ) ( 72 MPa)[cos (50 ) sin (50 )]

47.5706 MPa 47.6 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° ° + − ° − °

= − = − Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

12.16 The stresses shown in the figure act at a point in a stressedbody. Determine the normal and shear stresses at this point on the inclined plane shown.

Fig. P12.16

Solution The given stress values are: 2,150 psi, 860 psi, 1,460 psi, 40x y xyσ σ τ θ= − = = − = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 2,150 psi)cos ( 40 ) (860 psi)sin ( 40 ) 2( 1,460 psi)sin( 40 )cos( 40 )

531.4788 psi 531 psi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − − ° + − ° + − − ° − °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 2,150 psi) (860 psi)]sin( 40 )cos( 40 ) ( 1,460 psi)[cos ( 40 ) sin ( 40 )]

1,735.6620 psi 1,736 psi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − ° − ° + − − ° − − °

= − = − Ans.

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12.17 The stresses shown in the figure act at a point in a stressedbody. Determine the normal and shear stresses at this point onthe inclined plane shown.

Fig. P12.17

Solution The given stress values are: 18 MPa, 42 MPa, 30 MPa, 68.1986x y xyσ σ τ θ= = − = = ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(18 MPa)cos (68.1986 ) ( 42 MPa)sin (68.1986 )2(30 MPa)sin(68.1986 )cos(68.1986 )

13.0345 MPa 13.03 MPa (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + − °+ ° °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(18 MPa) ( 42 MPa)]sin(68.1986 )cos(68.1986 )(30 MPa)[cos (68.1986 ) sin (68.1986 )]

42.4138 MPa 42.4 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° °+ ° − °

= − = − Ans.

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12.18 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at thispoint on the inclined plane shown.

Fig. P12.18

Solution The given stress values are: 24 MPa, 80 MPa, 32 MPa, 33.6901x y xyσ σ τ θ= = = − = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(24 MPa)cos ( 33.6901 ) (80 MPa)sin ( 33.6901 )2( 32 MPa)sin( 33.6901 )cos( 33.6901 )

70.7693 MPa 70.8 MPa (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − °+ − − ° − °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(24 MPa) (80 MPa)]sin( 33.6901 )cos( 33.6901 )( 32 MPa)[cos ( 33.6901 ) sin ( 33.6901 )]

38.1538 MPa 38.2 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° − °+ − − ° − − °

= − = − Ans.

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12.19 The stresses shown in the figure act at a point in astressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

Fig. P12.19

Solution The given stress values are: 3,800 psi, 2,500 psi, 8,200 psi, 59.0362x y xyσ σ τ θ= − = − = = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 3,800 psi)cos ( 59.0362 ) ( 2,500 psi)sin ( 59.0362 )2(8,200 psi)sin( 59.0362 )cos( 59.0362 )

10,079.4185 psi 10,080 psi (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= − − ° + − − °+ − ° − °

= − = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 3,800 psi) ( 2,500 psi)]sin( 59.0362 )cos( 59.0362 )(8,200 psi)[cos ( 59.0362 ) sin ( 59.0362 )]

4,432.3424 psi 4,430 psi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − − ° − °+ − ° − − °

= − = − Ans.

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12.20 The stresses shown in the figure act at a point in a stressedbody. Determine the normal and shear stresses at this point on theinclined plane shown.

Fig. P12.20

Solution The given stress values are: 3.8 ksi, 9.4 ksi, 5.7 ksi, 38.6598x y xyσ σ τ θ= − = = = ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

( 3.8)cos (38.6598 ) (9.4 ksi)sin (38.6598 )2(5.7 ksi)sin(38.6598 )cos(38.6598 )

6.9122 ksi 6.91 ksi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + °+ ° °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[( 3.8 ksi) (9.4 ksi)]sin(38.6598 )cos(38.6598 )(5.7 ksi)[cos (38.6598 ) sin (38.6598 )]

7.6902 ksi 7.69 ksi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° °+ ° − °

= = Ans.

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12.21 The stresses shown in Fig.P12.21a act at a point on the freesurface of a stressed body.Determine the normal stresses σnand σt and the shear stress τnt at this point if they act on the rotatedstress element shown in Fig.P12.21b.

(a) (b) Fig. P12.21

Solution The given stress values are: 50 MPa, 15 MPa, 40 MPa, 36x y xyσ σ τ θ= = − = − = − ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(50 MPa)cos ( 36 ) ( 15 MPa)sin ( 36 ) 2( 40 MPa)sin( 36 )cos( 36 )

65.5853 MPa 65.6 MPa (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − − ° + − − ° − °

= = Ans. To find σt, add 90° to the value of θ used in Eq. (12-3):

2 2

2 2

2 2

cos sin 2 sin cos

(50 MPa)cos ( 36 90 ) ( 15 MPa)sin ( 36 90 )2( 40 MPa)sin( 36 90 )cos( 36 90 )(50 MPa)cos (54 ) ( 15 MPa)sin (54 ) 2( 40 MPa)sin(54 )cos(54 )

30.5853 MPa 30.6 MP

t x y xyσ σ θ σ θ τ θ θ= + +

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

= − = a (C) Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(50 MPa) ( 15 MPa)]sin( 36 )cos( 36 ) ( 40 MPa)[cos ( 36 ) sin ( 36 )]

18.5487 MPa 18.55 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − − ° − ° + − − ° − − °

= = Ans.

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12.22 The stresses shown inFig. P12.22a act at a point on thefree surface of a stressed body.Determine the normal stresses σnand σt and the shear stress τnt at this point if they act on therotated stress element shown inFig. P12.22b.

(a) (b) Fig. P12.22

Solution The given stress values are: 1,200 psi, 700 psi, 400 psi, 20x y xyσ σ τ θ= = = = ° The normal stress transformation equation [Eq. (12-3)] gives σn:

2 2

2 2

cos sin 2 sin cos

(1,200 psi)cos (20 ) (700 psi)sin (20 ) 2(400 psi)sin(20 )cos(20 )

1,398.6262 psi 1,399 psi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

= = Ans. To find σt, add 90° to the value of θ used in Eq. (12-3):

2 2

2 2

2 2

cos sin 2 sin cos

(1,200 psi)cos (20 90 ) (700 psi)sin (20 90 )2(400 psi)sin(20 90 )cos(20 90 )(1,200 psi)cos (110 ) (700 psi)sin (110 ) 2(400 psi)sin(110 )cos(110 )

501.3738 psi 5

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° + °+ ° + ° ° + °= ° + ° + ° °

= = 01 psi (T) Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt:

2 2

2 2

( )sin cos (cos sin )

[(1,200 psi) (700 psi)]sin(20 )cos(20 ) (400 psi)[cos (20 ) sin (20 )]

145.7209 psi 145.7 psi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − ° ° + ° − °

= = Ans.

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12.23 The stresses shown in Fig.P12.23 act at a point on the freesurface of a machine component.Determine the normal stresses σx and σy and the shear stress τxy at the point.

Fig. P12.23

Solution Redefine the axes, calling the rotated axes x and y. The angle from the rotated element to the unrotated element is now a positive value (since it is counterclockwise). Thus, the given stress values can be expressed as: 35 MPa, 27 MPa, 50 MPa, 30x y xyσ σ τ θ= = − = − = ° The normal stress transformation equation [Eq. (12-3)] gives σn, which is actually the normal stress in the horizontal direction (i.e., the original x direction) on the unrotated element:

2 2

2 2

cos sin 2 sin cos

(35 MPa)cos (30 ) ( 27 MPa)sin (30 ) 2( 50 MPa)sin(30 )cos(30 )

23.8013 MPa 23.8 MPa (C)

n x y xyσ σ θ σ θ τ θ θ= + +

= ° + − ° + − ° °

= − = Ans. To find σt, which is actually the normal stress in the vertical direction (i.e., the original y direction) on the unrotated element, add 90° to the value of θ used in Eq. (12-3):

2 2

2 2

cos sin 2 sin cos

(35 MPa)cos (30 90 ) ( 27 MPa)sin (30 90 ) 2( 50 MPa)sin(30 90 )cos(30 90 )t x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + − ° + ° + − ° + ° ° + °

2 2(35 MPa)cos (120 ) ( 27 MPa)sin (120 ) 2( 50 MPa)sin(120 )cos(120 )

31.8013 MPa 31.8 MPa (T)

= ° + − ° + − ° °

= = Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt, which is actually the shear stress on the horizontal and vertical faces of the unrotated element:

2 2

2 2

( )sin cos (cos sin )

[(35 MPa) ( 27 MPa)]sin(30 )cos(30 ) ( 50 MPa)[cos (30 ) sin (30 )]

51.8468 MPa 51.8 MPa

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° ° + − ° − °

= − = − Ans.

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12.24 The stresses shown inFig. P12.24 act at a point on thefree surface of a machinecomponent. Determine thenormal stresses σx and σy and the shear stress τxy at the point.

Fig. P12.24

Solution Redefine the axes, calling the rotated axes x and y. The angle from the rotated element to the unrotated element is now a negative value (since it is clockwise) . Thus, the given stress values can be expressed as: 18.2 ksi, 2.8 ksi, 5.0 ksi, 24x y xyσ σ τ θ= = = = − ° The normal stress transformation equation [Eq. (12-3)] gives σn, which is actually the normal stress in the horizontal direction (i.e., the original x direction) on the unrotated element:

2 2

2 2

cos sin 2 sin cos

(18.2 ksi)cos ( 24 ) (2.8 ksi)sin ( 24 ) 2(5.0 ksi)sin( 24 )cos( 24 )

11.9366 ksi 11.94 ksi (T)

n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − ° + − ° − °

= = Ans. To find σt, which is actually the normal stress in the vertical direction (i.e., the original y direction) on the unrotated element, add 90° to the value of θ used in Eq. (12-3):

2 2

2 2

2 2

cos sin 2 sin cos

(18.2 ksi)cos ( 24 90 ) (2.8 ksi)sin ( 24 90 )2(5.0 ksi)sin( 24 90 )cos( 24 90 )(18.2 ksi)cos (66 ) (2.8 ksi)sin (66 ) 2(5.0 ksi)sin(66 )cos(66 )

9.0634 ksi 9.06

t x y xyσ σ θ σ θ τ θ θ= + +

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

= = ksi (T) Ans. The shear stress transformation equation [Eq. (12-4)] gives τnt, which is actually the shear stress on the horizontal and vertical faces of the unrotated element:

2 2

2 2

( )sin cos (cos sin )

[(18.2 ksi) (2.8 ksi)]sin( 24 )cos( 24 ) (5.0 ksi)[cos ( 24 ) sin ( 24 )]

9.0679 ksi 9.07 ksi

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − − ° − ° + − ° − − °

= = Ans.