Mechanics of Materials Solutions Chapter12 Probs25 46

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.

12.25 Consider a point in a structural member that is subjected to planestress. Normal and shear stresses acting on horizontal and vertical planesat the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress

acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or

Fig. 12-16)

Instructors: Problems 12.25-12.28 should be assigned as a set. Fig. P12.25

Solution The given stress values are: 30 MPa, 10 MPa, 26 MPax y xyσ σ τ= = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(30 MPa) (10 MPa) (30 MPa) (10 MPa) ( 26 MPa)2 2

20.0000 MPa 27.8568 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠

= ± 1 47.86 MPapσ = and 2 7.86 MPapσ = − Ans.

max 27.86 MPa (maximum in-plane shear stress)τ = Ans.

avg 20.00 MPa (T) (normal stress on planes of maximum in-plane shear stress)σ = Ans.

1

26 MPatan 2 2.600( ) / 2 [(30 MPa) (10 MPa)] / 2

34.48 (clockwise from the axis to the direction of )

xyp

x y

p px

τθ

σ σ

θ σ

−= = = −− −

∴ = − ° Ans.




Fig. 12-16)


Solution The given stress values are: 16 MPa, 22 MPa, 28 MPax y xyσ σ τ= − = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 16 MPa) (22 MPa) ( 16 MPa) (22 MPa) ( 28 MPa)2 2

3.0000 MPa 33.8378 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠




2

28 MPatan 2 1.4737( ) / 2 [( 16 MPa) (22 MPa)] / 2

27.92 (counterclockwise from the axis to the direction of )

xyp

x y

p px

τθ

σ σ

θ σ

−= = =− − −

∴ = ° Ans.




Fig. 12-16)


Solution The given stress values are: 4 ksi, 20 ksi, 15 ksix y xyσ σ τ= − = = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 4 ksi) (20 ksi) ( 4 ksi) (20 ksi) (15 ksi)2 2

8.0000 ksi 19.2094 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠

= ± 1 27.21 ksipσ = and 2 11.21 ksipσ = − Ans.

max 19.21 ksi (maximum in-plane shear stress)τ = Ans.

avg 8.00 ksi (T) (normal stress on planes of maximum in-plane shear stress)σ = Ans.

2

15 ksitan 2 1.2500( ) / 2 [( 4 ksi) (20 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = = −− − −

∴ = − ° Ans.




Fig. 12-16)


Solution The given stress values are: 60 ksi, 10 ksi, 26 ksix y xyσ σ τ= = − = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(60 ksi) ( 10 ksi) (60 ksi) ( 10 ksi) (26 ksi)2 2

25.0000 ksi 43.6005 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ − − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠

= ± 1 68.60 ksipσ = and 2 18.60 ksipσ = − Ans.



1

26 ksitan 2 0.7429( ) / 2 [(60 ksi) ( 10 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = =− − −

∴ = ° Ans.


12.29 Consider a point in a structural member that is subjected to planestress. Normal and shear stresses acting on horizontal and verticalplanes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear

stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or

Fig. 12-16) (c) Compute the absolute shear stress at the point.


Solution The given stress values are: 35 MPa, 85 MPa, 30 MPax y xyσ σ τ= − = − = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 35 MPa) ( 85 MPa) ( 35 MPa) ( 85 MPa) ( 30 MPa)2 2

60.0000 MPa 39.0512 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − − − −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠

= − ± 1 20.95 MPapσ = − and 2 99.05 MPapσ = − Ans.


avg 60.00 MPa (C) (normal stress on planes of maximum in-plane shear stress)σ = Ans.

1

30 MPatan 2 1.2000( ) / 2 [( 35 MPa) ( 85 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = = −− − − −

∴ = − ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both negative,

2abs max

99.05 MPa49.5 MPa

2 2pσ

τ−

= = = Ans.


12.30 Consider a point in a structural member that is subjected toplane stress. Normal and shear stresses acting on horizontal andvertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear

stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15

or Fig. 12-16) (c) Compute the absolute shear stress at the point.


Solution The given stress values are: 16 MPa, 45 MPa, 10 MPax y xyσ σ τ= = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(16 MPa) (45 MPa) (16 MPa) (45 MPa) ( 10 MPa)2 2

30.5000 MPa 17.6139 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠

= ± 1 48.11 MPapσ = and 2 12.89 MPapσ = Ans.



2

10 MPatan 2 0.6897( ) / 2 [(16 MPa) (45 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = =− −

∴ = ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both positive,

1abs max

48.11 MPa 24.06 MPa2 2pσ

τ = = = Ans.






Solution The given stress values are: 66 MPa, 90 MPa, 114 MPax y xyσ σ τ= − = = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 66 MPa) (90 MPa) ( 66 MPa) (90 MPa) (114 MPa)2 2

12.0000 MPa 138.1304 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠




2

114 MPatan 2 1.4615( ) / 2 [( 66 MPa) (90 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = = −− − −

∴ = − ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 is positive and σp2 is negative, abs max max 138.13 MPaτ τ= = Ans.






Solution The given stress values are: 35 ksi, 15 ksi, 14 ksix y xyσ σ τ= = = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(35 ksi) (15 ksi) (35 ksi) (15 ksi) (14 ksi)2 2

25.0000 ksi 17.2047 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± +⎜ ⎟⎝ ⎠

= ± 1 42.20 ksipσ = and 2 7.80 ksipσ = Ans.



1

14 ksitan 2 1.4000( ) / 2 [(35 ksi) (15 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = =− −

∴ = ° Ans.


1abs max

42.20 ksi 21.10 ksi2 2pσ

τ = = = Ans.




or Fig. 12-16) (c) Compute the absolute maximum shear stress at the point.


Solution The given stress values are: 45 ksi, 15 ksi, 8 ksix y xyσ σ τ= = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(45 ksi) (15 ksi) (45 ksi) (15 ksi) ( 8 ksi)2 2

30.0000 ksi 13.0000 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠




1

8 ksitan 2 0.5333( ) / 2 [(45 ksi) (15 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = = −− −

∴ = − ° Ans.


1abs max

47.00 ksi 23.50 ksi2 2pσ

τ = = = Ans.






Solution The given stress values are: 12 ksi, 4 ksi, 14 ksix y xyσ σ τ= − = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 12 ksi) (4 ksi) ( 12 ksi) (4 ksi) ( 14 ksi)2 2

4.0000 ksi 16.1245 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠

= − ± 1 12.12 ksipσ = and 2 20.12 ksipσ = − Ans.


avg 4.00 ksi (C) (normal stress on planes of maximum in-plane shear stress)σ = Ans.

2

14 ksitan 2 1.7500( ) / 2 [( 12 ksi) (4 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = =− − −

∴ = ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 is positive and σp2 is negative, abs max max 16.12 ksiτ τ= = Ans.






Solution The given stress values are: 50 MPa, 35 MPa, 16 MPax y xyσ σ τ= − = − = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 50 MPa) ( 35 MPa) ( 50 MPa) ( 35 MPa) (16 MPa)2 2

42.5000 MPa 17.6706 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − − − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠




2

16 MPatan 2 2.1333( ) / 2 [( 50 MPa) ( 35 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = = −− − − −

∴ = − ° Ans.


2abs max

60.17 MPa30.09 MPa

2 2pσ

τ−

= = = Ans.






Solution The given stress values are: 70 MPa, 85 MPa, 56 MPax y xyσ σ τ= = − = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(70 MPa) ( 85 MPa) (70 MPa) ( 85 MPa) (56 MPa)2 2

7.5000 MPa 95.6151 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ − − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠

= − ± 1 88.12 MPapσ = and 2 103.12 MPapσ = − Ans.



1

56 MPatan 2 0.7226( ) / 2 [(70 MPa) ( 85 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = =− − −

∴ = ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 is positive and σp2 is negative, abs max max 95.62 MPaτ τ= = Ans.






Solution The given stress values are: 10 ksi, 50 ksi, 32 ksix y xyσ σ τ= − = − = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 10 ksi) ( 50 ksi) ( 10 ksi) ( 50 ksi) ( 32 ksi)2 2

30.0000 ksi 37.7359 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − − − −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠

= − ± 1 7.74 ksipσ = and 2 67.74 ksipσ = − Ans.


avg 30.00 ksi (C) (normal stress on planes of maximum in-plane shear stress)σ = Ans.

1

32 ksitan 2 1.6000( ) / 2 [( 10 ksi) ( 50 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = = −− − − −

∴ = − ° Ans.

(c) For plane stress, σz = σp3 = 0. Since σp1 is positive and σp2 is negative, abs max max 37.74 ksiτ τ= = Ans.






Solution The given stress values are: 30 ksi, 50 ksi, 14 ksix y xyσ σ τ= = = − The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(30 ksi) (50 ksi) (30 ksi) (50 ksi) ( 14 ksi)2 2

40.0000 ksi 17.2047 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± + −⎜ ⎟⎝ ⎠




2

14 ksitan 2 1.4000( ) / 2 [(30 ksi) (50 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

−= = =− −

∴ = ° Ans.


1abs max

57.20 ksi 28.60 ksi2 2pσ

τ = = = Ans.






Solution The given stress values are: 3.5 ksi, 6.0 ksi, 2.8 ksix y xyσ σ τ= = = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

(3.5 ksi) (6.0 ksi) (3.5 ksi) (6.0 ksi) (2.8 ksi)2 2

4.7500 ksi 3.0663 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± +⎜ ⎟⎝ ⎠




2

2.8 ksitan 2 2.2400( ) / 2 [(3.5 ksi) (6.0 ksi)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = = −− −

∴ = − ° Ans.


1abs max

7.82 ksi 3.91 ksi2 2pσ

τ = = = Ans.






Solution The given stress values are: 50 MPa, 90 MPa, 30 MPax y xyσ σ τ= − = − = The principal stress magnitudes can be computed from Eq. (12-12):

22

1, 2

22

2 2

( 50 MPa) ( 90 MPa) ( 50 MPa) ( 90 MPa) (30 MPa)2 2

70.0000 MPa 36.0555 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

− + − − − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠




1

30 MPatan 2 1.5000( ) / 2 [( 50 MPa) ( 90 MPa)] / 2


xyp

x y

p px

τθ

σ σ

θ σ

= = =− − − −

∴ = ° Ans.


2abs max

106.06 MPa53.03 MPa

2 2pσ

τ−

= = = Ans.


12.41 The principal compressive stress on a vertical plane through a point in a wooden block is equal tothree times the principal compression stress on a horizontal plane. The plane of the grain is 25°clockwise from the vertical plane. If the normal and shear stresses must not exceed 400 psi (C) and 90psi shear, determine the maximum allowable compressive stress on the horizontal plane.

Solution The principal compressive stress on a vertical plane (that is, the x face of a stress element) is equal to three times the principal compression stress on a horizontal plane (that is, the y face of a stress element). Thus, from the problem statement, we know that σx = 3σy. Since we are told that the stresses on the x and y faces are principal stress, we also know that τxy = 0. The plane of the wood grain is oriented 25° clockwise from the vertical plane; therefore, θ = −25°. We are told that the normal stress on the plane of the wood grain must not exceed −400 psi, or in other words, σn ≤ −400 psi. The normal stress transformation equation [Eq. (12-3)], which gives σn, can be rearranged to solve for σy:

2 2

2 2

2 2

2 2

cos sin 2 sin cos

400 psi 3 cos ( 25 ) sin ( 25 ) 2(0 psi)sin( 25 )cos( 25 )

400 psi [3cos ( 25 ) sin ( 25 )]

400 psi 400 psi 151.3546 psi[3cos ( 25 ) sin ( 25 )] 2.6428

n x y xy

y y

y

y

σ σ θ σ θ τ θ θ

σ σ

σ

σ

= + +

− ≥ − ° + − ° + − ° − °

− ≥ − ° + − °

− −∴ ≤ = = −− ° + − °

(a)

A second condition of the stresses acting on the plane of the wood grain is that the shear stress must not exceed 90 psi, or in other words, τnt ≤ 90 psi. The shear stress transformation equation [Eq. (12-4)], which gives τnt, can be rearranged to solve for τnt:

2 2

2 2

( )sin cos (cos sin )

90 psi [3 ]sin( 25 )cos( 25 ) (0 psi)[cos ( 25 ) sin ( 25 )]

90 psi 2 sin( 25 )cos( 25 )

90 psi 90 psi 117.4935 psi2sin( 25 )cos( 25 ) 0.7660

nt x y xy

y y

y

y

τ σ σ θ θ τ θ θ

σ σσ

σ

= − − + −

± ≥ − − − ° − ° + − ° − − °

± ≥ − − ° − °

± ±∴ ≤ = = ±− − ° − °

(b)

Since we are told that σy is a compressive normal stress, it is clear that we must choose the negative value for σy. Compare the two limits found in Eqs. (a) and (b) to find that the maximum compression stress that may be applied to the horizontal plane is 117.5 psiyσ ≤ − Ans.


12.42 At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an unknownpositive shear stress exist on a horizontal plane. One principal stress at the point is 8 MPa (C). Theabsolute maximum shear stress at the point has a magnitude of 95 MPa. Determine the unknown stresses on the horizontal and vertical planes and the unknown principal stress at the point.

Solution The absolute maximum shear stress can be found from Eq. (12-18)

max minabs max 2

σ στ −=

The absolute maximum shear stress at the point has a magnitude of 95 MPa. Suppose we assume that the given principal stress of −8 MPa is σmin. If this assumption is true, then max min abs max2 8 MPa 2(95 MPa) 182 MPaσ σ τ= + = − + = However, this assumption cannot be true because the normal stress on the horizontal plane is σy = −64 MPa, which is more negative than the given principal stress of −8 MPa. Therefore, we now know that the second principal stress must be negative and its magnitude must be greater than 64 MPa. The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore, 3 0 (since it is a free surface)z pσ σ= = Since both of the in-plane principal stresses must be negative, σmax = σp3 = 0. The minimum principal stress can now be determined from the absolute maximum shear stress: min max abs max2 0 MPa 2(95 MPa) 190 MPaσ σ τ= − = − = − Thus, the two in-plane principal stresses are: 1 8 MPapσ = − and 2 190 MPapσ = − Ans. Since σy is given, σx can easily be determined from the principal of stress invariance:

1 2

1 2 ( 8 MPa) ( 190 MPa) ( 64 MPa) 134 MPax y p p

x p p y

σ σ σ σ

σ σ σ σ

+ = +

∴ = + − = − + − − − = − Ans. The maximum in-plane shear stress can be found from

1 2max

( 8 MPa) ( 190 MPa) 91 MPa2 2

p pσ στ

− − − −= = =

Since σx, σy, and τmax are known, the magnitude of τxy can be found from the expression

22

max

22

2

( 134 MPa) ( 64 MPa)91 MPa2

84 MPa

x yxy

xy

xy

σ στ τ

τ

τ

−⎛ ⎞= +⎜ ⎟

⎝ ⎠

− − −⎛ ⎞= +⎜ ⎟⎝ ⎠

∴ = ± The problem states that a positive shear stress exists on a horizontal plane; therefore 84 MPaxyτ = Ans.


12.43 At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a verticalplane and 30 ksi (C) on a horizontal plane. An unknown negative shear stress exists on the verticalplane. The absolute maximum shear stress at the point has a magnitude of 32 ksi. Determine theprincipal stresses and the shear stress on the vertical plane at the point.

Solution Since σx and σy have opposite signs, the absolute maximum shear stress is equal to the maximum in-plane shear stress: max abs max 32 ksiτ τ= = Since σx, σy, and τmax are known, the magnitude of τxy can be found from the expression

22

max

22

2

(20 ksi) ( 30 ksi)32 ksi2

19.9750 ksi

x yxy

xy

xy

σ στ τ

τ

τ

−⎛ ⎞= +⎜ ⎟

⎝ ⎠

− −⎛ ⎞= +⎜ ⎟⎝ ⎠

∴ = ± The problem states that a negative shear stress exists on the vertical plane; therefore 19.98 ksixyτ = − Ans. From the principal of stress invariance:

1 2

1 2 (20 ksi) ( 30 ksi) 10 ksix y p p

p p

σ σ σ σσ σ

+ = +

∴ + = + − = − (a) The maximum in-plane shear stress is equal to one-half of the difference between the two in-plane principal stresses

1 2max

1 2 max

22 2(32 ksi) 64 ksi

p p

p p

σ στ

σ σ τ

−=

∴ − = = = (b) Add Eqs. (a) and (b) to find σp1: 1 p12 54 ksi 27 ksi 27 ksi (T)pσ σ= ∴ = = Ans. and subtract Eq. (b) from Eq. (a) to find σp2: 2 p22 74 ksi 37 ksi 37 ksi (C)pσ σ= − ∴ = − = Ans. The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore, 3 0 (since the point is on a free surface)z pσ σ= = Ans.


12.44 At a point on the free surface of a stressed body, a normal stress of 75 MPa (T) and an unknownnegative shear stress exist on a horizontal plane. One principal stress at the point is 200 MPa (T). The maximum in-plane shear stress at the point has a magnitude of 85 MPa. Determine the unknownstresses on the vertical plane, the unknown principal stress, and the absolute maximum shear stress at thepoint.

Solution Since σy = 75 MPa is less than the given principal stress, we will assume that σp1 = 200 MPa. If this assumption is true, then σp2 can be found from σp1 and τmax:

1 2max

2 1 max

22 200 MPa 2(85 MPa) 30 MPa

p p

p p

σ στ

σ σ τ

−=

∴ = − = − = Ans. The maximum in-plane shear stress at the point has a magnitude of 85 MPa. However, this assumption cannot be true because the normal stress on the horizontal plane is σy = −64 MPa, which is more negative than the given principal stress of −8 MPa. Therefore, we now know that the second principal stress must be negative and its magnitude must be greater than 64 MPa. From the principal of stress invariance:

1 2

(200 MPa) (30 MPa) (75 MPa) 155 MPax y p p

x

σ σ σ σ

σ

+ = +

∴ = + − = Ans. Since σx, σy, and τmax are known, the magnitude of τxy can be found from the expression

22

max

22

2

(155 MPa) (75 MPa)85 MPa2

75 MPa

x yxy

xy

xy

σ στ τ

τ

τ

−⎛ ⎞= +⎜ ⎟

⎝ ⎠

−⎛ ⎞= +⎜ ⎟⎝ ⎠

∴ = ± The problem states that a negative shear stress exists on the vertical plane; therefore 75 MPaxyτ = − Ans. The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore, 3 0 (since it is a free surface)z pσ σ= = Since both in-plane principal stresses are positive, the absolute maximum shear stress is found from

1 3 1abs max

0 200 MPa 100 MPa2 2 2

p p pσ σ στ

− −= = = = Ans.


12.45 For the state of plane stress shown, determine (a) the largest value of σy for which the maximum in-plane shear stress is equal to or less than 16 ksi and (b) the corresponding principal stresses.

Fig. P12.45

Solution Since σx, τxy, and τmax are known, the magnitude of σy can be found from the expression

22

max

22

2 2

2 2 2 2

2

(30 ksi)16 ksi (10 ksi)

2

(30 ksi)(16 ksi) (10 ksi)

230 ksi 2 (16 ksi) (10 ksi) 30 ksi 2 (16 ksi) (10 ksi)

x yxy

y

y

y

σ στ τ

σ

σ

σ

−⎛ ⎞= +⎜ ⎟

⎝ ⎠

−⎛ ⎞≥ +⎜ ⎟

⎝ ⎠−

− ≥

− − ≤ ≤ + −

5.02 ksi 54.98 ksi

54.98 ksiy

y

σ

σ

≤ ≤

∴ ≤ Ans.

The corresponding principal stresses are:

22

1, 2

22

2 2

(30 ksi) (54.98 ksi) (30 ksi) (54.98 ksi) (16 ksi)2 2

42.49 ksi 16.0 ksi

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ −⎛ ⎞= ± +⎜ ⎟⎝ ⎠



12.46 For the state of plane stress shown, determine (a) the largest value of τxy for which the maximum in-plane shear stress is equal to or less than 150 MPa and (b) the corresponding principal stresses.

Fig. P12.46

Solution Since σx, σy, and τmax are known, the magnitude of τxy can be found from the expression

22

max

22

2 2 2

2

(120 MPa) ( 70 MPa)150 MPa2

(150 MPa) (95 MPa)

x yxy

xy

xy

σ στ τ

τ

τ

−⎛ ⎞= +⎜ ⎟

⎝ ⎠

− −⎛ ⎞≥ +⎜ ⎟⎝ ⎠

≤ −

116.1 MPaxyτ∴ ≤ Ans.

The corresponding principal stresses are:

22

1, 2

22

2 2

(120 MPa) ( 70 MPa) (120 MPa) ( 70 MPa) (116.1 MPa)2 2

25 MPa 150 MPa

x y x yp p xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

+ − − −⎛ ⎞= ± +⎜ ⎟⎝ ⎠

= ± 1 175 MPa (T)pσ = and 2 125 MPa (C)pσ = Ans.

Mechanics of Materials Solutions Chapter12 Probs25 46

Documents