mechanics of materials solutions chapter12 probs47 74

7/25/2019 Mechanics of Materials Solutions Chapter12 Probs47 74

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12.47 Mohr’s circle is shown for a point in a

physical object that is subjected to plane stress.

(a) Determine the stresses σ x, σ y, and τ xy and show

them on a stress element.(b) Determine the principal stresses and the

maximum in-plane shear stress acting at the point

and show these stresses on an appropriate sketch

(e.g., see Fig. 12-15 or Fig. 12-16).

Fig. P12.47

Solution

2 2

1

2

max

avg

(35 ksi, 20 ksi)

(5 ksi, 20 ksi)

20 ksi

(15 ksi) (20 ksi) 25 ksi

20 ksi 25 ksi 45 ksi

20 ksi 25 ksi 5 ksi

25 ksi

20 ksi

p

p

x

y

C

R

C R

C R

R

C

σ

σ

τ

σ

=

=

=

= + =

= + = + =

= − = − = −

= =

= =

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress element) and σ p1 is foundfrom:

20 ksi 20 ksitan 2 1.3333 2 53.13 thus, 26.57

(35 ksi) (20 ksi) 15 ksi p p pθ θ θ = = = ∴ = ° = °

−

By inspection, the angle θ p from point x to σ p1 is turned clockwise.



Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposesonly 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.






and show these stresses on an appropriate sketch(e.g., see Fig. 12-15 or Fig. 12-16).

Fig. P12.48

Solution

2 2

1

2

max

avg

(90 ksi, 60 ksi)

(30 ksi, 60 ksi)

60 ksi

(30 ksi) (60 ksi) 67.08 ksi

60 ksi 67.08 ksi 127.08 ksi

60 ksi 67.08 ksi 7.08 ksi

67.08 ksi

60 ksi

p

p

x

y

C

R

C R

C R

R

C

σ

σ

τ

σ

=

=

=

= + =

= + = + =

= − = − = −

= =

= =


60 ksi 60 ksitan 2 2 2 63.43 thus, 31.72


−

By inspection, the angle θ p from point x to σ p1 is turned counterclockwise.











Fig. P12.49

Solution

2 2

1

2

max

avg

( 100 MPa, 30 MPa)

(20 MPa, 30 MPa)

40 MPa

(60 MPa) (30 MPa) 67.08 MPa

40 MPa 67.08 MPa 27.08 MPa

40 MPa 67.08 MPa 107.08 MPa

67.08 MPa

60 MPa

p

p

x

y

C

R

C R

C R

R

C

σ

σ

τ

σ

= −

=

= −

= + =

= + = − + =

= − = − − = −

= =

= =

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress element) and σ p2 is found

from:

30 MPa 30 MPatan 2 0.5 2 26.57 thus, 13.28

( 100 MPa) ( 40 MPa) 60 MPa p p pθ θ θ = = = ∴ = ° = °

− − −

By inspection, the angle θ p from point x to σ p2 is turned clockwise.











Fig. P12.50

Solution

2 2

1

2

max

avg

( 55 MPa, 25 MPa)

(15 MPa, 25 MPa)

20 MPa

(35 MPa) (25 MPa) 43.01 MPa

20 MPa 43.01 MPa 23.01 MPa

20 MPa 43.01 MPa 63.01 MPa

43.01 MPa

20 MPa (C)

p

p

x

y

C

R

C R

C R

R

C

σ

σ

τ

σ

= −

=

= −

= + =

= + = − + =

= − = − − = −

= =

= =


25 MPa 25 MPatan 2 0.7143 2 35.54 thus, 17.77


− − −

By inspection, the angle θ p from point x to σ p2 is turned counterclockwise.








them on a stress element.

(b) Determine the stresses σ n, σ t , and τ nt and show

them on a stress element that is properly rotated

with respect to the x-y element. The sketch mustinclude the magnitude of the angle between the x

and n axes and an indication of the rotation direction(i.e., either clockwise or counterclockwise).

Fig. P12.51

Solution

2 2

(60 MPa, 15 MPa) ( 20 MPa, 15 MPa)

( 10 MPa, 30 MPa) (50 MPa, 30 MPa)

20 MPa (40 MPa) (15 MPa) 42.72 MPa

x y

n t

C R

= = −

= − =

= = + =

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress element) and point 1 is

found from:

15 MPa 15 MPatan 2 0.3750 2 20.6

(60 MPa) (20 MPa) 40 MPa p pθ θ = = = ∴ = °

−

The magnitude of the angle β between point n and point 2 is found from:

30 MPa 30 MPatan 1 45

( 10 MPa) (20 MPa) 30 MPa β β = = = ∴ = °

− −





The angle α between point x and point n is thus180 20.6 45 114.4α = ° − ° − ° = °

Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half ofthis magnitude: 57.2°. By inspection, the 57.2° angle from point x to point n is turned in a

counterclockwise direction. The correct stresses on the n and t faces are shown in the sketch below.








them on a stress element.

(b) Determine the stresses σ n, σ t , and τ nt and show

them on a stress element that is properly rotated

with respect to the x-y element. The sketch mustinclude the magnitude of the angle between the x

and n axes and an indication of the rotation direction(i.e., either clockwise or counterclockwise).

Fig. P12.52

Solution

2 2

(10 ksi, 30 ksi) (50 ksi, 30 ksi)

(65 ksi, 10 ksi) ( 5 ksi, 10 ksi)

30 ksi (20 ksi) (30 ksi) 36.06 ksi

x y

n t

C R

= =

= = −

= = + =

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress element) and point 2 isfound from:

30 ksi 30 ksitan 2 1.5 2 56.3

(10 ksi) (30 ksi) 20 ksi p pθ θ = = = ∴ = °

−

The magnitude of the angle β between point n and point 1 is found from:

10 ksi 10 ksitan 0.2857 15.9

(65 ksi) (30 ksi) 35 ksi β β = = = ∴ = °

−





The angle α between point x and point n is thus180 56.3 15.9 107.8α = ° − ° − ° = °

Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half ofthis magnitude: 53.9°. By inspection, the 53.9° angle from point x to point n is turned in a clockwise

direction. The correct stresses on the n and t faces are shown in the sketch below.





12.53 Consider a point in a structural member that is subjected to

plane stress. Normal and shear stresses acting on horizontal andvertical planes at the point are shown.

(a) Draw Mohr’s circle for this state of stress.

(b) Determine the principal stresses and the maximum in-plane shearstress acting at the point using Mohr’s circle.

(c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15or Fig. 12-16).

Instructors: Problems 12.53-12.56 should be assigned as a set. Fig. P12.53

Solution

(b) The basic Mohr’s circle is shown.

2 2

1

2

max

avg

( 15 ksi) (5 ksi)5 ksi

2

(10 ksi) (12.5 ksi) 16.01 ksi

5 ksi 16.01 ksi 11.01 ksi

5 ksi 16.01 ksi 21.01 ksi

16.01 ksi

5 ksi

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− += = −

= + =

= + = − + =

= − = − − = −

= =

= = −

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress element) and point 2

(i.e., the principal plane subjected to σ p2) is found from:

12.5 ksi 12.5 ksitan 2 1.25 2 51.340 thus, 25.67

( 15 ksi) ( 5 ksi) 10 ksi p p pθ θ θ = = = ∴ = ° = °

− − −

By inspection, the angle θ p from point x to point 2 is turned clockwise.

(c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the

sketch below.











Solution


2 2

1

2

max

avg

(28 MPa) ( 50 MPa)11 MPa

2

(39 MPa) (44 MPa) 58.80 MPa

11 MPa 58.80 MPa 47.80 MPa

11 MPa 58.80 MPa 69.80 MPa

58.80 MPa

11 MPa

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+ −= = −

= + =

= + = − + =

= − = − − = −

= =

= = −

The magnitude of the angle 2θ p between point x (i.e.,the x face of the stress element) and point 1 (i.e., the

principal plane subjected to σ p1) is found from:

44 MPa 44 MPatan 2 1.1282 2 48.447 thus, 24.22

(28 MPa) ( 11 MPa) 39 MPa p p pθ θ θ = = = ∴ = ° = °

− −

By inspection, the angle θ p from point x to point 1 is turned counterclockwise.


sketch below.











Solution


2 2

1

2

max

avg

(16 ksi) ( 3 ksi)6.5 ksi

2

(9.5 ksi) (8 ksi) 12.42 ksi

6.5 ksi 12.42 ksi 18.92 ksi

6.5 ksi 12.42 ksi 5.92 ksi

16.01 ksi

6.5 ksi

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+ −= =

= + =

= + = + =

= − = − = −

= =

= =


(i.e., the principal plane subjected to σ p1) is foundfrom:

8 ksi 8 ksitan 2 0.8421 2 40.101 thus, 20.05

(16 ksi) (6.5 ksi) 9.5 ksi p p pθ θ θ = = = ∴ = ° = °

−



sketch below.











Solution


2 2

1

2

max

avg

( 12 ksi) (30 ksi)9.0 ksi

2

(21 ksi) (18 ksi) 27.66 ksi

9.0 ksi 27.66 ksi 36.66 ksi

9.0 ksi 27.66 ksi 18.66 ksi

27.66 ksi

9.0 ksi

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− += =

= + =

= + = + =

= − = − = −

= =

= =


(i.e., the principal plane subjected to σ p2) is foundfrom:

18 ksi 18 ksitan 2 0.8571 2 40.601 thus, 20.30

( 12 ksi) (9 ksi) 21 ksi p p pθ θ θ = = = ∴ = ° = °

− −



sketch below.








(b) Determine the principal stresses and the maximum in-plane shearstress acting at the point.

(c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15or Fig. 12-16).(d) Determine the absolute maximum shear stress at the point.


Solution

(b) The basic Mohr’s circle construction is shown.

2 2

1

2

max

avg

(6 ksi) (18 ksi)12 ksi

2

(6 ksi) (30 ksi) 30.59 ksi

12 ksi 30.59 ksi 42.59 ksi

12 ksi 30.59 ksi 18.59 ksi

30.59 ksi

12 ksi

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − = −

= =

= =



30 ksi 30 ksitan 2 5 2 78.690 thus, 39.35


−



sketch below.





(d) Since the point in a structural member is subjected to plane stress

3 0 z pσ σ = =

Three Mohr’s circles can be constructed to show stress combinations in the σ p1 – σ p2 plane, the σ p1 – σ p3

plane, and the σ p2 – σ p3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σ p1 – σ p2 plane (which is also the x-y plane). Therefore

abs max max 30.59 ksiτ τ = = Ans.











Solution


2 2

1

2

max

avg

( 35 MPa) ( 65 MPa)50 MPa

2

(15 MPa) (24 MPa) 28.30 MPa

50 MPa 28.30 MPa 21.70 MPa

50 MPa 28.30 MPa 78.30 MPa

28.30 MPa

50 MPa (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− + −= = −

= + =

= + = − + = −

= − = − − = −

= =

= =



24 MPa 24 MPatan 2 1.6 2 57.995 thus, 29.00


− − −



sketch below.






3 0 z pσ σ = =



In this case, the absolute maximum shear stress occurs in the σ p2 – σ p3 plane. Therefore

2 3

abs max

78.30 MPa 0 MPa39.15 MPa

2 2

p pσ σ τ

− − −= = = Ans.








(b) Determine the principal stresses and the maximum in-plane shear stress acting at the point.



Solution


2 2

1

2

max

avg

(0 MPa) ( 45 MPa)22.5 MPa

2

(22.5 MPa) (25 MPa) 33.63 MPa

22.5 MPa 33.63 MPa 11.13 MPa

22.5 MPa 33.63 MPa 56.13 MPa

33.63 MPa

22.5 MPa (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+ −= = −

= + =

= + = − + =

= − = − − = −

= =

= =



25 MPa 25 MPatan 2 1.1111 2 48.013 thus, 24.01

(0 MPa) ( 22.5 MPa) 22.5 MPa p p pθ θ θ = = = ∴ = ° = °

− −



sketch below.






3 0 z pσ σ = =




abs max max 33.63 MPaτ τ = = Ans.











Solution


2 2

1

2

max

avg

( 36 ksi) ( 18 ksi)27 ksi

2

(9 ksi) (12 ksi) 15 ksi



15 ksi

27 ksi (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− + −= = −

= + =

= + = − + = −

= − = − − = −

= =

= =

The magnitude of the angle 2θ p between point x (i.e., the x face of the stress

element) and point 2 (i.e., the principal plane subjected to σ p2) is found from:

12 ksi 12 ksitan 2 1.3333 2 53.130 thus, 26.57

( 36 ksi) ( 27 ksi) 9 ksi p p pθ θ θ = = = ∴ = ° = °

− − −



sketch below.






3 0 z pσ σ = =



In this case, the absolute maximum shear stress occurs in the σ p2 – σ p3 plane; therefore,

2 3

abs max

42 ksi 0 ksi21 ksi

2 2

p pσ σ τ

− − −= = = Ans.











Solution


2 2

1

2

max

avg

(60 MPa) (90 MPa)75 MPa

2

(15 MPa) (30 MPa) 33.54 MPa

75 MPa 33.54 MPa 108.54 MPa

75 MPa 33.54 MPa 41.46 MPa

33.54 MPa

75 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − =

= =

= =



30 MPa 30 MPatan 2 2 2 63.435 thus, 31.72

(60 MPa) (75 MPa) 15 MPa p p pθ θ θ = = = ∴ = ° = °

−


(c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on thesketch below.






3 0 z pσ σ = =




1 3

abs max

108.54 MPa 0 MPa54.27 MPa

2 2

p pσ σ τ

− −= = = Ans.











Solution


2 2

1

2

max

avg

(55 MPa) (4 MPa)29.5 MPa

2

(25.5 MPa) (28 MPa) 37.87 MPa

29.5 MPa 37.87 MPa 67.37 MPa

29.5 MPa 37.87 MPa 8.37 MPa

37.87 MPa

29.5 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − = −

= =

= =

The magnitude of the angle 2θ p between point x (i.e.,

the x face of the stress element) and point 2 (i.e., the


28 MPa 28 MPatan 2 1.0980 2 47.675 thus, 23.84

(55 MPa) (29.5 MPa) 25.5 MPa p p pθ θ θ = = = ∴ = ° = °

−


(c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on thesketch below.






3 0 z pσ σ = =















Solution


2 2

1

2

max

avg

( 250 psi) ( 500 psi)375 psi

2

(125 psi) (200 psi) 235.8 psi

375 psi 235.8 psi 139.2 psi

375 psi 235.8 psi 610.8 psi

235.8 psi

375 psi (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− + −= = −

= + =

= + = − + = −

= − = − − = −

= =

= =

The magnitude of the angle 2θ p between point x

(i.e., the x face of the stress element) and point 2

(i.e., the principal plane subjected to σ p2) is found from:200 psi 200 psitan 2 1.6 2 57.995 thus, 29.00

( 250 psi) ( 375 psi) 125 psi p p pθ θ θ = = = ∴ = ° = °

− − −



sketch below.






3 0 z pσ σ = =




2 3

abs max

610.8 psi 0 psi305.4 psi

2 2

p pσ σ τ

− − −= = = Ans.











Solution


2 2

1

2

max

avg

(4.5 ksi) (9.1 ksi)6.8 ksi

2

(2.3 ksi) (2 ksi) 3.05 ksi

6.8 ksi 3.05 ksi 9.85 ksi

6.8 ksi 3.05 ksi 3.75 ksi

3.05 ksi

6.8 ksi (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − =

= =

= =


(i.e., the principal plane subjected to σ p2) isfound from:

2 ksi 2 ksitan 2 0.8696 2 41.009 thus, 20.50

(4.5 ksi) (6.80 ksi) 2.3 ksi p p pθ θ θ = = = ∴ = ° = °

−



sketch below.






3 0 z pσ σ = =




1 3

abs max

9.85 ksi 0 ksi4.92 ksi

2 2

p pσ σ τ

− −= = = Ans.








(b) Determine the principal stresses and the maximum in-plane shearstress acting at the point. Show these stresses on an appropriate

sketch (e.g., see Fig. 12-15 or Fig. 12-16).(c) Determine the normal and shear stresses on the indicated planeand show these stresses on a sketch.

(d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.65-12.68 should be assigned as a set. Fig. P12.65

Solution


2 2

1

2

max

avg

( 90 MPa) ( 65 MPa)77.5 MPa

2

(12.5 MPa) (42 MPa) 43.82 MPa

77.5 MPa 43.82 MPa 33.68 MPa

77.5 MPa 43.82 MPa 121.32 MPa

43.82 MPa

77.5 MPa (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− + −= = −

= + =

= + = − + = −

= − = − − = −

= =

= =



42 MPa 42 MPatan 2 3.3600 2 73.426 thus, 36.71

( 90 MPa) ( 77.5 MPa) 12.5 MPa p p pθ θ θ = = = ∴ = ° = °

− − −

By inspection, the angle θ p from point x to point 2 is turned clockwise. The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.





(c) To determine the normal and shear stresses on the indicated plane, we must first determine the

orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 35° counterclockwise from the x axis. In

Mohr’s circle, all angle measures are doubled; therefore, point n (which represents the state of stress

on the n plane) on Mohr’s circle is rotated 2(35°) = 70° counterclockwise from point x. The angle between point n and point 1 is

180 73.426 70 36.574 β = ° − ° − ° = °

The σ coordinate of point n is found from:

cos

77.5 MPa (43.82 MPa)cos(36.574 )

42.31 MPa 42.31 MPa (C)

n C Rσ β = +

= − + °

= − = Ans.

The τ coordinate of point n is found from:

sin

(43.82 MPa)sin(36.574 ) 26.11 MPa

nt Rτ β =

= ° = Ans.

Since point n is below the σ axis, the shear stress acting on the plane

surface tends to rotate the stress element counterclockwise.


3 0 z pσ σ = =




2 3

abs max

121.32 MPa 0 MPa60.66 MPa

2 2

p pσ σ τ

− − −= = = Ans.











Solution


2 2

1

2

max

avg

(60 ksi) (24 ksi)42 ksi

2

(18 ksi) (16 ksi) 24.08 ksi

42 ksi 24.08 ksi 66.08 ksi

42 ksi 24.08 ksi 17.92 ksi

24.08 ksi

42 ksi (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − =

= =

= =


(i.e., the principal plane subjected to σ p1) isfound from:

16 ksi 16 ksitan 2 0.8889 2 41.634 thus, 20.82


−

By inspection, the angle θ p from point x to point 1 is turned counterclockwise. The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.






orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 56.31° clockwise from the x axis. In Mohr’s

circle, all angle measures are doubled; therefore, point n (which represents the state of stress on the n

plane) on Mohr’s circle is rotated 2(56.31°) = 112.62° clockwise from point x. The angle between point n and point 2 is

180 41.634 112.62 25.746 β = ° − ° − ° = °


cos

42 ksi (24.08 ksi)cos(25.746 ) 20.31 ksi (T)

n C Rσ β = −

= − ° = Ans.


sin

(24.08 ksi)sin(25.746 ) 10.46 ksi

nt Rτ β =

= ° = Ans.

Since point n is below the σ axis, the shear stress acting on the plane surface tends to rotate the stresselement counterclockwise.

(d) Since the point in a structural

member is subjected to plane stress

3 0 z pσ σ = =

Three Mohr’s circles can be

constructed to show stress

combinations in the σ p1 – σ p2 plane,

the σ p1 – σ p3 plane, and the σ p2 – σ p3 plane. These three circles are

shown on the right.


1 3

abs max

66.08 ksi 0 ksi33.04 ksi

2 2

p pσ σ τ

− −= = = Ans.











Solution


2 2

1

2

max

avg

(105 MPa) (45 MPa)75 MPa

2

(30 MPa) (35 MPa) 46.10 MPa

75 MPa 46.10 MPa 121.10 MPa

75 MPa 46.10 MPa 28.90 MPa

46.10 MPa

75 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − =

= =

= =



35 MPa 35 MPatan 2 1.1667 2 49.399 thus, 24.70


−







orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 40° clockwise from the x axis. In Mohr’s


plane) on Mohr’s circle is rotated 2(40°) = 80° clockwise from point x. The angle between point n and point 2 is

180 49.399 80 50.601 β = ° − ° − ° = °


cos

75 MPa (46.10 MPa)cos(50.601 ) 45.74 MPa (T)

n C Rσ β = −

= − ° = Ans.


sin

(46.10 MPa)sin(50.601 ) 35.62 MPa

nt Rτ β =

= ° = Ans.

Since point n is above the σ axis, the shear stress acting on the planesurface tends to rotate the stress element clockwise.

(d) Since the point in a

structural member is subjected

to plane stress

3 0 z pσ σ = =

Three Mohr’s circles can beconstructed to show stress

combinations in the σ p1 – σ p2

plane, the σ p1 – σ p3 plane, and

the σ p2 – σ p3 plane. These threecircles are shown on the right.


2 3

abs max

121.10 MPa 0 MPa60.55 MPa

2 2

p pσ σ τ

− − −= = = Ans.











Solution


2 2

1

2

max

avg

( 54 MPa) ( 28 MPa)41 MPa

2

(13 MPa) (15 MPa) 19.85 MPa

41 MPa 19.85 MPa 21.15 MPa

41 MPa 19.85 MPa 60.85 MPa

19.85 MPa

41 MPa (C)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− + −= = −

= + =

= + = − + = −

= − = − − = −

= =

= =



15 MPa 15 MPatan 2 1.1538 2 49.086 thus, 24.54


− − −







orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 36.87° counterclockwise from the x axis. In


on the n plane) on Mohr’s circle is rotated 2(36.87°) = 73.74° counterclockwise from point x. Theangle between point n and point 2 is

73.74 49.086 24.654 β = ° − ° = °


cos

41 MPa (19.85 MPa)cos(24.654 ) 59.04 MPa (C)

n C Rσ β = −

= − − ° = Ans.


sin

(19.85 MPa)sin(24.654 ) 8.28 MPa

nt Rτ β =

= ° = Ans.



member is subjected to plane

stress

3 0 z pσ σ = =



plane, the σ p1 – σ p3 plane, and the

σ p2 – σ p3 plane. These threecircles are shown on the right.


2 3

abs max

60.85 MPa 0 MPa30.43 MPa

2 2

p pσ σ τ

− − −= = = Ans.











Solution


2 2

1

2

max

avg

( 60 MPa) (100 MPa)20 MPa

2

(80 MPa) (80 MPa) 113.14 MPa

20 MPa 113.14 MPa 133.14 MPa

20 MPa 113.14 MPa 93.14 MPa

113.14 MPa

20 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

− += =

= + =

= + = + =

= − = − = −

= =

= =



80 MPa 80 MPatan 2 1.0 2 45 thus, 22.5

( 60 MPa) (20 MPa) 80 MPa p p pθ θ θ = = = ∴ = ° = °

− −

By inspection, the angle θ p from point x to point 2 is turned counterclockwise. The orientation of the

principal stresses and the maximum in-plane shear stress is shown on the sketch below.









180 45 60 75 β = − ° − ° = °


cos

20 MPa (113.14 MPa)cos(75 ) 49.28 MPa (T)

n C Rσ β = +

= + ° = Ans.


sin

(113.14 MPa)sin(75 ) 109.28 MPa

nt Rτ β =

= ° = Ans.

Since point n is above the σ axis, the shear stress acting on the plane surface tends to rotate the stresselement clockwise.



3 0 z pσ σ = =





shown on the right.













Solution


2 2

1

2

max

avg

(100 MPa) (0 MPa)50 MPa

2

(50 MPa) (210 MPa) 215.87 MPa

50 MPa 215.87 MPa 265.87 MPa

50 MPa 215.87 MPa 165.87 MPa

215.87 MPa

50 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − = −

= =

= =



210 MPa 210 MPatan 2 4.2 2 76.608 thus, 38.30


−

By inspection, the angle θ p from point x to point 1 is turned clockwise. The orientation of the

principal stresses and the maximum in-plane shear stress is shown on the sketch below.






orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 21.80° counterclockwise from the x axis. In


on the n plane) on Mohr’s circle is rotated 2(21.80°) = 43.60° counterclockwise from point x. Theangle between point n and point 2 is

180 76.608 43.60 59.792 β = − ° − ° = °


cos

50 MPa (215.87 MPa)cos(59.792 ) 58.61 MPa (C)

n C Rσ β = −

= − ° = Ans.


sin

(215.87 MPa)sin(59.792 ) 186.56 MPa

nt Rτ β =

= ° = Ans.


(d) Since the point in a

structural member is subjected

to plane stress

3 0 z pσ σ = =



plane, the σ p1 – σ p3 plane, and

the σ p2 – σ p3 plane. These threecircles are shown on the right.










(b) Determine the principal stresses and the maximum in-planeshear stress acting at the point. Show these stresses on an

appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16).(c) Determine the normal and shear stresses on the indicated planeand show these stresses on a sketch.


Solution


2 2

1

2

max

avg

(120 MPa) (40 MPa)80 MPa

2

(40 MPa) (180 MPa) 184.39 MPa

80 MPa 184.39 MPa 264.39 MPa

80 MPa 184.39 MPa 104.39 MPa

184.39 MPa

80 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − = −

= =

= =



180 MPa 180 MPatan 2 4.5 2 77.471 thus, 38.74


−







orientation of the plane relative to the x face of the stress element. Looking at the stress element, weobserve that the normal to the indicated plane is oriented 30.96° clockwise from the x axis. In Mohr’s


plane) on Mohr’s circle is rotated 2(30.96°) = 61.93° clockwise from point x. The angle between pointn and point 2 is

180 77.471 61.93 40.599 β = ° − ° − ° = °


cos

80 MPa (184.39 MPa)cos(40.599 ) 60.00 MPa (C)

n C Rσ β = −

= − ° = Ans.


sin

(184.39 MPa)sin(40.599 ) 119.99 MPa

nt Rτ β =

= ° = Ans.




3 0 z pσ σ = =





shown on the right.













Solution


2 2

1

2

max

avg

(25 MPa) (100 MPa)62.5 MPa

2

(37.5 MPa) (140 MPa) 144.94 MPa

62.5 MPa 144.94 MPa 207.44 MPa

62.5 MPa 144.94 MPa 82.44 MPa

144.94 MPa

62.5 MPa (T)

p

p

C

R

C R

C R

R

C

σ

σ

τ

σ

+= =

= + =

= + = + =

= − = − = −

= =

= =



140 MPa 140 MPatan 2 3.7333 2 75.00 thus, 37.50

(25 MPa) (62.5 MPa) 37.5 MPa p p pθ θ θ = = = ∴ = ° = °

−










100 75 25 β = ° − ° = °


cos

62.5 MPa (144.94 MPa)cos(25 ) 68.86 MPa (C)

n C Rσ β = −

= − ° = Ans.


sin

(144.94 MPa)sin(25 ) 61.25 MPa

nt Rτ β =

= ° = Ans.




3 0 z pσ σ = =




the σ p1 – σ p3 plane, and the σ p2 – σ p3 plane. These three circles are shown

below.







12.73 At a point in a stressed body, the principal

stresses are oriented as shown in Fig. P12.73. UseMohr’s circle to determine:

(a) the stresses on plane a-a.

(b) the stresses on the horizontal and vertical planes at the point.

(c) the absolute maximum shear stress at the point.

Fig. P12.73

Solution

The center of Mohr’s circle can be found from the two principal stresses:

1 2 ( 3 ksi) ( 20 ksi)11.5 ksi

2 2

p pC

σ σ + − + −= = = −

The radius of the circle is

1 2 ( 3 ksi) ( 20 ksi)8.5 ksi

2 2

p p R

σ σ − − − −= = =

(a) The stresses on plane a-a are found by rotating 270° counterclockwise from the σ p2 point on

Mohr’s circle. Therefore, the point at the top of the circle directly above the center corresponds to thestate of stress on plane a-a.

11.5 ksi 11.5 ksi (C)a a C σ −

= = − = Ans.

8.5 ksi (shear stress rotates the wedge element clockwise)a a Rτ −

= = Ans.





(b) The angle θ p shown on the problem statement sketch is

11tan (8 /15) 14.0362

2 pθ −

= = °

The σ p2 principal plane is rotated 14.0362° clockwise from the x face of the stress element. We needto find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we know

the location of σ p2 on Mohr’s circle, we can begin there and rotate 2θ p in the opposite direction to find

point x. Therefore, beginning at point σ p2, rotate 2(14.0362°) = 28.0724° counterclockwise to locate point x. The σ coordinate of point x is found from:

cos(2 )

11.5 ksi (8.5 ksi)cos(28.0724 ) 19.00 ksi (C)

x pC Rσ θ = −

= − − ° = Ans.

The τ coordinate of point x is found from:

sin(2 )

(8.5 ksi)sin(28.0724 ) 4.00 ksi (rotates element counterclockwise)

nt p Rτ θ =

= ° = Ans.

Similarly, the σ coordinate of point y is found from:

cos(2 )

11.5 ksi (8.5 ksi)cos(28.0724 ) 4.00 ksi (C)

x pC Rσ θ = +

= − + ° = Ans.

The τ coordinate of point y is also 4.00 ksi, and the shear stress on the y face rotates the stress elementclockwise.

The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both σ p1 and σ p2 are negative, the absolute maximum shear stress will be larger than themaximum in-plane shear stress. The radius of the largest Mohr’s circle gives the absolute maximum

shear stress. In this case, the absolute maximum shear stress occurs in the σ p2 – σ p3 plane; therefore,

2 3

abs max

20 ksi 0 ksi10.00 ksi

2 2

p pσ σ τ

− − −= = = Ans.





12.74 At a point in a stressed body, the principal

stresses are oriented as shown in Fig. P12.74. UseMohr’s circle to determine:

(a) the stresses on plane a-a.

(b) the stresses on the horizontal and vertical planes at the point.

(c) the absolute maximum shear stress at the point.

Fig. P12.74

Solution

The center of Mohr’s circle can be found from the two principal stresses:

1 2 (200 MPa) (50 MPa)125 MPa

2 2

p pC

σ σ + += = =

The radius of the circle is

1 2 (200 MPa) (50 MPa)75 MPa

2 2

p p R

σ σ − −= = =

(a) The stresses on plane a-a are found by rotating 270° counterclockwise from the σ p1 point onMohr’s circle. Therefore, the point at the bottom of the circle directly underneath the center

corresponds to the state of stress on plane a-a.

125 MPa 125 MPa (T)a a C σ −

= = = Ans.

75 MPa (shear stress rotates the wedge element counterclockwise)a a Rτ −

= = Ans.



(b) The angle θ p shown on the problem statement sketch is

11tan (3 / 4) 18.435

2 pθ −

= = °

The σ p1 principal plane is rotated 18.435° counterclockwise from the x face of the stress element. Weneed to find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we

know the location of σ p1 on Mohr’s circle, we can begin there and rotate 2θ p in the opposite direction

to find point x. Therefore, beginning at point σ p1, rotate 2(18.435°) = 36.87° clockwise to locate point x. The σ coordinate of point x is found from:

cos(2 )

125 MPa (75 MPa)cos(36.87 ) 185.0 MPa (T)

x pC Rσ θ = +

= + ° = Ans.

The τ coordinate of point x is found from:

sin(2 )

(75 MPa)sin(36.87 ) 45.0 MPa (rotates element counterclockwise)

nt p Rτ θ =

= ° = Ans.

Similarly, the σ coordinate of point y is found from:

cos(2 )

125 MPa (75 MPa)cos(36.87 ) 65.0 MPa (T)

x pC Rσ θ = −

= − ° = Ans.

The τ coordinate of point y is also 65.0 MPa, and the shear stress on the y face rotates the stresselement clockwise.

The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both σ p1 and σ p2 are positive, the absolute maximum shear stress will be larger than themaximum in-plane shear stress. The radius of the largest Mohr’s circle gives the absolute maximum

shear stress. In this case, the absolute maximum shear stress occurs in the σ p1 – σ p3 plane; therefore,

1 3

abs max

200 MPa 0 MPa100.0 MPa

2 2

p pσ σ τ

− −= = = Ans.

mechanics of materials solutions chapter12 probs47 74

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