Mechanics of Materials Solutions Chapter12 Probs47 74

12.47 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy 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 x = (35 ksi, 20 ksi) y = (5 ksi, 20 ksi) C = 20 ksi R = (15 ksi) 2 + (20 ksi) 2 = 25 ksi

σ p1 = C + R = 20 ksi + 25 ksi = 45 ksi σ p 2 = C − R = 20 ksi − 25 ksi = −5 ksi τ max = R = 25 ksi σ avg = C = 20 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and σp1 is found from: 20 ksi 20 ksi = = 1.3333 ∴ 2θ p = 53.13° tan 2θ p = thus, θ p = 26.57° (35 ksi) − (20 ksi) 15 ksi By inspection, the angle θp from point x to σp1 is turned clockwise.

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12.48 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy 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. P12.48

Solution x = (90 ksi, 60 ksi) y = (30 ksi, 60 ksi) C = 60 ksi R = (30 ksi) 2 + (60 ksi) 2 = 67.08 ksi

σ p1 = C + R = 60 ksi + 67.08 ksi = 127.08 ksi σ p 2 = C − R = 60 ksi − 67.08 ksi = −7.08 ksi τ max = R = 67.08 ksi σ avg = C = 60 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and σp1 is found from: 60 ksi 60 ksi = = 2 ∴ 2θ p = 63.43° tan 2θ p = thus, θ p = 31.72° (90 ksi) − (60 ksi) 30 ksi By inspection, the angle θp from point x to σp1 is turned counterclockwise.

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12.49 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy 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. P12.49

Solution x = (−100 MPa, 30 MPa) y = (20 MPa, 30 MPa) C = −40 MPa R = (60 MPa) 2 + (30 MPa)2 = 67.08 MPa

σ p1 = C + R = −40 MPa + 67.08 MPa = 27.08 MPa σ p 2 = C − R = −40 MPa − 67.08 MPa = −107.08 MPa τ max = R = 67.08 MPa σ avg = C = 60 MPa The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and σp2 is found from: 30 MPa 30 MPa = = 0.5 ∴ 2θ p = 26.57° thus, θ p = 13.28° tan 2θ p = (−100 MPa) − (−40 MPa) 60 MPa By inspection, the angle θp from point x to σp2 is turned clockwise.

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12.50 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy 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. P12.50

Solution x = (−55 MPa, 25 MPa) y = (15 MPa, 25 MPa) C = −20 MPa R = (35 MPa) 2 + (25 MPa) 2 = 43.01 MPa

σ p1 = C + R = −20 MPa + 43.01 MPa = 23.01 MPa σ p 2 = C − R = −20 MPa − 43.01 MPa = −63.01 MPa τ max = R = 43.01 MPa σ avg = C = 20 MPa (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and σp2 is found from: 25 MPa 25 MPa = = 0.7143 ∴ 2θ p = 35.54° tan 2θ p = thus, θ p = 17.77° (−55 MPa) − (−20 MPa) 35 MPa By inspection, the angle θp from point x to σp2 is turned counterclockwise.

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12.51 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy and show them on a stress element. (b) Determine the stresses σn, σt, and τnt and show them on a stress element that is properly rotated with respect to the x-y element. The sketch must include 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 x = (60 MPa, 15 MPa) n = (−10 MPa, 30 MPa)

y = (−20 MPa, 15 MPa) t = (50 MPa, 30 MPa)

C = 20 MPa

R = (40 MPa) 2 + (15 MPa) 2 = 42.72 MPa

The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 is found from: 15 MPa 15 MPa = = 0.3750 ∴ 2θ p = 20.6° tan 2θ p = (60 MPa) − (20 MPa) 40 MPa The magnitude of the angle β between point n and point 2 is found from: 30 MPa 30 MPa = =1 ∴ β = 45° tan β = (−10 MPa) − (20 MPa) 30 MPa 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.

The angle α between point x and point n is thus α = 180° − 20.6° − 45° = 114.4° Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half of this 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.

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12.52 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses σx, σy, and τxy and show them on a stress element. (b) Determine the stresses σn, σt, and τnt and show them on a stress element that is properly rotated with respect to the x-y element. The sketch must include 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 x = (10 ksi, 30 ksi) n = (65 ksi, 10 ksi)

y = (50 ksi, 30 ksi) t = (−5 ksi, 10 ksi)

C = 30 ksi

R = (20 ksi) 2 + (30 ksi) 2 = 36.06 ksi

The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 is found from: 30 ksi 30 ksi tan 2θ p = = = 1.5 ∴ 2θ p = 56.3° (10 ksi) − (30 ksi) 20 ksi The magnitude of the angle β between point n and point 1 is found from: 10 ksi 10 ksi tan β = = = 0.2857 ∴ β = 15.9° (65 ksi) − (30 ksi) 35 ksi 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.

The angle α between point x and point n is thus α = 180° − 56.3° − 15.9° = 107.8° Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half of this 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.

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12.53 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point using Mohr’s circle. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or 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. (−15 ksi) + (5 ksi) = −5 ksi C= 2 R = (10 ksi) 2 + (12.5 ksi) 2 = 16.01 ksi

σ p1 = C + R = −5 ksi + 16.01 ksi = 11.01 ksi σ p 2 = C − R = −5 ksi − 16.01 ksi = −21.01 ksi τ max = R = 16.01 ksi σ avg = C = −5 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

12.5 ksi 12.5 ksi = = 1.25 (−15 ksi) − (−5 ksi) 10 ksi

∴ 2θ p = 51.340°

thus, θ p = 25.67°

By inspection, the angle θp 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.

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12.54 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point using Mohr’s circle. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16).

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

Fig. P12.54

Solution (b) The basic Mohr’s circle is shown. (28 MPa) + (−50 MPa) = −11 MPa C= 2 R = (39 MPa) 2 + (44 MPa) 2 = 58.80 MPa

σ p1 = C + R = −11 MPa + 58.80 MPa = 47.80 MPa σ p 2 = C − R = −11 MPa − 58.80 MPa = −69.80 MPa τ max = R = 58.80 MPa σ avg = C = −11 MPa The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: tan 2θ p =

44 MPa 44 MPa = = 1.1282 (28 MPa) − (−11 MPa) 39 MPa

∴ 2θ p = 48.447°

thus, θ p = 24.22°

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

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12.55 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point using Mohr’s circle. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16).

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

Fig. P12.55

Solution (b) The basic Mohr’s circle is shown. C=

(16 ksi) + (−3 ksi) = 6.5 ksi 2

R = (9.5 ksi) 2 + (8 ksi) 2 = 12.42 ksi

σ p1 = C + R = 6.5 ksi + 12.42 ksi = 18.92 ksi σ p 2 = C − R = 6.5 ksi − 12.42 ksi = −5.92 ksi τ max = R = 16.01 ksi σ avg = C = 6.5 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: 8 ksi 8 ksi tan 2θ p = = = 0.8421 (16 ksi) − (6.5 ksi) 9.5 ksi

∴ 2θ p = 40.101°

thus, θ p = 20.05°

By inspection, the angle θp from point x to point 1 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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12.56 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point using Mohr’s circle. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16).

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

Fig. P12.56

Solution (b) The basic Mohr’s circle is shown. C=

(−12 ksi) + (30 ksi) = 9.0 ksi 2

R = (21 ksi) 2 + (18 ksi) 2 = 27.66 ksi

σ p1 = C + R = 9.0 ksi + 27.66 ksi = 36.66 ksi σ p 2 = C − R = 9.0 ksi − 27.66 ksi = −18.66 ksi τ max = R = 27.66 ksi σ avg = C = 9.0 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: 18 ksi 18 ksi tan 2θ p = = = 0.8571 (−12 ksi) − (9 ksi) 21 ksi

∴ 2θ p = 40.601°

thus, θ p = 20.30°

By inspection, the angle θp from point x to point 2 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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12.57 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.57-12.60 should be assigned as a set.

Fig. P12.57

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

(6 ksi) + (18 ksi) = 12 ksi 2

R = (6 ksi) 2 + (30 ksi) 2 = 30.59 ksi

σ p1 = C + R = 12 ksi + 30.59 ksi = 42.59 ksi σ p 2 = C − R = 12 ksi − 30.59 ksi = −18.59 ksi τ max = R = 30.59 ksi σ avg = C = 12 ksi The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: 30 ksi 30 ksi tan 2θ p = = =5 ∴ 2θ p = 78.690° (6 ksi) − (12 ksi) 6 ksi

thus, θ p = 39.35°

By inspection, the angle θp 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.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0

Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 30.59 ksi Ans.

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12.58 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.57-12.60 should be assigned as a set.

Fig. P12.58

Solution (b) The basic Mohr’s circle is shown. (−35 MPa) + (−65 MPa) = −50 MPa C= 2 R = (15 MPa) 2 + (24 MPa) 2 = 28.30 MPa

σ p1 = C + R = −50 MPa + 28.30 MPa = −21.70 MPa σ p 2 = C − R = −50 MPa − 28.30 MPa = −78.30 MPa τ max = R = 28.30 MPa σ avg = C = 50 MPa (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: tan 2θ p =

24 MPa 24 MPa = = 1.6 (−35 MPa) − (−50 MPa) 15 MPa

∴ 2θ p = 57.995°

thus, θ p = 29.00°

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

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane. Therefore σ p 2 − σ p3 −78.30 MPa − 0 MPa τ abs max = = = 39.15 MPa 2 2

Ans.

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12.59 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.57-12.60 should be assigned as a set.

Fig. P12.59

Solution (b) The basic Mohr’s circle is shown. (0 MPa) + (−45 MPa) = −22.5 MPa C= 2 R = (22.5 MPa) 2 + (25 MPa) 2 = 33.63 MPa

σ p1 = C + R = −22.5 MPa + 33.63 MPa = 11.13 MPa σ p 2 = C − R = −22.5 MPa − 33.63 MPa = −56.13 MPa τ max = R = 33.63 MPa σ avg = C = 22.5 MPa (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: tan 2θ p =

25 MPa 25 MPa = = 1.1111 (0 MPa) − (−22.5 MPa) 22.5 MPa

∴ 2θ p = 48.013°

thus, θ p = 24.01°

By inspection, the angle θp from point x to point 1 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 33.63 MPa Ans.

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12.60 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.57-12.60 should be assigned as a set.

Fig. P12.60

Solution (b) The basic Mohr’s circle is shown. C=

(−36 ksi) + (−18 ksi) = −27 ksi 2

R = (9 ksi) 2 + (12 ksi) 2 = 15 ksi

σ p1 = C + R = −27 ksi + 15 ksi = −12 ksi σ p 2 = C − R = −27 ksi − 15 ksi = −42 ksi τ max = R = 15 ksi σ avg = C = 27 ksi (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

12 ksi 12 ksi = = 1.3333 (−36 ksi) − (−27 ksi) 9 ksi

∴ 2θ p = 53.130°

thus, θ p = 26.57°

By inspection, the angle θp from point x to point 2 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane; therefore, σ p 2 − σ p3 −42 ksi − 0 ksi τ abs max = = = 21 ksi 2 2

Ans.

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12.61 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.61-12.64 should be assigned as a set.

Fig. P12.61

Solution (b) The basic Mohr’s circle is shown. (60 MPa) + (90 MPa) = 75 MPa C= 2 R = (15 MPa) 2 + (30 MPa) 2 = 33.54 MPa

σ p1 = C + R = 75 MPa + 33.54 MPa = 108.54 MPa σ p 2 = C − R = 75 MPa − 33.54 MPa = 41.46 MPa τ max = R = 33.54 MPa σ avg = C = 75 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: 30 MPa 30 MPa tan 2θ p = = =2 ∴ 2θ p = 63.435° (60 MPa) − (75 MPa) 15 MPa

thus, θ p = 31.72°

By inspection, the angle θp 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.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp3 plane; therefore, σ p1 − σ p 3 108.54 MPa − 0 MPa τ abs max = = = 54.27 MPa 2 2

Ans.

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12.62 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.61-12.64 should be assigned as a set.

Fig. P12.62

Solution (b) The basic Mohr’s circle is shown. (55 MPa) + (4 MPa) = 29.5 MPa C= 2 R = (25.5 MPa) 2 + (28 MPa) 2 = 37.87 MPa

σ p1 = C + R = 29.5 MPa + 37.87 MPa = 67.37 MPa σ p 2 = C − R = 29.5 MPa − 37.87 MPa = −8.37 MPa τ max = R = 37.87 MPa σ avg = C = 29.5 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

28 MPa 28 MPa = = 1.0980 (55 MPa) − (29.5 MPa) 25.5 MPa

∴ 2θ p = 47.675°

thus, θ p = 23.84°

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

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 37.87 MPa Ans.

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12.63 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.61-12.64 should be assigned as a set.

Fig. P12.63

Solution (b) The basic Mohr’s circle is shown. C=

(−250 psi) + (−500 psi) = −375 psi 2

R = (125 psi) 2 + (200 psi) 2 = 235.8 psi

σ p1 = C + R = −375 psi + 235.8 psi = −139.2 psi σ p 2 = C − R = −375 psi − 235.8 psi = −610.8 psi τ max = R = 235.8 psi σ avg = C = 375 psi (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: 200 psi 200 psi tan 2θ p = = = 1.6 (−250 psi) − (−375 psi) 125 psi

∴ 2θ p = 57.995° thus, θ p = 29.00°

By inspection, the angle θp from point x to point 1 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane; therefore, σ p 2 − σ p3 −610.8 psi − 0 psi τ abs max = = = 305.4 psi 2 2

Ans.

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12.64 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or Fig. 12-16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.61-12.64 should be assigned as a set.

Fig. P12.64

Solution (b) The basic Mohr’s circle is shown. C=

(4.5 ksi) + (9.1 ksi) = 6.8 ksi 2

R = (2.3 ksi) 2 + (2 ksi) 2 = 3.05 ksi

σ p1 = C + R = 6.8 ksi + 3.05 ksi = 9.85 ksi σ p 2 = C − R = 6.8 ksi − 3.05 ksi = 3.75 ksi τ max = R = 3.05 ksi σ avg = C = 6.8 ksi (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

2 ksi 2 ksi = = 0.8696 (4.5 ksi) − (6.80 ksi) 2.3 ksi

∴ 2θ p = 41.009°

thus, θ p = 20.50°

By inspection, the angle θp from point x to point 2 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

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(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp3 plane; therefore, σ p1 − σ p 3 9.85 ksi − 0 ksi τ abs max = = = 4.92 ksi 2 2

Ans.

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12.65 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and 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 (b) The basic Mohr’s circle is shown. (−90 MPa) + (−65 MPa) = −77.5 MPa C= 2 R = (12.5 MPa) 2 + (42 MPa) 2 = 43.82 MPa

σ p1 = C + R = −77.5 MPa + 43.82 MPa = −33.68 MPa σ p 2 = C − R = −77.5 MPa − 43.82 MPa = −121.32 MPa τ max = R = 43.82 MPa σ avg = C = 77.5 MPa (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

42 MPa 42 MPa = = 3.3600 (−90 MPa) − (−77.5 MPa) 12.5 MPa

∴ 2θ p = 73.426°

thus, θ p = 36.71°

By inspection, the angle θp 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.

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(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, we observe 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 β = 180° − 73.426° − 70° = 36.574° The σ coordinate of point n is found from: σ n = C + R cos β = −77.5 MPa + (43.82 MPa) cos(36.574°) = −42.31 MPa = 42.31 MPa (C)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (43.82 MPa) sin(36.574°) = 26.11 MPa Ans. Since point n is below the σ axis, the shear stress acting on the plane surface tends to rotate the stress element counterclockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane; therefore, σ p 2 − σ p3 −121.32 MPa − 0 MPa τ abs max = = = 60.66 MPa 2 2

Ans.

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12.66 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and 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. P12.66

Solution (b) The basic Mohr’s circle is shown. C=

(60 ksi) + (24 ksi) = 42 ksi 2

R = (18 ksi) 2 + (16 ksi) 2 = 24.08 ksi

σ p1 = C + R = 42 ksi + 24.08 ksi = 66.08 ksi σ p 2 = C − R = 42 ksi − 24.08 ksi = 17.92 ksi τ max = R = 24.08 ksi σ avg = C = 42 ksi (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: 16 ksi 16 ksi tan 2θ p = = = 0.8889 (60 ksi) − (42 ksi) 18 ksi

∴ 2θ p = 41.634°

thus, θ p = 20.82°

By inspection, the angle θp 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.

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(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, we observe 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 β = 180° − 41.634° − 112.62° = 25.746° The σ coordinate of point n is found from: σ n = C − R cos β = 42 ksi − (24.08 ksi) cos(25.746°) = 20.31 ksi (T)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (24.08 ksi) sin(25.746°) = 10.46 ksi Ans. Since point n is below the σ axis, the shear stress acting on the plane surface tends to rotate the stress element counterclockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp1–σp3 plane; therefore, σ p1 − σ p 3 66.08 ksi − 0 ksi τ abs max = = = 33.04 ksi 2 2

Ans.

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12.67 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and 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. P12.67

Solution (b) The basic Mohr’s circle is shown. (105 MPa) + (45 MPa) = 75 MPa C= 2 R = (30 MPa)2 + (35 MPa) 2 = 46.10 MPa

σ p1 = C + R = 75 MPa + 46.10 MPa = 121.10 MPa σ p 2 = C − R = 75 MPa − 46.10 MPa = 28.90 MPa τ max = R = 46.10 MPa σ avg = C = 75 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: tan 2θ p =

35 MPa 35 MPa = = 1.1667 (105 MPa) − (75 MPa) 30 MPa

∴ 2θ p = 49.399°

thus, θ p = 24.70°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 40° 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(40°) = 80° clockwise from point x. The angle between point n and point 2 is β = 180° − 49.399° − 80° = 50.601° The σ coordinate of point n is found from: σ n = C − R cos β = 75 MPa − (46.10 MPa) cos(50.601°) = 45.74 MPa (T)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (46.10 MPa) sin(50.601°) = 35.62 MPa Ans. Since point n is above the σ axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane; therefore, σ p 2 − σ p3 −121.10 MPa − 0 MPa τ abs max = = = 60.55 MPa 2 2

Ans.

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12.68 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and 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. P12.68

Solution (b) The basic Mohr’s circle is shown. (−54 MPa) + (−28 MPa) = −41 MPa C= 2 R = (13 MPa) 2 + (15 MPa) 2 = 19.85 MPa

σ p1 = C + R = −41 MPa + 19.85 MPa = −21.15 MPa σ p 2 = C − R = −41 MPa − 19.85 MPa = −60.85 MPa τ max = R = 19.85 MPa σ avg = C = 41 MPa (C) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

15 MPa 15 MPa = = 1.1538 ∴ 2θ p = 49.086° (−54 MPa) − (−41 MPa) 13 MPa

thus, θ p = 24.54°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 36.87° 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(36.87°) = 73.74° counterclockwise from point x. The angle between point n and point 2 is β = 73.74° − 49.086° = 24.654° The σ coordinate of point n is found from: σ n = C − R cos β = −41 MPa − (19.85 MPa) cos(24.654°) = 59.04 MPa (C)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (19.85 MPa)sin(24.654°) = 8.28 MPa Ans. Since point n is below the σ axis, the shear stress acting on the plane surface tends to rotate the stress element counterclockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp2–σp3 plane; therefore, σ p 2 − σ p3 −60.85 MPa − 0 MPa τ abs max = = = 30.43 MPa 2 2

Ans.

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12.69 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.69-12.72 should be assigned as a set.

Fig. P12.69

Solution (b) The basic Mohr’s circle is shown. (−60 MPa) + (100 MPa) = 20 MPa C= 2 R = (80 MPa) 2 + (80 MPa) 2 = 113.14 MPa

σ p1 = C + R = 20 MPa + 113.14 MPa = 133.14 MPa σ p 2 = C − R = 20 MPa − 113.14 MPa = −93.14 MPa τ max = R = 113.14 MPa σ avg = C = 20 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

80 MPa 80 MPa = = 1.0 (−60 MPa) − (20 MPa) 80 MPa

∴ 2θ p = 45°

thus, θ p = 22.5°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 30° 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(30°) = 60° clockwise from point x. The angle between point n and point 1 is β = 180 − 45° − 60° = 75° The σ coordinate of point n is found from: σ n = C + R cos β = 20 MPa + (113.14 MPa) cos(75°) = 49.28 MPa (T)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (113.14 MPa)sin(75°) = 109.28 MPa Ans. Since point n is above the σ axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 113.14 MPa Ans.

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12.70 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.69-12.72 should be assigned as a set.

Fig. P12.70

Solution (b) The basic Mohr’s circle is shown. (100 MPa) + (0 MPa) = 50 MPa C= 2 R = (50 MPa) 2 + (210 MPa) 2 = 215.87 MPa

σ p1 = C + R = 50 MPa + 215.87 MPa = 265.87 MPa σ p 2 = C − R = 50 MPa − 215.87 MPa = −165.87 MPa τ max = R = 215.87 MPa σ avg = C = 50 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to σp1) is found from: tan 2θ p =

210 MPa 210 MPa = = 4.2 (100 MPa) − (50 MPa) 50 MPa

∴ 2θ p = 76.608°

thus, θ p = 38.30°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 21.80° 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(21.80°) = 43.60° counterclockwise from point x. The angle between point n and point 2 is β = 180 − 76.608° − 43.60° = 59.792° The σ coordinate of point n is found from: σ n = C − R cos β = 50 MPa − (215.87 MPa) cos(59.792°) = 58.61 MPa (C)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (215.87 MPa) sin(59.792°) = 186.56 MPa Ans. Since point n is above the σ axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 215.87 MPa 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.

12.71 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.69-12.72 should be assigned as a set.

Fig. P12.71

Solution (b) The basic Mohr’s circle is shown. (120 MPa) + (40 MPa) = 80 MPa C= 2 R = (40 MPa) 2 + (180 MPa) 2 = 184.39 MPa

σ p1 = C + R = 80 MPa + 184.39 MPa = 264.39 MPa σ p 2 = C − R = 80 MPa − 184.39 MPa = −104.39 MPa τ max = R = 184.39 MPa σ avg = C = 80 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

180 MPa 180 MPa = = 4.5 (120 MPa) − (80 MPa) 40 MPa

∴ 2θ p = 77.471°

thus, θ p = 38.74°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 30.96° 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(30.96°) = 61.93° clockwise from point x. The angle between point n and point 2 is β = 180° − 77.471° − 61.93° = 40.599° The σ coordinate of point n is found from: σ n = C − R cos β = 80 MPa − (184.39 MPa) cos(40.599°) = 60.00 MPa (C)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (184.39 MPa)sin(40.599°) = 119.99 MPa Ans. Since point n is below the σ axis, the shear stress acting on the plane surface tends to rotate the stress element counterclockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown on the right.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 184.39 MPa 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.

12.72 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical 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 shear 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 plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems 12.69-12.72 should be assigned as a set.

Fig. P12.72

Solution (b) The basic Mohr’s circle is shown. (25 MPa) + (100 MPa) = 62.5 MPa C= 2 R = (37.5 MPa)2 + (140 MPa) 2 = 144.94 MPa

σ p1 = C + R = 62.5 MPa + 144.94 MPa = 207.44 MPa σ p 2 = C − R = 62.5 MPa − 144.94 MPa = −82.44 MPa τ max = R = 144.94 MPa σ avg = C = 62.5 MPa (T) The magnitude of the angle 2θp between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to σp2) is found from: tan 2θ p =

140 MPa 140 MPa = = 3.7333 (25 MPa) − (62.5 MPa) 37.5 MPa

∴ 2θ p = 75.00°

thus, θ p = 37.50°

By inspection, the angle θp 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.

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(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, we observe that the normal to the indicated plane is oriented 50° 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(50°) = 100° clockwise from point x. The angle between point n and point 2 is β = 100° − 75° = 25° The σ coordinate of point n is found from: σ n = C − R cos β = 62.5 MPa − (144.94 MPa) cos(25°) = 68.86 MPa (C)

Ans.

The τ coordinate of point n is found from: τ nt = R sin β = (144.94 MPa) sin(25°) = 61.25 MPa Ans. Since point n is above the σ axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

(d) Since the point in a structural member is subjected to plane stress σ z = σ p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the σp1–σp2 plane, the σp1–σp3 plane, and the σp2–σp3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the σp1–σp2 plane (which is also the x-y plane). Therefore τ abs max = τ max = 144.94 MPa Ans.

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12.73 At a point in a stressed body, the principal stresses are oriented as shown in Fig. P12.73. Use Mohr’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: σ + σ p 2 (−3 ksi) + (−20 ksi) = = −11.5 ksi C = p1 2 2 The radius of the circle is σ p1 − σ p 2 (−3 ksi) − (−20 ksi) R= = = 8.5 ksi 2 2 (a) The stresses on plane a-a are found by rotating 270° counterclockwise from the σp2 point on Mohr’s circle. Therefore, the point at the top of the circle directly above the center corresponds to the state of stress on plane a-a. σ a − a = C = −11.5 ksi = 11.5 ksi (C) Ans.

τ a − a = R = 8.5 ksi

(shear stress rotates the wedge element clockwise)

Ans.

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(b) The angle θp shown on the problem statement sketch is 1 θ p = tan −1 (8 /15) = 14.0362° 2 The σp2 principal plane is rotated 14.0362° clockwise from the x face of the stress element. We need to find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we know the location of σp2 on Mohr’s circle, we can begin there and rotate 2θp in the opposite direction to find point x. Therefore, beginning at point σp2, rotate 2(14.0362°) = 28.0724° counterclockwise to locate point x. The σ coordinate of point x is found from: σ x = C − R cos(2θ p ) = −11.5 ksi − (8.5 ksi) cos(28.0724°) = 19.00 ksi (C)

Ans.

The τ coordinate of point x is found from: τ nt = R sin(2θ p ) = (8.5 ksi) sin(28.0724°) = 4.00 ksi (rotates element counterclockwise) Similarly, the σ coordinate of point y is found from: σ x = C + R cos(2θ p ) = −11.5 ksi + (8.5 ksi) cos(28.0724°) = 4.00 ksi (C)

Ans.

Ans.

The τ coordinate of point y is also 4.00 ksi, and the shear stress on the y face rotates the stress element clockwise. The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both σp1 and σp2 are negative, the absolute maximum shear stress will be larger than the maximum 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 σp2–σp3 plane; therefore, σ p 2 − σ p 3 −20 ksi − 0 ksi = = 10.00 ksi Ans. τ abs max = 2 2

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12.74 At a point in a stressed body, the principal stresses are oriented as shown in Fig. P12.74. Use Mohr’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: σ + σ p 2 (200 MPa) + (50 MPa) = = 125 MPa C = p1 2 2 The radius of the circle is σ p1 − σ p 2 (200 MPa) − (50 MPa) = = 75 MPa R= 2 2 (a) The stresses on plane a-a are found by rotating 270° counterclockwise from the σp1 point on Mohr’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. σ a −a = C = 125 MPa = 125 MPa (T) Ans.

τ a − a = R = 75 MPa

(shear stress rotates the wedge element counterclockwise)

Ans.

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(b) The angle θp shown on the problem statement sketch is 1 θ p = tan −1 (3 / 4) = 18.435° 2 The σp1 principal plane is rotated 18.435° counterclockwise from the x face of the stress element. We need to find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we know the location of σp1 on Mohr’s circle, we can begin there and rotate 2θp in the opposite direction to find point x. Therefore, beginning at point σp1, rotate 2(18.435°) = 36.87° clockwise to locate point x. The σ coordinate of point x is found from: σ x = C + R cos(2θ p ) = 125 MPa + (75 MPa) cos(36.87°) = 185.0 MPa (T)

Ans.

The τ coordinate of point x is found from: τ nt = R sin(2θ p ) = (75 MPa) sin(36.87°) = 45.0 MPa (rotates element counterclockwise) Similarly, the σ coordinate of point y is found from: σ x = C − R cos(2θ p ) = 125 MPa − (75 MPa) cos(36.87°) = 65.0 MPa (T)

Ans.

Ans.

The τ coordinate of point y is also 65.0 MPa, and the shear stress on the y face rotates the stress element clockwise. The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both σp1 and σp2 are positive, the absolute maximum shear stress will be larger than the maximum 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 σp1–σp3 plane; therefore, σ − σ p 3 200 MPa − 0 MPa τ abs max = p1 = = 100.0 MPa Ans. 2 2

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