To determine The principal stresses at point A . The maximum in-plane shear stress developed at point A . The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A . Explanation Given information: Diameter of the shaft is 2 in. Calculation: Let N , V , and M be the normal force, shear force, and moment at the left cut segment of the shaft. Sketch the free body diagram of the left cut segment of the shaft as shown in Figure 1. Apply equations of equilibrium. Sum of forces in the x -direction is equal to 0. Σ F x = 0 N − 3 , 000 lb = 0 N = 3 , 000 lb Sum of forces in the y -direction is equal to 0. Σ F y = 0 − V + 75 lb = 0 V = 75 lb Sum of moments at C is equal to 0. Σ M C = 0 ( 75 lb ) ( 24 in . ) − M = 0 M = 1 , 800 lb ⋅ in . Find the cross sectional area ( A ) of the 1 in. radius of the shaft: A = π ( 1 in . ) 2 = 3.1416 in 2 Find the moment of inertia ( I ) of the 1 in. radius of the shaft: I = π 4 ( 1 in . ) 4 = 0.7854 mm 4 Find the first moment of area at point A ( Q A ) : The value of y ¯ ' is 0. Therefore, Q A = 0 Refer the cross section, The centroid distance of the section ( y ) to point A is 1 in . Find the normal stress at point A ( σ A ) using the relation: σ A = N A − M y I = 3 , 000 lb 3.1416 in 2 − ( 1 , 800 lb ⋅ in . ) ( 1 in . ) 0.7857 in 4 = − 1.336 × 10 3 psi = − 1.34 ksi There is no shear stress acting on the element. Therefore the value of the shear stress at point A ( τ A ) is 0. The value of normal stresses are σ x = − 1.34 ksi and σ y = 0 . The value of shear stress is τ x y = 0 . Sketch the state of stress at point A on the element as shown in Figure 2. Find the in-plane principal stresses ( σ 1 ) and ( σ 2 ) using the relation: The principal stress σ 1 = σ y = 0 . The principal stress σ 2 = σ x = − 1.34 ksi . Therefore, the principal stresses at point A ( σ 1 ) and ( σ 2 ) are 0 _ and − 1.34 ksi _ . Find the value of maximum in-plane shear stress ( τ max in-plane ) using the relation: τ max in-plane = ( σ x − σ y 2 ) 2 + τ x y 2 (1) Substitute − 1

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

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Chapter 9.3, Problem 9.34P

To determine

The principal stresses at point A.

The maximum in-plane shear stress developed at point A.

The orientation of the plane for maximum in-plane shear stress.

To sketch:

The results on an element located at point A.

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Chapter 9.3, Problem 9.33P

Chapter 9.3, Problem 9.35P

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Chapter 9 Solutions

Mechanics of Materials

Ch. 9.3 - In each case, the state of stress x, y, xy...Ch. 9.3 - Given the state of stress shown on the element,...Ch. 9.3 - Determine the normal stress and shear stress...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Also, find the corresponding orientation of the...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Determine the maximum principal stress at point B.Ch. 9.3 - Determine the principal stress at point C.Ch. 9.3 - Prove that the sum of the normal stresses x + y =...Ch. 9.3 - 9-2. The state of stress at a point in a member is...

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The principal stresses at point A . The maximum in-plane shear stress developed at point A . The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A .

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The principal stresses at point A. The maximum in-plane shear stress developed at point A. The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A.

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Chapter 9 Solutions

The principal stresses at point A.

The maximum in-plane shear stress developed at point A.

The orientation of the plane for maximum in-plane shear stress.

To sketch:

The results on an element located at point A.