Explanation Let the normal force in the x -direction be N and the torque be T . Let the force and moment in the z -direction be V z and M z . Let the force and moment in the y -direction be V y and M y . Sketch the free body diagram of the assembly as shown in Figure 1. Apply Equilibrium equations. Sum of forces in x -direction is equal to 0. Σ F x = 0 N − 450 = 0 N = 450 N Sum of forces in y -direction is equal to 0. Σ F y = 0 V y = 0 Sum of forces in z -direction is equal to 0. Σ F z = 0 V z + 300 = 0 V z = − 300 N Sum of moments in x -direction is equal to 0. Σ M x = 0 T + ( 300 N ) ( 200 mm ) = 0 T + ( 300 N ) ( 200 mm × 10 − 3 m 1 mm ) = 0 T = − 60 N ⋅ m Sum of moments in y -direction is equal to 0. Σ M y = 0 M y − ( 450 N ) ( 300 mm ) + ( 300 N ) ( 300 mm ) = 0 M y − ( 450 N ) ( 300 mm × 10 − 3 m 1 mm ) + ( 300 N ) ( 300 mm × 10 − 3 m 1 mm ) = 0 M y = 45 N ⋅ m Sum of moments in z -direction is equal to 0. Σ M z = 0 M z + ( 450 N ) ( 200 mm ) = 0 M z + ( 450 N ) ( 200 mm × 10 − 3 m 1 mm ) = 0 M z = − 90 N ⋅ m Sketch the assembly cross section at point B as shown in Figure 2. Refer Figure 2. The outer radius of the pipe cross section is 0.03 m and the inner radius of the pipe cross section is 0.02 m. Find the cross-sectional area of the assembly cross section ( A ) : A = π ( 0.03 2 − 0.02 2 ) = 1.5708 × 10 − 3 m 2 Find the polar moment of inertia of the assembly cross section ( J ) using the relation: J = π 2 ( 0.03 4 − 0.02 4 ) = 1.0210 × 10 − 6 m 4 Find the moment of inertia in y -direction and z -direction of the assembly cross section ( I y = I z ) : I y = I z = π 4 ( 0.03 4 − 0.02 4 ) = 5.1051 × 10 − 7 m 4 Calculate the first moment of area at point B in z -direction using Figure 2. The value of y ¯ ' is 0. Therefore, ( Q z ) B = Σ y ¯ ' A ' = 0 The centroid distance to point B ( z B ) is − 0.03 m . The expression for normal stress ( σ B ) due to axial and bending stress is shown below: σ B = N A + M y z B I y = 450 N 1.5708 × 10 − 3 m 2 + ( 45 N ⋅ m ) ( − 0.03 m ) 5.1051 × 10 − 7 m 4 = 0

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Combined Loading

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 9.4, Problem 9.67P

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The maximum in-plane shear stress at point B.

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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 B . The maximum in-plane shear stress at point B .

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The principal stresses at point B. The maximum in-plane shear stress at point B.

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The principal stresses at point B.

The maximum in-plane shear stress at point B.