Explanation Calculate the normal stress ( σ A ) acting at point A using the relation: σ A = M y z I y (1) Here, M y is the moment in y -direction, z is the distance in z -direction from centroid to point A , and I y is the moment of inertia. Sketch the internal forces and moment in free body diagram at point A as shown in Figure 1. Apply Equilibrium equations to find the value of moment at point A . Sum of moments in y -direction is equal to 0. Σ M y = 0 M y − ( 20 × 10 ) = 0 M y = 200 lb ⋅ in Sum of moments in x -direction is equal to 0. Σ M x = 0 T x + ( 20 × 12 ) = 0 T x = 240 lb ⋅ in Sum of forces in z -direction is equal to 0. Σ V z = 0 V z − 20 = 0 V z = 20 lb Find the moment of inertia of the section ( I ) : Outer radius of the pipe is 1.5 in. and the inner radius of the pipe is 1.375 in. I = π 4 [ ( 1.5 in ) 4 − ( 1.375 in ) 4 ] = 1.1687 in 4 Find the polar moment of inertia of the section ( J ) : J = π 2 [ ( 1.5 in ) 4 − ( 1.375 in ) 4 ] = 2.3374 in 4 Sketch the cross section at point A as shown in Figure 2. Find the first moment of area at point A ( Q A ) z using the relation: ( Q A ) z = Σ y ¯ ' A ' (2) Here, y ¯ ' is the centroid distance to point A and A ' is the area from centroid to point A . Refer to Figure 2. ( Q A ) z = 4 × 1.5 in . 3 π [ 1 2 × π × ( 1.5 in . ) 2 ] − 4 × 1.375 in . 3 π [ 1 2 × π × ( 1.375 in . ) 2 ] = 0.51693 in 3 . Substitute 200 lb ⋅ in for M y , 0 for z , and 1.1687 in 4 for I y in Equation (1). σ A = − ( 200 lb ⋅ in ) ( 0 ) 1

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ISBN: 2810014920922

Author: HIBBELER

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Chapter 9, Problem 9.91RP

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