To determine The state of stress ( σ A ) at point A of the post. The shear stress ( τ A ) at point A of the post. Explanation Given information: The outer radius of the cross section is 2.5 in. The inner radius of the cross section is 2 in. Calculation: Sketch a free body diagram of post as shown in Figure 1. Here, V z , V y shear in z , y direction, M z , M y moment in z , y direction, and T is the torque. Determine the internal loadings as follows: Refer to Figure 1, Apply equilibrium conditions in y direction. ∑ F y = 0 V y + 300 = 0 V y = − 300 lb Apply equilibrium conditions in z direction. ∑ F z = 0 V z + 400 = 0 V z = − 400 lb Take moment about x axis. ∑ M x = 0 T + 400 ( 1.5 ) = 0 T + 600 = 0 T = − 600 lb ⋅ ft Take moment about y axis. ∑ M y = 0 M y + 400 ( 5 ) = 0 M y + 2 , 000 = 0 M y = − 2 , 000 lb ⋅ ft Take moment about z axis. ∑ M z = 0 M z − 300 ( 5 ) = 0 M z + 1 , 500 = 0 M z = 1 , 500 lb ⋅ ft Sketch the free body diagram of circular cross section as shown in Figure 2. Refer Figure 2: Calculate the moment of inertia ( I ) about the y and z axes and the polar moment of inertia of the post cross section. I y = I z = π 4 ( R 4 − r 4 ) (1) Here R is the outer radius of the cross section and r is the inner radius of cross section. Substitute 2.5 in. for R and 2 in. for r in Equation (1). I y = I z = π 4 ( 2.5 4 − 2 4 ) = π 4 ( 39.0 − 16 ) = π 4 ( 23 ) = 5.7656 π in 4 Determine the polar moment of inertia ( J ) as follows: J = π 2 ( R 4 − r 4 ) (2) Substitute 2.5 in. for R and 2 in. for r in Equation (2). J = π 2 ( 2.5 4 − 2 4 ) = π 2 ( 23.0 ) = 11.53125 π in 4 Find the first moment area ( Q z ) A along z axis. Refer Figure 2, The first moment area Q A .it is located top of the area. Since A is zero. Thus, the moment area Q A along z axis is zero. ( Q z ) A = 0 . Find the first moment area ( Q y ) A along y axis. ( Q y ) A = 4 R 3 π [ π 2 ( R ) 2 ] − 4 r 3 π [ π 2 ( r ) 2 ] (3) Substitute 2.5 in. for R and 2 in. for r in Equation (3). ( Q y ) A = 4 ( 2.5 ) 3 π [ π 2 ( 2.5 ) 2 ] − 4 ( 2 ) 3 π [ π 2 ( 2 ) 2 ] = 10.4 − 5.3333 = 5.08333 in 3 Determine the normal stress at point A ( σ A ) for combined loadings: σ A = − M z y I z + M y z I y (4) Here, M z , M y moment in z , y direction, I z , I y moment of inertia about in z , y direction, y is the centroid of distance. Substitute 1 , 500 lb ⋅ ft for M z , − 2 , 000 lb ⋅ ft for M y , 0 for y , 5.7656 π in 4 for I z , I y , − 2.5 in . for z in Equation (4). σ A = − 0 + − 2 , 000 lb ⋅ ft ( 12 in 1 ft ) × ( − 2

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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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 8.2, Problem 8.42P

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The state of stress (σA) at point A of the post.

The shear stress (τA) at point A of the post.

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Chapter 8.2, Problem 8.41P

Chapter 8.2, Problem 8.43P

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The state of stress ( σ A ) at point A of the post. The shear stress ( τ A ) at point A of the post.

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The state of stress (σA) at point A of the post.

The shear stress (τA) at point A of the post.