Explanation Sketch a free body diagram of left segment of rod as shown in Figure 1. Determine the internal loadings as follows: Apply equilibrium conditions the force in x direction. ∑ F x = 0 N x − 100 = 0 N x = 100 kN Apply equilibrium conditions the force in y direction. ∑ F y = 0 V y − 10 = 0 V y = 10 kN Apply equilibrium conditions the force in z direction. ∑ F z = 0 V z + 20 = 0 V z = − 20 kN Take a moment about x axis. ∑ M x = 0 T x + 20 ( 0.03 ) + 10 ( 0.03 ) = 0 T x + 0.6 + 0.3 = 0 T x + 0.9 = 0 T x = − 0.9 kN .m Take moment about y axis. ∑ M y = 0 M y + 20 ( 0.2 ) + 100 ( 0.03 ) = 0 M y + 4 + 3 = 0 M y = − 7.00 kN .m Take a moment about z axis. ∑ M z = 0 M z + 10 ( 0.4 ) = 0 M z + 4 = 0 M z = − 4.00 kN .m Sketch the free body diagram of circular cross section as shown in Figure 2. Determine the area of cross section using the relation: A = π c 2 Here, c is the radius of circular diameter. Refer Figure 2. Substitute 0.03 m for c . A = π ( 0.03 ) 2 = 0.9 ( 10 − 3 ) π m 2 Determine the moment of inertia of rod using the relation: I y = I z = π 4 c 4 Substitute 0.03m for c . I y = I z = π 4 ( 0.03 ) 4 = 0.2025 ( 10 − 6 ) π m 4 Determine the polar moment of inertia of rod using the relation: J = π 2 c 4 Substitute 0.03m for c . J = π 2 ( 0.5 ) 03 4 = 0.405 ( 10 − 6 ) π m 4 Determine the second moment of inertia at point B about y axis: ( Q B ) y = 0 Area above the point B , the y axis is zero. Determine the second moment of inertia at point B about z axis: ( Q B ) z = z ¯ ′ A ′ = ( 4 r 3 π ) ( π 2 r 2 ) Substitute 0.03 m for r . ( Q B ) z = ( 4 ( ( 0.03 ) ) 3 π ) ( π 2 ( 0.03 ) 2 ) = 0.0127 ( 1.4137 × 10 − 03 ) = 18 × 10 -6 m 3 Determine the normal stress at point A for combined loadings: σ x = σ B = − N x A + M z y B I z + M y z B I y Centroid of the y axis is -0.03 m. Substitute 100 ( 10 3 ) N for N x , 0.9 ( 10 − 3 ) π m 2 for A , − 4.00 ( 10 3 ) for M z , -0.03 m for y B , 0.2025 ( 10 − 6 ) π m 4 for I z , 0.2025 ( 10 − 6 ) π m 4 for I y

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Stress transformation

The term stress indicates the measure of internal forces that develop due to an external force's application on an object. When a specific external force is applied to an object, internal stress generates in the material. The value of stress can obtain b…

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Chapter 13.2, Problem 48P

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Find the state of stress at point B.

Show the differential volume element at point B.

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Chapter 13.2, Problem 47P

Chapter 13.2, Problem 49P

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Find the state of stress at point B. Show the differential volume element at point B.

Solution Summary: The author shows the differential volume element at point B. Determine the internal loadings as follows: Apply equilibrium conditions the force in x direction.