Explanation Given information: The outer diameter of the steel pipe is 30 mm. The inner diameter of the steel pipe is 20 mm. Calculation: Sketch the Free Body Diagram of the A-36 steel pipe as shown in Figure 1. Refer to Figure 1. Find the internal loadings as shown below. Apply the Equations of Equilibrium as shown below. Summation of forces along y direction is Equal to zero. ∑ F y = 0 V y + 900 − 900 = 0 V y = 0 Take moment about point A along x direction is Equal to zero. ∑ M x = 0 T + 900 × 0.4 = 0 T + 360 = 0 T = − 360 N ⋅ m Take moment about point A along z direction is Equal to zero. ∑ M z = 0 M z + 900 × 0.15 − 900 × 0.25 = 0 M z − 90 = 0 M z = 90 N ⋅ m Find the moment of inertia about z axis as shown below. I z = π 64 ( d o 4 − d i 4 ) Here, d o is the outer diameter and d i is the inner diameter. Substitute 30 mm for d o and 20 mm for d i . I z = π 64 ( ( 30 mm × 1 m 1 , 000 mm ) 4 − ( 20 mm × 1 m 1 , 000 mm ) 4 ) = 10.15625 × 10 − 9 π m 4 Find the polar moment of inertia for the solid bar as shown below. J = π 32 ( d o 4 − d i 4 ) Substitute 30 mm for d o and 20 mm for d i . J = π 32 ( ( 30 mm × 1 m 1 , 000 mm ) 4 − ( 20 mm × 1 m 1 , 000 mm ) 4 ) = 20.3125 × 10 − 9 π m 4 The centroid of the bar c = d o 2 . Find the bending stress as shown below. σ Y = − M y A I Here, I is the moment of inertia, y A is the distance from the neutral axis, and M is the bending moment. Substitute 10.15625 × 10 − 9 π m 4 for I z , 90 N ⋅ m for M , and 0.015 m for y A . σ Y = − 90 × 0.015 10.15625 × 10 − 9 π = − 42.31 × 10 6 N / mm 2 × 1 MPa 10 6 N / mm 2 = − 42.31 MPa Find the distortion energy as shown below. τ = T c J Here, T is the torque and J is the polar moment of inertia. Substitute 20.3125 × 10 − 9 π m 4 for J , 360 N ⋅ m for T , and 0

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 10.7, Problem 10.78P

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The factor of safety against yielding of the material at point A according to maximum distortion energy theory.

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Chapter 10.7, Problem 10.77P

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