Explanation Given information: The applied force to the handles is 100 N. Calculation: Sketch the free body diagram of the handles as shown in Figure 1. Determine the reaction of the given support as shown in below: ∑ M D = 0 100 ( 0.25 ) − F C ( 0.05 ) = 0 25 − 0.05 F c = 0 − 0.05 F c = − 25 F C = 500 N Sketch the free body diagram of the segment as shown in Figure 2. Consider the equilibrium of the segment in the y direction as shown below: ∑ F y = 0 500 − V = 0 V = 500 N Take moment about C as shown in below: ∑ M C = 0 M − 500 ( 0.025 ) = 0 M − 12.5 = 0 M = 12.5 N .m Determine the moment of inertia of the cross section the centroidal axis ( I ) as shown in below: I = b d 3 12 Substitute 7.5 mm for b and 20 mm for d . I = ( 7.5 mm × 1 m 1 , 000 mm ) ( 20 mm × 1 m 1 , 000 mm ) 3 12 = 6 × 10 − 8 12 I = 5 × 10 − 9 m 4 Sketch a free body diagram of point A and point B as shown in below: Calculate section modulus at point B . Q B = y ¯ ′ A ′ Substitute 0.005 in for y ¯ ′ , ( 0.01 × 0.0075 ) in for A ′ . Q B = 0.005 ( 0.01 ) ( 0.0075 ) = 0.375 ( 10 − 6 ) m 3 Calculate section modulus at point A . Q A = 0 Determine the normal stress at point A of the smooth pipe as shown in below: σ A = M y I Substitute 0.01m for y , 12.5N.m for M , and 5 × 10 − 9 m 4 for I

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.2, Problem 30P

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Find the normal stress at point A and point B of the smooth pipe

Find the shear stress at point A and point B of the smooth pipe

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Chapter 13.2, Problem 29P

Chapter 13.2, Problem 31P

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Find the normal stress at point A and point B of the smooth pipe Find the shear stress at point A and point B of the smooth pipe

Solution Summary: The author calculates the normal stress and shear stress at points A and B of the smooth pipe. The applied force to the handles is 100 N.