Explanation Given information: The applied load is 150 kN. Calculation: Sketch the free body diagram of upper segment of the block as shown in Figure 1. Apply the equilibrium of force x in direction as shown in below: ∑ F x = 0 N + 150 = 0 N = − 150 kN Take a moment about y as shown in below: ∑ M y = 0 M y + 150 ( 0.05 ) = 0 M y = − 7.5 kN .m Take a moment about z as shown in below: ∑ M z = 0 M z + 150 ( 0.075 ) = 0 M z = − 11.25 kN .m Determine the rectangular cross section ( A ) as shown in below: A = b d Here, width of the rectangular cross section b and d is the thickness of the cross section. Substitute 100 mm for b and 150 mm for d . A = ( 100 mm × 150 mm ) × 1 m 1 , 000 mm = 0.015 m 2 Determine the moment of inertia of rectangular cross section as shown in below: I y = b d 3 12 Substitute 0.1 m for b and 0.15 m for d . I y = 0.15 × 0.1 3 12 = 1.5 × 10 − 4 12 = 12.5 × ( 10 − 6 ) m 4 Determine the moment of inertia of rectangular cross section ( I ) as shown in below: I z = b d 3 12 Substitute 0.1 m for b and 0.15 m for d . I z = 0.1 × 0.15 3 12 = 28.125 ( 10 − 6 ) m 4 Determine the normal stress at point A for combined loadings: σ A = N A − M z y I z + M y z I y (1) Refer Figure 1. For point A , y A = 0.075 m and z A = 0.05 m . Substitute − 150 × 10 3 N for N , 0.015 m 2 for A , 11.2 5 × 10 3 N

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 33P

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Find the normal stress developed at point A and point B for combined loadings.

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Chapter 13.2, Problem 32P

Chapter 13.2, Problem 34P

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