Explanation Given information: The weight of the block C is W = 25 lb . The modulus of elasticity is E = 29 × 10 6 psi . The maximum normal stress is σ m = 18 ksi . The length of the beam is L = 7.5 ft The section of the beam is W 5 × 16 . Calculation: Consider the acceleration due to gravity g = 32.2 ft / s 2 . Provide the section properties of the beam The beam is W 5 × 16 as shown below. The moment of inertia ( I y ) is 7.51 in . 4 . The elastic section modulus ( S x ) is 3 in . 3 . Sketch the deformation diagram as shown in Figure 1. Calculate the maximum bending moment ( M m ) as shown below. M m = σ m S x Substitute 18 ksi for σ m and 3 in . 3 for S x . M m = 18 × 3 = 54 kip ⋅ in . Calculate the equivalent force ( P m ) as shown below. P m = M m L Substitute 54 kip ⋅ in for M m and 7.5 ft for L . P m = 54 kip ⋅ in 7.5 ft × 12 in . 1 ft = 0.6 kips × 1 , 000 lb 1 kips = 600 lb Refer to Appendix D “Beam Deflections and Slope” in the text book, Calculate the maximum deflection ( y m ) as shown below. y m = P m L 3 3 E I Substitute 600 lb for P m , 7.5 ft for L , 7.51 in . 4 for I , and 29 × 10 6 psi for E

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ISBN: 8220100257063

Author: BEER

Publisher: YUZU

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Chapter 11.5, Problem 49P

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Find the largest speed v0 for which the maximum normal stress in the pipe does not exceed 18 ksi.

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Chapter 11.5, Problem 48P

Chapter 11.5, Problem 50P

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Find the largest speed v 0 for which the maximum normal stress in the pipe does not exceed 18 ksi .

Solution Summary: The author calculates the maximum normal stress in the pipe for the largest speed v_0 and the section properties of the beam as shown in Figure 1.

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Find the largest speed v0 for which the maximum normal stress in the pipe does not exceed 18 ksi.

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Chapter 11 Solutions