Explanation Given information: The failure stress in aluminium is σ f a i l = 400 MPa . The force applied is P = 60 kN . Factor of safety is 2 . For the rods to be in equilibrium, Consider the joint B . For the joint to be in equilibrium, ∑ F H = 0 ∑ F V = 0 Stress in rods is given by using the formula: σ a l l o w a b l e = σ f a i l F o S = F A r o d Rod area is given by: A r o d = π 4 d r o d 2 Calculation: The angle of rod A B with horizontal. tan θ = 3 4 θ = tan − 1 0.75 θ = 36.87 ° For the horizontal equilibrium of joint B . ∑ F H = 0 F A B cos θ − F B C = 0 F A B cos θ = F B C F A B cos 36.87 ° = F B C 0.8 F A B = F B C .............. 1 For the vertical equilibrium of joint B . ∑ F V = 0 F A B sin θ − P = 0 F A B sin θ = P F A B sin 36.87 ° = 60 kN 0.6 F A B = 60 kN F A B = 60 kN 0.6 F A B = 100 kN ............ 2 From (1) and (2), F B C = 0.8 F A B = 0

11th Edition

ISBN: 9780137605460

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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Chapter 1.7, Problem 78P

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The required diameters of the rods AB and BC .

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The required diameters of the rods A B and B C .

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