Explanation Given information: The force at point B is F B = 910 lb . The allowable shear stress for the segment is τ s a l l o w = 275 psi . Thickness of the segment is t = 1 2 in . . Free body diagram of body B . For the member to be in equilibrium, ∑ F H = 0 ∑ F V = 0 Shear stress in supporting material is given by using the formula of shear stress: τ s = F s h e a r A s h e a r Shear area of segment is given by: A s h e a r = b s e g m e n t × t Calculation: The angle of member B with the horizontal direction is: tan θ = 5 12 θ = tan − 1 5 12 = 22

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Chapter 1.7, Problem 86P

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The dimension of b the horizontal segment.

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Chapter 1.7, Problem 85P

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The dimension of b the horizontal segment.

Solution Summary: The author explains the dimension of b the horizontal segment and the allowable shear stress for the segment.

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