Ch.1 Problem 1.16 Solution | Michigan State University - Edubirdie

University:
Michigan State University
Course:
ERG 100 | Introduction to Engineering Design
Academic year:

2021
Views:

255

Pages:

1
Author:

esfredint7qo

PROBLEM 1.16 The wooden members A and B are to be joined by plywood splice plates, that will be fully glued on the surfaces in contact. As part of the design of the joint, and knowing that the clearance between the ends of the members is to be 14 in., determine the smallest allowable length L if the average shearing stress in the glue is not to exceed 120 psi. SOLUTION There are four separate areas that are glued. Each of these areas transmits one half the 5.8 kip force. Thus F = 1 1 P = (5.8) = 2.9 kips = 2900 lb. 2 2 Let l = length of one glued area and w = 4 in. be its width. For each glued area, A = lw Average shearing stress: τ= F F = A lw The allowable shearing stress is τ = 120 psi 2900 F = = 6.0417 in. τ w (120)(4) Solving for l, l= Total length L: L = l + (gap) + l = 6.0417 + 1 + 6.0417 4 L = 12.33 in. W

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