Ch.1 Problem 1.43 Solution | Study Guide - Edubirdie

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

2021
Views:

499

Pages:

1
Author:

gewoekerdodx

PROBLEM 1.43 Two wooden members shown, which support a 3.6 kip load, are joined by plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in the glue is 360 psi and the clearance between the members is 41 in. Determine the required length L of each splice if a factor of safety of 2.75 is to be achieved. SOLUTION There are 4 separate areas of glue. Let l be the length of each area and w = 5 in. its width. Then the area is A = lw. Each glue area transmits one half of the total load. § 1· F = ¨ ¸ (3.6 kips) = 1.8 kips ©2¹ Required ultimate load for each glue area: FU = ( F. S.) F = (2.75)(1.8) = 4.95 kips Required length of each glue area: FU = τ U A = τ U lw l = Total length of splice: FU τ Uw L=l+ = 4.95 × 103 = 2.75 in. (360)(5) 1 in. + l 4 L = 2.75 + 0.25 + 2.75 L = 5.75 in. W

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