Ch.1 Problem 1.18 Solution - Edubirdie

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

2021
Views:

104

Pages:

1
Author:

ilikemyicons0y5

PROBLEM 1.18 Two wooden planks, each 22 mm thick and 160 mm wide, are joined by the glued mortise joint shown. Knowing that the joint will fail when the average shearing stress in the glue reaches 820 kPa, determine the smallest allowable length d of the cuts if the joint is to withstand an axial load of magnitude P = 7.6 kN. SOLUTION Seven surfaces carry the total load P = 7.6 kN = 7.6 × 103. Let t = 22 mm. Each glue area is A = dt τ = P 7A A = P 7.6 × 103 = = 1.32404 × 10−3 m2 3 7τ (7)(820 × 10 ) = 1.32404 × 10 3mm 2 d = A 1.32404 × 10 = 22 t 3 = 60.2 d = 60.2 mm W

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