Ch.1 Problem 1.53 Solution | Study Guide - Edubirdie

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

2021
Views:

435

Pages:

1
Author:

Leilani Contreras

PROBLEM 1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section 41 in. thick and 1 in. wide, and is made of a steel with an ultimate strength in tension of 60 ksi. The pins at C and F each have a 21 -in. diameter and are made of a steel with an ultimate strength in shear of 25 ksi. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. SOLUTION Use member EFG as free body. Σ M E = 0 : 16 FCF − (26)(2) = 0 FCF = 3.25 kips Failure by tension in links CF. (2 parallel links) Net section area for 1 link: A = (b − d ) t = (1 − 12 )( 14 ) = 0.125 in 2 FU = 2 Aσ U = (2)(0.125)(60) = 15 kips Failure by double shear in pins. A = π 4 d2 = π § 1· 2 = 0.196350 in 2 4 ¨© 2 ¸¹ FU = 2 AτU = (2)(0.196350)(25) = 9.8175 kips Actual ultimate load is the smaller value. FU = 9.8175 kips Factor of safety: F. S . = FU 9.8175 = 3.25 FCF F. S . = 3.02 W

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