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

2021
Views:

24

Pages:

1
Author:

Alexa Livingston

PROBLEM 1.54 Solve Prob. 1.53, assuming that the pins at C and F have been replaced by pins with a 34 -in diameter. PROBLEM 1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section 14 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 12 -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. ΣM E = 0 : 16FCF − (26)(2) = 0 FCF = 3.25 kips Failure by tension in links CF. (2 parallel links) A = (b − d ) t = (1 − 34 )( 14 ) = 0.0625 in 2 Net section area for 1 link: FU = 2 Aσ U = (2)(0.0625)(60) = 7.5 kips Failure by double shear in pins. A= π 2 d2 = π § 3· 2 ¨ ¸ = 0.44179 in 4 4 © 4¹ FU = 2 AτU = (2)(0.44179)(25) = 22.09 kips Actual ultimate load is the smaller value. FU = 7.5 kips Factor of safety: F. S . = 7.5 FU = FCF 3.25 F. S . = 2.31 W