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

2021
Views:

241

Pages:

1
Author:

kundgetanufv

PROBLEM 1.46 For the support of Prob. 1.45, knowing that the diameter of the pin is d = 16 mm and that the magnitude of the load is P = 20 kN, determine (a) the factor of safety for the pin, (b) the required values of b and c if the factor of safety for the wooden members is the same as that found in part a for the pin. PROBLEM 1.45 A load P is supported as shown by a steel pin that has been inserted in a short wooden member hanging from the ceiling. The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 145 MPa in shear. Knowing that b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an overall factor of safety of 3.2 is desired. SOLUTION P = 20 kN = 20 × 103 N (a) Pin: A= Double shear: τ = π 4 d2 = π 4 (0.016) 2 = 2.01.06 × 10− 6 m 2 P P τU = U 2A 2A PU = 2 AτU = (2)(201.16 × 10−6 )(145 × 106 ) = 58.336 × 103 N F. S. = (b) Tension in wood: where w = 40 mm = 0.040 m b = 40.3 mm W PU = 58.336 × 10 3 N for same F.S. Double shear; each area is A = wc c= PU PU = A w(b − d ) 58.336 × 10 3 PU = 0.016 + = 40.3 × 10− 3m wσ U (0.040)(60 × 10 6 ) Shear in wood: F .S . = 2.92 W PU = 58.336 × 10 3 N for same F.S. σU = b =d + 58.336 × 103 PU = P 20 × 103 τU = PU P = U 2 A 2wc 58.336 × 103 PU = = 97.2 × 10− 3 m 2 wτ U (2)(0.040)(7.5 × 10 6 ) c = 97.2 mm W