Question

FOR PROBLEMS 4-6 (SHOW ALL PROCEDURE AND STEPS FOR CREDIT) A helical spring is fabricated using a steel wrapping wire 5/8-inch in diameter around a forming cylinder 5-inch in diameter (mean diameter). The spring cannot exceed a shearing stress of 20 ksi, while compressing 5 inches. Take G 12 x 106 psi PROBLEM 4 (T4A) The spring index is a) 5 b) 8 e) 10 d) None of the above PROBLEM 5 (T4A) If the spring index is 10, the most likely stiffness of the spring to meet the stress criteria and compression distance is a) 54 lb/in b) 64 lb/in c) 74 lb/in d) 84 lb/in (No information is missing. consider the shear stress equation!)

Answer 1

Given data for Helical Spring Wire diá 5/8 inch Mean diameter D = 5 inch Shearing stress'Is'= 20 ksi = 20x103 psi (y) Compression - 5 inches modulces of Rigidity G = 12X10psi ☺ Calculation of spring Index Spring Index 'd' = - Se - $*8 = 8 Hence option is correct. calculation of stiffness of the Spring Tmax= 8 KWEC kw= Wahl's Factor Here C= 10 kw=4c-1 +0.615 kw = 4x10-1 +0.615 1.145 4810-4 F = Imax rd 20x107x3.14 x (512 - 8kwe 8 x1,145 x 10 x (812 F = 267,8 lbf .: Stiffnese 'K' = Force (f), 267.8 y K=53.56~ 54eblin Hence optionca) is correct. © We wee add the required factor of safety, the diameter of helical spring 'will be increase. Hence optional es correct. 4c-u +0:615