Question

1- A helical compression spring is to be made of oil-tempered wire of 4-mm diameter with a spring index of C = 10. The spring is to operate inside a hole, so buckling is not a problem and the ends can be left plain. The free length of the spring should be 80 mm. A force of 50 N should deflect the spring 15 mm. (a) Determine the spring rate. (b) Determine the minimum hole diameter for the spring to operate in. (c) Determine the total number of coils needed. (d) Determine the solid length. (e) Determine a static factor of safety based on the yielding of the spring if it is compressed to its solid length

Answer 1

The detailed step by step calculations for all the sub-parts of the question are described in the images below:

Oriven Data : Page 1 4 mm 80 mm wire diameter, d = spring Index, c Free Length of the spring, Lf Applied force , F = 50 N Deflection of spring, s= 15mm, Now, the spring rate is given by, K = F 8 こ 50 15 K = 3.33 N/mm Ans a s الیا هلN , spring Index may be defined as, Spring Index, c = Mean diameter of the coil wire diameter d - 10 X 4 D 40 mm : The minimum hole diameter for the spring to operate in is D = 40 mm Ans - > Now for oil-tempered spring wire, the value of shear modulus CG) may be taken as, CT =80 Gla

-Let, Page - 2 n = Number of active coils n'= Total number of coils Deflection of the spring (8) is given by, 8 = 8 F. D?n God 4 ..15 - 8 x 50 (40)3 xn 80 x103 x (4)4 12 Assuming Plain ends of a helical compression spring, .. Total number of coils, n' = Number of active coils (n) n' = 12 Ans Now, Solid length of the spring, Ls =(n+1).d (12+1). (4) = وا - 52 mm Ams -