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

Solution! Cinen? - c=10 d=4mm Lo=80mm F = SON y = 15mm (a) K: F 50 y 15 → K=3.333 N/mm Answer. (6) D = c d 10242 40mm OD Dad =4044 OD = 44mm Answer © From Table of mechanical properties of come spring wires' G = 77.2 G Pa B Na dG SKD 49 (77.2) 10² 8 (3.333) 40 Na = 11.6 coils as the spring is plain ended Total number of coils, Nt = Na = 11.6 coils Answer

d) Le = d (Nt +1) Ls = 4(1):6 +1) solid Length 50.4 mm Answer = (e) From the table for oil tempered coine m=0.187 and A = 1855 Mpqimmy files. Sut А. am 40.184 Sut 1431 MPa Ssy = 0,50 Sut = 0.50 (1431) = 715.5MPa Ys Lo Ls Ys =80-50.4 ys 29.6 mm Fs kys (3.333) (296) 98.66 N KB = 4C +2. 4C-B L (D) +2 (0) -3 KB = 1.135

ar KB 8 FSD T ² 3 1.135 898.66) 40 X (43) で。 178.2 MPa Ms =. Ssu عر 715.5 178,2 ns 4.02 Answer