Question

A spring with end squared and ground, wire diameter d= 4 mm, outside diameter Do= 40 mm, 18 total coils, and free length Lf= 140 mm has been chosen for an application where the initial deflection of 15 mm and the working deflection is 50 mm. Determine minimum working length. shut height, clash allowance, spring index and spring rate for this spring.

Answer 1

sol:- Given :- 40 mm Wire diametex, de Umm Outside diameter, Do = No. of coils, N=18 Free length, 2f = 140 mm Initial deflection, Ni = 15 mm woruing deflection, rw = Calculate mean diameter as Thoron- D = Do-d = 40 - 4= 36 mm. 50 mm Now 36 9 spring Index is given by- 뭉 [C= g Lars spring inden Calculate ultimate tensile strength using relation - sut = Adb For spring with ende squared and ground- coefficient, A = 1753.3 mpa enpon ent, ba 1753.3 * (4-0-1822 -0.1822 sut = Sut 13 61.94918 MPa

Now, Yield shear strength is given by Sus=0.6 sut = 0.6 % 1361. 94918 Sus = 817.1695 MPa Assuming a factor of safety (f.0.3) of 2, working shear stress is given by- 817. 1695 Sus Los 2 Tworking - Ew = 408.5847 MPa / Working stress get Using evelation for Worring Shers, we (8XFXD To (1+004) * ( TT X d3 х -> 408.5847 = (8x F* 36 TX 43 9 408.5847 - 9.5% 9 288F HA 64 IT F = 270. 2091 N Working Force. Now, calculate spring rate using relation - k= F 270.2091 Hw working deflection 57 5.4042 wimm spring rate.

Shut height is given by- Ls = dxN =uX 18 72 mm Ils Ane. s shut height Free length is given by- Lst Leash & Hot Hinitial Similarly, 2f 140 = 72+ Leash + 50+ 15 Lelash 3mm Lane. G Clash allowaune. And shut height deflection is given by - Ushut L4 - Ls 140-72 - 68 mm nshut s shut height deflection. Minimum working length is given by - L - Lf - Kinitial - M working -140 - 15-50 Z = 75mm minimum working length.