Question

Mwenye 83 LO

A helical spring with stiffness k = 20 N / mm becomes locked when pressed with δB = X mm. The spring diameter ratio is Ck = D / d = 8 and in case of locking, the spring is dimensioned, locking and releasing, provided that the stresses do not exceed τb = 455 N / mm2.
Calculate the spring lengths in the state. The spring sliding module is given as G = 80800 N / mm2. The spring is flattened and ground for winding diameter (σk = 1350N / mm2 for spring material)

X=30 mm

Answer 1

- 8 shan stress To ssolution - Gwen Hat Spring stiffnes * = 20 N/nem, deflection of spring S8 = 30 mm ispring diameter "Ratio C= % 455 N/mm2 obpring siding Module G= 80 800 N/mm2, spring Material Ex=1350 N/mm2 stiffness is gurn by formula stiftness (k) Deflectim (se) For Load (P) 20 P पे 30 p = 600 N The permissible shear stress for the spring is calculated ap - T = 0.3 x GK = 0.3 x 1350 = 405 N/mm2 The permissible shear stress is acceptable ags TLTD. Now, Calculating diameter of spring wine . Wahl factor is calculated by formula K = 40-1 + 0.615 С. 46-4 K = (4x8-1) 0.615 + 8 (4x8-4) K = 1185 ने

→ T-K 8.P.C 7.02 405= 1.185 8x600 x 8 rx da a d = 35.78 mm by rounding off d = 36 mm # calculating activ number coils (N) wesing deflection formular of 88 = 8x P x C3 XN G x d 3 + 30 = 8 x 600 x 8 x N 80 800 x 36 N = 35.508 an Number of twens is always integen N=36 twins For flatned & ground end total number of given by formula Ne twins are = N + 1/2 36+ / & Nt N+= 36.5 twins in case # Length of spring of locking is given as Lo= Nexd Le = 36.5 x 36 9 LB - 1314 mm # For determining the free length (20) of the spring, we need to find the actual deflection for active coil CN=36 turns), we have -

Acutal deflection 8x P x C3 XN 11 a Sac Sac God = 8x3600 x 83 x 36 80800 x 36 Sac = 182.49 mm = Lot Sac Now, full ungth of spuing (Lo): Lo Lo 1314+ 182.49 . 1496-491 mm Answer

Please note that in free length of spring always actual deflection is added to the solid length .

Thank you.