Question

2. A helical compression spring is to be designed as shown in Fig Q2. The spring is required to exert a force of Po-102 N when compressed to a length of lo-44 mm. When its length is I-76 mm, it must exert a force of P-22 N. The spring will be cycled rapidly, with severe service required. Use ASTM A401 steel wire. The factor of safety, η, need to be larger than 1.5. Starting with a spring index of 9, and checking for stability is required. a) Compute the spring rate k and the free length Ir of the spring. (6 pts) b) Determine the wire diameter dw, of the spring to satisfy the conditions given, a a peened spring. Use the Goodman failure locus for the Zimmerli data. (11 pts) termine the solid length l, the force P, and stress corresponding to the solid length, the number of active coils, OD of the coil and the factor of safety guarding against statioc ssuming failure. (18 pts) (P = 0) 0 0 Fia. Q2.

Answer 1

From the required force condition,

when P_r = 102 N, L_i = 76 mm and P_o = 22 N, L_o = 44 mm

As we know force by the spring is directly proportional to the deflection of the spring form free length i. e. L_f.

Hence P_r = k (L_f - L_i)

P_o = k (L_f - L_o)

Putting values

102 = k (L_f - 76) .... (1)

22 = k (L_f - 44) .... (2)

Solving equation (1) and (2)

we get free length L_f = 35.2 mm and spring constant k = 2.5 N/mm .... Ans

From property data sheet of ASTM A401 steel,

Ultimate tensile strength S_ut = 1050 N/mm²

Modulus of rigidity G = 81370 N/mm²

Required FOS = 1.5 and from Goodman failure condition;

FOS

where S_sy is torsional yield strength

S_sy = 0.577 S_yt where S_yt is tensile yield strength

and S_yt = 0.75 S_ut

т=0.3 * Sut

τ = 315N/mm2

Spring index C = 9

Wahl factor is given as

4C-1 4C-4

Hence K = 1.093

Shear stress is give by

8 PC

8 102 9 315 = 1.093 *

$\therefore$Wire diameter d_w = 2.848 mm ..... Ans

Mean coil diameter D = C * d_w = 9 * 2.848 = 25.63 mm

Number of active coils N is given by below formula

$\delta = \frac{8 * P * D^{3} * N}{G * d_{w}^{4}}$

where $\delta$ is spring deflection for force P

$\delta$ = 22/2.5 = 8.8 mm

N is number of active coils

8 * N 22 * 25633 813702.844

$\therefore$Number of active coils N = 15.72 ~ 16 .... Ans

Assuming number of inactive coils = 2 (for peened spring)

Total coils N_t = N + 2 = 18

Solid length of the Spring = N_t * d_w = 18 * 2.84 = 51.12 mm .... Ans

Force required for compressing spring to solid length P_s = K * solid length

P_s = 51.12 * 2.5 = 127.8 N ..... Ans

Stress corresponding to solid length

$\tau = K *\frac{8*P_{s}*C}{\pi * d_{w}^{2}}$

$\tau = 1.093 *\frac{8* 127.8*9}{\pi * 2.84^{2}}$

$\tau$ = 396.91 N/mm² ..... Ans

Outer diameter of spring D_o = D + d_w/2

D_o = 25.63 + (2.84)/2 = 27.05 mm .... Ans