Question

Design a compression spring with an outside diameter of 11.00 inches from round type 302 stainless-steel wire to operate a plate dispenser in a cafeteria. When fully loaded, the spring is to be compressed 45 inches below the stop under a load of 80 lb, and when empty the spring should exert a force of 0.50 lb against the stop. The ends of the spring are closed. Specify the wire diameter, the number of active turns, the total number of turns, the solid length, and the outside diameter at the solid length. Use the solid length as the operating length.

Answer 1

Force when empty:

$F_1 = 0.5 = k*\Delta L_1$

Force at maximum Load:

$F_2 = 80 = k*\Delta L_2$

Given.

$\Delta L_2-\Delta L_1 = 45''$

Dividing the first two equations:

▲L2 80 ▲L1-0.5-160

$160\Delta L_1-\Delta L_1 = 45''$

$\Delta L_1= 45/159 = 0.283''$

$0.5=k*\Delta L_1=k*0.283$

k = 0.5/0.283 = 1.771b/in

Stress in the spring:

$S_s = \frac{8FD_m}{\pi D_w^3}$

Given D_out = 11''

F = 80lb

$11 = D_m+D_w \Rightarrow 11 - D_w = D_m$

$S_s = \frac{8*80*(11-D_w)}{\pi D_w^3}$

203.72(11- D

Yield strength of SS 302 is 30,000 psi

Taking Safety factor of 2

$30000*2 = \frac{203.72(11-D_w)}{ D_w^3}$

60000D3 = 203.72 (11-Dw) = 2240.92-203.72 * Du

Solving this, we get the wire diameter:

$\Rightarrow D_m = 11-0.331 '' = 10.669''$

Spring Constant is given by:

$k = \frac{G D_w^4}{8 D_m^3N_a}$

Shear Modulus of SS 302 is = 11200 ksi

$1.77 = \frac{11200*0.331^4}{8*10.669^3N_a}$

which gives:

Number of active turns: $N_a \approx 8$

Total number of turns = 8+2 = 10

Solid Length = 10*0.331 = 3.31''

Free Length : $L_f = \Delta L_2+Solid \;Length$

Equating volume of spring at full deflection and at no deflection

$\pi*D_w^3*45.59 = \pi*(D_o^2 - D_i^2)3*3.31$

$\pi*0.331^2*45.59 = \pi*((D_i+0.331)^2 - D_i^2)3*3.31$

1.509- 0.662D 0.331

$D_i = 2.114\Rightarrow D_o = 2.114+0.331 = 2.445$

Outside diameter at solid length = 2.445''