A compression spring (helical) of the exhaust valve mechanism is compressed with a pre-load of 350 N. When the spring is further compressed and the valve is fully opened, the torsional shear stress in the spring wire should not exceed 650 N/mm2. Due to space limitations, the outer diameter of the spring should not exceed 38 mm. The spring is to be designed for minimum weight. Calculate the wire diameter and the mean coil diameter of the spring.

Subject :- Machine Design -I

Topic :- Spring

Difficulty :- Medium

Assuming the outer diameter to be 38mm,

$D_o=D+d=Cd+d=d(C+1)$

i.e $d=\frac{D_o}{C+1}=\frac{38}{C+1}$ (a)

We have,

$\tau_max=K\frac{8P_maxD}{\pi d^3}=K\frac{8P_maxC}{\pi d^2}$

Therefore,

$650=K\frac{8*650*c*(c+1)^2}{\pi *38^2}$

$C(C+1)^2K=567.05$ (b)

The problem is solved by trial and error method. In practice, the spring index varies from 6 to 10. Considering values of C in this range, the results are tabulated as follows.

Value of K is,

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}$

Comparing (b) and the values in above table,

C=8

$d=\frac{38}{C+1}=\frac{38}{8+1}=4.22mm$

$d=5mm$ (i)

Mean coil diameter:

Since, $D_o=D+d$

$38=D+5$

D=33mm (ii)

Check for design:

$C=\frac{D}{d}=\frac{33}{5}=6.6$

$K=\frac{4C-1}{4C-4}+\frac{0.615}{C}=\frac{4*6.6-1}{4*6.6-4}+frac{0.615}{6.6}=1.227$

Also, $\tau_max=K\frac{8PC}{\pi d^2}=1.2\frac{8*700*606}{\pi 5^2}=470.58 N/mm^2\lt750 N/mm^2$

Therefore, dimensions are satisfactory.