Question 10.20: A helical compression spring of the exhaust valve mechanism ...

\text { Given } P_{\min .}=375 N \quad \tau_{\max .}=750 N / mm ^{2} .

D_{o}=42 mm .

Step I Wire diameter
For minimum weight,

P_{\max .}=2 P_{\min .}=2(375)=750 N .

Assuming the outer diameter to be 42 mm,

D_{o}=D+d=C d+d=d(C+1) .

or        d=\frac{D_{o}}{(C+1)}=\frac{42}{(C+1)}             (a).

From Eq. (10.6),

\tau=K\left(\frac{8 P D}{\pi d^{3}}\right)                   (10.6).

\tau_{\max .}=K\left(\frac{8 P_{\max .} D}{\pi d^{3}}\right)=K\left(\frac{8 P_{\max .} C}{\pi d^{2}}\right) .

Therefore,

750=K\left(\frac{8(750) C(C+1)^{2}}{\pi(42)^{2}}\right) .

C(C+1)^{2} K=692.72                   (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 in the following manner.

K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}                 (10.7).

Comparing Eq. (b) and the values in above table,

C = 8

d=\frac{42}{(C+1)}=\frac{42}{8+1}=4.67 mm .

∴        d = 5 mm               (i)
Step II Mean coil diameter
Since              D_{o}=D+d \quad 42=D+5

D = 37 mm                (ii)
Step III Check for design

C=\frac{D}{d}=\frac{37}{5}=7.4 .

K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(7.4)-1}{4(7.4)-4}+\frac{0.615}{7.4}=1.2 .

From Eq. (10.13),

\tau=K\left(\frac{8 P C}{\pi d^{2}}\right)               (10.13).

\tau_{\max .}=K\left(\frac{8 P C}{\pi d^{2}}\right)=(1.2)\left[\frac{8(750)(7.4)}{\pi(5)^{2}}\right] .

=678.38 N / mm ^{2} .

Therefore,

\tau_{\max }<750 N / mm ^{2} .

and the dimensions are satisfactory.