\text { Given } P=2000 N \quad G=81370 N / mm ^{2} .
Step I Force transmitted by each spring
Suppose suffix i and o refer to inner and outer spring respectively
D_{i}=64 mm \quad d_{i}=8 mm \quad N_{i}=12 coils .
D_{o}=80 mm \quad d_{o}=10 mm \quad N_{o}=8 coils .
\text { Since both springs have same deflection, } \delta_{i}=\delta_{o}
From Eq. (10.8),
\delta=\frac{8 P D^{3} N}{G d^{4}} (10.8)
\frac{8 P_{i} D_{i}^{3} N_{i}}{G d_{i}^{4}}=\frac{8 P_{o} D_{o}^{3} N_{o}}{G d_{o}^{4}} .
or \frac{P_{i} D_{i}^{3} N_{i}}{d_{i}^{4}}=\frac{P_{o} D_{o}^{3} N_{o}}{d_{o}^{4}}.
\frac{P_{i}(64)^{3}(12)}{(8)^{4}}=\frac{P_{o}(80)^{3}(8)}{(10)^{4}} .
\frac{P_{o}}{P_{i}}=1.875 (a).
Also, P_{o}+P_{i}=2000 N (b).
Solving equations (a) and (b) simultaneously,
P_{o}=1304.35 N \quad P_{i}=695.65 N (i).
Step II Maximum deflection of the spring
From Eq. (10.8),
\delta_{i}=\delta_{o}=\frac{8 P_{o} D_{o}^{3} N_{o}}{G d_{0}^{4}}=\frac{8(1304.35)(80)^{3}(8)}{(81370)(10)^{4}} (ii).
Step III Maximum shear stress
C=\frac{D_{o}}{d_{o}}=\frac{80}{10}=8 \quad C=\frac{D_{i}}{d_{i}}=\frac{64}{8}=8 .
From Eq. (10.7),
K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C} (10.7).
K=\frac{4 C-1}{4 C-4}+\frac{0.615}{C}=\frac{4(8)-1}{4(8)-4}+\frac{0.615}{8}=1.184 .
Outer spring
From Eq. (10.13),
\tau=K\left(\frac{8 P C}{\pi d^{2}}\right) (10.13).
\tau_{o}=K\left(\frac{8 P_{o} C}{\pi d_{o}^{2}}\right)=(1.184)\left[\frac{8(1304.35)(8)}{\pi(10)^{2}}\right] .
=314.61 N / mm ^{2} (iii).
Inner spring
\tau_{i}=K\left(\frac{8 P_{i} C}{\pi d_{i}^{2}}\right)=(1.184)\left[\frac{8(695.65)(8)}{\pi(8)^{2}}\right] .
=262.18 N / mm ^{2} (iii).