Question 10.17: A concentric spring is used as a valve spring in a heavy dut...

\text { Given } \quad P=6000 N \quad \delta=50 mm \quad \tau=800 N / mm ^{2} .

C = 6 G = 81370 N/mm²
Step I Axial force transmitted by each spring
The diametral clearance between the coils is equal to the difference between their wire diameters. From Eq. (10.26),

\frac{d_{1}}{d_{2}}=\frac{C}{(C-2)}                 (10.26).

\frac{d_{1}}{d_{2}}=\frac{C}{(C-2)}=\frac{6}{(6-2)}=1.5 .

From Eq. (10.23),

\frac{P_{1}}{P_{2}}=\frac{d_{1}^{2}}{d_{2}^{2}}           (10.23).

\frac{P_{1}}{P_{2}}=\frac{d_{1}^{2}}{d_{2}^{2}}=\left(\frac{d_{1}}{d_{2}}\right)^{2}=(1.5)^{2}=2.25                 (a).

\text { Also, } \quad P_{1}+P_{2}=P=6000 N             (b).

Solving equations (a) and (b) simultaneously,

P_{1}=4153.85 N \text { and } P_{2}=1846.15 N                       (i)

Step II Wire and mean coil diameters
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(6)-1}{4(6)-4}+\frac{0.615}{6}=1.2525 .

Outer spring
From Eq. (10.13),

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

\tau=K\left(\frac{8 P_{1} C}{\pi d_{1}^{2}}\right) \text { or } 800=(1.2525)\left[\frac{8(4153.85)(6)}{\pi d_{1}^{2}}\right] .

Outer spring
From Eq. (10.13),

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

\tau=K\left(\frac{8 P_{1} C}{\pi d_{1}^{2}}\right) \text { or } 800=(1.2525)\left[\frac{8(4153.85)(6)}{\pi d_{1}^{2}}\right] .

d_{1}=9.97 \text { or } 10 mm .

D_{1}=C d_{1}=6(10)=60 mm             (ii).

Inner spring
From Eq. (10.13),

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

\tau=K\left(\frac{8 P_{2} C}{\pi d_{2}^{2}}\right) \quad \text { or } \quad 800=(1.2525)\left[\frac{8(1846.15)(6)}{\pi d_{2}^{2}}\right] .

d_{2}=6.65 \text { or } 7 mm .

D_{2}=C d_{2}=6(7)=42 mm                  (ii).

Step III Number of active coils
From Eq. (10.8),

\delta=\frac{8 P D^{3} N}{G d^{4}}                 (10.8).

\delta=\frac{8 P_{1} D_{1}^{3} N_{1}}{G d_{1}^{4}} \quad \text { or } \quad 50=\frac{8(4153.85)(60)^{3} N_{1}}{(81370)(10)^{4}} .

N_{1}=5.67 \text { or } 6 \text { coils }                 (iii).

It is assumed that the springs have square and ground ends. Therefore,

\left(N_{t}\right)_{1}=N_{1}+2=6+2=8 \text { coils } .

Since the springs have the same solid length,

\left(N_{t}\right)_{1} d_{1}=\left(N_{t}\right)_{2} d_{2} \quad \text { or } \quad 8(10)=\left(N_{t}\right)_{2}(7) .

N_{2}=12-2=10 \text { coils }                     (iii).