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Chapter 10

Q. 10.19

A concentric spring consists of two helical compression springs one inside the other.
The free length of the outer spring is 15 mm greater than that of the inner spring. The wire diameter and mean coil diameter of the inner spring are 5 and 30 mm respectively. Also, the wire diameter and mean coil diameter of the outer spring are 6 and 36 mm respectively. The number of active coils in the inner and outer springs are 8 and 10 respectively. Assume same material for two springs and the modulus of rigidity of spring material is 81370 N/mm². The composite spring is subjected to a maximum axial force of 1000 N. Calculate:
(i) the compression of each spring;
(ii) the force transmitted by each spring; and
(iii) the maximum torsional shear stress induced in each spring.

Step-by-Step

Verified Answer

Given P = 1000 N G = 81370 N/mm²
Step I Compression of each spring
Suppose suffix i and o refer to inner and outer spring respectively.

D_{i}=30 mm \quad d_{i}=5 mm \quad N_{i}=8 \text { coils } .

D_{o}=36 mm \quad d_{o}=6 mm \quad N_{o}=10 \text { coils } .

Stiffness of springs
From Eq. (10.9),

k=\frac{G d^{4}}{8 D^{3} N}                   (10.9).

k_{i}=\frac{G d_{i}^{4}}{8 D_{i}^{3} N_{i}}=\frac{(81370)(5)^{4}}{8(30)^{3}(8)}=29.43 N / mm .

k_{o}=\frac{G d_{o}^{4}}{8 D_{o}^{3} N_{o}}=\frac{(81370)(6)^{4}}{8(36)^{3}(10)}=28.25 N / mm .

This type of spring is shown in Fig. 10.25. The free length of the outer spring is 15 mm greater than the inner spring. Therefore, the inner spring will not transmit any force till the outer spring is compressed by 15 mm. Suppose P is the axial force on the outer spring corresponding to this compression.

P=k_{o} \delta=28.25(15)=423.75 N .

After this load, both springs are active and each will transmit the force.
Remaining load shared by two springs = 1000 – 423.75 = 576.25 N
Concentric springs are parallel springs. From
Eq. (10.12),

k=k_{1}+k_{2}+\cdots           (10.12).

k=k_{o}+k_{i}=28.25+29.43=57.68 N / mm .

where k is the combined stiffness of the composite spring. Suppose x is the further compression of two springs.
Remaining load = kx or 576.25 = 57.68 x
x = 9.99 mm.

\text { Compression of outer spring }=\delta_{o}=15+9.99 .

= 24.99 mm             (i).

\text { Compression of inner spring }=\delta_{i}=9.99 mm               (i)

Step II Force transmitted by each spring
Force transmitted by outer spring

P_{o}=k_{o} \delta_{o}=28.25(24.99)=705.97 N               (ii).

Force transmitted by inner spring

P_{i}=k_{i} \delta_{i}=29.43(9.99)=294.01 N                 (ii).

Step III Maximum shear stress

C=\frac{D_{o}}{d_{o}}=\frac{36}{6}=6 \quad C=\frac{D_{i}}{d_{i}}=\frac{30}{5}=6 .

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. (10.6),

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

\tau_{o}=K\left(\frac{8 P_{o} D_{o}}{\pi d_{o}^{3}}\right)=(1.2525)\left[\frac{8(705.97)(36)}{\pi(6)^{3}}\right] .

=375.28 N / mm ^{2}                 (iii).

Inner spring

\tau_{i}=K\left(\frac{8 P_{i} D_{I}}{\pi d_{i}^{3}}\right)=(1.2525)\left[\frac{8(294.01)(30)}{\pi(5)^{3}}\right] .

=225.06 N / mm ^{2}                 (iii).

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