Question 2.15: A helical spring ‘B’ is placed inside the coils of a second ...

We know that the spring stiffness, k is

k=\frac{G d^4}{64 n R^3}

Now, the ratio of spring stiffness for springs A and B is

\frac{k_{ A }}{k_{ B }}=\left\lgroup \frac{d_{ A }}{d_{ B }} \right\rgroup^4 \times\left\lgroup \frac{R_{ B }}{R_{ A }} \right\rgroup^3

=\left\lgroup \frac{12}{7} \right\rgroup^4 \times\left\lgroup \frac{30}{45} \right\rgroup^3         (putting respective values)

= 2.56

If the load shared by spring A is P_{ A } \text { and that by } B \text { is } P_{ B } , we can write

\frac{P_{ A }}{k_{ A }}=\frac{P_{ B }}{k_{ B }}

as they suffer identical compression. Therefore, from above using addendo, we get

\frac{P_{ A }}{k_{ A }}=\frac{P_{ A }+P_{ B }}{k_{ A }+k_{ B }}

But P_{ A }+P_{ B }=210  N , therefore

P_{ A }=k_{ A } \times \frac{210}{k_{ A }+k_{ B }}=\frac{210}{\left[1+\left(k_{ A } / k_{ B }\right]\right.}\left\lgroup\frac{k_{ A }}{k_{ B }} \right\rgroup

Putting k_{ A } / k_{ B }=2.56 , we get

P_{ A }=\frac{2.56}{3.56} \times 210=151.01  N

Therefore, P_{ B }=(210-151.01)  N =58.99  N . Now from theoretical point of view,

\tau_{\max }=\frac{16 F R}{\pi d^3}\left(1+\frac{d}{4 R}\right)          [refer Eq. (2.14)]

whereas from practical point of view

\tau_{\max }=\frac{16 F R}{\pi d^3} \times k^{\prime}       [refer Eq. (2.15)]

where k’ is Wahl’s correction factor and is given by

k^{\prime}=\frac{4 S-1}{4 S-4}+\frac{0.615}{S}       \text { with } S=\frac{2 R}{d}         [refer Eq. (2.16)]

Thus, for spring A,

\left(\tau_{\text {max }}\right)_{\text {theoretical }}=21.36  MPa \text { and }\left(\tau_{\max }\right)_{\text {practical or actual }}=23.98  MPa

and for spring B,