A helical spring ‘B’ is placed inside the coils of a second helical spring ‘A’, having same number of coils, same free axial length and same material. The two springs are compressed by an axial load of 210 N which is shared by them. The mean coil diameters of A and B are 90 mm and 60 mm and the

Question

A helical spring ‘B’ is placed inside the coils of a second helical spring ‘A’, having same number of coils, same free axial length and same material. The two springs are compressed by an axial load of 210 N which is shared by them. The mean coil diameters of A and B are 90 mm and 60 mm and the wire diameters are 12 mm and 7 mm, respectively. Calculate the load shared by individual springs and maximum stress in each spring.

Accepted Answer

We know that the spring stiffness, k is

[latex] k=\frac{G d^4}{64 n R^3} [/latex]

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

[latex] \frac{k_{ A }}{k_{ B }}=\left\lgroup \frac{d_{ A }}{d_{ B }} ight group^4 imes\left\lgroup \frac{R_{ B }}{R_{ A }} ight group^3 [/latex]

[latex] =\left\lgroup \frac{12}{7} ight group^4 imes\left\lgroup \frac{30}{45} ight group^3 [/latex] (putting respective values)

= 2.56

If the load shared by spring A is [latex] P_{ A } ext { and that by } B ext { is } P_{ B } [/latex], we can write

[latex] \frac{P_{ A }}{k_{ A }}=\frac{P_{ B }}{k_{ B }} [/latex]

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

[latex] \frac{P_{ A }}{k_{ A }}=\frac{P_{ A }+P_{ B }}{k_{ A }+k_{ B }} [/latex]

But [latex] P_{ A }+P_{ B }=210 N [/latex], therefore

[latex] P_{ A }=k_{ A } imes \frac{210}{k_{ A }+k_{ B }}=\frac{210}{\left[1+\left(k_{ A } / k_{ B } ight] ight.}\left\lgroup\frac{k_{ A }}{k_{ B }} ight group [/latex]

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

[latex] P_{ A }=\frac{2.56}{3.56} imes 210=151.01 N [/latex]

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

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

whereas from practical point of view

[latex] au_{\max }=\frac{16 F R}{\pi d^3} imes k^{\prime} [/latex] [refer Eq. (2.15)]

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

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

Thus, for spring A,

[latex] \left( au_{ ext {max }} ight)_{ ext {theoretical }}=21.36 MPa ext { and }\left( au_{\max } ight)_{ ext {practical or actual }}=23.98 MPa [/latex]

and for spring B,

[latex] \left( au_{\max } ight)_{ ext {theoretical }}=27.81 MPa ext { and }\left( au_{\max } ight)_{ ext {actual }}=30.77 MPa [/latex]

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

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