Question

QUESTION 1 A compression coil spring made of an alloy steel is having the following specifications:- Mean diameter of coil = 50 mm; Wire diameter = 5 mm; Number of active coils = 20; If the spring is subjected to an axial load of 164 N. Calculate the maximum shear stress (Neglect the curvature effect) to which the spring material is subjected. (units of MPa).

Answer 1

Given,

Mean dia of the coil (D) = 50mm = 0.05m

Wire dia (d) = 5mm = 0.005m

Number of coils (N) = 20

Axial load (F) = 164N

When an axial force is applied to the spring torque is generated in the spring which results in shear stress. Now due to the applied load also there will be shear in the coils which is very less than the torque shear.

So, Maximum shear stress = Shear stress due to torque + Shear stress due to axial load

Now the torque acting on the spring T = F* (D/2)

  T = 164* (0.05/2)

T = 4.1Nm

the polar moment of inertia of the coil of dia d is Ip= ($\pi$d4) / 32

  Ip = ($\pi$*0.0054) / 32

Ip = 6.138*10-11 m4

Now shear stress dur to torque T is $\tau$ t = (T*r) / IP

  $\tau$t = {4.1* (0.005 / 2)} / 6.138*10-11

  $\tau$t = 1.67* 108 N/ m2

Now shear stress due to force F is $\tau$ f = F/ ($\pi$d2 / 4)

$\tau$f = 164 / ($\pi$* 0.0052 / 4)

$\tau$f = 8.35* 106 N/ m2

Therefore, Maximum shear stress $\tau$ max = $\tau$ t + $\tau$ f

$\tau$max = 1.67* 108 + 8.35* 106

  $\tau$max = 175.35* 106 N/ m2

  $\tau$max = 175.35 MPa