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= (d4) / 32
Ip = (*0.0054) / 32
Ip = 6.138*10-11 m4
Now shear stress dur to torque T is t = (T*r) / IP
t = {4.1* (0.005 / 2)} / 6.138*10-11
t = 1.67* 108 N/ m2
Now shear stress due to force F is f = F/ (d2 / 4)
f = 164 / (* 0.0052 / 4)
f = 8.35* 106 N/ m2
Therefore, Maximum shear stress max = t + f
max = 1.67* 108 + 8.35* 106
max = 175.35* 106 N/ m2
max = 175.35 MPa
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