Question

15. The free end of a torsional spring deflects through 90° when subjected to a torque of 4 N-m. The spring index is 6. Determine the coil wire diameter and number of turns with the following data: Modulus of rigidity 80 GPa ; Modulus of elasticity 200 GPa; Allowable stress = 500 MPa.

Answer 1

the equation for deflections of spring

Mtl 0 JG

where,
$\theta$= angle of twist (radians)

M_t = torsional moment = 4 N-m
l = length of bar (πDN)
J = polar moment of inertia of bar (πd⁴/32)
G = modulus of rigidity = 80 GPa

90° 2 (in radians)

  

C 6 d

D=6d

now putting values in equation

4TDN d 80 10 32 2

4 6d* N T 2 32 80 109

thus we get

  

163624617 37 d = N

now,

the equation for resultant stress, which includes torsional shear stress, direct shear stress and stress concentration due to curvature. This equation is given by,

8PD T=Ks Td3

here,

4С - 1 0.615 К, 1.2525 + 4С- 4 С

also

PD Mt 2

$\tau=K_{s}\left(\frac{16 M_{t}}{\pi d^{3}}\right)$

16 4 500 *10 = 1.2525

we get,

d0.0037 m 0.004 m 4 mm

thus,

D=24 mm

from

$163624617.37*d^3=N$

we get

$N = 10.27\approx 11\: turns$