Question 16.6: A close coiled helical spring has a stiffness of 4 N/mm. Its...
A close coiled helical spring has a stiffness of 4 N/mm. Its solid length is 160 mm when the adjecent coils touch each other. Determine the mean diameter of the spring and the diameter of the wire if their ratio is 10:1. Take modulus of rigidity equal to 80 GPa. If the gap between the coils is 5 mm, what is the load required to close it and what is the corresponding shear stress.
Stiffness, \frac{P}{\delta } =4 N/mm
Mean coil radius, R = 5d
Number of coils, n=\frac{160}{d}
Using Eq. (16.3), \frac{P}{\delta } =\frac{Gd^4}{64R^3n}
4=\frac{80000\times d^4}{64\times 125 d^3\times 160/d}or, Diameter of wire, d = 8 mm
Mean diameter of spring = 80 mm
Number of turns, n=\frac{160}{8} =20
Deflection required to close 5 mm gap between the coils
\delta =20\times5=100 mmForce required for 100 mm deflection,
P= 100 x 4 = 400 N
The corresponding shear stress,
\tau =\frac{16PR}{\pi d^3} \left\lgroup1+\frac{d}{4R} \right\rgroup= \frac{16\times400\times40}{\pi\times8^3} \left\lgroup1+\frac{8}{4\times40} \right\rgroup
= 158.75 MPa