Question 16.8: A 50 N weight is dropped on a close coiled helical spring fr...
A 50 N weight is dropped on a close coiled helical spring from a height of 150 mm. The spring has 10 coils of mean diameter 30 mm. If the diameter of the spring wire is 5 mm, determine the deflection and stress in the spring. Take modulus of rigidity equal to 80 GPa.
Work done by impact load = 50(150+\delta )
where,\delta = deflection due to impact load 50 N.
If P_e is the equivalent static load which will produce δ, then energy stored in the spring
=\frac{P_e\delta }{2}Equating,
Work done = Energy stored
50(150+\delta )=\frac{1}{2}P_e\deltaSubstitute P_e from Eq. (16.2)
\delta =\frac{64PR^3n}{Gd^4} (16.2)
or 50(150+\delta )=\frac{1}{2}\frac{\delta Gd^4}{64R^3n}\delta
=\frac{1}{2} \times\frac{80000\times5^4\times\delta ^2}{64\times15^3\times10}or, 23.148\delta ^2-100\delta -15000=0
or, Deflection, \delta = 27.7 mm
Equivalent static load
P_e=\frac{\delta Gd^4}{64R^3n} =\frac{27.7\times80000\times5^4\delta ^2}{64\times15^3\times10} =641.2 NShear stress, \tau =\frac{16PR}{\pi d^3}\left\lgroup1+\frac{d}{4R} \right\rgroup
=\frac{16\times641.2\times15}{\pi \times5^3} \left\lgroup1+\frac{5}{60} \right\rgroup =424.5 MPa