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.

Question

Accepted Answer

Work done by impact load = [latex]50(150+\delta )[/latex]

where,[latex]\delta[/latex] = deflection due to impact load 50 N.

If [latex]P_e[/latex] is the equivalent static load which will produce δ, then energy stored in the spring

[latex]=\frac{P_e\delta }{2} [/latex]

Equating,

Work done = Energy stored

[latex]50(150+\delta )=\frac{1}{2}P_e\delta [/latex]

Substitute [latex]P_e[/latex] from Eq. (16.2)

[latex]\delta =\frac{64PR^3n}{Gd^4} [/latex] (16.2)

or [latex]50(150+\delta )=\frac{1}{2}\frac{\delta Gd^4}{64R^3n}\delta [/latex]

[latex]=\frac{1}{2} imes\frac{80000 imes5^4 imes\delta ^2}{64 imes15^3 imes10} [/latex]

or, [latex]23.148\delta ^2-100\delta -15000=0[/latex]

or, Deflection, [latex]\delta[/latex] = 27.7 mm

Equivalent static load

[latex]P_e=\frac{\delta Gd^4}{64R^3n} =\frac{27.7 imes80000 imes5^4\delta ^2}{64 imes15^3 imes10} =641.2 N[/latex]

Shear stress, [latex] au =\frac{16PR}{\pi d^3}\left\lgroup1+\frac{d}{4R} ight group [/latex]

[latex]=\frac{16 imes641.2 imes15}{\pi imes5^3} \left\lgroup1+\frac{5}{60} ight group =424.5 MPa[/latex]

Question 16.8: A 50 N weight is dropped on a close coiled helical spring fr...

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