Question 16.2: A close coiled helical spring is made by 5 mm diameter steel...

A close coiled helical spring is made by 5 mm diameter steel wire having shear strength 500 MPa. The spring has mean diameter 50 mm. A prestress of 60 MPa causes the coils to be closed under no-load condition. Calculate the no load force in the spring and the maximum load that can be applied to the spring. Also calculate the maximum extension of spring having 30 number of coils of the modulus of rigidity is 80 GPa.

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Using Eq. (16.1)

\tau =\frac{16PR}{\pi d^3} \left\lgroup1+\frac{d}{4R} \right\rgroup

For no load condition force P is given by,

60=\frac{16P\times25}{\pi\times5^3} \left\lgroup1+\frac{5}{4\times25} \right\rgroup

or,              P=\frac{60\times\pi\times125}{16\times25\times1.05} =56.1  N

The maximum load is obtained when \tau= 500 MPa as follows:

P=\frac{500\pi\times125}{16\times25\times1.05} =467.5  N

Maximum deflection,

\delta =\frac{64nPR^3}{Gd^4}

 

=\frac{64\times30\times467.5\times25^3}{80000\times5^4} =109.57  mm

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