Question

A close-coiled helical spring has a stiffness of 10 N/mm. Its length when fully compressed,
with adjacent coils touching each other is 40 cm. The modulus of rigidity of the material
of the spring is 0.8x10 N/mm²
(i) Determine the wire diameter and mean coil diameter if their ratio is 1/10.
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Answer 1

The value of diameter is 20 cm

Explanation:

• We are given that:
• Stiffness of coil = 10 N/mm
• Length of coil = 40 cm
• Modulus of rigidity of the material  of the spring =  0.8x10 N/mm²
• To Find: Diameter of wire = ?

Solution:

Now

d / D = 1 /10

Length of solid = n x d

400 = n x d

n = 400 / d

Now d^2 = 0.4

d^2 = 0.4 x 10^3 = 400

d = √400 = 20 mm = 2 cm

D = 10 x 2 = 20 cm

Thus the value of diameter is 20 cm