Question

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subjected to an axial load of 200 N. Determine the deflection of the spring. C = 80 kN/mm2.

Answer 1

The deflection of the spring is 34.56 mm.

Step-by-step explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

`Radius_(mean) = 60\ mm`

`Diameter_(mean) = 120\ mm`

Load = 200 N

We need to calculate the deflection

Using formula of deflection

`\delta=(8pD^3n)/(Cd^4)`

Put the value into the formula

`\delta=(8*200*(120)^3*10)/(80*10^(3)*10^4)`

`\delta =34.56\ mm`

Hence, The deflection of the spring is 34.56 mm.