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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.

Respuesta :

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

[tex]Radius_{mean} = 60\ mm[/tex]

[tex]Diameter_{mean} = 120\ mm[/tex]

Load = 200 N

We need to calculate the deflection

Using formula of deflection

[tex]\delta=\dfrac{8pD^3n}{Cd^4}[/tex]

Put the value into the formula

[tex]\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}[/tex]

[tex]\delta =34.56\ mm[/tex]

Hence, The deflection of the spring is 34.56 mm.

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