A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?

required diameter is 7.65 mm

Explanation:

Given the data in the question;

F = 6660 N

l₀ = 380 mm = 0.38 m

E = 110 GPa = 110 × 10⁹ N/m²

Δl = 0.50 mm = 0.0005 m

So, lets assume the deformation is elastic;

d₀ = √(  [4l₀F] / [πEΔl] )

we substitute

d₀ = √(  [4 × 0.38 × 6660] / [π × (110 × 10⁹) × 0.0005]] )

d₀ = √( 10123.2 / 172787595.947 )

d₀ = √( 5.85875 × 10⁻⁵ )

d₀ = 0.007654 m

d₀ = ( 0.007654 × 1000 )mm

d₀ = 7.65 mm

Therefore, required diameter is 7.65 mm

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