Question

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?

Answer 1

Answer:

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