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
