A 1∕2-in.-diameter round steel rod 40 ft long supports a load of 4 kips. How much will it elongate?

Question

𝜀 = ΔL /L

f = P/A

E = (P∕A)/ (ΔL∕L) = (PL) / (A(ΔL))

Answer 1

E is modulus of elasticity

E of steel is 29000 ksi

Assuming all deformation is elastic, ΔL (elongation) = PL/AE

= ((4 kip)*(40 ft))/(pi*(1/4 in)^2)*(29000 ksi)

=0.028 ft or 0.337 in