Question

A solid cylinder ical steel coloumn is 4m long nd 9.0cm in diameter.wat will be its decrease its length wen carryin a lod of 80000 kg?Y=1.9*10^11pa.

Answer 1

we first find
  cross sectional area of the coloumn =πr² =(0.045m)²=6.36*10⁻³m²
then  from Y=(F/A)(ΔL/L)we have
              ΔL=FL/AY=8.00*10⁴*9.81N*4.0m/6.36*13⁻³m²*1.9*10¹¹Pa=2.6*10⁻³=2.6mm

Answer 2

Length of cylinder (L) = 4m
diameter = 9 cm = 0.09 m
radius = 0.09/2 = 0.045 m
cross sectional area(A) = πr² = π(0.045)² = 0.0063585 m²
weight (F)= 80000 kgf = 80000×9.8 = 7.84×10⁵ N
Young's modulus(Y) = 1.9×10¹¹ Pa
Let change in length = ΔL

According to hooke's law,

$Y= \frac{stress}{strain} = \frac{F/A}{ \Delta L/L } = \frac{F.L}{A. \Delta L }$

⇒$\Delta L= \frac{F.L}{A.Y}$

⇒$\Delta L = \frac{7.84*10^{5}*4}{0.0063585*1.9*10^{11}}$

⇒$\Delta L=0.002596m=2.596mm$