Answer to Question #123764 in Mechanical Engineering for Samuel

Question

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.

Answer 1

Given:

For aluminum bar

compressive Load (P)=180KN

modulus of elasticity (E_AL)=85GN/m²

Total Length(L) = 600 mm

Diameter aluminium bar (d_o) = 40 mm

Hole diameter (d_i)= 30 mm

Rest bar lenght (L₁)=(600-100)=500 mm

Hole length (L₂)= 100 mm

Find the total contraction on the bar due to a compressive load .

now, free body Diagram

Now,

$Area\,of\,crosssection\,without\,hole\,(A_{1})=\frac{\pi }{4}\times d_{i}^{2}=\frac{\pi }{4}\times 40^{2}=1256.63\,mm^{^{2}}$

$Area\,of\,crosssection\,with\,hole\,(A_{2})=\frac{\pi }{4}\times (d_{o}^{2}-d_{i}^{2})\\$

$\therefore$ $A_{2}=\frac{\pi }{4}\times (40^{2}-30^{2})=549.77\,mm^{^{2}}$

Now,

$Total\,contraction\,=\frac{P\times L_{1}}{A_{1}\times E}+\frac{P\times L_{2}}{A_{2}\times E}=\frac{p}{E}\times (\frac{L_{1}}{A_{1}}+\frac{L_{2}}{A_{2}})$

So,

$Total\,contraction\,=\frac{180\times 10^{3}N}{85\times 10^{3}N/m^{2}}\times \left ( \frac{500}{1256.63} +\frac{100}{549.77}\right )=1.228mm$