Question #123764

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.

Expert's answer

Given:

For aluminum bar

compressive Load (P)=180KN

modulus of elasticity (E_{AL})=85GN/m^{2}

Total Length(L) = 600 mm

Diameter aluminium bar (d_{o}) = 40 mm

Hole diameter (d_{i})= 30 mm

Rest bar lenght (L_{1})=(600-100)=500 mm

Hole length (L_{2})= 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"

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