n 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 modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN

Explanation:

It is to be noted here that the aluminium bar having some hole will not have any effect on the creation of strain in it due to stress .

length of bar L = 600 mm

= .6 m

cross sectional area = 3.14 x .02²

= .001256 m²

Compressive stress created =  180 x 10³ /  .001256

= 143312.1 x 10³ N / m²

Modulus of elasticity = stress / strain

85 x 10⁹ = 143312.1 x 10³  /  l / L

l / L = 143312.1 x 10³ / 85 X 10⁹

= 1686.02  x 10⁻⁶

l = L x 1686.02  x 10⁻⁶

= 600 x 10⁻³ x 1686.02  x 10⁻⁶

= .6 x 1686.02 x 10⁻⁶

= 1011.61 x 10⁻⁶

1.011 x 10⁻³ m

1.011 mm .

be subject to a force andchange its motion