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 .

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

they should be similar or the same