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

Answer:

$\delta = 0.385\,m$ (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

$\delta = \frac{P\cdot L}{A \cdot E}$

Where:

$P$ - Load experimented by the bar, measured in newtons.

$L$ - Length of the bar, measured in meters.

$A$ - Cross section area of the bar, measured in square meters.

$E$ - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ($D_{o} = 0.04\,m$, $D_{i} = 0.03\,m$)

$A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})$

Where:

$D_{o}$ - Outer diameter, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}]$

$A = 5.498 \times 10^{-4}\,m^{2}$

The total contraction of the bar due to compresive load is: ($P = -180\times 10^{3}\,N$, $L = 0.1\,m$, $E = 85\times 10^{9}\,Pa$, $A = 5.498 \times 10^{-4}\,m^{2}$) (Note: The negative sign in the load input means the existence of compressive load)

$\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}$

$\delta = -3.852\times 10^{-4}\,m$

$\delta = -0.385\,mm$

$\delta = 0.385\,m$ (Compression)