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

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

Step-by-step 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 = (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 = (\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 = (\pi)/(4)\cdot [(0.04\,m)^(2)-(0.03\,m)^(2)]`

`A = 5.498 * 10^(-4)\,m^(2)`

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

`\delta = ((-180* 10^(3)\,N)\cdot (0.1\,m))/((5.498* 10^(-4)\,m^(2))\cdot (85* 10^(9)\,Pa))`

`\delta = -3.852* 10^(-4)\,m`

`\delta = -0.385\,mm`

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