Question

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.

Answer 1

0.4757 mm

Step-by-step explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

`\Delta L = (PL)/(AE_(Al))`

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium
`E_(Al)` is:

`E_(Al)= 7*10^(10) \Nm^(-2)`

`\Delta L = (PL)/(AE_(Al))`

`\Delta L = (223000* 1.22)/(0.00817* 7*10^(10))`

ΔL = 4.757 × 10⁻⁴ m

ΔL = 0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm