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

Answer:

0.4757 mm

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:

$\Delta L = \dfrac{PL}{AE_{Al}}$

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:

$E_{Al}= 7*10^{10} \Nm^{-2}$

$\Delta L = \dfrac{PL}{AE_{Al}}$

$\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}$

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

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