Question

A straight bar of 500 mm length has its cross-sectional area of 500 mm2. Find the magnitude of the compressive load under which it will decrease its length by 0.2 mm. Take E for the bar material as 200 GPa.​

Answer 1

Answer:

40 Kn because 200* 0.2 = 40

Answer 2

Given: A straight bar of 500 mm length has its cross-sectional area of 500 mm^2. Decrease in length is 0.2 mm. E for the bar material is 200 GPa.

To find: Magnitude of compressive load

Explanation: E is called the Young's modulus of elasticity.

Value of E of this bar = 200 GPa

= 200 * 10^9 Pa

( 1 GPa= 10^9 Pa)

Area of the bar = 500 mm^2

= 500 * 10^-6 m^2

= 0.0005

( 1 mm^2 = 10^-6 m^2)

Length of the bar= 500 mm

= 0.5 m

( 1 mm = 0.001 m)

Decrease in length= 0.2 mm

= 0.0002 m

Strain in the bar

= Decrease in length/ Original length

= 0.0002/0.5

= 0.0004

Stress = Load / Area

= Load/ 0.0005

The formula that is used when load changes length is:

E = Stress/ Strain

$200 \times {10}^{9} = \frac{load}{0.0004 \times 0.0005}$

$load = 200 \times {10}^{9} \times 0.0004 \times 0.0005$

load = 40000 N

Therefore, the magnitude of compressive load is 40000 N.