Consider a rod of steel having radius of 8 mm and the length of 2 m. If a force of 150 kN stretches it along its lengths , then calculated the stress, percentage strain in the rod if the elongation in length is 7.46 mm.

Explanation

Radius of rod,` r=8 mm=8xx10^(-3)`m,length, L=2 m
Applied force, F=150 kN=`15xx10^(4)N`
Cross-section area of wire,
`A=pir^(2)=pixx(8xx10^(-3))^(2)=201xx10^(-6)m^(2)`
`DeltaL=7.46 mm=7.46xx10^(-3)m`, Percentage strain = ?
Stress in rod `=(F)/(A)=(15xx10^(4))/(201xx10^(-6))`
`=0.0746 xx 10^(10)Nm^(-2)`
`=7.46xx10^(8)Nm^(-3)`
Longitude strain `=(DeltaL)/(L)=(7.46xx10^(-3))/(2)=3.73xx10^(-3)`
Percentage strain `=3.73xx10^(-3)xx100=0.37 %`