Stress, Strain and YM

A tie bar of length 2.5m and diameter 10mm carries an axial load of 12 kN. The modulus of elasticity of the bar material is 180GPa. Determine the induced tensile stress, the tensile strain and the change in length that occurs.

Stress = F/A

Area = 10x2.5x10^-3 = 0.025mm2

Stress = 12 ÷ 0.025 = 480 GPa

YM = stress/strain

Strain=stress/YM = 480/180 = 2.7

Strain = change/original

Change in length = 2.7x2.5x10^-3 = 6.75x10^-3 mm

is this correct?

No you got the area wrong so messed the whole thing up

Area is pi r^2

And I recommend you to work in standard unit

Edit: I realised I shouldn't have posted the full solution but I can't delete it on mobile app
Posted from TSR Mobile

Man, I knew I made a stupid mistake somewhere. Good to know everything outside of that looks good though. The reason I didn't use standard unit is because it would make things confusing, and the question is set out so that it doesn't need converting like YM being in GPa and kN/mm2 gives stress in GPa. I think it would be fine though. Thanks for your help.