How to determine the value of young's modulus of elasticity if a steel bar, 40mm*10mm section and 2 m length is subjected to an axial compressive load of 40 kN. The contraction of the length is found to be 1 mm and the increase in 40 mm width is 0.006m?

Question

imtant · Answer

Given the formula is E= PL/ A delta.E for young's modulus of elasticity. P for 40 kN.

imtant · Answer

@Jack1

imtant · Answer

Okay, but may I know why delta L is in negative?

imtant · Answer

And for the A (cross section area) I got 0.0004 m^2.

imtant · Answer

So for the answer I got 200 Gpa. Is it same with you?

jack1 · Answer

so F = compressive load of 40,000N
length has reduces due to compression by 1 mm
width has increased by 6mm

using your equation
E= PL/ A delta.E
P = 40000 N
L = 2m ?
delta e = extension? 
A = ... initial or new now we have the different width... ?

jack1 · Answer

so ur equation was 40000 x 2 / 0.04 x 0.01 x ...0.001?

jack1 · Answer

what did u use as delta e?

imtant · Answer

Yes , it is. For delta or E?

jack1 · Answer

E= PL/ A delta.
so A delta = delta A ?
 = change in area?

imtant · Answer

$$E= \frac{ PL }{ A \delta } $$

imtant · Answer

I'm just using 0.0004m.

jack1 · Answer

0.04 x 0.01 = 0.0004 m^2 initial
new area = 0.039 x 0.046 = 0.001794 m^2

jack1 · Answer

so delta area = 0.001394m^2 ... yeah?

jack1 · Answer

sorry dude, i think i'm just muddying the waters, it's been too long since i learned physics i think

jack1 · Answer

@dan and @ganeshie8  
do you guys have a moment please?

imtant · Answer

It's okay Jack.

imtant · Answer

I think I will just using 0.001 for delta. And area are 0.0004 from (0.040*0.010).

imtant · Answer

Anyways Thanks Jack. :)

imtant · Answer

$$\delta = elongation.$$