# Young modulus (deriving formula)

1. Mar 22, 2013

### coconut62

Quoted from my book:
"The graph of ΔL against F has gradient L/EA, so the Young modulus E is equal to L(A x gradient)."

E is (stress/strain).

I dont understand that sentence. If I substitute the gradient into E= L(A x gradient), then

E= L(A)(L/EA)

E= L(L/E)

E= L^2 /E

E square = L square?

This doesnt seem to make sense. Can someone explain to me please?

2. Mar 22, 2013

### TSny

The last expression here is incorrect. If the gradient equals L/(EA), then what would be the correct expression for E in terms of L, A, and the gradient?

3. Mar 22, 2013

### Staff: Mentor

The stress σ is equal to the force per unit area F/A.

σ=F/A

The strain ε is equal to ΔL/L:

ε =ΔL/L

The stress is equal to the strain times Young's modulus:

σ = E ε

If we eliminate the stress σ and the strain ε from these equations, we get

$$\Delta L=(\frac{L}{AE})F$$

If you plot a graph of ΔL versus F, you will get a straight line through the origin with a slope (gradient) of L/AE. So E will be equal to L/A divided by the slope.

4. Mar 22, 2013

### coconut62

So does that mean the textbook is wrong?

5. Mar 22, 2013

### TSny

Yes. That's right. E = L/(A*gradient). Both A and the gradient are in the denominator.

6. Mar 22, 2013

### Staff: Mentor

I have trouble understanding your rendition of the algebra. The L is in the numerator, the A is in the denominator, and the gradient is in the denominator. E = L/(A x gradient). Is this what your textbook says?

7. Mar 23, 2013

### coconut62

No, my textbook doesn't have the "/" , just L (Ax gradient).

So the actual equation is E= L/(A*gradient), correct?

8. Mar 23, 2013

### Staff: Mentor

Do you honestly feel that you need to ask me this question? What does your knowledge of algebra tell you? Another way of figuring out which result is correct is to check the units. The units of E have to be force per unit area.