# Calculating young's modulus from data provided

## Main Question or Discussion Point

I am using one film mateial (Polyester Polyurethane) for some purpose. I have got data sheet from the manufacturer with following details:
Durometer 90Shore A D 2240
Specific Gravity 1.19 D792
Elongation @ Break Elast 515% D 412
Tensile Str. @ Break Elast 5,400 psi D 412
100% modulus 1500 psi D 412
300% modulus 2950 psi D 412

Here D.... is the ASTM test method. I have to determine Young modulus for this material, I don't have details of ASTM test method. I have uploaded PDF file of material spec too.
Can anyone help me here?

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## Answers and Replies

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What is your definition of Young's Modulus?

The information is right there in your list : What do you think the 100% & 300% Moduli are?

What is your definition of Young's Modulus?

The information is right there in your list : What do you think the 100% & 300% Moduli are?
Young modulus is slop of the stress vs strain curve in elastic region. The values which are given here as 100% & 500% moduli cannot be young modulus, because tensile strength is 5400psi which is generally much less than young modulus. Young modulus for this material should be around 150000 psi!

(Polyester Polyurethane)
You are looking at a plastic material.
This means that in a test to destruction it has a very small elastic region and a large plastic region.

$$YoungsModulus = stress\,x\,\frac{1}{{strain}} = stress\,x\,\frac{{{L_0}}}{{\Delta L}}$$

So at 100% extension

$$stress\,x\,\frac{{{L_0}}}{{2{L_0} - {L_0}}} = \frac{{stress}}{1} = 1500psi$$

and at 300%extension

$$stress\,x\,\frac{{{L_0}}}{{4{L_0} - {L_0}}} = \frac{{stress}}{3} = \frac{{2950}}{3} = 283psi$$

The fact that these two figures are different tell us that by the time you have drawn the material out to 300% extension you are well into the plastic region and require little extra force to achieve this. If you pulled the film out the 300% extension and let go it would remain well stretched.

It is true that you cannot be absolututely certain that the material is elastic up to 100% extension, from the figures given, but they are standard marker points on the curve.

Last edited:
You are looking at a plastic material.
This means that in a test to destruction it has a very small elastic region and a large plastic region.

$$YoungsModulus = stress\,x\,\frac{1}{{strain}} = stress\,x\,\frac{{{L_0}}}{{\Delta L}}$$

So at 100% extension

$$stress\,x\,\frac{{{L_0}}}{{2{L_0} - {L_0}}} = \frac{{stress}}{1} = 1500psi$$

and at 300%extension

$$stress\,x\,\frac{{{L_0}}}{{4{L_0} - {L_0}}} = \frac{{stress}}{3} = \frac{{2950}}{3} = 283psi$$

The fact that these two figures are different tell us that by the time you have drawn the material out to 300% extension you are well into the plastic region and require little extra force to achieve this. If you pulled the film out the 300% extension and let go it would remain well stretched.

It is true that you cannot be absolututely certain that the material is elastic up to 100% extension, from the figures given, but they are standard marker points on the curve.
Oops.
I got the point, thanks a lot.