Calculating young's modulus from data provided

[flyingbird]

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?

 

[flyingbird]

44 views so far and no reply! :(

 

[Studiot]

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?

 

[flyingbird]

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!

 

[Studiot]

(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.

 

[flyingbird]

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.