Calculating young's modulus from data provided

  • Thread starter flyingbird
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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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  • #2
44 views so far and no reply! :(
 
  • #3
5,419
7
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?
 
  • #4
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!
 
  • #5
5,419
7
(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.

[tex]YoungsModulus = stress\,x\,\frac{1}{{strain}} = stress\,x\,\frac{{{L_0}}}{{\Delta L}}[/tex]

So at 100% extension

[tex]stress\,x\,\frac{{{L_0}}}{{2{L_0} - {L_0}}} = \frac{{stress}}{1} = 1500psi[/tex]

and at 300%extension

[tex]stress\,x\,\frac{{{L_0}}}{{4{L_0} - {L_0}}} = \frac{{stress}}{3} = \frac{{2950}}{3} = 283psi[/tex]

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:
  • #6
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.

[tex]YoungsModulus = stress\,x\,\frac{1}{{strain}} = stress\,x\,\frac{{{L_0}}}{{\Delta L}}[/tex]

So at 100% extension

[tex]stress\,x\,\frac{{{L_0}}}{{2{L_0} - {L_0}}} = \frac{{stress}}{1} = 1500psi[/tex]

and at 300%extension

[tex]stress\,x\,\frac{{{L_0}}}{{4{L_0} - {L_0}}} = \frac{{stress}}{3} = \frac{{2950}}{3} = 283psi[/tex]

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. :smile:
 

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