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Tensioning Film with Correct Hanged Masses
Tensioning Film with Correct Hanged Masses
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Old 8th September 2008, 12:26 AM   #1
bshaw147 is offline bshaw147  United States
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Join Date: Aug 2008
Default Tensioning Film with Correct Hanged Masses

Hi All,

If we consider the free body diagram, tension = weight hanged. Now I need to relate the strain to the stress. For metals there is a linear relationship between stress and strain, Can someone better educate me on the applicability of the below work?


Young's Modulus = Stress/ Strain

Where Strain = (l-lo)/lo

l = final length element
lo = unstreched length element

and Stress = Force/(Change in Area)
= Tension/Change in area
=Mass hanged *9.8715(m/s^2)/ (Change in Area)

The Young modulus for a polyester film is on the order of

4GPa=4*10^7 Pa

I looked around in this forum and it is recommended that we create a 1 - 2 % elongation in the film. this will coorespong to a strain :

for a 2 % elongation will correspond to a Strain = 1/50

Change in area = 12x40.8" - 12"x40"=489.6-480=9.6 in^2 =6.19*10^-3 m^2

(Or do I only consider a smaller area element of the film?)

Stress = (Mass Hanged *9.8)/ (6.19*10^-3)

Young's Modulus=4*10^7= (50*Mass Hanged*9.8)*(6.19*10^3)

Solving for the hanged mass
Mass Hanged =13.1878 kg
=29.07 lbs

If we have 13 masses hanging along the width, then each mass hanged could be about a kg?

Happy Building,
Bryan
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Old 8th September 2008, 09:41 PM   #2
arend-jan is offline arend-jan  Netherlands
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Join Date: May 2006
Location: Netherlands
I think you're mixing up the dimensions. And 1GPa = 4e9, not 4e7.

Let's say you want to stretch a piece of film for a speaker of 1 meter high. You are going to stretch the film in the transverse direction (so the weights are hanging along the long side of the film which is 1 meter). The width of the film is of no concern.

Let's say the film is 6 microns thick (6e-6).
We take the Young's modulus for your Mylar to be about 4e9 (N/m^2).

F = force (N)
L = length of film = 1 (m)
t = thickness of film = 6 micron = 6e-6 (m)
e = change in width (dW/W) (W = width of film) = 2% = 0.02
E = 4e9 (N/m^2)
A = cross section of film that we apply the force to = L*t

Hooks Law:

F = E * A * e
= 4e9 * 1 * 6e-6 * 0.02
= 480 (N)

So that equals about 48kg spread over 1 meter of film on each side of the film (or one side of the film can be fixed). My advise would be to start with a little less tension.

Arend-Jan
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Old 9th September 2008, 12:34 AM   #3
bshaw147 is offline bshaw147  United States
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Join Date: Aug 2008
Hi AJ,

I am very happy that you replied to my posting. I saw your stretcher and I have to say that your work is extremely tidy and professional. That was a typo GPa is 10^9 agreed. Thank you for clarifying the area the force is applied to, I agree it is the length times thickness of film. Still 48kg is a lot to hang, i suppose if we hang several masses say 10 along the width of the panel so that they pull it along the symmetry axis (along the length) than it is more manageable. Thank you for your reply.
Bryan
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