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Force Required to Stretch a Rubber Band to a Certain Diameter

  • Thread starter rdijulio
  • Start date
  • #1
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Summary: Looking of someone to double check my work on this design issue.

I am posting this here because I am no longer a student, but it is similar to a simple homework problem

Problem: I am designing a rubber-band like product for a client. He wants to take a durable rubber band and stretch it over a circle that is 9.875" diameter and 1.75" wide. He wants it to be easily installed and removed, enough for a child or elderly person of 70 years. So I estimated 10 lbs of force would be ideal. The question I am trying to solve is: (At what un-stretched diameter is 10 lbs of force required to stretch this band to 9.875")

Relevant Equations: F =( [Ln-Lo]/Lo) * E * A

A=Cross Sectional Area (The band is not a rectangle)
E=Elastic Modulus
Ln=Length New (9.875"*π - 31.023" or .78798m)
Lo=Variable
F=Force (10lbs or 44.5N)

This is just Hooke's law, and I know that elastic bands don't behave under hooke's law at a certain point, but this was the best approximation I could find. Solving for Lo gives you:

Lo=Ln/(1 + (F / (E*A) ) )

The material I chose to calculate is Silicone 60A Durometer. I pulled the elastic modulus from this graph:
244379


Which gives me an E=3.6e6 pa.

The next step was to calculate the cross sectional area: Here is the drawing with the equations and variables:
244380


Plugging in gives 1.2834in2 or 8.28e-4m2

Now we have all our values:
Lo = .78798 / (1 + 44.5 / (3.6e6*8.28e-4) ) = .77638

D = .77638/pi = .2471m or 9.72inches


Just looking for someone to check my work. Provide possibly any more feedback, or an alternative way to calculate what I need. Thanks!
 

Answers and Replies

  • #3
rcgldr
Homework Helper
8,691
522
For latex rubber (used with fishing line to launch radio control gliders), an archived plot of the stress/strain curve looks like:

https://web.archive.org/web/20070808211048/http://www.hollyday.com/rich/hd/sailplanes/rubberdata.htm
This matches the data i got from my own measurements:

Code:
   strain versus tension / initial unit cross sectional area: (strain == pull distance)

     0% =   0 lb / in^2
    50% =  70 lb / in^2
   100% =  95 lb / in^2
   150% = 115 lb / in^2
   200% = 135 lb / in^2
   250% = 160 lb / in^2
   300% = 175 lb / in^2
   350% = 195 lb / in^2
   400% = 205 lb / in^2  (not recommended).
 
Last edited:
  • #5
3
0
For latex rubber (used with fishing line to launch radio control gliders), an archived plot of the stress/strain curve looks like:

https://web.archive.org/web/20070808211048/http://www.hollyday.com/rich/hd/sailplanes/rubberdata.htm
This matches the data i got from my own measurements:

Code:
   strain versus tension: (strain == pull distance)

     0% =   0 lb / in^2
    50% =  70 lb / in^2
   100% =  95 lb / in^2
   150% = 115 lb / in^2
   200% = 135 lb / in^2
   250% = 160 lb / in^2
   300% = 175 lb / in^2
   350% = 195 lb / in^2
   400% = 205 lb / in^2  (not recommended).
Good to know. Latex is an option we had been considering so this is helpful.
 
  • #6
20,231
4,264
For latex rubber (used with fishing line to launch radio control gliders), an archived plot of the stress/strain curve looks like:

https://web.archive.org/web/20070808211048/http://www.hollyday.com/rich/hd/sailplanes/rubberdata.htm
This matches the data i got from my own measurements:

Code:
   strain versus tension: (strain == pull distance)

     0% =   0 lb / in^2
    50% =  70 lb / in^2
   100% =  95 lb / in^2
   150% = 115 lb / in^2
   200% = 135 lb / in^2
   250% = 160 lb / in^2
   300% = 175 lb / in^2
   350% = 195 lb / in^2
   400% = 205 lb / in^2  (not recommended).
I assume those stresses are per initial cross sectional area (i.e., engineering stress), correct?
 
  • #7
rcgldr
Homework Helper
8,691
522
I assume those stresses are per initial cross sectional area (i.e., engineering stress), correct?
It's latex tubing, and the strain is the increase in length. Say the tubing is 1 meter long, then 50% strain is stretching the tubing so that total length becomes 1.5 meters long, 100% 2 meters long, ..., 300% 4 meters long. For a rubber band, the length would be the circumference.

For 10 lbs of tension at 300% strain, the cross sectional area would be (10/175) ~= 0.057 in^2.
 
  • #8
20,231
4,264
It's latex tubing, and the strain is the increase in length. Say the tubing is 1 meter long, then 50% strain is stretching the tubing so that total length becomes 1.5 meters long, 100% 2 meters long, ..., 300% 4 meters long. For a rubber band, the length would be the circumference.

For 10 lbs of tension at 300% strain, the cross sectional area would be (10/175) ~= 0.057 in^2.
So those stresses would be the "true stresses", not the "engineering stresses."
 
  • #9
rcgldr
Homework Helper
8,691
522
So those stresses would be the "true stresses", not the "engineering stresses."
I assume those stresses are per initial cross sectional area (i.e., engineering stress), correct?
Correct, the cross sectional area is the initial (zero tension) cross sectional area.

I missed the earlier post. I updated my original post to clarify this.

No attempt was made to measure the hysteresis. Each measurement was done after a stretch. A table of tensions versus return path from various stretches was not made.
 
Last edited:

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