# Force Required to Stretch a Rubber Band to a Certain Diameter

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:

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:

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

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rcgldr
Homework Helper
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).

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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.

Chestermiller
Mentor
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?

rcgldr
Homework Helper
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.

Chestermiller
Mentor
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."

rcgldr
Homework Helper
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.

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