Working out strain and stress of an elastic band (youngs modulus)

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Discussion Overview

The discussion revolves around calculating the stress and strain of an elastic band using Young's modulus. Participants explore the differences in behavior between elastomeric materials and metals, particularly in the context of large deformations and non-linear stress-strain relationships.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant seeks assistance in understanding how to calculate stress and strain for an elastic band under applied force in relation to Young's modulus.
  • Another participant questions the phrasing "in youngs modulus" and clarifies that stress requires knowledge of the cross-sectional area and applied force, linking it to Hooke's law.
  • Discussion includes various formulas for calculating strain based on the stretched and initial lengths of the band, highlighting different measures of stress (true stress vs. engineering stress).
  • Concerns are raised about the non-linear relationship between stress and strain for elastomers compared to metals, emphasizing the need for experimental data to establish this relationship.
  • Participants provide corrections regarding the dimensions and area of the elastic band, with one asserting the area of cross-section as 8 mm² and another correcting it to 80 mm².
  • There is mention of the Mooney-Rivlin model as a mathematical description of elastomer stress-strain behavior, referencing historical studies in statistical thermodynamics.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and interpretations of stress and strain for elastomeric materials, with no consensus reached on the best approach or formula to use.

Contextual Notes

Participants note the importance of unit consistency and the impact of stretching on the cross-sectional area, which complicates the calculations of stress and strain. There is also an acknowledgment of the need for experimental data to accurately determine the stress-strain relationship.

Gallows
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This is regarding youngs elastic modulus
I need help as I am trying to figure out the stress and strain of an elastic band when a force has been applied - in youngs modulus.

Does anyone know how to go about this?
 
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What do you mean by "in youngs modulus"?
I suppose you mean "in Young's modulus" but still not clear.
To calculate the stress you need to know the dimensions of the band (the area of the cross section) and the force.
Then you can find the strain from Hooke's law (this has Young's modulus in it).
 
nasu said:
What do you mean by "in youngs modulus"?
I suppose you mean "in Young's modulus" but still not clear.
To calculate the stress you need to know the dimensions of the band (the area of the cross section) and the force.
Then you can find the strain from Hooke's law (this has Young's modulus in it).

Hi, thank you for your reply, I think the dimensions of the elastic band are (height: 7cm) (width 4cm) and thickness (2mm)

So I think the area is (5.6nm)

I am still not sure how to work out the stress, strain and force
 
Gallows said:
This is regarding youngs elastic modulus
I need help as I am trying to figure out the stress and strain of an elastic band when a force has been applied - in youngs modulus.

Does anyone know how to go about this?

An elastomeric band is a different kind of animal from a metal bar, the latter of which is characterized in the region of elastic behavior (in extension) by Young's modulus.

An elastomer behaves differently. Its region of elastic behavior is much larger than that of a metal (in terms of the magnitude of the elastic deformations that can be sustained, followed by full recovery to the initial length). In addition, for an elastomer, the stress is not a linear function of the strain. Finally, because of the larger extensions that an elasomeric band suffers, its cross sectional area decreases significantly.

The real issue is how do you calculate the strain and the stress for large deformations. For uniaxial extensions like you are interested in, there are several strain measures that can be used and two different measures of stress:

ε =( l/l0 - 1 )

or ε =(( l/l0)2 -1)/2

or ε = (1-(l0/l)2)/2

where l is the stretched length of the band and l0 is the initial unstretched length.

The stress can be calculated in terms of the force per unit cross sectional area (the so-called true stress) or in terms of the force per unit initial cross sectional area (the so-called engineering stress):

σtrue = F/A = (F/A0) (l / l0)

σengineering = F/A0=(F/A) (l0/l)

If doesn't matter which measure of strain you choose to work with or which measure of stress you choose to work with. Whatever the case, the stress will be a unique non-linear function of the strain which you can measure experimentally (once and for all for the material in question).

σ = σ (ε)

The functionalities between all the various stress and strain measures are known, so you can convert from one strain measure to another and/or from one stress measure to another.

Elastomeric materials are typically cross linked polymers, and the fundamental deformational behavior of these types of materials was studied in the 1930's and 1940's using statistical thermodynamics. Various mathematical descriptions of the stress-strain behavior of elastomers are presented in the literature, based on these fundamental studies. The most well-known of these is the Mooney-Rivlin model.
 
Gallows said:
Hi, thank you for your reply, I think the dimensions of the elastic band are (height: 7cm) (width 4cm) and thickness (2mm)

So I think the area is (5.6nm)

I am still not sure how to work out the stress, strain and force

No, the volume is 56 cm^2. The area of cross section will be 4mmx2mm = 8mm^2. This is, assuming that you stretch the band along the 7 cm side.
You can calculate the initial stress as F/A. This will be the initial or "engineering" stress. However, as Chestermiller have explained in detail, the area changes significantly when you stretch the rubber band.
 
nasu said:
No, the volume is 56 cm^2. The area of cross section will be 4mmx2mm = 8mm^2. This is, assuming that you stretch the band along the 7 cm side.
You can calculate the initial stress as F/A. This will be the initial or "engineering" stress. However, as Chestermiller have explained in detail, the area changes significantly when you stretch the rubber band.

Watch those units. Actually the area is 80 mm2 = 0.8 cm2. This is the initial cross sectional area of the sample.

Do you have data on the measured force versus the measured length? If you do, then calculate the engineering stress by the equation I gave you, and calculate the strain by the first equation I gave you. Then plot a graph of the stress as a function of the strain, and see what it looks like. Then you can decide what kind of parametric relationship would be best for fitting to the data.
 
Chestermiller said:
Watch those units. Actually the area is 80 mm2 = 0.8 cm2. This is the initial cross sectional area of the sample.
Right. I did not see that the width is in cm. Thank you for correction.
 

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