# Stress-strain (area under curve)

• sassora
In summary, the conversation discusses the area under the stress-strain curve and its significance in determining work done and volume in a wire. The correct formula for calculating the area is given as Fe/AL, where F is the force, e is the distance the force moves, A is the cross-sectional area, and L is the unstressed length. The conversation also explores the concept of average force and how it relates to the calculation of work done. Ultimately, the area under the stress-strain curve can be thought of as work done per unit volume.
sassora

## Homework Statement

Deducing what the area under the stress-strain curve shows.

There are four option in the attached image. I can discount work done by considering the units. The remaining ones seem plausible, but only one is true.

## Homework Equations

stress = force / area; strain = proportional extension in length

## The Attempt at a Solution

stress units are N / m2
strain units are unitless

The units of the area could be N / m2 but Nm / m3 units are also consistent. Work done / Volume variants seem plausible.

If the line was y = mx then then area underneath would be (1/2)mx2. Here we have stress = young's modulus . strain.
I would have thought the area under the line is (1/2) young's modulus . strain2. Not sure that helps.

#### Attachments

• Stress-Strain area.png
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you are correct, look again at the units
Stress = F/A
strain = e/L
stress/strain = (F/A)/(e/L) = FL/eA FL is energy (work done) eA is a 'volume'...so your work done per unit volume makes sense !
sometimes cncelling units can 'hide' relevant physics information

Aren't we looking for the area rather than the gradient? Not sure why I would go straight to dividing stress by strain?

sassora said:
Aren't we looking for the area rather than the gradient? Not sure why I would go straight to dividing stress by strain?
Quite so. And FL would not be work done anyway.
Since we're looking at an area under a curve, we should consider the integral ##\int \sigma.d\epsilon ##. Assuming A remains constant, and L being the unstressed length always, how does that turn out in terms of F, A, e and L?

sassora said:
Aren't we looking for the area rather than the gradient? Not sure why I would go straight to dividing stress by strain?
sorry! you are quite right, area = stress x strain = (F/A) x (e/L) = Fe/AL...= work done in stretching/ volume of wire ...you were correct with your 'work done per unit volume'

haruspex said:
Quite so. And FL would not be work done anyway.
Since we're looking at an area under a curve, we should consider the integral ##\int \sigma.d\epsilon ##. Assuming A remains constant, and L being the unstressed length always, how does that turn out in terms of F, A, e and L?

it becomes Fe/AL

∫ σ dε = ∫ F / A dε
= F ε / A
= F (e / L) / A
= Fe / LA​

Fe is the work done and LA extends the cross-sectional area to the volume. So the area under the line is work done / volume.
If the cross-sectional A wasn't constant then I guess we'd be doing a second integral, which sounds quite complicated.

sassora said:
∫ σ dε = ∫ F / A dε
= F ε / A
Not quite. It is an integral, with σ varying: ∫ σ dε=##\frac {\int F.de}{AL}##, and ∫F.de is work done.
Note that this is valid whatever the shape of the curve. In the diagrammed example, F is proportional to e, so the work done is ½Fmaxemax.

σ varies with ε and I treated it as a constant? So you've rewritten the integral using ε = e/L to give ##\frac {\int F.de}{AL}##. I guess that shows that the area is work done over volume but how did you know to make that transformation? I would have thought that a F depends on ε but by extension on e also.

sassora said:
F depends on ε but by extension on e also.
Yes. Where have I indicated otherwise?

You haven't explicitly, I was trying to work out the reason for the change in variable. I understand the reason why my attempt wasn't quite right was by treating F as a constant although I'm not completely clear on that.

sassora said:
You haven't explicitly, I was trying to work out the reason for the change in variable. I understand the reason why my attempt wasn't quite right was by treating F as a constant although I'm not completely clear on that.
I changed the variable because the work done is related to the distance the force moves. That is given by e directly, and only indirectly by ε.
For both the area under a curve and the work done by a force, the integral is the correct form of the equation. (For work done by force, the completely correct form is the integral of a vector dot product, ##\int \vec F.\vec {dx}##.). You can only simplify to a simple product when the integrand is constant. In the case of a curve, that would be a horizontal line.

That makes sense - just because ε and e are correlated won't mean that the both give the work done, but something related to the work done.

Duplicate post

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I see from the diagram that the area is ½σmaxεmax.

If we assume A,L are constant then we have (1/2AL)(Fmaxemax)

For the work done are we then considering the average force applied is Fmax/2 and so the work done is (Fmax/2).emax?

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sassora said:
I see from the diagram that the area is ½σmaxεmax.

If we assume A,L are constant then we have (1/2AL)(Fmaxemax)

For the work done are we then considering the average force applied is Fmax/2 and so the work done is (Fmax/2).emax?
I have a purist objection to the use of the expression "average force" here, because that is defined to be an average over time, and we are not interested in how long the application of the stress took.
The linear increase of stress with strain shown is exactly like a spring. I assume you are familiar with the formula for stored energy in a spring, 1/2 kx2. The max force is kx, so it's the same result.

Yes I agree about the average force. I'd like to express this as:

Based on the triangle formed on the F-e graph: Work done = ½Fmax.emax
The area on the σ-ε graph will be the same but rescaled. σ will be 1/A of F and ε will be 1/L of e, the area under the σ-ε graph will be the area under the F-e graph divided by AL.

The area under than σ-ε graph = Work done / AL = Work done divided by volume

Thanks for your help. I found thinking about the various issues useful even when they weren't directly relevant to the original question.

sassora said:
Yes I agree about the average force. I'd like to express this as:

Based on the triangle formed on the F-e graph: Work done = ½Fmax.emax
The area on the σ-ε graph will be the same but rescaled. σ will be 1/A of F and ε will be 1/L of e, the area under the σ-ε graph will be the area under the F-e graph divided by AL.

The area under than σ-ε graph = Work done / AL = Work done divided by volume

Thanks for your help. I found thinking about the various issues useful even when they weren't directly relevant to the original question.
You are welcome.

## What is stress-strain curve?

The stress-strain curve is a graphical representation of the relationship between stress and strain in a material. It shows how a material behaves under different levels of stress, and can provide information about its strength, stiffness, and ductility.

## What is the significance of the area under the stress-strain curve?

The area under the stress-strain curve represents the energy required to deform a material. This is known as the material's toughness and can indicate its ability to withstand sudden impacts or loads.

## How is the area under the stress-strain curve calculated?

The area under the stress-strain curve is calculated by dividing the curve into smaller sections, calculating the area of each section, and then summing them together. This can be done using numerical integration techniques or by using a software program.

## What does the slope of the stress-strain curve represent?

The slope of the stress-strain curve represents the material's stiffness or modulus of elasticity. This is an important property as it indicates how much a material will deform under a given amount of stress.

## How does the stress-strain curve differ for different materials?

The stress-strain curve can vary greatly for different materials, depending on their composition and structure. Some materials, such as metals, have a gradual increase in stress and strain until they reach a breaking point, while others, such as polymers, may have multiple peaks and valleys in their curve due to their complex molecular structure.

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