Stress, strain, force

[Function]

Hi everyone

 

From my physics course in medicine:

 

[math]U=\int_{0}^{\Delta l}{Fdl}=\int_0^{\epsilon}{\sigma Old\varepsilon}=\cdots[/math]

 

With [math]F[/math] force applied on a beam, [math]l[/math] the length of the beam, [math]\Delta l[/math] the difference in length after appliance of the pulling force [math]F[/math] on the beam, [math]O[/math] the surface of the beam, [math]\sigma[/math] the stress [math]\frac{F}{O}[/math] on the beam and [math]\varepsilon[/math] the strain on the beam ([math]\frac{\Delta l}{l}[/math])

 

Can't seem to figure out why this is... Can someone help me?

 

Thanks.

 

F.

Edited September 22, 2014 by Function

[studiot]

O is the cross sectional area, not the surface area.

[Function]

O is the cross sectional area, not the surface area.

 

Well, yes.. Okay... Forgive me my linguistic ignorance.

[studiot]

This is a complicated way of saying that when an arm or wire etc is stretched in tension the work done = the strain energy stored = 1/2 stress x strain.

 

This arises because the force required to "pull out" the object rises from zero at the outset to the full value at the end of deltaL so the average value acting over the whole distance is 0.5(0+F) = F/2.

 

That is the value of the integral.

Edited September 22, 2014 by studiot

[Function]

Sorry, but I still can't seem to understand why the second integral is equal to the first one...

[elfmotat]

I'm going to change your variable O to an A, because I don't like it (it looks too much like zero). So we have:

 

[math]U=\int_{0}^{\Delta l} F dl[/math]

[math]\sigma =F/A[/math]

[math]\epsilon =\Delta l/l[/math]

 

We can first replace F with [math]F= \sigma A[/math]:

 

[math]U=\int_{0}^{\Delta l} \sigma A dl[/math]

 

Now we also have [math]\Delta l= l \epsilon [/math]. In the limit that [math]\Delta l[/math] becomes smaller and smaller it becomes [math]dl[/math], and [math]\epsilon[/math] becomes [math]d \epsilon[/math]. So our equation with differentials becomes:

 

[math]d l= ld \epsilon [/math]

 

Since we are changing variables we also need to change the limits of integration, which is where the strain in the upper bound comes from. (If change in length goes from 0 to Δl, then strain goes from 0 to ϵ.)

 

So after all of our substitutions, the integral becomes:

 

[math]U=\int_{0}^{\epsilon} \sigma A l d\epsilon[/math]

Edited September 22, 2014 by elfmotat

[Function]

I'm going to change your variable O to an A, because I don't like it (it looks too much like zero). So we have:

 

[math]U=\int_{0}^{\Delta l} F dl[/math]

[math]\sigma =F/A[/math]

[math]\epsilon =\Delta l/l[/math]

 

We can first replace F with [math]F= \sigma A[/math]:

 

[math]U=\int_{0}^{\Delta l} \sigma A dl[/math]

 

Now we also have [math]\Delta l= l \epsilon [/math]. In the limit that [math]\Delta l[/math] becomes smaller and smaller it becomes [math]dl[/math], and [math]\epsilon[/math] becomes [math]d \epsilon[/math]. So our equation with differentials becomes:

 

[math]d l= ld \epsilon [/math]

 

Since we are changing variables we also need to change the limits of integration, which is where the strain in the upper bound comes from. (If change in length goes from 0 to Δl, then strain goes from 0 to ϵ.)

 

So after all of our substitutions, the integral becomes:

 

[math]U=\int_{0}^{\epsilon} \sigma A l d\epsilon[/math]

 

Enlightening. Thanks.

("O" stands short for "oppervlakte", Dutch for "area", yet your irritation is agreeable ; I don't agree the choice of your word "variable", however...)

Edited September 22, 2014 by Function