The Student Room Group

Force extension graphs

For a force-extension graph where the loading curve has a different area under the graph (i.e. different energy) to the unloading curve, what quantities are meant by the area under the loading curve, area under the unloading curve, and area between the two curves respectively?
The area under the curve is the
a) work done on the material during loading
b) work done by the material during unloading
c) the difference between a) and b)
Reply 2
Original post by Stonebridge
The area under the curve is the
a) work done on the material during loading
b) work done by the material during unloading
c) the difference between a) and b)


Thanks.

I have another, separate question: in a stress-strain graph, how do we derive the formula Strain Energy per unit volume = (1/2) * F*x/(A*l)? It has to do with the graph being a straight line (within the limit of proportionality) and taking an integral. But besides that I don't know how it's defined.

Obviously it comes from Stress=F/A and Strain=x/l. Other than that I don't know. (F is force applied, A is area of material, x is extension, l is original length of material)
If I understand what you're asking

The area under the triangular stress strain line is

12(FA)(xL)\frac{1}{2}(\frac{F}{A})(\frac{x}{L})

½Fx is the work done by force. (½F is average force) and equal to the elastic potential energy EPE stored in the material.

so the area under the curve is
EPEAL\frac{EPE}{AL}

AL is the volume of the wire so this formula reduces to
Area under the line = epe (strain energy) per unit volume.
(edited 10 years ago)
Reply 4
Original post by Stonebridge
If I understand what you're asking

The area under the triangular stress strain line is

12(FA)(xL)\frac{1}{2}(\frac{F}{A})(\frac{x}{L})

½Fx is the work done by force. (½F is average force) and equal to the elastic potential energy EPE stored in the material.


Ah. So if it's not a straight line, then is the strain energy per unit volume = integral(0 to x){F(x)}dx/(A*l)? as elastic potential energy = integral(0 to x){F(x)}dx. For the straight line case F(x)=kx, so the integral is ½kx^2, which is the same as ½Fx, which translates to ½Fx/(A*l).
The area under the graph represents the same thing even if the graph isn't a straight line. Ii just simplifies the maths when it's straight.
Yes, it's that integral in the general case.
Reply 6
Original post by Stonebridge
The area under the graph represents the same thing even if the graph isn't a straight line. Ii just simplifies the maths when it's straight.
Yes, it's that integral in the general case.


OK, and is l in this case (since we want strain energy per unit volume) going to be the initial length (l0), or the total length (x+l0).
It's the initial length and initial volume.
The increase in length δL is very small compared with the original. So the increase in volume δV is also very small.
It can only be meaningful to talk about "per unit volume" when the change in volume is negligibly small, which it is in the case of stretched wires.
In calculus you take the limit as δL approaches zero.

Quick Reply

Latest