Calculating Yield Strength using a Load vs. Displacement Curve

[ginarific]

Hi,

This question came up on my midterm and I had no idea how to answer it without redrawing the entire curve as a stress vs. strain curve (which obviously took too long to do).

Anyway, I'm just requesting a general procedure, not a numerical answer. If you had a Load (y-axis) vs. Displacement (x-axis) curve, how do you calculate the yield strength? I know for a stress vs. strain curve, you have to have a 0.2% (0.002) offset, but my Professor told me that the offset is not of the same value in a Load vs. Displacement curve (which makes sense).

However ... what <i>is</i> the actual offset supposed to be? Is there a formula to calculate it?

Any help would be very much appreciated! :)

Thank you,
Gina

 

[Astronuc]

The 0.2% (0.002) is a value of strain, which is arbitrary, but it covers somewhat the uncertainty of when a given material actually departs from the purely linear (elastic) relationship between stress and strain.

Let l = length, then strain ε = (l - l_o)/l_o, where l_o = original length (usually the gauge length). Also, the displacement, d, is given by (l - l_o), so strain ε = d/l_o.

Similar the stress, σ, is just the load/force F divided by area A, i.e. σ = F/A, where A is constant until the specimen reaches UTS, which corresponds to the limit of uniform elongation, and necking begins.

Now back to length/displacement -

with ε = (l - l_o)/l_o, one rewrites the equation as ε = l/l_o - 1, and reorganizing the terms, l = l_o (1+ε),

so the length equivalent to the strain offset of 0.002 is just l = l_o*1.002, or d (0.002) = 0.002 l_o.

 

[General_Sax]

where A is constant until the specimen reaches UTS, which corresponds to the limit of uniform elongation, and necking begins.

So, if I have the following data:

A = 0.36m²
F = 9800N @ 0.2% offset

Then [sigma] = 9800/0.36 = 2.7*10⁴ Pa ??

I just have a problem regarding the constant area. Wouldn't the cross-sectional area be decreasing while the object is strained (stretched)?

 

[stewartcs]

So, if I have the following data:

A = 0.36m²
F = 9800N @ 0.2% offset

Then [sigma] = 9800/0.36 = 2.7*10⁴ Pa ??

That gives you the Engineering Stress (which is always a function of the original cross-sectional area).

I just have a problem regarding the constant area. Wouldn't the cross-sectional area be decreasing while the object is strained (stretched)?

Yes. The stress in this case is called the True Stress and is a function of the instantaneous minimum cross-sectional area of the specimen.

CS