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

 

[alphy]

Hello Gina,

Calculating the yield strength from a Load vs. Displacement curve is a little different from calculating it from a stress vs. strain curve. In a stress vs. strain curve, we use a 0.2% offset because it is a common standard in materials testing. However, in a Load vs. Displacement curve, the offset value can vary depending on the specific material and test conditions.

To calculate the yield strength from a Load vs. Displacement curve, you can follow these steps:

1. Find the point on the curve where the load starts to level off or plateau. This is the point where the material begins to yield.

2. Draw a straight line from the origin (0,0) to this point on the curve.

3. Extend this line until it intersects with the x-axis (displacement axis).

4. The value of displacement at this intersection point is your offset value.

5. The yield strength can then be calculated by dividing the yield load (the load at the plateau) by the cross-sectional area of the specimen.

It is important to note that the offset value may vary for different materials and testing conditions. It is always best to consult with your professor or refer to a materials handbook for the specific offset value to use for a particular material.

I hope this helps! Good luck with your studies.

Best,