Concept question on .2% strain

In summary, Chuck's professor went over how to calculate Youngs Modulus (E) using an offset and how to find the strain for a particular stress.
  • #1
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Hi all,

I am trying to wrap my head around this and for some reason I am blocked. I have taken Solid mechanics and strength of materials and am always used to calculating strain as ε=ΔL/L. I might be missing something easy as it has been a year since the class but in my Manufacturing Engineering class the professor went over an example that went like this:

During tensile test a specimen yields at 48KN. This is the .2% yield point.
Ao= .1m^2
Lo = .05m
Lf = .0523m

We calculate yield stress at normal with 48KN/.1m^2 = 480MPa

Then when we went to calculate Youngs Modulus (E = σ/ε), we used (ε - .002) for the ε. I feel like I am missing something. I think I know why we used this, I just wanted some clarification.

Basically E is the slope of the elastic curve. Dropping straight down on the stress strain curve gives the regular ε. If we offset this by .002 we can get the rise/run needed to get E. right?

Thanks,
Chuck
 
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  • #2
You have to pay attention to your decimal points. What is the area of the specimen again?

There is no such thing as a 0.2% yield point.
The yield point is where the specimen undergoes elongation without extra load. On the stress strain curve it looks like a blip. Not all materials have a unique yield point - low carbon steels being the most common that does.

A material without a yield point has what is called a yield strength, and there are different basis to determine that value. The 0.2 offset is one method used for steel and aluminium and some other metals.

To determine the 0.2% yield strength, one draws a line on the graph of the stress-strain curve parallel to the linear part of the graph that follows from the origin to the proportional limit. This drawn line is offset by a strain of 0.2% and the intersection of this line with the stress-strain curve gives the yield strength of the material.

Then when we went to calculate Youngs Modulus (E = σ/ε), we used (ε - .002) for the ε. I feel like I am missing something. I think I know why we used this, I just wanted some clarification.
The 0.002 would be the elongation at a particular stress. What stress did you use?

Basically E is the slope of the elastic curve. Dropping straight down on the stress strain curve gives the regular ε.
E is the slope of the elastic curve which is linear up to the proportional limit, where stress is proportional to strain, and you can calculate this from the yield stress and the elongation at that stress. Dropping down from the curve gives you the elongation.

If we offset this by .002 we can get the rise/run needed to get E. right?
NO
 
  • #3
256bits said:
You have to pay attention to your decimal points. What is the area of the specimen again?

Sorry, it was .0001 m^2

There is no such thing as a 0.2% yield point.

My text and professor for some reason refer to it as the .2% yield point in the examples.

The 0.002 would be the elongation at a particular stress. What stress did you use?
480Mpa

E is the slope of the elastic curve which is linear up to the proportional limit, where stress is proportional to strain, and you can calculate this from the yield stress and the elongation at that stress. Dropping down from the curve gives you the elongation.

Right, but what they are having us do is calculate E from the .2% yield strength, where the force, lengths, and area is given.

So we find the strain for that stress(which is past the proportional limit) with ΔL/Lo, subtract .002 from it, then divide the stress by the new strain.

I uploaded an example that is using true strain, but the same concept of subtracting .002.

Its just basically using the slope equation I thought. (y2-y1)/(x2-x1) which is in this case:

(480MPa - 0)/(.0046 - .002)...or am I just way off?
 

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  • #4
Its just basically using the slope equation I thought. (y2-y1)/(x2-x1) which is in this case:

(480MPa - 0)/(.0046 - .002)...or am I just way off?

That should be the equation for calculating E by the 0.2% offset. I do not see reason why that would not be so.

PS. I must have interpreted what you were explaining in your first post.
PS. 0.2% yield point is kind of arbitrary. On certain stress strain diagram one can see where the specimen does yield. Other materials are chosen to have other offsets, or other methods to determine E, etc. As long as we all agree on terminology and its description then we are all OK.

cheers
 
  • #5
Thanks a lot 256, I really appreciate the help!
 

1. What is meant by .2% strain?

.2% strain is a measure of the amount of deformation or change in shape of an object when subjected to stress or force. It is calculated as the change in length of the object divided by its initial length, multiplied by 100.

2. How is .2% strain different from other measures of strain?

.2% strain is a very small amount of deformation and is typically used to measure the elastic behavior of materials. Other measures of strain, such as 1% or 5%, are used for larger deformations and are often associated with plastic behavior.

3. Why is .2% strain important in materials testing?

.2% strain is important in materials testing because it is the point at which most materials begin to exhibit elastic behavior. This means that they can return to their original shape after the stress is removed. It is also a common standard for comparing the strength and stiffness of different materials.

4. How is .2% strain measured in experiments?

.2% strain is typically measured using a strain gauge, which is a device that can detect and measure small changes in length. The strain gauge is attached to the material being tested and as the material is subjected to stress, the gauge records the change in length. This data is then used to calculate the .2% strain.

5. Can .2% strain be exceeded without causing permanent damage to a material?

Yes, .2% strain is typically within the elastic range of most materials, meaning that they can be stretched or compressed beyond this point without causing permanent damage. However, if the strain is exceeded and the material is pushed into its plastic range, permanent damage may occur.

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