## Tensile Test vs. Bending Test Stresses

This is more of a theoretical question. Not sure if this should be under this section or Mechanical Engineering.

1. The problem statement, all variables and given/known data

For 1018 steel, I tested the UTS to be about 83 ksi for a dogbone specimen with a circular cross-section. However, when I determined the Ultimate Bending Stress from a 3 point bend test with 1018 steel cylindrical bar, the Ultimate bending stess was 204 ksi.

I've noticed the same differences in stress at the yield point. 73 ksi for tensile yield and 152ksi for bending yield.

I got similar values when running an FEA.

These values I got were what I was expected on seeing. I am just wondering why the stresses from a bending and tensile test vary so much? Bending fails in tension. They both also have the same fracture surface, microid covalesence.

Can anyone explain why there is a difference or recommend references?

Thanks!

2. Relevant equations
stress = force/area
stress = Mc/I

3. The attempt at a solution
Is some of the load absorbed by the compression component of bending?
Accuracy of equations? Mc/I is for elastic only.
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 Recognitions: Gold Member You are loading the specimen in two completely different ways. One is in pure tension and the other is in pure bending. Why would the results be the same? You are comparing apples to oranges. Also, the shear stress is usually about half the tensile stress for steels and other materials that obey Hooke's law. Thanks Matt
 Recognitions: Homework Help Science Advisor Hyrax: What was the specimen diameter and span length, and could you explain exactly how you applied and measured the transverse load, measured or calculated the moment, and measured or calculated the bending stress?

## Tensile Test vs. Bending Test Stresses

 Quote by CFDFEAGURU You are loading the specimen in two completely different ways. One is in pure tension and the other is in pure bending. Why would the results be the same? You are comparing apples to oranges. Also, the shear stress is usually about half the tensile stress for steels and other materials that obey Hooke's law. Thanks Matt
Matt,
can you point to a resource that explains that comparing them would be like comparing apples to oranges?

That is part of my question, Shouldn't the same material provide the same strength when tested under these two conditions especially considering that the failure mode at the fracture origin appears to be the same?

I've read up on bending theory and tension. I can't see why the two test conditions would yield different stresses. I've read that in bending, failure occurs in layers. But shouldn't a material fail or yield at the same stress regardless of testing conditions as long as grains are uniform?

 Recognitions: Gold Member Just because the fracture mode appears to be the same doesn't mean that it actually is. I will look more deeply into this and get back to you. Thanks Matt

 Quote by nvn Hyrax: What was the specimen diameter and span length, and could you explain exactly how you applied and measured the transverse load, measured or calculated the moment, and measured or calculated the bending stress?
The diameter was 0.236 inches. 36mm between supports of the 3 point bend setup. I used an MTS test machine. The MTS machine plotted the graph. From the graph, I found the ultimate load then calculated the stress:

32*((758 lb * 1.42 in)/4) /(pi*.236 in ^3 )

I got the yield load from the yield point.

Thanks nvm!

It's late. I'll check for updates tomorrow. Thanks again for all your assistance.
 Recognitions: Homework Help Science Advisor Hyrax: According to the PF rules, we aren't allowed to figure this out for you and tell you the answers. I think that is what they want you to figure out in this assignment. Hint 1: The main reason for the discrepancy is listed in post 1. Hint 2: As perhaps (?) another, minor, contributing factor, look up "engineering stress."

 Quote by nvn Hyrax: According to the PF rules, we aren't allowed to figure this out for you and tell you the answers. I think that is what they want you to figure out in this assignment. Hint 1: The main reason for the discrepancy is listed in post 1. Hint 2: As perhaps (?) another, minor, contributing factor, look up "engineering stress."
I looked up pure bending. How can there be no tensile stress in pure bending when there is a compression and tension component to pure bending?
 Recognitions: Homework Help Science Advisor In bending, there is tensile stress on the tension side, and compressive stress on the compression side. They probably mean there is no axial load, so there is no uniform tensile stress across the entire cross section. Hint 3: The main reason for the discrepancy is not because part of the load was absorbed by the compression component.

 Quote by nvn In bending, there is tensile stress on the tension side, and compressive stress on the compression side. They probably mean there is no axial load, so there is no uniform tensile stress across the entire cross section. Hint 3: The main reason for the discrepancy is not because part of the load was absorbed by the compression component.
argh.. I'll just take the possible explanations I found researching and present it to my boss. Thanks anyway.

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