Tensile Test vs. Bending Test Stresses

In summary, the conversation discusses the discrepancies in stress values obtained from tensile and bending tests on a 1018 steel specimen. The Ultimate Tensile Stress (UTS) and Ultimate Bending Stress (UBS) values were significantly different, as well as the stress at the yield point. The conversation also touches on the similarities in failure mode and the effect of shear stress. The main reason for the discrepancy is attributed to the different loading conditions in the two tests, with one being in pure tension and the other in pure bending. Other minor contributing factors, such as engineering stress and the compression component of bending, are also mentioned. Further research and discussion are needed to fully understand the differences in stress values between the two tests.
  • #1
Hyrax
5
0
This is more of a theoretical question. Not sure if this should be under this section or Mechanical Engineering.

Homework Statement



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!

Homework Equations


stress = force/area
stress = Mc/I

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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  • #2
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
 
  • #3
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?
 
  • #4
CFDFEAGURU said:
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?

Thanks for your input Matt!
 
  • #5
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
 
  • #6
nvn said:
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.
 
  • #7
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."
 
  • #8
nvn said:
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?
 
  • #9
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.
 
  • #10
nvn said:
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.
 

1. What is the difference between tensile test and bending test stresses?

The main difference between tensile test and bending test stresses lies in the direction of the applied force. In a tensile test, the force is applied in a linear direction to stretch the material, while in a bending test, the force is applied perpendicular to the material causing it to bend.

2. Which test is better for evaluating the strength of a material?

Both tensile test and bending test can provide valuable information about a material's strength. However, tensile test is typically considered more accurate and reliable as it measures the maximum load a material can withstand before breaking, while bending test only measures the maximum load that causes the material to bend.

3. What are the advantages of performing a tensile test?

Tensile test can provide a comprehensive view of a material's mechanical properties, including tensile strength, yield strength, and modulus of elasticity. It is also a widely accepted and standardized test method, making it easier to compare results from different studies.

4. When is a bending test preferred over a tensile test?

Bending test is often preferred over a tensile test when the material being tested is too brittle or too thin to withstand a tensile force. Bending test can also be used to evaluate the ductility and toughness of a material, which cannot be measured in a tensile test.

5. Can the results of a tensile test be used to predict the performance of a material under bending stress?

No, the results of a tensile test cannot accurately predict the performance of a material under bending stress. This is because the two types of stresses affect a material differently and may cause different types of failure. Therefore, both tests are necessary to fully understand a material's mechanical behavior.

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