Mechanics of Materials Torsion Test Help

In summary: Your Name]In summary, the conversation discussed a torsion test on mild steel and cast iron, and the required calculations for each material. The person asking the question was confused about the difference between the limit shear strain and the modulus of rupture, and sought clarification on the matter. The expert explained that the limit shear strain is the maximum amount of shear strain a material can withstand before failure, whereas the modulus of rupture is the maximum stress it can withstand. They also clarified that for brittle materials like cast iron, the limit shear strain would be zero.
  • #1
IllTry
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Homework Statement

Hello, I am writing up a lab for a torsion test that we performed on mild steel (ductile material) and cast iron (brittle material). For each of the materials, we must find:
• For Mild Steel:
- Torque vs. Angle of Twist per unit length.
- Shear Modulus
- Shear Yield Strength
- Torque Modulus of Resilience
- Torque Modulus of Toughness
- Limit Shear Stain
• For Cast Iron:
- Torque vs. Angle of Twist per unit length.
- Shear Modulus
- Modulus of Rupture
- Limit Shear Stain


Homework Equations


Ip = Polar MoI
We are using Shear Stress(tau)= Torsion*radius/Ip
and Shear Strain (gamma) = phi*radius/L



The Attempt at a Solution


This may be relativly simple to answer. For both materials I have an excel spread sheet with data from the lab. They gave me Torque and Twist angle, and from those two tables, I was able to create a second data set for Shear stress and strain. I put these tables into graphs and have a graph that looks similar to this for my cast Iron:
BzjABt3VOKD9wR8-PAvZkSBSJQ97uFO9J48lLLP_aJrrJkPPYEG0z33rhA7ylvIVgrHU-g8XstYtPhSDA?PARTNER=WRITER.jpg

I know shear modulus, or G, is just the slope of the line for the plastic behavioral part of the material. The modulus of rupture is defined in my lab as the shear stress calculated from the maximum torque found during the test. So It will just be my shear stress value at the very end of the graph (correct me if I am wrong). Then I come to the Limit shear strain and get confused. Is this the same as the modulus of rupture? It seems silly they would ask for two values which essentially mean the same thing. Do you know if this is another term for yield limit? if that's the case, then there should technically be none since it is a brittle material (please let me know if I am wrong here). Any Ideas or clarification will help.
 
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  • #2


Hello,

Thank you for sharing your question about the torsion test on mild steel and cast iron. From your description, it seems like you have a good understanding of the concepts and equations involved in this experiment.

To answer your question about the limit shear strain, it is not the same as the modulus of rupture. The limit shear strain is the maximum amount of shear strain that the material can withstand before it fails. This value is typically determined from the stress-strain curve for the material, and it is different from the modulus of rupture which is the maximum stress that the material can withstand before it fails.

In the case of brittle materials like cast iron, the limit shear strain would be zero since they do not exhibit plastic deformation and fail abruptly when the limit shear stress is reached. So, it would be correct to say that the limit shear strain for cast iron is essentially non-existent.

I hope this helps to clarify the concept of limit shear strain and its difference from the modulus of rupture. Let me know if you have any further questions or if I can assist you with anything else.


 
  • #3


I would first like to commend you on your thorough explanation of the problem and your understanding of the concepts involved in the torsion test. It is clear that you have put a lot of effort into your lab report and that you are seeking clarification on a few points. I will do my best to address your questions and provide some guidance for your analysis.

Firstly, you are correct in your understanding that the shear modulus, or G, is the slope of the linear portion of the stress-strain curve for a material. This is an important material property that describes the material's resistance to shearing forces. It is calculated by dividing the shear stress by the shear strain.

Next, you mention the modulus of rupture, which is defined as the maximum shear stress reached during the torsion test. This is also known as the ultimate shear strength and is an important value to determine the maximum amount of shear stress a material can withstand before it fails. This value is typically used to design structures and components to ensure they can withstand the expected loads.

The limit shear strain is a bit more complicated to define. It is not the same as the modulus of rupture, but it is related to it. The limit shear strain is the maximum amount of shear strain a material can withstand before it reaches its ultimate shear strength. In other words, it is the point at which the material starts to exhibit permanent deformation and is no longer able to return to its original shape once the load is removed. This value is important for understanding the ductility of a material, as materials with higher values of limit shear strain are able to withstand more deformation before failure.

In the context of your lab, it is possible that the limit shear strain is being used as another term for the yield limit. However, this would only be applicable for ductile materials, as brittle materials do not exhibit a yield point and therefore do not have a yield limit. It is important to clarify with your instructor or lab manual if this is the case.

In summary, for mild steel, you will need to calculate the shear modulus, shear yield strength, and the limit shear strain. For cast iron, you will need to calculate the shear modulus, modulus of rupture, and the limit shear strain. It is important to carefully review your data and calculations to ensure accuracy and to clearly label and explain your results in your lab report. I hope this helps clarify some of your questions and I wish you luck in your analysis.
 

1. What is the purpose of a torsion test in mechanics of materials?

A torsion test is used to determine the material's resistance to twisting or torsional forces. This helps engineers understand how a material will behave under load and how it can be used in different applications.

2. How is a torsion test performed?

In a torsion test, a sample of the material is clamped at both ends and a torque is applied at one end while the other end is fixed. The sample is twisted until it reaches its yield point or breaks. The amount of torque and the corresponding angle of twist are measured to determine the material's torsional strength.

3. What are the important parameters measured in a torsion test?

The two main parameters measured in a torsion test are torque and angle of twist. These values are used to calculate the shear stress and shear strain of the material, which are important indicators of its torsional strength.

4. How does a material's microstructure affect its torsional properties?

The microstructure of a material, such as the arrangement of its grains and the presence of impurities, can greatly impact its torsional properties. A material with a more uniform and fine grain structure will typically have higher torsional strength compared to a material with larger and more irregular grains.

5. What are some common applications of torsion testing in engineering?

Torsion testing is commonly used in engineering to test and select materials for applications that involve twisting or rotating forces, such as shafts, gears, and springs. It is also used to evaluate the quality of manufactured products and to study the effects of different processing techniques on a material's torsional properties.

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