- #1
IllTry
- 3
- 0
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:
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.