Calculate Modulus of Resilience with Grain Diameter Variation

[Sniperfx20]

Labratory measurements show that when a certain material has a grain diameter of 2.33x10-2 mm it's yield strength is 107 MPa. When the material's grain diameter decreases to 1.84x10-2 mm it's yield strength increases to 117 MPa.
When the average grain diameter is increased to 3.28x10-2 mm, what is the modulus of resilience (in J/m^3) if the modulus of elasticity is known to be 182 GPa?

any idea? I am stumped

 

[sumungo]

Hi,

There is two questions.

1: Grain size may or may not affect in elastic modulus properties. In work hardening metals (like in some copper alloys) decreasing grain size may increase E a bit, not to mention effects of microstructure and residual stresses. How the grain size was decreased, or if the sample is different one. Not patronizing You, but make sure, and more importantly, make it real that you are looking at the right things. At first, does it matter that much. If it does, and the more it does, the more there is a need for normalizing properties of the material.

2: Elastic resilience (U) is usually far more complex issue than solving it by using just a single linear formula. The concept of "resilience modulus" has very dynamic nature.

But:

U = σ²/2E, where σ is (elastic) yield stress.

Check out: http://www.brushwellman.com/alloy/tech_lit/april01.pdf
(it's the "Area" there)

One problem in this concept is that yield displacement should be defined somehow (and quite well, actually). Then, whatever would happen in the material during deformations having different displacements and speed (non-linear damping, fatigue), one should most preferably consider a bit more extensive study in kinematic properties of materials. Again, not patronizing You at all, just flashbackin' some bitter events in the past :).

Of course, if You are doing some rebound tests just for fun or class work, forget about it. Being too open-minded may cause that notorious brain spillage :)

 

[oswaler]

Based on the given information, we can calculate the modulus of resilience using the formula R = (σ^2)/2E, where R is the modulus of resilience, σ is the yield strength, and E is the modulus of elasticity.

First, we need to convert the given values to the appropriate units. The yield strength is given in MPa, so we need to convert it to Pa by multiplying by 10^6. This gives us a yield strength of 107x10^6 Pa for the grain diameter of 2.33x10^-2 mm and a yield strength of 117x10^6 Pa for the grain diameter of 1.84x10^-2 mm.

Next, we need to convert the modulus of elasticity from GPa to Pa by multiplying by 10^9. This gives us a modulus of elasticity of 182x10^9 Pa.

Now, we can plug these values into the formula for modulus of resilience:

R = (σ^2)/2E

For the grain diameter of 2.33x10^-2 mm:
R = ((107x10^6)^2)/2(182x10^9)
= 1.16x10^3 J/m^3

For the grain diameter of 1.84x10^-2 mm:
R = ((117x10^6)^2)/2(182x10^9)
= 1.59x10^3 J/m^3

Therefore, the modulus of resilience for the average grain diameter of 3.28x10^-2 mm would be the average of these two values:
R = (1.16x10^3 + 1.59x10^3)/2 = 1.38x10^3 J/m^3

In conclusion, the modulus of resilience for the average grain diameter of 3.28x10^-2 mm is 1.38x10^3 J/m^3. It is important to note that this value may vary depending on other factors such as the composition of the material and the testing conditions. Further experimentation and analysis may be needed to determine the true modulus of resilience for this material.