Solid Mechanics - Stress-Strain diagram - absorbed energy

[Feodalherren]

Homework Statement

1.The maximum energy (per unit volume) that can be absorbed by the steel alloy without
sustaining permanent deformation is _______ lb/in^2

1.4. The maximum energy (per unit volume) that can be absorbed bythe steel prior to fracturing is _______ kip/in^2.

.

Homework Equations

The Attempt at a Solution

If I remember correctly, stored energy is the area under the graph.
Since I don't know the equation I will work with estimates.

First, I used the 0.2 % offset method to find the Yield strength of the material and I found the yield strength to be about 65 ksi.
The area under the straight line is (1/2)(.002)(60E3)=60
I estimate the rest of the area as a rectangle: (0.002)(60)=120
which gives a grand total of 180 lb/in^2.

Yet the solution, which is not worked out, claims that the energy is only the first part, namely 60 lb/in^2. I thought a material wouldn't be deformed before it gets to the point of yield strength, and in the previous step I found the yield strength correctly. I'm thinking they did something wrong because the solution even goes on to say "Modulus of resilience"=60PSI.

1.4 should just be the same thing, total area under curve?

 

[SteamKing]

Homework Statement

1.The maximum energy (per unit volume) that can be absorbed by the steel alloy without
sustaining permanent deformation is _______ lb/in^2

1.4. The maximum energy (per unit volume) that can be absorbed bythe steel prior to fracturing is _______ kip/in^2.

.

Homework Equations

The Attempt at a Solution

If I remember correctly, stored energy is the area under the graph.
Since I don't know the equation I will work with estimates.

First, I used the 0.2 % offset method to find the Yield strength of the material and I found the yield strength to be about 65 ksi.
The area under the straight line is (1/2)(.002)(60E3)=60
I estimate the rest of the area as a rectangle: (0.002)(60)=120
which gives a grand total of 180 lb/in^2.

Yet the solution, which is not worked out, claims that the energy is only the first part, namely 60 lb/in^2. I thought a material wouldn't be deformed before it gets to the point of yield strength, and in the previous step I found the yield strength correctly. I'm thinking they did something wrong because the solution even goes on to say "Modulus of resilience"=60PSI.

1.4 should just be the same thing, total area under curve?

IDK why you are using the 0.2% offset method to determine yield strength, IIRC, 0.2% offset is used for various non-ferrous metals and alloys (like aluminum) which don't exhibit a clear linear stress-strain relationship in the stress-strain curve.

For the test results shown in the graph, there is clearly a linear stress-strain relationship up to a stress of approx. 60 ksi.

http://en.wikipedia.org/wiki/Yield_(engineering)

Once the material is stressed beyond yield, there will be a permanent set (it's kinda the definition of what yield is.)

The modulus of resilience is defined in this article:

http://en.wikipedia.org/wiki/Resilience_(materials_science)

 

[Feodalherren]

So now I'm thoroughly confused as to what the difference between yield strength and elastic limit is... And if I don't use the .2 % method, then how am I supposed to know where the yield point is? It's not given to me in the problem.

Nevermind, the information was in the wikipedia article, thanks.