Understanding Modulus of Toughness
Understanding Modulus of Toughness
(OP)
Hello all
I was hoping someone could help me understand the Modulus of Toughness for a timber compression experiment I did.
I tested a 45mm x 45mm x 45mmm timber cube under compressive load.
The load was applied over a cross sectional area of 45 * 45 = 2025mm^2
My stress strain curve is represented by a straight line function:-
Stress = 8.8771 * Strain
The area under my curve denotes the modulus of toughness which is the amount of energy the material can absorb per unit volume.
If I integrate my stress equation I get:-
Toughness = 4.43855ε2.
Integrate from 0 to 7.867 (which is the strain value of my material at failure) I get a value of approx 275.
If my units were:-
Stress = N/mm^2
Strain = %
Area = mm2
Then what units would my Toughness value be?
Also how is the volume factored into this, as I don't use the volume in any of the equations?
Can anyone shed any light?
Thank you.
I was hoping someone could help me understand the Modulus of Toughness for a timber compression experiment I did.
I tested a 45mm x 45mm x 45mmm timber cube under compressive load.
The load was applied over a cross sectional area of 45 * 45 = 2025mm^2
My stress strain curve is represented by a straight line function:-
Stress = 8.8771 * Strain
The area under my curve denotes the modulus of toughness which is the amount of energy the material can absorb per unit volume.
If I integrate my stress equation I get:-
Toughness = 4.43855ε2.
Integrate from 0 to 7.867 (which is the strain value of my material at failure) I get a value of approx 275.
If my units were:-
Stress = N/mm^2
Strain = %
Area = mm2
Then what units would my Toughness value be?
Also how is the volume factored into this, as I don't use the volume in any of the equations?
Can anyone shed any light?
Thank you.
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RE: Understanding Modulus of Toughness
1. How did you come up with this conclusion? Because, stress-strain curves for some typical woods I found on google definitely shows a plastic region.
2. Based on your straight line assumption (which I found highly doubtful), If I compare the failure strain of wood which is 7.88% with a measly 2% yield strain of 60 ksi steel, the question that comes to mind is, why the heck aren't we using wood as reinforcement in compression members such as columns?
3. Not only this, it also give a superior ultimate strength which according to following equation will be.
Stress = 8.8771 (assuming the unit of E is GPa) x 1000 x 7.88/100 = 700 MPa = 100 ksi approx.
One can say, steel has higher ductility than wood, then again, we can use this where ductility is not a requirement (being a devil's advocate here).
Again to reiterate, I found the intimated behavior in your point no. 1 to 3 highly doubtful.
4. Seems like, you forgot to divide the strain with 100 while calculating this.
5. Basically toughness is defined as the work done per unit volume and is equals to the area under the stress-strain curve. Which, in your case is equals to area of triangle = StessxStrain/2 = Magnitude (N/mm2)x(unitless) = N/mm2.
Above can also be expressed as = (N/mm2)x mm/mm = N-mm/mm3 = Workdone/Volume
Disclaimer: I've never worked on (or even studied) a wood structure, thus, my comments are solely based on common sense and knowledge of mechanical properties of materials other than wood.
RE: Understanding Modulus of Toughness