- #1
Synco
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- TL;DR Summary
- Well people i want to go from my steel/aluminum frame to the best material in the market. I have seen that exist a relation between Mpa(Yield strengh-elastic limit)/density but i get a very weird relation in a strange unit K*(m^2)/(s^2) which(K= any real number) is the expansion of an area in the time, but, what is it doing here if I'm searching the strongest material to reduce weight?
Well people i want to go from my steel/aluminum frame to the best material in the market. I have seen that exist a relation between Mpa(Yield strenght-elastic limit)/density but i get a very weird relation in a strange unit (m^2)/(s^2) which is the expansion of a growing area in the time, but, what is it doing here if I'm searching the strongest material to reduce weight?
Ok
1Mpa= 1000000 N/m^2
N=Kg*(m/s^2)
Density= Kg/m^3
so
Yield Strength/Density is:
((Kg*(m/s^2))/m^2)/(Kg/m^3)= m^2/s^2
according to this link
https://www.symbolab.com/solver/ste...}\right)/\left(K\cdot g/m^{3}\right)?or=input
Why Yield Strength(elastic limit)/Density?
If we think in total resistance( I'm not a mechanical engineer) we can imagine the following case:
Imagine that you have a hollow bar of X material with an L length with a volume V, using it as lever. If you fill the bar with the same material you can apply more force without breaking it or curve it and if the density tends to 0, for say something, you are filling the hollow bar without changing the weight so much, so we can think that's a benefit fill the bar without compromise the weight. In contrary case if the density tends to infinite(we have a black hole XD) you are adding so much weight without getting a benefit, so the correct "sense" of the density it is to go to 0.
Occurs exactly the contrary case with yield strength, if it goes to 0 it is useless if it goes to infinite it is awesome as a lever. Hence, the "correct" "sense" it is to grow. So the bigger number the better for this application.
But, if we are talking about volumes, resistance to pressure and weight, In what moment the acceleration of the expansion of the area it is representative to define what material it's better than the other to resist the same pressure by less weight?
Ok
1Mpa= 1000000 N/m^2
N=Kg*(m/s^2)
Density= Kg/m^3
so
Yield Strength/Density is:
((Kg*(m/s^2))/m^2)/(Kg/m^3)= m^2/s^2
according to this link
https://www.symbolab.com/solver/ste...}\right)/\left(K\cdot g/m^{3}\right)?or=input
Why Yield Strength(elastic limit)/Density?
If we think in total resistance( I'm not a mechanical engineer) we can imagine the following case:
Imagine that you have a hollow bar of X material with an L length with a volume V, using it as lever. If you fill the bar with the same material you can apply more force without breaking it or curve it and if the density tends to 0, for say something, you are filling the hollow bar without changing the weight so much, so we can think that's a benefit fill the bar without compromise the weight. In contrary case if the density tends to infinite(we have a black hole XD) you are adding so much weight without getting a benefit, so the correct "sense" of the density it is to go to 0.
Occurs exactly the contrary case with yield strength, if it goes to 0 it is useless if it goes to infinite it is awesome as a lever. Hence, the "correct" "sense" it is to grow. So the bigger number the better for this application.
But, if we are talking about volumes, resistance to pressure and weight, In what moment the acceleration of the expansion of the area it is representative to define what material it's better than the other to resist the same pressure by less weight?