So, $Y = 3K(1 - 2\sigma)$. The constants involved are the Young modulus, Bulk modulus and the Poisson ratio respectively. I don't know why this is true; nobody could tell me why (and I for some reason didn't think of asking you). But let us assume for a moment this is true.
I am writing out $dV/V$, yes. I have a wire of length $\ell$ and radius $r$, so $V$ here is $\pi r^2 \ell$ (unlike wiki where they do it with a cube). But w/e
now, if you can equate volumetric and tensile stress (as you did when you wrote your equations with that factor of $F/A$ in both) then that becomes $$\frac{Y}{K} = \frac{\text{tensile strain}}{\text{volumetric strain}}$$
Admittedly we weren't told, even by the good teachers we have, of any particular difference between them that should explain why I have this $1/3$ issue.
@0celo7 ...okay, i'm really getting tired of that attitude. not being fully rigorous does not mean being inconsistent.
and while i'm not versed enough in linear elasticity to actually give the arguments, I presume that when one actually uses them (i.e. elasticity tensors and the like) none of this is particularly strange or inconsistent.
If it's the sum of the stresses along the three orthogonal axes, then if the stresses are equal then "volumetric stress = 3 tensile stress" is pretty sensible
I agree, admittedly I have always tried to hand-wave past that issue: most objects (wires, cylinders) have a canonical cross section associated to it :P
Let us be more of a physicist than a mathematician and assume you're right for now. When another issue comes, I'll ping you and ask you to come up with a better definition :D
in fact, Lame's relation just corresponds to taking the trace of the following equation (stated there in index form): $\mathbf{\epsilon}=\frac{1}{E}\left[(1+\nu)\mathbf{\sigma}-(\text{tr }\mathbf{\sigma})1\right]$
...maybe. maybe i'm just being loopy :/
should've been a factor of $\nu$ in front of the trace in that last expression, woops
hmm, $K=\frac{\text{tr }\sigma}{\text{tr }\epsilon}$ is a nice statement @BalarkaSen
@SemiC No, wait a second. $dV/V = d\ell/\ell ( 1- 2\sigma)$. $F$ be volumetric stress. $\frac{F/A}{dV/V} = \frac{F/A}{d\ell/\ell} (1 - 2\sigma)^{-1}$. You're claiming $F$ = 3 x tensile stress (call it $F_\ell$)? That doesn't seem to give me the right thing. That says $1/3K =Y(1 - 2\sigma)^{-1}$.
So $F_\ell = 3F$? That does make sense in a way: if you apply the same amount of force by the x-axis, say, and symmetrically through all the three axes, the internal reaction in the first would be thrice the internal reaction of the latter.
Stress, after all, is internal reaction, and not the amount of force you apply to it.
@user1618033 i dont create math problems i just solve them
whilst, i create puzzles (logic is my focus)
meantime, i am getting accross an idle periode, my stoopitch mind getting rigid and i am more connected to a sensless uncreative software with that sour flavour of databases and computational-statistics
@balarka I just realized that when I said ratio of Young's modulus to bulk modulus earlier, I said it backwards i.e. I wrote Y/K on the LHS but K/Y on the RHS
