Thailand?
You know damn right when you no lift for 30 lbs instead half is 10.
Personally I do not nod back because I think that is some kind of homosexual communication.
Charmonica is at home without a Nautilus Gym machine. He is trying to figure out weight equivlants with free weights. He is wondering if he should count the weights on each end of the bar together or load each end of the bar with the weight that he would normally lift on the machine at the Gym.
“It’s alright, stewardess. I speak jive.”
It’s a random word generator right?
Just use Viagra and double everything.
...you ate SAND???
when as if you wieght the 40lb, the half men where can the lift 20lbs of floor, in the if but the 35 can for but to be women.
Only if you use 20lbs of chicken feathers..........
The answer is “49”
\epsilon_{ij} = \frac{1}{2} (u_{i,j} + u_{j,i}), (5.19)
the linear elastic isotropic relation between stress and strain
\sigma_{ij} = \lambda \delta_{ij} \epsilon{kk} + 2\mu\epsilon_{ij}, \end{displaymath} (5.20)
So again the answer is "49" and the equilibrium condition
\sigma_{ji,j} + f_i = 0. (5.21)
Substituting 5.19 into 5.20 into 5.21 gives the simplified Navier equations in terms of displacement $u_i$:
\mu u_{i,jj} + (\mu+\lambda) u_{j,ji} + f_i = 0. (5.22)
This can also be written in terms of the modulus of elasticity $E$ and Poisson's ratio $\nu$, where we have the relations
E = \frac{\mu(3\lambda + 2\mu)}{\lambda+\mu}; \nu = \frac{... ... \frac{\nu E}{(1-2\nu)(1+\nu)}; \mu = G = \frac{E}{2(1+\nu)}. (5.23)
But I like the simplicity of the Navier equations better, though in terms of $\mu$ and $\nu$, things aren't so bad:
\mu u_{i,jj} + \frac{\mu}{1-2\nu} u_{j,ji} + f_i = 0 (5.24)
So, this is trouble because the three equations are fully coupled. To separate them, we define the Galerkin vector $g_i$ such that
2\mu u_i = c g_{i,jj} - g_{j,ji}, (5.25)
where $c$ is a scalar constant whose value will be determined later. Substituting this into equation % latex2html id marker 1126 $\ref{eq:munu}$ gives
\left[\frac{c}{2}g_{i,kk} - \frac{1}{2} g_{k,ki}\right]_{,j... ...}{2} g_{j,kk} - \frac{1}{2} g_{k,kj} \right]_{,ji} + f_i = 0, (5.26)
and since $g_{k,kijj} = g_{j,kkji} = g_{k,kjji}$, this rearranges to
\frac{c}{2}g_{i,kkjj} + g_{k,kjji} \left[-\frac{1}{2} + \frac{c}{2(1-2\nu)} - \frac{1}{2(1-2\nu)}\right] + f_i = 0. (5.27)
Notice that if we choose $c=2(1-\nu)$, then an entire term disappears! So equation 5.25 becomes
2\mu u_i = 2(1-\nu) g_{i,jj} - g_{j,ji}, (5.28)
and the elastic equilibrium equation becomes
(1-\nu) g_{i,kkjj} + f_i = 0. (5.29)
And the original poster of this Undead Thread finally gets the zot.
i like cheese
You should print out this post and show it to your high school English teacher. He or she would be so proud.
