Fixed rectangular structure and gravitational force Suppose I have rectangular structure hanging from the ceiling.

It has dimension of height $H$, and square area $A$. The top is fixed to the ceiling, and gravitational force is pulling the structure downward.
In order to calculate how much height $H$ changes, I started with 
$$
[\epsilon]=[S][\sigma]
\longrightarrow
$$
$$
\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
{\frac{{\Delta x}}{L}}\\
{\frac{{\Delta y}}{W}}\\
{\frac{{\Delta z}}{H}}\\
 \vdots 
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{s_{11}}}&{{s_{12}}}&{{s_{12}}}& \cdots \\
{{s_{12}}}&{{s_{11}}}&{{s_{12}}}& \cdots \\
{{s_{12}}}&{{s_{12}}}&{{s_{11}}}& \cdots \\
 \vdots & \vdots & \vdots & \ddots 
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0\\
0\\
{{\sigma _z}}\\
 \vdots 
\end{array}} \right]\;\;\;\;\; \to \;\;\;\;\;\frac{{\Delta z}}{H} = {s_{11}}{\sigma _z} = \frac{{{c_{11}} + {c_{12}}}}{{\left( {{c_{11}} - {c_{12}}} \right)\left( {{c_{11}} + 2{c_{12}}} \right)}}\frac{F}{LW}\\
 \to \;\;\;\;\;\Delta x = \frac{{0.0077}}{{GPa}}\frac{H}{LW}F\;\;\;\;\; \to \;\;\;\;\;{K_{str}} = \frac{{130GPa}}{H}LW
\end{array}
$$
Here I assumed that the material is anisotropic silicon.
Then,
$$
\Delta x=\frac{F}{K_{str}}=\frac{mg}{K_{str}}
$$
If I set $H=100 mm$,$L=W=10 mm$, silicon density $\rho=2330kg$ and calculate
$\Delta x$, it is $1.7557nm$.
But, if I do the FEM simulation, I get $0.86895nm$

It is approximately half of my calculation!
Where did this discrepancy come from?
 A: The tension (stresses) is not constant along the bar, so you cannot use $F = k \delta$. 
Take a small segment ${\rm d}x$ and notice that the tension varies by $${\rm d}T = -\rho g A {\rm d}x$$
$\rho$ is density, $A$ is section area, $g$ is gravity
Here $x$ varies from the root to the end at $x=\ell$. The tension as a function of location is $$T(x) = \int {\rm d}T = T_0 - \rho g A x $$ and since there is no tension at the end (nothing pulling it), the end condition is $T(\ell)=0$ or $T_0 = \rho g A \ell$. Now the exact tension is
$$ T(x) = \rho g A \left(\ell -x\right) $$
The tension is known, and so is the stress $\sigma_x(x) = \frac{T(x)}{A} = \rho g (\ell-x)$. From this you get the strain along the x direction $\epsilon_x(x) = \sigma_x(x)/E^\star$ and the total deflection is
$$ \delta = \int \limits_0^\ell \epsilon_x(x) {\rm d}x = \int \limits_0^\ell \frac{\rho g (\ell -x)}{E^\star} {\rm d}x = \frac{\rho f \ell^2}{2 E^\star} = \frac{m g \ell}{2 A E^\star} $$

Stresses in the other directions should be zero $\sigma_y=0$, $\sigma_z=0$ and hence strains in the other directions are $\epsilon_y =-\nu \epsilon_x$ and $\epsilon_z =-\nu \epsilon_x$
