# 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?

• Do you get agreement when you use isotropic materials? Commented Sep 13, 2016 at 23:38
• No. for isotropic material, the result doesn't change much Commented Sep 14, 2016 at 0:06

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 $\epsilon_z =-\nu \epsilon_x$

• So, the last equation tells me, $\delta=\frac{mgl}{2AE}$, which $mg$ can be treated as $F$, right? then is it okay to say, $K_{stru}=\frac{2AE}{l}$? Commented Sep 14, 2016 at 0:05
• Except the force is not applied on the end, but throughout. The stiffness is equivalent to a bar of length $\frac{\ell}{2}$ with the load at the end $$K = \frac{A E}{\ell/2}$$ Commented Sep 14, 2016 at 1:06
• But I'm thinking of adding another force applied on the surface of the free bottom end, as $F_{2}$, and I want to calculate the deformation. When I apply force to merely at the bottom surface, stiffness constant is just $K=\frac{EA}{L}$. If I want to calcuate for deformation of gravity + $F_{2}$, shouldn't I use effective mass $m/2$ for gravitational force $F=mg$? Commented Sep 14, 2016 at 1:48
• Then change the tension to $T(x) = F_2 + \rho g a (\ell -x)$ and integrate again. Commented Sep 14, 2016 at 2:08