About
Forces Equilibrium
KU+\lambda B^T=F\\ B U=0
\begin{align}
\begin{bmatrix}
K & B^T \\
B & 0
\end{bmatrix}
\begin{bmatrix}
U \\
\lambda
\end{bmatrix}
=
\begin{bmatrix}
F \\
0
\end{bmatrix}
\end{align} How to get Displacement and React Force
\begin{bmatrix}
U \\
\lambda
\end{bmatrix}
=
\begin{bmatrix}
K & B^T \\
B & 0
\end{bmatrix}
^{-1}
\begin{bmatrix}
F \\
0
\end{bmatrix}Objective for Topology Optimization
In Topology Optimization problem, we are going to optimize the shape with respect to an objective, so called “Compliance”.
\min_x C(x) = F^T U(x)
Total Differential of the Equation
\frac{d}{dx} \left( K u + B^T \lambda - F \right) = 0\frac{dK}{dx} u + K \frac{du}{dx} + \frac{dB^T}{dx} \lambda + B^T \frac{d\lambda}{dx} = 0\\
\space \\
\begin{align}
K \frac{du}{dx} + B^T \frac{d\lambda}{dx} = -\left(\frac{dK}{dx} u + \frac{dB^T}{dx} \lambda \right)\space
\end{align} B \frac{du}{dx} + \frac{dB}{dx} u = 0\\
\space\\
\begin{align}
B \frac{du}{dx}=0 \space
\end{align} The (3) means that u is a kernel (in null space) of B. But this is already satisfied since the equation is derived from (4).
\begin{align}
Bu = 0
\end{align} So, (3) and (4) are summarized into (5)
\begin{align}
\begin{bmatrix}
K & B^T \\
B & 0
\end{bmatrix}
\begin{bmatrix}
\frac{du}{dx} \\
\frac{d\lambda}{dx}
\end{bmatrix}
=
-\begin{bmatrix}
\frac{dK}{dx} u + \frac{dB^T}{dx} \lambda \\
B \frac{du}{dx} + \frac{dB}{dx} u
\end{bmatrix}
=
\begin{bmatrix}
-\frac{dK}{dx} \\
0
\end{bmatrix}
\end{align} But, Lower Blocks does not mean anything, right? That just says that zero equals zero. Therefore,
\begin{align}
K \frac{du}{dx} + B^T \frac{d\lambda}{dx} = -\frac{dK}{dx} u\space
\end{align} What we are interested in is that how displacement varies with respect to x. So, the second term in the left is generally ignored.
\begin{align}
K \frac{du}{dx} = -\frac{dK}{dx} u\space
\end{align} As K is the second order form of X, it is differentiated as follows
K=XK_{org}X\\
\frac{dK}{dx}=2K_{org}XFinally, got the derivation of u.
\begin{align}
\frac{du}{dx} \simeq -K^{-1}\frac{dK}{dx} u = -2K^{-1}K_{org}X u \space
\end{align} \\
\text{where } K=XK_{org}X
