The Stiffness Method¶

One Spring Element

Two nodes: \(i, j\)
Nodal displacement: \(u_i, u_j\)
Nodal forces: \(f_i, f_j\)
stiffness constant: \(k\)

Element Equilibrium

Force from the spring: \(F = k (u_j - u_i)\)

\[ \left \{ \begin{aligned} f_i &= -F = k u_i - k u_j \\ f_j &= F = -k u_i + k u_j \end{aligned} \right. \implies \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \begin{bmatrix} u_i \\ u_j \end{bmatrix} = \begin{bmatrix} f_i \\ f_j \end{bmatrix} \]

\(\mathbf{k}\): Element stiffness matrix
\(\mathbf{u}\): Nodal displacement vector / DOF vector
\(\mathbf{f}\): Element nodal force vector

can also be derived from the Principle of Minimum Potential Energy

Spring System¶

Now consider a two-spring system:

\(f_i^m\): the internal force acting on node \(i\) from the spring element \(m\).

\[ \textcolor{skyblue}{ \begin{bmatrix} k_1 & -k_1 \\ -k_1 & k_1 \end{bmatrix}} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} f_1^1 \\ f_2^1 \end{bmatrix}, \, \textcolor{orange}{ \begin{bmatrix} k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} f_2^2 \\ f_3^2 \end{bmatrix} \]

Consider the equilibrium at each node,

\[ \left \{ \begin{aligned} & F_1 = f_1^1 = k_1 u_1 - k_1 u_2 \\ & F_2 = f_2^1 + f_2^2 = -k_1 u_1 + (k_1 + k_2) u_2 - k_2 u_3 \\ & F_3 = f_3^2 = k_2 u_3 - k_2 u_2 \end{aligned} \right. \]

which is

\[ \begin{bmatrix} k_1 & -k_1 & 0 \\ -k_1 & k_1 + k_2 & -k_2 \\ 0 & -k_2 & k_2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \end{bmatrix} \]

looks like the stiffness matrix of the system is the sum of the stiffness matrices of the elements!

Assemble the global stiffness matrix¶

Augment the element stiffness matrices to the size (DOF) of the system:

\[ \begin{aligned} & \begin{bmatrix} k_1 & -k_1 \\ -k_1 & k_1 \end{bmatrix} \implies \begin{bmatrix} k_1 & -k_1 & 0 \\ -k_1 & k_1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\[2ex] & \begin{bmatrix} k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix} \implies \begin{bmatrix} 0 & 0 & 0 \\ 0 & k_2 & -k_2 \\ 0 & -k_2 & k_2 \end{bmatrix} \end{aligned} \]

一般地，对于连接节点 \(i, j\) 的元素

Methods to Apply Fixed Displacements¶

当刚度矩阵奇异时，说明系统中存在刚体运动，需要施加边界条件。

Matrix partitioning¶

\[ \begin{bmatrix} K_{11} & K_{12} \\ K_{21} & K_{22} \end{bmatrix} \begin{bmatrix} d_{\text{fix}} \\ d_{?} \end{bmatrix} = \begin{bmatrix} F_{?} \\ F_{\text{fix}} \end{bmatrix} \]

1) Solve for \(d_{?}\): \(K_{22} d_{?} = F_{\text{fix}} - K_{21} d_{\text{fix}}\) 2) Solve for \(F_{?}\): \(F_{?} = K_{11} d_{\text{fix}} + K_{12} d_{?}\)

Note: for \(d_{\text{fix}} = 0\), we can delete the corresponding rows and columns

挪动固定位移到向量的上面

以 \(3 \times 3\) 的系统为例，假设 \(d_2 = \delta\) 是固定的

\[ \begin{bmatrix} K_{11} & K_{12} & K_{13} \\ K_{21} & K_{22} & K_{23} \\ K_{31} & K_{32} & K_{33} \end{bmatrix} \begin{bmatrix} d_1 \\ \delta \\ d_3 \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \end{bmatrix} \]

交换 \(d_1\) 和 \(d_2\)，需要交换刚度矩阵的 1,2 列：

\[ \begin{bmatrix} K_{12} & K_{11} & K_{13} \\ K_{22} & K_{21} & K_{23} \\ K_{32} & K_{31} & K_{33} \end{bmatrix} \begin{bmatrix} \delta \\ d_1 \\ d_3 \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \end{bmatrix} \]

再交换 \(F_1\) 和 \(F_2\)，需要交换刚度矩阵的 1,2 行：

\[ \begin{bmatrix} K_{22} & K_{21} & K_{23} \\ K_{12} & K_{11} & K_{13} \\ K_{32} & K_{31} & K_{33} \end{bmatrix} \begin{bmatrix} \delta \\ d_1 \\ d_3 \end{bmatrix} = \begin{bmatrix} F_2 \\ F_1 \\ F_3 \end{bmatrix} \]

Pros:

Fewer equations to solve
exactly satisfies constraints

Cons:

Complex implementation (interchanging rows and columns)

Row Substitution¶

Suppose we want to apply \(d_i = \delta\)

\[ \begin{bmatrix} & & & \cdots \\ K_{i1} & K_{i2} & \cdots & K_{ii} & \cdots\\ & & & \cdots \end{bmatrix} \begin{bmatrix} \\ \delta \\ \\ \end{bmatrix} = \begin{bmatrix} \\ F_i \\ \\ \end{bmatrix} \]

Save row \(i\), \(K_{ii} = 1, K_{ij} = 0\):

\[ \begin{bmatrix} & & & \cdots \\ 0 & 0 & \cdots & 1 & \cdots\\ & & & \cdots \end{bmatrix} \begin{bmatrix} \\ d_i \\ \\ \end{bmatrix} = \begin{bmatrix} \\ F_i \\ \\ \end{bmatrix} \]

Penalty Method¶

\[ \begin{bmatrix} & & & \cdots \\ K_{i1} & K_{i2} & \cdots & K_{ii} \textcolor{crimson}{+\text{BIG}} & \cdots\\ & & & \cdots \end{bmatrix} \begin{bmatrix} \\ \delta \\ \\ \end{bmatrix} = \begin{bmatrix} \\ F_i \textcolor{crimson}{+\text{BIG} \cdot \delta} \\ \\ \end{bmatrix} \]

then the matrix is approximately equivalent to

\[ K_{i1} d_1 + K_{i2} d_2 + \cdots + (K_{ii} + \text{BIG}) \delta + \cdots = F_i + \text{BIG} \cdot \delta \]

this is easy to program!