Exercises — FEM for structures — assembling global stiffness

The bar element stiffness is ${\mathbf{k}}^{(e)}=\frac{EA}{L}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right].$ Where this comes from: the bar force is $F=\frac{EA}{L}(u_j-u_i)$. The force on node $i$ is $\frac{EA}{L}(u_i-u_j)$, the force on node $j$ is $\frac{EA}{L}(u_j-u_i)$. Writing those two equations as a matrix gives the pattern above.

Row 1, column 2 = $-\frac{EA}{L}$. Meaning: the force appearing at DOF $i$ when we move DOF $j$ by one unit and hold DOF $i$ fixed. It is negative because pulling node $j$ away pulls node $i$ in the same direction — the bar drags it. That is the whole meaning of a stiffness entry $K_{ij}$: force at $i$ per unit displacement at $j$.

Only $K_{22}$. Look at the shared DOF: both elements list DOF 2. An entry $K_{ij}$ gets a contribution from an element only if both $i$ and $j$ are DOFs of that element. Element 1 owns DOFs $\{1,2\}$, element 2 owns $\{2,3\}$. The only pair $(i,j)$ living inside both sets is $(2,2)$. So $K_{22}$ is summed twice; every other entry comes from at most one element. (This is exactly why $K_{22}=2$ appears in the parent's two-bar truss.)

Let $k=\frac{EA}{L}$. Each bar contributes $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ into its two DOF rows/columns.

• Bar 1 → DOFs $[1,2]$
• Bar 2 → DOFs $[2,3]$
• Bar 3 → DOFs $[3,4]$

Scatter-add. Interior nodes 2 and 3 are each shared by two bars, so their diagonals become $1+1=2$: $\mathbf{K}=k\left[\begin{array}{cccc}1& -1& 0& 0\\ -1& 2& -1& 0\\ 0& -1& 2& -1\\ 0& 0& -1& 1\end{array}\right].$ Check the pattern: every row sums to zero. That is the rigid-body signature — slide the whole chain equally and no bar stretches, so no force. Good sign the assembly is correct.

Delete row/column 1 (fixed DOF). The reduced system, with $k=\frac{EA}{L}$: $k\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right]\left[\begin{array}{c}{u}_2\\ u_3\\ u_4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ F\end{array}\right].$ Row 1: $2u_2-u_3=0\Rightarrow u_3=2u_2$. Row 2: $-u_2+2u_3-u_4=0\Rightarrow u_4=-u_2+2u_3=-u_2+4u_2=3u_2$. Row 3: $-u_3+u_4=\frac{F}{k}\Rightarrow -2u_2+3u_2=u_2=\frac{F}{k}=\frac{FL}{EA}$. So $u_2=\frac{FL}{EA},u_3=\frac{2FL}{EA},u_4=\frac{3FL}{EA}.$ Physical check: three bars in series each carry the full force $F$, each stretches $\frac{FL}{EA}$, so displacements grow $1,2,3$ times down the chain. ✓

Every bar is $k\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$ on its two DOFs:

• A–C → DOFs $[2,1]$
• B–C → DOFs $[3,1]$
• D–C → DOFs $[4,1]$

C (DOF 1) sits in all three elements, so its diagonal collects $1+1+1=3$. Each outer node sits in one bar (diagonal 1) and couples only to C ($-1$): $\mathbf{K}=k\left[\begin{array}{cccc}3& -1& -1& -1\\ -1& 1& 0& 0\\ -1& 0& 1& 0\\ -1& 0& 0& 1\end{array}\right],K_{11}=3k.$ Insight: the diagonal entry at a node equals the count of stiffness contributions meeting there. A hub node connected to many elements gets a large diagonal — the more springs pull on you, the harder you are to move. Note $K_{23}=K_{24}=K_{34}=0$: outer nodes are not in the same element, so they never talk directly, only through C.

Symmetry: each element matrix is symmetric ($k_{ij}=k_{ji}$, the Maxwell–Betti reciprocity). Scatter-add of symmetric blocks into the same $(i,j)$ and $(j,i)$ slots preserves symmetry, so $\mathbf{K}={\mathbf{K}}^T$.

Why $K_{23}=0$: DOF 2 (node A) and DOF 3 (node B) never appear together in any single element. No element = no coupling = zero entry. See Sparse Matrix Storage — this is why $\mathbf{K}$ is mostly zeros.

General rule: $K_{ij}\ne 0$ iff DOFs $i$ and $j$ belong to at least one common element. Otherwise $K_{ij}=0$.

DOF ordering (state it before reading entries): the four rows/columns of the element matrix are, in order, $[u_{1x},u_{1y},u_{2x},u_{2y}],$ i.e. node 1's $x$-displacement, node 1's $y$-displacement, then node 2's $x$ and $y$. Every entry $k_{ab}$ below is "force in DOF $a$ per unit displacement of DOF $b$" using exactly this list.

Let $c=\cos \theta$, $s=\sin \theta$ where $\theta$ is the bar's angle from the $x$-axis. For a 45° bar, $c=s=\frac{1}{\sqrt{2}}$, so $c^2=s^2=cs=\frac{1}{2}$.

Why we need $c$ and $s$: the bar only resists stretching along its own axis. A global displacement $(u_x,u_y)$ contributes to axial stretch only through its projection onto the bar direction, and that projection uses $\cos$ and $\sin$ (this is Coordinate Transformations in FEM). The standard result, with rows/columns ordered $[u_{1x},u_{1y},u_{2x},u_{2y}]$: ${\mathbf{k}}_{\text{global}}^{(e)}=\frac{EA}{L}\left[\begin{array}{cccc}{c}^2& cs& -c^2& -cs\\ cs& s^2& -cs& -s^2\\ -c^2& -cs& c^2& cs\\ -cs& -s^2& cs& s^2\end{array}\right].$ Plug in $c^2=s^2=cs=\frac{1}{2}$: ${\mathbf{k}}_{\text{global}}^{(e)}=\frac{EA}{2L}\left[\begin{array}{cccc}1& 1& -1& -1\\ 1& 1& -1& -1\\ -1& -1& 1& 1\\ -1& -1& 1& 1\end{array}\right].$ Sanity check: each row still sums to zero (rigid translation → no force), and the $x$–$y$ entries are now coupled ($cs\ne 0$) precisely because the bar is tilted.

Let $g=\frac{EA}{2L}$.

Bar A (45°, node 1→2), DOFs $[1,2,3,4]$ ordered $[u_{1x},u_{1y},u_{2x},u_{2y}]$: the last $2\times 2$ block (the node-2 self-block, rows/cols $[u_{2x},u_{2y}]$) is $g\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$.

Bar B (−45°, node 2→3): here $c=\frac{1}{\sqrt{2}}$, $s=-\frac{1}{\sqrt{2}}$, so $c^2=s^2=\frac{1}{2}$ but $cs=-\frac{1}{2}$. Its element matrix, rows/cols ordered $[u_{2x},u_{2y},u_{3x},u_{3y}]$ → DOFs $[3,4,5,6]$: $g\left[\begin{array}{cccc}1& -1& -1& 1\\ -1& 1& 1& -1\\ -1& 1& 1& -1\\ 1& -1& -1& 1\end{array}\right].$ Its node-2 self-block (first $2\times 2$, rows/cols $[u_{2x},u_{2y}]$) is $g\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$.

Node 2 is shared → add the two self-blocks: $\mathbf{K}[3:4,3:4]=g\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]+g\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]=g\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\frac{EA}{L}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].$ So the four entries are $K_{33}=K_{44}=\frac{EA}{L}$ and $K_{34}=K_{43}=0$. Beautiful check: the cross-coupling cancels ($+1$ from bar A, $-1$ from bar B), leaving node 2 equally stiff in $x$ and $y$. The two symmetric diagonals brace it evenly — physically obvious once you see the figure.

(a) The un-reduced $\mathbf{K}$ has a 1-dimensional null space, so rank 3 (out of 4). The null vector is $\mathbf{u}=[1,1,1,1{]}^T$: sliding every node the same amount stretches no bar, gives zero strain energy, hence $\mathbf{Ku}=0$. This is the single rigid-body translation mode of a 1D chain — see Finite Element Method Overview.

(b) Energy: ${\mathbf{u}}^T\mathbf{Ku}=2U={\sum }_e\frac{EA}{L}(\Delta L_e{)}^2\ge 0$ always. Once node 1 is fixed, the only way to get zero strain in every bar is $u_2=u_3=u_4=0$ (each bar's stretch depends on a fixed neighbor chain back to the wall). So ${\mathbf{u}}^T{\mathbf{K}}_r\mathbf{u}>0$ for every nonzero $\mathbf{u}$ → positive definite → invertible → unique displacement solution. This is why solvers (Direct vs Iterative Solvers) only work after boundary conditions.

Determinant check: with $k=\frac{EA}{L}$, the reduced matrix is $k\left[\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 1\end{array}\right]$. Its determinant is $k^3\cdot 1=k^3>0$, and all leading minors ($2k>0$, $3k^2>0$, $k^3>0$) are positive → positive definite by Sylvester's criterion. ✓

Half-bandwidth $b$ = the largest $|i-j|$ over all nonzero $K_{ij}$. A bar couples the two DOFs at its ends, so $b$ is the largest label-gap between two physically-adjacent nodes.

(a) Consecutive numbering $1,2,3,4,5$. Each bar joins physically-adjacent nodes, whose global labels differ by exactly 1. So the worst nonzero off-diagonal has $|i-j|=1$ → $b=1$ (a tridiagonal matrix). Minimal storage.

(b) Scrambled labelling $1,3,5,4,2$ (physical node $1\to$ label 1, node $2\to$ label 3, node $3\to$ label 5, node $4\to$ label 4, node $5\to$ label 2). Now check each bar (physical neighbours) by its two labels:

• bar 1–2: labels $1,3$ → gap $2$
• bar 2–3: labels $3,5$ → gap $2$
• bar 3–4: labels $5,4$ → gap $1$
• bar 4–5: labels $4,2$ → gap $2$

The maximum gap is $2$, so $b=2$: the band is twice as wide as the consecutive case, even though the physics (the same 4 bars, same nonzero count) is unchanged.

Lesson: the same structure gives a wider band purely from worse numbering, and a wider band costs more memory and more solve time (direct factorization fills in within the band). Good node ordering — bandwidth-minimizing schemes like Cuthill–McKee — is therefore free performance, a real concern once meshes are refined (Mesh Refinement and Convergence, Sparse Matrix Storage, Direct vs Iterative Solvers).