Finite element method — nodes, elements, stiffness matrix: Exercises

This page is a self-test ladder for the parent topic. Every problem uses only the ideas built there: the 1D bar element stiffness $K^{e} = \frac{E A}{L} [1 - 1 - 1 1]$ , shape functions, assembly, and boundary conditions. Nothing new is assumed — if a symbol appears, it was defined in the parent.

Throughout, one bar element carries axial displacement only: $1$ degree of freedom (DOF) per node. Symbols reused from the parent:

$E$ = Young's modulus (stiffness of the material, in pascals).
$A$ = cross-section area.
$L$ = element length.
$k \equiv \frac{E A}{L}$ = the "spring constant" of one bar. Units: force / length (N/m).
$d_{i}$ = axial displacement of node $i$ . $F_{i}$ = axial force applied at node $i$ .

Figure — Finite element method — nodes, elements, stiffness matrix

The figure above is our reference structure for most problems: a chain of bars (springs) between numbered nodes, with the leftmost node pinned to a wall.

Level 1 — Recognition

Problem 1.1

Write the element stiffness matrix for a single bar with $E = 70 GPa$ (aluminium), $A = 2 \times 1 0^{- 4} m^{2}$ , $L = 0.5 m$ . Give the numeric value of $k$ .

Recall Solution 1.1

$k = \frac{E A}{L} = \frac{( 70 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 4} )}{0.5}$ . Numerator: $70 \times 1 0^{9} \times 2 \times 1 0^{- 4} = 1.4 \times 1 0^{7}$ N. Divide by $0.5$ m: $k = 2.8 \times 1 0^{7} N/m .$ So $K^{e} = 2.8 \times 1 0^{7} [1 - 1 - 1 1] N/m .$

Problem 1.2

A $3$ -node, $2$ -bar structure has been assembled into $K_{global} = k 1 - 1 0 - 1 2 - 1 0 - 1 1 .$ Which node connects to two elements, and how do you know just by reading the matrix?

Recall Solution 1.2

The diagonal entry $K_{ii}$ counts how much total stiffness pulls on node $i$ . Nodes $1$ and $3$ have diagonal $1$ (one bar each); node $2$ has diagonal $2$ (two bars add up). So node 2 connects to both elements. The off-diagonal $K_{25} = 0$ -style zeros also tell you node $1$ and node $3$ are not directly connected (no bar between them).

Level 2 — Application

Problem 2.1

A single aluminium bar ( $k = 2.8 \times 1 0^{7}$ N/m from Problem 1.1) is fixed at node $1$ and pulled with $F_{2} = 5600$ N at node $2$ . Find $d_{2}$ and the stress $σ$ if $A = 2 \times 1 0^{- 4} m^{2}$ .

Recall Solution 2.1

Fixed node $1$ means $d_{1} = 0$ . Keep only the free DOF (node $2$ ): $k d_{2} = F_{2} ⟹ d_{2} = \frac{5600}{2.8 \times 1 0 ^{7}} = 2 \times 1 0^{- 4} m = 0.2 mm .$ Strain (parent Step 2, with $d_{1} = 0$ ): $ϵ = \frac{d _{2} - d _{1}}{L} = \frac{2 \times 1 0 ^{- 4}}{0.5} = 4 \times 1 0^{- 4}$ . Stress (Hooke, parent Step 3): $σ = E ϵ = 70 \times 1 0^{9} \times 4 \times 1 0^{- 4} = 2.8 \times 1 0^{7}$ Pa $= 28$ MPa. Cross-check: $σ = F / A = 5600/ (2 \times 1 0^{- 4}) = 2.8 \times 1 0^{7}$ Pa. ✓ See stress & strain.

Problem 2.2

Two identical bars in series, each $k$ , nodes $1$ – $2$ – $3$ . Node $1$ fixed, load $P$ at node $3$ . Reproduce the reduced system and solve for $d_{2}, d_{3}$ .

Recall Solution 2.2

Delete row/column $1$ (fixed) from the global matrix: $k [2 - 1 - 1 1] [d_{2} d_{3}] = [0 P] .$ Row 1: $2 k d_{2} - k d_{3} = 0 \Rightarrow d_{3} = 2 d_{2}$ . Row 2: $- k d_{2} + k d_{3} = P \Rightarrow - k d_{2} + 2 k d_{2} = P \Rightarrow d_{2} = \frac{P}{k}$ , and $d_{3} = \frac{2 P}{k}$ . Physical check: two springs in series stretch twice as far at the free end. ✓

Level 3 — Analysis

Problem 3.1

Two bars in series but with different stiffnesses: element A ( $k_{A}$ ) between nodes $1$ – $2$ , element B ( $k_{B}$ ) between nodes $2$ – $3$ . Node $1$ fixed, load $P$ at node $3$ . Assemble, then solve for $d_{2}, d_{3}$ .

Recall Solution 3.1

Element A on DOFs (1,2), element B on DOFs (2,3). Node $2$ gets both diagonals: $K_{global} = k_{A} - k_{A} 0 - k_{A} k_{A} + k_{B} - k_{B} 0 - k_{B} k_{B} .$ Reduce (drop node 1): $[k_{A} + k_{B} - k_{B} - k_{B} k_{B}] [d_{2} d_{3}] = [0 P] .$ Row 1: $(k_{A} + k_{B}) d_{2} = k_{B} d_{3} \Rightarrow d_{2} = \frac{k _{B}}{k _{A} + k _{B}} d_{3}$ . Row 2: $- k_{B} d_{2} + k_{B} d_{3} = P$ . Substitute: $k_{B} d_{3} (1 - \frac{k _{B}}{k _{A} + k _{B}}) = P \Rightarrow k_{B} d_{3} \cdot \frac{k _{A}}{k _{A} + k _{B}} = P .$ $d_{3} = P \cdot \frac{k _{A} + k _{B}}{k _{A} k _{B}} = P (\frac{1}{k _{A}} + \frac{1}{k _{B}}) .$ That is the series-spring rule: total flexibility = sum of flexibilities. And $d_{2} = \frac{k _{B}}{k _{A} + k _{B}} d_{3} = \frac{P}{k _{A}}$ . Sanity: if $k_{A} = k_{B} = k$ , $d_{3} = 2 P / k$ , $d_{2} = P / k$ — matches Problem 2.2. ✓

Problem 3.2

Show the un-reduced $2 \times 2$ single-bar $K^{e}$ is singular and interpret the zero-determinant physically.

Recall Solution 3.2

$det [1 - 1 - 1 1] = (1) (1) - (- 1) (- 1) = 1 - 1 = 0.$ A zero determinant means $K^{e}$ has no inverse: $Kd = F$ has either no solution or infinitely many. Physically, with no support, sliding both nodes by the same amount ( $d_{1} = d_{2} = c$ ) produces zero strain ( $ϵ = (d_{2} - d_{1}) / L = 0$ ) and therefore zero force — a rigid-body translation costs no energy. The vector $[11]$ is the null-space direction. Boundary conditions exist precisely to kill this mode (parent's second [!mistake]).

Level 4 — Synthesis

Problem 4.1

A stepped rocket strut: bar A ( $E$ , $A_{1} = 4 \times 1 0^{- 4} m^{2}$ , $L = 1 m$ ) then bar B ( $E$ , $A_{2} = 2 \times 1 0^{- 4} m^{2}$ , $L = 1 m$ ), $E = 70$ GPa, in series, nodes $1$ – $2$ – $3$ . Node $1$ fixed, $P = 1.4 \times 1 0^{4}$ N at node $3$ . Find $d_{2}$ , $d_{3}$ , and the stress in each bar. Which bar is more stressed?

Recall Solution 4.1

$k_{A} = \frac{E A _{1}}{L} = \frac{70 \times 1 0 ^{9} \cdot 4 \times 1 0 ^{- 4}}{1} = 2.8 \times 1 0^{7}$ N/m. $k_{B} = \frac{E A _{2}}{L} = \frac{70 \times 1 0 ^{9} \cdot 2 \times 1 0 ^{- 4}}{1} = 1.4 \times 1 0^{7}$ N/m. Use Problem 3.1 results: $d_{2} = \frac{P}{k _{A}} = \frac{1.4 \times 1 0 ^{4}}{2.8 \times 1 0 ^{7}} = 5 \times 1 0^{- 4}$ m. $d_{3} = P (\frac{1}{k _{A}} + \frac{1}{k _{B}}) = 1.4 \times 1 0^{4} (\frac{1}{2.8 \times 1 0 ^{7}} + \frac{1}{1.4 \times 1 0 ^{7}})$ . $= 1.4 \times 1 0^{4} (3.5714 \times 1 0^{- 8} + 7.1429 \times 1 0^{- 8}) = 1.4 \times 1 0^{4} \times 1.0714 \times 1 0^{- 7} = 1.5 \times 1 0^{- 3}$ m. Stress bar A: $ϵ_{A} = (d_{2} - 0) / L = 5 \times 1 0^{- 4}$ , $σ_{A} = E ϵ_{A} = 70 \times 1 0^{9} \times 5 \times 1 0^{- 4} = 3.5 \times 1 0^{7}$ Pa $= 35$ MPa. Stress bar B: $ϵ_{B} = (d_{3} - d_{2}) / L = (1.5 \times 1 0^{- 3} - 5 \times 1 0^{- 4}) /1 = 1.0 \times 1 0^{- 3}$ , $σ_{B} = 70 \times 1 0^{9} \times 1.0 \times 1 0^{- 3} = 7.0 \times 1 0^{7}$ Pa $= 70$ MPa. Cross-check with $σ = F / A$ : same internal force $P = 1.4 \times 1 0^{4}$ N runs through both (series). Bar A: $1.4 \times 1 0^{4} /4 \times 1 0^{- 4} = 3.5 \times 1 0^{7}$ Pa ✓. Bar B: $1.4 \times 1 0^{4} /2 \times 1 0^{- 4} = 7.0 \times 1 0^{7}$ Pa ✓. Bar B is more stressed — the thinner section. This is why cutouts and thin walls drive design; see buckling.

Problem 4.2

For a single free-free bar of mass — imagine it discretised as one element with a lumped mass $m /2$ at each node. Without full modal machinery, argue from the singular stiffness why a free-free bar has a zero-frequency (rigid-body) mode, then state how many rigid-body modes a free unconstrained 3D structure has.

Recall Solution 4.2

A natural frequency satisfies $K ϕ = ω^{2} M ϕ$ (see modal analysis). The rigid-translation vector $ϕ = [11]$ gives $K ϕ = 0$ (from Problem 3.2's null space), so $ω^{2} M ϕ = 0$ , forcing $ω = 0$ . A zero-frequency mode = motion with no elastic restoring force = rigid-body mode. In free 3D space an unconstrained body has 6 rigid-body modes: 3 translations + 3 rotations. A valid analysis of a supported structure must remove all of them via boundary conditions.

Level 5 — Mastery

Problem 5.1

Three bars from a wall to a single free node arranged as a parallel bundle: nodes $1, 2, 3$ all fixed to the wall ( $d_{1} = d_{2} = d_{3} = 0$ ) each connected by its own bar to a common free node $4$ , with stiffnesses $k_{1}, k_{2}, k_{3}$ acting along the same axis. A load $P$ is applied at node $4$ . Assemble the reduced equation and find $d_{4}$ . Then evaluate for $k_{1} = k_{2} = k_{3} = k$ .

Recall Solution 5.1

Each bar contributes $+ k_{i}$ to the diagonal of the shared free node $4$ (all three pull on node $4$ ). Fixed nodes $1, 2, 3$ drop out. The single free DOF: $(k_{1} + k_{2} + k_{3}) d_{4} = P ⟹ d_{4} = \frac{P}{k _{1} + k _{2} + k _{3}} .$ Parallel springs add stiffnesses — contrast with the series result of Problem 3.1 where flexibilities added. For $k_{1} = k_{2} = k_{3} = k$ : $d_{4} = \frac{P}{3 k}$ — the bundle is three times stiffer than one bar, so it deflects one-third as much.

Problem 5.2

Convergence reasoning. A tapered bar (area shrinking linearly $A_{0} \to A_{0} /2$ over length $L$ ) is modelled by one constant-area bar of area $\overset{ˉ}{A} = 0.75 A_{0}$ (the average). The exact tip displacement under end load $P$ is $d_{exact} = \frac{P L}{E A _{0}} ln 2 \approx 0.6931 \frac{P L}{E A _{0}} .$ Compute the one-element FEM estimate and its percent error. Then state what h-refinement would do.

Recall Solution 5.2

One-element model: $k = \frac{E A ˉ}{L} = \frac{E ( 0.75 A _{0} )}{L}$ , so $d_{FEM} = \frac{P}{k} = \frac{P L}{0.75 E A _{0}} = 1.3333 \frac{P L}{E A _{0}} .$ Wait — compare like for like using $\frac{P L}{E A _{0}}$ as the unit. FEM coefficient $= 1/0.75 = 1.3333$ ; exact coefficient $= ln 2 = 0.6931$ . Percent error $= \frac{1.3333 - 0.6931}{0.6931} \times 100% \approx 92.4%$ . This is huge: a single average-area element badly mis-models a strong taper because our linear shape function forces constant strain, but a taper has strain rising toward the thin end. h-refinement (splitting into many short bars, each nearly constant area) makes each element's constant-strain assumption locally valid; the summed displacement converges to $0.6931 P L / (E A_{0})$ . See numerical methods and, for temperature-driven area/property changes, thermal analysis.

Recall Quick self-quiz

Single bar stiffness $k$ equals? ::: $E A / L$ Diagonal entry of assembled $K$ at a node tells you? ::: total stiffness (number/strength of bars) meeting there Why is un-constrained $K$ singular? ::: rigid-body motion costs zero strain energy → zero-determinant null space Series springs combine by adding? ::: flexibilities ( $1/ k$ ) Parallel springs combine by adding? ::: stiffnesses ( $k$ ) How many rigid-body modes for a free 3D body? ::: 6 (3 translations + 3 rotations) A linear bar element can represent only what kind of strain? ::: constant strain