Chapter 4: Finite Element Method

Force vector:

Displacement vector:

Stiffness matrix:

For element 1 (nodes 1, 3):

For element 2 (nodes 2, 4):

For element 3 (nodes 3, 4):

Now, writing global stiffness matrix by superposition:

Force-displacement relationship: $\{f\} = [K]\{u\}$

Eliminating rows and columns associated with $u = 0$ (boundary conditions), i.e., rows/columns 1 & 2 (since $u_1 = 0$ and $u_2 = 0$), and writing reduced matrix equation for unknown DOFs ($u_3$ and $u_4$):

Solving the 2×2 system using matrix inversion formula: $\{u\} = [K_r]^{-1}\{F_r\}$.

For a 2×2 matrix, the inverse is: $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

(c) Reaction forces: Back-substitute the displacements into the full global equation $\{f\} = [K]\{u\}$:

From Row 1: $R_1 = 1000 \times (1 \times 0 + 0 \times 0 - 1 \times \frac{15}{11} + 0 \times \frac{20}{11})$

From Row 2: $R_2 = 1000 \times (0 + 2 \times 0 + 0 \times \frac{15}{11} - 2 \times \frac{20}{11})$

Equilibrium check: $\sum F = R_1 + R_2 + P = -1363.63 - 3636.36 + 5000 \approx 0$   $\checkmark$

Forces in individual springs:

Element 1:

Element 2:

Element 3:

Given: 5 nodes, 4 spring elements (all with $k = 200$ kN/m), $u_1 = 0$ (fixed), $u_5 = 20$ mm $= 0.02$ m (known displacement), no external load on nodes 2, 3, 4.

Element connectivity:

• Element 1: nodes 1 – 2
• Element 2: nodes 2 – 3
• Element 3: nodes 3 – 4
• Element 4: nodes 4 – 5

Force vector: (Reactions at constrained nodes; no external load at free nodes)

Displacement vector: ($u_5 = 20$ mm $= 0.02$ m)

(a) Global Stiffness Matrix:

For each element, the stiffness matrix is given by:

Element 1 (nodes 1, 2):

Element 2 (nodes 2, 3):

Element 3 (nodes 3, 4):

Element 4 (nodes 4, 5):

Assembling the $5 \times 5$ global stiffness matrix by direct stiffness method (superimposing element contributions at corresponding DOFs):

Note: Diagonal entries at interior nodes are $k + k = 400$, and end nodes have just $k = 200$.

(b) Force-displacement relationship: $\{f\} = [K]\{u\}$

Eliminating rows/columns for known DOFs ($u_1 = 0$ and $u_5 = 0.02$ m) and writing equations for rows 2, 3, 4:

Row 2: $0 = 400 u_2 - 200 u_3$

Row 3: $0 = -200 u_2 + 400 u_3 - 200 u_4$

Row 4: $0 = -200 u_3 + 400 u_4 - 200 \times 0.02$

The term $-200 \times 0.02 = -4$ kN from the known displacement $u_5$ is moved to the left-hand side:

Solving the 3×3 system:

From Row 1: $400 u_2 = 200 u_3 \Rightarrow u_2 = 0.5\, u_3$   — (i)

From Row 2: $-200 u_2 + 400 u_3 - 200 u_4 = 0$

Substituting (i): $-200(0.5\,u_3) + 400 u_3 - 200 u_4 = 0 \Rightarrow 300 u_3 = 200 u_4 \Rightarrow u_3 = \frac{2}{3}u_4$   — (ii)

From Row 3: $-200 u_3 + 400 u_4 = 4$

Substituting (ii): $-200 \times \frac{2}{3}u_4 + 400 u_4 = 4 \Rightarrow \frac{800}{3} u_4 = 4 \Rightarrow u_4 = 0.015$ m

Back-substituting: $u_3 = \frac{2}{3} \times 0.015 = 0.01$ m   and   $u_2 = 0.5 \times 0.01 = 0.005$ m

(c) Global nodal forces (Reactions):

Using the original full equation $\{f\} = [K]\{u\}$ with all displacements now known: $\{u\} = \{0, 0.005, 0.01, 0.015, 0.02\}^T$ m.

Row 1: $R_1 = 200 \times 0 - 200 \times 0.005 = -1$ kN

Row 5: $R_5 = -200 \times 0.015 + 200 \times 0.02 = -3 + 4 = 1$ kN

Equilibrium check: $R_1 + R_5 = -1 + 1 = 0$   $\checkmark$ (No external loads applied, so net force must be zero.)

(d) Forces in individual springs:

Using $\{f^e\} = [K^e]\{u^e\}$ for each element:

Element 1 (nodes 1–2): $u_1 = 0$, $u_2 = 0.005$ m

Element 2 (nodes 2–3): $u_2 = 0.005$, $u_3 = 0.01$ m

Element 3 (nodes 3–4): $u_3 = 0.01$, $u_4 = 0.015$ m

Element 4 (nodes 4–5): $u_4 = 0.015$, $u_5 = 0.02$ m

Given: Two bar elements with $A_1 = A_2 = 250$ mm², $L_1 = L_2 = 150$ mm, $E = 20 \times 10^3$ N/mm². Load $P = 60$ kN at node 2. Gap of 1.2 mm between node 3 and the right wall. Node 1 is fixed ($u_1 = 0$).

Step 1: Identify unknowns and boundary conditions.

Initially, we assume the gap is not closed. Node 3 is free, so $F_3 = 0$.

Force vector:

Displacement vector:

Step 2: Element stiffness matrices.

Using $[K^e] = \frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$:

Step 3: Global stiffness matrix (assembling by direct stiffness method):

Step 4: Apply boundary conditions and solve.

$\{F\} = [K]\{u\}$:

Eliminating row/column 1 (since $u_1 = 0$), the reduced equations are:

From equation (ii): $u_3 = u_2$. Substituting into (i):

Step 5: Check if the gap closes.

The displacement at node 3 is $u_3 = 1.8$ mm, but the gap is only 1.2 mm. Since $u_3 > 1.2$ mm, the right end touches the wall and cannot displace further. The wall develops a reaction force $R_3$.

Step 6: Re-solve with gap constraint $u_3 = 1.2$ mm.

From the global equation, Row 2 gives:

Step 7: Back-substitute to find reactions.

Row 1: $R_1 = \frac{10^5}{3}(0 - u_2 + 0) = \frac{10^5}{3}(-1.5)$

Row 3: $R_3 = \frac{10^5}{3}(0 - u_2 + u_3) = \frac{10^5}{3}(-1.5 + 1.2)$

Equilibrium check: $R_1 + P + R_3 = -50 + 60 + (-10) = 0$   $\checkmark$

Step 8: Element stresses.

Using $\sigma = \frac{E}{L}(u_2 - u_1)$ for each element:

Element 1 (nodes 1–2):

Element 2 (nodes 2–3):

Force vector:

Displacement vector: $\{u\} = \begin{Bmatrix} 0 \\ u_2 \end{Bmatrix}$

Stiffness matrix:

$\{F\} = [K]\{u\}$:

Stress:

Verification from analytical relation at $x = 4000$:

At C.G. of element ($x = 2000$):

Load vector:

Stiffness matrix:

Reduced matrix equation:

Verification: At Node 2 ($x=2000$): $u_2 = 0.0075$ mm $\checkmark$   At Node 3 ($x=4000$): $u_3 = 0.01$ mm $\checkmark$

Stresses:

Verified at C.G. of element 1 ($x = 1000$): $\sigma = 0.075$ N/mm² $\checkmark$

Verified at C.G. of element 2 ($x = 3000$): $\sigma = 0.025$ N/mm² $\checkmark$

Given data:

$\Delta t = 80$°C, $P_1 = -60$ kN at node 2 (leftward), $P_2 = -75$ kN at node 3 (leftward). Both ends fixed: $u_1 = u_4 = 0$.

Step 1: Element thermal load vectors.

Using $\{f_t^e\} = AE\alpha \Delta t \begin{Bmatrix} -1 \\ 1 \end{Bmatrix}$:

Element 1: $A_1 E_1 \alpha_1 \Delta t = 2400 \times 83 \times 10^3 \times 18.9 \times 10^{-6} \times 80$

Element 2: $A_2 E_2 \alpha_2 \Delta t = 1200 \times 70 \times 10^3 \times 23 \times 10^{-6} \times 80$

Element 3: $A_3 E_3 \alpha_3 \Delta t = 600 \times 200 \times 10^3 \times 11.7 \times 10^{-6} \times 80$

Step 2: Assemble the global temperature load vector by adding element contributions at shared nodes:

Step 3: External force vector (forces at nodes 2 & 3 are leftward, i.e., negative):

Step 4: Global force vector = External + Temperature:

Displacement vector: $\{u\} = \begin{Bmatrix} 0 \\ u_2 \\ u_3 \\ 0 \end{Bmatrix}$

Step 5: Element stiffness matrices.

Using $k_i = \frac{A_i E_i}{L_i}$:

Step 6: Global stiffness matrix (assembling):

Step 7: Apply boundary conditions and solve.

$\{F\} = [K]\{u\}$. Eliminating rows/columns for $u_1 = 0$ and $u_4 = 0$:

Using matrix inversion: $\det = 10^{10}(3.89 \times 4.4 - 1.4 \times 1.4) = 10^{10}(17.116 - 1.96) = 15.156 \times 10^{10}$

Step 8: Support reactions.

Row 1 of the full equation: $F_1 = 10^5(2.49 \times 0 - 2.49 \times 0.22124) = 10^5 \times (-0.55089)$

Row 4: $F_4 = 10^5(-3 \times (-0.00406) + 3 \times 0) = 10^5 \times 0.01218$

Equilibrium check: $R_1 + (-60{,}000) + (-75{,}000) + R_4 = 246{,}101.4 - 60{,}000 - 75{,}000 - 111{,}102 \approx 0$   $\checkmark$

Step 9: Element stresses (with temperature effect).

Using $\sigma = E\left[\frac{1}{L}(u_j - u_i)\right] - E\alpha\Delta t$:

Element 1 ($u_1 = 0$, $u_2 = 0.22124$):

Element 2 ($u_2 = 0.22124$, $u_3 = -0.00406$):

Element 3 ($u_3 = -0.00406$, $u_4 = 0$):

Here $A_1 = 150 \times 25 = 3750$ mm², $A_2 = 75 \times 25 = 1875$ mm².

Load vector:

Stiffness matrix ($\bar{A} = 2812.5$ mm²):

Nodes at top, middle, and bottom: $A_1 = 3750$, $A_2 = 2812.5$, $A_3 = 1875$ mm².

Body force:

Stiffness matrices ($\bar{A}_1 = 3281.25$, $\bar{A}_2 = 2343.75$ mm²):

Reduced equation:

Stresses:

Strains:

Element forces: