Simply supported beam with two elements

Data

$EI = 200\times10^9 \times 8\times10^{-6} = 1.6\times10^6\ \text{N m}^2$, $L_e = 2$ m for each element.

$\dfrac{EI}{L_e^3} = \dfrac{1.6\times10^6}{8} = 2\times10^5$ N/m.

Nodes: 1 (left support), 2 (mid-span), 3 (right support). Degrees of freedom: $(v_1, \theta_1, v_2, \theta_2, v_3, \theta_3)$. Boundary conditions: $v_1 = v_3 = 0$.

   ^              10 kN
   |               |
   o======= (1) ===o======= (2) ===o
  1                2                3
  |<------ 2 m ---->|<------ 2 m -->|

Element matrix (each element)

With $L_e = 2$:

$$[k] = 2\times10^5\begin{bmatrix} 12 & 12 & -12 & 12 \\ 12 & 16 & -12 & 8 \\ -12 & -12 & 12 & -12 \\ 12 & 8 & -12 & 16 \end{bmatrix}$$

Reduced system

Free DOFs: $\theta_1, v_2, \theta_2, \theta_3$. By symmetry about mid-span, $\theta_2 = 0$ and $\theta_3 = -\theta_1$.

Row for $\theta_1$ (element 1, second row), with $v_1 = 0$, $\theta_2 = 0$:

$$2\times10^5\,(16\theta_1 - 12v_2) = 0$$

Row for $v_2$ (sum of element 1 third row and element 2 first row), with $v_1 = v_3 = 0$, $\theta_2 = 0$, $\theta_3 = -\theta_1$:

$$2\times10^5\,(-12\theta_1 + 24v_2 + 12\theta_3) = 2\times10^5\,(-24\theta_1 + 24v_2) = -10\,000$$

So $\theta_1 = 0.75v_2$, and $-18v_2 + 24v_2 = -0.05$ gives

$$v_2 = -0.008333\ \text{m},\qquad \theta_1 = -0.00625\ \text{rad}$$

Reactions

$R_1 = 2\times10^5(12v_1 + 12\theta_1 - 12v_2 + 12\theta_2) = 2.4\times10^6(-0.00625 + 0.008333) = 5000$ N, and by symmetry $R_3 = 5000$ N.

Comparison

Exact: $v_{max} = \dfrac{PL^3}{48EI} = \dfrac{10\,000(4)^3}{48(1.6\times10^6)} = 8.333$ mm; exact support rotation $\dfrac{PL^2}{16EI} = 6.25\times10^{-3}$ rad. The FE results match exactly because the cubic element is exact for a load applied at a node.

Answer: mid-span deflection = 8.33 mm downward; support rotations = 6.25 $\times10^{-3}$ rad (left clockwise, right anticlockwise); reactions = 5 kN each; the FE result equals the exact value.