Propped cantilever with UDL, two elements

Reading of the figure: total length 8 m, UDL 12 kN/m on the whole length, AB = 3 m, BC = 5 m, $EI$ constant. The figure shows a pin/roller support at A and a fixed end at C; the question asks for the rotations at B and C, so the results for the interchanged arrangement (A fixed, C pinned) are also given. Units: kN, m. Nodes: 1 = A, 2 = B, 3 = C.

Two-element model

$$[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T$$

(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) For AB, $L=3$ m; for BC, $L=5$ m. UDL equivalent nodal loads: AB: $-18$ kN, $-9$ kN m at A and $-18$ kN, $+9$ kN m at B; BC: $-30$ kN, $-25$ kN m at B and $-30$ kN, $+25$ kN m at C.

Case 1: A pinned ($v_A=0$), C fixed ($v_C=\theta_C=0$). Free DOFs: $\theta_A,v_B,\theta_B$.

$$EI\,\begin{bmatrix}1.3333 & -0.6667 & 0.6667 \\ -0.6667 & 0.5404 & -0.4267 \\ 0.6667 & -0.4267 & 2.1333\end{bmatrix}\begin{Bmatrix}\theta_A\\v_B\\\theta_B\end{Bmatrix}=\begin{Bmatrix}-9.000 \\ -48.000 \\ -16.000\end{Bmatrix}$$ $$\theta_A=-\frac{128}{EI},\qquad v_B=-\frac{262.5}{EI}\ \text{m},\qquad \theta_B=-\frac{20.0}{EI}\ \text{rad},\qquad \theta_C=0$$

Case 2: A fixed, C pinned (rotation at C is non-zero). Free DOFs: $v_B,\theta_B,\theta_C$.

$$EI\,\begin{bmatrix}0.5404 & -0.4267 & 0.2400 \\ -0.4267 & 2.1333 & 0.4000 \\ 0.2400 & 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}v_B\\\theta_B\\\theta_C\end{Bmatrix}=\begin{Bmatrix}-48.000 \\ -16.000 \\ 25.000\end{Bmatrix}$$ $$v_B=-\frac{202.5}{EI}\ \text{m},\qquad \theta_B=-\frac{72.0}{EI}\ \text{rad},\qquad \theta_C=+\frac{128.0}{EI}\ \text{rad}$$

Check with a single element (8 m)

Case 2 (A fixed, C pinned): free DOF is $\theta_C$ only. $k=\dfrac{4EI}{L}=\dfrac{EI}{2}$ and the equivalent moment at C is $+\dfrac{wL^2}{12}=64$ kN m, so

$$\theta_C=\frac{64}{EI/2}=\frac{128}{EI}\ \text{rad}$$

This equals the two-element value and the exact value $wL^3/48EI$. The deflection at B follows from Hermite interpolation of the nodal values plus the fixed-end beam deflection $\dfrac{wx^2(L-x)^2}{24EI}$ at $x=3$ m:

$$v_B=-\frac{90.0}{EI}-\frac{12\times9\times25}{24EI}=-\frac{90.0+112.5}{EI}=-\frac{202.5}{EI}\ \text{m}$$

which agrees with the two-element result. (For Case 1 the same single-element check gives $\theta_A=-128/EI$ and $v_B=-262.5/EI$, matching the exact value $wx(L^3-3Lx^2+2x^3)/48EI$ at $x=3$ m.)

Answer: Case 2: $v_B=-202.5/EI$ m, $\theta_B=-72/EI$ rad, $\theta_C=128/EI$ rad. Case 1: $v_B=-262.5/EI$ m, $\theta_B=-20/EI$ rad, $\theta_C=0$ (fixed).