--- title: "Propped cantilever with UDL, two elements" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:propped-cantilever-two-elements" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/propped-cantilever-two-elements site: BCT Engineering --- # Propped cantilever with UDL, two elements Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2074 Bhadra, 6+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/propped-cantilever-two-elements - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question A propped cantilever beam is loaded as shown in the figure below. Discretize the beam into two elements and find the deflection at point B and rotations at points B and C. Also check the result using a single element model. Take $EI$ as constant throughout the beam. [Figure: beam A-B-C of total length 8 m with 12 kN/m UDL over the whole length; A roller/pin support at the left end, B at 3 m from A (AB = 3 m, BC = 5 m), C fixed at the right end] ## Short answer For the given A pinned, C fixed case: vB=-262.5/EI m, thetaB=-20/EI rad, thetaC=0; the single-element model confirms these. If supports are interchanged (A fixed, C pinned): vB=-202.5/EI m, thetaB=-72/EI rad, thetaC=128/EI rad. ## Answer **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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 12 kN/m UDL on the 3 m element AB, what are the equivalent nodal loads? A. 18 kN and 9 kN m at each end, with opposite moment signs B. 36 kN at B only C. 12 kN and 4.5 kN m at each end D. 18 kN and 13.5 kN m at each end Answer: A. wL/2 = 18 kN and wL^2/12 = 9 kN m, the moment signs being opposite at the two ends. (Key point: Equivalent nodal loads) 2. In Case 2 (A fixed, C pinned), which DOFs are unknown? A. theta_A, v_B, theta_B B. v_B, theta_B, theta_C C. v_B, theta_B only D. theta_C only Answer: B. A fixed removes v_A and theta_A, the pin at C removes v_C only, so v_B, theta_B and theta_C remain. (Key point: Free DOFs for Case 2) 3. How does the single 8 m element check theta_C in Case 2? A. k = 4EI/L = EI/2 and the equivalent moment wL^2/12 = 64 kN m give theta_C = 128/EI B. k = 12EI/L^3 gives theta_C = 64/EI C. The check is impossible with one element D. It gives theta_C = 0 Answer: A. theta_C = 64/(EI/2) = 128/EI, matching the two-element result and the exact wL^3/(48EI). 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