--- title: "Beam: deflection, slope under load, BMD" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:beam-fem-s79" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-fem-s79 site: BCT Engineering --- # Beam: deflection, slope under load, BMD Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2079 Shrawan - Video: [Mechanics of Materials: Lesson 62 - Slope and Deflection Beam Bending Introduction](https://www.youtube.com/watch?v=vd5tpO3eRdg) — Jeff Hanson - 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/beam-fem-s79 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Determine the deflection and slope under the point load of the given beam using FEM. Also draw the BMD. Take $E = 210$ GPa and $I = 6\times10^6$ mm$^4$. [Figure: beam fixed at the left end; first segment $EI$ of 3 m then 2 m, with a 20 kN downward point load 3 m from the fixed end; then a roller support, followed by a $2EI$ segment of 5 m carrying 10 kN/m UDL, ending with a roller support] ## Short answer Use two-node beam elements with EI=1260 kN m^2; fixed and roller supports give v1=theta1=v3=v4=0. Solving the assembled system gives deflection under the 20 kN load as 4.33 mm downward and slope 0.00309 rad counter-clockwise. BMD values: -6.23, +8.82, -21.14, +21.57, 0 kN m at key sections. ## Answer **Data:** $E=210$ GPa $=210\times10^6$ kN/m$^2$, $I=6\times10^6$ mm$^4=6\times10^{-6}$ m$^4$, so $EI=1260$ kN m$^2$. Nodes: 1 = fixed end (x = 0), 2 = point load (x = 3 m), 3 = first roller (x = 5 m), 4 = last roller (x = 10 m). Elements: 1-2 ($L=3$ m, $EI$), 2-3 ($L=2$ m, $EI$), 3-4 ($L=5$ m, $2EI$, UDL 10 kN/m). Units: kN, m. #### Element stiffness $$ [k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T $$ (Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) Equivalent loads for the UDL element: $-25$ kN and $-20.833$ kN m at node 3; $-25$ kN and $+20.833$ kN m at node 4. The 20 kN load is a nodal load at node 2. #### Assembled equation Boundary conditions: $v_1=\theta_1=0$, $v_3=0$, $v_4=0$. Free DOFs: $v_2,\theta_2,\theta_3,\theta_4$. $$ \begin{bmatrix}2450 & 1050 & 1890 & 0 \\ 1050 & 4200 & 1260 & 0 \\ 1890 & 1260 & 4536 & 1008 \\ 0 & 0 & 1008 & 2016\end{bmatrix}\begin{Bmatrix}v_2\\\theta_2\\\theta_3\\\theta_4\end{Bmatrix}=\begin{Bmatrix}-20.000 \\ 0 \\ -20.833 \\ 20.833\end{Bmatrix} $$ (kN, m units.) Solving: $$ v_2=-4.329\ \text{mm},\quad \theta_2=0.00309\ \text{rad},\quad \theta_3=-0.00669\ \text{rad},\quad \theta_4=0.01368\ \text{rad} $$ **Deflection under the 20 kN load = 4.33 mm downward; slope = 0.00309 rad (3.09 x 10$^{-3}$, counter-clockwise).** #### End forces and bending moments | Element | $V_i$ | $M_i$ | $V_j$ | $M_j$ | |---|---|---|---|---| | 1-2 | 5.018 | 6.230 | -5.018 | 8.824 | | 2-3 | -14.982 | -8.824 | 14.982 | -21.140 | | 3-4 | 29.228 | 21.140 | 20.772 | 0.000 | Reactions: fixed end 5.018 kN and moment 6.230 kN m; first roller 44.210 kN; last roller 20.772 kN. Sum $=70$ kN = $20+10\times5$ (check). #### Bending moment diagram (sagging +) | Section | Moment (kN m) | |---|---| | Fixed end (x = 0) | -6.23 (hogging) | | Under 20 kN load (x = 3 m) | +8.82 | | Roller at x = 5 m | -21.14 (hogging) | | Max in last span (x = 7.92 m, 2.92 m from roller) | +21.57 | | Last roller (x = 10 m) | 0 | ```text x (m): 0 3 5 7.92 10 |--------|--------|-----------------|---------| BMD: -6.23 +8.82 -21.14 +21.57 0 (kN m) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is EI for E = 210 GPa and I = 6x10^6 mm^4 in kN m^2? A. 12600 B. 1260 C. 126 D. 126000 Answer: B. 210x10^6 kN/m^2 x 6x10^-6 m^4 = 1260 kN m^2. (Key point: Flexural rigidity) 2. Which DOFs are free given fixed node 1 and rollers at nodes 3 and 4? A. v2, theta2, theta3, theta4 B. v2, v3, v4 C. theta1, theta2, theta3 D. v2, theta2 only Answer: A. v1 = theta1 = 0, v3 = 0 and v4 = 0; what remains are v2, theta2, theta3, theta4. (Key point: Free DOFs) 3. For the 5 m span with 10 kN/m, the equivalent nodal loads are A. 25 kN and wL^2/12 = 20.833 kN m at each end B. 50 kN at each end and no moments C. 25 kN and wL^2/8 = 31.25 kN m D. 10 kN and 5 kN m Answer: A. wL/2 = 25 kN and wL^2/12 = 20.833 kN m (opposite signs at the two ends). (Key point: UDL equivalent loads) 4. How are the reactions checked? A. Sum of reactions equals total load 20 + 10x5 = 70 kN B. Sum of reactions equals 20 kN C. Reactions equal the shear at the fixed end only D. Sum of moments equals EI Answer: A. Vertical equilibrium: 5.018 + 44.210 + 20.772 = 70 kN. 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