--- title: "Rotation and deflection at free end, BMD" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:beam-fem-rotation-deflection-free-end" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-fem-rotation-deflection-free-end site: BCT Engineering --- # Rotation and deflection at free end, BMD Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2077 Chaitra, 10 marks - 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-rotation-deflection-free-end - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question Determine rotation and deflection at the free end of the given beam and hence draw BMD. Take $EI$ to be constant. [Figure: beam A-B-C, A fixed at the left end, span AB = 5 m with stiffness $2EI$ and a uniformly distributed load (value not clear in scan [?]); B is a roller support; span BC = 5 m with stiffness $EI$, free end C carrying a 50 kN point load and a 20 kNm moment] ## Short answer Two beam elements with constant EI are used. The free-end deflection is 3112/EI m downward and rotation is 880.73/EI rad clockwise. The bending moment diagram shows +103.75 kNm at fixed end A, zero at x=1.77 m, -270 kNm at roller B, and -20 kNm at free end C. ## Answer **Assumptions (figure partly unclear):** AB: $L=5$ m, stiffness $2EI$, UDL taken as $w=10$ kN/m; BC: $L=5$ m, stiffness $EI$; A fixed, B roller; at the free end C a 50 kN downward load and a 20 kN m clockwise moment. Results are in terms of $EI$ (kN m$^2$). Nodes: 1 = A, 2 = B, 3 = C. 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.) with $EI\to2EI$ for AB. Equivalent loads for AB: $-25$ and $-20.833$ at A; $-25$ and $+20.833$ at B. At C: $F_{v3}=-50$ kN, $F_{\theta3}=-20$ kN m (clockwise). #### Assembled equation Boundary conditions: $v_1=\theta_1=0$, $v_2=0$. Free DOFs: $\theta_2,v_3,\theta_3$. $$ EI\,\begin{bmatrix}2.4000 & -0.2400 & 0.4000 \\ -0.2400 & 0.0960 & -0.2400 \\ 0.4000 & -0.2400 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_2\\v_3\\\theta_3\end{Bmatrix}=\begin{Bmatrix}20.833 \\ -50.000 \\ -20.000\end{Bmatrix} $$ Solving: $$ \theta_B=-\frac{155.73}{EI}\ \text{rad},\qquad v_C=-\frac{3112.0}{EI}\ \text{m},\qquad \theta_C=-\frac{880.73}{EI}\ \text{rad} $$ **Free end C: deflection 3112.0/EI m downward; rotation 880.73/EI rad (clockwise).** #### End forces | Element | $V_i$ | $M_i$ | $V_j$ | $M_j$ | |---|---|---|---|---| | AB | -49.75 | -103.75 | 99.75 | -270.00 | | BC | 50.00 | 270.00 | -50.00 | -20.00 | Reactions: $R_A=-49.75$ kN (downward, since the overhang hogs the beam), $M_A=-103.75$ kN m (clockwise), $R_B=149.75$ kN. Check: $-49.75+149.75=100$ kN $=50+10\times5$. #### Bending moment diagram (sagging +) | Section | Moment (kN m) | |---|---| | A (fixed end) | +103.75 | | zero at x = 1.77 m from A | 0 | | B (support) | -270.00 (hogging) | | C (just left of the 20 kN m moment) | -20.00 | The moment at B is $-(50\times5+20)=-270$ kN m. In BC the moment varies linearly from $-270$ to $-20$. ```text A(fixed) B(roller) C(free) |-------------------------|------------------------| BMD: +103.75 (x=1.77 m: 0) -270 -20 (kN m) ``` **Answer:** $\theta_C=-880.73/EI$ rad, $v_C=-3112.0/EI$ m (downward). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the fixed-roller-free beam A-B-C, which degrees of freedom are left unknown after applying the boundary conditions? A. v1, theta1, v2 B. theta2, v3, theta3 C. v2, theta2, v3, theta3 D. theta1, theta2, theta3 Answer: B. A is fixed (v1 = theta1 = 0) and B is a roller (v2 = 0), so only theta2, v3 and theta3 remain. (Key point: Free DOFs) 2. For the UDL w on span AB (L = 5 m), what equivalent nodal loads are used at each end? A. wL/2 force and wL^2/8 moment B. wL force at A only C. wL/2 force and wL^2/12 moment (opposite signs at the two ends) D. wL^2/12 force and wL/2 moment Answer: C. Fixed-end forces of a UDL are wL/2 and wL^2/12, giving 25 kN and 20.833 kN m with opposite moment signs at A and B. (Key point: Equivalent nodal loads of the UDL) 3. Why is the bending moment at support B equal to -270 kN m? A. It is the UDL moment wL^2/8 on AB B. It is the 50 kN load times 5 m plus the 20 kN m moment, acting as hogging C. It is the fixed-end moment of AB only D. It is the 20 kN m moment alone Answer: B. Taking moments of the overhang BC about B gives 50 x 5 + 20 = 270 kN m, which hogs the beam. (Key point: Moment at B) 4. Which statement about the final results is correct? A. C deflects upward by 3112/EI B. C deflects downward by 3112/EI and R_A + R_B = 100 kN = 50 + 10 x 5 C. The reactions R_A and R_B are both upward and sum to 50 kN D. C rotates counter-clockwise because the overhang is loaded downward Answer: B. v_C is negative (downward) and theta_C is negative (clockwise); the vertical reactions -49.75 and 149.75 sum to the total load of 100 kN. (Key point: Result at free end and equilibrium check) ## Related in Computational Techniques in Civil Engineering - [Beam: deflection, slope under load, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-s79) — Computational Techniques in Civil Engineering, chapter 4 - [Beam BMD, SFD and mid-span deflection (FEM)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-bmd-sfd-midspan-concrete) — Computational Techniques in Civil Engineering, chapter 4 - [Beam reactions and mid-span deflection (2EI)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-support-reactions-midspan-k8) — Computational Techniques in Civil Engineering, chapter 4, asked 2× - [Beam reactions, deflection, BMD (30 kN, 20 kN/m)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-reactions-midspan-bmd-a2) — Computational Techniques in Civil Engineering, chapter 4, asked 2× - [Beam deflection at B, rotations (2 elements)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-two-elements-b-c) — Computational Techniques in Civil Engineering, chapter 4 - [Beam deflection and stresses at B (FEM)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-deflection-bmd-a) — Computational Techniques in Civil Engineering, chapter 4 ## Related in other subjects - [Fixed beam: end moments from support rotation or settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-end-moments-rotation-settlement) — Theory of Structures II, chapter 3, asked 4× - [Frame rotation and vertical deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-rotation-vertical-free-end-2073m) — Theory of Structures I, chapter 3 - [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3 - [Conjugate beam: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-slope-defl-free-end-2073) — Theory of Structures I, chapter 4