--- title: "Beam deflection at B, rotations (2 elements)" subject: "Computational Techniques in Civil Engineering" chapter: 4 key: "semester_8/computational_techniques_in_civil_engineering:beam-two-elements-b-c" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/beam-two-elements-b-c site: BCT Engineering --- # Beam deflection at B, rotations (2 elements) Computational Techniques in Civil Engineering · Chapter 4: Finite element method - Asked in: 2071 Bhadra, 10 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/beam-two-elements-b-c - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4 ## Question For the given beam find the deflection at point B and rotations at points B and C. Take $EI$ as constant throughout the beam. Discretise the beam into two elements. [Figure: beam A-B-C with 20 kN/m UDL over the whole length; AB = 5 m, BC = 5 m; A fixed at the left end, C roller supported] ## Short answer Discretise into two 5 m beam elements with fixed and roller supports; solve the reduced stiffness equations. Deflection at B is 1041.67/EI downward, rotation at B is 104.17/EI clockwise, and rotation at C is 416.67/EI anticlockwise. ## Answer The beam is divided into element 1 (A-B) and element 2 (B-C), each 5 m long, with nodes A, B, C. Dofs per node: (deflection $v$, rotation $\theta$). A is fixed ($v_A=\theta_A=0$) and C is a roller ($v_C=0$), so the unknowns are $v_B,\ \theta_B,\ \theta_C$. ```text 20 kN/m +++++++++++++++++++++++ |===========o==========o A fixed, C roller A B C |<--- 5 --->|<--- 5 -->| ``` #### Element stiffness ($L=5$ m) $$ [k]=\frac{EI}{125}\begin{bmatrix}12&30&-12&30\\30&100&-30&50\\-12&-30&12&-30\\30&50&-30&100\end{bmatrix} $$ #### Equivalent nodal loads (per element) $wL/2=20\times5/2=50$ kN, $wL^2/12=20\times25/12=41.667$ kNm. - Element 1: $(-50,\ -41.667,\ -50,\ +41.667)$ - Element 2: same. Loads on the free dofs $(v_B,\theta_B,\theta_C)$: $F_{v_B}=-50-50=-100$ kN, $F_{\theta_B}=+41.667-41.667=0$, $F_{\theta_C}=+41.667$ kNm. #### Assembled equations for the free dofs $$ EI\begin{bmatrix}0.192&0&0.24\\0&1.6&0.4\\0.24&0.4&0.8\end{bmatrix}\begin{bmatrix}v_B\\ \theta_B\\ \theta_C\end{bmatrix}=\begin{bmatrix}-100\\0\\41.667\end{bmatrix} $$ #### Solution $$ v_B=-\frac{1041.67}{EI},\qquad \theta_B=-\frac{104.17}{EI}\ \text{rad (clockwise)},\qquad \theta_C=+\frac{416.67}{EI}\ \text{rad (anticlockwise)} $$ #### Check with reactions and moments $\{R\}=[K]\{d\}-\{F\}$ gives $R_A=125$ kN, $M_A=250$ kNm and $R_C=75$ kN. $\Sigma F_y$: $125+75=200=20\times10$ and $\Sigma M_A$: $75\times10-200\times5+250=0$ (with $M_A$ anticlockwise). These agree with the propped-cantilever results $R_C=3wL/8=75$ kN, $R_A=5wL/8=125$ kN, $M_A=wL^2/8=250$ kNm. **Answer:** deflection at B $=1041.67/EI$ downward; rotations $\theta_B=104.17/EI$ (clockwise) and $\theta_C=416.67/EI$ (anticlockwise). Bending moment: $M(x)=-250+125x-10x^2$ (sagging +). $M_A=-250$, $M_B=+125$ kNm, zero at $x=2.5$ m, maximum $+140.6$ kNm at $x=6.25$ m, $M_C=0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With A fixed and C a roller, which degrees of freedom are the unknowns in the two-element beam? A. v_B, theta_B, theta_C B. v_B, v_C, theta_C C. theta_A, theta_B, v_C D. v_A, v_B, theta_B Answer: A. A fixed removes v_A and theta_A, and the roller removes v_C. What remains is v_B, theta_B and theta_C. (Key point: Unknown dofs) 2. For a 5 m element under 20 kN/m, what are the fixed-end forces used as equivalent nodal loads? A. wL/2 = 100 kN and wL^2/8 = 62.5 kNm B. wL/2 = 50 kN and wL^2/12 = 41.667 kNm C. wL = 100 kN and wL^2/12 = 41.667 kNm D. wL/2 = 50 kN and wL^2/8 = 62.5 kNm Answer: B. A UDL on a fixed-fixed beam gives shears wL/2 and end moments wL^2/12, which is 50 kN and 41.667 kNm here. (Key point: Equivalent nodal loads) 3. Why is the assembled load on theta_B equal to zero while the load on theta_C is +41.667 kNm? A. Because B is a roller support B. Because the UDL does not act on element 2 C. Because the two end moments at B (+41.667 from element 1 and -41.667 from element 2) cancel, but C gets only one D. Because the beam rotation at B is fixed Answer: C. Node B receives +41.667 from element 1 and -41.667 from element 2, so they cancel. Node C receives only element 2's +41.667. (Key point: Assembled load at free dofs) 4. How does the answer verify the finite element result? A. By comparing with a simply supported beam, giving R = wL/2 B. By checking that theta_B equals theta_C C. By repeating with one element only D. By getting R_A = 125, R_C = 75 and M_A = 250 matching the propped cantilever formulas 5wL/8, 3wL/8, wL^2/8 Answer: D. The reactions from {R} = [K]{d} - {F} satisfy equilibrium and match the propped cantilever results for fixed-roller beams. 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