--- title: "Stiffness matrix: continuous beam BMD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-continuous-beam-10kn-30kn-15knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-continuous-beam-10kn-30kn-15knm site: BCT Engineering --- # Stiffness matrix: continuous beam BMD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2070 Asar, 10 marks - Video: [Stiffness Matrix Method for Analysis of Beams ( With Overhanging )](https://www.youtube.com/watch?v=X9MJOI8XnOQ) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-continuous-beam-10kn-30kn-15knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Analyse the continuous beam and draw bending moment diagram which is loaded as shown in figure below. Use the stiffness matrix method. [Figure: beam ABC, A fixed; 10 kN at 2 m and 30 kN at 5 m from A; B support at 7 m (2 m + 3 m + 2 m); BC = 8 m (4 m + 4 m) with 15 kN/m UDL on the right 4 m; C roller.] ## Short answer Using theta_B and theta_C, solve EI[[1.0714,0.25],[0.25,0.5]]{theta}={-9.694,-55}. Final moments are M_AB=-17.07, M_BA=45.44, M_BC=-45.44, M_CB=0 kNm; maximum sagging in BC is 51.53 kNm at 5.38 m from B. ## Answer **Data:** A is fixed; AB = 7 m with 10 kN at 2 m and 30 kN at 5 m from A; B is a roller support; BC = 8 m with 15 kN/m on the right-hand 4 m (from 4 m to 8 m from B); C is a roller (free to rotate). EI is constant. #### Unknown displacements (coordinates) 1 = $\theta_B$, 2 = $\theta_C$ (clockwise +). $\theta_A=0$ and there are no support settlements. #### Fixed-end moments (clockwise +, kNm) - AB (point loads): load 10 kN ($a=2$, $b=5$): $\mp\dfrac{Pab^2}{L^2},\ \pm\dfrac{Pa^2b}{L^2}$ = $-10.204$, $+4.082$; load 30 kN ($a=5$, $b=2$): $-12.245$, $+30.612$. Total: $FEM_{AB}=-22.449$, $FEM_{BA}=34.694$. - BC (partial UDL, 15 kN/m over 4 m, from the right end): $FEM_{BC}=-25.000$, $FEM_{CB}=55.000$ (integration of the point-load formulas over the loaded length). #### Member stiffness equations - $M_{AB}=\frac{2EI}{7}\left(\theta_{B}\right) - 22.45$ - $M_{BA}=\frac{2EI}{7}\left(2\theta_{B}\right) + 34.69$ - $M_{BC}=\frac{2EI}{8}\left(2\theta_{B} + \theta_{C}\right) - 25.00$ - $M_{CB}=\frac{2EI}{8}\left(2\theta_{C} + \theta_{B}\right) + 55.00$ #### Stiffness matrix equation ($[K]\{d\}=\{P\}$) The roller C gives $M_{CB}=0$ and the joint B gives $M_{BA}+M_{BC}=0$: $$ EI\begin{bmatrix}1.0714 & 0.2500 \\ 0.2500 & 0.5000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-9.694 \\ -55.000\end{Bmatrix} $$ $K_{11}=\dfrac{4EI}{7}+\dfrac{4EI}{8}=1.0714EI$, $K_{12}=K_{21}=\dfrac{2EI}{8}=0.2500EI$, $K_{22}=\dfrac{4EI}{8}=0.5000EI$. $$ EI\theta_{B}=18.8140,\quad EI\theta_{C}=-119.4070 $$ #### Final end moments (kNm) | End | $M$ (kNm) | End | $M$ (kNm) | |---|---|---|---| | AB | -17.07 | BA | 45.44 | | BC | -45.44 | CB | 0.00 | Checks: Joint B: $45.44+(-45.44)=0.00$ ✓; roller C: $M_{CB}=0.00$ ✓. #### Bending moment diagram (sagging +, kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | -17.07 | -45.44 | 11.23 at 5.00 m | | BC | -45.44 | 0.00 | 51.53 at 5.38 m | BM under the 10 kN load: 6.25; under the 30 kN load: 11.23 kNm. Reactions: A: $H=0.00$, $V=11.66$, $M=17.07$; B: $H=0.00$, $V=49.02$, $M=0.00$; C: $H=0.00$, $V=39.32$, $M=0.00$. **Answer: $M_{AB}=-17.07$, $M_{BA}=45.44$, $M_{BC}=-45.44$ kNm, $M_{CB}=0$; maximum sagging in BC = 51.53 kNm at 5.38 m from B.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which displacements are the unknowns in the stiffness analysis of this beam? A. theta_B and theta_C, because A is fixed and there is no settlement B. theta_A and theta_B C. Vertical deflections at B and C D. Only theta_B Answer: A. A is fixed (theta_A = 0) and supports do not settle, so only rotations at B and C are free. (Key point: Unknown coordinates) 2. How is K11 = 1.0714EI obtained? A. 4EI/7 from AB plus 4EI/8 from BC B. 2EI/7 plus 2EI/8 C. 4EI/7 plus 2EI/8 D. 3EI/7 plus 3EI/8 Answer: A. For unit rotation of B, both members contribute 4EI/L: 0.5714EI + 0.5EI = 1.0714EI. (Key point: Stiffness coefficient K11) 3. What equation represents the roller at C? A. M_CB = 0 B. theta_C = 0 C. M_BC = 0 D. M_CB = M_BC Answer: A. A roller is free to rotate and cannot resist moment, so the end moment at C is zero; this gives the second equation. (Key point: Roller end condition) 4. What check does the final joint B result satisfy? A. M_BA + M_BC = 45.44 - 45.44 = 0 B. M_BA = M_AB C. M_BC = M_CB D. M_BA + M_AB = 0 Answer: A. No external moment acts at B, so the moments of the members meeting there must sum to zero. 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