--- title: "Flexibility method: beam reactions at B and C" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:flexibility-beam-200kn-100kn-reactions-b" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-beam-200kn-100kn-reactions-b site: BCT Engineering --- # Flexibility method: beam reactions at B and C Theory of Structures II · Chapter 2: Force method - Asked in: 2069 Asar, 10 marks - Video: [VE 11_3 Indeterminate Beam - Two Degrees - Force Method](https://www.youtube.com/watch?v=p26lL7OcxBQ) — Structural Analysis - Understanding Behavior - 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/flexibility-beam-200kn-100kn-reactions-b - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Use the force method (flexibility matrix) to find the reactions at supports B and C of the beam shown in figure below and also draw shear force and bending moment diagrams. [Figure: beam ABC, A fixed; 200 kN at 5 m from A; B support at 10 m; 100 kN at 5 m beyond B; C at 20 m total; spans 5 m, 5 m, 5 m, 5 m.] ## Short answer Take the cantilever fixed at A as the released structure, with redundants RB and RC. Compute flexibility coefficients and load deflections, then solve compatibility. Reactions: RB = 166.07 kN upward, RC = 28.57 kN upward. ## Answer **Assumptions (figure values read as given):** A is fixed at $x=0$; 200 kN acts at $x=5$ m; B is a roller support at $x=10$ m; 100 kN acts at $x=15$ m; C is a roller support at $x=20$ m. EI is constant. The beam is indeterminate to the second degree. Choose the reactions $R_B$ (coordinate 1) and $R_C$ (coordinate 2) as redundants, acting upward. The released structure is the cantilever fixed at A. #### Step 1: Flexibility matrix of the cantilever (L = 20 m) For a unit load at distance $a$ from the fixed end, the deflection at $x$ is $x^2(3a-x)/6EI$ for $x\le a$ and $a^2(3x-a)/6EI$ for $x\ge a$. $$ f_{11}=\frac{10^3}{3EI}=\frac{333.33}{EI},\quad f_{22}=\frac{20^3}{3EI}=\frac{2666.67}{EI},\quad f_{12}=f_{21}=\frac{10^2(3\times20-10)}{6EI}=\frac{833.33}{EI} $$ #### Step 2: Deflections due to the loads (downward) on the released beam At B (x = 10): the 200 kN load gives $\dfrac{200\times5^2(3\times10-5)}{6EI}=\dfrac{20833.3}{EI}$ and the 100 kN load (at 15 m, beyond B) gives $\dfrac{100\times10^2(3\times15-10)}{6EI}=\dfrac{58333.3}{EI}$. At C (x = 20): $\dfrac{200\times5^2(60-5)}{6EI}=\dfrac{45833.3}{EI}$ and $\dfrac{100\times15^2(60-15)}{6EI}=\dfrac{168750}{EI}$. $$ \Delta_1=\frac{79166.7}{EI},\qquad \Delta_2=\frac{214583.3}{EI} $$ #### Step 3: Compatibility (deflection at B and C must be zero) $$ \begin{bmatrix}333.33&833.33\\833.33&2666.67\end{bmatrix}\begin{Bmatrix}R_B\\R_C\end{Bmatrix}=\begin{Bmatrix}79166.7\\214583.3\end{Bmatrix} $$ $$ R_B=166.07\ \text{kN}\uparrow,\qquad R_C=28.57\ \text{kN}\uparrow $$ #### Step 4: Remaining reactions $$ R_A=200+100-166.07-28.57=105.36\ \text{kN}\uparrow,\qquad M_A=-200(5)-100(15)+166.07(10)+28.57(20)\;\Rightarrow\; M_A=-267.86\ \text{kNm (hogging)} $$ Check: moments of the loads and support reactions about A: $166.07\times10+28.57\times20-200\times5-100\times15=-267.86$ kNm, which is balanced by the fixed-end reaction moment of 267.86 kNm (anticlockwise) ✓. #### Shear force (kN) | Section | SF | |---|---| | A (right) | +105.36 | | 5 m (left / right of 200 kN) | +105.36 / -94.64 | | B (left / right) | -94.64 / +71.43 | | 15 m (left / right of 100 kN) | +71.43 / -28.57 | | C (left) | -28.57 | #### Bending moment (kNm, sagging +) | Section | BM | |---|---| | A | -267.86 | | 5 m (under 200 kN) | +258.93 | | B (10 m) | -214.29 | | 15 m (under 100 kN) | +142.86 | | C | 0 | The SF changes sign under the 200 kN load (maximum sagging BM +258.93 kNm), at B (hogging peak of -214.29 kNm, where the SF jumps) and under the 100 kN load. The BM is zero at C. **Answer: $R_B=166.07$ kN, $R_C=28.57$ kN (both upward); also $R_A=105.36$ kN, $M_A=267.86$ kNm (hogging).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are R_B and R_C chosen as redundants, and what is the released structure? A. The beam is 1st degree indeterminate; released structure is simply supported B. The beam is 2nd degree indeterminate; released structure is the cantilever fixed at A C. The beam is 3rd degree indeterminate; released structure is a simple beam D. The beam is determinate; no redundants are needed Answer: B. Fixed A plus rollers at B and C gives two extra reactions, so removing R_B and R_C leaves a cantilever fixed at A. (Key point: Degree of indeterminacy and redundants) 2. What compatibility condition is used to form [F]{R} = {Delta}? A. The moment at A is zero B. The deflections at B and C in the real beam are zero C. The shear force at B equals that at C D. The sum of vertical forces is zero Answer: B. The supports B and C do not settle, so the redundants must cancel the downward load deflections of the released cantilever at B and C. (Key point: Compatibility condition) 3. For the cantilever with L = 20 m, how are f11 and f22 obtained? A. f11 = 10^3/(3EI) for the unit load at B, f22 = 20^3/(3EI) for the unit load at C B. f11 = 10^3/(8EI), f22 = 20^3/(8EI) C. f11 = 10^2/(2EI), f22 = 20^2/(2EI) D. f11 = f22 = 20^3/(3EI) Answer: A. Each is the tip deflection of a cantilever of the length from A to the point, PL^3/3EI with P = 1. (Key point: Flexibility coefficients) 4. What are the redundant reactions found from the compatibility equations? A. R_B = 28.57 kN, R_C = 166.07 kN B. R_B = 105.36 kN, R_C = 28.57 kN C. R_B = 166.07 kN, R_C = 28.57 kN D. R_B = 94.64 kN, R_C = 71.43 kN Answer: C. Solving the 2x2 system gives R_B = 166.07 kN and R_C = 28.57 kN, both upward; R_A then follows from equilibrium. 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