--- title: "Moment area: slope and deflection under 50 kN" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-12-ss-defl-4m-d" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-12-ss-defl-4m-d site: BCT Engineering --- # Moment area: slope and deflection under 50 kN Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2077 Chaitra, 12 marks - Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=jkwPzvWsqIY) — AF Math & Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-12-ss-defl-4m-d - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Using moment area method, calculate the slope at supports, deflection at points C and D. Also calculate maximum deflection in the beam and its location. [Figure: simply supported beam AB with C and D; AC = CD = DB = 4 m (span 12 m); 50 kN at D; EI constant.] ## Short answer Using the moment-area method: θA = 355.56/EI clockwise, θB = 444.44/EI anticlockwise. Deflection at C is 1244.4/EI, at D is 1422.2/EI, both downward. Maximum deflection is 1548.3/EI downward at 6.532 m from A. ## Answer #### Given data Simply supported beam AB, $L = 12$ m. $AC = CD = DB = 4$ m. 50 kN at D (8 m from A). $EI$ constant. ```text 50 kN A ------- C ------- D v ----- B ^ 4 m 4 m 4 m ^ ``` #### Step 1: Reactions $R_{A} = 16.67$ kN, $R_{B} = 33.33$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 kN). #### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces | Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from A (m) | |---|---|---|---|---| | 0 to 8 | $\frac{50 x}{3}$ | EI | 533.33 | 5.333 | | 8 to 12 | $400 - \frac{100 x}{3}$ | EI | 266.67 | 9.333 | (Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.) #### Step 3: Tangent at A Deviation of the support at $x = 12$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support): $$ t = (533.33)(6.667) + (266.67)(2.667) = \frac{4266.67}{EI} $$ That support does not deflect, so $\theta_A = -t/12$: $$ \theta_A = \frac{-355.556}{EI}\ \text{(clockwise)} $$ **At C ($x = 4$ m):** $$ \theta = \theta_A + \sum A = -355.556 + [(133.33)] = \frac{-222.222}{EI}\ \text{(clockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -1422.222 + [(133.33)(1.333)] = \frac{-1244.444}{EI}\ \text{(downward)} $$ **At D ($x = 8$ m):** $$ \theta = \theta_A + \sum A = -355.556 + [(533.33)] = \frac{177.778}{EI}\ \text{(anticlockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -2844.444 + [(533.33)(2.667)] = \frac{-1422.222}{EI}\ \text{(downward)} $$ **At the point of maximum deflection ($x = 6.532$ m):** $$ \theta = \theta_A + \sum A = -355.556 + [(355.56)] = 0\ \text{(zero slope)} $$ $$ y = \theta_A\,x + t_{P/A} = -2322.489 + [(355.56)(2.177)] = \frac{-1548.319}{EI}\ \text{(downward)} $$ #### Slope at the supports $\theta_A = 355.56/EI$ (clockwise), as found above. At B ($x = 12$ m): $\theta_B = \theta_A + \text{total area} = -355.56 + 800 = 444.44$, so $\theta_B = \dfrac{444.44}{EI}$ (anticlockwise). (Total area of the $M/EI$ triangle $= \tfrac12\times12\times133.33 = 800$.) #### Maximum deflection The slope is zero in the portion AD, where $M = 16.667x$. Slope: $\theta(x) = -355.56 + 8.333x^2$. Setting it to zero: $$ x = \sqrt{\frac{355.56}{8.333}} = 6.532\ \text{m from A} $$ This lies in AD, so the assumption is valid. The deflection there, from the same formula, is $y_{max} = \dfrac{1548.3}{EI}$ downward. **Answer:** $\theta_A = 355.56/EI$ (clockwise), $\theta_B = 444.44/EI$ (anticlockwise); deflection at C $= 1244.4/EI$, at D $= 1422.2/EI$ (both downward); maximum deflection $= 1548.3/EI$ downward at $6.532$ m from A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 12 m beam with 50 kN at D (8 m from A), what are R_A and the M/EI diagram? A. R_A = 33.33 kN; triangle with peak 266.67/EI B. R_A = 25 kN; peak 200/EI C. R_A = 16.67 kN; triangle with peak 133.33/EI at D D. R_A = 16.67 kN; rectangle Answer: C. R_A = 50x4/12 = 16.67 kN, M_D = 16.67x8 = 133.33 kNm, so M/EI is a triangle peaking at D. (Key point: Reaction and M/EI shape) 2. How is theta_A = 355.56/EI obtained? A. Area of M/EI divided by 12 B. Total area (800) divided by 2 C. Moment of the area about A divided by 12 D. Moment of the M/EI area about B (4266.67) divided by 12 m Answer: D. B does not deflect, so theta_A = deviation of B from the tangent at A / L = 4266.67/12 = 355.56/EI. (Key point: Slope at A) 3. What is theta_B? A. -355.56 + 800 = 444.44/EI anticlockwise B. 355.56/EI clockwise C. 800/EI D. 444.44/EI clockwise Answer: A. Theorem I: slope at B = theta_A + total area of M/EI = -355.56 + 800 = +444.44/EI, opposite sense to theta_A (anticlockwise). (Key point: Slope at B) 4. Where is the slope zero (maximum deflection), and why is it not at mid-span? A. At x = 6 m, mid-span, as for any simply supported beam B. At x = 8 m under the load C. At x = 6.532 m, shifted toward the load because the load is off-centre; found from -355.56 + 8.333x^2 = 0 D. At x = 5.468 m, toward A Answer: C. The slope at D is positive (177.78), so zero slope lies in AD. In AD the slope is -355.56 + 8.333x^2, which is zero at x = 6.532 m. (Key point: Maximum deflection location) ## Related in Theory of Structures I - [Moment area: B, C, maximum deflection](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-200-300kn) — Theory of Structures I, chapter 4 - [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4 - [Moment area: slope and deflection at E and at 4 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-e-4m-a) — Theory of Structures I, chapter 4 - [Moment area: girder with 120 kN and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-14m-girder-two-loads) — Theory of Structures I, chapter 4 - [Moment area: slopes at A, B, deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-slopes-a-b-defl-d-2065) — Theory of Structures I, chapter 4 - [Moment area: slopes at A, B, deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-beam-60kn-12m) — Theory of Structures I, chapter 4 ## Related in other subjects - [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2 - [Slope deflection: beam with 50 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-30kn-bmd) — Theory of Structures II, chapter 3 - [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4 - [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7