--- title: "Moment area: B, C, maximum deflection" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-200-300kn" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-200-300kn site: BCT Engineering --- # Moment area: B, C, maximum deflection Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2080 Chaitra, 8 marks - 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-200-300kn - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Calculate deflection at points B and C of the given simply supported beam using moment area method. Also calculate the location and magnitude of maximum deflection. Take EI to be constant. [Figure: beam ABCD, A hinge, D roller; AB = 3 m with 200 kN at B; BC = 5 m with 300 kN at C; CD = 2 m.] ## Short answer Moment area method: deflection at B = 5550/EI downward, at C = 4300/EI downward. Maximum deflection occurs where slope zero between B and C, at 5.08 m from A, magnitude = 6852/EI downward. Analysis uses tangent at A, M/EI area moments, and condition for zero slope. ## Answer #### Given data Simply supported beam ABCD, span $AD = 10$ m: A hinge, D roller, $AB = 3$ m (200 kN at B), $BC = 5$ m (300 kN at C), $CD = 2$ m. $EI$ constant. $x$ is measured from A. ```text 200 kN 300 kN A ------v--------------v----- D ^ 3 m B 5 m C 2 m ^ ``` #### Step 1: Reactions $R_{A} = 200.00$ kN, $R_{D} = 300.00$ kN (positive = upward; check: $\sum F_y = 500.00$ kN equals the total load 500.00 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 3 | $200 x$ | EI | 900.00 | 2.000 | | 3 to 8 | $600$ | EI | 3000.00 | 5.500 | | 8 to 10 | $3000 - 300 x$ | EI | 600.00 | 8.667 | (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 $x = 10$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support): $$ t = (900.00)(8.000) + (3000.00)(4.500) + (600.00)(1.333) = \frac{21500.00}{EI} $$ That support does not deflect, so $\theta_A = -t/10$: $$ \theta_A = \frac{-2150.000}{EI}\ \text{(clockwise)} $$ **At B ($x = 3$ m):** $$ \theta = \theta_A + \sum A = -2150.000 + [(900.00)] = \frac{-1250.000}{EI}\ \text{(clockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -6450.000 + [(900.00)(1.000)] = \frac{-5550.000}{EI}\ \text{(downward)} $$ **At C ($x = 8$ m):** $$ \theta = \theta_A + \sum A = -2150.000 + [(900.00) + (3000.00)] = \frac{1750.000}{EI}\ \text{(anticlockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -17200.000 + [(900.00)(6.000) + (3000.00)(2.500)] = \frac{-4300.000}{EI}\ \text{(downward)} $$ **At the point of maximum deflection ($x = 5.08333$ m):** $$ \theta = \theta_A + \sum A = -2150.000 + [(900.00) + (1250.00)] = \frac{0.000}{EI}\ \text{(zero)} $$ $$ y = \theta_A\,x + t_{P/A} = -10929.167 + [(900.00)(3.083) + (1250.00)(1.042)] = \frac{-6852.083}{EI}\ \text{(downward)} $$ #### Maximum deflection The deflection is maximum where the slope is zero. The slope at B is $-1250/EI$ (clockwise) and at C it is $+1750/EI$, so the zero lies between B and C. In BC, $M/EI = 600/EI$ is constant, so the slope changes by $600/EI$ per metre: $$ -1250 + 600\,(x - 3) = 0 \;\Rightarrow\; x = 3 + 2.083 = 5.083\ \text{m from A} $$ The deflection there (Step 3 tangent offset, from the same formula) is $$ y_{max} = \theta_A x + t_{P/A} = \frac{-6852.08}{EI} $$ **Answer:** deflection at B $= 5550/EI$ downward; at C $= 4300/EI$ downward (kNm$^3$/EI). Maximum deflection $= 6852.1/EI$ downward at $5.083$ m from A (and 4.917 m from D). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the bending moment between B and C and why? A. Increases from 600 to 1000 kNm B. Constant 600 kNm, since the shear there is 200 - 200 = 0 C. Constant 300 kNm D. Decreases from 600 to 0 Answer: B. M at B = 200x3 = 600 kNm and the shear between B and C is zero, so the M/EI diagram is a rectangle of 600/EI. (Key point: Moment between the loads) 2. Why is theta_A = -t/10 with t = 21500/EI? A. Because the beam deflection at A is t B. Because the area of M/EI equals t C. Because D does not deflect, so the tangent at A must make up the deviation t over the span of 10 m D. Because the roller carries 300 kN Answer: C. The support at D has zero deflection, so tangent offset at D (theta_A x 10) must equal the deviation t, giving theta_A magnitude 2150/EI clockwise. (Key point: Slope at A from tangent deviation) 3. Slopes are -1250/EI at B and +1750/EI at C. Where does the maximum deflection occur? A. Under the 200 kN load B B. Between B and C, where the slope changes sign (zero slope) C. Under the 300 kN load C D. At mid-span, 5 m from A Answer: B. Maximum deflection is where slope is zero. The slope changes sign between B and C, so the point lies in BC. (Key point: Location of maximum deflection) 4. In BC, M/EI = 600/EI is constant, so slope changes by 600/EI per metre. Starting from -1250/EI at B, where is the slope zero? A. x = 4.0 m from A B. x = 5.083 m from A (3 + 1250/600) C. x = 5.5 m from A D. x = 6.0 m from A Answer: B. -1250 + 600(x - 3) = 0 gives x - 3 = 2.083, so x = 5.083 m from A. 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