--- title: "Moment area: slopes at A, B, deflection at D" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-beam-60kn-12m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-beam-60kn-12m site: BCT Engineering --- # Moment area: slopes at A, B, deflection at D Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2081 Chaitra, 8 marks - Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=FFv2pErDeBU) — Civil Engineering Exam - 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-beam-60kn-12m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine the slope at A and B and deflections at section D of the beam loaded as shown in figure below using moment area theorem. Take EI constant. [Figure: simply supported beam AB, span 12 m; 60 kN load at C, 4 m from A; section D between C and B, 6 m from B.] ## Short answer Using moment area theorems, slope at A = 533.3/EI clockwise, slope at B = 426.7/EI anticlockwise. Deflection at D = 1840/EI downward (units: rad and m for EI in kNm^2). The method uses tangent deviation from A and area moments of the triangular M/EI diagram. ## Answer #### Given data Simply supported beam AB, $L = 12$ m, 60 kN at C (4 m from A). D is 6 m from B (6 m from A). $EI$ constant. ```text 60 kN A ------v--------D-------- B ^ 4 m C 2 m 6 m ^ ``` #### Reactions and bending moments $$ R_A = \frac{60\times8}{12} = 40\ \text{kN},\qquad R_B = 20\ \text{kN} $$ $$ M_C = 40\times4 = 160\ \text{kNm},\qquad M_D = 20\times6 = 120\ \text{kNm} $$ The $M/EI$ diagram is a triangle A-C-B with peak $160/EI$ at C. #### Slope at A Deviation of B from the tangent at A (Theorem II): area of the triangle $= \tfrac12\times12\times160 = 960$; its centroid is $(0 + 4 + 12)/3 = 5.333$ m from A, i.e. 6.667 m from B. $$ \Delta_{B/A} = \frac{960\times6.667}{EI} = \frac{6400}{EI} $$ $$ \theta_A = \frac{\Delta_{B/A}}{L} = \frac{6400}{12\,EI} = \frac{533.33}{EI}\ \text{(clockwise)} $$ #### Slope at B (Theorem I) $$ \theta_B = \theta_A - \text{Area}_{AB} = \frac{533.33 - 960}{EI} = -\frac{426.67}{EI} $$ so $\theta_B = 426.67/EI$ (anticlockwise). #### Deflection at D Deviation of D from the tangent at A (moment of the $M/EI$ area between A and D about D): - Triangle A to C: area $= \tfrac12\times4\times160 = 320$, centroid 3.333 m from D: $320\times3.333 = 1066.67$ - Rectangle C to D: $2\times120 = 240$, centroid 1 m from D: $240$ - Triangle (extra height 40 at C, zero at D): $\tfrac12\times2\times40 = 40$, centroid 1.333 m from D: $53.33$ $$ \Delta_{D/A} = \frac{1066.67 + 240 + 53.33}{EI} = \frac{1360}{EI} $$ Tangent offset at D: $\theta_A\times6 = \dfrac{533.33\times6}{EI} = \dfrac{3200}{EI}$. $$ y_D = \frac{3200 - 1360}{EI} = \frac{1840}{EI}\ \text{(downward)} $$ **Answer:** $\theta_A = 533.3/EI$ (clockwise), $\theta_B = 426.7/EI$ (anticlockwise), deflection at D $= 1840/EI$ downward (kNm$^3$/EI, i.e. in metres with $EI$ in kNm$^2$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 12 m beam with 60 kN at C (4 m from A), what are R_A and the peak M/EI? A. R_A = 20 kN; peak 80/EI B. R_A = 30 kN; peak 120/EI C. R_A = 40 kN; peak 160/EI D. R_A = 40 kN; peak 240/EI Answer: C. R_A = 60x8/12 = 40 kN, so M_C = 40x4 = 160 kNm and the M/EI diagram is a triangle with peak 160/EI at C. (Key point: Reactions and peak M/EI) 2. How is theta_A found? A. Deviation of B from the tangent at A (moment of M/EI area about B) divided by L B. Area of the M/EI diagram divided by L C. Area of M/EI divided by EI only D. Moment of the M/EI area about A divided by L Answer: A. B does not deflect, so the tangent at A has to pass the deviation of B: theta_A = Delta_B/A / L = 6400/(12EI) = 533.33/EI. (Key point: Slope at A using Theorem II) 3. Given theta_A = 533.33/EI and total M/EI area 960, what is the slope at B? A. 533.33 + 960 = 1493.33/EI B. 533.33 - 960 = -426.67/EI, i.e. 426.67/EI anticlockwise C. 960/12 = 80/EI D. 533.33/EI, same as A Answer: B. Theorem I: slope change between A and B equals the M/EI area, so theta_B = theta_A - 960 = -426.67/EI (opposite sense to theta_A). (Key point: Slope at B using Theorem I) 4. How is the deflection at D obtained? A. Deviation of D from tangent at A minus the tangent offset theta_A x 6 B. Tangent offset theta_A x 6 (3200/EI) minus the deviation of D from the tangent at A (1360/EI) C. Add the tangent offset and the deviation: 3200 + 1360 D. Moment of the whole M/EI area about D Answer: B. The tangent at A is 3200/EI away from the original line at D, and D lies 1360/EI back from that tangent, so y_D = 3200 - 1360 = 1840/EI downward. 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