--- title: "AFD, SFD, BMD of frame with triangular load" subject: "Applied Mechanics" chapter: 6 key: "semester_1/applied_mechanics:frame-afd-sfd-bmd-10kn-per-m-triangular" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-10kn-per-m-triangular site: BCT Engineering --- # AFD, SFD, BMD of frame with triangular load Applied Mechanics · Chapter 6: Analysis of Beams and Frames - Asked in: 2074 Chaitra, 13 marks - Video: [Frame 2 Analysis// Draw AFD SFD and BMD// Assign Inclined Loads](https://www.youtube.com/watch?v=tqUsUQSGFM4) — Engineer's Home - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-bmd-10kn-per-m-triangular - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6 ## Question Draw axial force, shear force and bending moment diagram and indicate the salient features if any for the given frame loaded as shown in the figure below. [Figure: frame A-B-C-D with pin at A and roller at D; column A-B 5 m + 5 m with 30 kN horizontal load at mid-height; beam B-C 8 m + 2 m carrying a triangular load of up to 10 kN/m, a 20 kN point load, and a 15 kN·m moment at C; column C-D 4 m.] ## Short answer Reactions: A_x=30 kN left, A_y=19.83 kN up, D_y=40.17 kN up. AB column has constant moment 150 kN·m above mid-height. Beam BC: maximum sagging moment 171.7 kN·m at 2.32 m from B (shear zero), moment at C is 15 kN·m balanced by applied couple; column CD has no moment. ## Answer **Reading of the figure:** A is a hinge at (0, 0) and D is a roller at (10, 6). Column AB is 10 m high (5 + 5), with a 30 kN horizontal load to the right at mid-height (5 m). Beam B–C is 10 m long: a triangular load from 10 kN/m at B decreasing to zero at 8 m, a 20 kN point load at 8 m, and a 15 kN·m couple at C (taken counter-clockwise). Column CD is 4 m long, with the roller D at its foot. The roller gives a vertical reaction only. #### Support reactions Triangular load = ½ × 10 × 8 = 40 kN, acting at 8/3 = 2.667 m from B. Moments about A (anticlockwise positive): the 30 kN load at 5 m height pushes to the right, giving −30 × 5 = −150; the 40 kN triangular load gives −40 × 2.667 = −106.67; the 20 kN gives −20 × 8 = −160; the couple gives +15; D_y at 10 m gives +10 D_y. ```text ΣM_A = 0: 10 D_y − 150 − 106.67 − 160 + 15 = 0 D_y = 401.67/10 = 40.17 kN ↑ ``` ΣFx = 0: A_x + 30 = 0, so **A_x = 30 kN ←** ΣFy = 0: A_y + 40.17 − 40 − 20 = 0 → **A_y = 19.83 kN ↑** #### Member forces **Column AB (0 to 10 m)** - N = −19.83 kN (compression), constant. - V = 30 kN from A to mid-height, 0 from mid-height to B. - M = 30y from A to mid-height: 0 → **150 kN·m** at 5 m; constant 150 kN·m up to B (no shear). **Beam BC** (x measured from B; w(x) = 10(1 − x/8)) - N = 0 (no horizontal force). - V(x) = 19.83 − 10x + 0.625x², so V_B = **19.83 kN**; V = 0 at x = 2.32 m; just left of the 20 kN load V(8) = −20.17 kN; just right of it **−40.17 kN**, constant to C. - M(x) = 150 + 19.83x − 5x² + 0.2083x³ - M_B = **150.00 kN·m** - Maximum **M = 171.70 kN·m at x = 2.32 m** - M(4 m) = 162.67, M(8 m) = 95.33 kN·m - M_C (beam end) = 95.33 − 40.17 × 2 = **15.00 kN·m**, balanced by the applied 15 kN·m couple, so the column has no moment. **Column CD (4 m)** - N = **−40.17 kN** (compression); V = 0; M = 0 (the roller reaction acts along the column). #### Values at key points | Point | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | A | −19.83 | 30 | 0 | | Mid AB (5 m) | −19.83 | 30 → 0 | 150 | | B | −19.83 / 0 | 0 / 19.83 | 150 | | 2.32 m from B | 0 | 0 | 171.70 (max) | | 8 m from B | 0 | −20.17 / −40.17 | 95.33 | | C | 0 | −40.17 | 15 → 0 | | D | −40.17 | 0 | 0 | ```text BMD (kN·m) 150 ----(171.7)---- 95.3 \ 15 | C| | | (zero) 150 (constant) | | | 0 A D (0) ``` **Salient features:** the maximum bending moment is 171.7 kN·m in the beam at 2.32 m from B, where the shear is zero; the moment in the column is constant at 150 kN·m between mid-height and B because the shear there is zero; the shear in the beam changes sign once; there is a sudden jump of 20 kN in the SFD at the point load, and the 15 kN·m couple at C produces a drop of 15 kN·m in the moment from the beam end to the column; the BMD in the beam is cubic (triangular load) and linear under no load. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the resultant of the triangular load (10 kN/m to zero over 8 m) and where does it act? A. 40 kN at 2.667 m from B (the 10 kN/m end) B. 80 kN at 4 m from B C. 40 kN at 5.33 m from B D. 10 kN at 8 m Answer: A. Area = 1/2 x 10 x 8 = 40 kN, acting at 8/3 from the larger end. (Key point: Triangular load resultant) 2. What does the roller at D provide? A. Vertical and horizontal reactions B. Only a horizontal reaction C. Only a vertical reaction D. A moment reaction Answer: C. A roller on a horizontal surface resists only perpendicular (vertical) motion. (Key point: Roller reaction) 3. Why is the moment in AB constant at 150 kN·m from mid-height to B? A. The axial force is zero B. The shear there is zero C. The column is unloaded vertically D. The moment is zero Answer: B. dM/dy = V; with V = 0 the moment is constant. (Key point: Column AB moment) 4. Where does the beam moment peak? A. At B, 150 kN·m B. At the point load, 95.33 kN·m C. At C D. At 2.32 m from B where shear is zero, 171.7 kN·m Answer: D. Maximum M occurs where V = 0, at x = 2.32 m. 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