--- title: "Frame AFD, SFD, BMD (inclined column, triangular)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-30knm-20kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-30knm-20kn site: BCT Engineering --- # Frame AFD, SFD, BMD (inclined column, triangular) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2073 Shrawan, 12 marks - Video: [Frame Analysis || Shear Force & Bending Moment Diagram](https://www.youtube.com/watch?v=0pG1to_L44M) — Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-30knm-20kn - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw axial force, shear force and bending moment diagram of the frame shown in figure below. [Figure: frame with an inclined left leg A-B (A hinged at the base, 5 m of rise to the level of the 20 kN horizontal force, total height 10 m to B), a 20 kN horizontal force acting on the inclined leg at 5 m height, top member B-C of 10 m carrying a triangular load increasing from zero at B to 30 kN/m at C, right column C-D 10 m high with D fixed; horizontal dimensions 2 m and 10 m.] ## Short answer Assuming hinges at B and C for determinacy, reactions: H_A=0, V_A=50 kN up, H_D=20 kN left, V_D=100 kN up, M_D=200 kN·m. Diagrams show max bending moment 200.0 kN·m at D (tension left face of CD); zero at hinges B, C; AB peaks 50.0 kN·m under 20 kN load; BC local max 192.45 kN·m at 5.77 m from B. ## Answer **Assumptions (taken from the figure):** - Inclined leg A(0,0) to B(2,10) (length 10.2 m), A hinged; 20 kN horizontal (→) at mid-length (5 m height). - Beam B–C = 10 m, triangular load 0 at B to 30 kN/m at C (total 150 kN downward). Column CD = 10 m, D fixed. - With A hinged, D fixed and no internal hinge, the frame is indeterminate (5 reactions), so hinges are assumed at B and C to make it determinate. Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame. ```text Bo------------------------Co | | | | | | || | | | || | | | | | | | | | | | || | A D ``` #### Support reactions Hinge at B (forces on AB): $-2V_A + 10H_A + 20(5) = 0$. Hinge at C (forces on AB and BC): $-12V_A + 10H_A + 20(5) + 150(3.33) = 0$. Together with the equilibrium of the whole frame: - $H_A$ = 0.00 kN - $V_A$ = 50.00 kN ↑ - $H_D$ = 20.00 kN ← - $V_D$ = 100.00 kN ↑ - $M_D$ = 200.00 kN·m (anticlockwise) Check: the three equilibrium equations of the complete frame and both hinge conditions are satisfied. #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -49.03 | 9.81 | 0.00 | | AB at 20 kN load (mid-length) | -49.03 / -52.95 | 9.81 / -9.81 | 50.00 (right face) | | AB at B | -52.95 | -9.81 | 0.00 | | BC at B | -20.00 | 50.00 | 0.00 | | BC at C | -20.00 | -100.00 | 0.00 | | CD at C | -100.00 | 20.00 | 0.00 | | CD at D | -100.00 | 20.00 | 200.00 (left face) | #### Shape of the diagrams - AB: V constant on each half (it changes sign at the 20 kN load), M linear on each half (triangular, peak 50 kN·m at mid-length) - BC: V parabolic, M cubic (a simply supported span between the hinges B and C) - CD: V constant, M linear - AFD: constant between loads on a member, changing only where a load component acts along the member. - The BMD is drawn on the tension side of each member and is zero at hinges and at free ends. #### Salient features - Maximum bending moment = 200.00 kN·m at D (on CD), tension on the left face - V = 0 at 5.77 m from B along BC: M = 192.45 kN·m (bottom), the local maximum ($= wL^2/9\sqrt3$) - On AB the moment rises from 0 at A to $9.81\times5.10 = 50.0$ kN·m under the 20 kN load and falls back to 0 at B (tension on the right, inner face) - The bending moment is zero at the hinges B and C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are hinges assumed at B and C? A. A hinged and D fixed gives 5 reactions, so extra hinges are needed for a determinate frame B. To reduce the load on the beam C. Because the leg is inclined D. To make the axial force zero Answer: A. With 5 unknown reactions and only 3 equilibrium equations, assumed hinges (M = 0) supply the missing equations. (Key point: Indeterminacy) 2. The total load on BC (0 to 30 kN/m over 10 m) is: A. 300 kN B. 150 kN C. 30 kN D. 75 kN Answer: B. 1/2 x 10 x 30 = 150 kN. (Key point: Triangular load resultant) 3. Maximum bending moment is: A. 192.45 kN·m at B B. 150 kN·m at C C. 200 kN·m at fixed support D, tension on left face D. Zero everywhere Answer: C. M_D = 200 kN·m is the largest; the 192.45 value is only a local maximum on BC. (Key point: Maximum moment) 4. On BC, V = 0 at 5.77 m from B. What is the moment there? A. 0 B. 200 kN·m C. 50 kN·m D. 192.45 kN·m (local maximum, bottom) Answer: D. Moment is stationary where shear is zero; the answer gives M = 192.45 at that point. 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