--- title: "Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-80kn-30knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-80kn-30knm site: BCT Engineering --- # Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2071 Chaitra, 16 marks - 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-80kn-30knm - 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 loaded as shown in the figure below. [Figure: left column A-B, with A hinged at the base and B at 6 m above A (2 m and 6 m marked on the left); an 80 kN horizontal force acting on the left column at B; a 30 kNm couple at the top-left corner; a 50 kN point load on the beam; beam B-D-E with an internal hinge at D and a triangular load rising to 15 kN/m on the right part ending at E; right column E-F with a 10 kN/m distributed horizontal load and F hinged at the base; horizontal dimensions 2 m, 1 m and 9 m as marked.] ## Short answer Reactions: Ha=12.19→, Va=31.04↑, Hf=32.19←, Vf=86.46↑. AB: M linear 0→-73.13 kN·m left face. BC: M 18.96 bottom at C. Hinge D: M=0. DE: triangular load, M cubic, max -373.13 top at E. EF: horizontal distributed load, M parabolic from -373.13 to 0. Max BM=373.13 kN·m at E. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 6 m; 80 kN horizontal (→) and 30 kN·m clockwise couple at B. - Beam B–C–D–E: BC = 2 m (50 kN downward at C), CD = 1 m, internal hinge at D, DE = 9 m with triangular load from 0 at D to 15 kN/m at E (67.5 kN). - Right column EF = 6 m, F hinged, with 10 kN/m horizontal (←) over its full height. 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 B-------C--D---------------------------------E | | | | | | | | | | | | | | | | | | | | A F ``` #### Support reactions Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$; the internal hinge adds the condition $M_{hinge} = 0$ (moments of all forces on one side of the hinge), which makes the reactions determinate. - $H_A$ = 12.19 kN → - $V_A$ = 31.04 kN ↑ - $H_F$ = 32.19 kN ← - $V_F$ = 86.46 kN ↑ Check: the three equilibrium equations of the complete frame (and the hinge condition, where present) are satisfied to rounding (residuals below $10^{-9}$). #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -31.04 | -12.19 | 0.00 | | AB at B | -31.04 | -12.19 | -73.13 (left face) | | BC at B | -92.19 | 31.04 | -43.13 (top) | | BC at C | -92.19 | 31.04 | 18.96 (bottom) | | CD at C | -92.19 | -18.96 | 18.96 (bottom) | | CD at D | -92.19 | -18.96 | 0.00 | | DE at D | -92.19 | -18.96 | 0.00 | | DE at E | -92.19 | -86.46 | -373.13 (top) | | EF at E | -86.46 | 92.19 | -373.13 (right face) | | EF at F | -86.46 | 32.19 | 0.00 | #### Shape of the diagrams - AB: V constant (steps at point loads), M linear - BC: V constant (steps at point loads), M linear - CD: V constant (steps at point loads), M linear - DE: V parabolic, M cubic - EF: V linear, M parabolic - AFD: constant between loads on a member, changing only where a load component acts along the member. - BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends. #### Salient features - Maximum bending moment = 373.13 kN·m at E, tension on the right face - Point of contraflexure (M = 0) at 1.39 m from B along BC - Bending moment is zero at the internal hinge (value read from both sides: 0.00). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Total triangular load on DE (0 to 15 kN/m over 9 m) is: A. 135 kN B. 15 kN C. 67.5 kN D. 45 kN Answer: C. 1/2 x 9 x 15 = 67.5 kN. (Key point: Triangular load) 2. On DE (triangular load) and EF (uniform horizontal load) the shapes are: A. DE: V linear, M parabolic; EF: V constant, M linear B. DE: V constant, M linear; EF: V parabolic, M cubic C. Both: V constant, M linear D. DE: V parabolic, M cubic; EF: V linear, M parabolic Answer: D. Triangular load gives parabolic V; UDL gives linear V. (Key point: Shapes) 3. The maximum bending moment is: A. 373.13 kN·m at E, tension on the right face B. 73.13 kN·m at B C. 43.13 kN·m at B on the beam D. 18.96 kN·m at C Answer: A. The moment grows to 373.13 kN·m at the corner E. (Key point: Maximum moment) 4. Why is there a point of contraflexure at 1.39 m from B on BC? A. Shear force is zero there B. M changes from -43.13 (top) to +18.96 (bottom), so it passes through zero C. The hinge is there D. Axial force changes sign there Answer: B. A contraflexure point is where bending moment changes sign. 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