--- title: "Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-30knm-50knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-30knm-50knm site: BCT Engineering --- # Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2078 Kartik, 16 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-50knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw axial force diagram, shear force diagram and bending moment diagram for the frame shown, indicating the salient features. [Figure: left column of height 8 m, hinged at the base; a 50 kNm couple at the top left corner; horizontal top member of 10 m carrying a 30 kN/m UDL, with an internal hinge at its right end; an inclined member descending from the hinge (horizontal projection 3 m, vertical height 5 m) to a hinged support at the bottom right; a 15 kN force acting on the inclined member.] ## Short answer Reactions: H_A=68.75 kN→, V_A=200.00 kN↑; H_D=55.89 kN←, V_D=107.72 kN↑. Maximum bending moment 549.98 kN·m at B (left face). On BC, local maximum 166.67 kN·m at 6.67 m from B; contraflexure at 3.33 m. On CD, moment peaks at 21.87 kN·m at mid-length. A 50 kNm couple at B causes a moment jump. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 8 m; 50 kN·m couple at B (clockwise); beam B–C = 10 m with 30 kN/m downward, internal hinge at C (10, 8). - Inclined member CD: horizontal projection 3 m and vertical height 5 m, so D = (13, 3) is the hinged support (length 5.83 m). - The 15 kN force is taken perpendicular to CD at mid-length, pointing down and to the left. 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-------------------------------Co | \\ | \\ | \ | \\\ | \ | \\ | \\ | D | | | | A ``` #### 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. - $H_A$ = 68.75 kN → - $V_A$ = 200.00 kN ↑ - $H_D$ = 55.89 kN ← - $V_D$ = 107.72 kN ↑ Check: the three equilibrium equations of the complete frame and the hinge condition are satisfied to rounding. #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -200.00 | -68.75 | 0.00 | | AB at B | -200.00 | -68.75 | -549.98 (left face) | | BC at B | -68.75 | 200.00 | -499.98 (top) | | BC at C | -68.75 | -100.00 | 0.00 | | CD at C | -121.12 | 7.50 | 0.00 | | CD at mid-length (15 kN load) | -121.12 | 7.50 / -7.50 | 21.87 (lower-left face) | | CD at D | -121.12 | -7.50 | 0.00 | The moment drops from 549.98 to 499.98 kN·m across joint B because of the 50 kN·m couple applied there. #### Shape of the diagrams - AB: V constant, M linear - BC: V linear, M parabolic - CD: V constant on each half (it changes from +7.5 to -7.5 kN at the 15 kN load), M linear on each half (triangular, peak at mid-length) - 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; it is continuous at rigid joints (apart from the jump caused by an applied couple) and zero at hinges and at free ends. #### Salient features - Maximum bending moment = 549.98 kN·m at B (on AB), tension on the left face - V = 0 at 6.67 m from B along BC: M = 166.67 kN·m (bottom), the local maximum - Point of contraflexure (M = 0) at 3.33 m from B along BC - On CD the moment rises from 0 at C to $7.5\times2.92 = 21.87$ kN·m under the 15 kN load and falls back to 0 at D (tension on the lower-left, inner face) - The bending moment is zero at the internal hinge C (from both sides). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the bending moment at the internal hinge C at the end of beam BC? A. 549.98 kN·m B. 166.67 kN·m C. 499.98 kN·m D. Zero Answer: D. A hinge carries no bending moment, so M = 0 from both BC and CD. (Key point: Moment at hinge C) 2. In BC the shear is 200 kN at B and the UDL is 30 kN/m. Where does V = 0, and what is the moment there? A. At 3.33 m, M = 0 B. At 6.67 m from B, M = 166.67 kN·m (local maximum) C. At 10 m, M = 0 D. At 5 m, M = 549.98 Answer: B. V falls at 30 kN/m, so V = 0 at 200/30 = 6.67 m, where M is a local maximum. (Key point: Zero shear in BC) 3. The 15 kN force acts perpendicular to CD at mid-length. What shear does CD carry? A. +7.5 kN at C and -7.5 kN at D B. +15 kN at both ends C. 0 at both ends D. +100 kN at C Answer: A. The two hinged ends share the central load equally, giving V = +7.5 and -7.5 kN with M = 0 at both ends. (Key point: Load on inclined member) ## Related in Strength of Materials - [Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-80kn-30knm) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (12 kN, 25 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-12kn-25knm) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (30 and 20 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30-20) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (5 kN/m, 10 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10kn) — Strength of Materials, chapter 1, asked 2× - [Frame AFD, SFD, BMD (5 and 10 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-5knm-10knm) — Strength of Materials, chapter 1, asked 2× - [Frame AFD, SFD, BMD (60 kNm, 50 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-50kn-60knm) — Strength of Materials, chapter 1 ## Related in other subjects - [AFD, SFD, BMD of frame with hinge at D](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-8knm-hinge-50kn) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with hinge](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-hinge-4knm-20knm) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with hinge at C](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-hinge-5knm-10knm-triangular) — Applied Mechanics, chapter 6, asked 2× - [AFD, SFD, BMD for frame with internal hinge](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-internal-hinge-5kn-per-m) — Applied Mechanics, chapter 6