--- title: "Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-100kn-25knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-100kn-25knm site: BCT Engineering --- # Frame AFD, SFD, BMD (100 kN, 25 kN/m, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2075 Asoj, 16 marks - Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR - 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-100kn-25knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw axial force, shear force and bending moment diagrams for the frame. Indicate numerical values at salient points. [Figure: left column A-B-C with A hinged at the base, a 50 kN horizontal force at B (3 m above A), and C a further 2 m above B; top member C-D-E with a 100 kN point load at 2 m from C, an internal hinge at D (4 m from C) and a 25 kN/m UDL over D to E (4 m); an inclined member from E down to a hinged support F located 2 m inside the horizontal projection of E.] ## Short answer After computing support reactions, axial force, shear force and bending moment diagrams are plotted. Maximum bending moment is 433.33 kN·m at E (tension top in DE, right face of EF); zero at hinge D; maximum sagging moment 116.67 kN·m under the 100 kN load on CD. AB has N=-41.67 kN, V=26.67 kN constant. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 3 m, BC = 2 m; 50 kN horizontal (→) at B. Top member C–D–E: CD = 4 m (hinge at D), DE = 4 m. - 100 kN downward at 2 m from C; 25 kN/m downward over DE (total 100 kN). - The inclined member EF goes from E (8, 5) down to the hinged support F (6, 0) at ground level, 2 m inside E. 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 C--------------------D--------------------E | / | // | / | // B / | // | / | // | / | // | / | // 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. - $H_A$ = 26.67 kN ← - $V_A$ = 41.67 kN ↑ - $H_F$ = 23.33 kN ← - $V_F$ = 158.33 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 | -41.67 | 26.67 | 0.00 | | AB at B | -41.67 | 26.67 | 80.00 (right face) | | BC at B | -41.67 | -23.33 | 80.00 (right face) | | BC at C | -41.67 | -23.33 | 33.33 (right face) | | CD at C | -23.33 | 41.67 | 33.33 (bottom) | | CD at 100 kN load (2 m from C) | -23.33 | 41.67 / -58.33 | 116.67 (bottom) | | CD at D | -23.33 | -58.33 | 0.00 | | DE at D | -23.33 | -58.33 | 0.00 | | DE at E | -23.33 | -158.33 | -433.33 (top) | | EF at E | -138.34 | 80.47 | -433.33 (right face) | | EF at F | -138.34 | 80.47 | 0.00 | #### Shape of the diagrams - AB: V constant, M linear - BC: V constant, M linear - CD: V constant on each side of the 100 kN load (it steps by 100 kN there), M linear on each side - DE: V linear, M parabolic - EF: 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; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends. #### Salient features - Maximum bending moment = 433.33 kN·m at joint E (top of DE, right face of EF) - Maximum sagging moment = 116.67 kN·m under the 100 kN load on CD (2 m from C), tension on the bottom - The bending moment is zero at the internal hinge D (from both sides). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What extra condition does the internal hinge at D give? A. Shear force at D is zero B. Bending moment at D is zero, so reactions become determinate C. Axial force at D is zero D. Deflection at D is zero Answer: B. A hinge cannot transmit moment, so M = 0 about D for the forces on one side; this supplies the fourth equation for four reactions. (Key point: Internal hinge condition) 2. What is the total load of the 25 kN/m UDL on DE (4 m), and what are the V and M shapes there? A. 100 kN; V constant, M linear B. 50 kN; V linear, M parabolic C. 100 kN; V linear, M parabolic D. 100 kN; V parabolic, M cubic Answer: C. 25 x 4 = 100 kN. Under a UDL V varies linearly and M parabolically. (Key point: UDL resultant) 3. Where is the maximum bending moment and what is its value? A. 433.33 kN·m at D B. 80 kN·m at B C. 33.33 kN·m at C D. 433.33 kN·m at E, tension on the right face Answer: D. The BMD peaks at the joint E (-433.33 kN·m); the moment at the hinge D is zero. (Key point: Maximum moment) 4. Which set of reactions is correct for supports A and F? A. H_A = 26.67 kN ←, V_A = 41.67 kN ↑, H_F = 23.33 kN ←, V_F = 158.33 kN ↑ B. H_A = 50 kN, V_A = 100 kN, F carries no horizontal force C. V_A = 158.33 kN, V_F = 41.67 kN D. H_A = H_F = 25 kN, V_A = V_F = 100 kN Answer: A. Horizontal: 26.67 + 23.33 = 50 kN balances the 50 kN force; vertical: 41.67 + 158.33 = 200 kN equals 100 + 100 kN of loads. 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