--- title: "Frame AFD, SFD, BMD (50 kN, 15 kN/m)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-50kn-15knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-50kn-15knm site: BCT Engineering --- # Frame AFD, SFD, BMD (50 kN, 15 kN/m) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2076 Chaitra, 10 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-50kn-15knm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw axial force shear force and bending moment diagram for a given loaded frame. Also write the salient features. [Figure: frame with hinge support at A at the bottom of the left column (height 6 m to B); 50 kN horizontal force at B at the top left; beam B-C of 8 m span carrying 15 kN/m UDL; right column C-D-E with D and E 2 m apart vertically, a 20 kN horizontal force acting at D, and a roller/hinge support at E.] ## Short answer Reactions: HA=30 kN left, VA=27.5 kN up, VE=92.5 kN up. Maximum bending moment 205.21 kN·m occurs 1.83 m from B on BC (bottom tension). Point of contraflexure at 7.06 m from B on BC. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 6 m; 50 kN horizontal (→) at B; beam BC = 8 m with 15 kN/m downward; right column from C(8,6): CD = 4 m, DE = 2 m (E at ground level). - 20 kN horizontal force at D acts to the left; E is taken as a roller (vertical reaction only). 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 | | | | | | A E ``` #### Support reactions Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$. - $H_A$ = 30.00 kN ← - $V_A$ = 27.50 kN ↑ - $V_E$ = 92.50 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 | -27.50 | 30.00 | 0.00 | | AB at B | -27.50 | 30.00 | 180.00 (right face) | | BC at B | -20.00 | 27.50 | 180.00 (bottom) | | BC at C | -20.00 | -92.50 | -80.00 (top) | | CD at C | -92.50 | 20.00 | -80.00 (right face) | | CD at D | -92.50 | 20.00 | 0.00 | | DE at D | -92.50 | 0.00 | 0.00 | | DE at E | -92.50 | 0.00 | 0.00 | #### Shape of the diagrams - AB: V constant (steps at point loads), M linear - BC: V linear, M parabolic - CD: V constant (steps at point loads), M linear - DE: V constant (steps at point loads), M linear - 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 = 205.21 kN·m at 1.83 m from B on BC, tension on the bottom - V = 0 at 1.83 m from B along BC: M = 205.21 (bottom) kN·m (local maximum) - Point of contraflexure (M = 0) at 7.06 m from B along BC ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is H_A = 30 kN found? A. Taking moments about E B. From sum Fx = 0 using the 50 kN at B and the 20 kN at D (50 - 20 = 30), acting to the left at A C. From V_A and V_E D. From the UDL on BC Answer: B. The net horizontal load is 50 - 20 = 30 kN to the right, and E is a roller, so the hinge at A supplies 30 kN to the left. (Key point: Horizontal reaction) 2. In BC, V = 27.5 kN at B and the UDL is 15 kN/m. Where is V = 0 and what is M? A. At 1.83 m from B, M = 205.21 kN·m (maximum) B. At 8 m from B, M = -80 C. At 4 m from B, M = 180 D. At 0.5 m, M = 0 Answer: A. 27.5/15 = 1.83 m. The moment there is the local maximum, 205.21 kN·m, with the bottom in tension. (Key point: Zero shear point) 3. What are V and M in segment DE? A. V = 20 kN, M = 80 kN·m B. V = 0 and M = 0, because the roller reaction is vertical and passes along the axis of the column C. V = 92.5 kN, M = 0 D. V = 0 and M = -80 Answer: B. No transverse load acts below D, and the roller reaction acts along the member axis, so there is only axial force of 92.5 kN. (Key point: Segment DE) 4. What is the moment at C and which side is in tension? A. -80 kN·m, tension on the top of BC (right face of CD) B. 180 kN·m at the bottom C. 0 D. 205.21 kN·m Answer: A. At the right end of BC the moment is -80 kN·m (hogging), and it is carried around the rigid joint into column CD. 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