--- title: "Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-10knm-60kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-10knm-60kn site: BCT Engineering --- # Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2067 Asar (old course), 12 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-10knm-60kn - 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 following frame. [Figure: frame with a hinge at the top left; 10 kN/m UDL along the top member; left column height 2 m with a 60 kN horizontal force at its base support; right column 5 m + 3 m with a 2 kN vertical force at 5 m from the top; horizontal dimension 6 m; supports hinged.] ## Short answer Using the internal hinge for determinacy, reactions are H_A=60 kN ←, V_A=30 kN ↑, H_D=0, V_D=32 kN ↑. Maximum bending moment is 45 kN·m at 3 m from B on BC (sagging, bottom tension). Axial forces: AB and CD are in compression (30–32 kN). ## Answer **Assumptions (taken from the figure):** - Left column AB = 2 m, with the internal hinge at its top B (the top-left corner); A(0,6) hinged; the 60 kN horizontal force acts at the support A and is resisted directly by it. - Beam B–C = 6 m with 10 kN/m downward; right column C–D = 8 m, D(6,0) hinged; 2 kN vertical (down) at 5 m below C on CD. 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 | | | | A | | | | | | | | | | D ``` #### 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. - $H_A$ = 60.00 kN ← - $V_A$ = 30.00 kN ↑ - $H_D$ = 0.00 kN ← - $V_D$ = 32.00 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 | -30.00 | 0.00 | 0.00 | | AB at B | -30.00 | 0.00 | 0.00 | | BC at B | 0.00 | 30.00 | 0.00 | | BC at C | 0.00 | -30.00 | 0.00 | | CD at C | -30.00 | 0.00 | 0.00 | | CD at D | -32.00 | 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 - 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 = 45.00 kN·m at 3.00 m from B on BC, tension on the bottom - V = 0 at 3.00 m from B along BC: M = 45.00 (bottom) kN·m (local maximum) - 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. What does the hinge at the top B of the left column imply for the frame? A. The bending moment at B is zero, giving an extra equation for the reactions B. The shear at B is zero C. The column AB carries no axial force D. B can resist any moment Answer: A. A hinge transmits no moment, so M_B = 0 on either side; this extra condition makes the four reactions determinate. (Key point: Internal hinge at B) 2. Beam BC (6 m, 10 kN/m) behaves like a simply supported beam with 30 kN vertical end forces. What is the maximum moment and where? A. 30 kN·m at C B. 60 kN·m at B C. 45 kN·m at mid-span (3 m from B), tension at the bottom D. 90 kN·m at mid-span Answer: C. M_max = wL^2/8 = 10 x 36/8 = 45 kN·m where V = 0, at 3 m from B, causing sagging (bottom tension). (Key point: Maximum bending moment in BC) 3. Axial force in CD changes from -30 kN at C to -32 kN at D. Why? A. The 2 kN vertical load on CD adds 2 kN of compression below its point of application B. Moment at D changes it C. The hinge at D adds 2 kN D. Beam BC passes 32 kN into it Answer: A. Column CD carries 30 kN compression from the beam and the extra 2 kN downward load acts along the column, so below it the compression is 32 kN, equal to V_D. 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