--- title: "Frame AFD, SFD, BMD (20 kNm, 15 kN)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-20knm-15kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-20knm-15kn site: BCT Engineering --- # Frame AFD, SFD, BMD (20 kNm, 15 kN) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2078 Bhadra, 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-20knm-15kn - 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 given loaded frame. Also show the salient feature. [Figure: left column A-B, A hinged at the base, carrying a horizontal distributed load varying from 10 kN/m at A to 5 kN/m at B; beam B-C with a 5 kN/m UDL; a 20 kNm couple at C; right column from C down to a hinged support G; a 10 kN force at $30^\circ$ acting on the right column; a 15 kN downward force at the end E of the beam extension; horizontal dimensions 4 m and 2 m; vertical dimensions 3 m and 3 m on the right.] ## Short answer Reactions: H_A=39.67 kN←, V_A=39.00 kN↓; H_G=14.00 kN←, V_G=79.00 kN↑. Maximum bending moment 107.99 kN·m at C (top tension). On AB, local moment maximum 90.69 kN·m at 5.01 m from A. Contraflexure point on BC at 2 m from B. Axial forces constant on each member. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = 6 m; beam B–C = 4 m; column C–G = 6 m with G hinged; overhang C–E = 2 m carrying 15 kN downward at E. - Load on AB: horizontal, varying from 10 kN/m at A to 5 kN/m at B (acting →); 5 kN/m downward on BC; 20 kN·m couple at C (clockwise); 10 kN at 30° to the horizontal acts at mid-height of CG, to the right and downward. - Two hinged supports with no internal hinge are indeterminate, so a hinge is assumed at the mid-span of BC. 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 G ``` #### 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. - $H_A$ = 39.67 kN ← - $V_A$ = 39.00 kN ↓ - $H_G$ = 14.00 kN ← - $V_G$ = 79.00 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 | 39.00 | 39.67 | 0.00 | | AB at B | 39.00 | -5.33 | 87.99 (right face) | | BC at B | -5.33 | -39.00 | 87.99 (bottom) | | BC at C | -5.33 | -59.00 | -107.99 (top) | | CG at C | -74.00 | 5.33 | -57.99 (right face) | | CG at G | -79.00 | 14.00 | 0.00 | #### Shape of the diagrams - AB: V parabolic, M cubic - BC: V linear, M parabolic - CG: 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 = 107.99 kN·m at C (on BC), tension on the top - V = 0 at 5.01 m from A along AB: M = 90.69 (right face) kN·m (local maximum) - Point of contraflexure (M = 0) at 2.00 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. Why is a hinge assumed at the mid-span of BC? A. Two hinged supports give 4 reactions against 3 equations, so one extra condition (M = 0) is needed B. To remove the 20 kN·m couple C. To make AB a pinned bar D. Because BC has a UDL Answer: A. The moment at the hinge equals zero, supplying the fourth equation for H_A, V_A, H_G and V_G. (Key point: Reason for assumed hinge) 2. AB carries a horizontal load varying linearly from 10 kN/m to 5 kN/m. What are the shapes of V and M in AB? A. V linear, M parabolic B. V constant, M linear C. V parabolic, M cubic D. V zero, M constant Answer: C. A linearly varying (trapezoidal) load gives parabolic shear and cubic moment. (Key point: Shape on loaded column AB) 3. Where is the maximum bending moment of 107.99 kN·m and where is the tension? A. At B, bottom B. At C on BC, tension on top C. At G, right face D. At the hinge, bottom Answer: B. The rigid joint C, where the couple and overhang load act, has the largest moment, with the top fibre in tension. (Key point: Maximum bending moment) 4. Where does V = 0 occur in column AB and what does it give? A. At B, the moment is zero B. At 5.01 m from A, a local maximum moment of 90.69 kN·m C. At A, the maximum moment D. At G, the moment is 107.99 Answer: B. Because dM/dx = V, the moment has a local maximum where shear is zero, here 5.01 m from A. 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