--- title: "Frame AFD, SFD, BMD (12 kN, 25 kNm, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-12kn-25knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-12kn-25knm site: BCT Engineering --- # Frame AFD, SFD, BMD (12 kN, 25 kNm, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2079 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-12kn-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 shown in the figure below. Also indicate the salient features. [Figure: hinged support at A at the foot of the left column A-B (3 m) and B-C (3 m above); a 15 kN force at $30^\circ$ acting at B; beam C-D-E-F at the top with a triangular load increasing from zero at C to 5 kN/m at D and an internal hinge at D; a 25 kNm couple at E; a 12 kN downward point load at F; a column from E down to a hinged support G; horizontal dimensions 3 m, 3 m, 2 m; vertical dimensions 3 m and 2 m on the right.] ## Short answer Reactions: H_A=12.54 kN←, V_A=2.09 kN↓, H_G=0.45 kN←, V_G=29.09 kN↑. Maximum bending moment 51.26 kN·m at E (top tension); zero at hinge D. Overhang EF: constant shear -12 kN, moment 24 kN·m at E. Axial force constant on each member, e.g., AB 2.09 kN, EG -29.09 kN. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; AB = BC = 3 m; beam C–D–E–F with CD = DE = 3 m, EF = 2 m; internal hinge at D; column from E (6,6) down to hinged support G, taken 5 m long, so G = (6,1). - 15 kN at B acts at 30° to the horizontal, to the right and downward; triangular load on CD rises from 0 at C to 5 kN/m at D; 25 kN·m couple at E is clockwise; 12 kN downward at F (overhang EF). - The overhang EF is replaced at E by the 12 kN force plus a moment of $12\times2 = 24$ kN·m (hogging); moment in EF is $-12x$. 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 | | | | | | | | | | G | 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. - $H_A$ = 12.54 kN ← - $V_A$ = 2.09 kN ↓ - $H_G$ = 0.45 kN ← - $V_G$ = 29.09 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 | 2.09 | 12.54 | 0.00 | | AB at B | 2.09 | 12.54 | 37.62 (right face) | | BC at B | 9.59 | -0.45 | 37.62 (right face) | | BC at C | 9.59 | -0.45 | 36.26 (right face) | | CD at C | -0.45 | -9.59 | 36.26 (bottom) | | CD at D | -0.45 | -17.09 | 0.00 | | DE at D | -0.45 | -17.09 | 0.00 | | DE at E | -0.45 | -17.09 | -51.26 (top) | | EG at E | -29.09 | 0.45 | -2.26 (right face) | | EG at G | -29.09 | 0.45 | 0.00 | #### Shape of the diagrams - AB: V constant (steps at point loads), M linear - BC: V constant (steps at point loads), M linear - CD: V parabolic, M cubic - DE: V constant (steps at point loads), M linear - EG: 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 = 51.26 kN·m at E (on DE), tension on the top - Bending moment is zero at the internal hinge (value read from both sides: 0.00). - Overhang EF: V = −12 kN (downward, constant), N = 0, M = 0 at F rising linearly to 24 kN·m (hogging, tension on top) at E. At joint E the 51.26 kN·m of DE is balanced by 24 (overhang) + 25 (couple) + 2.26 (column EG) kN·m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the overhang EF with the 12 kN load at F replaced when analysing the frame at E? A. By a 12 kN force only B. By a 12 kN force plus a moment of 24 kN·m (hogging) C. By a moment of 12 kN·m only D. By a 25 kN·m couple Answer: B. Shifting the 12 kN force 2 m to E adds a moment 12 x 2 = 24 kN·m, and the moment in EF is -12x. (Key point: Overhang replaced at E) 2. At joint E, the 51.26 kN·m of DE is balanced by which moments? A. 24 (overhang) + 25 (couple) + 2.26 (column EG) B. 25 + 25 + 1.26 C. 12 + 25 + 14.26 D. Only the 25 kN·m couple Answer: A. The moments meeting at the rigid joint balance: 24 + 25 + 2.26 = 51.26 kN·m. (Key point: Joint E equilibrium) 3. On CD the load increases from zero at C to 5 kN/m at D. What are the shapes of V and M? A. V linear, M parabolic B. V constant, M linear C. V parabolic, M cubic D. V cubic, M parabolic Answer: C. A linearly varying load makes the shear parabolic and the moment cubic. (Key point: Triangular load shapes) 4. What is the bending moment at the internal hinge D, and why? A. Maximum, because the load is greatest there B. Zero, because a hinge cannot transmit bending moment C. 25 kN·m, equal to the couple D. 36.26 kN·m, equal to the moment at C Answer: B. A hinge allows free rotation, so M = 0 from both sides, and this gives the extra equation to find the reactions. 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