--- title: "Frame AFD, SFD, BMD (60 kNm, 50 kN)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-afd-sfd-bmd-50kn-60knm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-afd-sfd-bmd-50kn-60knm site: BCT Engineering --- # Frame AFD, SFD, BMD (60 kNm, 50 kN) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2072 Chaitra, 12 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-50kn-60knm - 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 given in figure below. [Figure: inclined member A-B-C rising from a hinged support at A to C (A to B horizontal projection 2 m, B to C 3 m); a 50 kN force and a 60 kNm couple at B; horizontal member C-D of 6 m carrying 5 kN/m UDL; right column D-E of height 4 m with E on a roller support.] ## Short answer Reactions: Ha=0, Va=43.64 kN↑, Ve=36.36 kN↑. AB: N=-30.86, V=30.86, M linear 0→87.27 kN·m. At B a 60 kN·m couple jumps moment to 147.27 kN·m (right face) on BC. CD: V linear -6.36→-36.36, M parabolic max 128.18 bottom at C. DE: N=-36.36, V=0, M=0. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; inclined member A–B–C at 45°: B = (2,2), C = (5,5); 50 kN downward and 60 kN·m clockwise couple at B. - Beam C–D = 6 m with 5 kN/m downward; column D–E = 4 m with E on 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 /C------------------------D / | /// | / | /// | / | /B/ | / | /// E / A/ ``` #### Support reactions Equilibrium of the whole frame gives $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$. - $H_A$ = 0.00 kN → - $V_A$ = 43.64 kN ↑ - $V_E$ = 36.36 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 | -30.86 | 30.86 | 0.00 | | AB at B | -30.86 | 30.86 | 87.27 (right face) | | BC at B | 4.50 | -4.50 | 147.27 (right face) | | BC at C | 4.50 | -4.50 | 128.18 (right face) | | CD at C | 0.00 | -6.36 | 128.18 (bottom) | | CD at D | 0.00 | -36.36 | 0.00 | | DE at D | -36.36 | 0.00 | 0.00 | | DE at E | -36.36 | 0.00 | 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 linear, M parabolic - 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 = 147.27 kN·m at B, tension on the right face ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is H_A = 0? A. Hinge A cannot carry horizontal force B. Roller E gives only a vertical reaction and there is no horizontal load, so sum Fx = 0 leaves H_A = 0 C. The inclined member cancels it D. The couple balances it Answer: B. The only horizontal support is A, and no horizontal external loads act. (Key point: Roller reaction) 2. Why does M jump from 87.27 to 147.27 kN·m at B? A. The 50 kN load causes a jump in M B. The member changes slope C. The 60 kN·m couple applied at B causes a jump equal to its value D. The hinge adds moment Answer: C. A concentrated couple produces a jump in BMD equal to its magnitude; a point load only changes the slope. (Key point: Couple effect) 3. On CD (5 kN/m UDL) V and M shapes are: A. V constant, M linear B. V parabolic, M cubic C. V zero, M constant D. V linear, M parabolic Answer: D. UDL gives linear V and parabolic M. (Key point: UDL on CD) 4. Reactions V_A = 43.64 and V_E = 36.36 kN sum to 80 kN. What loads do they balance? A. 50 kN load plus 5 x 6 = 30 kN UDL B. 50 kN load only C. The 60 kN·m couple plus 50 kN D. The UDL of 30 kN only Answer: A. The couple has no vertical force, so total vertical load is 50 + 30 = 80 kN. 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