--- title: "Frame SFD, BMD (100 kN, 20 kN/m, hinge)" subject: "Strength of Materials" chapter: 1 key: "semester_3/strength_of_materials:frame-sfd-bmd-100kn-hinge-60deg" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/frame-sfd-bmd-100kn-hinge-60deg site: BCT Engineering --- # Frame SFD, BMD (100 kN, 20 kN/m, hinge) Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments - Asked in: 2081 Baisakh, 12 marks - Video: [Frame Analysis || Shear Force & Bending Moment Diagram](https://www.youtube.com/watch?v=0pG1to_L44M) — Civil Engineering - 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-sfd-bmd-100kn-hinge-60deg - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1 ## Question Draw SFD and BMD for the following frame. Also indicate the salient points. [Figure: portal-type frame with a 4 m high left column carrying a 20 kN/m distributed load on its side and hinged at its base; a horizontal beam of two 2 m segments carrying a 100 kN point load at the middle of the top beam; an internal hinge at the right end of the horizontal beam; an inclined member going down from the hinge at $60^\circ$ to a hinged support, carrying a 30 kN load perpendicular to its length at mid-length; horizontal distance between the two supports 4 m.] ## Short answer Reactions from equilibrium and hinge condition: H_A = 36.08 kN left, V_A = 53.92 kN up, H_F = 17.94 kN left, V_F = 61.08 kN up. Maximum bending moment 92.15 kNm at C (bottom tension). Column AB peak 32.54 kNm at 1.80 m from A; inclined DF peak 34.64 kNm at mid-length. ## Answer **Assumptions (taken from the figure):** - A(0,0) hinged; column AB = 4 m; beam B–C–D = 2 m + 2 m; internal hinge at D (x = 4 m, y = 4 m). - The inclined member DF slopes down at 60° to the horizontal to a hinged support F at the level of A, so F = (6.31 m, 0) and DF = 4.62 m. - 20 kN/m acts horizontally (→) over the full 4 m of AB; 100 kN acts downward at C; 30 kN acts perpendicular to DF at its mid-length, pointing down and to the left. 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------------Do | \\ | \ | \\ | \\ | \ | \\\ | \ | \\ | \ | \\ | \\ | \ A \F ``` #### 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. - $H_A$ = 36.08 kN ← - $V_A$ = 53.92 kN ↑ - $H_F$ = 17.94 kN ← - $V_F$ = 61.08 kN ↑ Check: the three equilibrium equations of the complete frame and the hinge condition are satisfied to rounding. #### Axial force, shear force and bending moment at the key sections | Section | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB at A | -53.92 | 36.08 | 0.00 | | AB at B | -53.92 | -43.92 | -15.69 (left face) | | BC at B | -43.92 | 53.92 | -15.69 (top) | | BC at C | -43.92 | 53.92 | 92.15 (bottom) | | CD at C | -43.92 | -46.08 | 92.15 (bottom) | | CD at D | -43.92 | -46.08 | 0.00 | | DF at D | -61.87 | 15.00 | 0.00 | | DF at mid-length (30 kN load) | -61.87 | 15.00 / -15.00 | 34.64 (lower-left face) | | DF at F | -61.87 | -15.00 | 0.00 | #### Shape of the diagrams - AB: V linear, M parabolic - BC: V constant, M linear - CD: V constant, M linear - DF: V constant on each half (it changes from +15 to -15 kN at the 30 kN load), M linear on each half (triangular, peak at mid-length) - AFD: constant between loads on a member, changing only where a load component acts along the member. - The 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 = 92.15 kN·m at C (on BC), tension on the bottom - V = 0 at 1.80 m from A along AB: M = 32.54 kN·m (right face), the local maximum - On DF the moment rises from 0 at D to $15\times2.31 = 34.64$ kN·m under the 30 kN load and falls back to 0 at F (tension on the lower-left, inner face) - Point of contraflexure (M = 0) at 3.61 m from A along AB - Point of contraflexure (M = 0) at 0.29 m from B along BC - The bending moment is zero at the internal hinge D (from both sides). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What extra condition does the internal hinge at D give for finding the four support reactions? A. The shear force at D is zero B. The axial force at D is zero C. The bending moment at D is zero (taking moments of forces on one side) D. The reactions at A and F are equal Answer: C. A hinge cannot transmit moment. This fourth equation together with the three equilibrium equations makes the reactions determinate. (Key point: Hinge condition) 2. Where is the maximum bending moment of 92.15 kN·m and on which face is the tension? A. At C under the 100 kN load, tension at the bottom B. At D, tension on top C. At A, tension on the left D. At B, tension on the left face Answer: A. The BMD of the beam BCD peaks under the 100 kN point load at C, where the beam sags with tension at the bottom. (Key point: Maximum moment location) 3. The 30 kN load acts perpendicular to DF at mid-length. What shear does DF carry at D and at F, and what is M at the ends? A. V = +15 kN at D and -15 kN at F; M = 0 at both ends B. V = 30 kN at both ends; M = 0 C. V = 15 kN at both ends; M = 92.15 D. V = 0 at both ends; M = 15 Answer: A. A central load is shared equally by the two hinged ends, giving +15 and -15 kN, and the moment is zero at hinges. (Key point: Loading on inclined member) 4. Column AB carries a uniformly distributed horizontal load of 20 kN/m. What are the shapes of V and M in AB? A. V constant, M linear B. V parabolic, M cubic C. V linear, M parabolic D. V zero, M constant Answer: C. Under a UDL the shear varies linearly and the moment parabolically, with a local maximum where V = 0 (1.80 m from A). 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