--- title: "AFD, SFD, BMD for frame with internal hinge" subject: "Applied Mechanics" chapter: 6 key: "semester_1/applied_mechanics:frame-afd-sfd-internal-hinge-5kn-per-m" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-internal-hinge-5kn-per-m site: BCT Engineering --- # AFD, SFD, BMD for frame with internal hinge Applied Mechanics · Chapter 6: Analysis of Beams and Frames - Asked in: 2073 Chaitra, 14 marks - Video: [Three hinged frames, problem 001 [Structural Mechanics]](https://www.youtube.com/watch?v=73B-iN7mcWg) — Engineering Handbooks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/frame-afd-sfd-internal-hinge-5kn-per-m - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6 ## Question Draw the axial force, shear force and bending moment diagram for the given frame shown in figure below. Also show the salient features. [Figure: frame with pin support at A and roller support at F; internal hinge at C; column A-B-C of 3 m + 3 m with 15 kN load at 30° at B; beam C-D-E with triangular load rising to 5 kN/m, 20 kN·m moment at D and 10 kN downward at E; column D/E-F with 1 m dimension; horizontal dimensions 4 m and 2 m.] ## Short answer Reactions: A_x = 6.50 kN ←, A_y = 7.29 kN ↓, F_x = 6.50 kN ←, F_y = 34.79 kN ↑. Maximum hogging moment is 72.48 kN·m at D. Moment is zero at hinge C, supports A and F, and free end E. ## Answer **Reading of the figure:** A is a hinge support at (0, 0). Column A–B–C is 6 m high (3 + 3 m) with an internal hinge at C (top). The beam C–D–E has C–D = 4 m and D–E = 2 m. The right column D–F is 5 m long (F is 1 m above the level of A, at (4, 1)). Loads: 15 kN at B at 30° (down and to the right), a triangular load from zero at C to 5 kN/m at D, a 20 kN·m couple at D (clockwise), and 10 kN downward at E. **Support at F:** the figure description calls F a roller, but with the internal hinge at C a roller at F would give only 3 reactions for 4 conditions (3 equilibrium equations + M_C = 0). The column AC could then rotate about A, so the frame would be a mechanism. F is therefore taken as a hinge (pin) support, giving 4 reactions. #### Reactions (4 unknowns, 3 equations + hinge condition M_C = 0) Components: 15 kN at B = (12.99 →, 7.50 ↓). Triangular load = ½ × 5 × 4 = 10 kN at 2.667 m from C. **Hinge at C (left part, moments about C):** A_x × 6 + 12.99 × 3 = 0 → **A_x = −6.50 kN (6.50 kN ←)** ```text ΣFx = 0: A_x + F_x + 12.99 = 0 -> F_x = −6.50 kN (6.50 kN ←) ΣM_A = 0 (anticlockwise +): −12.99 × 3 (15 kN, horizontal part at B) = −38.97 −10 × 2.667 (triangular load) = −26.67 −10 × 6 (10 kN at E) = −60.00 −20 (clockwise couple at D) = −20.00 +4 F_y − 1 × F_x = 4 F_y + 6.50 (F is at (4, 1)) => 4 F_y = 139.14, F_y = 34.79 kN ↑ ΣFy = 0: A_y + 34.79 − 7.50 − 10 − 10 = 0 -> A_y = −7.29 kN (7.29 kN ↓) ``` **Reactions: A: 6.50 kN ← and 7.29 kN ↓; F: 6.50 kN ← and 34.79 kN ↑.** #### Member forces | Member | N (kN) | V (kN) | M (kN·m) | |---|---|---|---| | AB | +7.29 (tension) | 6.50 | 0 at A → 19.49 at B | | BC | +14.79 (tension) | −6.50 | 19.49 at B → 0 at C (hinge) | | CD (triangular load) | −6.50 (compression) | −14.79 at C → −24.79 at D | 0 at C → 72.48 at D (hogging); 31.24 at 2 m from C | | DE (overhang) | 0 | +10 | 20 at D → 0 at E (hogging) | | FD (column) | −34.79 (compression) | 6.50 | 0 at F → 32.48 at D | - Hinge forces at C: 6.50 kN horizontal and 14.79 kN vertical (the beam is held by the hinge). - Shear in CD: V(x) = −14.79 − 0.625x² (x from C); M(x) = 14.79x + 0.2083x³ (hogging). - Joint D check: 72.48 (beam) = 32.48 (column) + 20 (overhang) + 20 (couple) ✓. ```text BMD (kN·m) 0 C ______ 31.2 ____ 72.5 D --20-- E 0 | (hinge) | 19.5 .. 0 32.5 | | B 19.5 | | | 0 A 0 F ``` **Salient features:** the bending moment is zero at the internal hinge C, at the supports A and F, and at the free end E; the maximum moment is 72.48 kN·m, hogging at D; the maximum moment in the column AC is 19.49 kN·m at B, where the 15 kN load acts; the shear in the beam does not change sign between C and D; the column AC is in tension (the reaction at A is downward); the 20 kN·m couple at D causes no jump in the shear, only a jump of 20 kN·m in the BMD at D. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is A_x found? A. Taking moments about the hinge C for the left part, M_C = 0 B. From ΣFy = 0 C. By symmetry D. From ΣM about F Answer: A. With 4 unknown reactions, the hinge gives the fourth equation: moment about C of the left portion is zero. (Key point: Hinge condition) 2. What is the bending moment at the internal hinge C? A. Maximum B. Zero C. Equal to 19.49 kN·m D. Equal to the couple 20 kN·m Answer: B. A hinge cannot transmit moment. (Key point: Bending moment at hinge) 3. How does the 20 kN·m couple at D show on the diagrams? A. A jump in the SFD of 20 kN B. No effect C. A jump of 20 kN·m in the BMD, no jump in the SFD D. A jump in the AFD Answer: C. A couple causes a sudden step in moment only. (Key point: Effect of couple at D) 4. Which joint equilibrium check at D is correct? A. 72.48 = 32.48 + 20 + 20 B. 72.48 = 32.48 C. 72.48 = 20 + 20 D. 72.48 = 32.48 - 20 - 20 Answer: A. Beam moment equals column moment plus overhang moment plus couple. 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