--- title: "Plastic moment capacity of frame" subject: "Theory of Structures II" chapter: 5 key: "semester_5/theory_of_structures_ii:plastic-moment-capacity-frame-80kn-20knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/plastic-moment-capacity-frame-80kn-20knm site: BCT Engineering --- # Plastic moment capacity of frame Theory of Structures II · Chapter 5: Introduction to plastic analysis - Asked in: 2076 Chaitra, 10 marks - Video: [Plastic Analysis Problem No - 23 ( Finding Plastic Moment Capacity in a Frame )](https://www.youtube.com/watch?v=HsnNrS1vR3M) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/plastic-moment-capacity-frame-80kn-20knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5 ## Question Find the plastic moment capacity of the frame shown in figure below. [Figure: portal frame ABCD, A and D hinged; column AB (3I) with 20 kN/m UDL horizontally; beam BC (1.5I) with 80 kN vertical at 2 m from B, span 6 m (2 m + 4 m); column CD (2I) with 100 kN horizontal at 2 m above D, heights 2 m + 2 m marked.] ## Short answer Applying virtual work to beam, sway, and combined mechanisms, the largest required plastic moment capacity is from the sway mechanism: Mp = 120 kN·m. ## Answer #### Assumptions - Portal frame: A (0,0) and D (6,0) hinged, B (0,4), C (6,4); both columns are 4 m high, the beam is 6 m with the 80 kN at 2 m from B. - The plastic moments are in proportion to $I$: $M_{AB} = 3M_p$, $M_{BC} = 1.5M_p$, $M_{CD} = 2M_p$, where $M_p$ is the plastic moment capacity to be found. The loads are taken as ultimate (collapse) loads: 20 kN/m on AB and 100 kN on CD (2 m above D), both horizontal towards the right, and 80 kN vertical on the beam. - Method: virtual work for each mechanism, $M_p\sum(\text{factor}\times\theta) = \sum P\delta$; the required $M_p$ is the largest value obtained from all mechanisms. - Two hinged bases give 4 reactions, so the degree of indeterminacy is 1 and a mechanism needs 2 hinges (sway) or 3 hinges (beam). #### Mechanisms **Mechanism 1: beam mechanism** No sway. Hinges at B, under the 80 kN load (2 m from B) and at C, all in the beam ($1.5M_p$ is smaller than the column moments $3M_p$ and $2M_p$). The beam parts rotate $2\theta/3$ (hinge at B) and $\theta/3$ (hinge at C), with $\theta$ under the load. | Hinge | Rotation | Plastic moment | Internal work | |---|---|---|---| | under the 80 kN load | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ | | B (beam end) | $\frac{2}{3}\theta$ | $\frac{3}{2}M_p$ | $1\,M_p\theta$ | | C (beam end) | $\frac{1}{3}\theta$ | $\frac{3}{2}M_p$ | $\frac{1}{2}\,M_p\theta$ | Internal work $= 3\,M_p\,\theta$. | Load | Displacement along the load | External work | |---|---|---| | 80 kN | $\frac{4}{3}\theta$ | $\frac{320}{3}\,\text{kN·m}\,\theta$ | | 100 kN | $0\theta$ | $0\,\text{kN·m}\,\theta$ | | 20 kN/m on AB | - | $0\,\text{kN·m}\,\theta$ | **Equation:** $M_p\times 3\,\theta = \frac{320}{3}\,\theta$ (kN·m), so $M_p = \frac{320}{9} = 35.56$ kN·m. **Mechanism 2: sway mechanism** Both columns rotate $\theta$ about their hinged bases, the beam translates by $4\theta$ to the right. Hinges form at B and C in the beam, each rotating $\theta$. | Hinge | Rotation | Plastic moment | Internal work | |---|---|---|---| | B (beam end) | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ | | C (beam end) | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ | Internal work $= 3\,M_p\,\theta$. | Load | Displacement along the load | External work | |---|---|---| | 80 kN | $0\theta$ | $0\,\text{kN·m}\,\theta$ | | 100 kN | $2\theta$ | $200\,\text{kN·m}\,\theta$ | | 20 kN/m on AB | - | $160\,\text{kN·m}\,\theta$ | **Equation:** $M_p\times 3\,\theta = 360\,\theta$ (kN·m), so $M_p = 120$ kN·m. **Mechanism 3: combined mechanism** Beam and sway mechanisms combined so that the hinge at B cancels; hinges under the 80 kN load and at C. | Hinge | Rotation | Plastic moment | Internal work | |---|---|---|---| | under the 80 kN load | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ | | C (beam end) | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ | Internal work $= 3\,M_p\,\theta$. | Load | Displacement along the load | External work | |---|---|---| | 80 kN | $\frac{4}{3}\theta$ | $\frac{320}{3}\,\text{kN·m}\,\theta$ | | 100 kN | $\frac{4}{3}\theta$ | $\frac{400}{3}\,\text{kN·m}\,\theta$ | | 20 kN/m on AB | - | $\frac{320}{3}\,\text{kN·m}\,\theta$ | **Equation:** $M_p\times 3\,\theta = \frac{1040}{3}\,\theta$ (kN·m), so $M_p = \frac{1040}{9} = 115.56$ kN·m. #### Required plastic moment The largest required $M_p$ comes from the sway mechanism (the others give 35.56 and 115.56 kN·m); a frame with $M_p$ smaller than this would collapse in sway, while a larger value is safe against all mechanisms. **Answer: $M_p = 120$ kN·m** (so $M_{AB} = 360$, $M_{BC} = 180$, $M_{CD} = 240$ kN·m). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For finding the required plastic moment, which mechanism value is chosen? A. The smallest M_p B. The largest M_p from all mechanisms C. The mean value D. The beam mechanism value Answer: B. Loads are fixed, so the frame must resist all mechanisms; the largest required M_p governs. (Key point: Selection rule) 2. How are the plastic moments of members related? A. In proportion to I: AB = 3Mp, BC = 1.5Mp, CD = 2Mp B. All equal Mp C. Inversely proportional to I D. AB = Mp, BC = 3Mp Answer: A. Plastic moments are proportional to I. (Key point: Moment proportions) 3. In sway, hinges form at B and C in the beam. What is M_p? A. 35.56 kN.m B. 115.56 kN.m C. 120 kN.m D. 360 kN.m Answer: C. 3 Mp theta = 360 theta gives M_p = 120 kN.m. (Key point: Sway mechanism) 4. Two hinged bases give indeterminacy of 1, so a sway mechanism needs: A. 1 hinge B. 2 hinges C. 3 hinges D. 4 hinges Answer: B. Mechanism needs r+1 hinges = 2 hinges for sway, and 3 hinges for the beam. 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