---
title: "Collapse mechanism: continuous beam ABCD, BMD"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-continuous-beam-abcd-100kn-6knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-continuous-beam-abcd-100kn-6knm
site: BCT Engineering
---

# Collapse mechanism: continuous beam ABCD, BMD

Theory of Structures II · Chapter 5: Introduction to plastic analysis

- Asked in: 2069 Chaitra, 6 marks
- 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/collapse-continuous-beam-abcd-100kn-6knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

A prismatic continuous beam ABCD is fixed at A and simply supported at B, C and D. It is subjected to factored loads as shown in figure below. Find the collapse mechanism and draw BM diagram. [Figure: 100 kN at E (4 m from A), B at 8 m; 6 kN/m UDL on BC with 75 kN at F; C at 13 m; 50 kN at G, D; dimensions 4 m, 4 m, 5 m, 3 m, 2 m, 2 m as marked.]

## Short answer

Plastic analysis of three spans gives required Mp: AB: 100 kN·m (hinges at A, E, B), BC: 54 kN·m, CD: 33.33 kN·m. Collapse mechanism is span AB failing with Mp = 100 kN·m. Bending moment diagram at collapse shows moments A=-100, E=+100, B=-100, C=-39.31, F=+44.42, G=+30.35 kN·m.

## Answer

#### Assumptions (the figure is not fully clear)
- Continuous beam A-B-C-D, prismatic, with the same plastic moment $M_p$ in all spans: A fixed; B, C and D simple supports. AB = 8 m with 100 kN at E, 4 m from A; BC = 5 m with 6 kN/m over the whole span and 75 kN at F, 3 m from B; CD = 4 m with 50 kN at G, 2 m from C. The given loads are the factored (collapse) loads, so $M_p$ is the value required.
- Degree of indeterminacy $= 6 - 3 = 3$ (A fixed: 3; B, C and D: 1 each). Each span can fail on its own, and the largest $M_p$ among the spans governs.

#### Mechanism 1: span AB (hinges at A, E and B)
AB is fixed at A and continuous at B, so it acts as a fixed-ended beam. With rotation $\theta$ at A and at B and $2\theta$ at E, the deflection under the load is $4\theta$.
$$
M_p(\theta + 2\theta + \theta) = 100\times 4\theta \;\Rightarrow\; M_p = 100\ \text{kN·m}
$$
This is the usual $PL/8 = 100\times 8/8$ for a fixed beam with a central load.

#### Mechanism 2: span BC (hinges at B, F and C)
BC is also treated as a fixed-ended span. Part BF (3 m) rotates $2\phi$ and part FC (2 m) rotates $3\phi$ (since $3\times2\phi = 2\times3\phi$), so the deflection under the point load is $6\phi$. Hinge rotations: B $2\phi$, F $5\phi$, C $3\phi$.
- Internal work $= M_p(2\phi + 5\phi + 3\phi) = 10M_p\phi$.
- External work $= 75\times6\phi + 6\times\tfrac12\times5\times6\phi = 540\,\phi$.
$$
M_p = \frac{540}{10} = 54\ \text{kN·m}
$$

#### Mechanism 3: span CD (hinges at C and G)
CD has a simple end at D, so two hinges are enough. Part CG (2 m) rotates $\theta$ about C, so the deflection under the load is $2\theta$; part GD (2 m) then rotates $\theta$ about D, and the hinge at G turns $2\theta$.
- Internal work $= M_p(\theta + 2\theta) = 3\,M_p\theta$.
- External work $= 50\times2\theta = 100\theta$.
$$
M_p = \frac{100}{3} = 33.33\ \text{kN·m}
$$

#### Collapse mechanism and required $M_p$
| Span | Hinges | Required $M_p$ (kN·m) |
|---|---|---|
| AB | A, E, B | 100 |
| BC | B, F, C | 54 |
| CD | C, G | 33.33 |

The largest requirement is for span AB, so the collapse mechanism is the failure of AB with hinges at A, E and B (a beam mechanism), and **$M_p = 100$ kN·m**. Spans BC and CD stay elastic and carry the moments below.

#### Bending moment diagram at collapse
The hinges give $M_A = -100$, $M_E = +100$ and $M_B = -100$ kN·m. The hinge at B holds the moment $-M_p$ while BC and CD remain elastic as a two-span beam, so the three-moment equation at C (with $M_B = -100$, $M_D = 0$) gives the hogging support moment
$$
M_C = -39.31\ \text{kN·m}
$$
The sagging moments are then $M_F = +44.42$ kN·m under the 75 kN load in BC and $M_G = +30.35$ kN·m under the 50 kN load in CD. All are below $M_p$, so no other hinge forms and the diagram is statically admissible.

| Section | A | E | B | F | C | G | D |
|---|---|---|---|---|---|---|---|
| BM (kN·m) | -100 | +100 | -100 | +44.42 | -39.31 | +30.35 | 0 |

Shape (sagging plotted above the line): in AB, a triangle with a peak of $+100$ at E and $-100$ at A and B; in BC, a curve starting at $-100$ at B, rising to about $+44.42$ at F and falling to $-39.31$ at C; in CD, straight lines from $-39.31$ at C through $+30.35$ at G to 0 at D.

**Answer: $M_p = 100$ kN·m; the collapse mechanism is span AB, with plastic hinges at A, at E (under the 100 kN load) and at B.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the collapse mechanism of the continuous beam ABCD found?
   A. By averaging the Mp of all spans
   B. By finding Mp for each span's own mechanism and taking the largest
   C. By taking the span with the smallest load
   D. By using only the span with the UDL
   Answer: B. Each span can fail alone; the span needing the largest Mp governs the collapse. (Key point: Span-by-span mechanisms)

2. For span AB (fixed at A, hinges at A, E, B) with the 100 kN load at the centre, which equation gives Mp?
   A. Mp(theta + 2theta + theta) = 100 x 4 theta, giving Mp = 100 kN·m
   B. Mp(2theta) = 100 x 4 theta, giving Mp = 200 kN·m
   C. Mp(theta + 2theta + theta) = 100 x 8 theta, giving Mp = 200 kN·m
   D. Mp(3theta) = 100 x 4 theta, giving Mp = 133 kN·m
   Answer: A. Hinge rotations are theta, 2theta and theta, and the load moves 4theta, so 4Mp = 400, Mp = 100 kN·m (PL/8). (Key point: Mechanism of span AB)

3. Why do spans BC and CD stay elastic at collapse, and what check confirms the BM diagram?
   A. They need a larger Mp, and all moments equal Mp
   B. They have no load, so moments are zero
   C. They need less Mp (54 and 33.33) than AB, and all moments stay below Mp (statically admissible)
   D. They form hinges at the same time as AB
   Answer: C. AB needs 100 kN·m, more than BC and CD; the moments in BC and CD (e.g. -39.31, +44.42, +30.35) are all below Mp, so no further hinge forms. (Key point: Collapse span and BM check)

## Related in Theory of Structures II

- [Collapse mechanisms: continuous beam ABCD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-continuous-beam-abcd-80kn-5knm) — Theory of Structures II, chapter 5
- [Slope deflection: continuous beam, 75 kN/m, 190 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-75knm-190kn) — Theory of Structures II, chapter 3
- [Collapse load of fixed beam with UDL](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-fixed-beam-udl) — Theory of Structures II, chapter 5
- [Slope deflection: continuous beam, SFD and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-10knm-3ei) — Theory of Structures II, chapter 3
- [Slope deflection: beam with 50 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-30kn-bmd) — Theory of Structures II, chapter 3
- [Stiffness matrix: continuous beam BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-continuous-beam-10kn-30kn-15knm) — Theory of Structures II, chapter 3

## Related in other subjects

- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Lateral vibration of a simply supported beam](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Abeam-lateral-vibration-simply-supported) — Theory of Mechanism and Machine II, chapter 10
- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4