---
title: "Moment distribution: multi-bay frame, AFD, SFD, BMD"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-multi-column-frame-afd"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-multi-column-frame-afd
site: BCT Engineering
---

# Moment distribution: multi-bay frame, AFD, SFD, BMD

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2065 Shrawan, 20 marks
- Video: [Analysis of a Symmetrical Frame, Moment Distribution Method, Analysis of Frames - Problem No 12](https://www.youtube.com/watch?v=M4bD0ndvkpQ) — 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/moment-dist-multi-column-frame-afd
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Use the moment distribution method to analyze the frame shown in the figure. Draw Axial Force Diagram, Shear Force Diagram and Bending Moment Diagram for the system. [Figure: beam A-B-C-D-E (A and E cantilever ends) with spans of 1.5 m each; vertical loads 2 kN, 5 kN, 5 kN (upward) and 2 kN (upward); member stiffnesses EI, 2EI, 2EI, EI; columns BF (3EI), CG (4EI) and DH (3EI) with fixed bases, heights 2 m, 2 m, 1 m; 10 kN horizontal loads at B and D levels.]

## Short answer

Moment distribution is performed in two stages: non-sway (fixed-end moments from vertical loads and cantilevers) and sway (imposed lateral displacement). The sway correction factor is found from horizontal equilibrium. Final moments are the sum of non-sway and scaled sway moments.

## Answer

#### Assumptions (figure is not fully clear)
- Beam A-B-C-D-E: four members of 1.5 m each; AB and DE are cantilever overhangs ($EI$ each), BC and CD are $2EI$.
- Columns BF ($3EI$, 2 m), CG ($4EI$, 2 m) and DH ($3EI$, 1 m) have fixed bases at F, G, H.
- Vertical loads (upward, as stated): 2 kN at A, 5 kN at the middle of BC, 5 kN at the middle of CD and 2 kN at E. Horizontal loads: 10 kN to the right at B and at D. The beam is axially rigid, so all joints sway equally ($\Delta$): a **sway frame**.
- Clockwise end moments are positive (kN·m).

#### Distribution factors
Relative stiffness $K = EI/L$ (far ends fixed). The overhangs AB and DE do not resist rotation of the joint, so they only transfer fixed moments to B and D.

| Joint | Member | K | DF |
|---|---|---|---|
| B | BC | $2/1.5 = 1.333$ | 0.4706 |
| B | BF | $3/2 = 1.5$ | 0.5294 |
| C | CB | 1.333 | 0.2857 |
| C | CD | 1.333 | 0.2857 |
| C | CG | $4/2 = 2$ | 0.4286 |
| D | DC | 1.333 | 0.3077 |
| D | DH | $3/1 = 3$ | 0.6923 |

#### Stage 1: sway prevented (joints held against side movement)
Fixed-end moments from the vertical loads: for BC and CD, $PL/8 = 5\times1.5/8 = 0.9375$ (upward load, so $M_{BC} = M_{CD} = +0.9375$, $M_{CB} = M_{DC} = -0.9375$). The cantilevers apply $2\times1.5 = 3$ kN·m to joints B ($-3$) and D ($+3$). The 10 kN loads act at the joint levels and appear in the sway equation, not in the FEMs.

| Step | BC | CB | CD | DC | BF | FB | CG | GC | DH | HD | BA | DE |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| DF | 0.4706 | 0.2857 | 0.2857 | 0.3077 | 0.5294 | - | 0.4286 | - | 0.6923 | - | - | - |
| FEM | 0.938 | -0.938 | 0.938 | -0.938 | 0 | 0 | 0 | 0 | 0 | 0 | -3.000 | 3.000 |
| Dist 1 | 0.971 | 0 | 0 | -0.635 | 1.092 | 0 | 0 | 0 | -1.428 | 0 | 0 | 0 |
| C.O. 1 | 0 | 0.485 | -0.317 | 0 | 0 | 0.546 | 0 | 0 | 0 | -0.714 | 0 | 0 |
| Dist 2 | 0 | -0.048 | -0.048 | 0 | 0 | 0 | -0.072 | 0 | 0 | 0 | 0 | 0 |
| C.O. 2 | -0.024 | 0 | 0 | -0.024 | 0 | 0 | 0 | -0.036 | 0 | 0 | 0 | 0 |
| Dist 3 | 0.011 | 0 | 0 | 0.007 | 0.013 | 0 | 0 | 0 | 0.017 | 0 | 0 | 0 |
| C.O. 3 | 0 | 0.006 | 0 | 0 | 0 | 0.006 | 0 | 0 | 0 | 0.008 | 0 | 0 |
| **Final (converged)** | **1.895** | **-0.497** | **0.573** | **-1.590** | **1.105** | **0.553** | **-0.076** | **-0.038** | **-1.410** | **-0.705** | **-3.000** | **3.000** |

Column shears in stage 1: $\Sigma H_1 = \Sigma (M_{top}+M_{bot})/h = -1.344$ kN.

#### Stage 2: sway with no loads
Impose a sway with $6EI\Delta = 100$ (kN·m units). The fixed-end moments $-6(EI)\Delta/h^2$ are: BF $= 3\times100/4 = -75$, CG $= 4\times100/4 = -100$, DH $= 3\times100/1 = -300$ (both ends).

| Step | BC | CB | CD | DC | BF | FB | CG | GC | DH | HD |
|---|---|---|---|---|---|---|---|---|---|---|
| DF | 0.4706 | 0.2857 | 0.2857 | 0.3077 | 0.5294 | - | 0.4286 | - | 0.6923 | - |
| FEM | 0 | 0 | 0 | 0 | -75.00 | -75.00 | -100.00 | -100.00 | -300.00 | -300.00 |
| Dist 1 | 35.29 | 28.57 | 28.57 | 92.31 | 39.71 | 0 | 42.86 | 0 | 207.69 | 0 |
| C.O. 1 | 14.29 | 17.65 | 46.15 | 14.29 | 0 | 19.85 | 0 | 21.43 | 0 | 103.85 |
| Dist 2 | -6.72 | -18.23 | -18.23 | -4.40 | -7.56 | 0 | -27.34 | 0 | -9.89 | 0 |
| C.O. 2 | -9.11 | -3.36 | -2.20 | -9.11 | 0 | -3.78 | 0 | -13.67 | 0 | -4.95 |
| Dist 3 | 4.29 | 1.59 | 1.59 | 2.80 | 4.83 | 0 | 2.38 | 0 | 6.31 | 0 |
| C.O. 3 | 0.79 | 2.14 | 1.40 | 0.79 | 0 | 2.41 | 0 | 1.19 | 0 | 3.15 |
| **Final (converged)** | **38.19** | **27.31** | **56.26** | **96.10** | **-38.19** | **-56.60** | **-83.57** | **-91.79** | **-96.10** | **-198.05** |

Column shears in stage 2: $\Sigma H_2 = -429.22$ kN.

#### Sway correction
The columns must resist the 20 kN of applied horizontal load: $\Sigma H_1 + k\,\Sigma H_2 = -20$, so
$$
k = \frac{-20 - (-1.344)}{-429.22} = 0.04347
$$
Final moment $= M_{stage\,1} + k\,M_{stage\,2}$.

#### Final end moments (kN·m, clockwise +)
| End | BC | CB | CD | DC | BF | FB | CG | GC | DH | HD |
|---|---|---|---|---|---|---|---|---|---|---|
| M | 3.555 | 0.69 | 3.019 | 2.587 | -0.555 | -1.907 | -3.709 | -4.028 | -5.587 | -9.313 |

Joint checks: B: $3.555 - 0.555 - 3 = 0$; C: $0.690 + 3.019 - 3.709 = 0$; D: $2.587 - 5.587 + 3 = 0$.

#### Bending moment diagram (kN·m; + = sagging, tension at bottom)
| Point | BM |
|---|---|
| A | 0 |
| B (left, in AB) | +3.000 |
| B (right, in BC) | +3.555 |
| middle of BC | -0.443 |
| C (left) | -0.690 |
| C (right, in CD) | +3.019 |
| middle of CD | -1.659 |
| D (left) | -2.587 |
| D (right, in DE) | +3.000 |
| E | 0 |

Column moments at the top/base: BF 0.555 / 1.907; CG 3.709 / 4.028; DH 5.587 / 9.313. Each column bends in double curvature (moment changes sign along its height).

#### Shear force diagram (kN)
- Beam: AB +2.00 (constant); BC -5.33 up to the load, then -0.33; CD -6.24, then -1.24; DE -2.00 (constant).
- Columns: BF 1.23, CG 3.87, DH 14.90 (all resisting the horizontal load; $1.23+3.87+14.90 = 20$ kN).

#### Axial force diagram (kN)
- Beam: AB and DE zero; BC 8.77 compression; CD 4.90 compression (the horizontal load is shed to the columns at each joint: $10 - 1.23 = 8.77$, $8.77 - 3.87 = 4.90$, $4.90 + 10 - 14.90 = 0$).
- Columns: BF 7.33 tension, CG 5.91 tension, DH 0.76 tension (the upward loads of 14 kN in total are held down by the bases: $7.33+5.91+0.76 = 14.0$).

**Answer:** Sway factor $k = 0.0435$; final column-end moments: BF -0.555/-1.907, CG -3.709/-4.028, DH -5.587/-9.313 kN·m (top/base). Maximum moment 9.31 kN·m at the base of DH.

## Check yourself

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

1. How is the sway frame analysed?
   A. Stage 1 with sway prevented, Stage 2 with an arbitrary sway and no loads, then combined
   B. Single moment distribution with no sway
   C. Only a sway stage
   D. Using symmetry
   Answer: A. Joints are first held against sway; the sway stage then supplies the correction needed to satisfy the horizontal shear equilibrium. (Key point: Sway frame stages)

2. What is the role of the overhangs AB and DE?
   A. They have DFs like other members
   B. They only transfer their statically determinate moments (-3 and +3 kNm) to joints B and D
   C. They carry the sway
   D. They are removed
   Answer: B. A cantilever cannot resist joint rotation; its load moment 2x1.5 = 3 kNm is applied to the joint as a FEM. (Key point: Overhangs)

3. The sway factor k is obtained from...
   A. k = DF ratio of columns
   B. Sum of vertical loads
   C. Column shear equilibrium: Sum H1 + k Sum H2 = -20 kN (total horizontal load)
   D. Setting moments at A to zero
   Answer: C. The columns must resist the 20 kN applied horizontal loads, which fixes k = 0.0435. (Key point: Sway factor)

4. Why does column DH (1 m) carry the most shear (14.90 kN)?
   A. It has the heaviest load
   B. It is nearest the support
   C. It has the smallest EI
   D. It is the shortest column, so it is stiffest in sway (EI/h^2 effect)
   Answer: D. Sway FEM and shear stiffness vary as EI/h^2 and EI/h^3, so the short column attracts most of the horizontal force. (Key point: Share of shear)

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