---
title: "Moment distribution: portal frame, AFD, SFD, BMD"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-frame-60kn-15knm-afd"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-60kn-15knm-afd
site: BCT Engineering
---

# Moment distribution: portal frame, AFD, SFD, BMD

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2067 Asar, 15 marks
- Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR
- 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-frame-60kn-15knm-afd
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyze the frame shown in figure by the moment distribution method. Also draw AFD, SFD and BMD for the structure. [Figure: portal frame, columns 2EI of 6 m, beam 4EI of 7 m carrying 15 kN/m UDL and 60 kN at 2 m from B; fixed bases.]

## Short answer

Perform moment distribution for sway-prevented case, then apply a unit sway case and combine using k = -R1/R2 = 0.04882. Final moments: M_AB=26.40, M_BA=57.69, M_BC=-57.69, M_CB=54.43, M_CD=-54.43, M_DC=-29.66 kNm. Draw AFD, SFD and BMD from these.

## Answer

**Data:** portal frame with fixed bases A and D, columns AB and CD of height 6 m ($2EI$), beam BC of span 7 m ($4EI$) carrying 15 kN/m and a 60 kN load at 2 m from B. Because the load on the beam is not symmetric about the centre line, the frame sways slightly. The analysis is done in two cases.

#### Step 1: Distribution factors ($k=EI/L$)
- Joint B: relative stiffness $\Sigma k$ = 0.9048; BA: $k=EI/L$ = 0.3333, DF = 0.3684; BC: $k=EI/L$ = 0.5714, DF = 0.6316
- Joint C: relative stiffness $\Sigma k$ = 0.9048; CB: $k=EI/L$ = 0.5714, DF = 0.6316; CD: $k=EI/L$ = 0.3333, DF = 0.3684

#### Step 2: Fixed-end moments (clockwise +)
Beam BC: UDL $\mp\dfrac{15\times7^2}{12}=\mp61.25$ kNm; point load ($a=2$, $b=5$): $\mp\dfrac{60\times2\times5^2}{7^2}=-61.22$ at B and $+\dfrac{60\times2^2\times5}{7^2}=+24.49$ at C. Total $FEM_{BC}=-122.47$ kNm, $FEM_{CB}=85.74$ kNm.

#### Case I: sway prevented (imaginary support at B)
| | AB | BA | BC | CB | CD | DC |
|---|---|---|---|---|---|---|
| DF | - | 0.368 | 0.632 | 0.632 | 0.368 | - |
| FEM | 0.00 | 0.00 | -122.47 | 85.74 | 0.00 | 0.00 |
| Balance | 0.00 | 45.12 | 77.35 | -54.15 | -31.59 | 0.00 |
| Carry-over | 22.56 | 0.00 | -27.08 | 38.68 | 0.00 | -15.79 |
| Balance | 0.00 | 9.98 | 17.10 | -24.43 | -14.25 | 0.00 |
| Carry-over | 4.99 | 0.00 | -12.21 | 8.55 | 0.00 | -7.12 |
| Balance | 0.00 | 4.50 | 7.71 | -5.40 | -3.15 | 0.00 |
| Carry-over | 2.25 | 0.00 | -2.70 | 3.86 | 0.00 | -1.58 |
| Balance | 0.00 | 0.99 | 1.71 | -2.44 | -1.42 | 0.00 |
| Carry-over | 0.50 | 0.00 | -1.22 | 0.85 | 0.00 | -0.71 |
| Further cycles (converged) | 0.30 | 0.61 | 0.61 | -0.35 | -0.51 | -0.25 |
| **Final M** | **30.60** | **61.20** | **-61.20** | **50.91** | **-50.91** | **-25.46** |

Reaction of the imaginary support: $R_1=-2.571$ kN.

#### Case II: sway without external load
Assume FEM $=-100$ kNm at the ends of both columns (equal EI, equal height, equal $\Delta$).
| | AB | BA | BC | CB | CD | DC |
|---|---|---|---|---|---|---|
| DF | - | 0.368 | 0.632 | 0.632 | 0.368 | - |
| FEM | -100.00 | -100.00 | 0.00 | 0.00 | -100.00 | -100.00 |
| Balance | 0.00 | 36.84 | 63.16 | 63.16 | 36.84 | 0.00 |
| Carry-over | 18.42 | 0.00 | 31.58 | 31.58 | 0.00 | 18.42 |
| Balance | 0.00 | -11.63 | -19.94 | -19.94 | -11.63 | 0.00 |
| Carry-over | -5.82 | 0.00 | -9.97 | -9.97 | 0.00 | -5.82 |
| Balance | 0.00 | 3.67 | 6.30 | 6.30 | 3.67 | 0.00 |
| Carry-over | 1.84 | 0.00 | 3.15 | 3.15 | 0.00 | 1.84 |
| Balance | 0.00 | -1.16 | -1.99 | -1.99 | -1.16 | 0.00 |
| Carry-over | -0.58 | 0.00 | -0.99 | -0.99 | 0.00 | -0.58 |
| Further cycles (converged) | 0.14 | 0.28 | 0.72 | 0.72 | 0.28 | 0.14 |
| **Final M** | **-86.00** | **-72.00** | **72.00** | **72.00** | **-72.00** | **-86.00** |

Reaction of the imaginary support: $R_2=52.667$ kN.

#### Step 3: Combination
$k=-\dfrac{R_1}{R_2}=0.04882$ and the final moments are $M=M_I+kM_{II}$:

| End | Case I | k × Case II | Final M (kNm) |
|---|---|---|---|
| AB | 30.60 | -4.20 | 26.40 |
| BA | 61.20 | -3.52 | 57.69 |
| BC | -61.20 | 3.52 | -57.69 |
| CB | 50.91 | 3.52 | 54.43 |
| CD | -50.91 | -3.52 | -54.43 |
| DC | -25.46 | -4.20 | -29.66 |

Joint check: B $0.00$, C $0.00$.

#### Step 4: BMD, SFD and AFD
Bending moment (sagging/inside tension +, kNm):
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 26.40 | -57.69 | - |
| BC | -57.69 | -54.43 | 105.09 at 2.39 m |
| CD | -54.43 | 29.66 | - |
Under the 60 kN load: 103.96 kNm; mid-span: 95.82 kNm.

Shear force and axial force:
| Member | Axial N (+ tension) | Shear at start | Shear at end |
|---|---|---|---|
| AB | -95.82 | -14.01 | -14.01 |
| BC | -14.01 | 95.82 | -69.18 |
| CD | -69.18 | 14.01 | 14.01 |

Shear sign: positive when the force on the start end acts along the member's local y axis (90° anticlockwise from the direction A→B, B→C, C→D). Axial: negative = compression. The vertical reactions are 95.82 kN at A and 69.18 kN at D (total 165.00 kN = 15×7+60), the horizontal reactions are 14.01 kN and -14.01 kN (equal and opposite).

SFD of the beam: 95.82 kN at B, falling by 15 kN/m to 65.82 kN just left of the 60 kN load, then 5.82 kN, and reaching -69.18 kN at C.

**Answer: $M_{AB}=26.40$, $M_{BA}=57.69$, $M_{BC}=-57.69$, $M_{CB}=54.43$, $M_{CD}=-54.43$, $M_{DC}=-29.66$ kNm (clockwise end moments +).**

## Check yourself

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

1. Why is DF_BC (0.632) bigger than DF_BA (0.368)?
   A. Beam stiffness 4EI/7 = 0.571EI exceeds column 2EI/6 = 0.333EI
   B. Beam is shorter
   C. Column is hinged
   D. Because it carries the load
   Answer: A. DF is proportional to EI/L: 0.5714/0.9048 = 0.632 versus 0.3333/0.9048 = 0.368. (Key point: Distribution factors)

2. Why does this frame sway?
   A. Load on the beam is not symmetric about the centreline
   B. Bases are hinged
   C. Columns are unequal
   D. UDL always causes sway
   Answer: A. The 60 kN load at 2 m makes the loading unsymmetric. (Key point: Why sway)

3. How is k found?
   A. k = -R1/R2 from the imaginary support reactions of Case I and II
   B. k = R1 x R2
   C. k = R2/R1
   D. k = FEM ratio
   Answer: A. The correction makes the total reaction at the imaginary support zero. (Key point: Sway factor)

4. Total vertical reaction should be:
   A. 15x7 + 60 = 165 kN
   B. 105 kN
   C. 60 kN
   D. 225 kN
   Answer: A. Vertical equilibrium: UDL 105 kN plus point load 60 kN equals 165 kN. (Key point: Vertical reactions)

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