---
title: "Moment distribution: frame with 20 kN, 6 kN/m"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-frame-20kn-6knm-5kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-20kn-6knm-5kn
site: BCT Engineering
---

# Moment distribution: frame with 20 kN, 6 kN/m

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2069 Asar, 10 marks
- Video: [Moment Distribution Method - Analysis of Frames - Problem No 10](https://www.youtube.com/watch?v=wzQPEz-5Kcw) — 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-frame-20kn-6knm-5kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the frame loaded as shown in figure below using the moment distribution method and draw bending moment diagram. Take EI is constant. [Figure: beam A-B-C; A hinged; 20 kN at 2 m from A; B at 5 m (2 m + 3 m); BC = 9 m with 6 kN/m UDL, C roller; column below B of 4 m (2 m + 2 m) fixed at D with 5 kN horizontal at the mid-level.]

## Short answer

Moment distribution with no sway gives M_BA=29.66, M_BC=-53.60, M_BD=23.94, M_DB=8.22 kNm. M_AB=M_CB=0. EI constant.

## Answer

**Assumptions:** the beam A-B-C is horizontal: A is a hinge, AB = 2 m + 3 m = 5 m with 20 kN at 2 m from A, BC = 9 m with 6 kN/m, C is a roller. The column BD is 4 m long (2 m + 2 m), fixed at D, with a 5 kN horizontal load (to the right) at its mid-height. EI is constant. The hinge at A prevents horizontal movement of the beam, so there is **no sway**; only joint B rotates.

#### Fixed-end moments (clockwise +)
- AB ($a=2$, $b=3$): $FEM_{AB}=-\dfrac{20\times2\times3^2}{5^2}=-14.40$, $FEM_{BA}=+\dfrac{20\times2^2\times3}{5^2}=9.60$ kNm; the far end A is hinged, so the modified $FEM_{BA}'=FEM_{BA}-\tfrac12FEM_{AB}=16.80$ kNm
- BC: $\mp\dfrac{6\times9^2}{12}=\mp40.5$ kNm; C is a roller, so $FEM_{BC}'=-40.5-\tfrac12(40.5)=-60.75$ kNm
- BD (5 kN at mid-height, $L=4$): $FEM_{BD}=2.50$, $FEM_{DB}=-2.50$ kNm ($\pm PL/8$ with the sign for a horizontal load to the right)

#### Distribution factors at B ($k=EI/L$; AB and BC use 3/4 because their far ends are hinged)
- Joint B: relative stiffness $\Sigma k$ = 0.4833; BA: $k=EI/L$ = 0.1500 (far end hinged: 3/4 factor), DF = 0.3103; BC: $k=EI/L$ = 0.0833 (far end hinged: 3/4 factor), DF = 0.1724; BD: $k=EI/L$ = 0.2500, DF = 0.5172

#### Moment distribution
| | BA | BC | BD | DB |
|---|---|---|---|---|
| DF | 0.310 | 0.172 | 0.517 | - |
| FEM | 16.80 | -60.75 | 2.50 | -2.50 |
| Balance | 12.86 | 7.15 | 21.44 | 0.00 |
| Carry-over | 0.00 | 0.00 | 0.00 | 10.72 |
| **Final M** | **29.66** | **-53.60** | **23.94** | **8.22** |


Joint B: $29.66+(-53.60)+(23.94)=0.00$ ✓; $M_{AB}=M_{CB}=0$ (hinge and roller).

#### Bending moment diagram
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -29.66 | 12.13 at 2.00 m |
| BC | -53.60 | 0.00 | 36.90 at 5.49 m |
| BD | 23.94 | -8.22 | - |
BM under the 20 kN load: 12.13 kNm. Mid-span BM of BC: 33.95 kNm.

Reactions: A: $H=-10.54$, $V=6.07$, $M=0.00$; C: $H=0.00$, $V=21.04$, $M=0.00$; D: $H=5.54$, $V=46.89$, $M=-8.22$.

**Answer: $M_{BA}=29.66$, $M_{BC}=-53.60$, $M_{BD}=23.94$, $M_{DB}=8.22$ kNm; $M_{AB}=M_{CB}=0$.**

## Check yourself

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

1. Why is only the rotation of joint B considered?
   A. The hinge at A prevents horizontal movement of the beam, so there is no sway
   B. Column BD is infinitely stiff
   C. The 5 kN load is negligible
   D. C is a roller, which prevents rotation
   Answer: A. The hinge at A restrains horizontal translation of the beam, so the frame cannot sway and only B rotates. (Key point: Why no sway)

2. How are the FEMs at B for spans AB and BC modified?
   A. Each is adjusted by half the FEM at the released far end (FEM' = FEM_B - 1/2 FEM_far)
   B. They are doubled
   C. They are unchanged and the stiffness is also unchanged
   D. They are set to zero
   Answer: A. Because A is a hinge and C a roller, their end moments are released, so half of those FEMs are added back at B: 16.80 and -60.75 kNm. (Key point: Modification for hinge and roller ends)

3. Why does BD get the largest DF (0.517)?
   A. It is fixed at D with k = EI/4 = 0.25, larger than the 3/4-reduced stiffness of BA and BC
   B. It carries the horizontal load
   C. It is the shortest member only
   D. DFs are equal for all members
   Answer: A. k_BD = 0.25EI while BA is 0.15EI and BC 0.0833EI, so BD has the largest share 0.25/0.4833. (Key point: Distribution factor of column)

4. What moment is carried over to D?
   A. Half of the 21.44 distributed to BD, i.e. 10.72 kNm
   B. The full 21.44 kNm
   C. Zero since D is fixed
   D. Half of 12.86 kNm
   Answer: A. D is fixed so COF = 1/2: 0.5 x 21.44 = 10.72 kNm, none is carried to the hinged A or roller C. (Key point: Carry-over to column base)

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