---
title: "Moment distribution: frame with cantilever 15 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-frame-50knm-15kn-cantilever"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-50knm-15kn-cantilever
site: BCT Engineering
---

# Moment distribution: frame with cantilever 15 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2071 Shrawan, 10 marks
- Video: [SA40: Moment Distribution Method: Frames (No Sidesway)](https://www.youtube.com/watch?v=BjbVlb6HvMc) — Dr. Structure
- 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-50knm-15kn-cantilever
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the frame shown in figure below by using the moment distribution method. Draw bending moment diagram. [Figure: beam of 5 m (2I) fixed at the left end with 50 kN/m UDL; 1 m overhang with 15 kN at the end; column (I) of height 4 m below the junction, fixed base.]

## Short answer

Using moment distribution: overhang BC contributes a fixed moment of -15 kNm at B. Fixed-end moments for AB are ±104.17 kNm. After distribution with DFs 0.615 (BA) and 0.385 (BD) and carry-over, final moments are M_AB=-131.60, M_BA=49.29, M_BC=-15.00, M_BD=-34.29, M_DB=-17.15 kNm.

## Answer

**Assumptions:** the beam AB = 5 m (2I) is fixed at A and carries 50 kN/m; it continues as a 1 m overhang BC (2I) with 15 kN at C; the column BD (I, 4 m) runs down from the junction B to its fixed base D. The fixed support A prevents sway, so only joint B rotates.

#### Step 1: Overhang BC
$M_{BC}=15\times1=15$ kNm (hogging), a known fixed moment on joint B.

#### Step 2: Distribution factors at B ($k=EI/L$)
- Joint B: relative stiffness $\Sigma k$ = 0.6500; BA: $k=EI/L$ = 0.4000, DF = 0.6154; BC: overhang, DF = 0 (its moment is a known fixed value); BD: $k=EI/L$ = 0.2500, DF = 0.3846

#### Step 3: Fixed-end moments (clockwise +)
- AB: $\mp\dfrac{wL^2}{12}=\mp\dfrac{50\times5^2}{12}=\mp104.17$ kNm
- BC (overhang): $-15$ kNm at B
- BD: 0

#### Step 4: Distribution
Unbalanced moment at B: $+104.17-15=89.17$ kNm. The balancing moment is $-89.17$ kNm, divided 0.615 : 0.385 between BA and BD, and half of each share is carried over to the fixed ends A and D.

| | AB | BA | BC | BD | DB |
|---|---|---|---|---|---|
| DF | - | 0.615 | 0.000 | 0.385 | - |
| FEM | -104.17 | 104.17 | -15.00 | 0.00 | 0.00 |
| Balance | 0.00 | -54.87 | 0.00 | -34.29 | 0.00 |
| Carry-over | -27.44 | 0.00 | 0.00 | 0.00 | -17.15 |
| **Final M** | **-131.60** | **49.29** | **-15.00** | **-34.29** | **-17.15** |


#### Step 5: Results and BMD (kNm)
Joint B: $49.29+(-15.00)+(-34.29)=0.00$ ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -131.60 | -49.29 | 68.51 at 2.83 m |
| BC | -15.00 | 0.00 | - |
| BD | -34.29 | 17.15 | - |

BM sign: sagging positive for the horizontal members; for the column, positive = tension on the right-hand side when travelling from B down to D. The span AB has hogging at both ends and a sagging span moment; the overhang has a straight-line BMD from 0 at C to 15.00 kNm (hogging) at B.

**Answer: $M_{AB}=-131.60$, $M_{BA}=49.29$, $M_{BD}=-34.29$, $M_{DB}=-17.15$ kNm, $M_{BC}=-15.00$ kNm (overhang, clockwise end moments +).**

## Check yourself

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

1. How is the 1 m overhang BC with the 15 kN end load handled in the moment distribution?
   A. It is given a distribution factor equal to its stiffness share
   B. It is replaced by a known hogging moment of 15 kNm at joint B, and BC gets DF = 0
   C. It is ignored because it carries no UDL
   D. It is replaced by a vertical force of 15 kN at B
   Answer: B. The overhang is statically determinate, so its moment 15 x 1 = 15 kNm is a fixed known moment on joint B and it takes no share of the balancing moment. (Key point: Overhang treated as known moment)

2. What is the fixed-end moment magnitude for span AB (5 m, 50 kN/m UDL)?
   A. 50 x 5^2 / 8 = 156.25 kNm
   B. 50 x 5^2 / 12 = 104.17 kNm
   C. 50 x 5 / 8 = 31.25 kNm
   D. 50 x 5^2 / 24 = 52.08 kNm
   Answer: B. For a fixed-fixed span with a full UDL the FEM is wL^2/12 = 104.17 kNm. (Key point: Fixed-end moment of AB)

3. Why is the unbalanced moment at B equal to 89.17 kNm?
   A. 104.17 + 15, since both moments act the same way
   B. 104.17 - 15, since the overhang moment opposes the FEM of BA
   C. 104.17 - 34.29, using the column moment
   D. 104.17 / 2 + 15
   Answer: B. FEM_BA = +104.17 and the overhang moment is -15, so the sum at B is 104.17 - 15 = 89.17 kNm. (Key point: Unbalanced moment at B)

4. After the balancing moment of -89.17 kNm is found, how is it used?
   A. Split equally between BA and BD, and fully carried over to A and D
   B. Split 0.615 : 0.385 between BA and BD, and half carried over to the fixed ends A and D
   C. Split 0.615 : 0.385 between BA and BC, with no carry-over
   D. Given wholly to BD because the column is stiffer
   Answer: B. The balancing moment is shared in proportion to DF (BA 0.615, BD 0.385) and half of each share is carried over to the far fixed ends. (Key point: Distribution and carry-over)

## Related in Theory of Structures II

- [Castigliano: fixed-end moment, 15 kN/m, L = 5 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-propped-udl-15knm-5m) — Theory of Structures II, chapter 2, asked 2×
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## Related in other subjects

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