---
title: "Moment distribution: frame with 50 kNm"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-frame-50knm-30knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-50knm-30knm
site: BCT Engineering
---

# Moment distribution: frame with 50 kNm

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2078 Kartik, 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-50knm-30knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Draw the bending moment diagram of the frame given below using the moment distribution method. [Figure: horizontal member A-B-C with A fixed; AB (1.5I) carries a 50 kNm moment, dimensions 2 m and 4 m as marked; BC (2I) of 5 m carries 30 kN/m UDL; column BD (I) of height 6 m, fixed at D.]

## Short answer

Overhang BC fixed-end moment -375 kNm. Moment distribution at joint B with DF_BA=0.6, DF_BD=0.4, carry-over yields final clockwise end moments: M_AB=107.50 kNm, M_BA=231.67 kNm, M_BC=-375.00 kNm, M_BD=143.33 kNm, M_DB=71.67 kNm.

## Answer

**Assumptions:** AB = 2 m + 4 m = 6 m (A fixed) carries a clockwise couple of 50 kNm at 2 m from A; BC = 5 m is a free-ended overhang with 30 kN/m; the column BD = 6 m is fixed at D. Relative stiffness: AB = 1.5I, BC = 2I, BD = I (E constant). No sway occurs because the fixed support at A prevents the horizontal movement of the girder.

#### Step 1: Overhang BC
The overhang is statically determinate: $M_{BC}=\dfrac{30\times5^2}{2}=375$ kNm (hogging). It is a fixed moment on joint B and takes no part in the distribution (DF = 0).

#### Step 2: Stiffness and distribution factors (only joint B is free)
- Joint B: relative stiffness $\Sigma k$ = 0.4167; BA: $k=EI/L$ = 0.2500, DF = 0.6000; BC: overhang, DF = 0 (its moment is a known fixed value); BD: $k=EI/L$ = 0.1667, DF = 0.4000

#### Step 3: Fixed-end moments (clockwise +)
For a couple $M$ at distance $a$ from A: $FEM_{AB}=\dfrac{Mb(2a-b)}{L^2}$, $FEM_{BA}=\dfrac{Ma(2b-a)}{L^2}$ with $a=2$, $b=4$, $L=6$:
$FEM_{AB}=\dfrac{50\times4\times(4-4)}{36}=0$, $FEM_{BA}=\dfrac{50\times2\times(8-2)}{36}=16.67$ kNm.
Overhang moment on B: $-375$ kNm. Column BD has no load.

#### Step 4: Moment distribution
Unbalanced moment at B = $16.67-375=-358.33$ kNm, so the balancing moment is $+358.33$ kNm, shared 0.6 : 0.4. Half of each share is carried over to the far fixed ends.

| | AB | BA | BC | BD | DB |
|---|---|---|---|---|---|
| DF | - | 0.600 | 0.000 | 0.400 | - |
| FEM | 0.00 | 16.67 | -375.00 | 0.00 | 0.00 |
| Balance | 0.00 | 215.00 | 0.00 | 143.33 | 0.00 |
| Carry-over | 107.50 | 0.00 | 0.00 | 0.00 | 71.67 |
| **Final M** | **107.50** | **231.67** | **-375.00** | **143.33** | **71.67** |


#### Step 5: Bending moment diagram (kNm)
- Joint B: $231.67+143.33-375=0$ ✓
- Column BD: top moment 143.33, base moment 71.67 (carry-over of one half)
- Overhang: 375 kNm at B (statics)

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 107.50 | -231.67 | jump at 2.00 m: -22.22 just left, +27.78 just right of the couple |
| BC | -375.00 | 0.00 | - |
| BD | 143.33 | -71.67 | - |

BM sign: positive = tension on the right-hand side when travelling A to B to C and B to D (sagging for the horizontal beam). In AB the shear is constant, $V_A=-(107.50+231.67+50)/6=-64.86$ kN, so the BMD is two parallel straight lines: from 107.50 at A to -22.22 just left of the couple, then a jump of +50 (the clockwise couple) to +27.78, then down to -231.67 at B. The overhang BC has a parabolic diagram from 0 at C to 375 kNm (hogging) at B.

**Answer: $M_{AB}=107.50$, $M_{BA}=231.67$, $M_{BC}=-375.00$, $M_{BD}=143.33$, $M_{DB}=71.67$ kNm (clockwise end moments +).**

## Check yourself

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

1. What is the moment at B from the 5 m overhang BC with 30 kN/m, and how is it treated in the distribution?
   A. 75 kNm; distributed to BA and BD
   B. 375 kNm hogging; a fixed moment on B, with DF of BC = 0
   C. 187.5 kNm; carried over to C
   D. 375 kNm; distributed 50:50
   Answer: B. The overhang is determinate: M = wL^2/2 = 30 x 25/2 = 375 kNm. It is a known moment on joint B and takes no part in the distribution, so its DF is 0. (Key point: Treatment of overhang)

2. With stiffness k = EI/L, BA = 1.5I/6 and BD = I/6, what are DF_BA and DF_BD?
   A. 0.5 and 0.5
   B. 0.4 and 0.6
   C. 0.6 and 0.4
   D. 0.75 and 0.25
   Answer: C. k_BA = 0.25 and k_BD = 0.1667, so the sum is 0.4167. The DFs are 0.6 and 0.4. (Key point: Distribution factors)

3. What is the balancing moment at B in the distribution?
   A. -358.33 kNm
   B. +16.67 kNm
   C. +358.33 kNm
   D. -375 kNm
   Answer: C. The unbalanced moment is 16.67 - 375 = -358.33 kNm. The balancing moment is the same with reversed sign, +358.33, split 0.6 : 0.4. (Key point: Unbalanced moment)

4. Why is no sway correction required?
   A. Because the column BD is very stiff
   B. Because there is no horizontal load
   C. Because the carry-over factor is zero
   D. Because the fixed support at A prevents horizontal movement of the girder
   Answer: D. Fixed support A holds the girder horizontally, so joint B cannot translate, and a non-sway moment distribution is enough. (Key point: No sway)

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