---
title: "Moment distribution: rectangular frame, two horizontal loads"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:moment-dist-rectangular-frame-25kn-50kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-rectangular-frame-25kn-50kn
site: BCT Engineering
---

# Moment distribution: rectangular frame, two horizontal loads

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2066 Jestha, 20 marks
- Video: [Frame Analysis Example - Shear and Moment Diagram (Part 1) - Structural Analysis](https://www.youtube.com/watch?v=Pa9d98iukV0) — structurefree
- 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-rectangular-frame-25kn-50kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

A single storey rectangular frame of span 6 m is fixed at the bases and has a beam of sectional stiffness 4 EI and columns of 2 EI with storey heights 4 m. Two horizontal concentrated forces of magnitude 25 kN and 50 kN, directed towards right, are acting at the beam-column joint and at the middle of the left column respectively on the left side. Use the moment distribution method to draw bending moment diagram for the frame.

## Short answer

Frame analyzed by moment distribution with sway. Case I: no sway, Case II: sway with assumed FEMs. Combined with factor k=0.625. Final moments: AB=-82.92, BA=-28.33, BC=28.33, CB=38.33, CD=-38.33, DC=-50.42 kNm.

## Answer

**Data:** single-storey frame with fixed bases A and D, span 6 m, storey height 4 m, columns $2EI$, beam $4EI$. Horizontal loads (to the right): 25 kN at the top-left joint B and 50 kN at the mid-height of the left column AB. Because the loads are lateral, the frame sways and the analysis is in two cases.

#### Step 1: Stiffness and distribution factors ($k=EI/L$)
- Joint B: relative stiffness $\Sigma k$ = 1.1667; BA: $k=EI/L$ = 0.5000, DF = 0.4286; BC: $k=EI/L$ = 0.6667, DF = 0.5714
- Joint C: relative stiffness $\Sigma k$ = 1.1667; CB: $k=EI/L$ = 0.6667, DF = 0.5714; CD: $k=EI/L$ = 0.5000, DF = 0.4286

#### Step 2: Fixed-end moments
The only member load is the 50 kN on column AB ($L=4$, load at mid-height): $\mp\dfrac{PL}{8}=\mp\dfrac{50\times4}{8}=\mp25$ kNm; sign for a load to the right on the column: $FEM_{AB}=-25.00$, $FEM_{BA}=25.00$ kNm.

#### Case I: sway prevented by a horizontal support at B
| | AB | BA | BC | CB | CD | DC |
|---|---|---|---|---|---|---|
| DF | - | 0.429 | 0.571 | 0.571 | 0.429 | - |
| FEM | -25.00 | 25.00 | 0.00 | 0.00 | 0.00 | 0.00 |
| Balance | 0.00 | -10.71 | -14.29 | 0.00 | 0.00 | 0.00 |
| Carry-over | -5.36 | 0.00 | 0.00 | -7.14 | 0.00 | 0.00 |
| Balance | 0.00 | 0.00 | 0.00 | 4.08 | 3.06 | 0.00 |
| Carry-over | 0.00 | 0.00 | 2.04 | 0.00 | 0.00 | 1.53 |
| Balance | 0.00 | -0.87 | -1.17 | 0.00 | 0.00 | 0.00 |
| Carry-over | -0.44 | 0.00 | 0.00 | -0.58 | 0.00 | 0.00 |
| Balance | 0.00 | 0.00 | 0.00 | 0.33 | 0.25 | 0.00 |
| Carry-over | 0.00 | 0.00 | 0.17 | 0.00 | 0.00 | 0.12 |
| Further cycles (converged) | -0.04 | -0.08 | -0.09 | -0.02 | 0.02 | 0.01 |
| **Final M** | **-30.83** | **13.33** | **-13.33** | **-3.33** | **3.33** | **1.67** |

Reaction of the imaginary support: $R_1=-46.875$ kN.

#### Case II: sway with no loads
Assume FEM $=-100$ kNm at both ends of each column.
| | AB | BA | BC | CB | CD | DC |
|---|---|---|---|---|---|---|
| DF | - | 0.429 | 0.571 | 0.571 | 0.429 | - |
| FEM | -100.00 | -100.00 | 0.00 | 0.00 | -100.00 | -100.00 |
| Balance | 0.00 | 42.86 | 57.14 | 57.14 | 42.86 | 0.00 |
| Carry-over | 21.43 | 0.00 | 28.57 | 28.57 | 0.00 | 21.43 |
| Balance | 0.00 | -12.24 | -16.33 | -16.33 | -12.24 | 0.00 |
| Carry-over | -6.12 | 0.00 | -8.16 | -8.16 | 0.00 | -6.12 |
| Balance | 0.00 | 3.50 | 4.66 | 4.66 | 3.50 | 0.00 |
| Carry-over | 1.75 | 0.00 | 2.33 | 2.33 | 0.00 | 1.75 |
| Balance | 0.00 | -1.00 | -1.33 | -1.33 | -1.00 | 0.00 |
| Carry-over | -0.50 | 0.00 | -0.67 | -0.67 | 0.00 | -0.50 |
| Further cycles (converged) | 0.11 | 0.22 | 0.44 | 0.44 | 0.22 | 0.11 |
| **Final M** | **-83.33** | **-66.67** | **66.67** | **66.67** | **-66.67** | **-83.33** |

Reaction of the imaginary support: $R_2=75.000$ kN.

#### Step 3: Combination
$k=-\dfrac{R_1}{R_2}=0.62500$

| End | Case I | k × Case II | Final M (kNm) |
|---|---|---|---|
| AB | -30.83 | -52.08 | -82.92 |
| BA | 13.33 | -41.67 | -28.33 |
| BC | -13.33 | 41.67 | 28.33 |
| CB | -3.33 | 41.67 | 38.33 |
| CD | 3.33 | -41.67 | -38.33 |
| DC | 1.67 | -52.08 | -50.42 |

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

#### Bending moment diagram (kNm; positive = inside tension for the frame traversed A-B-C-D)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -82.92 | 28.33 | - |
| BC | 28.33 | -38.33 | - |
| CD | -38.33 | 50.42 | - |
Under the 50 kN load: 22.71 kNm.

Reactions: A: $H=-52.81$, $V=-11.11$, $M=82.92$; D: $H=-22.19$, $V=11.11$, $M=50.42$. Check: horizontal reactions sum to -75.00 kN, balancing 25 + 50 = 75 kN.

**Answer: $M_{AB}=-82.92$, $M_{BA}=-28.33$, $M_{BC}=28.33$, $M_{CB}=38.33$, $M_{CD}=-38.33$, $M_{DC}=-50.42$ kNm.**

## Check yourself

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

1. Why is the analysis done in two cases (Case I and Case II)?
   A. The beam has two different stiffnesses
   B. There are two loads
   C. The lateral loads make the frame sway, so sway is first prevented and then corrected
   D. The bases are fixed
   Answer: C. Case I holds the frame against sway with an imaginary support; Case II applies a sway to remove that artificial restraint. (Key point: Why two cases)

2. In Case II (sway with no loads), the assumed FEMs in the columns are...
   A. -100 at both ends of each column, equal because sway rotates both ends equally
   B. Zero, because there are no loads
   C. +100 at the top and -100 at the bottom
   D. PL/8 as for a point load
   Answer: A. A relative sway of the column ends induces equal moments of the same sign at both ends, so an arbitrary value (-100) is assumed. (Key point: Case II fixed-end moments)

3. How is the factor k that scales Case II obtained?
   A. k = R1/R2
   B. k = -R1/R2, so the imaginary support reaction cancels (k = 0.625)
   C. k = DF_BA / DF_BC
   D. k = 25/50
   Answer: B. R1 + k R2 = 0 removes the force in the imaginary support, giving k = 46.875/75 = 0.625. (Key point: Sway correction factor)

4. How is the final moment at AB found?
   A. Case I only (-30.83)
   B. Case II only (-83.33)
   C. Average of the two cases
   D. Case I + k x Case II = -30.83 - 52.08 = -82.92 kNm
   Answer: D. The final moments are the superposition of Case I and k times Case II. (Key point: Final combination)

## Related in Theory of Structures II

- [Moment distribution: multi-bay frame, AFD, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-multi-column-frame-afd) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame, 25 kN, 100 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-25kn-30knm-100kn) — Theory of Structures II, chapter 3
- [Moment distribution: frame with 80 kN, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-80kn-10knm-60kn) — Theory of Structures II, chapter 3
- [Moment distribution principle; frame stiffness matrix](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aprinciple-moment-dist-stiffness-matrix-u1v1) — Theory of Structures II, chapter 3
- [Stiffness matrix: frame with 50 kNm moment](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-40kn-10knm-50knm) — Theory of Structures II, chapter 3
- [Moment distribution: frame with 20 kN, 6 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-20kn-6knm-5kn) — Theory of Structures II, chapter 3

## Related in other subjects

- [Beam with two symmetrical loads by moment-area](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Asymmetrical-beam-two-loads-moment-area) — Strength of Materials, chapter 8
- [Truss member forces with 25, 50, 10 kN loads](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Atruss-forces-25kn-50kn-10kn-hinge-roller) — Applied Mechanics, chapter 7