---
title: "Stiffness matrix: BMD of frame"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-60kn-10knm-2ei"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-60kn-10knm-2ei
site: BCT Engineering
---

# Stiffness matrix: BMD of frame

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2076 Chaitra, 12 marks
- Video: [Direct Stiffness Method - Frame -1](https://www.youtube.com/watch?v=TYAKMvDHlrI) — The Classroom
- 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/stiffness-frame-60kn-10knm-2ei
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Draw BMD of the given frame using the stiffness matrix method. [Figure: frame; beam A-B-C with A fixed, AB (2EI) with a 60 kN point load, BC (EI) of 4 m with 10 kN/m UDL, C roller; column BD (EI) of 4 m height down to D; dimensions 2 m, 2 m, 4 m as marked.]

## Short answer

No sway due to fixed support A. Unknowns θ_B and θ_C; M_CB=0 equilibrium. Fixed-end moments and slope-deflection equations yield 2×2 matrix. Solution: EIθ_B=-2.667, EIθ_C=-12.0. Final end moments: M_AB=-32.67, M_BA=24.67, M_BC=-22.00, M_CB=0, M_BD=-2.67, M_DB=-1.33 kNm.

## Answer

**Assumptions:** the girder A-B-C is horizontal with A fixed and C on a roller; AB = 4 m (2EI) with the 60 kN load at mid-span; BC = 4 m (EI) with 10 kN/m downward; the column BD (EI, 4 m) hangs below B and is fixed at D. The fixed support A prevents sway, so only $\theta_B$ and $\theta_C$ are unknown.

#### Fixed-end moments (clockwise +)
- AB: $\mp\dfrac{PL}{8}=\mp\dfrac{60\times4}{8}=\mp30$ kNm
- BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
- BD: no load

#### Stiffness (slope-deflection) equations
- $M_{AB}=\frac{2(2EI)}{4}\left(\theta_{B}\right) - 30.00$
- $M_{BA}=\frac{2(2EI)}{4}\left(2\theta_{B}\right) + 30.00$
- $M_{BC}=\frac{2EI}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
- $M_{CB}=\frac{2EI}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
- $M_{BD}=\frac{2EI}{4}\left(2\theta_{B}\right)$
- $M_{DB}=\frac{2EI}{4}\left(\theta_{B}\right)$

#### Equilibrium
- Joint B: $M_{BA}+M_{BC}+M_{BD}=0$
- Roller C: $M_{CB}=0$

Writing these in terms of $\theta_B,\theta_C$ gives the stiffness matrix equation:
$$
EI\begin{bmatrix}4.0000 & 0.5000 \\ 0.5000 & 1.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-16.667 \\ -13.333\end{Bmatrix}
$$
$$
EI\theta_{B}=-2.6667,\quad EI\theta_{C}=-12.0000
$$

#### Final end moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -32.67 | BA | 24.67 |
| BC | -22.00 | CB | 0.00 |
| BD | -2.67 | DB | -1.33 |

Checks: Joint B: $24.67+(-22.00)+(-2.67)=0.00$ ✓; Roller C: $M_{CB}=0.00$ ✓.

#### Bending moment diagram ordinates (sagging + for the girder; column: tension on the right when going down, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -32.67 | -24.67 | 31.33 at 2.00 m |
| BC | -22.00 | 0.00 | 10.51 at 2.55 m |
| BD | -2.67 | 1.33 | - |

Reactions: A: $H=1.00$, $V=32.00$, $M=32.67$; D: $H=-1.00$, $V=53.50$, $M=1.33$; C: $H=0.00$, $V=14.50$, $M=0.00$.

**Answer: $M_{AB}=-32.67$, $M_{BA}=24.67$, $M_{BC}=-22.00$, $M_{CB}=0$, $M_{BD}=-2.67$, $M_{DB}=-1.33$ kNm.**

## Check yourself

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

1. Why are only θB and θC unknown in this frame?
   A. The columns have no stiffness
   B. A is fixed, which prevents sway of the girder
   C. The loads are symmetric
   D. Roller C stops all rotation
   Answer: B. The fixed support at A prevents sway, so only the joint rotations remain. (Key point: Number of unknowns)

2. The FEM for AB (60 kN at mid-span of 4 m) is:
   A. PL/12 = 20 kNm
   B. PL/4 = 60 kNm
   C. PL/8 = 30 kNm
   D. wL^2/12 = 13.33 kNm
   Answer: C. For a central point load on a fixed-fixed span, FEM = PL/8 = 60 x 4 / 8 = 30 kNm. (Key point: Fixed-end moment)

3. What boundary condition gives the second joint equation at C?
   A. θC = 0
   B. MCB equals MBC
   C. Vertical sway at C is zero so MCB = 0 and θC = 0
   D. MCB = 0 because a roller cannot carry moment
   Answer: D. A roller end is free to rotate and carries no moment, so MCB = 0. (Key point: Roller condition)

4. Equilibrium of joint B involves which members?
   A. BA, BC and BD moments summing to zero
   B. BA and BC only
   C. BC and BD only
   D. Only the column BD
   Answer: A. Three members meet at B (AB, BC and the column BD), so MBA + MBC + MBD = 0. (Key point: Joint B equilibrium)

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