---
title: "Stiffness matrix: continuous beam reactions and BMD"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-continuous-beam-10kn-6knm-fixed"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-continuous-beam-10kn-6knm-fixed
site: BCT Engineering
---

# Stiffness matrix: continuous beam reactions and BMD

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2068 Baishakh, 10 marks
- Video: [Stiffness Matrix Method for Analysis of Beams  ( With Overhanging )](https://www.youtube.com/watch?v=X9MJOI8XnOQ) — 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/stiffness-continuous-beam-10kn-6knm-fixed
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Using the stiffness matrix method, find support reactions and draw bending moment diagram for the given loaded continuous beam. [Figure: beam, left end hinged; 10 kN at the middle of the first span (1.5 m + 1.5 m); middle support; second span 4 m with 6 kN/m UDL, right end fixed.]

## Short answer

Set theta_A and theta_B as unknowns; solve EI[[1.333,0.667],[0.667,2.333]]{theta_A,theta_B}={3.75,4.25} to get EI theta_A=2.2188, EI theta_B=1.1875. End moments: M_AB=0, M_BA=6.81, M_BC=-6.81, M_CB=8.59 kNm. Reactions: R_A=2.73, R_B=18.83, R_C=12.45 kN, M_C=8.59 kNm.

## Answer

**Data:** AB = 3 m (1.5 m + 1.5 m) with 10 kN at mid-span, A is a hinge; BC = 4 m with 6 kN/m, C is fixed; B is the middle support. EI is constant.

#### Unknown displacements
1 = $\theta_A$ (rotation of the hinged end), 2 = $\theta_B$ (clockwise +). $\theta_C=0$; no settlements.

#### Fixed-end moments (clockwise +)
- AB: $\mp\dfrac{PL}{8}=\mp\dfrac{10\times3}{8}=\mp3.75$ kNm
- BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{6\times4^2}{12}=\mp8$ kNm

#### Member stiffness equations
- $M_{AB}=\frac{2EI}{3}\left(2\theta_{A} + \theta_{B}\right) - 3.75$
- $M_{BA}=\frac{2EI}{3}\left(2\theta_{B} + \theta_{A}\right) + 3.75$
- $M_{BC}=\frac{2EI}{4}\left(2\theta_{B}\right) - 8.00$
- $M_{CB}=\frac{2EI}{4}\left(\theta_{B}\right) + 8.00$

#### Stiffness matrix equation
Equilibrium: $M_{AB}=0$ (hinge) and $M_{BA}+M_{BC}=0$ (joint B).
$$
EI\begin{bmatrix}1.3333 & 0.6667 \\ 0.6667 & 2.3333\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B}\end{Bmatrix}=\begin{Bmatrix}3.750 \\ 4.250\end{Bmatrix}
$$
with $K_{11}=\dfrac{4EI}{3}$, $K_{12}=K_{21}=\dfrac{2EI}{3}$, $K_{22}=\dfrac{4EI}{3}+\dfrac{4EI}{4}$.
$$
EI\theta_{A}=2.2188,\quad EI\theta_{B}=1.1875
$$

#### End moments (kNm)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 0.00 | BA | 6.81 |
| BC | -6.81 | CB | 8.59 |

Checks: $M_{AB}=0.00$ ✓; $M_{BA}+M_{BC}=0.00$ ✓.

#### Support reactions
Using the end moments and the equilibrium of each span:
- Span AB (taking moments about B): $R_A=\dfrac{10\times1.5-6.81}{3}=2.73$ kN (the hogging moment at B reduces the reaction at A)
- $R_B=18.83$ kN (sum of the shears of AB and BC at B)
- $R_C=12.45$ kN, $M_C=-8.59$ kNm (anticlockwise +), i.e. fixed-end moment 8.59 kNm (hogging)

Check: $R_A+R_B+R_C=34.00$ kN $=10+6\times4=34$ kN ✓.

#### Bending moment diagram (sagging +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -6.81 | 4.09 at 1.50 m |
| BC | -6.81 | -8.59 | 4.31 at 1.93 m |
BM under the 10 kN load: 4.09 kNm.

**Answer: $M_{BA}=6.81$, $M_{BC}=-6.81$, $M_{CB}=8.59$ kNm, $M_{AB}=0$; $R_A=2.73$ kN, $R_B=18.83$ kN, $R_C=12.45$ kN, $M_C=8.59$ kNm.**

## Check yourself

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

1. Which displacements are the unknowns?
   A. theta_A (hinged end rotation) and theta_B
   B. theta_B and theta_C
   C. Only theta_B
   D. Vertical deflections
   Answer: A. A hinge rotates freely, C is fixed (theta_C = 0), so theta_A and theta_B are unknown. (Key point: Unknowns)

2. What are FEMs of spans AB and BC?
   A. 3.75 kNm (PL/8) and 8 kNm (wL^2/12)
   B. 7.5 and 16
   C. 3.75 and 12
   D. PL/4 and wL^2/8
   Answer: A. 10x3/8 = 3.75 and 6x16/12 = 8. (Key point: Fixed-end moments)

3. Which two conditions give the stiffness equation?
   A. M_AB = 0 (hinge) and M_BA + M_BC = 0 at B
   B. M_AB = M_BA
   C. M_BC = M_CB
   D. M_CB = 0
   Answer: A. The hinge carries no moment and the joint B moments balance. (Key point: Equilibrium conditions)

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