---
title: "Stiffness matrix: frame with coordinates 1, 2, 3"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-matrix-frame-120kn-coords"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-matrix-frame-120kn-coords
site: BCT Engineering
---

# Stiffness matrix: frame with coordinates 1, 2, 3

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2074 Chaitra, 10 marks
- Video: [How To Make a Stiffness Matrix || Most Imp Topic Of Structural Analysis](https://www.youtube.com/watch?v=b63PB97uXJY) — THE CIVIL BANDA
- 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-matrix-frame-120kn-coords
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Generate the stiffness matrix for the frame shown in the figure below and determine the end moments and horizontal reactions at supports due to the load given. [Figure: frame ABCD; A fixed (column AB: 1.5I, 10 m); beam BC (2I) of 10 m; column CD (I) of 5 m, D fixed; 120 kN horizontal at B; coordinates 1 (rotation at B), 2 (rotation at C) and 3 (horizontal sway) marked.]

## Short answer

The 3×3 stiffness matrix for rotations θB, θC and sway Δ is EI[[1.4,0.4,-0.09],[0.4,1.6,-0.24],[-0.09,-0.24,0.114]]. Solving yields end moments M_AB=-128.96, M_BA=-118.21, M_BC=118.21, M_CB=193.43, M_CD=-193.43, M_DC=-282.99 kNm and horizontal reactions H_A=24.72 kN, H_D=95.28 kN to the left.

## Answer

**Assumptions:** A(0,0), B(0,10), C(10,10), D(10,5): AB = 10 m (1.5I), BC = 10 m (2I), CD = 5 m (I); A and D fixed; the 120 kN load acts to the right at B. Axial deformation is neglected. Coordinates: **1 = rotation of B, 2 = rotation of C, 3 = horizontal sway $\Delta$** (right +). EI denotes the stiffness of I.

#### Generation of the stiffness matrix (column by column)
Each $k_{ij}$ is the force needed at coordinate $i$ to give a unit displacement at coordinate $j$ with the other coordinates held at zero. Member stiffnesses: $4EI/L$, $2EI/L$, $6EI/L^2$, $12EI/L^3$.

| Member | EI | L (m) | $4EI/L$ | $2EI/L$ | $6EI/L^2$ | $12EI/L^3$ |
|---|---|---|---|---|---|---|
| AB | 1.5EI | 10 | 0.600 | 0.300 | 0.0900 | 0.01800 |
| BC | 2EI | 10 | 0.800 | 0.400 | - | - |
| CD | EI | 5 | 0.800 | 0.400 | 0.2400 | 0.09600 |

- $k_{11}=\dfrac{4(1.5EI)}{10}+\dfrac{4(2EI)}{10}=1.400EI$, $k_{21}=k_{12}=\dfrac{2(2EI)}{10}=0.400EI$
- $k_{22}=\dfrac{4(2EI)}{10}+\dfrac{4EI}{5}=1.600EI$
- $k_{13}=-\dfrac{6(1.5EI)}{10^2}=-0.0900EI$, $k_{23}=-\dfrac{6EI}{5^2}=-0.2400EI$
- $k_{33}=\dfrac{12(1.5EI)}{10^3}+\dfrac{12EI}{5^3}=0.11400EI$

(The negative signs in the sway column follow from the sign convention: clockwise rotation positive, sway to the right causes clockwise chord rotation.)

$$
[K]=EI\begin{bmatrix}1.4000 & 0.4000 & -0.0900 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0900 & -0.2400 & 0.1140\end{bmatrix}
$$

#### Load vector and solution
Only the horizontal load at coordinate 3: $\{F\}=\{0,\;0,\;120\}^T$.
$$
EI\begin{bmatrix}1.4000 & 0.4000 & -0.0900 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0900 & -0.2400 & 0.1140\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 0.000 \\ 120.000\end{Bmatrix}
$$
$$
EI\theta_{B}=35.8209,\quad EI\theta_{C}=223.8806,\quad EI\Delta_{1}=1552.2388
$$

#### End moments from the member equations (kNm, clockwise +)
- $M_{AB}=\frac{2(1.5EI)}{10}\left(\theta_{B} - (0.3000)\Delta_1\right)$
- $M_{BA}=\frac{2(1.5EI)}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right)$
- $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right)$
- $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right)$
- $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -128.96 | BA | -118.21 |
| BC | 118.21 | CB | 193.43 |
| CD | -193.43 | DC | -282.99 |

#### Horizontal reactions
Column shear = (sum of end moments)/height:
$H_A=\dfrac{M_{AB}+M_{BA}}{10}=-24.72$ kN, $H_D=\dfrac{M_{CD}+M_{DC}}{5}=-95.28$ kN (both resist the load, so they act to the left); check: $120.00\approx120$ kN ✓.

Full reactions: A: $H=-24.72$, $V=-31.16$, $M=128.96$; D: $H=-95.28$, $V=31.16$, $M=282.99$.

**Answer: $M_{AB}=-128.96$, $M_{BA}=-118.21$, $M_{BC}=118.21$, $M_{CB}=193.43$, $M_{CD}=-193.43$, $M_{DC}=-282.99$ kNm; $H_A=24.72$ kN and $H_D=95.28$ kN (to the left).**

## Check yourself

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

1. k11 equals:
   A. 2(2EI)/10
   B. 4(1.5EI)/10 + 4(2EI)/10 = 1.4EI
   C. 1.5EI only
   D. 0.4EI
   Answer: B. Sum of 4EI/L of members at B. (Key point: k11)

2. k33 is:
   A. 6EI/L^2
   B. Sum of 4EI/L
   C. 0.3EI
   D. 12EI/L^3 summed over columns = 0.114EI
   Answer: D. Sway stiffness comes from the columns: 12(1.5EI)/1000 + 12EI/125. (Key point: k33)

3. Horizontal reaction at a column base is:
   A. Sum of end moments divided by column height
   B. Moment times height
   C. Half the load
   D. Zero
   Answer: A. Column shear = (M_top + M_bottom)/h. (Key point: Reactions)

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