---
title: "Generate stiffness matrix of frame"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-matrix-frame-3i-2i-i"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-matrix-frame-3i-2i-i
site: BCT Engineering
---

# Generate stiffness matrix of frame

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2066 Jestha, 10 marks
- Video: [How to Compile Stiffness Matrix](https://www.youtube.com/watch?v=8hddX-cOKfw) — Engineering Edu
- 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-3i-2i-i
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Generate stiffness matrix for the frame shown below. [Figure: frame ABCD; AB (2I), BC (3I) of 4 m, CD (I); heights 3 m and 1 m; A and D fixed.]

## Short answer

Stiffness matrix for coordinates θB, θC, Δ (sway) is K = EI [[5.6667, 1.5, -1.3333], [1.5, 7.0, -6.0], [-1.3333, -6.0, 12.8889]]. Axial deformations neglected; chord rotations included.

## Answer

**Assumptions:** A(0,0), B(0,3), C(4,3), D(4,2): column AB is 3 m high ($2EI$), beam BC is 4 m ($3EI$), column CD is 1 m high ($EI$, short column), A and D are fixed. Axial deformations are neglected. Coordinates (positive directions): **1 = rotation of B, 2 = rotation of C (both clockwise), 3 = horizontal sway of the beam (to the right)**. The chord rotations are $\psi_{AB}=\Delta/3$ and $\psi_{CD}=\Delta/1$.

#### Member constants
| Member | EI | L (m) | $4EI/L$ | $2EI/L$ | $6EI/L^2$ | $12EI/L^3$ |
|---|---|---|---|---|---|---|
| AB | $2EI$ | 3 | 2.6667EI | 1.3333EI | 1.3333EI | 0.8889EI |
| BC | $3EI$ | 4 | 3.0000EI | 1.5000EI | - | - |
| CD | $EI$ | 1 | 4.0000EI | 2.0000EI | 6.0000EI | 12.0000EI |

#### Generation of the stiffness matrix
- **Unit rotation of B (coordinate 1):** $k_{11}=\dfrac{4(2EI)}{3}+\dfrac{4(3EI)}{4}=5.6667EI$; $k_{21}=\dfrac{2(3EI)}{4}=1.5000EI$; $k_{31}=-\dfrac{6(2EI)}{3^2}=-1.3333EI$
- **Unit rotation of C (coordinate 2):** $k_{22}=\dfrac{4(3EI)}{4}+\dfrac{4EI}{1}=7.0000EI$; $k_{12}=1.5000EI$; $k_{32}=-\dfrac{6EI}{1^2}=-6.0000EI$
- **Unit sway (coordinate 3):** $k_{33}=\dfrac{12(2EI)}{3^3}+\dfrac{12EI}{1^3}=12.8889EI$; $k_{13}=-1.3333EI$; $k_{23}=-6.0000EI$

$$
[K]=EI\begin{bmatrix}5.6667 & 1.5000 & -1.3333 \\ 1.5000 & 7.0000 & -6.0000 \\ -1.3333 & -6.0000 & 12.8889\end{bmatrix}
$$

The matrix is symmetric ($k_{ij}=k_{ji}$). The short column CD (1 m) is laterally very stiff: it contributes $12EI$ to the sway stiffness $k_{33}$ and $-6EI$ to $k_{23}$, so most of the sway resistance, and most of the coupling between sway and the rotation of joint C, comes from CD. The equation of the frame is $\{P\}=[K]\{d\}$ with $\{d\}=\{\theta_B,\theta_C,\Delta\}^T$.

## Check yourself

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

1. What are the three coordinates used for the stiffness matrix?
   A. Rotation of B, rotation of C and horizontal sway of the beam
   B. Rotation of A, B and C
   C. Vertical deflections of B, C and D
   D. Axial elongations of AB, BC, CD
   Answer: A. Axial deformation is neglected, so the freedoms are theta_B, theta_C and the common sway. (Key point: Coordinates)

2. k11 (unit rotation of B) is the sum of which member terms?
   A. 12EI/L^3 of AB and CD
   B. 4EI/L of AB and 4EI/L of BC
   C. 2EI/L of AB and BC
   D. 6EI/L^2 of AB and BC
   Answer: B. k11 = 4(2EI)/3 + 4(3EI)/4 = 5.6667EI, the rotational stiffness of the members meeting at B. (Key point: Diagonal term k11)

3. Why does the short 1 m column CD dominate k33 (12.89 EI)?
   A. It is the longest column
   B. It has the largest EI
   C. Its shear stiffness 12EI/L^3 grows rapidly as L decreases, giving 12EI
   D. It is hinged at D
   Answer: C. Sway stiffness of a fixed column is 12EI/L^3, so a 1 m column gives 12EI against only 0.889EI for the 3 m column. (Key point: Sway stiffness)

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