---
title: "Stiffness matrix: frame with 40 kN/m, 60 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-matrix-frame-40knm-60kn-1m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-matrix-frame-40knm-60kn-1m
site: BCT Engineering
---

# Stiffness matrix: frame with 40 kN/m, 60 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2068 Chaitra, 8 marks
- Video: [FRAME ANALYSIS- STIFFNESS METHOD: Previous Year University solved (KTU)](https://www.youtube.com/watch?v=tfHAOJYfp98) — 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-matrix-frame-40knm-60kn-1m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Determine the stiffness matrix for the frame shown in figure below. [Figure: portal frame, EI constant; beam of 4 m carrying 40 kN/m UDL; left column 2 m + 1 m (60 kN horizontal at the 1 m level), right column 2 m + 2 m (60 kN horizontal at mid-height, pointing left); both bases fixed.]

## Short answer

Use coordinates theta_B, theta_C, sway Delta; axial deformation neglected. Member stiffnesses from EI/L give K = EI [[2.333, 0.5, -0.667], [0.5, 2.0, -0.375], [-0.667, -0.375, 0.6319]] for (theta_B, theta_C, Delta). Solving gives EI theta_B=26.34, EI theta_C=-9.85, EI Delta=44.80.

## Answer

**Assumptions (figure partly illegible):** A(0,0), B(0,3), C(4,3), D(4,-1): the left column is 3 m high (2 m + 1 m) and the right column is 4 m high (2 m + 2 m), with the beam BC = 4 m horizontal at the top; EI is constant; the bases A and D are fixed. Loads: 40 kN/m on BC (downward); 60 kN horizontal (to the right) on AB at 2 m above A; 60 kN horizontal (to the left) at mid-height of CD. Coordinates: **1 = $\theta_B$, 2 = $\theta_C$ (clockwise), 3 = horizontal sway $\Delta$ of the beam (right +)**. Axial deformation is neglected.

#### Chord rotations
$\psi_{AB}=\Delta/3$ and $\psi_{CD}=\Delta/4$.

#### Stiffness matrix (column by column)
Member stiffnesses: AB ($L=3$): $\dfrac{4EI}{3}=1.3333EI$, $\dfrac{2EI}{3}=0.6667EI$, $\dfrac{6EI}{9}=0.6667EI$, $\dfrac{12EI}{27}=0.4444EI$; CD ($L=4$): $EI$, $0.5EI$, $0.375EI$, $0.1875EI$; BC ($L=4$): $EI$, $0.5EI$.
- $k_{11}=\dfrac{4EI}{3}+\dfrac{4EI}{4}=2.3333EI$, $k_{12}=\dfrac{2EI}{4}=0.5000EI$, $k_{13}=-\dfrac{6EI}{3^2}=-0.6667EI$
- $k_{22}=\dfrac{4EI}{4}+\dfrac{4EI}{4}=2.0000EI$, $k_{23}=-\dfrac{6EI}{4^2}=-0.3750EI$
- $k_{33}=\dfrac{12EI}{3^3}+\dfrac{12EI}{4^3}=0.6319EI$

$$
[K]=EI\begin{bmatrix}2.3333 & 0.5000 & -0.6667 \\ 0.5000 & 2.0000 & -0.3750 \\ -0.6667 & -0.3750 & 0.6319\end{bmatrix}
$$

#### Load vector (for completeness)
Fixed-end moments: BC $\mp\dfrac{40\times4^2}{12}=\mp53.33$; AB (60 kN at 2 m from A, $L=3$): $FEM_{AB}=-13.33$, $FEM_{BA}=26.67$; CD (60 kN to the left at 2 m from C): $FEM_{CD}=-30.00$, $FEM_{DC}=30.00$ kNm. The equivalent loads are
$$
\{P\}=\begin{Bmatrix}26.667\\-23.333\\14.444\end{Bmatrix}
$$
and $[K]\{d\}=\{P\}$ gives
$$
EI\theta_{B}=26.3388,\quad EI\theta_{C}=-9.8519,\quad EI\Delta_{1}=44.7970
$$

#### End moments (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -25.64 | BA | 31.92 |
| BC | -31.92 | CB | 56.65 |
| CD | -56.65 | DC | 8.28 |

Checks:
- Joint B: (31.92) + (-31.92) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (56.65) + (-56.65) = 0.00 kNm (applied clockwise moment 0.00) ✓

**Answer: $[K]=EI\begin{bmatrix}2.333&0.500&-0.667\\0.500&2.000&-0.375\\-0.667&-0.375&0.6319\end{bmatrix}$ (coordinates $\theta_B,\theta_C,\Delta$).**

## Check yourself

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

1. Which coordinates are used for the stiffness matrix of the frame?
   A. theta_B, theta_C and horizontal sway of the beam
   B. Vertical displacements of B and C
   C. Rotations at A and D
   D. Only the sway
   Answer: A. With axial deformation neglected, the three DOFs are the two joint rotations and the common sway. (Key point: Coordinates)

2. What is k33 (stiffness for unit sway)?
   A. 12EI/27 + 12EI/64 = 0.6319EI
   B. 4EI/3 + 4EI/4
   C. 6EI/9 + 6EI/16
   D. EI/3 + EI/4
   Answer: A. Each column gives 12EI/L^3: 12EI/3^3 + 12EI/4^3 = 0.4444EI + 0.1875EI. (Key point: Sway stiffness k33)

3. k11 is the moment at joint B for unit rotation of B. What is it?
   A. 4EI/3 + 4EI/4 = 2.3333EI
   B. 2EI/3 + 2EI/4
   C. 4EI/4 only
   D. 6EI/9
   Answer: A. Both members meeting at B (column AB and beam BC) contribute 4EI/L. (Key point: Stiffness k11)

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