---
title: "Stiffness matrix: frame with 20 kN/m, 10 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-20knm-10kn-2i"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-20knm-10kn-2i
site: BCT Engineering
---

# Stiffness matrix: frame with 20 kN/m, 10 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2078 Bhadra, 12 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-frame-20knm-10kn-2i
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the frame given below using the stiffness matrix method. [Figure: portal frame ABCD; beam BC (2I) of 8 m carrying 20 kN/m UDL with 10 kN horizontal at C; column AB (1.5I) of height 6 m, column CD (I) of height 4 m; A and D fixed.]

## Short answer

Unknowns θ_B, θ_C, and sway Δ. Fixed-end moments and slope-deflection equations, with storey-shear equilibrium, form a 3×3 stiffness matrix. Solving gives EIθ_B=71.60, EIθ_C=-70.12, EIΔ=5.93. Final end moments (kNm): M_AB=34.32, M_BA=70.12, M_BC=-70.12, M_CB=72.35, M_CD=-72.35, M_DC=-37.28.

## Answer

**Assumptions:** A(0,0) and D are fixed; the girder BC (2I, 8 m) is horizontal at the top, AB = 6 m (1.5I) and CD = 4 m (I), so D is 2 m above A. The UDL of 20 kN/m acts downward on BC and the 10 kN load at C acts to the right. Axial deformations are ignored. Take $EI$ as the stiffness of I.

#### Unknown displacements
$\theta_B$, $\theta_C$ and the sway $\Delta$ of the girder. Chord rotations: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/4$, $\psi_{BC}=0$.

#### Fixed-end moments
$FEM_{BC}=-\dfrac{20\times8^2}{12}=-106.67$ kNm, $FEM_{CB}=+106.67$ kNm. Columns carry no load.

#### Member stiffness equations
- $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B} - (0.5000)\Delta_1\right)$
- $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right)$
- $M_{BC}=\frac{2(2EI)}{8}\left(2\theta_{B} + \theta_{C}\right) - 106.67$
- $M_{CB}=\frac{2(2EI)}{8}\left(2\theta_{C} + \theta_{B}\right) + 106.67$
- $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

#### Equilibrium equations
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=0$
- Storey shear: $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{4}+10=0$

Combining gives the stiffness matrix equation:
$$
EI\begin{bmatrix}2.0000 & 0.5000 & -0.2500 \\ 0.5000 & 2.0000 & -0.3750 \\ -0.2500 & -0.3750 & 0.2708\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}106.667 \\ -106.667 \\ 10.000\end{Bmatrix}
$$
Typical terms: $K_{11}=\dfrac{4(1.5EI)}{6}+\dfrac{4(2EI)}{8}=2EI$, $K_{12}=\dfrac{2(2EI)}{8}=0.5EI$, $K_{13}=-\dfrac{6(1.5EI)}{6^2}=-0.25EI$ (sign convention of the rows above, positive sway to the right).

#### Solution
$$
EI\theta_{B}=71.6049,\quad EI\theta_{C}=-70.1235,\quad EI\Delta_{1}=5.9259
$$

#### Final end moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 34.32 | BA | 70.12 |
| BC | -70.12 | CB | 72.35 |
| CD | -72.35 | DC | -37.28 |

Checks:
- Joint B: (70.12) + (-70.12) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (72.35) + (-72.35) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Storey shear: $\dfrac{104.44}{6}+\dfrac{-109.63}{4}+10=0.00$ ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 34.32 | -70.12 | - |
| BC | -70.12 | -72.35 | 88.77 at 3.99 m |
| CD | -72.35 | 37.28 | - |

Reactions: A: $H=17.41$, $V=79.72$, $M=-34.32$; D: $H=-27.41$, $V=80.28$, $M=37.28$.

**Answer: $M_{AB}=34.32$, $M_{BA}=70.12$, $M_{BC}=-70.12$, $M_{CB}=72.35$, $M_{CD}=-72.35$, $M_{DC}=-37.28$ kNm.**

## Check yourself

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

1. Which unknowns are solved in this frame (axial deformation ignored)?
   A. Only θB and θC
   B. θB, θC and the sway Δ of the girder
   C. Only the sway Δ
   D. θA, θB, θC and θD
   Answer: B. A and D are fixed so they do not rotate, but joints B and C rotate and the girder can sway sideways. (Key point: Unknown displacements)

2. Why is the chord rotation of beam BC taken as zero while those of the columns are not?
   A. BC is the stiffest member
   B. BC carries a UDL
   C. The girder moves horizontally as a whole, so its ends have no relative vertical displacement
   D. BC is fixed at both ends
   Answer: C. With axial deformation neglected, B and C move equally sideways, so only the columns get chord rotation (Δ/6 and Δ/4). (Key point: Chord rotations)

3. The third equilibrium equation (storey shear) states that:
   A. The sum of joint moments at B is zero
   B. Vertical reactions equal the UDL
   C. The moment at A equals the moment at D
   D. The column shears (sum of end moments / height) plus the 10 kN load balance horizontally
   Answer: D. Horizontal force equilibrium of the girder gives (MAB+MBA)/6 + (MCD+MDC)/4 + 10 = 0. (Key point: Sway equation)

4. The fixed-end moment magnitude for BC (20 kN/m, 8 m) is:
   A. 106.67 kNm
   B. 160 kNm
   C. 213.33 kNm
   D. 80 kNm
   Answer: A. wL^2/12 = 20 x 64 / 12 = 106.67 kNm, negative at B and positive at C. (Key point: Fixed-end moment)

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