---
title: "Stiffness matrix: frame, flexural deformations only"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-15knm-10kn-15knm-flexural"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-15knm-10kn-15knm-flexural
site: BCT Engineering
---

# Stiffness matrix: frame, flexural deformations only

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2072 Kartik, 10 marks
- 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-15knm-10kn-15knm-flexural
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the frame shown in figure using the stiffness matrix method. Consider only flexural deformations. [Figure: frame; left column 2EI with 15 kN/m UDL horizontally; beam EI of 3 m + 2 m with 10 kN downward load and 15 kNm moment; right column EI of height 2 m + 1 m, fixed base.]

## Short answer

Stiffness matrix method considering flexural deformations: unknowns θ_B, θ_C and sway Δ; final end moments are M_AB = -41.70, M_BA = -7.18, M_BC = 7.18, M_CB = 20.34, M_CD = -5.34, M_DC = -13.29 kNm.

## Answer

**Assumptions (figure partly illegible):** A(0,0), B(0,3), C(5,3), D(5,0): both columns are 3 m high (left column AB = 2EI, right column CD = EI), the beam BC = 5 m (EI). The 15 kN/m load acts horizontally to the right on AB, the 10 kN load acts downward at 3 m from B and the 15 kNm moment at C is clockwise. A and D are fixed. Only flexural deformation is considered (axial and shear deformations are neglected).

#### Unknowns
$\theta_B,\theta_C$ and the sway $\Delta$ of the beam: $\psi_{AB}=\psi_{CD}=\Delta/3$.

#### Fixed-end moments (clockwise +)
- AB: $\mp\dfrac{wL^2}{12}=\mp\dfrac{15\times3^2}{12}=\mp11.25$ kNm, with the sign for a horizontal load on a vertical member: $FEM_{AB}=-11.25$, $FEM_{BA}=11.25$ kNm
- BC ($a=3$, $b=2$): $FEM_{BC}=-\dfrac{10\times3\times2^2}{5^2}=-4.80$, $FEM_{CB}=+\dfrac{10\times3^2\times2}{5^2}=7.20$ kNm

#### Stiffness equations of the members
- $M_{AB}=\frac{2(2EI)}{3}\left(\theta_{B} - (1.0000)\Delta_1\right) - 11.25$
- $M_{BA}=\frac{2(2EI)}{3}\left(2\theta_{B} - (1.0000)\Delta_1\right) + 11.25$
- $M_{BC}=\frac{2EI}{5}\left(2\theta_{B} + \theta_{C}\right) - 4.80$
- $M_{CB}=\frac{2EI}{5}\left(2\theta_{C} + \theta_{B}\right) + 7.20$
- $M_{CD}=\frac{2EI}{3}\left(2\theta_{C} - (1.0000)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{3}\left(\theta_{C} - (1.0000)\Delta_1\right)$

#### Equilibrium
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=15$
- Storey shear: the 15 kN/m load on AB is replaced by its fixed-end reaction at the girder level (the fixed-end shear $wL/2=22.5$ kN at B), giving the sway equation with right-hand side 22.50 kN.

$$
EI\begin{bmatrix}3.4667 & 0.4000 & -1.3333 \\ 0.4000 & 2.1333 & -0.6667 \\ -1.3333 & -0.6667 & 1.3333\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}-6.450 \\ 7.800 \\ 22.500\end{Bmatrix}
$$
$$
EI\theta_{B}=9.0133,\quad EI\theta_{C}=11.9186,\quad EI\Delta_{1}=31.8476
$$

#### Final end moments (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -41.70 | BA | -7.18 |
| BC | 7.18 | CB | 20.34 |
| CD | -5.34 | DC | -13.29 |

Checks:
- Joint B: (-7.18) + (7.18) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (20.34) + (-5.34) = 15.00 kNm (applied clockwise moment 15.00) ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -41.70 | 7.18 | 8.46 at 2.59 m |
| BC | 7.18 | -20.34 | - |
| CD | -5.34 | 13.29 | - |

Reactions: A: $H=-38.79$, $V=-1.50$, $M=41.70$; D: $H=-6.21$, $V=11.50$, $M=13.29$.

**Answer: $M_{AB}=-41.70$, $M_{BA}=-7.18$, $M_{BC}=7.18$, $M_{CB}=20.34$, $M_{CD}=-5.34$, $M_{DC}=-13.29$ kNm.**

## Check yourself

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

1. What does considering only flexural deformation mean?
   A. Only bending moments are zero
   B. Only axial deformation is included
   C. Rotations are neglected
   D. Axial and shear deformations are neglected
   Answer: D. Members are treated as axially rigid, so joints sway as a whole with one Delta. (Key point: Flexural only)

2. How is the 15 kN/m on AB handled in the sway equation?
   A. Replaced by its fixed-end shear wL/2 = 22.5 kN at girder level
   B. Ignored
   C. As a moment at A
   D. As 15 kN
   Answer: A. Member loads are converted to equivalent joint loads; the fixed-end shear wL/2 = 22.5 kN. (Key point: Lateral UDL)

3. Both columns are 3 m high, so?
   A. psi = Delta
   B. psi_AB = psi_CD = Delta/3
   C. psi_AB = Delta/5
   D. psi_CD = 0
   Answer: B. Rotation is sway divided by the column height. (Key point: Chord rotation)

4. Joint C equation?
   A. M_CB = 15
   B. M_BC + M_CB = 15
   C. M_CB + M_CD = 15
   D. M_CB + M_CD = 0
   Answer: C. The applied 15 kNm clockwise moment at C must be balanced by member end moments. (Key point: Joint C)

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