---
title: "Stiffness matrix: frame, end moments"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-60kn-50kn-10knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-60kn-50kn-10knm
site: BCT Engineering
---

# Stiffness matrix: frame, end moments

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2074 Asoj, 10 marks
- Video: [CH5 Stiffness Matrix (Frame) Part 2/2](https://www.youtube.com/watch?v=PAV07r5Tz7U) — DrM4S4
- 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-60kn-50kn-10knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Generate the stiffness matrix of the frame loaded as shown in figure below. Also determine the end moments considering stiffness equations of each member. [Figure: portal frame ABCD; beam BC (2I) with 50 kN at mid-span (3 m + 3 m) and 10 kNm moment at C; column AB (I) 5 m, column CD (2I) 5 m; 60 kN horizontal at B; A and D fixed.]

## Short answer

The structure stiffness matrix is EI[2.1333 0.6667 -0.2400; 0.6667 2.9333 -0.4800; -0.2400 -0.4800 0.2880] for {θ_B, θ_C, Δ}. End moments: M_AB = -53.27, M_BA = -36.76, M_BC = 36.76, M_CB = 103.45, M_CD = -93.45, M_DC = -116.51 kNm.

## Answer

**Assumptions:** columns AB = 5 m (I) and CD = 5 m (2I), girder BC = 6 m (2I); A and D fixed. The 60 kN horizontal load at B acts to the right, the 50 kN load acts downward at mid-span of BC and the 10 kNm moment at C is clockwise. Axial deformation is neglected.

#### Coordinates
1 = $\theta_B$, 2 = $\theta_C$ (clockwise +), 3 = $\Delta$ (sway to the right). $\psi_{AB}=\psi_{CD}=\Delta/5$.

#### Fixed-end moments
$FEM_{BC}=-\dfrac{PL}{8}=-\dfrac{50\times6}{8}=-37.5$ kNm, $FEM_{CB}=+37.5$ kNm. Other members unloaded.

#### Stiffness equations of each member
- $M_{AB}=\frac{2EI}{5}\left(\theta_{B} - (0.6000)\Delta_1\right)$
- $M_{BA}=\frac{2EI}{5}\left(2\theta_{B} - (0.6000)\Delta_1\right)$
- $M_{BC}=\frac{2(2EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 37.50$
- $M_{CB}=\frac{2(2EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 37.50$
- $M_{CD}=\frac{2(2EI)}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
- $M_{DC}=\frac{2(2EI)}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

#### Stiffness matrix of the frame
Joint B: $K_{11}=\dfrac{4EI}{5}+\dfrac{4(2EI)}{6}$, joint C: $K_{22}=\dfrac{4(2EI)}{6}+\dfrac{4(2EI)}{5}$, coupling $K_{12}=\dfrac{2(2EI)}{6}$, sway terms $K_{13}=-\dfrac{6EI}{5^2}$, $K_{23}=-\dfrac{6(2EI)}{5^2}$, $K_{33}=\dfrac{12EI}{5^3}+\dfrac{12(2EI)}{5^3}$.
$$
[K]=EI\begin{bmatrix}2.1333 & 0.6667 & -0.2400 \\ 0.6667 & 2.9333 & -0.4800 \\ -0.2400 & -0.4800 & 0.2880\end{bmatrix}
$$
Load vector (applied loads minus fixed-end effects): $\{F\}=\{37.500,\ -27.500,\ 60.000\}^T$ (rows 1 and 2: $-\sum FEM$ plus the applied moment at B and C; row 3: the 60 kN lateral force).

$$
EI\begin{bmatrix}2.1333 & 0.6667 & -0.2400 \\ 0.6667 & 2.9333 & -0.4800 \\ -0.2400 & -0.4800 & 0.2880\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}37.500 \\ -27.500 \\ 60.000\end{Bmatrix}
$$
$$
EI\theta_{B}=41.2829,\quad EI\theta_{C}=28.8240,\quad EI\Delta_{1}=290.7758
$$

#### End moments using the member stiffness equations (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -53.27 | BA | -36.76 |
| BC | 36.76 | CB | 103.45 |
| CD | -93.45 | DC | -116.51 |

Checks:
- Joint B: (-36.76) + (36.76) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (103.45) + (-93.45) = 10.00 kNm (applied clockwise moment 10.00) ✓
- Storey shear: $\dfrac{-90.03+(-209.97)}{5}+60=0.00$ ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -53.27 | 36.76 | - |
| BC | 36.76 | -103.45 | 41.65 at 3.00 m |
| CD | -93.45 | 116.51 | - |

Reactions: A: $H=-18.01$, $V=1.63$, $M=53.27$; D: $H=-41.99$, $V=48.37$, $M=116.51$.

**Answer: $M_{AB}=-53.27$, $M_{BA}=-36.76$, $M_{BC}=36.76$, $M_{CB}=103.45$, $M_{CD}=-93.45$, $M_{DC}=-116.51$ kNm.**

## Check yourself

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

1. Which coordinates are used for the frame?
   A. Rotations of B and C and one sway of the beam
   B. Rotations of A and D
   C. Vertical displacement of B and C
   D. Only the sway
   Answer: A. A and D are fixed; B and C rotate and the frame sways sideways (axial deformation neglected), giving three coordinates. (Key point: Coordinates)

2. What is K11, the moment at B for unit rotation of B?
   A. 6EI/5^2
   B. 4EI/5 (column AB) + 4(2EI)/6 (beam BC)
   C. 2EI/5 + 2(2EI)/6
   D. 12EI/5^3
   Answer: B. K11 is the sum of 4EI/L of members meeting at B: column AB (I, 5 m) and girder BC (2I, 6 m). (Key point: Stiffness term)

3. Fixed-end moments in BC for the 50 kN mid-span load?
   A. -50 and +50
   B. 0 and 0
   C. -37.5 at B and +37.5 at C kNm
   D. -25 and +25
   Answer: C. PL/8 = 50 x 6/8 = 37.5 kNm, clockwise at C, anticlockwise at B. (Key point: Fixed-end moment)

4. Joint C is checked as M_CB + M_CD = 10 kNm. Why?
   A. The sum must always be 10
   B. Because CD is 10 m
   C. The load vector is 10
   D. The 10 kNm clockwise moment is applied at C
   Answer: D. The joint must balance the applied external moment, so the member moments at C add up to the applied 10 kNm. (Key point: Final moment check)

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