---
title: "Stiffness matrix: frame, 25 kN, 100 kN"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-25kn-30knm-100kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-25kn-30knm-100kn
site: BCT Engineering
---

# Stiffness matrix: frame, 25 kN, 100 kN

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2071 Chaitra, 10 marks
- Video: [FRAME ANALYSIS BY USING STIFFNESS METHOD](https://www.youtube.com/watch?v=S2pzmCxag0I) — 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-25kn-30knm-100kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the frame shown in figure below by using the stiffness matrix method and draw bending moment diagram. [Figure: frame; left column 2I (3 m + 7 m) with 100 kN horizontal; 25 kN vertical at top-left; beam (2I) of 8 m with 30 kN/m UDL; right column I of 5 m; fixed bases.]

## Short answer

Using the stiffness matrix method with joint rotations and sway as unknowns, the equilibrium equations are solved to give the final end moments: M_AB = -122.33 kNm, M_BA = 110.27 kNm, M_BC = -110.27 kNm, M_CB = 174.70 kNm, M_CD = -174.70 kNm, M_DC = -169.28 kNm.

## Answer

**Assumptions:** A(0,0), B(0,10), C(8,10), D(8,5). Left column AB (2I) is 10 m high (3 m + 7 m) with the 100 kN horizontal force (to the right) acting 7 m above A; beam BC (2I) is 8 m and carries 30 kN/m (the '30 kNm' of the figure is read as a UDL of 30 kN/m); right column CD (I) is 5 m; A and D are fixed. The 25 kN vertical load at B only produces axial force in AB (it acts through the column axis) and no bending moment. Axial deformation is neglected.

#### Unknowns
$\theta_B,\theta_C$ and the sway $\Delta$: $\psi_{AB}=\Delta/10$, $\psi_{CD}=\Delta/5$.

#### Fixed-end moments (clockwise +)
- AB (100 kN at $a=7$ m from A, $b=3$ m): $FEM_{AB}=-63.00$, $FEM_{BA}=147.00$ kNm (from $Pab^2/L^2=100\times7\times9/100$ and $Pa^2b/L^2=100\times49\times3/100$)
- BC: $\mp\dfrac{30\times8^2}{12}=\mp160$ kNm

#### Stiffness equations
- $M_{AB}=\frac{2(2EI)}{10}\left(\theta_{B} - (0.3000)\Delta_1\right) - 63.00$
- $M_{BA}=\frac{2(2EI)}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right) + 147.00$
- $M_{BC}=\frac{2(2EI)}{8}\left(2\theta_{B} + \theta_{C}\right) - 160.00$
- $M_{CB}=\frac{2(2EI)}{8}\left(2\theta_{C} + \theta_{B}\right) + 160.00$
- $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

#### Equations of equilibrium
Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$; sway ($\sum H=0$, with $H_A=\dfrac{M_{AB}+M_{BA}-100\times3}{10}$ and $H_D=\dfrac{M_{CD}+M_{DC}}{5}$): $\dfrac{M_{AB}+M_{BA}}{10}+\dfrac{M_{CD}+M_{DC}}{5}+\dfrac{100\times7}{10}=0$. When the fixed-end moments are moved to the right-hand side, the load term of this row becomes the fixed-end reaction of the 100 kN load at B, $\dfrac{Pa^2(L+2b)}{L^3}=\dfrac{100\times49\times16}{1000}=78.40$ kN $\left(=70+\dfrac{147-63}{10}\right)$.

$$
EI\begin{bmatrix}1.8000 & 0.5000 & -0.1200 \\ 0.5000 & 1.8000 & -0.2400 \\ -0.1200 & -0.2400 & 0.1200\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}13.000 \\ -160.000 \\ 78.400\end{Bmatrix}
$$
$$
EI\theta_{B}=56.5023,\quad EI\theta_{C}=-13.5535,\quad EI\Delta_{1}=682.7287
$$

#### Final end moments (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -122.33 | BA | 110.27 |
| BC | -110.27 | CB | 174.70 |
| CD | -174.70 | DC | -169.28 |

Checks:
- Joint B: (110.27) + (-110.27) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (174.70) + (-174.70) = 0.00 kNm (applied clockwise moment 0.00) ✓

#### Bending moment diagram ordinates (sagging/inside tension +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -122.33 | -110.27 | 96.11 at 7.00 m |
| BC | -110.27 | -174.70 | 98.59 at 3.73 m |
| CD | -174.70 | 169.28 | - |

Reactions: A: $H=-31.21$, $V=136.95$, $M=122.33$; D: $H=-68.79$, $V=128.05$, $M=169.28$.

**Answer: $M_{AB}=-122.33$, $M_{BA}=110.27$, $M_{BC}=-110.27$, $M_{CB}=174.70$, $M_{CD}=-174.70$, $M_{DC}=-169.28$ kNm.**

## Check yourself

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

1. Why does the 25 kN vertical load at B not appear in the moment analysis?
   A. It is balanced by D
   B. It is a moment
   C. It acts through the column axis and gives only axial force
   D. It is too small
   Answer: C. Axial deformation is neglected and the load causes no bending. (Key point: 25 kN vertical)

2. The 100 kN on AB is replaced for sway by F_eq = 78.4 kN. What is it?
   A. P/2
   B. P x a/L
   C. 100 x 0.3
   D. Fixed-end shear at B, Pa^2(L+2b)/L^3
   Answer: D. Member loads are converted to joint loads using fixed-end reactions. (Key point: Equivalent force)

3. FEM_BA for the 100 kN at 7 m from A (L = 10)?
   A. +Pa^2b/L^2 = 147 kNm
   B. 63 kNm
   C. -147
   D. 250
   Answer: A. 100x49x3/100 = 147, clockwise at the near-far end. (Key point: FEM_BA)

4. UDL 30 kN/m on 8 m beam BC gives?
   A. 30
   B. wL^2/12 = 160 kNm
   C. 240
   D. 120
   Answer: B. 30x64/12 = 160. (Key point: Beam FEM)

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