---
title: "Stiffness matrix: frame with 100 kN, 15 kNm"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-100kn-20knm-15knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-100kn-20knm-15knm
site: BCT Engineering
---

# Stiffness matrix: frame with 100 kN, 15 kNm

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2072 Chaitra, 12 marks
- Video: [Stiffness Method Example: Part 1](https://www.youtube.com/watch?v=O1LwyvdZdCc) — Simmy Sigma
- 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-100kn-20knm-15knm
- 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 ABCD; A fixed; column AB (EI) of 10 m; beam BC (2EI) of 10 m with 20 kN/m UDL and 15 kNm moment at C; column CD (EI) of 5 m, D fixed; 100 kN horizontal at B.]

## Short answer

Stiffness matrix method incorporating sway: final end moments are M_AB = -27.18, M_BA = 10.02, M_BC = -10.02, M_CB = 256.78, M_CD = -241.78, M_DC = -249.64 kNm; maximum positive moment in BC is 131.82 kNm at 3.76 m.

## Answer

**Assumptions:** A(0,0), B(0,10), C(10,10), D(10,5): AB = 10 m (EI), BC = 10 m (2EI) with 20 kN/m, CD = 5 m (EI); A and D fixed; the 100 kN load at B acts to the right and the 15 kNm moment at C is clockwise. Axial deformation is neglected.

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

#### Fixed-end moments
$FEM_{BC}=-\dfrac{20\times10^2}{12}=-166.67$ kNm, $FEM_{CB}=+166.67$ kNm.

#### Stiffness equations
- $M_{AB}=\frac{2EI}{10}\left(\theta_{B} - (0.3000)\Delta_1\right)$
- $M_{BA}=\frac{2EI}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right)$
- $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 166.67$
- $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 166.67$
- $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

#### Equations of equilibrium
Joint B: $M_{BA}+M_{BC}=0$; Joint C: $M_{CB}+M_{CD}=15$; storey shear: $\dfrac{M_{AB}+M_{BA}}{10}+\dfrac{M_{CD}+M_{DC}}{5}+100=0$.

$$
EI\begin{bmatrix}1.2000 & 0.4000 & -0.0600 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0600 & -0.2400 & 0.1080\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}166.667 \\ -151.667 \\ 100.000\end{Bmatrix}
$$
$$
EI\theta_{B}=185.9848,\quad EI\theta_{C}=19.6496,\quad EI\Delta_{1}=1072.9167
$$

#### Final end moments (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -27.18 | BA | 10.02 |
| BC | -10.02 | CB | 256.78 |
| CD | -241.78 | DC | -249.64 |

Checks:
- Joint B: (10.02) + (-10.02) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (256.78) + (-241.78) = 15.00 kNm (applied clockwise moment 15.00) ✓
- Storey shear: $\dfrac{-17.16}{10}+\dfrac{-491.42}{5}+100=0.00$ ✓

#### Bending moment diagram ordinates (sagging/inside + , kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -27.18 | -10.02 | - |
| BC | -10.02 | -256.78 | 131.82 at 3.76 m |
| CD | -241.78 | 249.64 | - |

Reactions: A: $H=-1.72$, $V=75.32$, $M=27.18$; D: $H=-98.28$, $V=124.68$, $M=249.64$.

**Answer: $M_{AB}=-27.18$, $M_{BA}=10.02$, $M_{BC}=-10.02$, $M_{CB}=256.78$, $M_{CD}=-241.78$, $M_{DC}=-249.64$ kNm.**

## Check yourself

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

1. Columns AB (10 m) and CD (5 m) share the same sway Delta. Chord rotations are?
   A. Both Delta/10
   B. Both Delta/5
   C. Zero
   D. psi_AB = Delta/10, psi_CD = Delta/5
   Answer: D. Chord rotation is Delta divided by the column height. (Key point: Sway relation)

2. Which equation gives the sway?
   A. Storey shear: column shears plus 100 kN = 0
   B. Joint B moment sum alone
   C. Vertical equilibrium
   D. theta_B = theta_C
   Answer: A. Horizontal equilibrium of the girder: column shear forces balance the 100 kN lateral load. (Key point: Third equation)

3. FEM_BC for 20 kN/m on a 10 m beam?
   A. -250 and +250
   B. -166.67 and +166.67 kNm
   C. -100 and +100
   D. -200 and +200
   Answer: B. wL^2/12 = 20x100/12 = 166.67. (Key point: FEM)

4. Joint C equation?
   A. M_CB = 15
   B. M_CD = 15
   C. M_CB + M_CD = 15 (applied clockwise moment)
   D. M_CB + M_CD = 0
   Answer: C. The 15 kNm moment at C is clockwise, so the member moments sum to 15. (Key point: Joint C)

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