---
title: "Stiffness method: frame, flexural deformations only"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-10kn-40knm-flexural"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-10kn-40knm-flexural
site: BCT Engineering
---

# Stiffness method: frame, flexural deformations only

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2067 Asar, 15 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-10kn-40knm-flexural
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyze the frame shown in figure using the stiffness method (displacement method). Consider only flexural deformations and take EI as constant throughout. [Figure: frame; left column of 5 m, hinged at the base A (with 3 m dimension); 10 kN horizontal at B; beam BC of 3 m; C fixed with 40 kNm moment applied.]

## Short answer

Use stiffness method with no sway; unknowns are rotations at A and B. Solve equilibrium to get EI theta_A = -10.3448, EI theta_B = 20.6897. Final moments: M_AB=0, M_BA=12.41, M_BC=27.59, M_CB=13.79 kNm.

## Answer

**Assumptions (figure partly described):** the frame is A-B-C: column AB is 5 m high with a hinge at A; the beam BC is 3 m long, fixed at C. At joint B a 10 kN horizontal force (to the right) and a 40 kNm clockwise moment act. EI is constant and only flexural deformation is considered (axial and shear deformations neglected). Since C is fixed, B cannot translate horizontally; the vertical displacement of B is prevented by the column (axially rigid), so no sway occurs and the unknowns are the rotations $\theta_A$ (hinge) and $\theta_B$ (clockwise +).

#### Member stiffness equations (no member loads)
- $M_{AB}=\frac{2EI}{5}\left(2\theta_{A} + \theta_{B}\right)$
- $M_{BA}=\frac{2EI}{5}\left(2\theta_{B} + \theta_{A}\right)$
- $M_{BC}=\frac{2EI}{3}\left(2\theta_{B}\right)$
- $M_{CB}=\frac{2EI}{3}\left(\theta_{B}\right)$
The 10 kN force acts at the joint B and is carried directly to the supports by axial force in BC (it produces no bending moment).

#### Equilibrium
- Hinge A: $M_{AB}=0$
- Joint B: $M_{BA}+M_{BC}=40$ (applied clockwise moment)

#### Stiffness matrix equation
$$
EI\begin{bmatrix}0.8000 & 0.4000 \\ 0.4000 & 2.1333\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 40.000\end{Bmatrix}
$$
$K_{11}=\dfrac{4EI}{5}=0.8EI$, $K_{12}=\dfrac{2EI}{5}=0.4EI$, $K_{22}=\dfrac{4EI}{5}+\dfrac{4EI}{3}=2.1333EI$.
$$
EI\theta_{A}=-10.3448,\quad EI\theta_{B}=20.6897
$$

#### End moments (kNm, clockwise +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 0.00 | BA | 12.41 |
| BC | 27.59 | CB | 13.79 |

Checks: $M_{AB}=0.00$ ✓; Joint B: $12.41+27.59=40.00$ kNm = applied 40 kNm ✓.

#### Reactions
A: $H=2.48$, $V=-13.79$, $M=0.00$; C: $H=-12.48$, $V=13.79$, $M=-13.79$ (M anticlockwise +). Horizontally the supports give 2.48 kN at A (equal to the column shear caused by the end moments) and -12.48 kN at C; their sum balances the 10 kN load. The column shear is $(M_{AB}+M_{BA})/5=2.48$ kN.

#### Bending moment diagram
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | -12.41 | - |
| BC | 27.59 | -13.79 | - |

**Answer: $M_{AB}=0$, $M_{BA}=12.41$ kNm, $M_{BC}=27.59$ kNm, $M_{CB}=13.79$ kNm; $\theta_B=20.690/EI$ (clockwise).**

## Check yourself

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

1. Why are the unknowns only theta_A and theta_B?
   A. C is fixed and the column is axially rigid, so B cannot translate (no sway)
   B. A is fixed
   C. The 10 kN force is zero
   D. Because EI is constant
   Answer: A. B cannot move, so only the rotations of the hinge A and joint B remain. (Key point: No sway)

2. What does the 10 kN horizontal force at B do?
   A. Carried directly to supports by axial force in BC; no bending moment
   B. Produces sway
   C. Adds to the joint moment
   D. Changes theta_A
   Answer: A. It acts at the joint and is resisted axially, so it does not enter the moment equations. (Key point: 10 kN force)

3. What is the joint B equilibrium?
   A. M_BA + M_BC = 40 (applied clockwise moment)
   B. M_BA + M_BC = 0
   C. M_BA = M_BC
   D. M_BA + M_BC = 10
   Answer: A. The sum of member end moments equals the applied 40 kNm. (Key point: Joint B equation)

4. Condition at hinge A?
   A. M_AB = 0
   B. theta_A = 0
   C. M_BA = 0
   D. M_AB = 40
   Answer: A. A hinge cannot carry moment, so 2EI/5 (2 theta_A + theta_B) = 0. (Key point: Hinge condition)

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