---
title: "Stiffness matrix: frame with 50 kNm moment"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:stiffness-frame-40kn-10knm-50knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-frame-40kn-10knm-50knm
site: BCT Engineering
---

# Stiffness matrix: frame with 50 kNm moment

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2079 Baishakh, 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-40kn-10knm-50knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Analyse the given frame using the stiffness matrix method. [Figure: portal frame ABCD, columns AB and CD of height 4 m with EI, beam BC span 8 m with EI carrying 10 kN/m UDL; 40 kN horizontal at B; 50 kNm moment applied at C; A and D fixed.]

## Short answer

Using stiffness matrix for unknowns θ_B, θ_C and sway Δ, solving joint and storey equilibrium gives end moments: M_AB = –44.92, M_BA = –1.08, M_BC = 1.08, M_CB = 96.42, M_CD = –46.42, M_DC = –67.58 kNm (clockwise positive); EIΔ = 236.67 kNm³.

## Answer

**Assumptions:** the 10 kN/m UDL acts downward on BC, the 40 kN load at B acts to the right and the 50 kNm moment at C is clockwise. EI is constant, axial deformation is neglected, A and D are fixed.

#### Degrees of freedom
$\theta_B,\theta_C$ (joint rotations, clockwise +) and $\Delta$ (horizontal sway of beam BC, right +). Since BC is axially rigid, B and C move equally, so $\psi_{AB}=\psi_{CD}=\Delta/4$ and $\psi_{BC}=0$.

Member relation: $M_{ij}=\dfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+FEM_{ij}$ (clockwise end moments +).

#### Fixed-end moments
$FEM_{BC}=-\dfrac{wL^2}{12}=-\dfrac{10\times8^2}{12}=-53.33$ kNm, $FEM_{CB}=+53.33$ kNm. The columns carry no load between joints, so their FEMs are zero.

#### Stiffness equations for the members
- $M_{AB}=\frac{2EI}{4}\left(\theta_{B} - (0.7500)\Delta_1\right)$
- $M_{BA}=\frac{2EI}{4}\left(2\theta_{B} - (0.7500)\Delta_1\right)$
- $M_{BC}=\frac{2EI}{8}\left(2\theta_{B} + \theta_{C}\right) - 53.33$
- $M_{CB}=\frac{2EI}{8}\left(2\theta_{C} + \theta_{B}\right) + 53.33$
- $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

#### Equilibrium conditions
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=50$ (the applied clockwise moment)
- Storey shear: the column shears resist the 40 kN load, $\dfrac{M_{AB}+M_{BA}}{4}+\dfrac{M_{CD}+M_{DC}}{4}+40=0$

Substituting the member equations gives $[K]\{d\}=\{F\}$:
$$
EI\begin{bmatrix}1.5000 & 0.2500 & -0.3750 \\ 0.2500 & 1.5000 & -0.3750 \\ -0.3750 & -0.3750 & 0.3750\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}53.333 \\ -3.333 \\ 40.000\end{Bmatrix}
$$
(First two rows: joint B and C. Third row: sway equation. The load vector holds the applied loads minus the fixed-end effects.)

#### Solution
$$
EI\theta_{B}=87.6667,\quad EI\theta_{C}=42.3333,\quad EI\Delta_{1}=236.6667
$$

#### Final end moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | -44.92 | BA | -1.08 |
| BC | 1.08 | CB | 96.42 |
| CD | -46.42 | DC | -67.58 |

Checks:
- Joint B: (-1.08) + (1.08) = 0.00 kNm (applied clockwise moment 0.00) ✓
- Joint C: (96.42) + (-46.42) = 50.00 kNm (applied clockwise moment 50.00) ✓
- Storey shear: $\dfrac{(-46.00)+(-114.00)}{4}+40=0.00$ ✓

#### Bending moment diagram ordinates (sagging/inside tension +, kNm)
| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | -44.92 | 1.08 | - |
| BC | 1.08 | -96.42 | 39.76 at 2.78 m |
| CD | -46.42 | 67.58 | - |

Reactions: A: $H=-11.50$, $V=27.81$, $M=44.92$; D: $H=-28.50$, $V=52.19$, $M=67.58$ (M anticlockwise +).

**Answer: $M_{AB}=-44.92$, $M_{BA}=-1.08$, $M_{BC}=1.08$, $M_{CB}=96.42$, $M_{CD}=-46.42$, $M_{DC}=-67.58$ kNm; sway $EI\Delta=236.67$.**

## Check yourself

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

1. Which unknowns are used, and what is the chord rotation of the columns?
   A. theta_B, theta_C and sway; psi_AB = psi_CD = Delta/4 and psi_BC = 0
   B. theta_A, theta_D and sway; psi_BC = Delta/8
   C. Only theta_B and theta_C; no sway
   D. theta_B, theta_C and sway; psi_AB = Delta/8
   Answer: A. A and D are fixed. Beam BC is axially rigid, so B and C move equally by Delta, rotating each 4 m column by Delta/4 while BC does not rotate. (Key point: Degrees of freedom)

2. What are the fixed-end moments on beam BC (10 kN/m, 8 m)?
   A. -80 and +80 kNm
   B. -53.33 at B and +53.33 kNm at C
   C. -40 and +40 kNm
   D. +53.33 at B and -53.33 kNm at C
   Answer: B. FEM = wL^2/12 = 10 x 64/12 = 53.33 kNm, negative at B and positive at C with the clockwise-positive convention. (Key point: Fixed-end moments)

3. What are the joint equilibrium equations at B and C?
   A. M_BA + M_BC = 0 and M_CB + M_CD = 0
   B. M_BA + M_BC = 50 and M_CB + M_CD = 0
   C. M_BA + M_BC = 40 and M_CB + M_CD = 50
   D. M_BA + M_BC = 0 and M_CB + M_CD = 50 (the applied clockwise moment)
   Answer: D. The sum of member end moments at a joint equals the applied clockwise moment there: 0 at B and 50 kNm at C. (Key point: Joint equilibrium equations)

4. Why is a third equation, the storey shear equation, needed?
   A. Because sway Delta is a third unknown and the column shears must resist the 40 kN horizontal load
   B. Because the joint equations are not independent
   C. To check the fixed-end moments
   D. Because the beam carries a UDL
   Answer: A. There are three unknowns, theta_B, theta_C and Delta. The sway equation expresses horizontal equilibrium: the column shears balance the 40 kN load. (Key point: Sway equation)

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