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

# Stiffness matrix: frame with 100 kN, end reactions

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2075 Asoj, 8 marks
- Video: [Direct Stiffness Matrix Method for Analysis of Frames - Problem No 1](https://www.youtube.com/watch?v=_10iV0yrUa0) — Stan Academy
- 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-matrix-frame-100kn-end-reactions
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Generate the stiffness matrix for the frame shown and determine the end reactions at the support. [Figure: portal frame ABCD; 100 kN horizontal at B; beam BC (EI) of 6 m; columns EI, AB 6 m high and CD 2 m high; A hinged, D fixed.]

## Short answer

The frame's 4×4 stiffness matrix is assembled from member stiffness equations ignoring axial deformation. Solving for joint rotations and sway gives end moments M_BA=-19.63, M_BC=19.63, M_CB=53.27, M_CD=-53.27, M_DC=-140.19 kNm and horizontal reactions H_A=3.27 kN, H_D=96.73 kN to the left.

## Answer

**Assumptions:** A(0,0) is a hinge, B(0,6) and C(6,6) are the top corners (AB = 6 m, BC = 6 m, CD = 2 m high, so D is 4 m above A and fixed). EI is the same for all members; the 100 kN load at B acts to the right. Axial deformations are ignored.

#### Degrees of freedom (coordinates)
1. $\theta_A$ (hinge rotation) 2. $\theta_B$ 3. $\theta_C$ 4. $\Delta$ (horizontal sway of the girder). Chord rotations: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/2$, $\psi_{BC}=0$.

#### Member stiffness equations ($M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)$, no member loads)
- $M_{AB}=\frac{2EI}{6}\left(2\theta_{A} + \theta_{B} - (0.5000)\Delta_1\right)$
- $M_{BA}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{A} - (0.5000)\Delta_1\right)$
- $M_{BC}=\frac{2EI}{6}\left(2\theta_{B} + \theta_{C}\right)$
- $M_{CB}=\frac{2EI}{6}\left(2\theta_{C} + \theta_{B}\right)$
- $M_{CD}=\frac{2EI}{2}\left(2\theta_{C} - (1.5000)\Delta_1\right)$
- $M_{DC}=\frac{2EI}{2}\left(\theta_{C} - (1.5000)\Delta_1\right)$

#### Equilibrium equations
- Hinge A: $M_{AB}=0$
- Joint B: $M_{BA}+M_{BC}=0$
- Joint C: $M_{CB}+M_{CD}=0$
- Storey shear: $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{2}+100=0$

#### Stiffness matrix of the frame $[K]\{d\}=\{F\}$
$$
EI\begin{bmatrix}0.6667 & 0.3333 & 0.0000 & -0.1667 \\ 0.3333 & 1.3333 & 0.3333 & -0.1667 \\ 0.0000 & 0.3333 & 2.6667 & -1.5000 \\ -0.1667 & -0.1667 & -1.5000 & 1.5556\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 0.000 \\ 0.000 \\ 100.000\end{Bmatrix}
$$
The matrix is symmetric. The only non-zero load is the 100 kN at the sway coordinate.

#### Solution
$$
EI\theta_{A}=44.8598,\quad EI\theta_{B}=-14.0187,\quad EI\theta_{C}=86.9159,\quad EI\Delta_{1}=151.4019
$$

#### End moments (kNm, clockwise on the member end +)
| End | $M$ (kNm) | End | $M$ (kNm) |
|---|---|---|---|
| AB | 0.00 | BA | -19.63 |
| BC | 19.63 | CB | 53.27 |
| CD | -53.27 | DC | -140.19 |

Checks: $M_{AB}=0.00$ ✓; Joint B: $-19.63+(19.63)=0.00$ ✓; Joint C: $53.27+(-53.27)=0.00$ ✓.

#### End reactions
Column shears: $H_A=\dfrac{M_{AB}+M_{BA}}{6}$ and $H_D=\dfrac{M_{CD}+M_{DC}}{2}$ (magnitudes), and the vertical reactions from the girder moments.
- Support A: $H_A=3.27$ kN (to the left), $V_A=-12.15$ kN, $M_A=0$
- Support D: $H_D=96.73$ kN (to the left), $V_D=12.15$ kN, $M_D=140.19$ kNm (anticlockwise +)
- Check: $\sum H=100-3.27-96.73=0.00$ ✓, $\sum V=0.00$ ✓

| Member | BM at first end | BM at second end | Max within member |
|---|---|---|---|
| AB | 0.00 | 19.63 | - |
| BC | 19.63 | -53.27 | - |
| CD | -53.27 | 140.19 | - |

**Answer: $EI\Delta=151.402$; $M_{BA}=-19.63$, $M_{BC}=19.63$, $M_{CB}=53.27$, $M_{CD}=-53.27$, $M_{DC}=-140.19$ kNm; reactions $H_A=3.27$ kN, $H_D=96.73$ kN (both to the left).**

## Check yourself

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

1. How many coordinates does this frame have?
   A. Three
   B. Four: θA, θB, θC and Δ
   C. Two
   D. Six
   Answer: B. The hinge at A adds its own rotation to B, C and sway. (Key point: Coordinates)

2. Why do ψAB = Δ/6 and ψCD = Δ/2 differ?
   A. Columns have different EI
   B. Loads differ
   C. Columns differ in height (6 m and 2 m)
   D. D is hinged
   Answer: C. ψ = Δ/h, and the shorter column rotates more. (Key point: Chord rotations)

3. The sway equation is:
   A. MBA + MBC = 0
   B. MAB = 0
   C. Sum of all moments = 0
   D. Sum of column shears plus 100 = 0
   Answer: D. Horizontal equilibrium: (MAB+MBA)/6 + (MCD+MDC)/2 + 100 = 0. (Key point: Storey shear)

4. Why is the load vector {0,0,0,100}?
   A. The 100 kN acts at the sway coordinate only
   B. Moments are zero everywhere
   C. Loads act on all coordinates
   D. The hinge takes the load
   Answer: A. No joint moments are applied, only the horizontal force corresponding to sway. (Key point: Load vector)

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