---
title: "Flexibility matrix for coordinates 1 and 2"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:flexibility-matrix-propped-beam-coordinates-1"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-propped-beam-coordinates-1
site: BCT Engineering
---

# Flexibility matrix for coordinates 1 and 2

Theory of Structures II · Chapter 2: Force method

- Asked in: 2070 Asar, 6 marks
- Video: [Flexibility Co-Efficient Problem 1](https://www.youtube.com/watch?v=pbnd0M0OgUQ) — Firoz Nadaf
- 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/flexibility-matrix-propped-beam-coordinates-1
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Generate the flexibility matrix for the coordinates shown in figure below. [Figure: beam of two spans of 10 m each, EI constant, left end fixed with coordinate 1 (rotation at the fixed end), interior support with coordinate 2 (rotation), right end roller.]

## Short answer

Flexibility matrix for rotations at fixed end A and interior support B: [F] = (1/EI)[3.333 1.667; 1.667 6.667] or (L/6EI)[2 1; 1 4]. Matrix is symmetric with positive diagonals; inverse gives stiffness matrix.

## Answer

Reading of the figure: a beam of two spans AB and BC, each $L = 10$ m, with constant $EI$; A is a fixed end, B an interior support and C a roller. Coordinate 1 is the rotation at the fixed end A, and coordinate 2 is the rotation at the interior support B. To define flexibility coefficients, the restraints against these rotations are released: A becomes a hinge and the beam is cut into two simply supported spans by a hinge over B, so the primary structure is made of the two simply supported spans AB and BC. A unit moment at coordinate 1 acts at A and a unit moment at coordinate 2 acts at B (on both spans).

```text
 (1) A o------------o B (2)------------o C
        |<-- 10 -->|     |<-- 10 -->|
```

#### Flexibility coefficients
The flexibility coefficient $f_{ij}$ is the rotation at coordinate $i$ due to a unit moment at coordinate $j$.

**Unit moment at coordinate 1 (A):** it acts on span AB only. For a simply supported span of length $L$ with a unit moment at one end, the rotation at that end is $\dfrac{L}{3EI}$ and the rotation at the far end is $\dfrac{L}{6EI}$:

$$
f_{11} = \frac{L}{3EI} = \frac{10}{3EI} = \frac{3.333}{EI},\qquad f_{21} = \frac{L}{6EI} = \frac{10}{6EI} = \frac{1.667}{EI}
$$

**Unit moment at coordinate 2 (B):** it acts on both spans, so each span turns by $\dfrac{L}{3EI}$ at B, and span AB rotates $\dfrac{L}{6EI}$ at A:

$$
f_{22} = \frac{L}{3EI} + \frac{L}{3EI} = \frac{20}{3EI} = \frac{6.667}{EI},\qquad f_{12} = \frac{L}{6EI} = \frac{1.667}{EI}
$$

($f_{12} = f_{21}$, as required by Maxwell's theorem.)

#### Flexibility matrix
$$
[F] = \frac{1}{EI}\begin{bmatrix} 3.333 & 1.667 \\ 1.667 & 6.667 \end{bmatrix} = \frac{L}{6EI}\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}
$$

The matrix is symmetric with positive diagonal terms. Its inverse is the stiffness matrix for the same coordinates:

$$
[K] = [F]^{-1} = \frac{6EI}{7L}\begin{bmatrix} 4 & -1 \\ -1 & 2 \end{bmatrix}
$$

## Check yourself

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

1. For coordinates here, f_ij is:
   A. Moment at i due to unit rotation at j
   B. Rotation at coordinate i due to unit moment at coordinate j
   C. Deflection at j due to unit load at i
   D. Shear at i due to unit moment at j
   Answer: B. Coordinates are rotations, so flexibility is rotation per unit moment. (Key point: Definition of f_ij)

2. Why is f22 = 20/(3EI), twice f11?
   A. Coordinate 2 has a fixed end
   B. The unit moment at B acts on both spans, each turning L/3EI
   C. B is a roller
   D. Because EI doubles
   Answer: B. Both spans adjoin B, so contributions L/3EI + L/3EI add. (Key point: f22)

3. A unit moment at A on span AB gives the rotation at B equal to:
   A. L/3EI
   B. L/12EI
   C. L/6EI
   D. 0
   Answer: C. The far-end rotation of a simply supported span with an end moment is L/6EI = 1.667/EI. (Key point: f21)

4. The matrix [F] = (L/6EI)[[2,1],[1,4]] is symmetric because of:
   A. Equilibrium
   B. Maxwell's reciprocal theorem
   C. Compatibility
   D. Castigliano's first theorem
   Answer: B. f12 = f21 by Maxwell's theorem. (Key point: Symmetry)

## Related in Theory of Structures II

- [Fixed beam: end moments from support rotation or settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-end-moments-rotation-settlement) — Theory of Structures II, chapter 3, asked 4×
- [Flexibility matrix: reactions at support D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-frame-2knm-5kn-coords) — Theory of Structures II, chapter 2
- [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4
- [Stiffness matrix: frame with coordinates 1, 2, 3](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-coords-123-15knm-10kn) — Theory of Structures II, chapter 3
- [Generate stiffness matrix of frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-matrix-frame-coordinates-123) — Theory of Structures II, chapter 3
- [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4

## Related in other subjects

- [Frame: horizontal displacement and rotation at roller](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-horizontal-rotation-roller-2079) — Theory of Structures I, chapter 3
- [Rotation and deflection at free end, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-rotation-deflection-free-end) — Computational Techniques in Civil Engineering, chapter 4
- [Frame: roller displacement and hinge rotation](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-roller-hinge-rotation-2071b) — Theory of Structures I, chapter 3
- [Frame rotation and vertical deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aframe-rotation-vertical-free-end-2073m) — Theory of Structures I, chapter 3