---
title: "Flexibility matrix: reactions at D, BMD"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:flexibility-matrix-frame-30kn-reactions-d"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-frame-30kn-reactions-d
site: BCT Engineering
---

# Flexibility matrix: reactions at D, BMD

Theory of Structures II · Chapter 2: Force method

- Asked in: 2071 Chaitra, 10 marks
- Video: [Beam Analysis: Flexibility Method](https://www.youtube.com/watch?v=blRCb4y_DYU) — Captain Matrix
- 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-frame-30kn-reactions-d
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Generate the flexibility matrix to determine the reactions at support D for the frame loaded as shown in the figure below. Also determine the reactions at support D and draw the bending moment diagram. Show all the steps. [Figure: frame; column AE-B (A fixed) with 4 m + 4 m... marked dimensions 2 m, 4 m, 4 m on the left; 30 kN horizontal at E; beam BFC with 4 m + 2 m; column CD with D hinged, 4 m high.]

## Short answer

Releasing hinge at D and taking horizontal X1 and vertical X2 as redundants, flexibility matrix [F] = (1/EI)[[160,72],[72,360]]; load vector {Δ0} = (1/EI){640, -1440}. Solving yields X1 = -6.37 kN, X2 = 5.27 kN. Thus support D reactions: H_D = 6.37 kN left, V_D = 5.27 kN up.

## Answer

Reading of the figure (dimensions as assumed): column ABE with A fixed at the base, E 4 m above A (30 kN horizontal, towards the right) and B 8 m above A; beam BFC with $BF = 4$ m and $FC = 2$ m; column CD, 4 m high, so D is 4 m above the level of A and is hinged. EI is constant.

$r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take $X_1$ = horizontal reaction ($\rightarrow$) and $X_2$ = vertical reaction ($\uparrow$). The primary structure is the cantilever fixed at A.

#### Step 1: Coordinates and unit-load moments
$M_0$ is due to the 30 kN load only: $M_0 = -30\times4 = -120$ kNm at A, zero above E. The unit moments $m_1$ and $m_2$ are those caused by a unit horizontal force and a unit vertical force at D (see the table below, + tension on the inner face).

#### Step 2: Flexibility matrix $[F]$ and load vector $\{\Delta_0\}$
$$
\begin{aligned}
f_{11} &= \frac{160.000}{EI} \\
f_{12} &= \frac{72.000}{EI} \\
f_{22} &= \frac{360.000}{EI} \\
\Delta_{10} &= \frac{640.0}{EI} \\
\Delta_{20} &= \frac{-1440.0}{EI}
\end{aligned}
$$

$$
[F] = \frac{1}{EI}\begin{bmatrix} 160.00 & 72.00 \\ 72.00 & 360.00 \end{bmatrix},\qquad
\{\Delta_0\} = \frac{1}{EI}\begin{Bmatrix} 640.0 \\ -1440.0 \end{Bmatrix}
$$

#### Step 3: Compatibility, $[F]\{X\} = -\{\Delta_0\}$
$$
\begin{aligned}
160.000\,X_1 + 72.000\,X_2 &= -640.0 \\
72.000\,X_1 + 360.000\,X_2 &= 1440.0
\end{aligned}
$$

$$
X_1 = H_D = -6.37\ \text{kN},\qquad X_2 = V_D = 5.27\ \text{kN}
$$

#### Step 4: Bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)
| Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ |
|---|---|---|---|---|
| A | -120.00 | -4.00 | 6.00 | -62.86 |
| E | 0.00 | 0.00 | 6.00 | 31.65 |
| B (column top) | 0.00 | 4.00 | 6.00 | 6.15 |
| F | 0.00 | 4.00 | 2.00 | -14.95 |
| C | 0.00 | 4.00 | 0.00 | -25.49 |
| D | 0.00 | 0.00 | 0.00 | 0.00 |

#### Step 5: Shear force (kN)
| Member | SF |
|---|---|
| AE | +23.63 |
| EB | -6.37 |
| BF, FC | -5.27 |
| CD | +6.37 |

#### Reactions
- D: $H_D = 6.37$ kN (towards the left), $V_D = 5.27$ kN ($\uparrow$)
- A: $H_A = 23.63$ kN (towards the left), $V_A = 5.27$ kN ($\downarrow$), $M_A = 62.86$ kNm

Check: $\Sigma H$: $30 - 23.63 - 6.37 = 0$ and $\Sigma V$: $-5.27 + 5.27 = 0$.

**Answer: reactions at D: $H_D = 6.37$ kN (towards the left) and $V_D = 5.27$ kN (upward).**

## Check yourself

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

1. With r = 5, how many redundants and which?
   A. One; vertical at D
   B. Three; moments
   C. Two; horizontal and vertical reactions at hinge D
   D. Five; all reactions
   Answer: C. D_s = 5 - 3 = 2, released at hinge D. (Key point: Redundants)

2. In [F] the terms f12 and f21 are both 72. Why?
   A. Coincidence
   B. The load is symmetric
   C. Both are zero
   D. Maxwell's reciprocal theorem makes [F] symmetric
   Answer: D. f12 = f21 follows from Maxwell's theorem, as both equal the integral of m1 m2/EI. (Key point: Flexibility matrix property)

3. X1 = H_D = -6.37 kN where X1 was assumed to the right. What is the real reaction?
   A. 6.37 kN to the right
   B. 6.37 kN to the left
   C. Zero
   D. 30 kN to the left
   Answer: B. A negative value means the reaction acts opposite to the assumed direction. (Key point: Sign of solved redundants)

4. Which check holds for the horizontal forces?
   A. 30 + 23.63 = 6.37
   B. 30 = 23.63 - 6.37
   C. 30 - 23.63 - 6.37 = 0
   D. 23.63 = 6.37
   Answer: C. The 30 kN load is balanced by H_A = 23.63 and H_D = 6.37, both to the left. (Key point: Equilibrium check)

## Related in Theory of Structures II

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