---
title: "Flexibility matrix: frame reactions at D, SFD, BMD"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:flexibility-matrix-frame-50kn-30knm-d"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-frame-50kn-30knm-d
site: BCT Engineering
---

# Flexibility matrix: frame reactions at D, SFD, BMD

Theory of Structures II · Chapter 2: Force method

- Asked in: 2073 Shrawan, 7 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-50kn-30knm-d
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Using the flexibility matrix method, determine the reactions at support D of the frame loaded as shown in figure below. Also draw SFD and BMD. Take EI = constant. [Figure: frame ABCD; A fixed, column AB 5 m high with 50 kN horizontal at B; beam BC with 30 kNm moment at 2 m from B, total span 5 m; column CD 3 m high, D on roller/hinge.]

## Short answer

Flexibility matrix method with redundants H_D and V_D at hinge D. Primary structure is cantilever fixed at A. Compatibility equations yield H_D = -23.44 kN (leftward) and V_D = 31.72 kN (upward). Reactions at A: H_A=26.56 kN left, V_A=31.72 kN down, M_A=74.50 kNm.

## Answer

Reading of the figure: A (0, 0) fixed; column AB is 5 m high with 50 kN horizontal at B (towards the right); beam BC is 5 m long with a couple of 30 kNm (taken clockwise) applied at F, 2 m from B; column CD is 3 m high (D is 2 m above the level of A) and D is hinged. EI is constant.

$D_s = 3 + 2 - 3 = 2$. Release the hinge at D: $X_1$ = horizontal reaction ($\rightarrow$), $X_2$ = vertical reaction ($\uparrow$).

#### Flexibility matrix and load vector (flexibility method)
The primary structure is the cantilever fixed at A. The loads (50 kN at B and the couple at F) give $M_0$, and unit forces at D give $m_1$, $m_2$ (the table of moments below lists their values; + means tension on the inner face).

$$
\begin{aligned}
f_{11} &= \frac{65.667}{EI} \\
f_{12} &= \frac{50.000}{EI} \\
f_{22} &= \frac{166.667}{EI} \\
\Delta_{10} &= \frac{-46.7}{EI} \\
\Delta_{20} &= \frac{-4115.0}{EI}
\end{aligned}
$$

$[F]\{X\} = -\{\Delta_0\}$:

$$
\begin{aligned}
65.667\,X_1 + 50.000\,X_2 &= 46.7 \\
50.000\,X_1 + 166.667\,X_2 &= 4115.0
\end{aligned}
$$

$$
X_1 = H_D = -23.44\ \text{kN},\qquad X_2 = V_D = 31.72\ \text{kN}
$$

#### Final 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 | -280.00 | -2.00 | 5.00 | -74.50 |
| B (column) | -30.00 | 3.00 | 5.00 | 58.28 |
| F (left of couple) | -30.00 | 3.00 | 3.00 | -5.16 |
| F (right of couple) | 0.00 | 3.00 | 3.00 | 24.84 |
| C | 0.00 | 3.00 | 0.00 | -70.33 |
| D | 0.00 | 0.00 | 0.00 | 0.00 |

#### Shear force (kN)
| Member | SF |
|---|---|
| AB | +26.56 |
| BF, FC (beam) | -31.72 |
| CD | +23.44 |

#### Reactions
- D (hinge): $H_D = 23.44$ kN (towards the left), $V_D = 31.72$ kN ($\uparrow$)
- A (fixed): $H_A = 26.56$ kN (towards the left), $V_A = 31.72$ kN (downward), $M_A = 74.50$ kNm

Check: $\Sigma H$: $50 - 26.56 - 23.44 = 0$; $\Sigma V$: $-31.72 + 31.72 = 0$ (no vertical load).

The BMD is linear in each member, with a jump of 30 kNm at F from the couple.

**Answer: at D, $H_D = 23.44$ kN (towards the left) and $V_D = 31.72$ kN (upward).**

## Check yourself

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

1. D_s = 3 + 2 - 3 = 2. Which redundants are taken?
   A. Moments at A and B
   B. Horizontal and vertical reactions at the hinge D, leaving a cantilever fixed at A
   C. Only the vertical reaction at D
   D. Reactions at A, with D left as is
   Answer: B. Releasing the hinge at D gives a determinate cantilever fixed at A, with X1 = H_D and X2 = V_D as redundants. (Key point: Redundants chosen)

2. The equations are [F]{X} = -{D0}. Why is the right side 46.7 and 4115.0 (positive) when D10 and D20 are negative?
   A. Because the loads are reversed
   B. Because the matrix is negative
   C. Because the load displacements are moved to the other side with a minus sign
   D. Because the flexibility terms are subtracted
   Answer: C. Total displacement at D is zero: D0 + F X = 0, so F X = -D0, which turns negative D0 into positive terms. (Key point: Compatibility equations)

3. With H_D = 23.44 kN (left) and H_A = 26.56 kN (left), which check confirms the horizontal equilibrium?
   A. 50 = 26.56 + 23.44
   B. 50 + 26.56 = 23.44
   C. 26.56 = 23.44
   D. 50 = 26.56 - 23.44
   Answer: A. The 50 kN load to the right is resisted by the two leftward reactions: 26.56 + 23.44 = 50. (Key point: Result and equilibrium check)

## Related in Theory of Structures II

- [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
- [Flexibility matrix: reactions at E and A, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-frame-75knm-20kn-e) — Theory of Structures II, chapter 2
- [Flexibility matrix: reactions at D, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-frame-30kn-reactions-d) — Theory of Structures II, chapter 2
- [Stiffness matrix: frame with 100 kN, end reactions](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-matrix-frame-100kn-end-reactions) — Theory of Structures II, chapter 3
- [Stiffness matrix: BMD of frame, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-frame-60kn-a-fixed-d) — Theory of Structures II, chapter 3
- [Flexibility matrix method: portal frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-portal-12000) — Theory of Structures II, chapter 2

## Related in other subjects

- [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6
- [Frame AFD, SFD, BMD (50 kN, 15 kN/m)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-50kn-15knm) — Strength of Materials, chapter 1
- [Frame SFD, BMD (100 kN, 20 kN/m, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-sfd-bmd-100kn-hinge-60deg) — Strength of Materials, chapter 1
- [Frame AFD, SFD, BMD (30 kN/m, 50 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-30knm-50knm) — Strength of Materials, chapter 1