---
title: "Flexibility matrix: frame with 48 kNm moment"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:flexibility-frame-48knm-hinged-a"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-frame-48knm-hinged-a
site: BCT Engineering
---

# Flexibility matrix: frame with 48 kNm moment

Theory of Structures II · Chapter 2: Force method

- Asked in: 2068 Chaitra, 8 marks
- Video: [Problem 7 Analyse Frame By Force Method / Flexibility Method [ FRAME ]](https://www.youtube.com/watch?v=Mlmp9s0tlgc) — Ganesh Shegar
- 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-frame-48knm-hinged-a
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Use the force method (flexibility matrix) to solve the frame as shown in figure below. [Figure: portal frame; left column EI of 6 m, hinged at the base A; 48 kNm moment at the top-left joint; beam EI of 4 m (1 m + 3 m); right column EI of 4 m, fixed base.]

## Short answer

Using force method with hinge reactions (horizontal X1, vertical X2) as redundants, solve flexibility equations. X1 = 2.95 kN, X2 = 9.63 kN downward. Key bending moments: M_B = -17.70 kNm (AB), 30.30 kNm (BC); M_C = -8.21 kNm; M_D = 3.59 kNm.

## Answer

**Assumptions:** A(0,0) is hinged at the base of the 6 m left column AB; the 4 m beam BC is horizontal at the top of AB; the 4 m right column CD is fixed at D, whose base is 2 m above A. The 48 kNm moment acts clockwise at joint B. EI is the same for all members. Axial deformation is ignored.

#### Degree of indeterminacy
Reactions: hinge A (2) + fixed D (3) = 5; equations = 3. So the frame is indeterminate to the 2nd degree. Redundants: the reactions at the hinge A, **coordinate 1 = horizontal $X_1$ (right +)** and **coordinate 2 = vertical $X_2$ (up +)**. Released structure: cantilever fixed at D with A free.

#### Moment diagrams of the released structure (positive = inside tension)
| Case | AB (A to B) | BC (B to C) | CD (C to D) |
|---|---|---|---|
| Load: 48 kNm at B ($M_L$) | 0 | 48 | 48 |
| Unit horizontal force at A ($m_1$) | 0 to -6 | -6 | -6 to -2 |
| Unit vertical force at A ($m_2$) | 0 | 0 to 4 | 4 |

(In case 1 the moment equals the height of the section above A; in case 2 it equals the horizontal distance from A.)

#### Flexibility coefficients ($\int m_im_j\,dx/EI$)
$$
f_{11}=\frac{1}{EI}\left[\frac{6^3}{3}+6^2(4)+\frac{6^3-2^3}{3}\right]=\frac{285.33}{EI},\quad
f_{22}=\frac{1}{EI}\left[\frac{4^3}{3}+4^2(4)\right]=\frac{85.33}{EI},\quad
f_{12}=f_{21}=\frac{-112.00}{EI}
$$
$$
\Delta_{1L}=\frac{1}{EI}[(-6)(48)(4)+(-1)(48)(16)]=\frac{-1920}{EI},\qquad
\Delta_{2L}=\frac{1}{EI}\left[\frac{4}{2}(48)(4)+4(48)(4)\right]=\frac{1152}{EI}
$$

#### Compatibility (A cannot move)
$$
\begin{bmatrix}285.33&-112.00\\-112.00&85.33\end{bmatrix}\begin{Bmatrix}X_1\\X_2\end{Bmatrix}=-\begin{Bmatrix}-1920\\1152\end{Bmatrix}
\;\Rightarrow\; X_1=2.95\ \text{kN},\quad X_2=-9.63\ \text{kN}
$$
So the hinge A develops a horizontal reaction of 2.95 kN to the right and a vertical reaction of 9.63 kN downward.

#### Reactions at D
$H_D=-2.95$ kN (left), $V_D=9.63$ kN (up), $M_D=3.59$ kNm (anticlockwise). Check, moments about A: $V_D(4)+|H_D|(2)+M_D=48.00$ kNm, equal to the applied 48 kNm ✓.

#### Final bending moments ($M=M_L+X_1m_1+X_2m_2$, inside tension +)
| Joint | Value (kNm) |
|---|---|
| A | 0 |
| B (on AB) | -17.70 |
| B (on BC) | 30.30 |
| C | -8.21 |
| D (on CD) | 3.59 |

Joint B: 30.30 - (-17.70) = 48 kNm, equal to the applied moment ✓. Negative values mean tension on the outside face. The BM varies linearly in every member (no loads between joints).

**Answer: $X_1=H_A=2.95$ kN and $X_2=V_A=9.63$ kN (downward); $M_B=-17.70$ kNm in AB and $30.30$ kNm in BC, $M_C=-8.21$ kNm, $M_D=3.59$ kNm.**

## Check yourself

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

1. Why is the frame 2nd degree indeterminate?
   A. 5 reaction components (hinge 2 + fixed 3) against 3 equations
   B. 6 reactions against 3 equations
   C. 3 reactions against 3 equations
   D. 4 reactions against 2 equations
   Answer: A. Five unknown reactions minus three equilibrium equations leaves two redundants, the horizontal and vertical reactions at A. (Key point: Degree of indeterminacy)

2. What condition is imposed to find X1 and X2?
   A. Displacement of A in the released frame plus redundant effects equals zero
   B. The moment at A equals 48 kNm
   C. The reaction at D is zero
   D. Axial deformation equals bending deformation
   Answer: A. Support A is a hinge that cannot move, so [F]{X} = -{Delta_L}. (Key point: Compatibility)

3. What reactions develop at hinge A?
   A. X1 = 2.95 kN to the right, X2 = 9.63 kN downward
   B. X1 = 9.63 kN right, X2 = 2.95 kN up
   C. X1 = 48 kN, X2 = 0
   D. X1 = 2.95 kN left, X2 = 9.63 kN up
   Answer: A. Solving the 2x2 system gives X1 = 2.95 kN and X2 = -9.63 kN (downward). (Key point: Result)

## Related in Theory of Structures II

- [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
- [Force method: fixed portal frame, 30 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aportal-fixed-force-method-4x4-30knm) — Theory of Structures II, chapter 2
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