---
title: "Consistent deformation: chair-frame with couple"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:consistent-deformation-chair-frame-couple"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-chair-frame-couple
site: BCT Engineering
---

# Consistent deformation: chair-frame with couple

Theory of Structures II · Chapter 2: Force method

- Asked in: 2066 Bhadra, 10 marks
- 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/consistent-deformation-chair-frame-couple
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Use the consistent deformation method to draw bending moment diagram of the chair-frame loaded with a couple as shown. Take $E = 2\times10^{4}$ MPa, $\ell = 3$ m, $M = 50$ kNm and $I = 4.5\times10^{8}\ \text{mm}^4$. Also draw shear force and normal thrust diagrams corresponding to the bending moment diagram. [Figure: chair-shaped frame; fixed top support, vertical member (2I) of height $\ell$, horizontal member (I) of length $\ell/3$ with the couple M applied at its end, then vertical member (2I) of height $\ell$ and horizontal span $2\ell/3$ down to a fixed base.]

## Short answer

Release E; redundants H_E=9.22 kN, V_E=5.32 kN, M_E=-8.24 kNm. Bending moments: A=13.03, C left=-19.95, C right=30.05 kNm. Shear and axial constant per member; AB and CD 5.32 kN compression, BC and DE 9.22 kN tension.

## Answer

**Assumptions (the figure is only described):** the chair frame is A(0,6) - B(0,3) - C(1,3) - D(1,0) - E(3,0) with $\ell=3$ m: AB vertical $2I$ (3 m), BC horizontal $I$ ($\ell/3=1$ m), CD vertical $2I$ (3 m) and DE horizontal $I$ ($2\ell/3=2$ m). A (top) and E (base) are fixed. The couple $M=50$ kNm acts clockwise at joint C. Axial deformation is neglected.
$E=2\times10^4$ MPa $=2\times10^7$ kN/m², $I=4.5\times10^8$ mm⁴ $=4.5\times10^{-4}$ m⁴, so $EI=9000$ kNm² and $2EI=18000$ kNm². Only the relative stiffnesses matter for the redundants, so EI is a common factor.

#### Degree of indeterminacy
$D_s=6-3=3$. Release E and take $X_1=H_E$ (right +), $X_2=V_E$ (up +), $X_3=M_E$ (anticlockwise +) as redundants. The primary structure is the cantilever A-B-C-D-E fixed at A.

#### Moments in the primary structure (kNm)
| Section | $m_1$ (H=1) | $m_2$ (V=1) | $m_3$ (M=1) | $M_L$ |
|---|---|---|---|---|
| A (AB) | 6.00 | 3.00 | 1.00 | -50.00 |
| B (AB) | 3.00 | 3.00 | 1.00 | -50.00 |
| B (BC) | 3.00 | 3.00 | 1.00 | -50.00 |
| C (BC) | 3.00 | 2.00 | 1.00 | -50.00 |
| C (CD) | 3.00 | 2.00 | 1.00 | 0.00 |
| D (CD) | 0.00 | 2.00 | 1.00 | 0.00 |
| D (DE) | 0.00 | 2.00 | 1.00 | 0.00 |
| E (DE) | 0.00 | 0.00 | 1.00 | 0.00 |

The couple at C acts on the part between the support A and C, so in the primary structure it produces a constant moment of 50 kNm on AB and BC (the sign is negative here) and no moment on CD and DE. Sign: positive = tension on the right-hand side when travelling A, B, C, D, E.

#### Flexibility matrix ($\int m_im_j\,ds/EI$, shown in units of 1/EI where $EI$ is the stiffness of $I$)
$$
[F]=\frac{1}{EI}\begin{bmatrix}45.00&32.25&12.00\\32.25&28.50&12.00\\12.00&12.00&6.00\end{bmatrix},\qquad
\{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}-487.50\\-350.00\\-125.00\end{Bmatrix}
$$

#### Consistent deformation (E is fixed)
$$
[F]\{X\}=-\{\Delta_L\}\Rightarrow X_1=9.220\ \text{kN},\quad X_2=5.319\ \text{kN},\quad X_3=-8.245\ \text{kNm}
$$

#### Reactions at A
$H_A=-9.220$ kN, $V_A=-5.319$ kN, $M_A=-13.032$ kNm (anticlockwise +). Check: $\sum H=0$, $\sum V=0$ ✓; moment equilibrium holds with the applied 50 kNm couple.

#### Bending moment diagram (kNm, sign as above)
| Section | BM |
|---|---|
| A | 13.03 |
| B | -14.63 |
| C (left, on BC) | -19.95 |
| C (right, on CD) | 30.05 |
| D | 2.39 |
| E | -8.24 |

The jump at C (50.00 kNm) equals the applied couple. Every member has a straight-line BM because there are no loads along the members.

#### Shear force and normal thrust diagrams
| Member | Axial N (+ tension) | Shear at start | Shear at end |
|---|---|---|---|
| AB | -5.32 | -9.22 | -9.22 |
| BC | 9.22 | -5.32 | -5.32 |
| CD | -5.32 | -9.22 | -9.22 |
| DE | 9.22 | -5.32 | -5.32 |

Shear and axial force are constant along each member. Axial: negative = compression. AB and CD carry 5.32 kN compression, while BC and DE carry 9.22 kN tension. The shear in each member equals the slope of its BM diagram, i.e. the change in BM along the member divided by the member length.

**Answer: $H_E=9.22$ kN, $V_E=5.32$ kN, $M_E=-8.24$ kNm; $M_A=13.03$ kNm, $M_C=(-19.95, 30.05)$ kNm.**

## Check yourself

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

1. The chair frame is fixed at both A and E. What is its degree of static indeterminacy, and what redundants are taken?
   A. 1; only H_E
   B. 3; H_E, V_E and M_E released at E
   C. 6; all reactions at A and E
   D. 2; V_E and M_E only
   Answer: B. Six reactions minus three equilibrium equations gives D_s = 3. Support E is released and H_E, V_E, M_E are the redundants, leaving a cantilever fixed at A. (Key point: Degree of indeterminacy)

2. In the primary structure (cantilever fixed at A), which members carry bending moment due to the couple at C?
   A. All four members equally
   B. Only CD and DE
   C. Only AB and BC, as a constant moment of 50 kNm
   D. Only BC
   Answer: C. The couple acts on the part between support A and C, so it gives a constant 50 kNm on AB and BC and nothing on CD and DE, which lie beyond the load. (Key point: Effect of the couple in the primary structure)

3. Which condition gives the redundants X1, X2, X3?
   A. The sum of all moments at C is zero
   B. The flexibility matrix times X equals minus the load displacements, so E has zero displacement as it is fixed
   C. The shear force in every member is zero
   D. The redundants are equal to the reactions of a simply supported frame
   Answer: B. Since E is actually fixed, the displacements at E from the loads and the redundants must cancel: [F]{X} = -{Delta_L}. (Key point: Compatibility equation)

4. What does the jump in bending moment at C (from -19.95 to 30.05 kNm) represent?
   A. The shear force at C
   B. The axial thrust in member CD
   C. The error in the iteration
   D. The applied couple of 50 kNm
   Answer: D. A concentrated couple causes a sudden jump in the BMD equal to its magnitude; here 30.05 - (-19.95) = 50 kNm. (Key point: Jump in BMD at C)

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