The chair frame is fixed at both A and E. What is its degree of static indeterminacy, and what redundants are taken?

3; H_E, V_E and M_E released at E

In the primary structure (cantilever fixed at A), which members carry bending moment due to the couple at C?

Only AB and BC, as a constant moment of 50 kNm

Which condition gives the redundants X1, X2, X3?

The flexibility matrix times X equals minus the load displacements, so E has zero displacement as it is fixed

The sum of all moments at C is zero

The shear force in every member is zero

The redundants are equal to the reactions of a simply supported frame

2066 Bhadra · 10 marks

Consistent deformation: chair-frame with couple

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.

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.]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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