Consistent deformation: bent frame with support movement

Assumptions: A(0,0), C(0,4), B(5,4). Column AC (3EI) is 4 m high with the 30 kN horizontal load (to the right) at mid-height; beam CB (5EI) is 5 m with 20 kN/m downward; B is fixed. Support A moves: 10 mm down, 10 mm to the left and rotates 0.002 rad clockwise. The value of EI is not given in the question, so the answer is derived in terms of EI (kNm² units, E in kN/m², I in m⁴), and a numerical example with $EI=20\,000$ kNm² is added. Axial deformations are ignored.

Step 1: Degree of indeterminacy and primary structure

Reactions = 3 + 3 = 6, equilibrium = 3, so $D_s=3$. Release B. Redundants: $X_1=H_B$ (right +), $X_2=V_B$ (up +), $X_3=M_B$ (anticlockwise +). Primary structure: cantilever B-C-A fixed at A (which moves with the support).

Step 2: Displacement of B in the primary structure

(a) Due to the loads (using the moment diagrams of the primary structure): the 20 kN/m UDL on CB and the 30 kN force on AC give

$$\{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}733.33\\-2079.17\\-436.67\end{Bmatrix}$$

(b) Due to the movement of A. The primary structure moves as a rigid body with A: for A displaced $(u_A,v_A)=(-0.010,-0.010)$ m and rotated $\theta_A=-0.002$ rad (anticlockwise +), the point B at $(5,4)$ from A moves by

$$u_B=u_A-\theta_A\,y=-0.010+0.002\times4=-0.002\ \text{m},\quad v_B=v_A+\theta_A\,x=-0.010-0.002\times5=-0.020\ \text{m},\quad \theta_B=-0.002\ \text{rad}$$ $$\{\Delta_S\}=\{-0.002,\;-0.020,\;-0.002\}^T\ (\text{m, m, rad})$$

Step 3: Flexibility matrix of the primary structure (unit forces at B)

$$[F]=\frac{1}{EI}\begin{bmatrix}7.111&-13.333&-2.667\\-13.333&41.667&9.167\\-2.667&9.167&2.333\end{bmatrix}$$

(e.g. $f_{11}=\dfrac{1}{3EI}\displaystyle\int_0^4(4-y)^2dy=\dfrac{7.111}{EI}$; $f_{33}=\dfrac{4}{3EI}+\dfrac{5}{5EI}=\dfrac{2.333}{EI}$.)

Step 4: Consistent deformation (B is fixed)

$$[F]\{X\}+\{\Delta_L\}+\{\Delta_S\}=\{0\}\quad\Rightarrow\quad \{X\}=\{X_L\}+EI\{x_S\}$$ $$\{X_L\}=\begin{Bmatrix}-19.286\ \text{kN}\\54.571\ \text{kN}\\-49.286\ \text{kNm}\end{Bmatrix},\qquad \{x_S\}=\begin{Bmatrix}0.003978\\0.004157\\-0.010929\end{Bmatrix}\ (\text{per unit }EI)$$

So $H_B=-19.29+0.00398\,EI$ kN, $V_B=54.57+0.00416\,EI$ kN and $M_B=-49.29-0.01093\,EI$ kNm. The settlement effects grow in proportion to EI.

Step 5: Results for the numerical example ($EI=20\,000$ kNm²)

$H_B=60.27$ kN, $V_B=137.71$ kN, $M_B=-267.86$ kNm.

Bending moments (sagging/inside tension +, kNm):

Shear and axial forces (EI = 20 000; axial: + tension):

The BMD is drawn from the table: AC has a straight-line moment with a kink at the 30 kN load, CB is parabolic (UDL) between the C and B values.

Answer: redundants at B: $H_B=-19.29+0.00398EI$ kN, $V_B=54.57+0.00416EI$ kN, $M_B=-49.29-0.01093EI$ kNm; for EI = 20 000 kNm²: $H_B=60.3$ kN, $V_B=137.7$ kN, $M_B=-267.9$ kNm.