Consistent deformation: frame, BMD, SFD, normal thrust

Assumptions: A(0,0) and D are fixed bases. Column AB (2I) is 6 m high, beam BC (2I) is 6 m and column CD (I) is 3 m high, so D is 3 m above the level of A. The 1 kN/m UDL acts horizontally to the right on AB; the 3 kN load acts downward at 3 m from B on BC. Axial deformation is ignored. Positive bending moment = tension on the inside of the frame.

Degree of indeterminacy

Reactions: 3 + 3 = 6; equations = 3, so $D_s=3$. Release the fixed support D. The redundants are $X_1=H_D$ (right +), $X_2=V_D$ (up +) and $X_3=M_D$ (anticlockwise +). The primary structure is a cantilever fixed at A.

Moments in the primary structure (kNm; $m_1,m_2,m_3$ for unit $X_1,X_2,X_3$)

Flexibility matrix and load displacements ($\int m_im_j\,ds/EI$)

$$[F]=\frac{1}{EI}\begin{bmatrix}45.00&27.00&13.50\\27.00&144.00&27.00\\13.50&27.00&9.00\end{bmatrix},\qquad \{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}6.75\\-303.75\\-51.75\end{Bmatrix}$$

(For example $f_{33}=\sum \dfrac{L}{EI}=\dfrac{6}{2}+\dfrac{6}{2}+\dfrac{3}{1}=9$.)

Consistent deformation (D is fixed: all three displacements are zero)

$$[F]\{X\}=-\{\Delta_L\}\;\Rightarrow\; X_1=-2.404\ \text{kN},\quad X_2=1.842\ \text{kN},\quad X_3=3.831\ \text{kNm}$$

So $H_D=2.40$ kN to the left, $V_D=1.84$ kN upward, $M_D=3.83$ kNm anticlockwise.

Reactions at A (equilibrium)

$H_A=6-2.40=3.60$ kN (left), $V_A=3-1.84=1.16$ kN (up), $M_A=4.90$ kNm (anticlockwise). Check: $\sum F_x=6-3.60-2.40=0$ ✓, $\sum F_y=3-1.16-1.84=0$ ✓.

Bending moment diagram ($M=M_L+\sum X_im_i$, kNm; + inside tension)

AB bends in a parabola (UDL); BC is two straight lines with a peak under the load; CD is a straight line with a change of sign.

Shear force and normal thrust

Shear sign: positive when the force on the start end of the member acts along its local y-axis (90° anticlockwise from the direction A to B, B to C, C to D). Axial: negative = compression. Normal thrust in AB, BC, CD is therefore 1.16, 2.40 and 1.84 kN compression. The shear in AB goes from 3.60 kN at A to -2.40 kN at B (linear, due to the UDL), and in BC it is 1.16 kN before the 3 kN load and -1.84 kN after it.

Answer: $H_D=2.40$ kN, $V_D=1.84$ kN, $M_D=3.83$ kNm; $M_A=-4.90$, $M_B=-1.33$, $M_C=-3.38$ kNm (inside tension +).