Force method: frame with support settlement

Reading of the figure: A (0, 0) fixed, column AB 4 m ($EI$), beam BC 4 m ($2EI$) with 10 kN/m, column CD 4 m ($EI$) with D hinged at the level of A (the extra 2 m dimension is not needed). Support D settles by $200/EI$ (downward), where $EI$ is the stiffness of the columns.

$D_s = 2$. Release the hinge at D; redundants: $X_1$ = horizontal reaction ($\rightarrow$ positive) and $X_2$ = vertical reaction ($\uparrow$ positive).

Primary structure moments (sign: + tension on the inner face)

• $M_0$ (UDL on the cantilever A-B-C): $-80$ kNm at B and constant down the column; in the beam it falls parabolically from $-80$ at B to 0 at C.
• $m_1$ (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column falling from 4 at B to 0 at A.
• $m_2$ (unit vertical force at D): zero in CD, falling linearly from 4 at B to 0 at C in the beam, and constant at 4 in the column.

Flexibility coefficients (in $1/EI$ units)

$$\begin{aligned} f_{11} &= \frac{74.667}{EI} \\ f_{12} &= \frac{48.000}{EI} \\ f_{22} &= \frac{74.667}{EI} \\ \Delta_{10} &= \frac{-853.3}{EI} \\ \Delta_{20} &= \frac{-1440.0}{EI} \end{aligned}$$

Compatibility with settlement

The displacement of D along $X_1$ is zero. The displacement along $X_2$ (upward) is $-200/EI$ because D moves down. The right side of each equation is $-\Delta_{i0} + \delta_i$ (second equation: $1440 - 200 = 1240$):

$$\begin{aligned} 74.667\,X_1 + 48.000\,X_2 &= 853.3 \\ 48.000\,X_1 + 74.667\,X_2 &= 1240.0 \end{aligned}$$ $$X_1 = 1.28\ \text{kN},\qquad X_2 = 15.78\ \text{kN}$$

Final bending moments (kNm)

Reactions

• At D: $H_D = 1.28$ kN ($\rightarrow$), $V_D = 15.78$ kN ($\uparrow$)
• At A: $H_A = 1.28$ kN ($\leftarrow$), $V_A = 24.22$ kN ($\uparrow$), $M_A = 16.87$ kNm (anticlockwise)

Check: $V_A + V_D = 40.00$ kN $= 10\times4 = 40$ kN.

Answer: reactions at the hinged support D: horizontal $1.28$ kN (to the right) and vertical $15.78$ kN (upward).