Force method: portal frame, 50 kN/m

Portal frame: span $L = 6$ m, height $h = 5$ m, both bases hinged, EI constant, beam load $w = 50$ kN/m. A hinged portal has $r = 4$, so $D_s = 1$. Take the horizontal reaction $H$ at D as the redundant (D is released into a roller).

   50 kN/m
 B vvvvvvvvvvvv C
 |              |
 | 5 m          | 5 m
 |              |
 A o            o D  <- H
 |<---- 6 m --->|

Primary structure (D on a roller)

Reactions: $V_A = V_D = 50\times6/2 = 150$ kN.

• $M_0$: columns 0; beam $M_0 = 150x - 25x^2$ (maximum $225$ kNm at mid-span).
• $m$ (unit horizontal force at D): column CD: $m = y$ (from D); beam: $m = 5$; column AB: $m = y$ (from A).

Flexibility coefficient and load term

$$\begin{aligned} f_{11} &= \frac{1}{EI}\left[2\int_0^5 y^2dy + 5^2(6)\right] = \frac{1}{EI}\left[83.33 + 150\right] = \frac{233.33}{EI} \\ \Delta_{10} &= \frac{1}{EI}\int_0^6 M_0\,(5)\,dx = \frac{5}{EI}\cdot\frac{wL^3}{12} = \frac{5\times900}{EI} = \frac{4500}{EI} \end{aligned}$$

(The loads push D outwards by $\Delta_{10}$ in the primary structure, and the inward thrust $H$ brings it back by $f_{11}H$.)

Compatibility (D does not move horizontally)

$$\Delta_{10} - f_{11}H = 0 \ \Rightarrow\ H = \frac{4500}{233.33} = 19.286\ \text{kN} \left(= \frac{135}{7}\right)$$

The horizontal reaction acts inwards (towards the frame) at each support.

Final results

Bending moments: $M_A = M_D = 0$; $M_B = M_C = -H h = -19.286\times5 = -96.43$ kNm (tension on the outside); mid-span $M = 225 - 96.43 = 128.57$ kNm (sagging).

Answer: at each hinge, $V = 150$ kN (upward) and $H = 19.286$ kN (towards the other support).