Force method: L-frame reactions at A

L-frame: A hinged (end of horizontal member AB, length $l$, UDL $w = 10$ kN/m), B rigid corner, vertical member BC of height $l$ down to C fixed. EI is constant.

  w = 10 kN/m
 vvvvvvvvvvvvv
 A o----------B
               |
               | l
               |
              /// C (fixed)
 |<---- l ---->|

The frame is statically indeterminate to the 2nd degree ($r = 2 + 3 = 5$). Release the hinge at A to get the cantilever C-B-A, and take $X_1$ = horizontal reaction at A and $X_2$ = vertical reaction at A as redundants.

Moment expressions

In AB, $x$ measured from A; in BC, $y$ measured down from B.

Flexibility coefficients

$$\begin{aligned} f_{11} &= \frac{1}{EI}\int_0^l y^2\,dy = \frac{l^3}{3EI} \\ f_{12} &= \frac{1}{EI}\int_0^l y\,l\,dy = \frac{l^3}{2EI} \\ f_{22} &= \frac{1}{EI}\left[\int_0^l x^2\,dx + \int_0^l l^2\,dy\right] = \frac{4l^3}{3EI} \end{aligned}$$

Load displacements

$$\begin{aligned} \Delta_{10} &= \frac{1}{EI}\int_0^l y\left(-\frac{wl^2}{2}\right)dy = -\frac{wl^4}{4EI} \\ \Delta_{20} &= \frac{1}{EI}\left[\int_0^l x\left(-\frac{wx^2}{2}\right)dx + \int_0^l l\left(-\frac{wl^2}{2}\right)dy\right] = -\frac{5wl^4}{8EI} \end{aligned}$$

Compatibility (zero displacement at the hinge A)

$$\begin{aligned} \tfrac13 X_1 + \tfrac12 X_2 &= \frac{wl}{4} \\ \tfrac12 X_1 + \tfrac43 X_2 &= \frac{5wl}{8} \end{aligned}$$

Solving: $X_2 = \dfrac{3wl}{7}$ and $X_1 = \dfrac{3wl}{28}$.

With $w = 10$ kN/m:

$$V_A = \frac{30\,l}{7} = 4.29\,l\ \text{kN}\ (\uparrow), \qquad H_A = \frac{15\,l}{14} = 1.07\,l\ \text{kN}\ (\rightarrow)$$

(with $l$ in metres). Check from the support at C: $V_C = 10l - 4.29l = 5.71\,l$ kN up, $H_C = 1.07\,l$ kN to the left, $M_C = 5l^2/14 = 0.357\,l^2$ kNm.

Answer: at A, $H_A = 3wl/28 = 1.07\,l$ kN (to the right) and $V_A = 3wl/7 = 4.29\,l$ kN upward.