Circular arch: BM, thrust, shear at D, BM diagram

Assumptions (details partly unclear in the figure): three-hinged circular arch, $A$ and $B$ hinged, crown hinge $C$ at $x = 8$ m with rise $f = 5$ m; $AD = 4$ m, $DC = 4$ m, $CB = 8$ m, so the span is $L = 16$ m and $D$ is at $x = 4$ m. UDL 20 kN/m on the left part $AC$ ($x = 0$ to 8 m). The 10 kN horizontal load acts to the right at the arch point 3 m (horizontally) to the right of the crown ($x = 11$ m).

Radius: $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{64+25}{10} = 8.9$ m. Equation: $y = \sqrt{8.9^2-(x-8)^2} - 3.9$. At $D$: $y_D = 4.05$ m, $\tan\theta = \dfrac{8-4}{\sqrt{8.9^2-16}} = 0.5031$, $\theta = 26.71^\circ$. At the 10 kN load, $y(11) = 4.479$ m.

Reactions

Let $H_A$ act to the right at $A$ and $H_B$ to the left at $B$.

Horizontal equilibrium: $H_A - H_B + 10 = 0 \Rightarrow H_B = H_A + 10$.

Moments about $A$:

$$V_B\times 16 = (20\times 8)\times 4 + 10\times 4.479 \Rightarrow V_B = \frac{684.79}{16} = 42.8\ \text{kN}, \qquad V_A = 160 - 42.8 = 117.2\ \text{kN}$$

Hinge condition (left part, moments about $C$; the 10 kN load is on the right part):

$$V_A\times 8 - H_A\times 5 - 160\times 4 = 0 \Rightarrow H_A = \frac{117.2\times 8 - 640}{5} = 59.52\ \text{kN}, \qquad H_B = 69.52\ \text{kN}$$

Forces at D

$$\begin{aligned} M &= V_A\times 4 - 20\times 4\times 2 - H_A\,y_D = 468.8 - 160 - 59.52\times 4.05 = 67.71\ \text{kN m}\\ V_b &= V_A - 20\times 4 = 37.2\ \text{kN}\\ V &= V_b\cos\theta - H_A\sin\theta = 6.48\ \text{kN}\\ N &= V_b\sin\theta + H_A\cos\theta = 69.89\ \text{kN} \end{aligned}$$

Bending moment diagram

$M = V_A\,x - H_A\,y - \text{(moments of the loads)}$ at various sections:

The bending moment is continuous at the 10 kN load (the load has no moment arm about its own point); only the slope of the curve changes there. $M = 0$ at $A$, $C$ and $B$ (hinges). The largest sagging moment is about $+70.1$ kN m at $x \approx 4.6$ m (under the UDL), and the largest hogging moment is about $-113.1$ kN m at $x \approx 12.7$ m (in the right half).

 BMD (positive = sagging, scale approx.)
 x (m)   0    4     8   11    12.7    16
 M       0   67.7  0  -97.4  -113.1   0

Answer: at $D$: bending moment $= 67.71$ kN m, normal thrust $= 69.89$ kN, radial shear $= 6.48$ kN.