Arch 18 m: forces at 3 m intervals, diagrams

Data and assumptions: symmetrical parabolic three-hinged arch, $L = 18$ m, $f = 3$ m, $y = \dfrac{4\times 3\,x(18-x)}{18^2} = \dfrac{x(18-x)}{27}$. 80 kN at 4.5 m from the right support ($x = 13.5$ m); UDL 5 kN/m over half the span (taken as the left half, $x = 0$ to 9 m).

Reactions and thrust

$$V_B\times 18 = 80\times 13.5 + (5\times 9)\times 4.5 = 1282.5 \Rightarrow V_B = 71.25\ \text{kN}, \qquad V_A = 125 - 71.25 = 53.75\ \text{kN}$$ $$H\times 3 = V_A\times 9 - 45\times 4.5 \Rightarrow H = \frac{53.75\times 9 - 202.5}{3} = 93.75\ \text{kN}$$

Forces at 3 m intervals

Simple-beam values: $M_b$, $V_b$ at each section; $\tan\theta = \dfrac{dy}{dx} = \dfrac{18-2x}{27}$.

$$M = M_b - H\,y, \quad V = V_b\cos\theta - H\sin\theta, \quad N = V_b\sin\theta + H\cos\theta$$

(At the supports $x=0$ and $18$, the values are those just inside the span; $M = 0$ at $A$, $C$ and $B$. The shear is the radial shear and $N$ the normal thrust.)

Diagrams drawn on the horizontal x-axis

Bending moment (+ sagging, scale not exact):

    0 m |  0.0
    3 m | ------- -17.5
    6 m | ------- -17.5
    9 m |  0.0
   12 m | ++++++++++++++++++++++++ 57.5
   15 m | ++++++++++++++++++++++++ 57.5
   18 m |  0.0

Radial shear:

    0 m | ------ -7.3
    3 m | -- -2.7
    6 m | ++ 2.8
    9 m | +++++++ 8.8
   12 m | ++++++++++++++++++++++++ 28.9
   15 m | ---------------------- -27.0
   18 m | ------ -7.3

Normal thrust (always compressive):

    0 m | ++++++++++++++++++++++ 107.8
    3 m | +++++++++++++++++++++ 101.4
    6 m | ++++++++++++++++++++ 96.7
    9 m | +++++++++++++++++++ 93.8
   12 m | ++++++++++++++++++ 89.6
   15 m | +++++++++++++++++++++++ 114.6
   18 m | ++++++++++++++++++++++++ 117.5

Maximum sagging moment $\approx 109.7$ kN m at $x \approx 13.5$ m, and maximum hogging moment $\approx -19.7$ kN m at $x \approx 4.5$ m (in the left half, under the UDL).