Vertical force on stepped dam

The number of steps is not legible, so this reading is used. Assumed: water depth equals the dam height $H=3$ m, with 10 steps, each 0.3 m high and 0.3 m deep, and the dam is $w=3$ m wide. The water stands against the upstream stepped face.

 water ____
        |_|__          each step 0.3 x 0.3 m
        |   |_|__
        |       |_|__

The horizontal force acts on the vertical projection and does not matter here. The vertical force equals the weight of water above the stepped surface up to the free surface. Only the horizontal treads carry vertical pressure.

Tread $k$ ($k=1,\ldots,10$) lies at depth $0.3k$ m below the surface, has horizontal area $0.3\times3=0.9\ \text{m}^2$, and carries pressure $\gamma(0.3k)$:

$$F_V=\gamma\,w\,(0.3)\sum_{k=1}^{10}0.3k=9810(3)(0.3)(0.3)(55)$$ $$F_V=9810\times3\times4.95=145.7\ \text{kN}$$

(Equivalently, the volume of water above the steps is $3\times4.95=14.85\ \text{m}^3$, so $F_V=9.81\times14.85=145.7$ kN.)

Answer: vertical force on the dam $\approx146$ kN downward (for 3 m depth and 10 steps). For $n$ steps, $F_V=\gamma w(0.3)^{2}\dfrac{n(n+1)}{2}$.