Question 3.P.28: Determine the resultant vertical force on the curved structu......
Determine the resultant vertical force on the curved structure AB and also the line of action. At any section the height, x = 0.5 Z² where z is the width. Consider a width of 1 m. Water stands upto x = 2 m. (note the constant 0.5 should be dimensional, 1/m).
The area and centre of gravity are to be determined by integration. Considering a strip at x and width dx,
A=\int_{0}^{x}Z\,d x\,{\mathrm{where}}\ Z=(x/0.5)^{0.5} \\ ∴ A = 0.9428 x^{1.5}To obtain the line of centre of gravity from z = 0 line
A \overline{Z}\,=\,\int_{0}^{x}Z\,d x\,Z/2=\int_{0}^{x}x\,d x=x^{2}/2 \\ ∴ \overline{Z} = (x^{2} / 2) (1/0.9428 x^{1.5} ) = x^{0.5} / 1.8856The vertical force due to gas pressure = vertical projected area × pressure
Top width, Z = (3/0.5)^{0.5} = 2.45 \mathrm{m}. Considering 1 m width.
P = 0.8 \mathrm{bar} = 0.8 × 10^{5} \mathrm{N/m^{2}}
Pressure force = 2.45 × 1 × 0.8 × 10^{5} \mathrm{N} = 195959 \mathrm{N}. This acts along 1.225 m from Z = 0 position. The vertical force due to the weight of water which stands upto x = 2 m over the surface.
Weight = volume × sp. weight = area × depth × sp. weight
A = 0.9428 x^{1.5}, x = 2, A = 2.6666 \mathrm{m^{2}}
Force due to water = 2.6666 × 9810 = 26160 N. This force acts at
\overline{Z} = x^{0.5} / 1.8856 = 0.75 \mathrm{m} from the vertical at Z = 0
Total force = 195959 + 26160 = 222119 N, To determine the line of action:
195959 × 1.225 + 26160 × 0.75 = Z (222119),
z = 1.169 m.
