Question 3.P.16: Determine the resultant force and the direction of its actio......
Determine the resultant force and the direction of its action on the segmental gate shown in Fig. P. 3.16.
Height, h = 4 × 2 \sin 30 = 4 m. Considering 1 m width, the horizontal force = 9810 × 2 × 4 = 78480 N. The line of action (centre of pressure) is 1.333 m from the bottom.
Vertical force (upwards) equals the weight of displaced. Volume of segment of circle
= [(π R^{2} × 60/360) – (2 R \sin 30 × R \cos 30/2)] × 1 = 1.4494 m^{3}
Weight = 1.449 × 9810 = 14218 N upwards direction.
\tan θ = 78480/14218, θ = 79.73° from vertical
The net force is (78480^{2} + 14218^{2})^{0.5} = 79757.5 N
The centre of gravity of a segment of a circle from centre is given by
(2/3) R \sin^{3} θ/(\mathrm{Rad} θ – \sin θ \cos θ) \mathrm{where} θ is half of the segment angle.
Substituting the values
h_{CG} = (2/3) × 4 × \sin^{3} 30/[(π / 6) – \sin 30 \cos 30] = 3.67974 m
taking moments about the centre,
(14218 × 3.67974) – (2 – 1.33) × 78477 = 0.
The quantities are equal. So the resultant passes through the centre and as the resulting moment about the centre is zero. The line of action will pass through O and its direction will be (90 – 79.73)° with horizontal, as shown in figure.