LA DIMENSIÓN CLÁSICA

Consider first a thin, flat, submerged, rectangular plate of any width w and length L and inclined at any angle  to a liquid surface, as shown in Figure 3.31. C denotes the centroid of the plate (lies at the centre of area), and P the centre of pressure.

The resultant net pressure force F_{p net}acts at the centre of pressure, P, and, since the liquid is at rest, F_{p net}must be perpendicular to the plate (as there is no shear force). Now, F_{p net}can be resolved into vertical and horizontal components given by

F_v Fp netcos and F_h Fp netsin

The components of F_hand F_valong the plate must cancel as the resultant force is perpendicular to the gate.

冕_y1^y2wy2dy 冕_y1^{y2wy dy} 冕_A^y2dA

冕_A^{y dA}

Figure 3.31Forces acting on inclined, submerged plate View in x-direction

(projection onto vertical plane)

F_{p net}F_v

F_v(reaction) b F_h

O θ

O O, O

Volume V of liquid above ab

Plate length (a to b) = L Area A =wL

Liquid surface Liquid surface

h_p h_c

Area = A_v

L sinθ C

w P

X

a

G

Vertical component F_v

For equilibrium, the vertical component F_vmust be balanced by an equal and opposite reaction F_v. This reaction must support the weight of all the liquid lying directly above the plate (between the dotted lines). If the volume of this liquid is V, then

Weight of liquid mg  (V)g  Fv (3.44)

The weight of the liquid acts at the centre of gravity G of the volume V. Thus, the line of action of F_vmust pass through G. Equation (3.44) applies to any shape of flat, submerged surface and also to curved as well as flat surfaces.

Horizontal component, F_h

For a flat, submerged surface of any shape at any , the horizontal component of force (the resolved component from equation (3.39)) is

F_h gAhCsin ^(3.45)

Consider an end view of the inclined plate on a vertical plane (direction X, illustrated in Figure 3.31). The view is the vertical projection of the plate and forms an imaginary vertical surface of width w, height L sin and area Av wL sin (hCand h_pare the same as for the inclined surface). The net (horizontal) force on the vertical surface, using equation (3.45), is

F_h gAvh_C

 g(wL sin)hC

 gAhCsin

which is identical to equation (3.30). Therefore, F_hcan be found by considering the horizontal force on the projected area A_vof the inclined plate.

The same also applies for the horizontal force on curved surfaces and can be explained by considering a submerged body of uniform width w and in a state of equilibrium, as shown in Figure 3.32. One end of the body is curved and the other end is flat and vertical.Whether the body is viewed in direction X or in direction X exactly the same rectangular shape is seen.

The net force on the vertical end is F_h2and must be horizontal. F_h2acts at the centre of pressure P₂. The horizontal component of the net force on the curved end is F_h1acting through P₁(which does not lie on the curved surface).

For equilibrium

F_h1 F_h2

otherwise the body would move in one direction or the other, and h_p1 h_p2

otherwise the body would rotate.

The vertical force on the plate is equal to the weight of water above the plate.

Figure 3.32Submerged body

F_n2 F_n2

Liquid surface Liquid surface

h_p2 h_p2

h_p1

p₂

P₁ p₂

X’ w X

F_h2 F_h1

To summarize, the horizontal force on any curved or inclined surface can be determined in the following way:

Surfaces can be flat or curved (concave or convex) with the fluid above or below them. In all cases the vertical component of the hydrostatic force is equal to the weight of water above the gate (or the weight of the water that would be there if the gate were not), as shown in Figure 3.33. The line of action of the vertical force passes through the centre of gravity of the fluid above the gate.

The horizontal component of force is the same as that on the vertical projection of the vertical gate.

Two other comments need to be made about curved surfaces:

▪ On the surface of a curved plate, the lines of action of the horizontal and vertical

components of force will not meet on the surface of the plate, and the point of intersection of the forces is not referred to as the centre of pressure as it is not on the surface of the plate.

▪ When the surface is split into small elements, the force from each element is locally perpendicular to the surface of the plate. This means that, for a plate made in the shape of a circular arc, the resultant force will pass through the centre of the arc.

To find F_hand h_pfor any curved or inclined flat surface, find F_hand h_pfor the projection of that surface onto a vertical plane.

Figure 3.33Illustration of vertical force acting on curved gates

A rectangular sluice gate, 1 m wide and 3 m long, is submerged in water with the upper edge hinged to a wall at a depth of 1 m below the water surface. The gate is inclined at 30° to the horizontal. Taking  as 1000 kg m^ⴚ3and g as 9.81 m s^ⴚ2, find the net force on the gate that is due to gauge pressure, the depth of the centre of pressure and the normal force F that must be applied at the lower edge of the gate to open it. Neglect friction at the hinge.

F_pacts perpendicular to the gate.

To find the weight of water above the gate, consider the shape of the volume of water to be made up of a cuboid and triangular prism.

The horizontal force is found by considering the vertical projection of the gate.

Figure 3.34

F_v

F_v G

G

3 m Gate is 1 m wide

F_open F_p

30°

h 1 m

Figure 3.35

F_p 30°

h=1 m 3 sin 30 m

1 m

y F_V (weight of water above the gate)  Vg

FV [(3 cos 30)(1)  (3 cos 30)(3 sin 30)] (1)(g)

 44.60 kN



1 2

dF_h ghw dh

F_h gw冕₁^{1 3 sin 30h dh 1g}冤^h22冥^2.51 2.625g  25.75 kN

The total force on the gate is found by resolving the vertical and horizontal forces F_p 兹F苶F²v苶²h

 兹44.6苶 25.75²苶²

 51.5 kN

The line of action of the horizontal force is found by considering the force on the vertical projection of the gate (as shown in the previous worked example) and taking moments about the water surface:

dM ghw dhh

M gw冕₁^{2.5h2dh 1g}冤^h33冥^2.51

 4.875 g M Fhh_p h_p 4

2 . . 8 6 7 2 5

  1.86 m5

To find the force required to open the gate, take the moments for all the forces about the hinge:

F_py_p Fopen 3 sin30 hp

y



p

1

 y_p 1.72 m

⇒ Fopen 51.5 1 3

0³ 1.72

 29.5 kN

Figure 3.36

Figure 3.37

A sluice gate has a width 2 m and a radius of 6 m. The gate pivot is level with the water surface. Find the horizontal and vertical components of force on the gate and the line of action of both. Also find the resultant force.

Find the horizontal component of the force by considering the projected vertical gate:

F_h gAvh_C 1000  9.81  (6  2)  3  353.2 kN or

Fh gh(2h)

 F_h 2g冕₀^{6hh  2g}冤^h22冥⁶0 36g  353.2 kN F_p

F_v

F_h h_p= 1.86 m

F_p y_p

F_open h

3

In document TÉRMINO EXCLUSIÓN SOCIAL (página 36-40)