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CHAPTER II.

HYDROSTATIC THRUST.

27. Let us apply immediately the mechanical Axiom or Corollary of § 5, derived from the Definition of a Fluid, "The Stress on any plane in a Fluid at rest is a Normal Pressure," to the solution of an important Hydrostatical Question, the determination of the

Thrust of Water against a Reservoir Wall.

First, suppose the wall of the reservoir a masonry dam with a vertical face, AB representing the elevation of the face in a plane perpendicular to its length (fig. 20); and draw any plane BC through B the foot of the face AB to meet in C the surface of the water AC (which we have shown to be a horizontal plane).

Consider the equilibrium of the water in ABC; and, merely to fix the ideas, it is convenient to suppose this water solidified or frozen; we shall often make use of this supposition hereafter.

The forces maintaining the equilibrium of ABC are the force W of its weight acting vertically downwards, the thrust P of the wall acting horizontally, and the thrust R of the water on the plane BC, acting perpendicular to BC.

42

HYDROSTATIC THRUST

Then, denoting the angle ABC by 0, and resolving parallel to the plane BC,

P sin 0 W cos 0, or P= W cot 0.

=

We notice that if 0=45°, then P= W, so that the thrust P on the wall is given by the weight of the isosceles prism ABC.

Now if AB=h, the depth of the water in feet, then AC=h tan 0; and if l denotes the length of the wall, and w the weight in lb of a cubic foot of water or the liquid, then W = {wlh2 tan 0,

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an expression independent of 0, as it should be.

The average pressure on AB is P/hl=wh, the pressure in lb/ft2 at depth th feet.

But the pressure at any point of AB being proportional to the depth, as represented by the ordinate of the straight line Ab, the thrust P on AB is not uniformly distributed over AB; but the distribution of thrust P, as represented by the pressure p, is uniformly varied, as represented by the ordinate of Ab; the thrust P being represented by the area ABb.

or

Resolving vertically to determine R,

R sin 0 W=wlh2 tan 0,

=

R={wlh2 sec 0;

also if the vertical through G, the centre of gravity of ABC, meets BC in K, then K will be the point of application of the resultant thrust R; and CK=3CB.

Therefore also H is the point of application of the thrust P on AB, when HK is horizontal, and therefore AH AB; the three forces P, R, W which maintain the equilibrium of ABC meeting in K.

ON A RESERVOIR WALL.

43

The points H and K are called the centres of pressure

of AB and BC.

28. Next suppose the reservoir is bounded by a parallel earthwork dam, forming a wall sloping at an angle a suppose; to determine the thrust on this sloping wall DE. Again draw through the foot of this wall E a plane EF in any direction; but let us for simplicity choose the direction perpendicular to DE, and let DE=b=h cosec a.

[graphic][subsumed][merged small][merged small][merged small]

Then as before, from the conditions of equilibrium of the liquid in DEF, we find the normal thrust on DE Q=wlb2 sin awlh2 cosec a ;

so that the horizontal component of this thrust is

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as it should be; and the three forces which maintain the equilibrium of DEF now intersect in its C.G.

The average pressure over DE is

Q/blwb sin a=wh,

the pressure at the depth of the middle point of DE.
The same formulas will give the thrust on the walls of
the water reaching to any depth h short of the full depth ;
and thence we can infer the distribution of stress in the
interior of the solid material of the walls, and the condi-
tions of stability to be satisfied so that the walls shall
not fail by upsetting or crushing.

44

THRUST ON A RESERVOIR DAM.

This is a very important problem in practical Engineering with the high reservoir walls in existence or course of construction, such as the Vyrnwy dam of the Liverpool Water Works in North Wales, 120 feet high, or the projected Quaker Bridge Dam of the New York Water Works, to be made 260 feet high.

The height of these dams is reckoned from the foundation, which is carried down through the alluvial soil to the solid rock; it is assumed in the design of the dam that the alluvial soil is porous, so that the water pressure is propagated through it.

29. Now if the vertical wall AB in fig. 20 is continued down to the rock foundation at O, an extra depth BO=a feet, then the hydrostatic thrust on the part BO under ground, being the difference of the thrusts on AO and AB, will be given by

{wl{(a+h)2 — h2} = wl({a2+ah); so that the average pressure over BO is w(ja+h), the pressure at the depth of the C.G. of BO.

Again, by taking moments round A, the moments of the thrusts on AB and AO being }wlh3 and wl(a+h)3, we find that the resultant thrust on BO will act at a depth below A, given by

} (a+h)3 — } h3 __ }a2+ah+h2
(a+h)2 — 1 h2

and therefore at a height above B

a+h_3a2+ah+h2

a+h

a+h

;

a+3h

a

a+2h

In earthwork dams a wall of puddled clay must be carried down to the rock foundation, to prevent percolation of water; but the great danger to avoid in an earthwork dam is water running over the crest; the water

THEORY OF EARTH PRESSURE.

45

cuts a channel which increases in size and saws the dam in two, as at the Conemaugh dam failure, which caused the Johnstown flood in America; this dam was strong and safe enough till the water was allowed to overflow the

crest.

*30. The Theory of Earth Pressure.

Substances in a finely divided, pulverised, or granular state, such as sand, loose earth, grain, meal, or a mass of spherical granules, large or small, such as lead shot or cannon balls, imitate to a certain extent the behaviour of liquids, and require to be restrained by walls; and it is important to determine the thrust which may be expected to be exerted on a retaining wall; the usual procedure is as follows.

Suppose AB (fig. 21) is an end elevation of a vertical wall, which supports a mass of the loose substance, filled up to the level AC of the top of the wall; and drawing any plane BC through the foot of the wall, consider the equilibrium of ABC, supposed solidified.

The wall AB being supposed smooth, the thrust P on it will be horizontal; and supposing the wall AB to yield ever so little horizontally, the prism ABC will slip on BC, and a stress R across BC will be called into play, which now will not be normal to BC, but will make an angle, e suppose, with the normal, in the direction resisting motion.

This angle e, the limiting angle of friction, is taken to be the greatest angle of slope of the loose substance at which it will stand when heaped up; it is also called the angle of repose of the substance.

* Articles which may be omitted at a first reading are marked with an asterisk_*.

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