Ex. 450. There is a river wall of Aberdeen granite 15 ft. high and having a rectangular section; the water comes to the distance of one foot from the top of the wall; find its thickness when the line of resistance cuts the base 6 in. within the extrades. Ans. 5.34 ft.

Ex. 451. In the last Example if the wall had a section of the form shown fig. 124, where AB is 1 ft. long, the vertical face of the wall being towards the water; determine the width at the bottom when the line of resistance cuts the base 6 in. within the extrados. If the walls in this Example and the last are 200 ft. long, determine the solid contents of each.

Ans. (1) 5.86 ft. (2) 10,290 and 16,020 cub. ft. Ex. 452. In each of the last Examples determine the distance from the extrados of the point at which the line of resistance cuts a horizontal joint 8 ft. below the surface of the water. Ans. (1) 1.98 ft. (2) 1.75 ft.

[The point will, of course, be that round which the moment of the weight of the incumbent portion of the wall equals the moment of the pressure of the water on the eight feet.]

Ex. 453. A river wall whose section is a right-angled triangle just supports the pressure of water when its surface is on a level with the top of the wall; show that the thickness of the base

if the hypothenuse of the triangle is turned towards the water; but when the perpendicular is turned towards the water the thickness of the base

where w is the weight of a cubic foot of water, and w1 that of a cubic foot of the material of the wall. And show from hence that in the former case the thickness of the base is greater or less than in the latter according as the specific gravity of the wall is greater or less than 2.

Ex. 454. A wall of brickwork is to be built round a reservoir 20 ft. deep; its slope is inward; it is one foot thick at top; what must be its thickness at the bottom, that when the reservoir is full, the line of resistance may cut the base 6 in. within the extrados? Ans. 10.74 ft.

Ex. 455. The wall of a reservoir full to the brim is of brickwork and is 20 ft. high and 2 ft. thick; it is supported by props at intervals of 6 ft.; the length of each is 20 ft., and its inclination to the horizon 30°: determine the thrust on each prop, its weight being neglected. Ans. 54,632 lbs.

Ex. 456. In the last Example determine the thickness of the wall that would just support the pressure of the water if the props were removed. If the wall stand on its lowest section without the aid of cement, what must be the coefficient of friction between the surfaces?

Ans. (1) 8.6 ft. (2) 0.65.

Ex. 457. A reservoir is divided by a brickwork wall 12 ft. high 2 ft. thick, the water on one side is 10 ft. deep; what must be the depth on the Ans. 10-4 ft.

other side if the wall is just overthrown?

Ex. 458. A cofferdam sustains the pressure of 26 ft. of water, and is

supported at intervals of 10 ft. by props DE and CF; given that BC and BD are respectively 4 ft. and 18 ft. and that DE and cF are respectively 30 ft. and 18 ft.; find the thrust on each prop. And what must be the weight of the struts, and of the cofferdam, that the whole be not overthrown? The thickness of the cofferdam, and the adhesion at B, are to be neglected.

Ans. (1) Thrust on DE=88,020 lbs.; on EC=144,400 lbs. (1) 84,900lbs.

E

F

FIG. 125.

B

A

D

C

95. The Pressure of Earth. - Let a B represent a section of a wall supporting earth, whose surface is a c, it is required to determine the pressure produced

FIG. 126.

A

C

B

On AB by the earth. Now, it must be remembered that two extreme cases may come under consideration: the first arises when the earth is thoroughly penetrated with water, in which case the pressure is the same as would result from hydrostatic pressure; the second arises when the cohesion of the earth is so considerable that it would stand with its face vertical even if the wall were removed. Dismissing these two extreme cases, let us suppose the wall A B removed, the following result will then ensue: the earth being friable will weather and break away until its surface has taken a slope вc, inclined to the horizon at an angle equal to the limiting angle of resistance; when reduced to this state it will have no further tendency to break away, and, unless washed down by rain, or removed by some other extrinsic cause, will remain permanently at rest at that slope, which is therefore called

N

its natural slope. Hence, in the case we are considering, the wall is required to give a certain degree of support to the wedge of earth ABC; this wedge is generally supported in some degree by the cohesion of its parts with each other and with the earth below BC, so that the wall will be sufficiently strong if it will support the earth, on the supposition that the cohesion is quite destroyed, unless (which is not contemplated) the earth should be saturated with water. The angle of the natural slope of fine dry sand is about 35°; of dry loose shingle about 40°; of common earth, pulverised and dry, about 45°.*

Proposition 20.

If w is the weight of a cubic foot of earth, and its natural slope, the pressure produced on the vertical face of a retaining wall by earth which does not rise above its summit, and which has a horizontal surface, is the same as that produced by a fluid the weight of a cubic foot of which is w tan2

FIG. 127.

A

Y

Let AB be the section of the wall, BAC of the earth; take any portion AX equal to x of the wall, and suppose its length to be 1 foot; draw xy, making C an angle o with the horizon greater than ; then the weight w of the wedge AXY equals w x2 cotan 0, and acts vertically through a point Pwhere XP= xx, and is supported by the reaction R1 of XY and by the reaction R of the wall: the latter

N

R1

X

W

B

PR

reaction is equal and opposite to

the pressure produced by the earth on the wall, and its direction is perpendicular to AX: also, since the surface X Y

* See Mr. Moseley's Mechanical Principles of Engineering, p. 441.

will not exert a greater pressure than is just necessary to support AXY, the direction of R1 must be inclined to the normal to xy at an angle equal to ; also, the directions of R and R1 must pass through the point p, in which w's direction cuts xy, so that Nx will equal of AX; moreover,

R: W :: Sin R1PW : Sin R1PR :: Sin (0-4) : cos (0-4)

... R=w tan (0-4)= w x2 cotan 0 tan (0-4) Now, according as o has different values R will have different values, and if we determine the value of 6 for which Ris greatest, the wall cannot be called on to supply a greater reaction, and this must therefore equal the pressure which AX actually sustains. But

cot Otan (0-4)=

cos e sin (0-4)_sin (20-)-sin sin e cos (0-4) sin (2014) + sin

1

2 sin

sin (20-4) + sin

which is manifestly greatest when the fractional part of the expression is least, i. e. when 20-6 equals, so that the

Φ required value of is+ is + 42'

value of the pressure is

and, therefore, the required

x2

ww2 cotan(+) tan (-) = ww2 tan2(-)

4

W

acting through a point n which is below a by a distance equal tox; but this is the same as the pressure that would be produced by a fluid each cubic foot of which

Ex. 459. A mass of earth the specific gravity of which is 1.7, whose surface is horizontal, presses against a revêtement wall whose top is on the level of the ground and height 20 ft., the natural slope of the earth being 45°; determine the pressure of the earth on each foot of the length of the wall. Ans. 3646 lbs.

Ex. 460. If the wall in the last Example is of brickwork and has a rectangular section, determine its thickness to enable it to sustain the pressure of the earth. Ans. 4.65 ft.

Ex. 461. The vertical face of a revêtement wall of brickwork sustains the pressure of 20 ft. of earth, the surface of which is horizontal and 2 ft. below the summit of the wall; the thickness of the wall at top is 1 ft.: what must be its thickness at bottom if it just sustains the earth, the specific gravity of the earth being 2 and its natural slope 45°? Also determine the thickness that would enable the wall to sustain the pressure if the earth were thoroughly permeated with water.*

Ans. (1) 5.47 ft. (2) 9.6 ft.

Ex. 462. If a pressure P is applied against a wall supported on the opposite side by earth with its surface horizontal; show that when p is on the point of causing the earth to yield, the resistance of the earth is the same as that of a fluid the weight of a cubic foot of which equals (weight of cubic foot of earth)

X

tan2

Φ + 4

2

[The reasoning in this case is step by step the same as that given in Prop. 20, except that now the wedge of earth is on the point of being forced up, so that the direction of R1 will be on the other side of the perpendicular

to x Y.

Ex. 463. The wall of a reservoir of brickwork is 4 ft. thick and 15 ft. above the surface of the ground; the foundations are 15 ft. deep; the natural slope of the earth is 45° and it weighs 100 lbs. per cubic foot; when the reservoir is full (so that the water presses against the whole 30 ft. of wall), will the wall stand, supposing the adhesion of the cement perfect?

B

C

Ans. Yes; excess of the moment of the greatest pressure that could support the wall over that of the pressure of the water 73,480.