Adding H2. B2 to both sides of the equation, and then

H2. C2. w

taking the square roots, ť2 + 2 H. t. B + H.2 B2 = W

If 2 be taken as the factor of safety, then_t=H

12. C2. w B2
3. W

+ - H. B.

3

Taking the latter formula,

let it be applied to a wall 10 feet high, the weight of the wall per cubic foot being equal to that of the earth, which is to be taken as compact earth, the batter 1 in 8,

thick at the top; adding on the batter th of the height,

the thickness at the ground level becomes 1.8 +

10

= 3.05 8

feet. This formula is commonly used where the ground is not treacherous, but of course the ratios of w to W are filled in when they are not the same. Thus a masonry wall will weigh 130 lbs. per foot, and rammed earth 100 lbs. per cubic foot, which would reduce the above thickness at the top to 1.45 feet.

The foregoing investigations refer to retaining walls carrying banks having a horizontal top surface, and it is interesting to notice that the formula for moment of horizontal thrust will apply to water. The formula arrived at for water was M = 10·416 D3, where D = H: that for the

water has no angle of repose, C vanishes; w = 62.5; hence

M=

H3 x 62.5
6

expression.

10.416 H3, the same as the previous

When the wall a b d c is loaded, as shown in Fig. 81, it is said to be surcharged, and in this case not only is there a greater load upon the wall, but the centre of pressure acts higher up than one

third its height: in this case it is simplest to determine the horizontal thrust by the graphic method, and then equate its moment with the moment of resistance of the wall.

Sometimes retaining walls are made with counterforts in front, as shown in the plan A B, where C Care the counterforts; these have the

effect of increasing the

leverage of the wall, and therefore its moment of resistance, as it must to upset turn over on the edges hh, instead of kk, provided the masonry does not give way at the joint of the counterforts with the wall. In order to strengthen the wall between the counterforts, it is sometimes made arched in plan, as shown by the dotted lines; this will materially assist in preventing it from bulging between the counterforts, and in fact such walls should always be so built.

Counterforts behind the wall are not nearly of so much use, as they do not increase the moment of resistance of the other material composing it, though because their own centres of gravity are carried back further from the edge, they are in themselves of more value than the same quan

tity of material distributed equally on the back of the wall.

Walls requiring a great resistance may sometimes be made hollow, but care must be taken that the weight of material is sufficient to resist the horizontal thrust acting to slide the wall on its base, or these hollow places-pockets or voids, as they are termed-should not be carried completely down through the wall, but left with a sufficient bottom for them to be filled up with earth or concrete, preferably the latter, which by its weight will add to the stability of the structure.

If a retaining wall carry a superposed weight, its stability will be proportionately increased, and this is the case in retaining walls carrying bridges, and also in the abutments of arched bridges. It is a common practice to call the supports of bridges abutments, but this is inaccurate except for arches and certain trusses that abut on and thrust against their supports; in the case of girders the bridge merely puts a weight or vertical pressure on the supports. I therefore avoid applying the term abutments to these supports, piers being a more suitable expression. In arched and trussed bridges the piers are intermediate supports, on which there is, under the full load, only vertical weight, the thrusts from the arches on either side balancing each other.

Retaining walls, when calculated, invariably look heavy both on paper and on the ground, and there is therefore an inclination to cut them down in size: no doubt they form a very heavy class of work compared with structures of strength, and it is necessary they should be so, for in making them we are opposing weight to weight, and the resisting mass must bear some proportion to that endeavouring to overthrow it. I call attention to this point especially to caution students against this inclination to diminish the section, for that which is calculated will be

practically requisite, and in no case would I let the factor of safety be less than 2. There have been many instances of retaining walls falling down, and more of others which, although they have not actually fallen, have bulged and shown signs of weakness, requiring to be cobbled and patched up either temporarily or permanently, so far as anything connected with such a structure can be said to be permanent. The way in which the temporary support saves the work is by holding it up until the earth behind it, which had been disturbed by building operations, becomes settled down and hardened again, so as to be more in a position to stand by itself than it was when the wall was first completed.

For places where material is of no consequence, dry retaining walls are frequently built, having only the coping in mortar. They must be much heavier than mortared walls, and although good enough for small heights, are not to be recommended for large or heavily loaded walls.

CHAPTER XVI.

ARCHES-ABUTMENTS—BUTTRESSES.

CONDITIONS Similar to those attending the stability of a retaining wall are imposed in the case of an arch: no part of the arch must turn upon one or other of its edges, and the joints must make a proper angle with the line of thrust, which throughout the length of the arch should lie in the middle third of the depth or thickness of the arch.

As has been shown in a previous part of this work, the tangential force due to a radial force = w × r, where w is the radial force per lineal foot, and r the radius in feet of the element on which it acts, at the point of its action; and this formula will give the horizontal thrust at the crown of an arch, carrying at that point a load w per lineal foot, and having a radius r at the crown : let p = wr.

In Fig. 82 let the arrow P represent the direction of the force p acting at the crown of the arch. Through the centre of gravity of each arch stone, or voussoir, and its accompanying load, draw a vertical line, as at ef, gh. This for each voussoir will pass through the centre of gravity of a mass comprising the voussoir itself, spandrels and internal bearing walls, arches or flags upon the voussoir, also superposed pavement, ballast, and other load.

Produce the direction of the arrow P to intersect eƒ at a, and make a b=p; mark off a d equal to the load on the voussoir D, complete the parallelogram a bed, ac will be