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If these forces-the weight of the dam and the pressure of the waterwere just equal, the dam would yet remain in place against sliding, because of the friction of the masonry on its base, which is dependent on its weight and the surface of contact at CD, and has an established value for stone masonry resting on a rock bed of six-tenths of 1 per cent., which is practically equivalent to increasing the weight of the masonry by that proportion. If a good quality of cement mortar be used in bedding the masonry on its foundation, the shearing resistance of the mortar may, properly, be considered.

OVERTURNING.

The probability of the dam sliding on its base is not the only manner of failure to be considered. A more important problem is to so proportion the dimensions and profile that it may not be turned over down stream by the pressure of the water in the reservoir.

Referring to the last figure (2) we find the forces acting on the dam to be its weight, represented by its center of gravity G, acting in the vertical line G V, and the water pressure F acting in the horizontal line FP, at two-thirds the depth of the water. If from the intersection of the lines of action of these forces at I there be laid off, by scale, the proportionate values of the forces in the direction in which they act, we shall have the lines I W and I P, respectively, each scaled to represent 312,500 units (or 31), the weight and water pressure being equal. If from W and P the lines W R and P R be drawn parallel, respectively, to IP and I W, there results a (square) parallelogram I PW R, called in ongineering language the parallelogram of forces, in which the diagonal IR is the "resultant" of the forces, representing the direction and value of the combined action of the two forces I W and IP. This value is readily found in this case by the well-known geometrical relation that in a right-angled triangle the square of the hypothenuse (or diagonal) is equal to the sum of the squares of the sides.

This resultant represents two principles, the first that if its equivalent force were applied to the masonry, and acting in that line, it would cause the masonry to slide on its base, disregarding friction of the base or the resistance of the mortar; the second, that as the line of the resultant passes out of, or cuts the down-stream face of the dam at a point above the down-stream toe D, its action is to overturn the dam by turning it around the point D. To resist and prevent such an overturning it has been demonstrated that the width of the masonry on its base C D must be enough to bring the resultant IR to cut the base CD inside of the point D.

To determine this width the moment of the forces W and F must be considered. If the masonry, considered as a column, were supported on a point at V (referring again to Fig. 2), vertically under the center of gravity G, the downward action would be equal to its weight, and a small force would hold it in equilibrium (upright) in the absence of the water pressure, but if the column were supported at the point D (as shown), the downward action would be much greater, so that a great force acting somewhere on the face A C would be required to hold the column in a vertical position. The force of the downward action of the weight W is found, or represented, by multiplying the weight by the perpendicular distance from the line of action of the force, or weight, to the assumed point of support; in this case this distance is VD equal half the width of the base.

PROBLEMS INVOLVED IN THE SWEETWATER DAM. 365

This product is said to be the moment of the weight, or force, W about the point D, minus in this case 312,500 × 10.41+ = 3,253,125 foot-pounds moment. The force that, acting against the face A C, would hold the column in equilibrium, must have an equal moment about D, to be found by multiplying the force by its lever arm with reference to the same point D minus the perpendicular distance from the line of action F P of the force F to the point D, which will be ED. In the case of the water-pressure force applied at one-third the height h, F C = ED=33} feet, the moment is therefore (312,500×333) 10,416,666 foot-pounds-31 times the moment of the masonry weight, so that about one third of the water-pressure force would support the column in an upright position, and any greater force would certainly overturn it on the toe D.

The moment of the force F (water pressure) about the toe D being 3 the moment of the weight about the same point, proves conclusively that the column, or dam, would be overturned about D. To find the width of base C H, such that the column shall be stable against overturning, the moment of the weight must equal the moment of the water pressure (10,416,666 foot-pounds). If the base to be found be represented by x, the area (and cubic feet) of the column will be 100 multiplied by x (100x) and its weight that product by 150 pounds to the cubic foot (100x × 150-15,000x), and this multiplied by the lever arm

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must be the moment to equal 10,416,666 foot-pounds. To solve this, to find the value of x, we must divide 10,416,666 by 7,500 = 1,388.8888+ for the value of x2, and the square root of this-37.2678+=x, and x = 18.6339+. Substituting these values for x, and x, in the last computations the results will be for the area and cubic feet of the profile (100x37.2678+)3,726.78 feet, for the weight (3,726.78 by 1.50) 559,017+, and for the moment (559,017 × 18.6339+) 10,416,667 foot-pounds, proving that if the base C H be taken as 37.27 feet the moment of the weight will equal the moment of the water pressure. If now a new parallelogram of the forces be constructed (the center of gravity of the column having transferred to O, the center of AM H C, from K the intersection of the forces), we have K P W R, and the direction of the resultant K R cutting the toe H, indicating stability.

If the last found base be increased by its one-half to T = 55.90 feet, to secure stability, the weight of the masonry will be increased one-half, or to 838,525 pounds. One-half this weight is in excess of the weight found to resist sliding in respect of the water pressure (312,500 pounds), an excess of more than one-third (419,262-312,500=106,762), which would secure stability against sliding, so that if the profile of the dam should be divided in half by a diagonal line from A to T, thus dividing the masonry into equal parts, we yet have weight enough to prevent sliding, and finding the center of gravity of the triangle ACT, at G, at one-third the height and base from C, and constructing a new parallelogram of forces G P W R, Fig. 3, we will find that the resultant cuts the base inside the toe T, indicating stability with referenc to overturning about the toe T. Now, as there is no material press of water at the top A of the dam, indicating very little thicknes masonry there, and finding that width of base is the requirement

stability against overturning, we are led to the conclusion that the the oretic profile of a dam of stable equilibrium is that of a right-angled triangle, with the water resting against the perpendicular side.

Argument is not necessary to show that this form will bring the quantity of masonry to the least practicable weight.

Considering this triangle form of profile with reference to the equality of moments of the forces about the toe T (Fig. 3), we shall have the

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center of gravity of the dam (G) at one-third of its height above the base, and vertically above V, one-third of the width of the base C D, from E, and two-thirds of the base from D, this last distance being the lever arm for the weight W. Calling, again, the height of the

PROBLEMS INVOLVED IN THE SWEETWATER DAM 367

dam and depth of water h (100 feet), and its base x equal C D, the area and cubic of the profile will be ha its weight ha by 150

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300hx2 plication we get = 50 hx2 for the moment of the weight about 6 the toe D. Putting this equal to the moment of the water pressure, as before, we have, substituting 100 for h, 5,000 x2 = 10,416,666, and dividing x2 10,416,666 by 5,000 we have for the value of x2 2,083.33, and extracting the square root of this number we find for the value of x (the base C D of the dam) 45.6435+ feet, which value will, if substituted for x in the last equation, make the moments of the two forces acting about D equal, 100 × 45.6435+ for the weight of the profile of the dam will be × 150 2 342,326.25 pounds, which multiplied by its lever arm, two-thirds of 45.6435+ (30.429+) gives 10,416,666 foot-pounds, equal to the moment of the water pressure.

The weight of the masonry thus found (342,326.25 pounds) is nearly 10 per cent. in excess of the weight determined to guard against sliding on the base (312,500 pounds), so that the profile just determined is heavy enough to prevent failure by sliding.

As in the case of guarding against sliding, so in this case the base of the dam is increased in width to guard against overturning, and the members of the engineering profession have agreed that this increase, called co-efficient of stability, shall be one-half of the base, as above determ ned, making the base in this case 68.465+feet =CT., ouly 13.5 feet more than was found for the final base of the masonry column dam. By this increase of the bottom width of the dam, its profile area and weight are increased one-half, as shown in the triangle C A T, and its center of gravity is now found at Q, at one-third of the height and base from C, as for C A D, and acts in the vertical QY. For a new parallelogram of forces from Q we shall have for the water pressure, as before, 312,500 pounds, and for the weight of the masonry W (100 × 68.46 × 150 ÷ 2) 513,489 pounds. Completing the parallelogram of

NOTE. The algebraic expression of these relations is as follows: When h represents the height of the dam and the contained water, calling the weight of water 1, and of the masonry 2.4 (624×2.4=150)=w, the moment of the water pressure will be expressed by

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forces, we find that the resultant cuts the base somewhat within the point D. Now, as C Y was taken one-third of the base, C T for line of

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center of gravity Q, and D T is onehalf C D, it is one-third the whole base, and equals C Y, and there remains Y T as the middle third of the base and it is cut by the resultant of the acting forces. The result of this disposition of the forces with reference to the profile, is the generally accepted maxim. A masonry dam is stable when the resultant of the forces acting upon it cuts the base within the middle third.

If the upstream face A C of the dam be built with a batir or slope, CE, and the bottom width be kept the same as before, bringing T back to N, there will be no change in the area or weight of the profile. The center of gravity will be transferred upstream two-thirds of C E, and the intersection of the line of the resultant with the base will be changed by an equal distance, but will be nearer the down-stream toe by one-third of C E, reducing the co-efficient of stability by that amount, but at the same time the weight of the water resting vertically over the upstream batir C E may be considered as increasing the weight acting on the base E N, and also of the force W.

FIG. 5-Ban Dam, France.

If there should be such a condition of over-fullness of the reservoir as would cause the water to over

flow to a depth of 10 feet, the pressure F would be increased, and the point of action raised accordingly, reducing the stabil ity of the dam proportionately. To avoid this effect a spill or waste-way is provided, generally at one end of the dam, through which the surplus water is allowed to flow away. The waste-way should have a capacity equal to the greatest known flood flow of the stream to be obstructed to form the reservoir.

The dam may receive shocks from waves, or masses of ice floating on the surface of the water of the reservoir, which would to some extent affect the stability of the narrow top of the triangular dam. To guard against this effect, and to,

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FIG. 6.-Terrasse Dam.

at the same time, provide a convenient road-way from end to end of the dam, the top is given a width, A B, which is added to the down stream

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