38. From equations 6 and 7, and the same values as above of H=210; h=10; T=10; g=2.4; b=0; B 174.84, there results tan .4919 and R=54717.6, intersecting base 87.10 feet from toe, resulting in 10.55 tons per square foot maximum oblique compression at the toe if upthrust should not occur, and if shear (= to force 5) be uniformly distributed; or 12.71 tons if shear increases uniformly from zero at the heel.

39. It is needless to deduce equations for obliquity and magnitude of maximum pressure at the heel, for it must be nearly vertical, and its amount can but little exceed the weight of a column of masonry a foot square and as high as the dam, or 210 × 150 ÷ 2000=15.75 tons per square foot.

40. It appears then that in either event the pressures are all safe, being well within the 20 tons which is the usually accepted limit in modern practice.

41. It also appears that in the event of upthrust of the assumed amount and distribution, making tan → = .8395, something more than the friction of plane surfaces is required to prevent sliding, and reliance must be in part on cohesion of masonry, and on anchorage in foundations and abutments. At the San Carlos Dam site the downstream convergency of the abutments would alone insure against sliding.

42. Let it be determined what would be the effect on the proposed dam of 184.84 feet base if upthrust instead of being full at the heel, diminishing uniformly to zero at the toe, should continue uniformly full to the toe. Force 6 on page 8 would then become (H+h) (B+T+b) with lever arm of (B+T+b); and the resultant of all vertical forces would be forces 1+2+3+4-6; and its lever arm would be

Moments 1+2+3+4−6
Forces 1+2+3+46°

Assuming H=210; h=10; T=10; g=24; b=0; B=174.84; the above vertical resultant becomes

2.4 X 174.84 x 210

2

+2.4 x 10 x 210 (210 + 10) (174.84 + 10 + 0) =

8434.88 units of 62 pounds; and the lever arm becomes.

2.4 X 174.84 × 210 /2 174.84)

2

(3174.84)

+(2.4 × 10X210) (174.84+5) (210+

10) (174.84 +10) (174.84+10)÷8434.88=270.75 feet.

Force 5-24150, with lever arm of 73.04 feet.

8434.88 24150 :: 73.04: (209.12), which is 270.75 -209.12 61.63 from toe; which is the one-third point as before.

But now tan = 2.863, and friction must receive great help from anchorages, or cohesion, or arch action, or downstream convergency of abutments, to prevent sliding. The oblique compression at the toe is now 8.64 tons per square foot.

The moment of stability, if the dam were straight, is now but 9.4 per cent in excess of the overturning moment, but the resultant falls on the one-third point, and the dam will stand if sliding be prevented.

43. Let it be found how much, if any, tension per square foot would exist across the bottom of a vertical plane parallel to the length of the dam and situated at a distance d from the toe, assuming full upthrust at the heel, diminishing uniformly to zero at the toe, and that H=210; h=10; T=10; g=2.4; b=0; B 174.84; which assumptions, as heretofore shown, make resultant intersect at the limit of the middle third of the base. In this case, assuming absolute rigidity of both dam and foundation, the vertical reaction of the foundation would increase uniformly from zero at the heel to 9.729 tons per square foot at the toe.

44. Now, considering the dam as a bracket or cantilever bearing the reaction as a load, increasing from zero at the heel to 9.729 tons per linear foot at the toe, and neglecting for the time the horizontal shear along the foundation plane-a vertical shear as well as a moment of rupture resulting in tension occurs at vertical planes throughout a considerable distance upstream from the toe. Referring to the diagram below and the nomenclature thereon, the vertical shear along any plane A B situated a distance d upstream from the toe, equals in amount the difference between that part of the total reaction represented by the trapezoid A C D E acting upward and that portion of the weight of the dam represented by triangle A B C acting downward, or

which reduced gives 9.729d-.0713d = total shear in tons, and

6

5

dividing by d gives (9.729.0713d) = shear per square feet in tons. d=136.45 feet.

To make shear = 0

Therefore shear per square feet on vertical planes, if uniformly distributed over such planes, diminishes uniformly from 8.11 tons at the toe to zero at 136.45 feet upstream from toe.

45. Regarding now the dam as a cantilever beam, the moment of rupture at section A B situated at a distance d upstream from the toe, is equal to the difference of moments of trapezoid ACDE of reactions, and of mass of triangle A B C of the dam, with respect to a point in line A B.

Making P=9.73 tons, and T+B=184.84, say 185.

555-d

Moment of rupture = 9.73d2
-9.73d (551) - .015d foot-tons.

1110

Eq. 9.

To determine moment of resistance of beam at cross-section A B, assume that modulus of elasticity in cyclopean masonry is the same in tension and compression, which will locate the neutral

1 6

axis at ✈ of the depth of the beam above its bottom, or of d=.6d above point A. Now letting t the tension at the lowest point A, the moment of resistance of the beam on its tension side alone

will equal 7.6d

Xt.

1

"Sxxdx = (60) and on the

.6d'

same, making

3

(.6d)t and on the compression side the

Eq. 10.

Moment of resistance = .24d't

Placing this equal to the amount of rupture given above:

.24dt = 9.73d (555) — .015а3

1110

= 20.27 − .099d very nearly.

The above is the tension in tons per square feet in "farthest fiber" or at point A. The total tension across joint A B is the sum of the progression diminishing uniformly from the above maximum at A to zero at the neutral axis, which is of A B above A or at of 6/5 d.6d, making total tension = maximum tension X.3d. It is 2/3 of this latter amount which must be overcome by horizontal shear along the base between the points A and C, if tension is to be prevented.

46. The above formulæ give for assigned values of d, vertical shear, maximum tensions, total tensions, etc., as follows:

47. Now the total horizontal shear along the base is the water pressure against the upstream face of the dam, which is

If this were uniformly distributed over the base it would amount to 754.69÷185=4.08 tons per foot, which is entered in the last column of the table above. Comparison of the last two columns indicates that, under the above assumption of uniform distribution of horizontal shear, no tension would exist at lower edge of any vertical longitudinal section. But to guard against the possibility of tension to result, should the horizontal shear toward the toe fall short of a uniform distribution, as well as to reduce the larger vertical shears shown in column two, some thickening is desirable at the toe, which would be fully furnished by the reversed curve required for overfall, as indicated on the diagram on page 8, by making that curve begin as much as 35 or 40 feet back from the toe.

48. The distribution of horizontal shear at base of dam, though assumed in the above calculation to be uniform, is not necessarily so. Different authors have assumed various laws of distributionparabolic, rectangular, trapezoidal, triangular, and compounds of parabolic and triangular. But in reality the distribution at the base of any given dam is dependent upon special conditions obtaining at that base, such as relative compressibility of masonry and foundation; presence or absence of friction; presence or absence of adhesion; and presence and location of anchorages. For example: (A) If the masonry and foundation rock were both nonyielding, and if their junction were a horizontal plane with only friction to prevent sliding, then with resultant intersecting at the downstream one-third point, making the normal pressures between dam and foundation increase uniformly from zero at the heel to twice the mean at the toe, the friction, which opposes the horizontal shear, being proportional to load or pressure, would likewise increase uniformly from zero at the heel to twice the average shear at the toe. The same would still be true if there be vertical yielding in either dam or foundation, or both, provided this yielding or compression be proportional to the load, and provided further that it be limited to only the vertical direction, leaving the dam still rigid as a cantilever beam. But in fact the masonry is compressible horizontally also, whereby the toe of the dam bends up, relieving itself of a portion of the vertical reaction it would otherwise need to withstand, but at the expense of somewhat increasing the normal pressures at and near the downstream one-third point. Thus, just as the diagram of vertical pressures. may depart from the triangular form, so also would vary the diagram of active friction as an opponent of sliding. (B) If sliding be prevented, not by friction, but by adhesion, as of mortar between dam and foundation, or else by a large number of equal and equidistant notches constituting anchorages, then in either case the distribution of horizontal shear will be dependent on relative compressibilities of masonry and foundation rock. If both be equally compressible in the downstream direction, as in certain models in which dam and foundation were made of one piece of yielding material, the shear

would be distributed uniformly. The same would be true if both were absolutely noncompressible or unyielding. If the material of the dam were compressible and that of the base noncompressible, then the shear would be a maximum at the heel, diminishing uniformly toward the toe, making its diagram a trapezoid; and an inclined fracture due to tension would be likely to occur near the heel. But if the foundation be compressible and the dam noncompressible, the shear would increase uniformly from heel to toe, making its diagram a trapezoid with its wider end downstream. (C) If there be a single large anchorage at or near the heel, as is often employed for the additional purpose of intercepting water, then the shear would be largely concentrated at that point, a condition tending strongly to oblique rupture near the heel. But if the single large anchorage be at the toe instead, without the intercepting wall, if any, at the heel, being an integral part of the dam, the resulting partial concentration of shear at the toe, if not in excess of the shearing strength, will be beneficial in insuring against tension, which, in the absence of horizontal shear at and near the toe, must exist across the lower parts of vertical planes in that vicinity.

49. In the proposed San Carlos Dam on stratified rock dipping some 30° downstream, its junction with the foundation would naturally be in a succession of steps or notches throughout the distance. from heel to toe, a condition which, if taken alone, is, as stated above, conducive to a uniform distribution of horizontal shear. The foundation rock, however, being of varying degrees of hardness from that of a grindstone to hard quartzite, though containing many incipient seams in various directions is, as a whole, less compressible than ordinary dam concrete, which, as also stated above, would tend toward making the shear next the base decrease somewhat from heel to toe. But it is known that concrete curing under constant moisture, as would be the case at the bottom of the dam, is inclined to expand a little, which, on account of the confinement due to the interlocking with the steps in the foundation, would result in an initial compressive stress in the concrete, making it bear hardest against the downstream faces of notches near the toe and the upstream faces of notches near the heel, which in turn would result in transmitting a greater proportion of the horizontal pressure of water to be borne by the more distant notches, making the shear greater toward the toe; and this may offset the reverse effect of the greater rigidity of bedrock and leave the shear not greater at the heel than at the toe. In any event it is possible in construction to preordain a greater shear toward the toe by providing that the principal anchorages shall be in that vicinity, and to that end the natural notches in the bedrock at and near the heel would be obliterated by filling them with concrete, which, together with top of intercepting wall, if any, would be first finished either level or with slight downstream slopes. Such procedure would insure against the oblique fracture at the heel and would provide near the toe sufficient shearing tendency to prevent any tension across vertical planes, for, as shown on page 14, not more than the average of the shear throughout the base is required for such purpose.

50. In theory, analysis and calculation can be carried to greater refinement, but the physical data do not warrant it. Some parts of the foundation may be more yielding than others, rendering even