21. We can therefore calculate for a straight dam with full confidence that curving the dam will not only introduce the arch action ever ready to assist gravity, but will by virtue of the curvature alone increase the efficiency of the gravity section and make the dam, though severed from the abutments, harder to overturn. 22. What must be the top width of a trapezoidal dam section to accommodate parabolic curve of discharge for water h feet above its crest? Mean velocity = }√2gh, y = }√√2ght, and x = 4gť, whence stream slope of dam will be in the neighborhood of 6 on 5, let 23. Now, what shall the depth of overflow, or h, be assumed to be? It, of course, depends on length of overfall and quantity of water to be discharged. The quantity can not be definitely known; we can only judge by the past. By far the greatest flood ever known was that of 1905. It has been stated by an engineer that high-water marks left by that flood between Florence and the Buttes indicated a maximum flow there of 190,000 cubic feet per second. The writer has found driftwood lodged 26 feet above low water in the river several hundred feet above the dam site, which would mean a flow at the time of about 150,000 cubic feet per second, which is not out of harmony with the reported flow of 190,000 at Florence, for the San Pedro is said to have been in flood at the same time. But it can be argued that the chance for a repetition of such flood is small and that the chance of its occurring on a full reservoir is much smaller, and that the chance of its continuing so long that the storage room even on top of a full reservoir would not reduce the maximum flow at the dam would be very much smaller still, making of the combination a chance so remote that it would not be necessary to provide against so great an overfall as 150,000 cubic feet per second. The writer would not expect 50,000 cubic feet per second to ever pass the dam, and were only the cost of dam subject to risk, he would consider the question of insurance and would not expend much money toward providing against an overfall greater than 50,000 cubic feet per second. But in a case like this, where a failure of the dam would be so appalling in loss of life and in loss of property many times the cost of dam, the engineer is not warranted in assuming less than 150,000 cubic feet per second as a possible maximum flow. 24. To pass 150,000 cubic feet per second would require a spillway 500 feet long with water nearly 20 feet above its crest, or else 300 feet of spillway with 20 feet of water together with 500 feet of dam with 10 feet of water over it. 25. The board having decided on a dam storing to an elevation of 180 feet, and having proposed spillways past its ends with crests at 180 feet on which movable gates may be added at some future time, and having suggested that the dam be 10 feet higher, or to elevation 190, but with top and downstream face of form suitable to pass possible floods in excess of the capacity of the spillways-it is assumed as a limit for purpose of calculation, a flow of 150,000 cubic feet second, of which about 95,000 would be passing over spillways 300 feet long and 55,000 over 500 feet of dam, requiring a head of about 20 feet above the spillways and 10 feet above the dam, or water to an elevation of 200 feet. per 26. While there would be room on the dam and in the excavations at its ends for more than 500 feet of overfall and 300 feet of spillways, yet there would be towers and, in time, possible bridge piers also, on the dam and its approaches to obstruct the flow. 27. From the above considerations it will be assumed in the calculations of the dam that water may rise 10 feet above its crest, making the h on page 6 (par. 22) equal to 10 feet. Hence, necessary top width to accommodate the parabolic curve suitable for a head of 10 feet is 8/15X10=5 feet. But to the 5 feet should be added 4 feet or more to provide for chamfer of upstream edge, making 10 feet in all. If bridge piers on the dam be needed, they could spring from an elevation 12 feet lower, where the width is 10 feet greater, making 20 feet in all. A narrower width at and near the top would make an arch that would deflect so much that it would not be of help to the gravity section. 28. It may be argued with reason, that inasmuch as a height of water 10 feet above the dam is so improbable, it would be regarded sufficiently safe to make a cross-section which, with water only 5 feet or even less above the crest, would give a resultant intersecting the base at the one-third point, and let the resultant, if the water should go higher, fall a little outside the middle third; for the dam would not fail until the resultant had reached nearly to the toe, unless the masonry or rock foundation should fail through crushing, or unless the cohesion of masonry be insufficient to withstand the tension that would then occur in the upstream face, and along the base toward the toe. But it is better to make a design absolutely safe and to assume water 10 feet above the crest or 200 feet above datum, or 220 feet above the foundation, which will be done in the numerical work to follow. 29. In the figure on page 8 the single equivalent of all the vertical forces is in magnitude equal to Forces (1+2+3+4−6), and its point of application is distant horizontally from the toe of the dam= Moments (1+2+3+4−6) Forces (1+2+3+4-6) The above equivalent, when combined with force 5, will have an oblique resultant which, under the conditions imposed, is to pass through the one-third point next the toe. The triangle of forces being similar to the triangle formed with the intercept of the base by the vertical and oblique force lines, prolonged if necessary, there results the following proportion: Forces (1+2+3+4-6): Force 5: H (H+3h) from which, after reduction, the following general equation results: 30. In the same figure, page 8, when reservoir is empty and wind is blowing upstream, we have again an oblique resultant, which by the conditions imposed is to pass through the one-third point next the heel. Here the equivalent of the vertical forces is in magnitude = Forces (1+2+3), and its point of application is horizontally distant from the heel Moments (1+2+3) the lever arms in this case being the whole base (B+T+b), less the lever arms given on page 8. This equivalent, combined with force 7, gives the required resultant; and from similar triangles, as before, the following proportion obtains: 31. The above equations, 1 and 2, are perfectly general, each containing the two unknown quantities B and b; and the two are simultaneous for any given values of g, H, h, T, and W. The two equations can be, of course, solved for values of B and b in the general terms of g, H, h, T, and W, but it is simpler to first introduce the assigned numerical values of g, II, h, T, and W into both equations and solve for B and b afterwards. 32. Eliminating B from equations 1 and 2, there results. 2h (1-3) T2] + [[3 W (1 − 3) + (II +3h) } II +2 (2—1 +3 −g): (II+3h)}H+2( +3−9) T2 4h 4 H T] + [(1 - 3H ·(H+h) W T H g + 841 'b=0 Eq. 3 T Assigning numerical values: H=210; h=10; T=10; g=2.4; and W=.8 (which is 0.8 of 6250 pounds per square foot); and solving, gives b=.5367 feet, and B=174.42 feet, making the whole base= 0.5367 +10 +174.42 184.96 feet. Had W been 8/21 (or 23.8 pounds. per square foot) b would have become zero. In view of the fact that wind pressure would be omitted by the board, and that it usually is omitted in dam design, in the following discussion b will be made zero (which provides for about 24 pounds wind pressure) unless batter shall be required toward the base to limit stress in masonry. 33. To determine necessary batter for different heights: Assume H=210; h=10; T= 10; g=2.4; b=0. Whence by Eq. 1 284 B2+10,720 B+28,400-1,058,400,000=0, from which B=174.84, and whole base = 10 +174.84184.84. Assume H=200; h=10; T=10; g=2.4; b=0. 27 B2+1,020 B+2,700-920,000=0. B=166.40, and whole base = 10+166.40 176.40. Assume H=150; h=10; T=10; g=2.4; b=0. 200 B2+7,600 B+20,000 4,050,000=0. B = 124.22 and whole base = 10 +124.22 134.22. = Assume H=100; h=10; T=10; g=2.4; b=0. 130 B2+5,000 B+13,000 - 1,300,000=0. B=82.11, and whole base = 10 +82.11 = 92.11. Assume H=50; h=10; T=10; g=2.4; b=0. 60 B2+2,400 B+6,000-200,000=0. B=40.28, and whole base = 10 +40.28 50.28. With uniform batter for the 210 feet we have 210 of 174.84 166.52 which is 0.12' greater than the 166.40; = of 174.84 124.89 which is 0.67' greater than the = 200 for 200'B of 174.84 = 83.26 which is 1.15' greater than the 82.11; 210 for 50'B= of 174.84 = 41.63 which is 1.35' greater than the 40.28. 50 210 34. It appears therefore that a straight batter will not result in much excess strength at any height in this case, which assumes similar upthrust at every height within the masonry. It is impossible that masonry would be so poor as to permit so much upthrust within itself, but it must be remembered that each height has its ends on the sloping walls of the canyon where upthrust is just as likely to occur as on the very lowest part, hence requiring that the ends, if not the center, of each zone in height shall be of cross sections as calculated above. To reduce only the central portions of each zone, leaving its ends of larger size, would introduce a complexity impracticable to follow in dam construction. 35. But we must examine whether the straight batter figured above as 174.84 in 210 (practically 6 on 5), will result in excessive compression in either the masonry or the rock foundation, and whether the direction of the resultant is within the angle of friction. 36. Referring to figure on page 8 R = Force 5÷sin 0 or √(Forces 1+2+3+4 −6)2 + (force 5)2 Reducing gives: = Eq. 5. Assume H=210; h=10; T=10; g=2.4; b= ; B=174.84; and there results tan 0=0.8295 and R 37562.7; which latter, since in the formula water-unity, must be multiplied by 62, making 2347669 lbs. 1173.835 tons. Combining vertical component and uniform shear gives 10.55 tons per square foot for maximum oblique compression at the toe, or 12.71 tons if shear increases uniformly from 0 at the heel. = 37. The above includes upthrust which operates to increase tan and to diminish R. If through good fortune the upthrust should not prevail, then R = Force 5÷sin 0 or √(Forces 1 + 2 + 3 + 4)2 + (force 5)2 |