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MILLER CANDIDATED FOR THE DEGREE OF BACHELOR OF SCIENCE J'NE l 9 3 O THESIS C C) i): Oi") UTLINE A. INTRODUCTION B. DIRECT FLOW SYSTEE l. Wading Pool a. Estimate of attendance b. Design of wading pool 0. Piping system d. Selection of chlorination asparatus 2. Dam a. Investigation of Stream Flow b. Design of Dam 0. Cost of Dam “‘3. Cost of Direct Flow SyStem O. 'INDIRECT F CW SYS EM 1. Design of System a. Design of piping for recirculation b. Selection of chlorination apparatus 0. Design of Filtration Plant 2. Cost of Indirect Flow System D. DRAIXAGE SYSTEM AND GRADING 1. Design of System 2. Cost of Drainage System and Grading E. OKCLUSIOX. 94103 PREFAC This thesis was chosen, because, first of all, it presented a real problem. A problem that had been pro- posed and was likely to be carried out in the near future. Health an be app iness are the two things which stand foremost in our minds. When it is possible, to change existing conditions, to safeg: uard health an ad estab- liSh a better means for recreation,gshould be considered a great accomplishment. Because of the limited amount of time avail- able it was impossible to go into detail in the design of as the wish of the authors :3 all parts of the system. It ‘ to go into such detail as to ffect a.souno es ti ate of (D costs of different systems. It is difficult to mention all the sources frpm which have cozne help and inspiration for the prepara- . ééé . . a . .. ” tion of th the81s. fie are indebted to hr. Taylor, Engineer in char: e of Pontiac Water Works for inforxation and advice concerning the investi ation of the Clinton River; Professor Theroux at nichi gm nState College for advice on design on wading pool and filtration plant; Professor Cade at hicniian State College for advice concerning design of dam and drainage System; Professor Mallmann at Kichigan state 0011986 for advice concerning analysis and method of purification of the river water. bin t‘t“ rm 0 k INTRODUCTION 5 Dodge Brothers State Park No. 8 is located in Macomb County two miles south of Utica or nine miles northwest of fit. Clemens. This park lies either side of Clinton River and extends along the river a distance of 3,400 feet. It contains 31 acres of ground. Playground equipment has been installed and additional equipment in the way of tables, benches and garbage cans have bee. added. At the present time many children are allowed to bathe in Clinton River at this park. Due to the fact that raw sewage is emptied in Clinton River at Utica a distance of less than two miles upstream, the river is in such a sanitary condition that it is unfit for bathing purpdses. A sample was taken April 19, 1930. From this sample 5 tubes of lactose broth each of which had been inoculated with 10 0.0. of the sample shows from 20 to 55% of gas production in 24 hrs. A count of 13,000 bacteria per 0.0. was obtained from agar plates. The presence of Escherichia coli was confirmed by use of eosin methylene-blue and Endo's media. Typical colonies were fished from Endo's medium and was placed in lactose broth which showed a gas production of from 15 to M5§ in #8 hours. The presence of Escherichia coli, an intestinal bacteria indicates fecal pollution and waters with fecal pollution are dangerous from a standpoint of thB - ”‘- L- Idisease, typhoid fever. Warning can be given to bathers of this water as to its dangerous condition,but until such time that a suitable place for bathing is constructed there is a grave danger of disease from this source. The purpose of this thesis is to compare economically the estimated costs of the supply for a wad— ing pool. The supply will be furnished either by a direct flow or recirculation system. In the direct flow system a dam will be designed to furnish the head. This dam will support a footbridge. The supply in this system will only be chlorinated as a means of purification. In the recirculation system the supply can either be taken from Clinton River or a well which is in operation at present time. The supply in this system will be filtered through a rapid sand filter and treated with chlorine as a means of purification. Pumps will be installed to furnish required head. This thesis also deals with design of a drainage system for Dodge Brothers Park #8. An Estimate of the Cost of an Installation of a Direct Flow Wading Pool. Design Data On the following page a table shows total weekly attendance at Dodge Park #8, also attendance on the Sunday of that w ek and the per cent of a total weekly attendance visiting park on Sunday. Following is a table showing the attendance at Dodge Park #8 in previous years. Table No./ Year No Visiting Park % increase during year 1927 245,700 1928 431,400 43 1929 763,108 u3.5 From this above table no.’ assuminp' a Q #3% increase for year 1930 the total yearly attendance would be 1,000,000 peOple. “4- Table No. Date Total Weekly Sunday % of Total Attendance Attendance Weekly as: ttendance visiting park on Sunday 5/4 150 5/11 3300 2170 66 5/18 10000 8500 85 5/25 5300 2900 53 6/1 #5000 5500 12 6/8 8000 M150 52 6/15 19000 9600 50 6/22 79000 30000 38 6/29 53000 98000 90 7/6 88000 31000 35 7/13 #6000 37000 80 7/20 63000 55000 87 7/27 87000 50000 58 3/3 74000 61500 83 8/10 28000 25000 89 8/17 39000 37000 95 8/24 #0000 36500 89 8/31 29000 28000 96.5 9/7 38000 6000 16 9/1h 4000 3400 85 9/21 2600 2250 87 9/28 1000 650 65 10/5 190 100 53 10/12 , 510 520 90 Total£I63,llO 1,555 5 The average 0 of a total weekly attend- ance visiting park on Sunday is 67.6 with a maximum of 96.5% and a minimum of 12%. We will use 75% as a value for design data. With 75,000 as an estimated value for average maximum weekly attendance for year 1929 we have: 763,110 : 75,000 :: 1,000,000 : X X : 100,000 (approx) average maximum weekly attendance for year 1930. Using 75% as referred to above we have 100,000 x .75 = 75,000 people; the maximum number attending park on Sunday for maximum year. Assuming that 1% of the total number in a maximum day will be children using wading pool we have: 75,000 x .01 : 750 children using pool on a maximum day. From a report of the Joint Committee on Bathing Places of the American Public Health Associa- tion and the Conference of State Sanitary Engineers we have stated that, "at large outdoor pools where a con- siderable pr0portion of the water in shallow water, we may assume that 50 per cent of the non—swimmers would be on a shore. The average space allowance for each non- swimmer in the water is approximately one-half that of the swimmer in deep water. Combining these factors an allowance of 10 square feet per bather should be allowed for this portion of the pool." In consideration of the fact that this is a wading pool and only children using it we will use 5 square feet per child. Then 750 x 5 : 3,750 sq. ft. of water surface required. Maximum depth of wading pool 24" Assume that 500 children will be the maximum number using the pool per hour. Again from the Report of the Joint Committee as before referred to it states in re— gard to the frequency of changing water that "the total number of bathers using a swimming pool during any period of time shall not exceed 20 persons for each 1,000 gallons of clean water added to the pool during that period." Therefore we have 5%% x 1,000 = 25,000 gal,/ hr. 25000 : 416 gal,/ min. (water to be supplied to pool) 00 Also the Report of the Joint Committee states that "the slope of the bottom of any part of a pool where the water is less than 6 feet deep must not be more than 1 foot in each 15 feet: summing up these Specifications for the pool we have: Haximum number of children using pool in one day 750 haximum number using pool per hour 500 Total water surface area to be not less than 3,750 sq. ft. There shall be 416 gal/ min. of fresh water supplied at time of maximum load. Maximum depth of water to be 24". Inlets and outlets to be so arranged that there will be no dead ends, no short circuiting and even distribu- tion. Haximum width to be 40' and pool shall be located in a natural oI-bow which exists in park. Pool shall slope to a central outlet and can be drained for cleaning. Pool shall be lined with a concrete slab 6" in thickness and to be reinforced with eXpanded metal reinforcing. 'With this design data in mind we will design pool as follows: The pool will be semi—circular, 40' wide with a center line onglO' radius. The inner edge a semi-circle with 190' radius and the outer edge a semi-circle with 230' radius. The pool will be 150' long with semi-circular corners of 12.5' radii. The ‘ pool will have a vertical wall around outsice j \ height the top of which is 3" above surface of water. The elevation of the surface of the water in the pool will be 602.40. At one nd there will be three inlets, each 1'x3", discharging from a distribution chamber. From this end of pool the bottom will slope down to a point at elevation o00./0 which is 25' along center line from end of pool. Thence the bottom of the will $10pe to a point 25' from Opposite end which is at elevation 600.40. At this point there will be located a main outlet with an auxiliary outlet at opposite end from inlet near the vertical wall. For details see plate Ho. 4. -C/'- At inlet end of pool there will be located a distribution chamber with three outlets int pool. This is to provide even distribution of inlet water over entire length of in- let end of pool See plate Ifo.4-for details. At a distance of 150' from distribution chamber toward river these will be located the chlorine apparatus inclosed in a small building which will also provide for a Q storage of chlorine cylinders etc. Hear this will be located a small chrmber in which there will be a float valve which will keep the water surface at a constrnt elevation of 602.40. The building will be constructed of waterproof concrete as shown on plate So. 4. The pool thus specified has an area of 5,858 sq. ft of water surface, a maximum depth of 24", width of 40', maximum slope of bottom of 1' in 13% feet and a circulation system such that there will be even distri- bution, no short circuiting or dead ends. The velocity of flow is Q = AV #16 z 50 X V .0185 ft./ sec. or thich is eoual to 1.11 ft./ min. This velocity is low but it is satisfactory. Capacity of tank equals 6,2 8 cu. ft. or 47,085 gallons. This concludes the design of wading pool, chlorination plant and piping system. -9- a..- Stream Flow Investigation of Clinton River In consideration of drawing the supply for the wading pool from the river, and supplying the necessary gate head by means of a dam; it will be necessary to investi the variation of stream flow in the Clinton Fiver. It is hoped to accurately determine the maximum stream flow that will probably occur once in a given number of years. An exhaustive search for stream flow records was made through the publications of the United States Weather Bureau, and the office of the State Highway Depart- ment, but without success. The only exiSting records were taken at the Water Works in Pontiac; but due to the large difference in the characteristics of the drainage area at Pontiac and that at Utica, hr. Taylor, Engineer in charge of the Pontiac Water Works advised us against their use in this inveStigation. hr. Taylor suggested that we ob~ tain the stream flow records for the Huron Fiver at Ann Arbor as the characteristics of the Huron River drainage 0 area and that of the Clinton hiver are very similar. The suggestion of hr.Taylor was followed. Monthly average stream flow of the Huron River at Borton Dam were obtained from the State Board of Health. The records were complete and extended from 1904 to 1928 inc. The records are tabulated on the following page, and are also plotted on a hydrograph. 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I. .rllilI-J‘J _ u . 4 1111'. ‘ . , . . . ”LL LVI‘IFL.LHH?)FF»1I 'l‘... ’0 ‘ ID. ‘I. l.‘ -i bLblL _ ._TL1_- a ,_._ pin 5 5"! ”yr. . ~ W _ . r» . » b ~ ‘ _ ~ 7 . .. . _ , _ I__1. , . . _ , _ . MILuQ'» .L. ..» . » »-¥.LI»H¥PNLo .. .,._,., . ”LLLI u n . , w _ . _ u . _ . . . H __ ~ . - . . ., _ _ N . . . - h ¢ w _ . p . . . ~ . . _ 0 §‘ ' I I ' VA . a .‘A ‘ O o a. o 9 Q I I “A a - . fl . . _ . — O . . .. t . _ .. /, . .. r .5 . . _ .. . .9127 . N2 1.. , , 2 J , 1 . .1 . . 11.}.15 3.1 ./0. .;z4 ., . I. o: b. \J t x .. V. :1... . . u, a) In . I; .. . o .. . u b. ~ .. n J ’ Mr w. o{ ‘I v l..fl( [w 1; IN/a *Iwn.../a JIU,(I- M .~AI/HIIVUHI..|..1L I} WUprflflaW I“ II, P, YOU . ”a“ [\a/ / , . . k . \ .a . I . . . . . . I \l . ‘9. 'A A a r) . \.. \‘J .m\ 3 D. . . .J \1 ,4. \ . - C -1 -P ‘. f‘ L‘ . n .. - _ ‘ , , .. . . . . - .r «K . - . ti- . _ . . . r . . t ‘ 0 ll ! 11 n - w .\. 1. Jim. - I ,_ .. \\. _ I | h \I. . . v .. . , AH. - ‘ I \. A!“ . - . _ Ill... ., . a‘nl m . . Ilium. I r....4.r;n/,..,.:.u. an I) .17) . . J... v I:. . ,3» V ,.‘Iz . ~13-. In the following table, the percentages of the various types of soil as they are measured in each drainage are listed. The areas used in computing these percentages of soil was measured by a planimeter from the soil map published by the Iichigan State Department of Agriculture. The relative perviousness of the various soils are expressed in whet one might term as perviousness indices. After reviewing texts on soils it was decided that the perviousness of the soils were related in about the proportion indicated by the Lerviousness index in the table. s The pervioueness ratio oetween the two .0 as is determined by dividing the sum oi (D drainage ar the moments (the product of the percent of a soil in Clinton area, times the perviousness index, times the ratio of soil in Clinton area to soil in Huron area) by the average of the indices. By way of explanation, it is quite logical to assume that the perviousness ratio should be the average of the individual soil ratios and directly prOportional to the percent of their existence in the Clinton area, and directly proportional to the perviousness index. The results of the following calcula- tions indicate that the Clinton drainage area above Utica is 1.18 times as pervious as the Huron River drainage area above Ann Arbor. Hence, in view of these calculations, the -ih. discharge of the Clinton Fiver in c.f.s. per square mile at Utica is probably 1.00 of the discharge of the Huron River, 0 considering for tLe present that the rainfall upon the two areas is the same. u oflpmh mmmnm20fibnmm La: . om.a u m u vanfi_mMmHm>< ~>p< 0 .III m H mvmm mxmq hvcmw mac. m:.o ma mm M\: madman Hana amm. mm.H m: m.mm M\m manages. mm”. mo.H mm m.em m gmma pso mmha coazm ea HHom mo pnmommm on «pea cepcfiao wmya mmmcflmam mma< mmnnfiwam CH HHom mo capsfiflo. coasm MmecH pcoaou pcmoamm mo ofipmm Mo unmoemm mo pamoamm mmmnmdofl>amm HHom mnm>flm Cowmfifio new noadm msp mo _ mofiumwuwpomnmgo mmm« mmmcflmam -15- The next factor to be considered is the rain- fall over the two drainage areas. From the United States Weather Bureau Report of 1920 (1930 report is not avail- able) the average yearly raingall over the Huron Fiver drainage area from the year 1880 to 1920 is 31.69". This record is complete. From the same report the average rain— fall over the Clinton Hiver Drainare area from the year _. v 1887 to 1920 is 30.01”. The records over this area are fl rather incomplete. For this reason, and because or the fact that there exists very little difference between the two averages, no correction will be made for rainfall in 'estimating the discharge of the Clinton River at Utica- There remains only the simple Operation’ of dividing each monthly discharge as plotted on the Huron River h'drograph by the constant 1.18, and th result will be the probable discharge for the Clinton River at that time, eXpressed in c.f.s per sq. mile. These results are listed on the following table. linton River. -17- Tabulation of Average second per square mile. ‘determined from discharge Apr. Drainage area = 1904 to Oct. 1928 on drainage area. 320 square miles. the probable stream discharge of the monthly discharge in cubic feet per These values are records covering the period from the Huron River and a study of its Year Jan Feb Mar April taxi June July Aug. Sept.Qct. Nov.3ec. 1904 1.198 .452 .322 .146 .156 .210 .281 .183 .180 05 .228 .190 .836 .746 .677 1.212 .446 .333 .291 .287 .510 .546 06 .768 .476 .578 .693 .509 ..361 .165 .174 .092 .156 .305 .501 <17:L098.434 .806 .961 .899 .459 .268 .155 .248 .354 .405 .506 08 .754 1420 2334 .980 .701 .298 .108 .302 .161 .243 .168 .265 09 .338 .921 .741 .553 1262 .544 .196 .128 .128 .140 .390 .453 10 .523 .554 .92 .546 .855 .430 .137 .119 .170 .180 .195 .199 11 .342 .706 .475 .631 .299 .188 .092 .092 .135 .371 .503 .535 12 .281 .251 .935 L870 .688 .278 .129 .309 .394 .566 .856 .522 131.389 .809 L6042.042 1173 .612 .258 .178 .206 .273 .430 .527 14 .409 .448 .739 l£Dl.l517 .402 .436 .176 .461 .394 .338 .338 15 .4091.260 .840 .559 .431 .454-.413 .531 .856 .616 .468 .451 16 L040 .949 lHOSTLSBBTLISS ..893 .388 .186 .158 .242 .288 .357 17 .452 .326 .800 1201 .703 .789 .460 .164 .242 .293 .370 .265 18 .1911.475za705 .746 .465 .189 .114 .086 .157 .168 .283 .571 19 .403 .323 L142 L640 L167 .388 .226 .187 .190 .268 .387 .387 20 .223 .241.1420 .916 .561 .296 .162 .162 .155 .149 .292 .542 21 .439 .280 .912 .895 .497 ..266 .214 .187 .428 .484 .654 .800 22 .470 .565 .9601.891 .776 .306 .164 .137 .176 .197 .218 215 23 .236 .220 .964 .536 .395 .430 .146 .109 .147 .162 196 376 24 .324 .3201.008 .885 .549 .439 .284 .131 .146 147 .137 .196 25 .153 .444 .665 .378 .219 111 .070 .131 .182 .480 .798 .521 26 .418 .5411.4621.897 .500 .283 .138 .212 .480 .459 .515 .468 27 .322 .811 .815 .501 .464 .474 .249 .125 .172 .244 .326 .924 28 .667 .682 .710 .784 .419_.422 .247 .205 .153 .181 -19- From the above table a typical yearly dis- charge curve is plotted which shows the probably discharge of the Clinton River. The ordinates are obtained by find- ing the average for each month in the preceding table. The above data is also arranged into a frequency distribution which shows the number of times each rate of flow would have probably occured during tne period covered by the data. The third column in the frequency distribution shows the number of times t1 (0 p particular flow is equal to or greater than its self. The fourth column shows the percent of time and the fifth and sixth the frequency of occurance of each particular rate of flow. The percent of time and frequency in years is plOtted on probability paper. The resulting probability curve will be used to determine the size of flood to be expected once in a given number of years. 1 1_ 1... . _ . 11 . I 41, 1 v.1? P 1 {J 1 1 . 11 _ 1 11 ... _ 1 1 1 1. .rl- 1. 1 . 1|.. 1 . 1 L 1 1 1 1 1 1 1 1 A 1 1 I . _ 1 1 1 vi‘b'fl v1! )PIQ’IIJhn ' . 1 _ 1 1. n 1 _ 1 1 1 1 1 1 1 1. l 1 _ 1 1 _ 1 114 1 . _ 1 1 1 1 ‘1- It'll d'l'a.‘ll..l|llli. ‘07!!! 1 1 1. 1 1 1 1 1 1 . . . 1 . 1 . . 1 4 11] .I [In 1.11. 0"! .0! slot: 1. 1 11111 1111111111 -11 1‘ IILA ITI-thiv 1.).- l. v .7.»ll?lv IP’ ictuIIui31316111 it. In." \“JI-- (O .4.’I'I" ‘4’!" ‘t‘I‘ "a‘it"i' U.‘I"-ii:!. 4. I. l .1 'l: 'vlll ti»! vE 1» .3: . n 1'1". [finzfzagcmiy ZQASfVZowébn' cfl/.C7£m&un lavek.4541M»mn/3éaa Average Lumber Summatiin of Percent Frequency Lonthly of Occurrences of time Months Years Discharge Occurrenc- c.f.s./ es sq.mi. .100 5 295 99.9 1.001 .0835 .150 23 290 98.1 1.02 .085 .200 42 267 90.4 1.11 .0925 .250 21 225 76.1 1.31 .109 .300 24 204 69.0 1.45 .121 .350 15 180 60.9 1.65 137 .400 15 165 55.7 1.80 .150 .490 21 150 ,0.7 1.97 70* ,00 27 129 43.5 2.30 .192 .550 15 102 34.4 2.91 .242 .600 8 87 29.3 3.42 .285 .650 3 79 26.6 3.76 .313 .700 6 76 25.6 3.91 .326 .750 8 70 23.6 4.24 .353 .800 8 62 20.8 4.80 .401 .850 6 54 18.1 5.52 .460 .900 8 48 16.1 6.21 .518 .950 7 40 13.4 7.46 .621 1.000 5 33 11.0 9.09 .755 1.050 ‘3 28 . 9.32 10.72 .903 1.100 2 25 8.31 12.03 1.003 1.150 1 23 7.6 13.10 1.09 1.200 4 22 ‘ 7.30 13.70 1.14 1.250 2 18 5.94 16.85 1.40 Average tumber of Summation Percent F1e0uency Ionthly Cccurrences of of time Ionths Years Discharges Occurrences 'c.f.s./sq.mi 1.300 2 16 5.26 19.00 1.58 1.350 O 1.400 1 14 4.58 21.85 1.82 1.450 2 13 4.24 23.60 1.97 1.500 2 11 3.56 28.10 2.34 1.550 1 9 2.88 34.85 2.90 1.600 0 1.650 2 8 2.54 39.40 3.28 1.700 0 1.750 0 1.800 0 1.850 1 6 1.86 53.80 4.48 1.900 2 5 1.53 6.54 5.44 1.950 0 2.000 O 2.050 1 3 0.848 118.0 9.83 2.100 O 2.150 0 2.200 6 o 2.250 0 2.300 . 0 2.350 1 2 0.509 96.7 16.3 2.400 0 2.450 0 1231 Averag lumber of Summation Percent Xonthly Cccurr nces of of time Discharges Occurrences c.f.s./sq.mi 2.500 2.550 2.600 2.650 2.700 2.750 31400000 .171 0"" 16: G. fitxksb trunk g {Wmflttwmx i - wkxxk 48 waUXWHK 3.1.4 2. §V§L~Y~UUQ “no \mutfixfin [\‘k . 3.0 N 0 oi '5 Q h F. a- '3 a/W ”Dagnzao’ I’M-19094327} .27qu aJMJag; 1.96,». 1v]. ’0. floral/AU 1101/) a.) $6.005me $46.. in. gown... all? 00.11.3119.-- HM. Him/Mr 20161.; 10 .0216 v. 7.0.. «56.0? \ v. v .flgfliguuo :10 «manly/2W Ill).-- .‘w W .\ S 9%.».1‘“. 022/14 10 4250..» Ufl o. .0 d f o h. c! G" M 5.0 wart- °§\1c» v o y .44. , , b— /EL.;. —__i ‘ . L.‘2_lll-l. :92 7: ¢.¢~I. i. Q)! .‘a . ..I'I .f. If}- Lily 11:19.?! l’il10.v . h.- ‘I‘II v. _ . . , M . w _. _ m _ . ‘ u m rm h _ . u. ._, . . _ a a a p; «___ flw-~ Qfik , Q1, ,- a ....__._ -__._ - _‘ ~- --.__.‘_ -39- Design of Dam The dam, as shown on plate No. 2 will be a crest ' gravity type with epillway and flash boards in the center section. The flash boards and supporting structure will be so arranged as to enable an operator to instantly open the spillway section when such an occasion arises. The flash boards and their supporting structure may also be removed during that part of the year when the park is Then; closed. -The33-need not remain in place,any thing during the winter except thecresc section and the epillway piers. In addition to the above provisions, there exists three controlling factors; the minimum water surface elevation to be maintained in the wading pool and the mini- mum stream flow to be expected over the crest of the dam, the careful consideration of these two factors will deter- mine the correct crest elevation; the third factor, upon which depends the dimensions of the Spillway, is the maxi- mum stream flow. Crest Elevation Water surface elevation of wading pool=602.40 Min. " " " of pond =602.4O (Head loss between pond and wading pool is negligible) The probable minimum monthly flow was selected from the probable average monthly flow table on page (Z Probable minimum flow 0.086 c.f.s. per sq.mi. 01‘ .086 X 320 2 2705 C.f.So -30- As there are two crest dams over which the water will flow, 5 x 27.5 = 3.75 c.f.s. over each dam tetermination of depth on crest frgmgFrancis' formula Q = cl (h } hv) Q = discharge in c.f.s. : 15.75 length of crest in ft : 10' h : head on crest v2 . hv : a; : velocity head C = Coefficient of discharge From curve 77, Hydroelectric Handbook, by Creager and Justin C = 3.95 Velocity head, hv Area as measured from cross-section of stream : 217 sq. ft. Velocity of stream The velocity head may be neglected _ 3/2 13.79 : 3-95 I 10 (h) h - The above results would be correct if there were no leakage around the flash boards. But due to the numerous small Open ngs around the flash boards, there will be considerable leakage. It is quite probable, and for the want of more accurate data, that 20% of the minimum flow ;31_ will leak through the flash boards. Fence 80% of 13.75 c.f.s . 11.30 c.f.s. dis- lcharge over each crest dam. by Francis' formula JL Q *6// ‘ //.00 = 3 75‘): /0€é/% — 4132.'7- .476' A " J-iJ‘XIo) “ /27’)§ Xinimum elevation of water surface 602.40 " head on crest dam .MB 051:09; Let crest elevation at 602.0' Capacity of Soillwey 1st. Trial. The capacity of the spillway depends upon the maximum flow that will probably occur once in 50 years a frequency consistant with a develOpment of this type. From the stream flow probability curve on page 4511, the average monthly flow that will probably occur once in .39 years is 2.7 c.f.s. per sq. mile Or 320 x 2.7 = 864 c.f. s. It is also desired to limit the elevation of the water surface when 8o4 c.f.s. are flowing to 60%.00, as this elevation is only sli: 5htly lower than the adjoin- By Creager & Justin formula (43) for spillways Q : ld'vzgvzfiyhv _ where in d would ecual 2/3 (h l hv) if the conditions were such that the hydraulic v jump would Operate. But in this case the down stream slope throu5h the saillwry is not sufficient to cause the hydraul jump. Hen e there remains the possibility o: determining d from the probable elevation of the water surface below the soillway. From the gage Carve on page the water ‘ surface elevation Should oe 603.5 when the river dischar5e Spa c.f.s. hence h : 004 - 598.5 2 5.5. sown Stream head h: o03.5—598.5 : 5.0' 598.5 = elevation of spillway floor. Also assume shillway width : 26' therefore d : 731-5 1' % I; z .9074 ér ___ 7" I a = 3'08 From C/OJJ -.Se'c//'a)v ,7 7~= 3‘7 2 2,3 F/fl J“ 303 z 57—2 JXJZZ Q -- d/4V72/17+/1,—:7 j 62 = . 907 X51)" x26 Vi¢.y/étr+./éy—v907lfif Q Q .._. , gay X 575x 2‘ Véy,y/.o9/x.§.r+./6¢' 6.15.5. —- ‘7°7X 5‘,J"x2 C X6.J"/ = Xl‘l F/aw over Cresf JOWJ a =' C‘Cz//4%ér) j/‘L /:V'flv 755V} 425’ 53V¢Vfl7"'*'uflmr/717 d: 3'8 ; (1".2/7 3 62 = 191-?” ,r/o/z/érj A: 3:8x.6’/7X/0x3./:- 74.3 C‘.£J. Z)’ 76:? = /7,?. 6 auel’ 6/(5/ /om.5‘ )s‘l » 7am pr/wa, /o #3 ‘4'},‘5. 7‘4/4/ flycla/der -33.. The above calculations were based upon the assump- d tion that sporoximately sea c.f.s. were flowing and the the elevation 0* the tail water as taken from the gaging curve was 603.5 The results indicate that the elevation of the tail water should be higher. Capacity of ¢pillway 2nd Trial Assume 925 c.f.s. Elevation of tail water from gaging curve o03.65 6!: In — .. *4: .521—{Z : .7JJ/ é Joya !—-—-————-———7 Q = y.7;J;///};77/m~4 7‘ 4,, Velocity head 3 o/ I z r‘- . -=—4—* = a - g, 91373357/26 Val/[ocnnl‘fi/36) Q ; 0'77J-xj‘ofx26/Y 5357’:- 7J'J Cf}. Over crest of dams From table 24 Creager and Justin C = 3.5; C = .739 Q = 3.8 x .739x10 (2.136) 3/2 = 57.5 Over crest dams 2 x 87.5 = 175.0 Through spillway : 753 Total discharge 925 c.f.s. The Open spillway and the drest dams will discharge 928 c;f.s. at a surface elevation of 604.0 which discharge exceeds 864 ccf.s. Hence the assumed spillway dimensions will be edOpted. -34... ro Spillwey vidth 5: Floor elevetion 598.5 flex. woter surfece33+,9 Too of spillwsy piers 605.0 Spillwey Design The flesh bosrds will be held in place by vertic- al studding secured at the bottom by iron shoes, end at the tOp by a horizontal steel truss. The truss will be 2 five paneled Pratt, 25 ft. long end 5' wide. Selection of flesh boards. Too elevation of boards : 603.0 Floor elevation = ngdi head on botzom board 4.5 Water pressure at bottom 4.5x62.5 : 252 lb/ft Scan of board \f' U Do‘ a 1105 lb ft - 13300 lb. in 22x516X576 3 xpf/Afif 7‘ l... é‘ W/z ’2 / Extreme fiber stress of Douglas Fur from corsegie ook 1200 lb/in2 p4: ;;z ;- /7300 c "D :3 :2. OJ ,1: — 147193. : //,0)- ”.73 ’ C /Zoo for 3'IF/4vk {Midsure 2 7/y) _1_: 1. /.7.7Jx/Z : /_>j//;7 :2 C L375" 3" Douglas Fur timber will be used for Flash boards. Flash board studdin ”v —’ C.) I000- : 3.07 in 3 Carnegie Cross Tire 1-25 :.l = 5.5 C . , .3 +-,\ _ Too re.cticn - 1.5 35:3 : $20? 6.5 Bottom " = 3550 «820 = 2730? Steel Truss This truss will be rather unusual in that it U must support its own weight in a horizontal position, a deed load of a temporary foot bridge, a live load and impact, and the horizontal panel point load, the loading which makes the truss necessary. 0. Us. or 25,1017, 4“! a {V k p gk k Q ‘1- Z‘ Ila/o 5’10 310 ' ’1” [:0 /64¢o / -30 Length of diagonal : Influence line for shear in panel 1-2 4 .7 s I I. J 4V \ - ‘ J- 2 -L I J Influence line for shear in panel 2-3 3 3: I'\zJ Influence line for shear in panel 3-4 . , J 3‘33“‘--~._~_“‘_ Full load stresses (horizontal) V»; m an: filn \(u (by index method) hember Index Stress Factor Stress v, -v1 -16l+o M —is37 V, -v, ~2460 533 4755 (5 -v, ~2uso £551 43799 v, -v,- -2l'+éo 5:: -27» 5 v, -v, 4640 £6 48537 L, -L, o 0 L1 -L, 14640 E'é'é' $1837 L, -L., ~ l21+60 :53 4.0755 L, .. Lr {£13140 K:-6 1'13}? L, ~L, o ‘5— o v, -L -16‘~+o 1 4640 -37- Hember Index Stress Factor Stress“ V, -L,L 4320 l -820 Y,-L, 0 1 0 m,-L“ 0 l o v, —L, —1o~+o 1 -161+o Q -L, iisuo .51 sense . 5 V',_-LJ $320 7.51 ll23l 5 V;-Lq O 0 V;- L, O 0 z--Lv l$20 7551 $1231 17‘ -L , 11549 .Rl 9.2462 Determination of Maximum Stresses The maximum stress occurs in top and bottom chord when all panel points are fully loaded. The quantity for the chord members on the preceding page is maximum. Maximum shear in panel 1-2 occurs when all panel points are fully loaded, as shown by influence line. Hence values for U,L,; U‘Lk Y’Lz; and V L on preceding page are a maximum. Maximum shear occurs in panel 2—3, with points 3,4 and 5 loaded. V=__/_f2+3 720:?f5' J. Z4. : —- 771-; l/,-- 1.,- = - 95! /‘//IX. K4, .— 7.__,7 7;,- = + uric #7". r’ ’4‘"lr : _____7,'{'/ 5' 73.} :v+/a?O ”Ix, -3g- haximum shear occurs in panel 3—4, with points 4 and 5 loaded. V: /:'2_ 370 5192‘ i” ”.3": ””4, “1‘19 : ~y¢z "/ 4V ”3 ‘41] 411/ Uv‘Lc '1 16:” #72 = 7‘ 76/7 Vy‘Z-q /.J’ Cows/'Jelleal 67.1 4 71?”.5/027 founffl’ Maximum stresses Tap chord _2755 Bottom chord ;2755 V’-L,and g L, _1540 V; ’1‘; " YrL; - 985 V; -L. " Vqu _ .492 V, - L. " V, - 12462 V2' L " V, V, Hugo Y: ‘Lv " V. L: 1 7'49 Chord Hembers. The chord members are subject to two loadinr ’- b \J s . A vertical load causing bending and tension due to horizon— tal loading , Vertical load Foot bridge (see drawing No.2) Rafters 2"x6"x8' 12 21 Flooring2"x60"xl' .835 c.f. .381 c.f. Rail & Bracing 2"x#"x6' .333 c.f. Floor bracing l"x4"x5' .139 c.f Amountof bridge timber per foot 1.688 c.f. 3-39- Weight of yellow pine a; lb per cu.ft. 44 x 1.688 - 75#/ft 37.5 per ft for each I Beam. Live Load Assume 100 lbs per lin.foot 50 lb per foot per I Beam Live Moment = W12 _ ’8'" - 1) x 255218 = L@900 lb.ft ‘3 Impact : :00 - , c - ‘ . poo l sexes - ”90 X 979 - fllég IOO Live homent 1 Impact 9680 lb.ft Assume weight of I Beam as 25 lb per ft. 'Total dead load 25 i 37.5 = 62.5 lb per ft. Dead moment = 62.5; 28x28 = 6120 lb ft 8 Live Moment :g9680 Total Moment 15S00 lb ft " " 189,500 lb in. Design of Lower Chord .Aloow 4 - 3/8" rivet holes in The design of this member is exactly similar to that of an escentrically loaded column for which Spofford in Theonr bf Structures, gives the formula (28) P S - v “—... i 1.1 2 A I ; PL CE -ho- in which bi v—‘ I Bending moment due to vertical load. a Axial load Distance from c.g to extreme fiber , l = constant, C— = +3 for uniform load. C) *4 ’U ll : Karimum stoss Unsupported length : Modulus of elasticity H [1:] L“ (D II = Homent of Inertia Assuming 18.4" 8" I Beam . . 2 A area from Carnegla A z 5.34 - 1.5 = 3.8M ln . 2 as l 189509 x4 ° 56.9 4 2755 x55 x144 x28 x28 2SX29000OUO J 7" 20# I A : 5.83 -l.5 : 4.33 - 27R" ; 189500x3.5 ; . 41.9“;2755x5xluux28x28 43x29,ooo,ooo i r 66 000 , , . I = A 30 4 uijg—g—I;12 = 630 l 15430 = 10060 18.48# 8" I Beam will be used for lower chord. U) 717 I cacao = 717 $13050 = 13767 lb/in2 50. 1.12 4¥l "Design of 613‘; er Shard The formula is the sure for a ccmlgle 831.121 :1er1bel with the excel:- tin of change in Sign, thus S I ..g... + 3.1 A “Dry 18.4 {.1 8" I beam by formula 22 bpofford's Structures : 16000 - 50 L T o H’b‘ P+w H II £8 1: 12 : 105 5.26 5.6 X 12 = 80 .64 P = 16000 - 60 x 103 . 10,850 e per in2 A s : 2755 .+,189500 x 4 5.34 56.3 - 2755 x 5 x 144 x 28 x 28 46 x 29000.000 5 = 517.L#25800Q#4 ALL : 517 13580 :36 .9 " 1012 s = 14097 -f—/in 2 Try 21.8 lb 9" I beam 1 = 16000 - 28 x 12 50 = 11320 lb yer in 2 A 5.67 S I 2755 + 189500 I 4. 5 fine-2 04:09 " 3:55 1: 5 X 14"! X 28 x 28g 48 I 29, 000, 000 6 ;- 426+ 853000 a 436 +10160 34.9 - 1.12 ; 10616 lb per in 2 21.8 9" I Beams will be used for both tOp and bottom chords. Design of mailers 71 1.2 and V6 L5 load 2462 Required Area :1 ___2_fi-.£___2___ -_-: .154 sq. in. 16000 Try 2% x 2 x 1/4" L .11.06 - l/d'x 1/2 = *015 .98 sq. in. Try 2 x 1 1/4 x 1/4 .-. .75 - 1/4 x 1/2 =___.13 .62 Ski. in. Use 2 x 1 1/4 x 1/4 L3 for diagonals. Design of Liembers V1 L1 am 176 1‘6 load 0 153.0 cam-wression. Use Channel for stability of I beams. Use 6" 6.2 / ft channel a ,‘x- 0 9 a 11:40 : Euc lb yer 1n“ adeqlately sale. 2.79 1112 Assign of members 132 L2 - U5 L5 £35 111. octagressi on. Use 4115.4; channel 985 = .651 111/1:12 1.66 Also use s-.1e channel for {.73 11., ; V4 ’ 492 : .315 lb yer in2 U Design of joints cn upper chord. bse hitch azg les 4" x 3" x 1/4" L to Connect [to I been 1 - 3/8 " rivet in s. S. 's 1221;} 1 - 3/8" " " Bearin- (ID I H C) O 5) ..J (T) .f . C) II H o N (.31 N 0 However use 4 rivets in each leg. Use 174" gas set plate to connect angle to channel. Load in 71 12 : 2462 lbs. 2462 = 1.87 1320 Use 5 rivets in both channel ani 1:311: connection. 7‘5 Design of lower chord joint , Use hitch angle 4 x 4 x 1/4 L Shear in rivets of channel leg of hitch 3115 la .-. 8201‘, However use 4 rivets as are usediin uglier joints. Rivets for I been leg of hitch angle . 039 4 rivets for shear. 3/8" rivets. in tens ion .11 X 16000 n 1760?} Tells ion transferred from I beam to connectian equals 820:}. L‘se 2 rivets for tension vhich nukes a to tal of 6 15/8" rivets in horizontal leg of hitch angle. Use the same glsset plate as is used for up} er joints. #4 Desi-en or bpil lway I-iers. Brac’rst to taize horizontal thrust of truss. § % ¥ _r—" #5” l 1'12“. it “215.» 'T a . 0.); D :. to .. -o ‘J 0‘ Balding moment .-.- 16420 x 4.5 = 7550 lb. in. v . 60 for shear then web is reinforced éjdv 9 x .875 x 60 4t However it would be advisable to use 6" :-.- d 9” - D As I 11 I 7380? - .088 sq. in. fsjd 18000 x .875 x 6 .0 Use 1/4" 0 YBars for moment. 2 x .049 -_- 098 sq. in. 'J .-. v = 1540 = 199 lb "per in2 gojd 2 x .785 x .875 x 6 Extend bars 40 bar diameters 10" Web reinforcing; Use 1/4 0 K bars. If vertical steel were used, load c:.-rried by steel would be'(60-40) 9 x 6: 1080 "; at 45° 1080 = 765 lb. 1.41 Cepqcity of :1: Bar . 2 x .049 x 16000 .- 1670 lbs. Ali Eiers After testing piers of various design for overturn'ng, sliding and soil pressnre, the following desibn was found satisfactory. .7)’ - .5712" .7) j 1' 4 7C __ '_'* :27 1 ‘Ir 0\ 7.77 7F 4L Lionents Lioznent Par 1; l) imensi ons t . 1: Plus Iiinu s ; 2x5x9.77x150 8800 4.89 43000 b 9x6.4xlx150 8650 5.05 46600 c l.35x4.5xlx150 911 1.40 1280 d .87x4.5xlx150 588 8.53 5100 a .75x. 5x.75x150 65 2.37 150 f 6.41:1.5xlx150 1440 5.05 7270 Vert.tmlss 7152 5.50 4180 2‘21; Elorz. Truss 1640 12.37 20500 :2:- 'ffater %e2.5x.56 1120 6 6710 iressure 2 I—resssre on 2.5521/2262.Z“1'"i;e.6 _ 1510 5.5 564.0 Fladlboard _ __ 4370 82650 104590 Resulting moment = 104590 7195.0 10. ft. " 32650 I 71940 21219 '2 Ch . Vol-O Llomtnt distance of Resultant -_- 9.77 - c0238 :3 1.51 2 l‘ccentrici by C 9.77 a 1.63 6 Iesultart falls vithin middle t1-ird of base. h ., ‘- actor of safety agaimt wm‘turning 104530 : 5.2 f.S. for overturning "I £8.th of safety against sliding; from table :11 p 132 Z-ool's Sonorete Construction 701 2, the V friction for wet clay :- 0.33 cctfi‘ is lent 6‘7 sliding force _-.- 4570 Iesistirg force = 21214 x 0.55 = 7071 m_ = 1062 £05. for Sliding d p1 . 2121_4_ ( 1+ 6151.51 )-_- 7w. (1+.922) = 13.92 51/ ft2 5X 9.77 9.77 P2 = 724 (1 - .922) = 56 1/ ft 2 From table on p 1:",- 'Eiool's Concrete Comtmct 11;.11 Vol 2 allowable soil pressure for soft clay is 2C00 lbs, lyer 51;. ft. ;>esi;n of Spillway Floor. 14 // F 'o I'fidth between flash board studding 5.6' Bottom reacti on of flash board Stud .2 2730;} The floor will be a 6" slab reinforced with wire :;1esh,.:1th a 2' cut off wall on line of up stream edge of 1: ier. Llorizcntal thrust of flesh board staddizg to be carried into soil by cantilever wall. Desi)“. of ccntilever tell n (I; box-mule for passive earth 1r; ssqre from 3001's Concrete Cons‘; F Pricti on angle for clay on rage 10 : 2o degrees. P: 1/2whz 1+st 1 - sinfi‘ 1.1+ sin 25 :- 1+.‘25 : 2.117 1 - Sill 25 l " .523 In as much as a factor of ea: ety against 5110114.; of 1.5 if ad V girl‘essm‘e 13' used in formula will be multiplied by 1.5 1. ( 7—100 h‘ x 2.47 ) 5.6 (.31 h) .41 C. ‘1 O H X (C __fi"4100 1; 2.44 3" “Fr‘; 5 Cr " QOMO‘I 1A..) .0 h LLeL-Le depth of wall 2.5' Bending moment = 4100 x 1.25 x 12 :.- 615-00 1b in. 8 d 3 II __ II V 61500 : 2.9%" Use 3" Allomble v: 40 1‘01er sq in v-—-E——-; d: V = 4100______ = 1.78” bid ij 653.074xé0 P .0077 As : .0077 x 3 x 66 a 1.55 Sq in. Use 11 - 15/6,". 95bars 3, 6" Specin: 11 X .14: = 1054 Sula 111. I'dct icn Vol 2. Design of entire normal flow of be in ace- cdd‘ooo}; by 01 for unit used in an attenget to arrive at c. as it will be necessary to find 47 01‘~;st Gravity 353:1. "rue ends of tie dim 17111 be g'avity secti 0213 over 1111011 the _“I The desi;;,11 oi tine crest 21111 (+4 (I. "3 H d g 1, Fl |...J *4 {J C O or lance with tat suggested (11 gage ZOE-9 of l’ydro Llectric ~ ger 3; Just in. Tile coordinates of time curve of a crest head as listed on gage 220 of the ifydro Electric Zigzdbozk, are a. '1 fairly accurate ferrule 111‘ the curve; the area under the curve ani its centroid. such values v5.11 be necessary in detera;1inin~: tie weigit and moment of the d 2.13. Clrve of crest ( ‘o'p stream face vertical). de Representative coordinates of curve for unit head and upstream face vertical x - 0.7; Y = 0.257 x = 2.7; Y = 2.32 The curve is of the form X(1 2 AY .711 = 0257 a 2.7n : 2.82 a N lose 37 = 10g. .257 + 10g. a N 10g. 207: 10g. 2.32 + log. a Subtracting N (logo 2.7 - logo -7) log. 2.82 - 10g. .257 - 10g 20 82 N ' 0557 z 1.04 = 1.773 10g §;%_ '5 (2'7)l.773 3 2.82 a A =2 2.07 Formula Xl°773 = 2.07 y A comparison between the coordinates given in the handbook and those computed by the formula will be made below. Coordinates from handbook : by formula 8 x . 0 yo 5 8.1 .82 8:; 2112 '05? .7 .257 02 22 a; 57 1:4 :870 '572 1.7 1.22 2.2 1.96 1.96 2'7 2°33 81 3.2 E. 3. . .9 3.; 6.23 6.18 . . .57 The usual head under which this dam will operate is 2 feet, and both coopdinates of the above curve should be multiplied by the head (2 feet) to obtain the proper curve of the crest. Hence the formula will be changed to comply with the 2 ft. head. Let x. of unit curve equal 5 x of new curve. Let y. of unit curve eoual % y of new curve. hence 1%)1.773 : 2'07 (£9 x1’773 _ 3712—7 ‘ 2491-? 11'773 : 3.54 y Q NT- 01’3'17‘ ‘02-0 J’T’fi -t 4p 48 _J 3: s A J Coordinates x Y .5 0'-6" .0825 o‘-1" 12 3'2 . 1:... n . I- n 2.0 gu-on .5ES o'-11§" 2'5 2'-5" 1.43 1'-5 1/8 " 3.0 3'-0" 1.98 l'-11%" Z-S 3'-6" 2.61 2'- 7 3/8" .0 '-o" 3.31 3'- 3%" 1.13 u'-1&" 3.5 3'-6" Area under curve. 11'773 = 3.5H y X = 2.04 y'504 3-5 3-5 , 3-5 A ={ 3‘ dY : 2°04 0/ f5"LL dy = 2.01 rib—gt; 33°56” o _ 3‘5 ——-1. en - 1030AL 3,1056); 0 = 1.30“- x 3.5 5 = 9026 SQofto Centroid 3.5 ~ 3-5 _ 2 1 _ -U¢a 2 (. 6#) A x - ~/0 % X QY - O ‘3' 20 y I 5 dy _ 2 08/..5 1.128 2.08 y2.128 3'5 - ' 0 Y dy = 2.123 0 2.128 - 1&00 I : 7&6 :- 1052 Determination of the dimensions of the footing under the dam so that the resultant will pass through the middle third. Up lift will be considered as being effective under the main portion of the dam only. A moment equation will be set up, and the equation will be solved for additional width of base (x) Md : Moment of dam M, = " .” water Wd 3 Weight of dam C = Unit wt. of Concrete 1 = “ ” " water b = base width of crest section Y = depth of base. 2 Uplift - i- wheb ‘3 b. _1" 2 <51? Md : Wd x + % by x2 - Mw- {heb 3b uplift = 1/3 (Wd + cxy) (b . x) 1 Ma + de + % 0y 12 - MW - 3'Wh26 -- %.(de + de + bcyx 4 cyxe) (Hy-hm 12- <%wd-wd+%bcy> x + (Md - Mi ' %- de - é'whebz) = 0 The equation is now in quadratic form Valuation of the known quanities. Wt. i’ Moment a 9 26 x 150 1390 2.61 36g0 b .6 x 3. 5 x 150 315 n.3g c 4.73 x 3 x 150 21130 2.3 % w (h: -h2) : i 62-h (9.52 - 6252. 1500# Centroid of Water Pressure Static Water Pressure P Y-3h-oh4'h2 : n316fi*6fi1605 = 2068 5 n' * 3 h 6 x 3 + 3 x 6.5 Impact Pl : hvw v2 3 62.u g 6.52; n x u : 202 lb. 8 32.2 Considered to act % h. 202 x 1.75 = 35% 1b. ft. MI Md Substituting the above values in formula and solving for Ix! (250x3--50xnx-(3wm-3%54§4oanWx3no + (10025 - a; h --% 3835 x n.73 -3-62. .4 x 9 5 x n.73 x 4. 73): 75 X2 + lane 1 - M809 . any x -6f‘2.. - - J/g*fifl h 30 _ l8u6 .yfiglgéfl: thqu809 2a 150 x . ~12.3 r 3Ju10i000 . 1,uu2,000 150 X .- -1203 I 14.7 x a 204 Or -2700 The width of the base will be increased 2.4 feet Resistance to sliding N = Weight of dam — uplift N = 3835 + 3x2.ux150 - 3.5x262.u n.73 : 3515 lb. Coefficient of friction on wet clay = 0.33 Total horizontal force = 1500 + 202 = 1702 : llll_. g 0.69 Not safe 1702 Cantilever cut off wall will be used to prevent sliding. As it is desirable to have factor of safety against sliding off 1.5, the horizontal force will be multipled by 1.5 To be taken by cantilever wall = 1.5 x 1702 - 1171 - 1389 lb. Passive earth pressure P = % Wh2 1 + sin 1 - sin Angle of internal friction 2 25° 1 + sin 22° = 1.u2 : 2.u l - 8111 O -' o 23 7 p = (% 100 h2 x 2.h7 ) h : \FEffiflifff—‘R - . 6, use h = 3.5' Win 33 7 51f Bending Moment : 175 x 12 x 1389 = 29200 lb.in. d= $2 723200 fl :_- 11.76- 12 x 107.7 d = V = 1389 = .70" b j v 12 x.'7 x 3 J d If 5" 3 D = 8" p = .0077 A5 = .0077 X 12 I 5 =00u63 BGoiDo Use 3 " at 6" u = V : 1389 = 177 lb. per in 25j d 3. x - x 3 Extend bars 50 bar diameter 25" Design of Retaining wall. Gravity Type After various trials, the following dimensions were adopted. 5 w d :1 Moment Part dimension Wt. E. '9 Moment a ' .5n x 6.5 150 273 .82 22% b 1 x 6.5 x 150 975 1.5 1&63 C 105 X 605 150 732 205 1830 2 d 1 x u x 150 600 2.0 1200 e 105 g 605 100 “88 3.0 1H6} f .5 x 6.5 x 100 {3g5 3.75 1220 3393 7&00 576 P : i 100 x 7752 1 - sin 25° 1 7 sin 25° 50 x 7752 x.402 = 1130# Overturning moment 3 1130 x 2.5 = 2820 lb. ft. F.S. of overturning 3 1399 = 2,62 . 2820 Earth Pressure x — ZHOO - 2820 -- 1.35 3393 0 2 - 1.35 = 0.65 ft. P 2 'p2 = 898 x .025 = 21 lb. per ft. Slidin . Coe%ficient of sliding friction for wet clay Resisting force = 3390 x .33 ‘ 1131 lb. Sliding force a 1130 lb. Factor of Safety : 11 1 ; 1130 1 A cantilever wall will be used to resist sliding. force will be increased 150%. 1130 x 1.50 - 1131 = 564 lbs. Earth pressure P : % Wh2 %% + sin ¢L) - sin 1% I 1 : gain E5 = 2.u7 56h . 50 x 2.47 n2 h .yf‘6fi""_—“‘ = 2.14; 2.5 ft. 5 x 2047 # B.M. - 56H x 2.5 x 12 = 17000 in. d = YJiidaaf : 3.6 use 9 ” 12 X 10707 D n. 7" 33Z3—-( 1 + é_§¢;éi.) = 848 x 1.975 . 1675 lb.ft. '33 The sliding Jri' v = _y__. = 564 . 13.5 1b in2 b j d 123x .874 x 4 - P = ~0077 AS - .0077 x 4 x 12 - 0.37 sq. in. Use 5" ¢ bars at 6" = .392 sq. in. Cost of Dam (exclusive of foot bridge) Spillway 2 Piers 10.1 c.y. @ 820.00 = 2202.00 Steel Truss 762 1b. 0 .06 = 45.72 Flash Board Studding 4-(M25) Carnegie Cross Ties 6' - 11%" Long 406 lb c .05 = 24.36 Douglas Fir Flash Boards 645 bd.ft. @ 90.00 per M I 58.05 Spillway Floor 6.8 c.y. @ 20.00 136.00 Crest Gravity Dam f Dam 26.4 c y @ 20.00 = 528.00 Apron 1.5 c y @ 20.00 a 30.00 Retaining Wall 39.2 c.y. @ 20.00 a _]84.00 $1,808.13 COst of the different factors in the installation of a Direct Flow Wading Pool. Cost of dam (not including foot bridge) ------ $1,808.10 135.7 cu.yd.concrete in place & 10.70 ------ 1,450.00 260 sq.ft. of roof surface & .25 ------ 65.00 8 sheets of 3-9-30,in 12' widths of expanded metal @ $4.30 ----- 34.40 1 Type D.B.M Paradon direct feed chlorinator 250.00 2 - 6' x 2' —6" x 3/8" cover plates 0 10.50-- 21.00 1-8" globe tank float value -------- 115.00 era 1 - 8“ Standard wedge gate valve ------ $40.00 1 -l2" Standard wedge gate valve ------ 80.00 2 - Floor stands for valves @311.00 — - - - 22.00 1 - 22" manhole frame & cover -------- 12.00 190' of 8" C.I. pipe in place@ $1.70 — - - - 323.00 1183' of 12" drain tile 0 90¢ per ft. -- - - 1,064.70 2 - Door casing & door @ 38.00 - - -- - - -- 16.00 3 Windows complete with shash & glass @ $9.00- 27.00 t 5,328.20 An Estimate of the Cost of Installation of a Recirculation Flow Wading Pool Design Data For Pool The report of the Joint Committee on Bathing Places of the American Health Association and the Conference of State Sanitary Engineers states that, "In a recirculation or flowing through pool in which the dirty or used water is continually being withdrawn and replaced by fresh or filtered water, purification of the pool water proceeds by consecutive dilution. The first portion withdrawn from the pool will all be dirty water, but, owing to the constant admixture of entering clean water with dirty water remaining in the pool, each succeeding portion of water withdrawn will consist of a decreasing proportion of dirty water mixed with an increasing proportion of clean water. In proportioning the rate at which fresh water should be added to a flowing through pool or the capacity of pumps, filters, etc., for a recirculation pool, this law must be taken into consideration." In view of this fact and assuming as we did in direct flow pool that 500 children will be maximum per hour, but in this case adding only 500 gal. for each 20 children using pool in an JF7' hour, we have 500 x 500 - 12,500 gal./2hr., the amount of water 2 to be added to pool per hour. 12,500 gal. hr. = 208.33 gal. min. Inlets for fresh or re-purified water should be located at points so as to produce as far as possible a uniform cir- i culation of water throughout the entire pool. Eight inlets I will be used; located at intervals entirely around the perimeter of the pool as shown on plate No. 5. Three outlets will be used which will be located on center line of pool, one at midpoint between the two ends and the others 50' in opposite directions along center line from the one at center of pool. Total water surface to be not less than 3,750 sq.ft. as was obtained in direct flow data of design. Max. depth; 21" ' Max. wide of pool 40' but there shall be a 6' side wall with low curb on outside of walk all of the way around pool. This walk will be not less than 4 ' in thickness. With this design data in mind, the pool will be semi- cirnlar, 40' wide. The center line will a semi-circle of 210' radius and inner edge of pool to be a semi-circle of 190‘ radius and outer edge a semi-circle of 230' radius. This pool will have circular corners of 12.5' radius, also about the outside there will be a 6' sidewalk with low curb on outside of walk. This walk will be 4" in thickness. The bottom of pool will slope to central outlet at center of pool and this will be connected to 12" drain pipe. This is to serve the purpose of draining pool for cleaning without pump— ing the water back to filters. The location and size of inlet 419 and outlet pipes are shown on Plate No.£; The pool will be 6" in thickness reinforced with expanded metal reinforcing. The elevation of the water surface will be 602.67'. Rapid Sand Filter Assuming the filter will take care of 2 gal. per minute for each square foot of surface, the area required would be 208.33 : 104.17 sq. ft. of filter bed surface. We will use a 9' x 12' bed. The sand bed will be 27“ thick and will have a uniformity coefficient of not greater than 1.7 and an effective size of 35 mm. The upper edge of the wash water gutters will be 30" above top of sand. There will be 18" of gravel varying in size,the larger stones being at the bottom. The collection system will consist of galvanized iron pipe 3" in diameter with two rows of 3/8" holes spaced 4" center to center along under side. These will connect with a 8" central drain pipe which will empty into the storage tank. To wash the filter the flow will be reversed through the collection system and upward velocity of water will be 18" per min. To maintain this velocity it will require 1326 gal. min. The pipe and the throat of the tube is proportional to the square of the rate of flow through the tube. As the foat A drops, it opens the balanced valve so that the level of water in compartment A is the same as in B. The effective head forcing chemical solution through the control valve is, therefore, proportional to the difference of pressure causing the flow through the venturi tube. The rate of flow of chemical is, therefore, proportional to the flow through the Venturi tube. There will be two pumps,the smaller one of 250 gal.per min. A#__ ' "W‘W‘r A .., ,___, w .‘V'H‘ —‘__ é/' capacity to be operated whenever filter is supplied with water. The larger pump of 1200 gal. per min. capacity will be Operated wheneVer filter is washed. When water is added to pool after it has once been filled with either be taken from river or from a well which is in operation at the park.as their supply is not reliable and, using a rate of 1326 gal. per min. it will be possible to wash filter for a period of 6 min. before supply of stored water is exhausted. There will'be 28 holes of 3/8" diameter in each piece of pipe in collection system. Then 28 x 16 x 3.1416 x .1875 2 _ 108 x 144 .0318%. orifice ratio. The walls and floor will be 8“ in thickness and it will be possible to obtain a head of 5% ft. above sand before the filter must be washed. There will be an apparatus for dosing supply to filter with a sufficient amount of alum to effect a layer of schmutzdecke. The successful operation of the filter is dependent upon this layer. A detail cross section of this dosing apparatus is shown on plate No. 5, and it operates as follows: The water to be treated passes through the venturi tube from right to left. The difference in pressure between may contain objectional minerals. The supply from river is the better to use. The pool will occupy relatively the same position as direct flow pool did as shown on plage No. IV. The building will be located at any convenient place near outlet to pool. For general arrangement of recirculation system, see plate No. v. Chlorination. In either the direct flow or recirculation system, a type D B M Paradon Direct Feed Chlorinator will be used. The chlorine éuz will be applied in the small chamber where the float valve is located. In case of direct flow system, the residual chlorine content must be considerably greater than in the recirculation system. Also the dose of chlorine of direct flow must be greater in most cases because of the condition of water than in the recirculation system. Estimated Cost of a Recirculation System Wading Pool 206 Cu. Yd. of concrete in place 0 $11.50 @2369.00 1170 Sq.Ft. of roof surface in place@ .25¢ 292.50 1 Type D B M Paradon Direct Feed Chlorinator 250.00 1 Proportional chemical feed apparatus including venturi meter 150.00 Expanded metal reinforcing 34.40 1,250 ft. of 12” drain tile in place@ 90¢ 1125.00 1 8" globe float valve 115.00 9.34 cu. yd. of sand @ $5.00 - 46.70 5.73 cu. yd. of gravel @ $2.00 11.50 1 - 250 gal. per min. centrifugal pump,including motor 1000. 00 l - 1200 gal. per min. centrifugal pump, including motor 2,000.00 20% ft. of 2' x 3/8" C.I.coverplate 28.90 9 - 8" gate valves 0 840.00 360.00 1 - 12” gate valve @ $80.00 80.00 4 Windows including gass & frame @ 39.00 36.00 3 Doors including frames @ $8.00 24.00 2650' of 8" C.I. Pipe 100' of e" C.I. Pipe 83' of 3" perforated pipe . I ,2 4 - 6" bend * *“‘ 2 00 10- 8” bend 7 5' l - 8" x 8" Cross 8 - 8" x 3” crosses 3 0 8" x 8" Tees Suitable intake constructed of concrete, lump sum 50.00 TOtal $8,698.00 63 Drainage It is desired to lay out an underdrain system that will drain the entire park. The low land along the river will be drained by a system of drain tile placed 50 feet apart, the high ground will be drained by tile placed 100 feet apart. This spacing of drains as chosen should be entirely adequate as the top 2 or 3 feet of the low land is a sandy loam and quite porous. The higher ground is mostly sand and gravel through which percolation is very rapid. The low land drainage system is designed to work with in very narrow limits of elevation. It will be noted on the profile of the main drains which is also used to drain the wading pool, that there is less than a foot difference between the invert elevation of the outlet and the elevation of the bottom of the wading pool. The plan of the drainage system is shown on the topo- graphical map, drawing No. 1. To determine the size of the mains, it was necessary to determine the rate at which the water is to be removed. It is the usual practice on large drainage systems to arrive at this value by predicting the largest storm that will probably occur once in a predetermined number of years. But in the case of a drainage system of so small an extent and where a rise in the elevation of the river of only two feet would render the system practically useless, a thorough investigation of past rainfall records and large storms is entirely unnecessary. According to Pickle's Drainage and Flood Control, the average drainage modulus based on years of record is 3/8" of rain to be removed in 24 hours in the State of Michigan. This 4? value is equivalent to 0.0151 cubic feet per second per acre. By the use of this value, the discharge and slope of the mains will be computed. é;J’ An Estimate of the Cost of a Drainage System for Dodge Park #8. A run—off modulus of .0151 was used for the design of this drainage system. This value was obtained from the study of the Clinton River drainage area. For the main drain, (A b C H I J K L M), a 12" main was used. This main had a slope of .055' / 100' and was designed to carry .984 cu. ft. / sec. To obtain size of all mains a table on page 161 of the 1925 edition of Drainage and Flood Control Engineering by Pickels, was used. The elevation of 599.25' was chosen as lowest elevation for outlet. 5" drain tile was used for all laterals. For general layout of drainage system see plate No. l, and for elevations, size, and slope of all mains, see cross section Sheet No. 5 Standard drain tile of clay will be used and following is lengths of the different sized drains with cost of same. 1,183 ' of 12" main a-90Q per ft. in place 31,06u.70 212' of 10" main 0 70¢ per ft. in place 148.40 3,694' of 6" main @ 45¢ per ft. in place 1,653.30 17,830' of 5" lateral 9 40¢ per ft. in place 2,132.00 Total $9,998.40 The above prices in place include all wyes, reducers, and other connections, also with a suitable outlet to be made of concrete, in place. Grading. It will be necessary to fill over pipe lines D S P 0 R, J 0 and part of the natural ox-bow; also some filling and grading underneath pool. The average length of haul will be about 500 445 feet as the earth used in the fill can be obtained from the higher portion of park. To make these proper fills it will be necessary to haul approximately 5,000 cu. yd. At 80¢ per cu. yd. the cost would be $4,000. Conclusion. In conclusion maywle present a comparison of the costs of the two different types of wading pools and their respective water supplies. Also a summary of the cost of drainage and grading. Cost of Wading Pool with direct flow system 85,328.20 Cost of Wading Pool with indirect flow system 8,698.00 Cost of drainage system 9,993.ho Cost of Grading 4,000.00 In view of the above tabulations the Direct Flow System is the most economical to construct. It is also the most economical to maintain, because no pumping is necessary as the dam furnishes the head. 0n the other hand, a more pure water is obtained with the indirect Flow System, because recirculation necessarily requires filtration as well as chlorination. 1. .Epzw .1111 s 9' ,1? 111111 1 111“ 3 12930 96 30 097