ZEN'V BI ’OhIIiEiIH fiIV‘cidC-l A DESIGN OF A P'RlMARY TANK AND AN AEREATION TANK FOR A SEWAGE TREATMENT PLANT Thesis for the Degree of M. S. MICHIGAN STATE COLLEGE E. Rlbeiro I. 1945 “1 115.315 This is to certify that the thesis entitled A Design of a Primary Tank and an Aeraation Tank for a Sewage Treatment Plant presented by Efrain Ribeiro-Ibanez . has been accepted towards fulfihnent of the requirements for H. S. degree hi 01711 Engineering Chester L. Allen Major professor Date June 8, 1945 _ _ , .__,__,_ ' I — . -_-- Adu- hi.“ 1'5 I so I 1K} " i" k ‘ ‘_; i p o L ‘i _. tr ,. lie. 1. U 0 all t! . ow “L . #2 IL .r‘L ‘ . 7 .t ,\ 0 JAN. 5. , ,. \. _ .6-Ii..k‘1p.1ni 1‘. This-AS;- féi gt. i L! 1‘ 6- 1 (.‘1’ .. 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TM“ V’ )"'_ ‘ ’ k \. 1’ ‘J’ [\‘I"\ ) X 0C ll '1’ :4; A f\‘ I .— . 'Jo' " Ell. 1Y1. 't r‘» . 3' tm :11 MIC O .f‘ .~ 0 ‘— 1 .. : ,‘ f I l'u . x, i t f: -.- in . uj r P. r". I 3.’ c L hm" J THAI" ‘II 10 x J x d m" 1. 1 {JC' I; trial garln - 13 _ 1. Design of the walls. In this case all the walls of the aeration tank are of long spans and must be desingned then like a cantilevers walls, for this re- ason we are going to consider the four walls of our tank like a cantilevers and design it in that wav. If we assume athickness of 1 ft 3 in for the walls, we have: A "f° The maximun presshg. of the water is in the bottom of the wall and is equal then to: w x h p 62.4 x 13 p p a 811.2 lbs Now we know the resultant of the oressume of the water is going to be aoolied at the 1/3 of the wall, and its value is then: P.: i x w x h2 s x 811.2 x 13 P - 5,272.8 lbs. P The maximun moment of this force is at the bottom of the wall and is equal to: n : 2 x h/3 x 12 5272.8 x 13/} x 12 M : 2745185 in lbs - 14 - The minimun thickness for the wall is then: we have: 274,185 in lbs. 12 in 7:68 I: 157 ( From O'Rourke ) Then we have: d 3 274:185 157x 2 d 12. 06 in Practicallv we can considered 12 in for the mlnimun thickness, and if we assume 1.5 in for covering at each side we have a net thicness of 12 olus 2 x 1.5 equal to 15 in. for the thickness. Now we must check this value for the shear: V': 5,272 lbs then: v : V bx 3x d v : 5272 . 123778xl2 v = 42.5 lbs psi. The allowable shear is: I 0.12 x to Va“: v'= 0.12 x 900 v'= 240 lbs psi. Then our section is safe. This shear is in the bottom of the wall but all the other sections ere also safe because the value of "d" varies with the value of "h" - 15 - and in this way the value for the shear is each when the value for "d" is smallest. Quantity of steel. For the steel we have: As 3 H _ T, x 3*x d in which: Ms 274,185 in lbs. :3: 20,000 lbs psi. J : 7/8 6 : 12 in. then: time smallest A 3 274 18 ‘ 20,505sfsvéxre' ‘ 12 A8 =~ 1.14 sq in. We select round steel of 7/8 in and 6 in of space between centers with a total area of 1.20 sq in and a perimeter of 5.5 in. We have to check the bond for this steel: u : V ES‘x J x d u': 273 5. x. x 91.5 lbs psi. u The allowable bond is: u = 0.05 r; u 3 0.05 x 2,000 = 100 lbs p81. then in our case is safe. Cut-Off Bars. We are going to have at the middle of the wall a moment which is going to be half of the maximun moment at the bottom, so the: steel required in that section is only the half of the steel used in the lower third of the wall, for this season at a heigth of i of 13 or 6.5 ft. we out every other bar. Anchorage. The steel wil be bent down in the toe of the footing at a distan- ce beyond the too of the footing or base slab suficient to dvelop their strength in bond or: distance = 20,000 6 x 100 x 7/8 distance a 30 in. Temperature Steel. The outside of the wall is exoosed to the full temperature varia- tion while the inside somewhat insulated, but we are going to pla- ce the same quantity of steel in both sides, so we have: A, = p x'b x d Ag - 0.00125 x 12. x 12 A3 = 0.18 sq in. We place round steel of s in, and spaced 12 in between centers gi- ven a total area of 0.20 sq in. Considerations. We had designed the walls considering only the interior side of it supporting the pressuvn of the sewage, but also the exterior side is _ 17 - going to support the pressmnn of the soil and for the ground water, so providing we are going to have the tank sometimes empty, and in that case we are going to have areversal of efforts with the exte~ rior side supporting all the present-1, so we are going to put the same reinforcement in both sides of the wall. Also the length of our four walls is very close, so we adoot the same dimensions and reinforcements for all of them are going the same providong the efforts which are going to support are very close and is unnecesary to repeat the same calculus. - 18 - 2. Design of the Beamg. These beams are going to support the motors employed in the aera~ tion of the sewage, this motors has a weigth of 12200 lbs each one, also the beams are going to be used to walk so we are ging to assume 100 lbs per linear foot for the weigth of the people, the weight in the center of the beam is 1,200 lbs plus 500 move conssidering accesories and impact of the motor. Then for our design we have a uniform load of 100 lbs per linear foot, and a concentrated load of 1,700 lbs at the center of the beam. We adopt for a preliminary calculus the following dimensions, for our design we are going to consider each beam composed by two Tee beams, so our tentative section and dimensions is the following: Iwootbe. _ I I "‘ 3'2. ‘ I as: f ‘_... .....«._ g , _ p g, . ‘ " f 1 u ' . .___._.._- -11“, _ __ - _ . g E F \ ' «'f— 1": i 5 ' $ . l : i ‘ I. It“ ‘ 1.. F | ' 1 1 1 1 1 £ ‘ A”\J . j J é— ‘0“ )1. s — an» —, Then the weigth of each Tee beam is: w = 18 x 4 1:3 10 x 12 x 150 w a 199.50 lbs per linear foott this is considering each linear foot of the beam, and 150 lbs per each cubic foot of reinforced concrete. -19... We have a uniform load of 100 lbs per linear foot, so the total uniform load is : w a 200 l 100 W : 300 lbs p.1.f. The shear caused bv this uniform load is then: v a 300 x s x 29.33 V = 4,400 lbs. For the moment cause by this uniform load we are going to consider the beams like partially continuos beams and for this reason we assume a moment of 1/10 w12, so we have: M = 1/10 x w x 12 = 1/10x300x257332 M = 309,600 in. lbs. Concentrated load. We know the concentrated load is: p 1200 K 500 1700 lbs. P of this concentrated load, half of this is for each of the tee beams so the concentrated load is: P = #11700 = 850 lbs. The shear due to the concentrated load is: V = s x 850 V = 425 lbs For the maximun moment due to the concentrated load, if we can si— dered the beam like a partially continues, we have to make a re- duction in the moment which is & P1, and this reduction mus be _ 20 _ in the same proportion like the uniform load, so our moment is go— ing to be: 8/10 x 1 x P x l a u 8/10x%1850x29.33112 59,832 111-le. H M Total moment and total shear. Making the sume of the values we got before we have: Total shear: v 4,400 / 425 = 4825 lbs H Total moment: M : 309,600 x 59,832 3 369,432 in-lbs. With this total moment we can check our first tentative section. Assuming that web reinforcement is to be provided, the allowable unit shearing stress is : v = 0.06.19: f}, v e 0.006 x 2,500 : 120 lbs p.s.i. So we have: bsd . V’ 77126 4 824 ‘0'“ ‘W b'd = 46 sq in. But we have a tentative section dd 16 x 10 or 160 sq in, this sec- tion is to big for the value we obtained before, so we must redu- ce our section and check again. For this pur posse we are going to ad0pt another dimensions and proceed to the calculus. - 2 1 - New Section For The Tee Beams.- We adOpt the following dimensions for our new tentative sectidn for the tee beams: cc ". ”- Q“ —- -— —-—---1 n“ i v i I ¢ a“ 5 Then the weigth of the beam is in this case: . = lgxalfia 8x10 x 150 w - 158 lbs per linear ft. then the total uniform load is: w : 158 / 100 = 258 lbs p.1.ft. The shear for this load is then: 2581éx2933 V = 3,780 lbs. V The maximun moment for the uniform load is considering like before the beams partially continuos: l/lO xw x l2 0.1 x 258 x 29.352 x 12 266,000 in-lbe [I 3?. I! u -22.. Concentrated Load. The concentrated load is the same like in the other case or: P : 850 lbs Then the shear for the concentrated load is: v = '2: x 850 :: 425 lbs The maximun moment for the concentrated load is then: BZIOxixle E II M 8/10x1241850x29.33x12 M' 59.772 in-lba Then the total moment and the total shear are: 3780 I 425 4,205 lbs Total Shear: V V Total moment: M = 266,000 / 59.772 M = 325,772 in-lbs. So we can check our new section with the values we got, assuming like in our first section that web reinforcement is to be provided the allowable unit shearing stress is 120 lbs p.s.i., and the mini- mun section is then: ’ : V b d .__3_E,T§O b'd a 4 205 00875XI§6 bfid It 40 sq in. Our new tentative section is 8 x 14 in. or 112 so. in, and like in - 23 _ the first case we have a biggest section than that required, but we are going to adnt this second section for our definite section , be- cause is imposible to reduce more our section. Steel for the Tee Beam. To get the steel for the beams we must compute first in which portion of the beam is the neutral axis. First we must get the value of "k", we know: in which: n a 15 f3- 20,000 lbs psi. tbs 900 lbs psi. so we have: R : 15 i5 7 20,000 900 k-= 0.55 The value of "kd" isthen: kd 0.55(1# - 1.5) kd : 6.875 in. .We put 14 - 1.5 for the value of "d" considering 1.5 in for covering the steel, and also we are considering one row of bars. The value obtained for "kd" of 6.875 in indicates that the neutral axis is on the stem of the beam and then the Tee-beams formulas can be apply to obtain the steel necesary for our case. To obtain the quantity of steel required we must get first the va- lues of ”a" and "J” : - 24 _ The value of "z" is : 8:3kd‘2tx2 2kd ~ t 3 3 a 5 3x6.875 — 2xh )x 4 2x 9 75-1X37x3 8 - 1072 To obtain the value of "3" we know: 3d = d - a then: . J x 1205 3 12050 " 1072 3: 10.78 12050 J = 0.862 Then with this value for "J“ and the maximun moment we got before. we can get the value of the quantity of steel required: As: M fax JR 6 And replacing with the values we have: = 352 772 ‘3 267666§f882§i275 As : 1.63 sq in. We select round steel , 4 bars, of Q in of diameter given a total area of 1.76 sq in. and a perimeter of 9.424 in. Now we must check if all this steel can be put in one row like we assume before. if the separation of the oars between centers is 2-5 times the diameter of the bars , and if we considered for tthe ex~ ter’or covering of the steel 1 in at each side we have: 3 x i x 2.5 / 2x1 = 7.6 in _ 25 - Then we obtain a value of 7.7 in, and we have a width of 8 in son in this case is right andfwe can put all of our bars ill on- lv one row. We Check for the bond: 9:: V EoXJXd 4 205 u : 9.525x.862x1§.5 u 43 lbs p.s.i. The allowable bond is: 0.05 x 2,000 : 100 lbs psi. ll 11 then we are safe. Width of flange of the tee beam. To Check the width of flange of the beam we use the nexte formula: t Me 3 r6 x ( 1 ”‘3 )x,£ x b x J x d2 '"2E——“' .d in which: He . 352,772 in-lbs fc : 900 lbs psi. t : 4 in d : 12.5 in J : 8.62 b : width of flange. Then replacing this values in the formula we obtain: b 3 352L772 23.895 b = 10.5 in. -25.. So the munimun width of flange we can use is 10.5 in but we ha- ve assumed a value of 18 in, son our case is safe. Bent of bars. The number of the bars in the beams is four, and we can bent two of them. To know the distance at which we must do it, we drew the parabola of moments of the beam, and then divided the vertical maximun distance( corresponding to the maximun moment), in four parts and we assume each of this parts are absorbed for each of the bars. In this way we know we can bent two of the bars at 4.25 ft of the extreme, the others two bars are prolongued until the end of the beams. M ~z~~~~o, no l'm;.!"r ‘,,L 'l , h .4 A z;m*“w_ ‘ 4i .5“ )- d ’2‘5‘71“ ‘3— 5 t1 I'mps . Making the investigacion for stirruns we noticed that the shear is verv small for the section of the beams and so theoretically we do no need to put stirrups. But considering we need some su- pport for the horizontal bars and for prevention of any extra stress we are going to place vertical stirrups of e in and spa- ced one ft. all along the beams. - 27 _ Anchorage and extensions of bars. For end anchorage we are going to use a hook in the end of the bars , then the bars are bent in a full semicircle with a radius of four diameters, then 4 x % = 3 in, and a free end of eight diameters or 8 x g = 6 in. The extension of the bars in the beams in the middle subport is nrescribed by a minimun of 12 diameters or 12 x g = 9 in, but we are going to give an extension of 18 in. BIBLIOGRAFY Design Of Concrete Structures, by Urquhart and Q’Rourke. Fourth Edition. mac Graw Hill Rook "ompany. Reinforced Concrete Construction, by Hool. Mac Graww Hill nomnany. Reinforced ficncrete Structures, by Dean Peabody. John Wiley and Fons. Reinforced Concrete Design Handbook. American Goncrete Institute. Sewerage and Sewage Discosal, by Metcalf and Eddy Vac Gravy and Hill Company. American “ivil Engineer's Hanbook, by Merriman and wiggin. John wiley and Sons. Bulletins in Reinforced Concrete Design of the Portland Cement Association. Al. - ,5 (b J 4 < ‘2 - . ... 1 9%“ .‘a‘ ”1' O. ’3' 3" ‘r wufrwd u .:. mutt .s .. .K L (Q L\ \NH‘ \2 .. u. “on. ...\_s. .c ). .3: rucixb zit . ._ .. :3 o. . GK :3: ,, Eff: ... 3) .0 0! re... _ 8-8 41011)); E: h r . I .. ._._ '57" .— 33 .0 .9! " .0 70’ . J ‘ l3“ .Qn ‘Is‘ ..3..x. ..... . .s ,. .., .. . . .. .1 .. .w. .. to) $3.... . .. .‘ ... A u. k V. \ -\\. Va 3 ‘1 y o .. I. O . o . s . V. . ' .0 . I .o v. 8' ¢ .u c n v. I C - . . ' ‘ e I- .h o n. ' up .. . .. n .0.“ 2:; kw, wort. ‘ 1» a“ §°\8 v3 I O t G. I ‘ , B u '3." a... . .... . -.L . 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