Design of Water Tank
A Project Submitted
In Partial Fulfillment of the Requirements
For the Degree of
Bachelor of Technology
In Civil Engineering
By
Nibedita Sahoo
10401010
DEPARTMENT OF CIVIL ENGINEERING
NATIONAL INSTITUTE OF TECHNOLOGY ROURKELA
Design of Water Tank
A Project Submitted
In Partial Fulfillment of the Requirements
For the Degree of
Bachelor of Technology
In Civil Engineering
By
Nibedita Sahoo
10401010
Under the Guidance of
Prof. S.K. Sahoo
DEPARTMENT OF CIVIL ENGINEERING
NATIONAL INSTITUTE OF TECHNOLOGY ROURKELA
NATIONAL INSTITUTE OF TECHNOLOGY
ROURKELA
CERTIFICATE
This is to certify that the project entitled “DESIGN OF WATER TANK” submitted by Miss Nibedita Sahoo [Roll no. 10401010] in partial fulfillment of the requirements for the award of bachelor of technology degree in Civil engineering at the National Institute of Technology Rourkela (deemed University) is an authentic work carried out by her under my supervision and guidance.
To the best of my knowledge the matter embodied in the project has not been submitted to any other university/institute for the award of any degree or diploma.
Date: Prof. S.K Sahu
Department of Civil Engineering
National Institute of Technology Rourkela - 769008
ACKNOWLEDGEMENT
I would like to express my profound sense of deepest gratitude to my guide and motivator Prof. S.K. Sahu, Professor, Civil Engineering Department, National Institute of Technology, Rourkela for his valuable guidance, sympathy and co-operation for providing necessary facilities and sources during the entire period of this project.
I wish to convey my sincere gratitude to all the faculties of Civil Engineering Department who have enlightened me during my studies. The facilities and co-operation received from the technical staff of Civil Engineering Department is thankfully acknowledged.
I express my thanks to all those who helped me one way or other.
Last, but not the least, I would like to thank the authors of various research articles and books that I referred to.
Nibedita Sahoo Roll No 10401010 B. Tech 8th Semester
CONTENTS
Chapter No.
Title
Page
No.
CERTIFICATE i ACKNOWLEDGEMENT ii CONTENTS iii ABSTRACT iv LIST OF FIGURES v LIST OF TABLES vi 1 INTRODUCTION 1 1.1 OBJECTIVE 1 2 THEORY 22.1 DESIGN REQUIREMENT OF CONCRETE 4 2.2 JOINTS IN LIQUID RETAINING STRUCTURES 5 2.3 GENERAL DESIGN REQUIREMENTS 11 2.4 FLEXIBLE BASE CIRCULAR WATER TANK 19 2.5 RIGID BASE WATER TANK 21 2.6 UNDER GROUND WATER TANK 23
2.7 PROGRAMS 25
3 RESULTS AND DISCUSSION 42
3.1 DESIGN OF CIRCULAR TANK 43 3.2 DESIGN OF UNDERGROUND WATER TANK 46
4 CONCLUSION 48
ABSTRACT
Storage res ervoi rs and o verh ead t ank are us ed to st ore wat er, liq uid pet rol eum , pet ro leu m p rod uct s an d sim i lar li quid s. Th e fo rce an al ys is of the reserv oi rs or tanks is ab out th e s ame i rresp ecti ve of t he ch emi cal nat ure of th e p ro du ct. Al l t ank s are desi gned as crack free s tru ctu res t o elimi nat e an y l eak age.
This project giv es in b rief, th e th eo r y b eh ind t he d esi gn o f liq uid ret ai nin g struct ure (circular wat er tank with flex ibl e and ri gi d bas e and rect an gul ar u nd er ground wat er tank) u s in g wo rki n g st ress meth od. This repo rt als o i ncl ud es com put er su brouti nes to anal yze and d es ign ci rcul ar wat er t ank with fl ex ible and ri gid b as e and rect an gul ar u nder groun d wat er tank. Th e pro gram h as b een writt en as M acro s in Mi croso ft Ex cel usin g Vis u al Basi c pro grammi n g l an gu age. In th e end , th e pro grams are vali dat ed wi th th e results o f m anu al calculation giv en i n “Concrete Stru ctu re” book.
LIST OF FIGURES
Figure No. Title Page
No.
2.1(a) Contraction joint with discontinuity in steel 6 2.1(b) Contraction joint with continuity in steel 6
2.2 Expansion joint 7
2.3 Sliding joint 7
2.4 Construction joint 8
2.5(a) Temporary open joint with prepared joint surface 8 2.5(b) Temporary open joint with joint sealing compound 9 2.5(c) Temporary open joint with mortar filling 9 3.1 Flexible base circular tank 44
LIST OF TABLES
Table No. Topic Page
No.
1 Permissible Concrete Stresses 12 2 Design of Circular Tank 43 3 Design of Under Ground Tank 46
CHAPTER 1
INTRODUCTION
Storage res ervoi rs and o verh ead t ank are us ed to st ore wat er, liq uid pet rol eum , pet ro leu m p rod uct s an d sim i lar li quid s. Th e fo rce an al ys is of the reserv oi rs or tanks is ab out th e s ame i rresp ecti ve of t he ch emi cal nat ure of th e p ro du ct. Al l t ank s are desi gned as crack free s tru ctu res t o elimi nat e an y l eak age. W at er o r raw p etrol eum retainin g sl ab and walls can be o f reinforced con cret e with ad eq uat e co v er to th e reinfo rcement. Water and p et rol eu m and react with co ncrete and , th erefo re, no sp ecial treatm ent to t he s urface is requi red. Ind ust rial wast es can also be coll ect ed and pro cessed in co ncret e tanks wit h few ex ception s. The pet rol eum prod u ct s u ch as p et rol, di es el o il, etc. are lik el y to l eak t hrou gh the co n cret e wall s, therefore su ch t an ks n eed s peci al m emb ran es t o prev ent l eak age. R eservoi r i s a co mmo n term ap plied to li quid s to rage stru ct ure and it can be bel ow or abov e th e grou nd l ev el . R eservoi rs belo w the grou nd level are normal l y built to sto re l arge q uantit ies o f wat er wh ereas thos e o f o verh ead t yp e are bu ilt for d irect dist ribu tio n b y gravit y flo w and are u su all y of smaller cap acit y.
1.1 OBJECTIVE
1. To m ak e a stud y a bout th e an al ys is an d desi gn o f wat er t an k s.
2. To make a stud y abo ut th e guid elin es fo r the d esi gn of li q uid ret ainin g stru ct ure accordin g t o IS Cod e.
3. To kn ow abo ut t he d esi gn phil oso ph y fo r the s afe an d econo mical desi gn of wat er t ank .
4. To d ev elop p ro grams for th e d esi gn of water tank of flex i ble b as e an d ri gi d b as e and th e un dergrou nd t ank t o av oid t he tedio us calcu latio ns. 5. In th e end , th e pro grams are v alidated with th e res ults of m an ual calculation gi ven in “Concrete Structure” book.
CHAPTER 2
2.1 DESIGN REQUIREMENT OF CONCRETE (I. S. I)
In wat er ret aini n g stru ct ure a dens e i mperm eabl e con cret e is requi red therefore, p roport ion of fin e an d cou rs e aggregat es t o cem en t sho uld be such as t o giv e hi gh qualit y co n cret e.
Concret e mix weak er th an M2 0 is n ot used. The mi nimum quantit y of cement i n the co n crete mix s hall b e not less th an 30 k N/m3.
The desi gn of t he co ncrete mix sh all be s uch t hat th e resul tan t con crete is suffi ci entl y imp ervi ous. Effi ci ent comp action preferab l y b y vib ration is ess ent ial . Th e p erm eabilit y of th e th o rou ghl y comp act ed co n cret e i s dep en dent on wat er cement ratio. In creas e i n wat er cement ratio in creas es perm eabilit y, whil e con crete wit h l ow wat er cem ent rati o is diffi cul t to com pact. Ot her causes o f l eakage in con crete are d efects s uch as segregatio n an d ho n e y combi n g. All joi n ts shoul d b e made wat er-ti ght as thes e are p otenti al s ources of leakage.
Desi gn of l iqui d retaini n g s tru ct ure is different from o rd inar y R .C.C, stru ct ures as it requ ires th at con cret e s h ould not crack and hen ce t en sil e stress es in concret e s hould b e withi n p ermissibl e limits .
A rei nforced con cret e m emb er o f liquid ret ainin g st ru ct ure is desi gn ed on the usu al prin cipl es i gno rin g t en sil e resistan ce of con cret e in b en din g. Addit ion all y it shoul d be ens ured th at ten sile st ress on th e liq uid ret aini n g face o f th e equi valent co n cret e s ection does not ex ceed th e permiss ibl e tens ile stren gth of concret e as giv en in t abl e 1. Fo r cal cul ati on pu rp os es the cov er is als o t ak en int o con crete area.
Crack in g m a y be caus ed du e to res traint to sh rin k age, ex pan sion an d con tractio n o f concret e du e to t emp eratu re o r sh rin k age and swellin g due to moi stu re effect s. Such rest raint m a y b e caus ed b y –
(i) The int eracti on b etween rein fo rcemen t and co ncret e d uri n g sh rin kage due to dr yin g.
(ii ) The bo undar y co nditio ns.
(iii ) The di fferenti al co nditi ons p revailin g th rou gh th e l arge thickn ess of massi ve concret e.
Us e o f sm all size bars p laced prop erl y, l eads t o clos er crack s b ut o f small er widt h. Th e risk of cracki n g du e t o t emp erat ure an d sh rin kage effects ma y be mi ni mized b y lim itin g t h e ch an ges in moi stu re co nt ent and temp erature to which the st ru ctu re as a whol e is subj ected. The ri sk of cracki n g can also b e mini mized b y red uci n g th e restrai nt on th e free ex pan sio n of t h e st ruct ure with lo n g walls o r slab fo und ed at or b elo w gro und level , restrai nt can b e minim ized b y t he p rovi sion of a sli din g la yer. This can b e p rovi ded b y foundi n g the st ru ct ure o n a flat l a yer of con crete wit h i nterpositi on of som e mat eri al to break t he bon d and facili tat e movem ent .
In cas e l en gth of st ruct ure i s l arge it sh ould b e su bdi vid ed i nto s uit abl e lengths separated b y m ov em ent j oint s, esp eci all y wh ere sectio ns are ch anged th e m ov em ent joi nts sho uld b e p rovi ded.
Where struct ures h ave t o sto re h ot liq ui ds, st ress es caus ed b y di fferen ce in temp eratu re b et ween i nsid e an d o utsid e of t h e res ervoi r sho uld b e t ak en into accoun t.
The co effi ci ent of ex pansio n du e to t em peratu re ch an ge is t ak en as 11 x 10- 6/° C and coefficient of shrinkage may be taken as 450 x 10- 6 for i niti al shri nk age an d 2 00 x 10- 6 for d r yi n g s hrin kage.
2.2 JOINTS IN LIQUID RETAINING STRUCTURES
2.2. 1 MOVE ME NT JOINTS . There are t hree t yp es o f m ov em ent join ts.
(i)C ontra cti on J oint . It is a mo vem ent j oint with delib erat e disconti nuit y witho ut i niti al gap b etween th e co ncrete on eit her sid e of th e j oint. The purpos e of this joi nt is to accomm od ate contractio n o f th e con cret e.
Fi gure 2.1(a)
A cont raction joi nt ma y b e eith er com p let e con tractio n join t or p arti al con tractio n join t. A com plet e co ntractio n joint is on e in whi ch both st eel and con cret e are int errupt ed and a parti al con tractio n joi nt i s o ne in whi ch onl y th e con cret e is int erru pted, t he rei nfo rcin g st eel run ni n g th ro u gh as shown i n Fi g.2. 1(b).
Fi gure 2.1 (b )
(ii )Exp ansion Joint . It is a j oint wit h co mplet e dis co ntin uit y in bo th rei nforci n g steel and co ncret e and it i s to accommo date eith er ex pansi on or co ntractio n o f th e stru ct ure. A t ypi cal ex pansio n j oint is s ho wn i n Fi g.2. 2
Fi gure 2.2
This t yp e of joi nt requires th e provisi on of an init ial gap bet ween th e adj oinin g parts of a s tru ctu re whi ch b y cl osin g o r o penin g accommod at es the ex p ansi on or con tracti on of th e struct ure.
(iii ) Slid ing Joint. It is a j oint wit h comp let e di sconti nuit y in both rei nforcem ent and co ncrete and with speci al provisi on t o facil i tat e mov ement i n pl an e o f th e join t. A t yp i cal joint is sho wn in Fi g. 2 .3
Fi gure 3.3
This t yp e o f jo int i s provided b et ween wall an d floo r in s om e c yl ind ri cal tank desi gns .
2.2. 2. CO NTRACT I ON JO INTS
This t yp e o f jo int i s provided for con ven ien ce in cons tructi on .
mov ement. One appli cation o f th es e join ts is b et ween su ccessi ve lift s in a reserv oir wall . A t yp ical jo int is s ho wn i n Fi g.3 .4.
Fi gure 3.4
The nu mb er of jo ints sho uld be as sm all as po ssibl e an d t hese joints shoul d b e k ept from possi bilit y o f p ercol ation o f water.
2.2. 3 TE MPO RARY JOI NTS
A gap is som etim es l eft t empo raril y between t he con cret e of adjoini n g parts of a st ru ctu re whi ch aft er a suit abl e i nterv al and befo re th e st ru ctu re is pu t to us e, is fill ed wit h m ort ar o r co n cret e com pl etel y as i n Fi g. 3.5 (a) or as sh own in Fi g. 3. 5 (b) and (c) wit h s u itabl e join tin g materi als. In t he first cas e width o f th e gap sh oul d b e suffi ci ent t o allo w t h e si d es t o b e prep ared b efore fil li n g.
Fi gure 3.5 (b )
Fi gure 3.5 (c)
2.2. 4 S PACING O F JOINTS
Unl ess alt ernativ e effecti ve mean s are taken to avoi d cracks b y allo win g fo r th e additi on al stress es th at m a y be ind uced b y t em peratu re o r shri nk age chan ges o r b y u nequ al s ettl ement , mov em ent joi nts sho uld b e pro vid ed at th e foll o win g spacin g:-
(a) In rein fo rced co n cret e fl oo rs, mov em ent joi nts sho uld b e spaced at n ot more th an 7.5 m apart in two di recti ons at ri gh t an gl es. Th e wall and fl oor joints sho uld b e i n l ine ex cept wh ere sli din g j oints o ccur at the base of the wall i n whi ch correspo nd en ce is not s o imp ort an t.
(b )For floo rs wit h o n l y nomin al p ercentage o f rein fo rcem ent (s mall er th an the mini mum sp ecifi ed ) the con crete flo or sho uld b e cast in pan els wit h sides n ot m ore t han 4.5m.
(c) In con cret e walls, the mov em ent join t s should n orm all y b e placed at a maximum s paci n g o f 7 .5m. in rein fo rced walls and 6m in unrein fo rced wal ls. The m ax imu m len gth d esi rabl e bet ween verti cal mo vem ent j oints will depend upo n t h e t en sile st ren gt h o f th e wall s, an d m a y be in creas ed b y suit abl e rein fo rcement. When a s lidin g l a yer is p laced at the fou nd atio n of a wall, the len gth o f th e wall that can be k ept free o f cracks dep en ds o n the capacit y of wall s ection to resis t th e fri ctio n indu ced at the pl an e o f slid in g. App rox imat el y t he wal l has to st and th e effect o f a fo rce at th e pl ace o f sli din g equ al t o wei ght o f h al f t he len gth of wall multipli ed b y th e co -effi ci ent of fri ctio n.
(d )Am on gst th e mov em ent joi nts in flo ors and walls as menti oned above ex pan sio n join ts s ho uld n orm all y b e p ro vid ed at a sp acin g of n ot mo re than 30 m b et ween s uccessiv e ex p ansi on joint s o r b et ween t he en d o f t he stru ct ure and the n ex t ex pan sion j oint ; all oth er join ts b ein g of the con struction t yp e.
(e)When, ho wever, the t emp erature chan ges to be accom mod ated are abn orm al or occu r m ore frequ en tl y th an usu al as in t he cas e of st orage o f warm liqui ds or in unins ul ated roo f sl abs, a smaller sp acin g th an 30m shoul d b e adopt ed th at i s greater propo rti on o f mov em ent joi n ts sh oul d be of th e ex pansi on t ype). Wh en th e ran g e of t emp eratu re i s small, fo r ex ampl e, i n cert ain cov ered st ructu res, or wh ere res train t i s smal l, for ex ampl e, i n certai n elevated structu res non e of t he mo v ement j oints pro vid ed in sm all st ru ctu res up to 45m l en gth need be o f th e ex pansio n t yp e. Wh ere sl idin g joints are provi ded bet ween th e wall s and eith er th e floo r o r roo f, th e p rovisi on o f mov em en t joints in each el ement can b e con sid ered independ entl y.
2.3 GENERAL DESIGN REQUIREMENTS (I.S.I)
2.3. 1 Pl ain Concrete Stru ctu res. Pl ai n con crete mem ber o f rein fo rced
con crete liq uid ret ainin g st ru ctu re m a y be d esi gn ed agai n st st ru ctu ral fai lu re b y allo win g t ensi on i n pl ain con cret e as p er th e p ermi ssibl e limit s fo r tensio n in bendin g. Th is wi ll autom at icall y t ak e care o f failure d ue t o cracki n g. However, nomin al rei nforcem ent sh all b e provid ed, fo r pl ain con crete st ru ctu ral m emb ers.
2.3. 2. Permiss ible S tres ses in Con crete.
(a ) Fo r resi stance to cra cking. Fo r cal culati ons relat in g to t he res ist an ce
of mem bers to crackin g, th e p ermi ssibl e s tress es in tensi o n (direct and due to b endin g) and shear s hall co nfi rm t o th e v alu es specifi ed in Tabl e 1. The p ermissi ble t en sile stres ses d ue to ben din g appl y to th e face of t he memb er i n cont act with th e liqu id. In m embers less t han 2 25 mm. th ick and in co ntact with li qu id on on e si de t hese permissi bl e stres ses in ben din g app l y als o to th e face remot e from t he liq uid.
(b) For s tren gth calcul ati ons. In stren gth cal cu lati ons th e permi ssibl e
con crete st ress es s hall b e in acco rd ance wit h Table 1 . Where th e calculated s hear st ress in con cret e al on e ex ceeds th e p ermis sibl e v alu e, rei nforcem ent act in g in co njun cti on with diagon al com pression in t he con crete shall b e p ro vid ed t o t ak e t he wh ole of th e sh ear.
Table 1. Permissib le con crete s tress es in ca lcul ati ons rela tin g to resis tan ce to cra cki ng
Perm issi ble st ress i n KN/m^2 tensi on Grad e of con cret e
Direct Bend in g shear M15 1.1 1.5 1.5 M20 1.2 1.7 1.7 M25 1.3 1.8 1.9 M30 1.5 2.0 2.2 M35 1.6 2.2 2.5 M40 1.7 2.4 2.7
2.3. 3 Permissibl e S tress es in Steel (a ) For res istance to cra ckin g.
When st eel an d co ncrete are ass um ed to act to get h er for ch eckin g t he
tens ile st ress in co n cret e for avoi dance of crack, t h e t ens ile st ress i n s teel will b e limit ed b y th e requi rem ent th at th e permi ssibl e t ensil e stress in the con crete is not ex ceeded s o th e tensil e st ress in s teel s hal l be equ al to t he pro du ct of modul ar rat io o f st eel and con crete, and th e correspo ndin g allo wabl e t ensile st ress in co ncret e.
(b) Fo r strength cal cula tions .
In stren gth cal cul ati ons t he p ermissi bl e s tress sh all be as foll ows:
(i) Tensil e st ress in memb er in di rect t en sion 1000 k g/ cm2 (ii ) Ten sil e st ress in memb er in b en din g o n
l iqui d retai nin g face o f mem bers o r face awa y from
l iqui d fo r m emb ers l ess th an 225 mm thick 1000 k g/ cm2 (iii )On face awa y from liq uid fo r mem bers 225mm o r m ore
i n t hickness 1250 k g/ cm2 (iv )Tensil e st ress in sh ear rein fo rcem en t,
Fo r m em bers 22 5mm or m ore in t hickness 1250 k g/ cm2 (v )Comp ressi ve st res s in co lumn s s ubj ect ed to direct l oad 1250 k g/ cm2
2.3. 4. S tress es du e to drying Sh rinka ge or T emperatu re Ch ange.
(i)St ress es du e to d ryi n g s hri nk age o r temperatu re ch an ge m a y b e i gno red pro vid ed th at –
(a) Th e p ermi ssib le stres ses sp eci fi ed abo ve in (ii) and (iii) are not oth erwis e ex ceeded.
(b ) Ad equ at e precau tions are tak en to avoid crackin g o f co ncrete du ri n g the const ru cti on peri od and un til t he res ervoi r is p ut i nto us e.
(c) Reco mmend atio n regardi n g join ts giv en in arti cl e 8. 3 an d for su itabl e sliding l a yer beneat h the res ervoi r are compli ed wit h, o r the res ervoi r is to b e us ed onl y fo r the s to rage of wat er or aq u eous liqu i ds at or n ear ambi ent temp eratu re and th e circum st an ces are s u ch th at t he con crete will nev er d r y ou t.
(ii )Shrink age st ress es ma y ho wev er b e required t o be calculat ed in sp eci al cases, wh en a sh rin k age co -effi cient of 3 00 x 10- 6ma y b e as su med .
(iii ) When t he s hri n kage st ress es are al lowed, t he p ermiss i ble st ress es, tens ile stres ses to concret e (d irect and ben din g) as gi ven in Tab le 1 ma y be in creas ed b y 33. 3 3 p er cen t.
2.3. 5. Flo ors
(i)Provi sion of mov emen t join ts.
Movem ent joi nts sh ould b e p ro vid ed as discuss ed in arti cl e 3 .
(ii ) Fl oo rs of tanks restin g on g round.
If t he t ank is res tin g di rectl y ov er gro u nd, fl oo r ma y b e co nstructed of con crete wit h n omi nal percent age o f rein fo rcem en t p rov id ed t hat it i s cert ain t hat the grou nd will carr y t he l o ad with out app reciabl e su bsid en ce in an y p art and t hat the co ncret e fl oo r is cast i n p an els wi th sid es not more th an 4.5m . wi th co nt racti on o r ex pan sion joint s b etween. In s uch cases a screed or concrete l a yer l ess t han 75mm t hi ck shall first be placed
on t he grou nd and covered wit h a slidi n g l a yer o f bitum en p ap er o r oth er suit abl e m at eri al t o des tro y th e bon d bet ween th e s creed an d floo r con crete.
In no rm al ci rcums tances th e s creed la yer s hal l be o f grade not weak er than M 10, where in juri ous so ils o r agg ressiv e wat er are ex pect ed , th e screed la yer s hall b e o f grad e no t weak er th an M 15 an d if necess ar y a sulph at e resist in g or oth er sp ecial cem en t shoul d b e us ed.
(iii ) Floo r of tanks restin g on suppo rts
(a) If t he t an k is su pp ort ed o n walls o r ot h er s imil ar s upp ort s t he floo r sl ab shal l b e desi gn ed as flo or in buil din gs for bendin g m oments du e t o wat er load and s elf wei ght.
(b )When th e flo or i s ri gidl y co nn ected to th e wall s (as i s gen erall y t he case) the ben din g m oments at th e jun cti on b etween th e wall s and flo ors shal l be t ak en into accoun t in the d esi gn of fl oo r togeth er wit h an y di rect fo rces t rans ferred to the floor from th e wal ls o r from th e fl o or to th e wall due to sus pensio n o f the fl oo r from th e wall.
If t he wal ls are no n-mo nolit hi c wit h the floo r sl ab, su ch as in cas es , wh ere m ov ement jo i nts h av e been provi ded b etween t he flo or sl abs and wal ls, th e floor sh all be d esi gn ed onl y for th e v erti cal lo ads on th e floo r. (c) In conti nuous T-b eams and L-b eams with rib s on th e sid e remot e from the liqui d, the t ensi on in concret e on t he l iqui d si de at th e face o f t he support s sh all n ot ex ceed th e p erm issi ble stress es fo r cont roll i n g cracks in con crete. Th e width of th e slab s hall b e determin ed in us ual mann er for calculation o f the resist ance to crackin g o f T-b eam, L-b eam section s at support s.
(d )Th e fl oo r slab ma y b e suit abl y ti ed to the walls b y rod s pro perl y embed ded in bot h th e sl ab and th e walls . In su ch cases no s eparate b eam (curved o r st rai ght ) is necess ar y under t he wal l, p ro vid ed th e wall of t he tank its el f i s d esi gn ed to act as a b eam o v er th e s upp orts u nd er it.
(e)Som etim es it ma y be econom ical t o provid e th e floo rs o f ci rcul ar t an ks, in the s hape of dom e. In such cas es the dom e s hall be desi gned for the
verti cal l o ads o f th e liquid o v er it an d t he rat io o f its ris e t o its di am eter shal l be so adjus ted that th e st res ses in the dom e are, as far as pos sibl e, wholl y com pres siv e. The d ome sh all b e s u ppo rted at its bott om on th e rin g beam whi ch sh all b e des i gn ed for res u ltant circu mferenti al tensio n i n add ition to v erti cal l oad s.
2.3. 6. Wall s
(i)Provi sion of joints
(a)Where it i s d esi red to al low th e walls to ex pan d o r co nt ract sep arat el y from t he floo r, or t o p rev ent mom ents at t he b as e of th e wall owi n g to fix it y to th e flo or, sl idin g joi nts m a y b e emplo yed .
(b )Th e sp acin g of verti cal m ov ement j oints sh oul d be as discuss ed in art icl e 3 .3 while t h e maj ori t y o f these joints m a y b e of t he p artial or com plet e con tractio n t yp e, su ffici en t join ts of th e ex pan sio n t yp e sho uld be provi ded to s atis fy th e requi rem ents gi ven in arti cl e
(ii )Pressu re on Wall s.
(a) In liqu id ret aini n g st ru ctu res wit h fi x ed or fl oat in g cov ers the gas press ure d ev elop ed abo ve li quid s urface sh all b e ad ded to the liq uid press ure.
(b )When th e wall of liqui d ret ai nin g st ruct ure i s bu ilt i n ground , o r has eart h em bank ed agai nst it, t he effect o f eart h pres su re sh all be t ak en i nto account .
(iii ) W alls o r Tanks Rectan gula r o r Pol ygonal in Plan.
While d esi gnin g th e wal ls of rect an gul ar or pol ygo nal con cret e tan ks, th e foll owi ng point s s ho uld b e born e i n mi nd .
(a) In pl ane wal ls, t he li quid pres sure i s resist ed b y b oth verti cal an d horizo nt al b end ing moments. An est imat e sh ould be mad e o f the pro po rtio n of th e pressure resist ed b y b en din g m om ents in t he verti cal and horizo nt al pl an es. The di rect h orizo nt al tens ion caus ed b y the direct pull due t o water pressu re on th e end wall s, shoul d b e ad ded to t hat resulti n g from horizont al bending mom ent s. On l iquid ret aini ng faces, t he t ensil e
stress es du e to th e combin ati on o f direct horizont al t en sion and b endi n g action sh all s atis f y t he fol lo win g condi ti on:
(t’/t )+ ( óc t’ / óc t ) ≤ 1
t’ = calcul ated d irect t ensil e st ress in co ncrete
t = permis sibl e d irect tens ile st ress i n con cret e (Table 1 )
ó ′c t = calcul ated t ens ile s tress du e t o b en din g in con cret e.
óc t = permis sibl e t en sil e stres s d ue to bendi n g in con cret e.
(d )At th e verti cal ed ges where th e walls of a res ervoi r are ri gi dl y joi ned, horizo nt al rei nforcement and h au nch b ars sho uld be provi ded to resi st th e horizo nt al b endi n g moments ev en i f th e wall s are desi gned to with st and the wh ol e load as v erti cal b eams or cant i lev er witho ut l at eral supp ort s. (c) In t he cas e o f rect an gul ar or pol ygo nal tan ks, t h e sid e walls act as t wo -wa y sl abs , whereb y the -wall is co ntin ued or rest rain ed i n th e ho rizo ntal direction , fixed or hi n ged at t he b ottom and hin ged or free at the to p. The wal ls th us act as thin pl at es s ubj ect ed tri an gul ar lo adin g an d wi th boun dar y condit ions var yi n g b et ween full rest raint an d free ed ge. The an al ys is o f mo ment and forces m a y be m ad e o n th e b asis of an y reco gniz ed m et hod .
(iv ) W alls of Cylind ri cal T anks.
While desi gni n g walls of c yl ind ri cal tan ks th e follo win g poi nts sh ould be borne in mind:
(a)Walls of c ylin dri cal t an ks are eith er cast m ono lithi call y with th e base or are set in gro ov es and k e y wa ys (mov ement jo ints ). In eit her cas e defo rm ation of wall und er i nflu en ce of l i quid p ress ure is rest ricted at an d abo ve th e b as e. C on seq uentl y, o nl y p art of th e t ri an gular h yd rost ati c lo ad will b e carri ed b y ring t ens ion and p art of th e lo ad at bot tom wi ll be support ed b y cantil ever action .
(b )It is di fficult to rest ri ct rot ati on o r settlem ent o f th e base slab and it is adv isabl e to provi de verti cal rein fo rcem ent as i f the walls were fu ll y fix ed at t he bas e, i n addi t ion t o the reinforcement required to res ist horizontal
rin g t en sion for hin ged at base, co nditi on s of walls , unl ess th e app rop ri at e amount o f fix it y at the b as e is est ablish ed b y an al ys is with due con sid erati on to th e dimensio ns of t he b ase sl ab th e t yp e of j oint b etween the wall and sl ab , and , wh ere appli cabl e, th e t yp e o f soil s uppo rti n g the bas e slab.
2.3. 7. Ro ofs
(i) Pro vision of Mov emen t join ts.
To av oid t he po ssib ilit y o f s ym p ath eti c cracki n g it is im port ant t o ens ure
that mo vem en t jo int s in th e ro of corres pond wi th t hos e in the walls , i f roo f an d walls are m onolit hi c. It, ho wever, p ro visi on i s m ad e b y m eans of a sli ding join t for m ov ement b et ween th e roo f and th e wal l co rres pon dence of jo ints i s n ot s o im portant.
(ii )Loadin g
. Fi eld co v ers o f li qu id ret ai nin g st ru ctu res sh ould b e desi gn ed fo r grav it y loads, such as the wei ght of roo f slab, eart h co ver if an y, li ve loads and mech ani cal equi pm ent. Th e y s ho uld als o be d esi gn ed fo r u p ward lo ad if the liqu id retainin g s tru ctu re is su bject ed to int ern al gas p ress ure.
A sup erfi ci al lo ad su ffi ci ent t o en su re s afet y with th e un eq ual intensit y o f loadin g which occurs du ring th e pl aci n g o f t he earth cov er s houl d be allo wed fo r i n d esi gn in g roo fs .
The engin eer should speci f y a lo adi n g u nder th es e t emp orary co nditi ons whi ch sho uld not be ex ceeded. In desi gn i n g th e ro of, allo wan ce sh ould b e made fo r th e t empo rar y cond itio n o f so me sp ans lo ad ed an d oth er sp ans unlo ad ed , even tho u gh i n t he fi n al st at e the lo ad ma y b e sm all and evenl y distribut ed .
(iii )Wa ter ti ghtn ess . In cas e o f t an ks i ntend ed fo r th e st orage o f wat er
fo r dom esti c p urpos e, th e roo f mu st b e mad e wat er-ti ght. This m a y be achi ev ed b y limiti n g the st ress es as fo r t he rest o f th e t ank , or b y th e us e of th e cov erin g of t he wat erp roo f m em bran e o r b y p ro vidi n g slo pes to ens ure ad eq u ate d rai nage.
(iv ) Pro tection agai nst co rrosion . Pro tectio n meas ure s hal l be p ro vid ed
to the un dersi d e of the roo f to prev ent it from corro s ion du e to con dens ation.
2.3. 8. Mini mu m Rei nforcemen t
(a)The mi nimum rei nfo rcement in walls, floo rs an d ro ofs in each of two direction s at ri gh t an gl es sh all hav e an area o f 0 .3 p er cent o f the co n cret e sectio n in th at d irect ion for s ectio ns up t o 100mm , thi ck ness. Fo r s ection s of t hickn ess great er than 10 0mm, and less th an 4 50mm th e min imum rei nforcem ent in each of th e two di recti o ns shall b e lin earl y redu ced from 0.3 percent fo r 10 0 mm thi ck s ection t o 0. 2 percent fo r 4 50mm, thi ck sectio ns. Fo r secti ons o f thickn ess greater th an 45 0m m, minimum rei nforcem ent i n each of th e t wo directio ns sh all b e k ept at 0 .2 p er cent. In con cret e s ectio n s of th ickn ess 2 25 mm or great er, t wo la yers o f rei nforcem ent st eel shal l b e p laced on e near each face o f th e s ectio n to make up t he mini mu m rein fo rcem ent.
(b )In sp ecial ci rcu mstances floo r sl abs ma y be co nst ruct ed with percen tage o f rei nforcement less th an speci fi ed abov e. In no case t he percen tage of rein fo rcem ent i n an y m em ber b e l ess th an 0°15% of gross sectio nal area o f t he member.
2.3. 9. Mini mu m Co ver to Reinforcement.
(a)Fo r liq uid faces of p art s of m emb ers eith er in co ntact wi th the li quid (such as in ner faces or roo f sl ab) t he mi n imum cov er to all reinfo rcement shoul d b e 25mm o r the di am et er of t he main b ar wh ichever is grat er. In the p resence o f t he sea water an d soils and wat er o f corros i ve characters the cov er should be i ncreas ed b y 12mm but this add ition al co ver sh all n ot be taken in to accoun t fo r desi gn cal cu lat i ons.
(b )For faces awa y from liq uid and for parts o f the st ru ctu re nei th er in con tact with the liq uid o n an y face, nor en closi n g t he sp ace abo ve t he liquid , t he cov er sh al l b e as fo r o rd in ar y c oncret e m emb er.
2.4 FLEXIBLE BASE CIRCULAR WATER TANK
Fo r smaller cap aciti es rect an gular tanks are us ed and for bi gg er capaciti es circular t ank s are u sed . In ci rcul ar t an k s with fl ex ible j oint at th e b as e tanks wall s are s ubj ected to h yd ro stat ic press ure . so t he tank wall s are desi gn ed as th in c yl i nder. As the ho op t ension gradu all y redu ces t o zero at top, t he rein fo rcem ent is gradu all y red uced t o minim um rei nforcement at top. The m ain rei nfo rcement co nsist s of ci rcul ar h oop s. Verti cal rei nforcem ent equ al to 0.3% o f concret e are is p rov id ed and h oop rei nforcem ent is ti ed to t his rein fo rcem en t.
STEP 1
DETERM INAT ION OF D IAMETER OF THE WATER TANK Diamet er=D=√(Q * 0.004) / ((H - Fb) * 3.14)
Where Q=cap acit y of th e wat er tank H=hei ght o f th e wat er t ank Fb =free b oard o f t he water tank
STEP 2
DES IGN OF DOM E SHAPED R OOF Thi ckn es s o f d om e = t=100mm Li v e lo ad = 1 .5KN/m2
Self wei ght of dom e = (t / 10 00 ) * uni t wei ght o f co ncret e Fin ish es lo ad = 0 .1 KN/m2
Tot al l oad = li ve lo ad + s elf wei ght + fi n ishes lo ad Cent ral ris e= r =1m
Radi us o f dom e = R = ((0. 5 * D) ^ 2 + r ^ 2) / (2 * r) cos A = ((R - r) /R)
Meridi on al t hru st = (total lo ad * R ) / (1 + co sA)
Circum ferenti al th ru st = tot al lo ad * R * (cos A - 1 / (1 + co sA)) Meridi on al st ress = meridio nal th ru st / t
Hoop st ress = ci rcum ferenti al th rust / t
Reinfo rcem en t in bo t h di recti on = 0.3 * t* 10 Hoop t ensi on = meri dional t hrust * cos A * D * 0.5
Reinfo rcem en t in to p rin g beam =As_t op ringb eam h oop t ensi o n / Ts Cros s s ection area o f top ri n g beam
= (h oop t ensi on / PS T di rect ) - (m - 1) * As_t op rin gbeam
STEP 3
DETERM INAT ION OF HOOP RE INFOR CEMENT HTi = 0. 5 * (w * (H - i ) * D)
Asi = HTi / Ts
Where, HTi =h oop tensio n at a d ep th of i from t he top
Asi =ho op rein fo rcem ent at a d ept h o f i from th e t op
STEP 4
DETERM INAT ION OF THIC KNESS OF CYLINDR IC AL WALL HT = 0.5 * (w * H * D)
t = 0. 001 * (HT1 / P STdirect - (m - 1) * As) Where, t=thi ck n ess of th e wall
HT=ho op tensio n at t he base o f t an k
PSTdirect =p ermissi b le s tress du e to direct tensio n As =ho op rein fo rcem ent at b as e
STEP 5
DETERM INAT ION OF VERT IC AL R EINFORC EMENT Asv = (0.3 - 0.1 * (t - 1 00 ) / 350 ) * t * 10
Where, Asv = verti cal rein forcement o f th e wall t=thi ck n ess of th e wall
STEP 6
DES IGN OF BAS E
Thi ckn es s o f b as e =1 50mm
2.5 RIGID BASE CIRCULAR TANK
The d es i gn o f ri gid bas e ci rcul ar t ank can be d on e b y th e app rox imat e meth od. In t his m eth od it is ass um ed th at some po rtio n of th e tan k at base acts as cantil ever an d thu s s ome lo ad at bottom are t ak en b y the canti lever effect. Lo ad i n th e top po rtio n is taken b y th e ho op t ensi on. Th e cantil ev er effect wi ll depend o n the d imen sion of th e t ank and t he thickn ess of th e wal l. For H2/Dt b et ween 6 to 12 , th e canti l ev er po rtion ma y b e ass um ed at H/3 or 1m from b as e whi ch ev er i s mo re. Fo r H2/Dt bet ween 6 to 12 , th e cantil ev er po rtio n ma y b e assum ed at H/ 4 or 1m from bas e whi ch ever i s m ore.
STEP 1
DETERM INAT ION OF D IAMETER OF THE WATER TANK Diamet er=D=√(Q * 0.004) / ((H - Fb) * 3.14)
Where Q=cap acit y of th e wat er tank H=hei ght o f th e wat er t ank Fb =free b oard o f t he water tank
STEP 2
DES IGN OF DOM E SHAPED R OOF Thi ckn es s o f d om e = t=100mm Li v e lo ad = 1 .5KN/m2
Self wei ght of dom e = (t / 10 00 ) * uni t wei ght o f co ncret e Fin ish es lo ad = 0 .1 KN/m2
Tot al l oad = li ve lo ad + s elf wei ght + fi n ishes lo ad Cent ral ris e= r =1m
Radi us o f dom e = R = ((0. 5 * D) ^ 2 + r ^ 2) / (2 * r) cos A = ((R - r) /R)
Meridi on al t hru st = (total lo ad * R ) / (1 + co sA)
Circum ferenti al th ru st = tot al lo ad * R * (cos A - 1 / (1 + co sA)) Meridi on al st ress = meridio nal th ru st / t
Hoop st ress = ci rcum ferenti al th rust / t
Hoop t ensi on = meri dion al t hrust * cos A * D * 0.5
Reinfo rcem en t in to p rin g beam =As_t op ringb eam h oop t ensi o n / Ts Cros s s ection area o f top ri n g beam
= (h oop t ensi on / PS T di rect ) - (m - 1) * As_t op rin gbeam
STEP 3
Assu me th e t hickn es s o f the wall =t = 0.1 5m Fin d t he v alue of H ^ 2 / (D * t )
(i) 6 < H ^ 2 / (D * t ) < 1 2
Cantil ever hei ght =H/ 3 o r 1m (whi ch ev er is mo re) (ii ) 12 < H ^ 2 / (D * t) < 30
Cantil ever hei ght =H/ 4 o r 1m (whi ch ev er is mo re)
STEP 4
DETERM INAT ION OF RE INFORC EMENT IN WA LL Max imum hoo p t ens i on=pD/ 2
Where, p=w* (H-cantil ev er hei ght) w=unit wei ght o f water
Area o f steel req uired=m ax imum hoo p t ension /ós t STEP 5
DETERM INAT ION OF RE INFORC EMENT IN CANT ILEVER HE IGHT Max imum bendi n g m oment = 0 .5 * (w * H) * (cantil ev erht ^ 2) / 3 Effecti ve d ept h=t -4 0 mm
Area o f steel req uired=m ax imum bendin g mom ent/ (j* effectiv e depth * ós t) STEP 6
DETERM INAT ION OF D ISTR IBUT ION STEE L IN WALL
Dist ribu tion st eel p rovid ed = (0.3 - 0. 1 * (t - 100 ) / 3 50) * t * 10
STEP 7
DES IGN OF BAS E
Thi ckn es s o f b as e =1 50mm
Minimum rei nforcem ent req uired =(0. 3/1 0 0)* 150 *10 00mm2
2.6 UNDER GROUND WATER TANK
The tanks l ik e pu ri ficati on t an ks, Imh off tanks , s epti c t anks, an d gas hold ers are bu ilt un dergrou nd. The d esi gn p ri nci pl e o f und ergrou nd tank is sam e as fo r t ank s are su bject ed to i nt ernal water p ressu re and o utsid e eart h p ress ure. Th e b as e is s ubj ected to wei ght of wat er and s oil p ress ure. These t ank s m a y be cov ered at t h e to p.
Whenev er there is a pos sibili t y o f water t ab le to ris e, s oil beco mes satu rat ed and eart h p ressu re ex ert ed b y s aturat ed s oil sh ould b e t ak en int o con sid erati on.
As th e ratio o f the l en gth of t ank to its breadth is great er th an 2, th e lon g wal ls wi ll b e d esi gn ed as canti levers an d the t op p ort ion of t h e sh ort walls will b e desi gned as s lab su pp ort ed b y lon g wal ls. Bot tom on e met er o f the sho rt wal ls will be d esi gned as canti lever slab.
STEP 1
DETERM INAT ION OF D IM ENS ION OF THE TANK Assu min g len gt h is equal to th e t hree tim es of b readt h. Area o f t he t ank = Q / H
B =√ (area of tank / 3) L=3 B
STEP 2
DES IGN OF LONG WALLS
1.fi rst cons id eri n g t h at press ure of s atu rat ed soil acti n g from o utsid e an d no wat er p ress ure fro m insi de, calculate t he depth and ov er all depth o f the walls .
2. C al cul at e th e max imum b en din g mom ent at b as e of lon g wall.
3. C al cul at e th e area of st eel and p rovi de it o n th e out er face of t he wall s. 4. Now cons id eri n g wat er p ress ure actin g from ins id e and no earth
press ure acting from outsi de, cal culat e th e m ax imum water press ure at bas e.
6. C al cul at e th e area of st eel and p rovi de it o n th e inn er face of t he wall s. 7. Dist ribu tion st eel pro vided = (0.3 - 0. 1 * (t - 1 00 ) / 35 0) * t * 1 0
STEP 3
DES IGN OF SHORT WALLS
1. Bottom 1m acts as cantil ev er an d rem aini n g 3m act s as s l ab su ppo rt ed on lo n g walls. Cal culat e t he wat er p res sure at a depth of (H-1) m from top.
2. C al cul at e th e max imum b en din g mom ent at s upp ort and cen tre.
3. Calcul ate th e correspo ndin g area o f s teel req ui red and p rovid e o n the out er face o f s ho rt wall respectiv el y.
4. Th en th e short walls are desi gned fo r cond ition p res su re of s at urat ed soil actin g fro m o uts ide and no eart h p res sure from i nsi de.
STEP 4
Bas e sl ab is check agai nst upli ft.
STEP 5
2.7 PROGRAMS
2.7.1 Design of Flexible Base and Rigid Base Circular Tank
Sub circular_fl ex ible_ri gid ()
Dim Q As Dou bl e 'capacit y o f th e tank i n lt. Dim H As Dou bl e 'd ept h o f t he wat er t ank in m. Dim Fb As Doub le 'free bo ard o f th e t ank in m . Dim D As Int eger 'di am eter of th e t an k i n m. Dim d sq r As Dou bl e
Dim HTi As Doubl e ' max imum hoo p t ensi on at im fro m to p in N/m^2 Dim HT1 As Dou ble 'm ax imum ho op t en si on at 1m from t op
Dim Asi As Do ubl e ' area o f st eel req ui red at im from top Dim w As Dou bl e 'd ensi t y of wat er in N/ m^3
Dim Ts As Doubl e 't he permi ssibl e stres s in rein fo rcemn t Dim AST As Do ubl e 'all owabl e stres s in t ensi on
Dim fck As Do ubl e 'the comp ressiv e stren gth of con cret e
Dim PSTdi rect As Doubl e 'p ermi ssib le tension st ress direct i n N/mm ^2 Dim PSTb endi n g As Dou ble 'p ermis sibl e tens ion st ress b endin g
Dim m As Doubl e 'm odul ar rati o of co n cret e Dim t As Doubl e 'thi ckn es s o f wall
Dim Asv As Doubl e 'verti cal rei nforcem ent
Dim Asv f As Do ubl e 'v ertical rein fo rcem ent on each face Dim n As Int eger 'n o .of rows
Dim d iarein f As Int eger 'di ameter of th e rein fo rcem en t in mm Dim Sv As Int eger 's pacin g p ro vided per m.
Dim v erti calSv As In teger 's p aci n g provi d ed per m o n each face (v ertical ) Dim AraSv As Doub l e 'area o f o ne b ar
Dim ass um edt As Do ubl e 'ass um ed t hi ckn ess o f th e tank Dim Hsq rb yd t As Do ubl e
Dim m ax ht As Dou bl e 'max imum ho op t ension in ri gid base t ank desi gn Dim m ax htast As Do ubl e 'area of st eel required d ue to hoo p st ress
Dim s paci n g As In teger 'sp acin g o f st eel on b oth face du e t o h oop stres s Dim m ax bm As Dou ble 'm ax imum b en di ng mo ment i n cantil ever p ort ion Dim m ax bm ast As Doubl e 'area o f st eel requ ired for cant ilever p ort ion Dim d st As Doubl e 'distribut ion st eel
Dim p ercentrei nf As Dou ble 'p ercentage o f di stributi on st eel Dim d st_s paci n g As In t eger
Dim t _roo f As Dou bl e 'thi ckn es s o f ro of Dim li velo ad As Do u ble 'liv e lo ad on dom e Dim s el f_ wt As Do u ble 's el fwt o f d ome Dim fini sh es As Do u ble 'wt o f fini sh es
Dim t otal_lo ad As Doubl e 'tot al l oad o n dom e Dim r As Int eger 'central ris e of th e dom e Dim rad_ dom e As Doubl e 'radi us of dom e Dim cos A As Do ubl e
Dim m er_thrust As Dou ble 'm eridio nal t rust
Dim ci rc_th rust As Dou ble 'ci rcumferent ial t hrust Dim m er_st ress As Doubl e 'meridion al stress Dim h oop _st ress As Dou ble 'h oop st ress
Dim ast _dom e As Do ubl e 'area of st eel in dom e Dim h oop _t ensi on As Do ubl e 'ho op t ensi on in do me
Dim ast _top ringb eam As Doubl e 'area o f steel in top ri n g beam Dim d _b as e As Do ub le 'd epth o f b as e slab
Dim ast _sl ab As Do u ble 'min imum rein fo rcement p ro vid ed in s lab Dim d eff_wall As Doubl e 'effecti ve d ept h o f th e wall
Dim ast _cant As Dou ble 'area of steel p ro vid ed on cantil ev er p orti on w = 1 000 0
Sheet2 .Cel ls.Cl ear Sheet3 .Cel ls.Cl ear Q = Sh eet1. Cell s(2, 2) H = Sh eet1. Cell s(2, 3) Fb = Sh eet1.C ells (2 , 4) fck = Sheet1 .Cells (2, 5) m = Sh eet 1.C ells (2 , 6)
diarein f = Sh eet1 .Cells(2 , 7) AraSv = (3.1 41 * di arei nf ^ 2) / 4 If fck = 15 Th en PSTdirect = 1.1 PSTbendi n g = 1.5 Els eIf fck = 20 Th en PSTdirect = 1.2 PSTbendi n g = 1.7 Els eIf fck = 25 Th en PSTdirect = 1.3 PSTbendi n g = 1.8 Els eIf fck = 30 Th en PSTdirect = 1.5 PSTbendi n g = 2 Els eIf fck = 35 Th en PSTdirect = 1.6 PSTbendi ng = 2.2
Els eIf fck = 40 Th en PSTdirect = 1.7 PSTbendi n g = 2.4 End If
'desi gn o f fl exible base
dsq r = (Q * 0.004) / ((H - Fb ) * 3 .14 ) D = Sq r(dsq r)
Sheet2 .Cel ls(1, 1).Valu e = "D IAM ETER in m " Sheet2 .Cel ls(2, 1).Valu e = D
i = 0 n = 2 Asi = 0 increment: If i < H Th en HTi = 0. 5 * (w * (H - i ) * D) Asi = HTi / Ts Sv = AraSv * 1000 / Asi
Sheet2 .Cel ls(1, 3).Valu e = "AT DEPTH IN m FR OM TOP " Sheet2 .Cel ls(n, 3).Valu e = (H - i )
Sheet2 .Cel ls(1, 4 ). Valu e = "SPAC ING OF RE INFORCEM ENT PER 1m ON EAC H FACE in mm"
Sheet2 .Cel ls(n, 4).Valu e = S v * 2 i = i + 1
n = n + 1
GoTo in crement: Els eIf i >= H Th en
t = 0. 001 * (HT1 / P STdirect - (m - 1) * Asi ) Asv = (0.3 - 0.1 * (t - 1 00 ) / 350 ) * t * 10 verti calS v = AraSv * 100 0 / As v
End If
Sheet2 .Cel ls(1, 2).Valu e = "TH IC KNESS in mm " Sheet2 .Cel ls(2, 2).Valu e = t
Sheet2 .Cel ls(1, 5).Valu e = "VERT ICA L REINFORCEM ENT SPAC ING ON EAC H FACE in mm^2"
Sheet2 .Cel ls(2, 5).Valu e = verti calSv
Sheet3 .Cel ls(1, 1).Valu e = "D IAM ETER in m " Sheet3 .Cel ls(2, 1).Valu e = D
'desi gn o f ri gi d bas e tan k ass umedt = 150
Sheet3 .Cel ls(1, 2).Valu e = "TH IC KNESS in mm " Sheet3 .Cel ls(2, 2).Valu e = as sum edt
Hsq rb yd t = H ^ 2 / (D * assum edt ) If 6 < Hs qrb yd t < 12 Th en If H / 3 > 1 Then cantil ev erht = H / 3 Els eIf H / 3 <= 1 Th en cantil ev erht = 1 End If Els eIf 1 2 < Hsq rb yd t < 30 Th en If H / 4 > 1 Then cantil ev erht = H / 4 Els eIf H / 4 <= 1 Th en cantil everht = 1
End If End If
maxht = w * 2 * (H / 3) * (D / 2) maxhtast = m ax ht / Ts
spacin g = AraSv * 1 000 / m ax htas t
Sheet3 .Cel ls(1, 3).Valu e = "AT DEPTH IN m FR OM TOP " Sheet3 .Cel ls(2, 3).Valu e = (H - cantil ev erht )
Sheet3 .Cel ls(1, 4 ). Valu e = "SPAC ING OF RE INFORCEM ENT PER 1m ON EAC H FACE in mm"
Sheet3 .Cel ls(2, 4).Valu e = sp aci ng * 2 st = 15 0 cb c = fck / 3 m = 28 0 / (3 * cb c) k = (m * cbc) / (m * cb c + s t) j = 1 - k / 3 qcrack = 0.5 * k * j * cbc
maxbm = 0 .5 * (w * H) * (cantil ev erht ^ 2) / 3 deff_wall = assum ed t - 40
ast_ cant = m ax bm * 10 ^ 6 / (j * st * d eff_wall)
Sheet3 .Cel ls(5, 1).Valu e = "Ast in cant il ev er po rti on i n mm ^ 2" Sheet3 .Cel ls(5, 2).Valu e = ast _cant
'dist rib utio n st eel
percen t_ rei nf = 0. 3 - 0.1 * (ass umedt / 10 00 - 0 .1 ) / 0.3 5 dst = p ercent_rein f * 0.1 5 * 10 00 * 1 000 / 100
dst_s paci ng = AraSv * 1 000 / d st
Sheet3 .Cel ls(7, 1).Valu e = "D ISTR IBUTION S TEEL" Sheet3 .Cel ls(7, 2).Valu e = dst
Sheet3 .Cel ls(8, 1 ). Valu e = "SPAC ING OF RE INFORCEM ENT PER 1m ON EAC H FACE in mm"
Sheet3 .Cel ls(8, 2).Valu e = dst_ spacin g * 2 'desi gn o f d om e s hape roo f
t_ro of = 10 0 livel oad = 1.5
sel fwt = (t_ roo f / 10 00) * 24 finis h es = 0.1
total _lo ad = liv elo ad + selfwt + fini sh es r = 1
rad_d om e = ((0.5 * D) ^ 2 + r ^ 2) / (2 * r) cos A = ((rad _dom e - r) / rad_ dom e)
mer_th ru st = (tot al_ l oad * rad_ dom e) / (1 + co sA)
circ_th rust = tot al _lo ad * rad_dom e * (co sA - 1 / (1 + cosA)) mer_st ress = mer_t hrust / t_ roo f
hoop _st ress = ci rc_t hru st / t_ roof ast_ dom e = 0.3 * t_roof * 10
Sheet2 .Cel ls(10, 1 ). Val ue = "DES IGN OF R OOF" Sheet2 .Cel ls(11, 1 ). Val ue = "CENTRAL R ISE in m " Sheet2 .Cel ls(11, 2 ). Val ue = r
Sheet2 .Cel ls(12, 1 ). Val ue = "TH ICKNES S in mm " Sheet2 .Cel ls(12, 2 ). Val ue = t_ ro of
'desi gn o f t op ring
hoop _t ensi on = m er_ thru st * cos A * D * 0.5 ast_ top rin gb eam = h oop_t ension / Ts
ac_to pri n gbeam = (h oop_t ensio n / PS Tdi rect) - (m - 1 ) * As _t opri n gb eam Sheet2 .Cel ls(13, 1 ). Val ue = "RE INFORC EMNET IN DOM E i n mm^ 2 " Sheet2 .Cel ls(13, 2 ). Val ue = ast_ dom e
Sheet2 .Cel ls(16, 1 ). Val ue = "c/s AREA OF R ING BEAM in mm^2 " Sheet2 .Cel ls(16, 2 ). Val ue = ac_top ri n gb eam
Sheet2 .Cel ls(17, 1 ). Val ue = "R EINFORCEMENT IN R ING BEAM in mm^2 "
Sheet2 .Cel ls(17, 2 ). Val ue = ast_t op rin gb eam 'DES IGN OF BASE
d_b as e = 1 50
ast_ sl ab = (0 .3 / 10 0 ) * 15 0 * 10 00
Sheet2 .Cel ls(20, 1 ). Val ue = "DES IGN OF BASE" Sheet2 .Cel ls(21, 1 ). Val ue = "DEPTH OF S LAB in m " Sheet2 .Cel ls(21, 2 ). Val ue = d _b as e
Sheet2 .Cel ls(22, 1 ). Val ue = "RE INFORC EMENT in mm ^2 " Sheet2 .Cel ls(22, 2 ). Val ue = ast_ slab
2.7.2 Design of Underground Tank
Sub undergroun d_t an k() Dim Q As Dou bl e
Dim H As Dou bl e Dim an gle As Doub l e Dim d ensit y As Dou ble
Dim w_ wat er As Do ubl e 'unit wt of wat er Dim w_s oil As Do ub le 'unit wt o f s oil Dim area_t an k As Doubl e
Dim Fck As In teger ' ch aract eri sti c st ren gt h o f con crete Dim cb c As Int eger
Dim m As In t eger Dim k As Double Dim j As Doubl e
Dim q crack As Dou b le Dim L As Dou ble Dim B As Doubl e
Dim p As Double 'earth pres su re
Dim Ka As Doub le 'coeff o f eart h p ressu re
Dim m ax BM_lo n gwall As Doubl e 'm ax m B. M at b as e of lo n g wal l Dim m ax BM_lo n gwall_soil As Doub le
Dim d eff As Dou ble 'effectiv e depth req uired for wal l Dim av gd As Do ubl e 'average t hi ckn ess o f wall
Dim d As Int eger 'p rovid ed d epth o f th e wal l
Dim s teel_ lon g_in ner As Dou bl e 'area of steel provid ed on in n er sid e of lon g wall
Dim s teel_ lon g_o uter As Dou bl e 'area of steel provid ed on ou t er sid e of lon g wall
Dim m ax BM_s ho rt_ cent re As Dou ble 'b en din g mom ent at centre i n sh ort wal l
Dim m ax BM_s ho rt_s upport As Do ubl e 'b ending mom en t at s up port in sho rt wal l
Dim t _sh ort As Doub le Dim t _avlbl e As Do u ble
Dim T As Dou ble 't ension in sho rt wall
Dim s teel_ sho rt As Dou ble 'area of st eel al on g sho rt wall
Dim s teel_ sho rt_ sup port As Doub le 'area of st eel at s upp ort s hort wall Dim s teel_ sho rt_ cent re As Doub le 'area o f st eel at cent re s ho rt wall Dim d rct_ com prsn As Doubl e 'di rect com pressi on du e t o lo n g wal l Dim Leff As Dou ble
Dim Beff As Doubl e Dim wt_l on g As Dou ble Dim wt_s ho rt As Do ubl e Dim wt_b as e As Dou ble
Dim wt_ eart h_p roj ection As Do ubl e Dim u pward_pr As Doubl e
Dim d ownward_ pr As Do ubl e Dim fric_res As Dou ble
Dim s ubm rgd_ earth p r As Doubl e Dim t ot_ fri c_res As Dou ble Dim u p_p r_1m As Doubl e Dim s lf_ wt As Doubl e Dim n et _up _p r As Doubl e Dim wt_ wall_p roj As Doubl e Dim R As Doubl e 'reaction
Dim d _b as e As Do ub le 'thi ck nes s o f base
Dim s teel_ base_su pp ort As Do ubl e 'st eel in b as e Dim BM _ed ge As Do ubl e
Dim a As Do ubl e
Dim t ot_p r_ 1mwall As Dou bl e
Dim ass um ed_ d_ roo f As Do ubl e 'thi cknes s of roo f s lab Dim s el fwt As Dou bl e 'selfwt o f ro of sl ab
Dim li vewt As Do ubl e 'liv e l oad o n roo f s lab Dim fini sh es As Do u ble 'finis hes lo ad on roo f Dim t otal_lo ad As Doubl e
Dim m ax BM_ roo f As Do ubl e 'm ax m BM on roof slab Dim ast _roo f As Dou ble 'rei nforcem ent of ro of sl ab
Dim d st_ roof As Do ubl e 'dist ribu tion rei nforcement o f ro of sl ab Dim d _roo f As Dou b le
Dim d eff_roo f As Do ubl e
Dim b m_s ho rt_s upp o rt As Dou bl e Dim b m_s ho rt_ centre As Dou bl e
Dim as_ sho rt_sup po rt_out er As Dou bl e Dim as_ sho rt_cent re_out er As Dou ble Sheet2 .Cel ls.Cl ear
Q = Sh eet1. Cell s(2, 1). Valu e H = Sh eet1. Cell s(2, 2). Valu e an gle = Sh eet1.C ells (2, 3 ). Val ue w_so il = Sh eet1 .Cell s(2 , 4 ). Valu e w_wat er = Sh eet1 .Cells(2 , 5).Valu e Fck = S heet 1.C ells (2 , 6 )
If Fck = 15 Then PSTdirect = 1.1 PSTbendi n g = 1.5 Els eIf Fck = 20 Then
PSTdirect = 1.2 PSTbendi n g = 1.7 Els eIf Fck = 25 Th en PSTdirect = 1.3 PSTbendi n g = 1.8 Els eIf Fck = 30 Th en PSTdirect = 1.5 PSTbendi n g = 2 Els eIf Fck = 35 Th en PSTdirect = 1.6 PSTbendi n g = 2.2 Els eIf Fck = 40 Th en PSTdirect = 1.7 PSTbendi n g = 2.4 End If st = 15 0 cb c = Fck / 3 m = 28 0 / (3 * cb c) k = (m * cbc) / (m * cb c + s t) j = 1 - k / 3 qcrack = 0.5 * k * j * cbc area_t an k = Q / H B = (area_t an k / 3 ) ^ 0.5 L = 3 * B
Sheet2 .Cel ls(2, 1).Valu e = L
Sheet2 .Cel ls(1, 2).Valu e = "BREADTH" Sheet2 .Cel ls(2, 2).Valu e = B
'l on g wall
't ank full and no soil pres su re
maxBM _lo n gwall = (w_wat er * H ^ 3 ) / 6
deff = Sq r((m ax BM _ lon gwall * 6 * 1 0 ^ 6 ) / (1 000 * PS Tb end i n g)) x yz :
d = d eff + 10
steel_l on g_in ner = m ax BM _lon gwall * 10 ^ 6 / (j * d eff * s t) av gd = (d + 15 0) * 0 .5
distr_lo ng = (0 .3 - 0. 1 * (av gd - 1 00 ) / 35 0) * 1 000 * av gd / 1 00 'soil pres su re on l y n o wat er p ressu re
a = 3 .14 * an gl e / 18 0
Ka = (1 - Sin (a)) / (1 + Si n(a))
p = w_wat er * H + (w_so il - w_wat er) * Ka * H maxBM _lo n gwall _so il = (p * H ^ 2 ) / 6
steel_l on g_outer = m ax BM _lon gwall _soi l * 1 0 ^ 6 / (j * (d - 5 0) * st ) Sheet2 .Cel ls(1, 3).Valu e = "TH IC KNESS "
Sheet2 .Cel ls(2, 3).Valu e = d
Sheet2 .Cel ls(1, 4).Valu e = " LONG WA LL"
Sheet2 .Cel ls(2, 4).Valu e = "S TEEL ALO NG INNER S IDE" Sheet2 .Cel ls(2, 5).Valu e = steel _lon g_i n ner
Sheet2 .Cel ls(3, 4).Valu e = "S TEEL ALO NG OUTER S IDE" Sheet2 .Cel ls(3, 5).Valu e = steel _lon g_o u ter
Sheet2 .Cel ls(4, 5).Valu e = dist r_l on g 'sho rt wall
't ank full no eart pressure
maxBM _sh ort_ centre = (w_wat er * (H - 1 ) * B ^ 2 ) / 16 maxBM _sh ort_su ppo rt = (w_wat er * (H - 1) * B ^ 2 ) / 12 t_sho rt = Sqr((m ax BM_sh ort _su ppo rt * 6 * 1 0 ^ 6 ) / (1 000 * PSTbendi n g))
t_avlbl e = 150 + (d - 150 ) * (H - 1) / H If t_s ho rt > t _avlbl e Then
GoTo x yz
Els eIf t _sh ort < t_ av lble Th en
steel_ sho rt = (m ax BM_sh ort _su ppo rt * 1 0 ^ 6) / (st * j * t _sh ort ) T = w_wat er * (H - 1 )
steel_ sho rt_ sup po rt = (m ax BM _sho rt _su ppo rt * 10 ^ 6 - T * 0.25 * t_sho rt ) / (s t * j * t_ sho rt) + (T * 10 ^ 3) / st
steel_ sho rt_ cent re = (max BM_sho rt_ cent re * 10 ^ 6 - T * 0.2 5 * t_s ho rt) / (st * j * t _sh ort ) + (T * 10 ^ 3) / st
End If
Sheet2 .Cel ls(6, 4).Valu e = "S HORT W ALL"
Sheet2 .Cel ls(7, 4).Valu e = "S TEEL ALO NG INNER S IDE" Sheet2 .Cel ls(8, 4).Valu e = "AT SUPP OR T"
Sheet2 .Cel ls(8, 5).Valu e = steel _sh ort _su ppo rt Sheet2 .Cel ls(9, 4).Valu e = "AT CENTR E" Sheet2 .Cel ls(9, 5).Valu e = steel _sh ort _centre 't ank emp t y & earth press ure outsi de
drct_com prsn = w_ wat er * H + (w_soi l - w_wat er) * Ka * H bm_sh ort _su ppo rt = (d rct _ comp rs n * B ^ 2) / 12
bm_sh ort _ centre = (drct_ com prsn * B ^ 2) / 16
as_ sho rt_ cent re_o uter = b m_sh ort_ centre * 10 ^ 6 / (j * s t * t_ sho rt) Sheet2 .Cel ls(7, 6).Valu e = "S TEEL ALO NG OUTER S IDE"
Sheet2 .Cel ls(8, 6).Valu e = "AT SUPP OR T"
Sheet2 .Cel ls(8, 7).Valu e = as _sh ort _su pp ort_ out er Sheet2 .Cel ls(9, 6).Valu e = "AT CENTR E"
Sheet2 .Cel ls(9, 7).Valu e = as _sh ort _cen t re_out er Sheet2 .Cel ls(10, 6 ). Val ue = "DIS TR IBU TION STEEL" Sheet2 .Cel ls(10, 7 ). Val ue = di str_lon g
'as sum e 3 0cm p roj ection an d 4 0cm as b as e t hickn ess 'ch eck agai nst upli ft
ab c: prj = 0. 3
Leff = L + 2 * d / 1 0 00 + 2 * prj Beff = B + 2 * d / 1000 + 2 * prj
wt_lo n g = 2 * (Leff - 2 * 0. 3) * (av gd / 10 00) * 24 * H wt_s hort = 2 * B * (av gd / 10 00 ) * 24 * H
wt_b as e = Leff * Beff * 0. 4 * 24
wt_ eart h_p roj ectio n = 2 * (Leff + B + 2 * av gd / 1 000 ) * w_so il * H * 0. 3 upward_ pr = Leff * Beff * (H + 0. 4) * 10
downward_p r = wt _l on g + wt_ sho rt + wt _base + wt _ earth _projectio n fri c_ res = 0.1 5 * (up ward _p r - down ward _pr)
subm rgd _earth pr = (w_wat er + (w_soil - w_wat er) * Ka) * (H + 0.4 ) tot_pr_ 1mwall = su bmrgd _earth pr * (H + 0.4 ) * 0. 5
tot_ fri c_ res = 2 * ( Leff + B + 2 * av gd / 1000 ) * to t_p r_ 1mwall If tot _fri c_ res > fri c_res Th en
Sheet2 .Cel ls(1, 6).Valu e = "PR OJ ECT ION" Sheet2.Cel ls(2, 6).Value = prj
Els eIf t ot_ fri c_ res <= fri c_res Th en prj = prj + 0 .1 GoTo ab c End If ‘ d e s i g n o f b a s e up_p r_ 1m = (H + 0. 4 ) * w_wat er net _up _p r = u p_p r_ 1 m - 0.4 * 25
wt_ wal l_p roj = avgd * H * 25 + H * w_s oil
R = 0.5 * (n et _up _p r * (B + 2 * av gd / 1 0 00) - 2 * (av gd / 1 00 0 * H * 2 5 + H * w_soil * prj ))
BM _ed ge = 0 .5 * (n et_up_ pr * prj ^ 2) + (w_soil - w_wat er) * H * (H / 0.3 + 0.2) * 0 .5 - 0.5 * w_so il * H * prj ^ 2
d_b as e = Sq r(BM_ ed ge * 1 0 ^ 6 / (q crack * 1 000 ))
steel_ base_su ppo rt = BM_ ed ge * 1 0 ^ 6 / (j * d_ b ase * st) distr_b as e = (0 .3 - 0. 1 * d_b as e / 3 50 ) * 1 000 * d _b as e / 10 0 Sheet2 .Cel ls(6, 1).Valu e = "BASE THIC KNESS "
Sheet2 .Cel ls(6, 2).Valu e = d_b as e
Sheet2 .Cel ls(7, 1).Valu e = "RE INFORCEMENT" Sheet2 .Cel ls(7, 2).Valu e = steel _b as e_s u ppo rt
Sheet2 .Cel ls(8, 1).Valu e = "D ISTR IBUTION S TEEL" Sheet2 .Cel ls(8, 2).Valu e = dist r_b as e
'Desi gn o f ro of
ass umed_d _roo f = 1 00
sel fwt = assu med_d _ roo f * 2 5 / 100 0 livewt = 1.5
finis h es = 0.1
total _lo ad = s elfwt + liv ewt + fi nis hes maxBM _roof = t otal _load * B ^ 2 / 8
d_roo f = Sq r(max BM _ro of * 10 ^ 6 / (q crack * B * 1 000 )) If assum ed _d_ro of / 2 > d_ ro of Th en
ast_ ro of = m ax BM _roof * 10 ^ 6 / (j * d _ roo f * st ) dst_ roo f = 0.1 5 * 10 * d_ ro of
Sheet2 .Cel ls(10, 1 ). Val ue = "DES IGN OF R OOF" Sheet2 .Cel ls(11, 1 ). Val ue = "TH ICKNES S in mm " Sheet2 .Cel ls(11, 2 ). Val ue = d _roof + 20
Sheet2 .Cel ls(12, 1 ). Val ue = "RE INFORC EMNET IN R OOF in mm^2 " Sheet2 .Cel ls(12, 2 ). Val ue = ast_ ro of
Sheet2 .Cel ls(13, 1 ). Val ue = " D ISTR IBUTION STEEL IN R OOF” Sheet2 .Cel ls(13, 2 ). Val ue = ds t_ roo f
Els eIf ass um ed_ d_ ro of < d_ roo f Th en deff_roo f = ass umed _d_ roo f - 20
ast_ ro of = m ax BM _roof * 10 ^ 6 / (j * d eff_roo f * st ) dst_ roo f = 0.1 5 * 10 * assum ed _d_ ro of
Sheet2 .Cel ls(10, 1 ). Val ue = "DES IGN OF R OOF" Sheet2 .Cel ls(11, 1 ). Val ue = "TH ICKNES S in mm " Sheet2 .Cel ls(11, 2 ). Val ue = assu med_d _ roof
Sheet2 .Cel ls(12, 1 ). Val ue = "RE INFORC EMNET IN R OOF in mm^2 " Sheet2 .Cel ls(12, 2 ). Val ue = ast_ ro of
Sheet2 .Cel ls(13, 1 ). Val ue = " D ISTR IBUTION STEEL IN R OOF in mm^2 "
Sheet2 .Cel ls(13, 2 ). Val ue = ds t_ roo f End If
CHAPTER 3
RESULTS
3.1 Design of Circular Tank with Flexible and Rigid Base
Cap acit y= 5000 00 Li tres . Depth of th e t an k = 4m
Compres siv e s tren gt h o f con crete= M20 Free bo ard= 0.2m Diamet er o f bars us ed= 16mm
Tab le 2 THEOR IT ICA L VA LUES PROGR AM VA LUES Diamet er i n m 13 13 Thi ckn es s o f walls i n mm 260 212. 767 Thi ckn es s o f roo f in mm 100 100 Cent ral ris e o f ro of i n m 1 1 Reinfo rcem en t in do me i n mm^ 2 300 300 Cros s s ection area o f top ri n g beam 228. 73 Reinfo rcem en t in ri n g b eam 2744 .71 1 Depth of b as e sl ab in mm 150 150 Reinfo rcem en t in b as e sl ab 450 450 Spacin g o f hoo p rei n fo rcem ent p er
1m at d ept h m from t op in mm
4 140 154
3 200 206
2 300 310
1 618
Spacin g o f verti cal rein fo rcem en t per 1m in bot h faces in mm
R IG ID BAS E
Theoreti cal Pro gram v al ues
Thi ckn es s in mm 150 150
Reinfo rcem en t in cantilever po rtio n 1284 823. 6 Hoop rei nforcem ent spacin g on each
face i n mm
130 232
Spacin g o f dist ribu ti on st eel on both face in mm
429 428. 57
Flexibl e bas e ci rcul ar tank
Rigid b as e ci rcul ar tank
3.2 Design of Underground Tank
Cap acit y= 192m3 Depth of th e t an k = 4m Compres siv e s tren gt h o f con crete= M20 Free bo ard= 0.2m Diamet er o f bars us ed= 16mm An gl e of rep os e o f s oil= 30 d egree Unit wei ght o f soil = 16KN/mm3 Unit wei ght o f water= 10KN/ mm3
DESCR IPT ION THEOR IT ICA L VA LUE PROGR AM VA LUE Length (m ) 12 12 Bread th (m ) 4 4 Thi ckn es s o f wall (m m) 650 624 Steel alon g i nn er sid e (mm2) 1390 .52 1325 .84 6 Steel alon g out er sid e(m m2) 1777 .7 1700 .87 5 Lo n g
wal l
Dist ribu tion st eel(m m2) 867. 34 843. 66 Steel alon g i nn er sid e
at s upp ort(mm2)
1145 .45 1011 .85 44 steel alo n g inn er si d e
at cent re(mm2)
995. 453 808. 876 Steel alon g out er sid e
at s upp ort(mm2)
1367 .32 5 1299 .19 Steel alon g out er sid e
at cent re(mm2)
1050 .47 8 974. 394 Short
wal l
distribut ion st eel (m m2) 967. 45 843. 66 Bas e t hickn ess (mm ) 400 373. 38 Reinforcem ent in bas e (mm2) 3547.56 3289.62
Dist ribu tion st eel in bas e (mm2) 834. 59 721. 82 Proj ection in bot h si de of wall (m ) 0.3 0.3 Roof thi ck ness (mm ) 100 62.1 25 Reinfo rcem en t in roo f (mm2) 433 1484 .57 Dist ribu tion st eel in roo f(mm2) 150 63.1 87
CHAPTER 4
CONCLUSION
Storage of wat er in the form o f tanks for drinki n g and was hi n g pu rpos es , swimmi ng pool s for ex ercise and en jo yment , and s ewage s edim ent ati o n tanks are gaini n g in creasi n g i mpo rt an ce in th e p resent da y li fe. Fo r sm all cap acit ies we go fo r rectan gu lar water t ank s whil e fo r bi gg er capaciti es we p rovi de circul ar wat er t an ks.
Desi gn o f water t an k is a v er y t edi ous met hod . P articul arl y d esi gn of und er gro und wat er tank in volv es l ots of mat hemati cal formu lae and calculation . It is als o time co nsum in g. Hence pro gram giv es a sol utio n to the abov e probl ems .
There is a littl e di fference b et ween th e d esi gn v alu es o f pro gram to that o f manual calcul ation. The p ro gram giv es the l east v alu e for the d esi gn. Hence desi gn er s hou ld no t p rov id e l ess t han th e v al ues we get from t he pro gram. In case of theoreti cal calcul at ion d esi gn er init iall y add s om e ex tra v alu es to th e o btai ned v al ues to b e in s afer si d e.
REFERENCES
Da yaratn am P . Des i gn o f R ein fo rced Co n cret e St ru ctu res. New
Del hi. Ox fo rd & IBH publ icati on. 2000
Vazi rani & Ratwani . Concret e St ru ctu res. New Delhi. Khann a
Publish ers. 199 0.
Sa yal & Goel .R ein forced Co n cret e St ru ctures. New Del hi. S . Chan d
publi catio n.2 004 .
IS 45 6-200 0 CODE FOR P LA IN AND R EINFORC ED CONC RETE IS 33 70 -19 65 C ODE FOR C ONCR ETE S TRUCTURES FOR HE
