22 --LIQUID RETAINING STRUCTtJ~s

-,_:.1 l~TRODliCTION

Liquid retaining structure is a general term applied to underground · d d tanks tank:-. reservotrs. aqueducts an even ams. They are used to st ' overhead · A 1· ·d · · ore Wate petroleum and chemtcals etc. tqu t retammg structure can have a · r, liquid rtcmn<~ular shape in plan. It can be bui lt either below or above the grctrcular shape or a

t e ·t I h ound lev I capacity tanks are usually but t be ow t e ground level. Over head c· 1 e · Large . trcu ar W

l:' a common sight in Indta. An over head water tank is usually sup ater tanks . d . . h . 'd ported on a of columns etther aroun tts penp ery or m a gn . The height of the 1 number lso referred to as the height of staging may vary from about 7 m t co umns Which is • · . o say 25

structures not only should have suffictent strength but should also b ti m. Such _cracks. Water and liquid petroleum do not react w ith concrete. Th e, ree from any . . . d. h . 'd c: f ere.ore no . treatment ts requtre on t e mst e sur1ace o the reservoirs. Typic 1 r : sp.ectal structures are shown in Figs. 22.1 a-h. a tqutd retaming

VENTILATOR

PLAN PLAN

(a) RECTANGULAR (b) CIRCULAR

(c) UNDER GROUND

Fig. 22.1 Typical liquid retaining structures (cont.)

)TAIRS ~ TWO WLUMNS

INlk()DIJr '1\(JN

BEAM

STAIRS

RAFT

(d) RECTANGULAR

STAIRS ON SINGLE COLUMN

(e) CIRCULAR 'MTH DOMES

PILES .-~

(f) INTZE TANK

Fig. 22.1 Typical liquid ~taining structurfl (coot)

694 LIQUID RETAINING STRUCl URES

COLUMN I 1 SHAFT •

STAIRS

CONICAL DOME

~~

(g) INTZE TANK ON SHAFT

I I --- I I

(h) CONICAL TANK ON SHAFT

Fig. 22.1 Typical liquid retaining structures

1gn Ph1losophy

A \\-ater retaining structure may be designed using either limit state method or . • . d . d · n to ensure orkmg stress method. All relevant limit states must be cons1dere m estg J' 't

adequate degree of safety and serviceability in accordance with IS : 456. The ~~ .state of serviceability in deflection is likely to be critical only in exceptional cases t not

· . · d flexure mus axunum calculated surface width of cracks for direct tenswn an . 013

y d 0 . . . . d fl ural tensJOn excee .2 mm. The crack Width for members m d1rect tenswn an ex . . does not

be deemed to be satisfactory if the stress in steel under service condttJOnS exceed 150 MPa in high strength deformed bars.

Jmoermeabilitv of Concrete

triJCtures. - . . . . . . · · d retaining 5• - ctl)' The lmpermeabthty of concrete JS a bas1c requirement for hqut because 11 dire ,e

llus is 1mponant not only for its direct effect on the leakage but also . n frost damagd - · abrasto , . ['111 an ~rr ..... , .. Aurab1hty, resistance to leaching, chemical attack, eroswn_,. f anY un1fo rhe

protection from corrosion of embedded steel. The. per~cabJht~ 0 dependent ~~able th ... rn .. ~"-lv compacted concrete of given mix proporttons tn Jarg'bly with the av !tall'

· · · · · ompatl e h · re~u ement rat1o_ It ts essenttal to select a concrete m1X c 11 as t e . . . b'J'ty as we tcle shape and gradtng to have a tugh degree of worka 1 1 · e

concrete ts sufficientlv impervious higher'~. -· ever. ( t"t

· M25 J--foW dinS 0 d ncretc used in such structures 15 · nderstan IY all d 1 the u 1 on ' . f concrete .. t.~et us. fir~t ~ve ~!".tina momefl ·

hu>nt .. ,l- •- ·

MEMBERS SUBJEC1Eo·10 BENOJN .

(1 MOME'NT l

·al tension and bending rnornent The d . · d ax · '""" of combm~ f over head water tanks are not diseu,!td '""- <irtut~...., .,d stagmg o ~Pre...,, MEMBERS SUBJECTED To AXIAL TtNSJON 22

.2 b must satisfy the following conditions: h mem ers sue fi . . ti

h U

ld be suf Ictent rem orcement to resi• an ~. lensiie , . There s o "'"~ (1) . h

.. The calculated tensile stress tn concrete s oufd not"'""~. P<nnissible..., (11)

d that the concrete and steel act together, and COntltft is unaa<kot. It is assume

fi t condition : From the '" T ~ a., x A, where, T - force of tension

5

cr51 = permissible stress in tension in steel reinforcement

A1 = area of reinforcement

<22.1)

From the second condition :

Equivalent area of cross-section is given by :

A -

Therefore, (j <

bD+(m-I)A1

T

where, b - ct - bD+(m-I)A1

width of the cross-section D ~

depth of the cross-section

d I . 280 mo u ar raho, --3crcb

computed tensile stress in concrete

• m "'

0 c1 -

0 ct' ~ permissible tensile stress in concrete

crcb "" permissible stress in concrete in bending compression lbus, dirne . 051

0ns of the section may be selected. 22.3 ME ,

~BERs SUBJECTED TO BENDING MOMENT Such

(i} lh fllernb d' · · ers must satisfy the following con 1t1ons ·

22.2)

..2J)

e cone . tli) lh rete 1s not perrn itted to crack. c:- tenstle satSs 10

e corn . . . crete and th teint; Press•ve stress in bendmg '" eon · sible values

orcern dinl' penms. . ~~ . enr should not exceed the correspon :. _. expresswn e bt>ntt· . I , the flllk,\\108 -1

llg str - · · · ~ g" en n ess m compression or rens1on 1• - •

t22.4)

6 LIQUID RETAINING STRUCTURES

If 8 sectilm is designed strictly as a homogeneous section on no cracf.. ba, . 1 . . 'bl . . b d' . '·' · t ten tt neutral axis will be nt 0.5 D. If pemusst e stress m en mg tension in cone . s d

. . . . . . rete 1s 1 7 1Pa for M 20 c'-"ncrete, the correspon mg stress m tenswn steel wtll be m x 1 7 h : l .h. · I r- t I l 'h · · · t at 1s

14 " 1.7::: 23.S MPa. ts stress ts too .ow 10r s c-e . e pemHSstble stress in steel · the liquid retaining fnce of the structu~e ts taken as 115 MPa for Fe 250 grade and 1 ~~ MPa for Fe 415 grade steel. lnus. 1t can be seen that although the design of • . water retaining structures ts ~upp_osed to be do.ne on no crack bas1s, some tension is permitted. It means a slight crnckmg 1s acceptable m concrete such that the crack. width is less th 02 mm. It should be remembered that the entire force of tension is resisted by steel an~ not by concrete.

The face of the concrete away from the water face can be designed on cracA basis provided a minimum of 115 mm uncracked section ts available on the water face, that is, the depth of neutral a.xi~ from the water face is at least 115 mm in compression.

If the section is treated as a composite section, the depth of neutral axis will be different than 0.5 D. Let the depth of neutral a.xis from the extreme compression edge 1s

x as shown in Fig. 22.2. Let us take moment of areas about the neutral axis : -x · (d-x)

bx.:.. = b(d- x) + (m - l)A (d-x) 2 2 t

(22.5)

I • b T~ ·I

D_t_

1 T x > D/2

. - · - . - . - d- · -1-•

At • • • I I ( '• T

SECllON STRESS

Fig. ll.l Concrete uncracked in tension 't is easY . . um steel. I . is For an assumed value of area of tension steel near about the mmlm .

1 01 secuon

h . . f th, equlva e to compute t e depth of neutral axis. The moment of mcrtta o c . g1ven as:

I = bD3 . 2·

12 + bD (x - 0.5 0)2 + (m - 1) At (d- x)

(22.6)

where d • effective depth of the section h re y"' d ~ ~~ . • 22.4, w e J•''s . Thus, the tensile stress in concrete can be computed usmg Eq.S. ce the neotf&

The tensi le stress in concrete surrounding the steel will be Os/m. 10

the mid depth, therefore,

o -/nt <;. rr th

MEMB l:iRS ~U8JEClf:n To Bf: NDr .

. .NG ~IO~tcN

t;97

1 , 1 Section CraCII tt . .

depth of neutral ax1s from the extreme co Let the t' t" . 111pressJon e-'"e I

. 1

,1

us take moment o e '' "'" ""' of ''• . "' s • ~ lho-. • ., J ~ . • k . l.u, ~I ton about th fig. ~.. . . 1

. 'tel'! mav be ta ·en equivalent tom times its n,... f e neutrallt.~IS l)e ore a o ' - "'' o '"'~ete.

X

b x 1 == m A1 (d - x) -

A At = p bd or. p - I - ----Let

bd , b .~ · ,.. 111 p bd (d - x) or, or. , ')

x- == 2 m p d (d - x}

vr,

~

I· b ~

TI .. ____ -··-0

0cb

N.A. - ·-1

l • I / '· T e • (Jst/m

SECTION STRESS

Fig. 22.3 Concrete c-racked in tension

0 \~ -+· 2 111 p d x - 2 m pdl

If ' '"' Nd, N a -rnp+Jm2p2+2mp ~l!er1 •

la[J, .. I 1 . . • . . •

~here

~ <r~

'

.

') · lle depth of neutral ux1s 1s gl\'l!n as ·

nst

n Cb :::0

Nh ..

N

N (1 Sl

1 + ,;1 a~~~

computed stress in steel in bending tensillO

corn . . nl'l'ssion puted stress in concrete in benJmg t.'OIIlt·-

In case I'T ... 0 1 'and o,.b

· ·' "'sl s • •

(\~

then N N11

i · '-"'I j(lll . ' b I811Ct'• ~" Coefllcient of the neutral AXIS fl\r a 11

coefl1c1enr of the. nc.·utral axis

<22.1)

(22.8)

X< D/2

(,, '"' __ .,,.

(~19bl

(21. IIJ\

Ll<)lJ ID RETAINING S fRUCTURES

p nni . ible tn:'is in ·teel in bending tens1on

m1i ' ib l .; tres ~ in concrete in bending ~.:ompress ion

F r e of com pre ~ ion

For e of tension T

I - crst b Nd ") -

1 ment of resL tance \\ ith respect to concrete

here K

10 R = ( ~crcbbNd) x jd

MO R - _!_ crcb Nj bd2 - K bd2 2

j - coefficient of lever arm

or I

- I- _• J ~ _,

1om nt of re istance \ 1th respect to steel

or

M - (cr51 A1) x jd

A = t M

cr st jd

(22.11)

(22.12)

(22.13)

(22.14)

(22.15)

(22. 16)

(22.17)

· fi of tensiOn Eq. ::!2 17 give the quantity of steel required to resist the enttre orce k the d ·r · d'd not crac , However. if there \\ as a fibre of concrete surrounding the steel, an I It 1 th ·treme

. . t cr at e ex ress 111 concrete would be cr5l m. The compressive stress 111 co~cre e cb . this case, ibre will Still be in the ratio of the distance from the neutral roos. Ho~ever, lll since no the poruon of the stress diagram below the neutral axis is only an imagmarybon: the mid

. . . . ually a ov e e are res1sted by concrete. In thts case, the neutral axts IS us dep h of the section. and

22.-l Mf.:\1B£RS SUBJECTED TO COMBINED BE 'Dit 'G MOMENT

A~v AXIAL TENSION

f ncra lr.ed Sect ton

On 1Jqu1d retaimng face ~ llo mg condrt•on :

· ses shoU of the plane walls, the tensile stres.

here o

0 • bt

o,, Obt ..­- + .....:;;.:.. .;:a 0 I

lJ t l 0' ht

comp ted tress m concrete in bending tension

permi 'i tble tress In con rete in bending tension

f tht ld sa"s )

MEMBERS SUBJECTED fOCOM BfNEo ""I 'V\JAL ·~

These stresses may be cornp 1

. · )(>!'-iANoar)Jr. u ed usmg f ••ufNG Mc

Cracked Section -q~ 22.3 and 22

JMf}q 699 4 thr()ugh 22 6

(i} When eccentricity is small th · at If ten .1 ' 31 e /ore In this case, the line of acr1· e 1S large 1 on of th ,e .. At'T

layer_s ~f steel .as shown in Fig. 22.4. e force lies Within 1

area 1s meffecttve. The Whole seer . the section betw ton IS in tension een the two

The total tensile force acts at d' and the concrete . a 1stance fi

cr512 are stresses m stee I areas A d e rom the c . , tl an At2, respective) . g. of steel areas.

Takmg moment of tensile force b }. If 0 stt and s a out the bottom steel

crstl x All x (dl + d2) '= T (d2- e) '

I· b

, 1 • dl

- · - dt AXIS THROUGH · - · -

e C.G. OF STEEL

• T

d'

T Fig. 22.4 Section under small emntnctt)

or (d , -e)T -

Hen ostJ All + crst2 At2 = T ce cr

. st2 can be computed It/) ~~'/

len ecc, · . \tTl I (Jifncuy ts Iaroe. that is, tl!nsilc}orcc IS small (t! · ' 1

n th ' ~ ~~ L case h · 'Je the •u 0 g. 22 5 • 1 e hne of action of the force lies (11!!~ 1 h

Cq The d · . d ant H~n~. t n be loca 11 ecr and bend in•• strc ses are equally omm ted by trial and error. o

~l)f~e Of rompr' . c 'iS Jon in concrett'

Force of . I '11411 tens1on in stl•el

tbr,utn of filr .. ... gJ\.es :

r~2 I )

(.1- )

sh'''ln I

~utrJI a.1 ;S

IIIIJIIHUI ININ<• 'lfllt ' fllll•'

b

- N,1 _.._;,N.A.

t C./l

t

_ l -T

h~. 12.'\ s( dion unclt·r lnr~:t l'l'('t•nh tdly

( Nd d')

h h Nd < 111 I ) t\ , Nd n, h

I or 11111111 nt JUthhtnun,

T •km. llllllll\~lll Ill • lithe lor~l'S olhoul the tension steel,

f22.2l)

( Nd d'J 1)/1., Nd olh(d d') 'f ( c ~ 1 d') (22.22)

JtHI N (22 .2 J)

I he 11 r 111 111 lfl.1y he t.ur inl out m lht• lollowrng \IL'P>:

p I A umc .1 v.rl111.: nl N

t ·n.! < nn1p111c Ire~ cs 1n onl'rctc anti \ll'cl ~'\hand o~, usmg .qs. . 17 • 22 21 and 22 22 p1

p •• 1 mpu1 wrr ctcd value of N ustng f·.q 22.2 3.

t o1np 1rc th I\\ n 'aluc of N and 11cr.1tc 11 the: dJIIL•r cncc IS u • . . , · II'ICI;CplnbJc.

22. I' I H \11'\SIIJLJ S'T HFSSJ.;S IN CON IU.TE

· n 111d d . I tt•nsiO , • f '' r' It m ( 11 <Jud;~n • The pcrmi\\thlt trcssc 111 concrete 111 lrcc bcntfll18 'I · se~ due to n u~ It 'II I n r 'I\' •n Ill l.thlc )) I I he p I miS~Ihk len I sllc~. . Jnnl2i "'"1 I

I n her , less I I c pp I 1 lh' I.J. 'ol the Ill 'lllht•r In t:OillJl t Wllh the htlllld II ITICI bcntflll!' Jl

. ., I ' I l'SSCS '" lh1 K n J trl 0111 ·t With the hqul(j 1111 tJill' ~Ide, thLSC pCIIlii.~SI l ~ '>I I IJ ph 1bl 11 lh 1 lllnl • 1.1 't'

T hit ll.l l'rrmh,thl tnt~ilf• ln·s~,. In l'ntH'rt'h' on wulcr ll f: ce ( Ml'lt)

S1re s -( 1r;J';Je ;.,j , nmrt:tc ---

M20 M25 'M'10 Ml'i Duct.:tten~Hm 1.2 I 3 1.5 I 6 -l.kndrng ICBSIOII 1.7 I II ~Ul :.z.:z

IIC''·l''• 111

I~ ,J56

lor fl'\1\lmlc ' lo eroding I Itt' cone 1

. J ., I . rc e Is a limed ~-llrm 111 Mec WI I >c lrmitcd hy lht· r 11u 1 to"'" uncra"'-' lb

· trement thatrh '"W cnnnck 1\ 1101 C)(( t dcd, that 1 . tiJ 1 fl't'l e perm, lbi: tenSJ)e le e!tnlbe .I • e Ire tn I ·I h"ll L mouular rat to .md the torrt·spond10., P'"r

1• bl a •Jttqualtorh

b ' n ISS I C ICO\IIe 1,...,

for .1/n•ngth 'ulculalton I ablt· 22.3

··• In COncrete P'oducr •

1 he fltrm isstble tre,s In steel remfon:emenr are g1~en m

Tahfc: 22.J l'crml• iblc •t . · · ·• ·• rts\e rn rtmfnrrtment IMI'II

Strc~s lligt, lrength f ~ -- deformed bars

f em1Je \Ires~ in direct ten ion , ISO hendrng .JOd shcur

Cornprcs\ivc Mre\s '" columM ~uhjcctcd to direlt load m

ll.? MINIMUM IU:JNFOI{CEMENT

1· 'lhc lllinlll 1 • d t · h ofth 1111

11 11gh strength rcmlorconcnt m wull , noo•> an roo m ca. 1Wo u 11 t•q . cross- ei"tlon a I .

1011\ .at nght <to 'les should be 0 35°o olthe urtace 1 re ,\ 10•"11 f "

.,. Ill ' IJ;.22 6

I

D

I o -J .. f () 5 D of co~cr Soo rnm, each rein for ment face conrro s250 mm d pth of

boo . 1 control rnrn, ach r mforcement ace f d pth to i d thts sur ac ' Ononng < ny ccntrul cora beyon

I 'I drd lab' • J:•Jtt• 22.6u Surflu:t• J'cm• \- \utll and ~usptn

LIQUID RETAINING STRUCTURES

_L D/2 T

NOBOTIOM REINFORCEMENT T 0 DUNDER

_l 30Dmm

_L D/2 T I

0 DFROM300 T0500mm

Tl_ 10Dmm

j_ 250 T I D

Tj_ 100mm

Fig. 22.6b Surface zones in ground slabs

o > 500mm

f inforcement may 2 In walls of less than 200 mm thickness, the calculated amount o re all be placed in one face.

. J shall f . forcmg stee 3. \\'hen reinforcement is placed in two layers, the two layers o rem th minimum be placed one near each face of the section to make up e reinforcement.

. . . d or enclosing 4. For hquid faces of parts of members either in contact wtth the hqut t should be the space above the ltquJd, the mm1mum clear cover to a rem r should · · · 11 · forcemen be 25 mm or diameter of the bars, whichever is greater. The cove determmed based on durability criteria as discussed m Chapter 9.

22.8 A USES OF CRACKING AND CONTROL

. · reasons· Crack 1n Water retammg ~tructures may occur due to Jollowmg J

1 loads an . ~wma te I) Exec 1ve direct or flexural tension m concrete due to n concre · · ackmg 1 temperature gradtent5 due to olar radiations may cause cr

DOtvtE

(l) Changes tn th~ moisture content of

703 and may result tn cracking of co concrete may c . ncrete ause dun . (J) Heat ts evolved as cement hydrat ens1ona1 chan

more after casting and then fall tes and the temPeratu . ges . · owards b' re W1l1 TJ r, this ttme whtle the concrete is still w k am lent. Crack· se or a day or

ea . tng usually """ ·""urs at llle cracking of concrete is controlled using

one or more of th . . . e following methods· (I) By avotdmg or reducmg the gradie t f ·

moisture of especially the early age en ° steep changes in temperat . oncrete. Tvn.. f ure and procedure and cunng method may al'C '"" o shuttermg, desh•"'· . . c . h uect the chan . w=rmg mo1sture. unng s ould be done for a rn· . ges In temperature d h · k f k · lllllllum of 14 da an (2) T e ns o crac tng due to overall tern tur. ys

minimized by limiting the changes in rnofstura c and shnnkage effects may be h re content and tem t e structure as a whole is subjected. perature to which

. (3) Cracking may be controlled by reducing the r ·tr · . ~--~h contraction of the structure as well as provis· f•L expans,on or (4) Th d T . lon o u•e movement Joints e etat mg of remforcement should be done very carefully. ·

12.9 DOME

A dome may be used in circular tanks as a roof or as a floor A dome is a shell !lnerated by the revol t' f 1 · ~ . u ton o a regu ar geometncal curve about one of its axis It rna) ,~obtamed by the revolution of a circular curve parabtJiic curve elliptical curve or a n~,ttr ' I , ' ' ~ 1 lange about its hypotenuse. The latter gives a COI!Ical dome. Domes CarT) loads

'· amY through the development of membrane forces. The bending moments and shear fees are negligible.

~. ~m, .m•y bo ~"umed to eons;st of, numb<cofhnri"'"' "'~ '""' '"'~ ~h ~Y . qulllbnum ts maintained by each ring independently of the rmgs abole 11

.-\t section th .1 aJ thrust and

loa ' ere are two forces at right angles to each other · m.:rtuton P compres · th · fe ence of the dome lliile .. Stan or tension. The hoop force acts along e meum r

I lllendlonaJ h be e~1!1uated usmg the liowin t rust acts along a meridian. These forces can g expressions.

vnifo~#t/ ' ~ dtstributed loads

let IV b f dome (edge stmpl) ~Poned). e the Vertical load per unit area of the surface 0 ·~er,d ·

'onar th rust at any point in the dome

wR ' '.I) \-- .. ~ .. 1 +cose

rad ' IUs of the dome h d >me

. vft e ' angl . h ·trticala\1) co~ e of the section measured trom t e v lPrcsslon at any point tn the dome

0

..

If

tll.: UID RE r INING STRUC1 URES

(- I + cos9 + cos

2 9)

wR 1 + cose (2225)

lu of the right hand e pr~ssion 1s positive, the force is compressive otherwise it

t n it

lfH mpr . 1on is zero, that is,

- l + cos e + cos~ e - 0

or e = st o 48'

· 1 1 9 15 1nore than 51° 48' the dome will develop hoop tension. lf m1 centra ang e •

LOt. d, fJ m, uniformly towards the base

It refers to the h~ drostatic pressure act~ng radially from zero at the crown to maximum n ar d_e_ a ho,,n tn Fig 22 .1 (edges s1mply supported).

Meridional thrust T 1 -

I I I I I I I 18/ n I I I I II

0

Fig. 22.7

wR2 (l+cos9-2cos2

9)

6(1 +cosO)

(22 26)

wR2 (4cos2 S+cosS-S) HoopcompressionT2 = - 6(l+cos9) . rovid(d

. bearn IS p o balance the horizontal component of force T 1 at edges, a nng

whtch wtll develop hoop tension .

22.10 0[ ICN Of' TA KS

.. ore , S I•• Roof e tarter 1

. 1

l!l1d d me Th r3d1a ~e

The roof lab •>f a tank may be etther a flat slab or a . 0 '100g the ·fall}' It

ec omic I s it r ~ists the load through membrane acu~n h 11

surchal'Se 1

etrcumferen tal directwns, 'I he roof~~ designed for the !iclf weag t,

DESIGN OF 1' ANKS

load and mechanical equipment if . • any. A provtded for mamtenance and a ventilat . lllanhole 1\ltb

or ts provided for a !eel ladder Rmg beam atr Circulation generauy

In case of over head circular wate ta . b h. h . r nks the I" Clfcular nng earn w IC m tum is support d , wa uflaor ma b

bending moment, shear force, torsional mo~ on columns. The nn~ ~- PP<Irted on a ent and hoop tens· uqm thu develop Jon or corn-... Floors .. ..,, "'"ten.

If a tank is resting directly over ground th fl a nom mal remforcement provtded that th; s e.1 oor may be constructed m concrttc th

b 'd · OJ can carry the '"""· 1\1 su s1 ence m any part. In case the water tab! . VG<Q v.ttbout apprectab be • h e IS close to the fl e

anng capaCity s ould be modified accordmgly If th oor or above 1 the nng beam, the floor should be designed as fl ·. b .e

1 tank 15 supported on wan or

h ~ d oors m Ul dmgs for bend s ear 10rce ue to load of water and self weight G

11 mg moment and

to th 11 Th h enera y, the floor IS ng~dly COnnected e wa s. us, t e direct forces transferred to the floor£ th . Its

should be duly accounted. rom e I\ a and vn:e ~ersa

It is economical to provide Circular tanks with a floor m the shape of a dome In ·uc ~ases, the dome should be designed for the vertical load of the liqmd. The ~of tbt o~e and rts diameter should be so adjusted that the stresses tn the dome arecompresSI\e

: ar as possible. The dome is supported at its bonom on the ring beam v.luch 1 eslgned for resultant circumferential tension in addition to vertical loads

Walls of rectangular tanks

h In plane walls, the liquid pressure is resisted by bending moment in v~cal a:!d th:ll:~~tal ?lanes. The horizontal tension caused by direct pull dut 10 "~ter pressure: mo

111 ~ornmg end walls should be added to that resulting from honzontal bend! g

ent as sho · · · · d · d ta·lrng t e f, the vert' wn m F1g. 22. 8. Extra care rs reqwre tn e 1

real edges where the walls are rigidly joined together

In recta I I b or two way slab (F•~ 22 9J It ~!u ar or square tanks, the wall may act as on~" a) ;ra fret at the top. In the ~orrz011 tal d/ b~ fi~ed or hinged at the bottom. and hinged The wall thus act as thin Plates sub ectJon, 1t may be either continuous or res~·amed dar) ~dit1ons ~~'Ill~ bttween tl 11~cted to triangular hydrostatic pressure With bo~ be earned out ng ~e theo u restraint and free edge. The analysis of such ."al s ma ent and shelf ·orce cOeflicie~s ~f elasticity. IS : 33 70 . Part 4 - 1965 gMS mom

or some common cases. Walls or the u . ed for the foJiowang load a:o . nderground water tank should be de 1gn _ .. ..A

h) 1' to earth IIV"'-<lnk full arth pressure dut

II. • hydrostatic pr~ssure due to '"atl•r and e ,. •lr ~ 1 ~ro

~ ank enlpty d it In cast die~ llblllerg - earth pres ure due to e rth aroun . ld be pJOdll~u

ed due to water table, the oil properties shoO

LIQUlO RETAINING STRUCTURES

DEFLECTED SHAPE ""'

I

TENSION I \ \

I

'

TENSION r f\

WATER / PRESSURE

r

\ -

\.. ,)

PLAN

I

4 ~

J ~

4

Fig. 22.8 Deformations in tank walls in plan

D~' ------,c·

BASE

A'

T h

_L

Fig. 22.9 Slab action in tank walls

c·

Wa/11 of circular tank.s f the wnil

· n o · re defonnatiO fherefO ' These are generally cast monolithically with the base The the base. d at th' under the triangular hydrostatic pressure i restricted at and ab~ve art of the loa part of the hydrostatic pressure is carried by hoop tensaon· an P bottom by the vertical cantileve, act1on (Fig. 22. I 0)

VERTICAL WALL

BOTTOM DOME

-

707

-(a) CANTILEVER BEAM ACTION

(b) HOOP TENSION

Fig. 22.10 Forces in walls of a circular tank 22.11 ILLUSTRATIVE EXAMPLES

Example 22.1

Design a section to resist a direct tensile force of 150 kN!m width. Use M~O concrete and Fe 415 grade steel.

Solution

crc, "' 1.2 MPa in direct tension, m - 13, cr51 =ISO MPa

Area of tension steel A1

_ __!_= 150xiOO~=IOOOmm1 !50

1.2 = 150000 or D = 113 mm

~ Adopr thkkness of the section = Ill mm. . _ IOOO mm'J Ovtdeg I A = ~ ~ ,oo - ' 111

111 bars @I 00 mm c/c on each face (Total stee · 0.35 1000

~ 1

I OOOD+(JJ-1) IOOO

e-- ( 1 Minimum steel = 0.35 ~o of surface zon - 100 ~K

1 ~ . = 202 mml I 000 mm en o,,de 8 ture reioforcem

'llrn bars.@Joo mm c/c on each face as tempera

- 2 X 50 I 000 300 Total steel - -,

_ 333 mm·

t~n... > 202 mm2 '·''PI - -

F\ e 22.2 dth proJu 111 ve · ~~,.,~, 1.. s'&n nt.'r me '~~sto a se . t I . ~~~~m t· I non th ct,oll to re i t a bendmg moment o . r114 t~l

e Water fa . U eM 25 con rete and Fe 41. g

0

!uti n

r 1 2 on ret ,

a . 1.8 MP , m I I, o t r:: I 50 MPa.

My

I

h re, I

Let D = 225 mm

10 and d = 225-25-- 195 mm for lOmm bar

2

Let At - 0.35% of surface zone of concrete

A = t 0

·35

X (225

X I ooo) = 394 mm2 say 400 mm2 100 2

Let u calculate depth of the neutral axis

2 = b(d-x) +(m-I)A

1(d-x)

• 2

x2 1000- =

2 _I o_o_o (225 - x)2 + I 0 X 400 )" (225 - x)

2

or x = 150 mm.

I = IOOOz 2253

+ JOOOx225 (150- 225 )2

+ 10x540(l95-ISO~ 12 2

• 1276 x J06 mm4

x (195- 150)- 0 53 MPa

< 1.8 MPa QJ(

1 ensile stre s in steel m crbt .- 1 1 "' 0.53 = 5.8 MPa oK

< J 50 MPa It is

125 mm c/c. Hence, dopt a thickne of 225 mm and provide I 0 mm bars @ po stblc to reduce 1he wall thickness.

·, ""'',. ll.J

Reck tgn the ection in example 22.2 on crack basi . • ,,fution

l.et u dhpt ' d pth f) -

d -

I 50 mm. 10

150-25-- • 120 mm . 2

ftrsl mal

Let At = 0.24% .... 0 35

o/. = 0.175% of. • of rface Zor:

&To s are llf Let us calculate the depth of neutral . . crete ~IS USJOg tq

22 _ 9aor22911 N - -mp+ f22---

'/ rn 4 pl +2m;

=-I I 0.24 ((j~~--~ /-+ (I 100 '"'00024 2

N . J +2,1l ... or.m4 - 0.2046

Coefficient of lever arm j = 1 _ I == 0.93

. Area of tensile steel At = M _ lhJ06

crsdd 150xO 93-<120

- 894 mm2

894 Pr - I 000 x I 50 == 0 6 o/o

Second tria/ > 0.24 % of gross area assumed

ltr A == 0 6 0 / r . / 0 = 900 mm2

N - -II x0.006+J(IIx0006)2+2..-.IJ 0.006

N - 0.303

. N J - I--= 0.90

3 15x 106

A1 = =926mm2•0.6'• He 150xO 9xl20 nee d

~~~~ 1 ' a Opt a th · kn Pr rdt 10 mm bars w c c. tc ess of the section equal to 150 mm. ov fr,

"'Pit 22.4 Oesr

•1dt~ gn a seer 1 to " • ~ rn m ~ Producin ton to resist a pull of 30 kN and a bending moment ~IS .-J.. I "'' g tens· te andfe.. i'"Y• ~11on '0 " on the water face Use M20 concrr

or ltt ~s 0 4 " ~'oss design of 0 em and ~ I . area ( .... Oan tJncracked ection. Let us assume a thJckn

' 1~ lfte '17 5%) st el on both the face (Ftg. ~-· 11

SS IIJ

concrete in a ial ten ion

0 c1 =:: T

bD+(m-J)A1

LIQUID RETATNING STRUCTURES

T • 0 - .

l • -I •

_i 30

~ . ___. 1 C/L • 30

T

• • • • 1000 ·I

Fig. 22.11

30x 1000

1000 x 300+(13-l)x ~·~; xiOOOx300

cr,, = 0.097 MPa

Tensile stress in concrete in bending My

crbt - I

. 1. t D/2 Effective cover th t . ic symmetrically reinforced, neutral axts tes a . Smce e sec ton ..., . 0

m to tenston or compression reinforcement ts 3 m .

I

I

r

bD3

D d')2 __ + (m- I) A1

(0.5 -12

3003 - 1000 X

12 X 0.24 X 1000 X 300 (!50- J0)2 + (13- I) 100

_ 2374 x 106 mm4

7.5 x 106

x 150 = 0.4 7 MPa 2374 x i06

The interaction equation gives :

1 or 0.097 + 0.47 = 036

1.2 1.7

< 1

It is possible to reduce the thickness of the section.

A = bD + (m - I) A1

-22 5 em· k ss- . Let us try thic ne

' 500 ,,. 0 24 225 "' 2JI .. 1000 )( 225 + (13 - I))( ,·00 X 1000 )( )2

2253 0 24 X I 000 )( 225 J .. 1 000 " 12

+ ( 13 - I ) :..: I OO (~22~-30 f .. 99300"' 10 mm4

oK

or

ILLUSTRA TlVE EX AMPL£s

T 3oooo

- A == 231500 == 0 129 MPa

7.5x 106 x 112.5

99300x 104 - "' 0 85 MPa

0.129 0.85

1.2 +J:7"'061<J.o

Provide a thickness of 225 mm and 8 mm bars @ 180

m

1 Ill c c each face. woo Total steel = 2 x 50 x == 556 nun2 180

or.

Example 22.5

Pr -556x 100 ---- == 0.247 o/o 1000x225

7JJ

OK

OK

The wall of a water tank is subjected to a direct pull of30 kN and a bending moment of 10 kNm in the horizontal plane as shown in Fig. 22. 12. Find the maximum stresses in concrete and steel. Use M20 concrete.

T

10¢ 0

C/L -Pol- -PoJ-

.

• I •

• I • e 1000

•

• 100 C/C EACH FACE

I· 1so ~

F 2 Vertical ection throua ig. 22. t It will

fhe

' II , rccentrj . . The dtrc·t ) doll). Crty he outside the ect10n d

'llaru. fhe section i · as umed to b era~ k

1 ...

L t

or

or

LIQUID R TAINJNG STRUClUR '

• o o. J .. 1 0- 0 "" 120 mm. :. Nd 36 mm

_2 t e ,

:- , b h +-~d) (m -I) A" (N~~d') "'• (d- d') ~ T (•-T•d') "' ['~ 36(120-

336)•t2x7s.sxtoxe63~3~}ctzo-3o>]

- 30 x 103 (334- 75 + 30)

acb - 1.24 MPa

Eq 22 2 I gives,

I (Nd-d') _ CJst At- 2 acb b Nd - (m- 1) AS\: Nd O'cb T

(78 5 X 10) 0' - ! X 1 24 X 1000 X 36- 12 X 78.5 X 10 ( 36 - 30) 1.24 = 30 X 103 st 2 · 36

or o 51 - 69 MPa

1 = 0. 19 < 0.30 assumed earlier

1+-6_9_ N -

13x 1.24

Second trial

Let N = 0.25, :. Nd - 30 mm Eq 22.22 gi'leS,

""' [ 1~, 3o(tzo-3

3° )•o] - 30 x to' (334 -75 + 30)

or -8670

= 5.25 MPa 1650

Eq 22 21 gi'le

I 785 0 t- - If 5.25 Y 1000 X )0- Q = 30000 2

1rd trw/

or o 1t = 138.5 MPa

N • 1

- 0.275

1 + 138.5 J) X 5.25

.. 0.33 > 0.25

· Nd .. £q . 22.22 gives; 71J

[~x33(120 a,b 2 33) 3 + 12 ... 785xJoy(~ - 30)

33- "020- 33) 1

or, "" 30 >< I Ol (3 ..

34 -7h 30) crcb "" 4.63 MPa < 7.0 MPa

Eq 22.2 I gives, OK

I )85 (jst- - X 4.63 X J000 X 33 - J2 X 78S J( 33-30

2 33 .>:463==30x J()l

or 0 st - 140 MPa < 150 MPa

I OK

140 "'OJO,.. 0275 1+--

N -o·

lfd · 13x4.63 esJTed, one more trial may be carried out.

Exam pie 22.6

The floor of a w t t k · . 05kNm/m in a e~ an IS subjected .toadirectpull~f50kN~dbendlngmomenrof max· the verttcaJ plane. A sectwn of the floor, . shown m ftg 22 13 find tbt tmum stres ·

ses rn concrete and steel. Use M25 concrete

1ot~1 . to11 b

'~o rn

T 12Q-

l •

8¢ @ 75 C/C EACH FACE

• • • - · - · ·--

• • •

1000 1

l 30 I eli l 30 T

Fig. 22. JJ Vertical section through noor

Ill, At 1 "" A 12 : 8 mm bar @ 75 mm c c.

1 I QUID RETAINING STRUCTURES

a,

or = 25 MPa < 150 MPa

E: _2 _o •v .

a t I All + a t2 A t2 T

or

1 dul r rat10 for M25 concrete m

D re ten 10n m concrete,

B m t n 10n m c.onc rcte

I w th Ill '"a Crilcked <,ectum.

F: x mpfe 22.7

[soooo-2s, so(' ;~o) J ·"'------- · 50 MPa

SOx I ~O • 75

< 150 MPa

280 280 --=- = II 3obc 3x8.5

SOx 103

- - - - 1000 120 x l000+( 11 - J) x 2 x 50 x 75

0.375 MPa < 1.2 MPa

50

I I '

4.54 MPa > 1.7 MPa

OK

OK

2.65 m h•gh D 1 n qu re Y. ater tan~ having inner dimensions of 7.5 m " 7 S m rtcd on the w o w II ti eo at the bottom and free at tha ~op . ·r he tank i<; d1rcc~ly suppo !)se M20 n~ Th n r I b I mrmollthlc with the 'walls. ·r he free hoard · ~ I 5 em.

on rm nd Fe 41 ~ •nl II S IJ bar

( p Clly of water tank 140m3 140kllolitre 7.5/7.5 ~ 2.5 atth~

to 280.mrll I t h1e; ' e of th v rt1 al w·dt he 23 () mm at the top and mcr~;a <,e ""''''' f H ta nY. l..et centre to (.. ntre linn nsion of the tank he

rn b

L ~5 rn t f) ,23 m 7.7J m

L

II 7 71

:L 5 3 10 ()

4 Jol L/11 of IS 1!70 . r'MI

n t1 b ndrn~ llltHnent J11 11 t 11 IIHJ o n the w.lli ' r J,, .

I)

ILLUSTRATIVE f:XAMPtf<.S

·mum S.M. in vert1cal direction at th bQ "ax• e ttom of wall where

Yw - density of water == 10 kN!m3

Moment coefnci ots

X y 0 L -At vertical centre of y"'-H 2 the wall At Vertical

Vertical edge Depth Horizontal Horizontal below top moment moment

Mx moment Mv Mv 0 0 O.D25 - 0 082 0.25 0.01 0.019 - 0071 0.50 0.005 0.010 - 0055 0.75 - 0.033 - 0004 - 0.028 1.0 0.126 -0,025 0

\lax•m um BM m hon zontal direction at ends of wall

B.M. at m1d he1ght of the wall 8 . M = -0.126 x 10 x2.53 - 19 70k mm

+ 0 005 ... I 0 X 2 53 - I 49 Nm m

= - 0.082 X 10 X 2.53 = - 12 81 kNrnm

B.M at the m id height of wall - - 0.055 x 10 x 2 53 - B.bO lNm.m For no crack. .

mg m the wall, ncar bottom

or t ~

In ltknc\s f 0 Wa ll rcqutrcd, t "'

for 00 cr, -1 -

"<'~'ng. . •n lhc wall ncar top, lhltk

Oc~\ Of . Waif required, t

I) 6 d 9. 70 10 :6~ mm

/000 I 7

h b 1~.8 10 21~ rum

1000 I. 7 '~ . ~ ,,, ii(J d t r r II ro. I

4 ht opt <! w I I . I h ' n ' 1 ht 01 a lh" J..ncs~ of ., 80 mm uJ tJc

1 -· 1 ..

! 6 ~ , ... I 1- . IHl\111111 I -~ 1' t, '' )uncn 1ons of thl' IJn" ,tr /Jt {"ll

~ I) l

''' l l7o t nd . tli ,ent~ t,•r ~dt '•t• '"I do ' nor 'I ~ hl''ll lt~rle l'' . tol • '' r,., >. r cr • 1 tt 11~11 •II tJt , ' I! •llfopt tft 11111 · Ul Jlt I 'II tu11 1 . •

lt Ill l"Ol)\ 'I 1(1\'t:

l!llll 11h I

11 R 1 1 b l RU fURl!

ant tt m

0 10 2.52 a 28 .12kN/m

_i 230

T 7500

1 230 -T 50

50 I f3~(..... _..:...7.::..:.50_0 ----,•13~ ~

(a) PLAN

2650

3750

C/L

I . I

100

l L_j_ ___ ---=:!:l t 50

720 1:4:8 P.C.C.

(b CTlONAL L VA TION

I , 21.1 uar w t r tank

ILLUSTRATIVE tv .,..A fiLES

. umS.F. atmidpointandtopoftheven .1 !a. tm tea edge of

::: 0. 4(){; '( 2

.. o 406 " IQ , 2 s2 2 • s 37 The s.F. in the wall o~ the vertical edge wiU ~ me S.f. at bottom edge will cause tension in the base 51~ ' et us check the section of wall on the vertical edge L -~

2537xto3

1000x230

1.2

t- 230mm

+ 6Y)2.8QyJ06

1000; 23ol

1.7

095 I

717

Let us check the section of base slab near the bottomed t The llllcl:t.."Sl of 1 wund the edges is 280 mm

28.12x 103

l000x280

I. 2

6.1" l9.70x 106 + =089<1

1000x2802

l'mtca/ remforcement in wall

fi)At base of wall

1.7

BM. at bottom causes tension on the water face= 19. 0 mm 0

=b"' 7

MPa, cr5t : 150 MPa, m = 13, D = 280 mm, d = 250mm

For a balanced section • N = --'-:-::- :: 0.3 8

150 1+--

13)( 7

J- 1- /3=08

. .. _M_ "' _.:.:~--:::; O~tjd

'br Ill. "' 604 mm2 m

lnillJu 0 35 Ill steel required i · 1000 280 "' 490 mm -'l.-f1 I 00

quattc h . eAt ,. etglu of wall

tau tc · 0

ron on th uter Ia e of'' II

B.M. 0 01

•

~~ m 1 ~ mt11

mm andd • 775mm lrum top D•24 a 2. mm r .. I so MPa inl d • 1 li mm

ornm m t

o·

I

25 17 z I

LH ... ll l R :~ INl Is IRUC' rtJRl·S

u l utred 1 0 ' 011

0 .• --100

1000 "'4~ -;-=- = 430 mm2/m -

1 mm a 1 ... 5 mm d e on ' ') 2 n m• m tl04 nun•

the water face at bottom or Wall

J pm nt I ngth 50 4> - . 00 mm Ot<

b urt il d at I 000 mm above the base and the rernain1·

. ng at n the outer face of the wall, provide 10 mm bars@ 12S rnrn

.:__ - _ mm~ m > 430 mm2/m) above the base.

j_ 230

WALL COMPRESSION ,.. I FACE I

~ c L - · - . .

T r \ e

"'

T TENSI~ FACE

WALL

• ....

(a) fig. 22.J5a Force in walls at joint in plan at top

M = 12.81 kNm/m, T = 25.37 kN/m

!::!,= 12.8JxJOOO =SOOmm T 25.37

e

abc)ut the c g of compression zone,

(5 0 115 200 +- 0 87 )( 200) = A,)( 150 0.87 X 200

25 37 x 103

:-:589 = 570

mm2fm 26100

(II) e r 1 'P of ...,all m th m dd!e .35% ( 400 mm2

M '"' 7.5 ~· m/rn

ILLIJS'fRJ\'r IVe c . XAMptF..s

·' rnmurJUm reinforcement. prOvlue us provide I 0 mm bars r;;, 2SO

1 et \"'/, mrn c/ . JO mtn (a) 125 mm c/c (- 628 .... 2 c stagger"·'

71'1

tee! IS b ... rn /rn 67 "" on ea h 1 1000 rnm he1ght a ovc the base. Beyond 0 rn 012101) 'f ~ face so that

1 10

the outer face to take care of tens· 1 rn he1ght pr . hts steel rnay b he total c'C on •on on the , OVtde I 0 e Provided

provide I 0 mm bars @ 125 rnrn c/c o h outer face. rnrn bars ra 12S rnrn 1,m > 570 mm2/m clc). Curtail 50 ~/t e Water face ne

nun 0 /Q bars ar ends f maming bars at 20 0 mm from the ends at IOoo rnrn c o the Wall (• 628

re . 'rom the Base slab ( F1g 2 2. 15 b) ends, and the

WALL

BASE T

TENSION SLAB FACE

--;4-T e

.~ - -·- .

COMPRESSION FACE

(b)

Fig. 22. J Sb Forces in wall and base slab at joint in elevation

M , 19 70 kNmtm, T = 28.!2 kN m

D -' 280 mm. d "" 250 mm

e = M = 19 70xl000 = ]OOmm lakin T 28.12

g moment ab h. 11.12 )( 101 7

out t e e.g. of compression zone.

( 00

+ 140-250 + 0.87 X 250) = A, 150 X 0 87 X 250 Or

28.12 103 8075 700 mm· m A = _,:.:..:..:_:_:._....:--

Prov ·d 1 150 x 217 5 ~4y" I e I 0 rnn b I b r II ed e ~{I o ~ b n

'>e cun· 1 1

<trs ~~) I 00 mm clc at top face of the s a nea 1

b a I ed at 1 O<JO zooo mm trlliTl th eu ~Cduc mm and the remaining bars at d I c the lh. k fi n the nd l'r''' • c

ll)lll b •c ness of the base slab to 100 mm at 1()00 nun ro• •• ar, fitl ..,00 2 m l in ea h dlr t• n

the b .. rnrn c/c ( • 50 x 1000 "" 2 50 mml/m . I 75 mm otr<,Jtt ot th I 200

e s ab .

/L-

A

2~ C/L-

B

l !Ql ID Rl I I I (, S IRUCTURFS

r shtn n 111 Fig. - 2. 1 thl. b. c and d.

T 1000

3750

10¢ c 125 C/C (V) EACH FACE

10¢ @ 250 C/C (H) EACH FACE

1_,ooo.r

(a) PLAN NEAR BOTTOM OF TANK

3750

C L

i3750

280

C/LT

C L

1-r"" • - - • - • - • -

~10¢@

125 C/C (H) BOTHWAYS

' T ' 1000

I I i3750

. : + 10¢ @ 125 C/C (H) I 4 1000 10¢

1@ 250 C/C (H~

lo4 t • ~ I 1&,~~-=~ .• n,,w~t=.~.------~. 230

, ...., • • a a a

(b) PLAN NEAR TOP OF TANK

- A

- B

fi • Z2.16 k inforcement details in square water tank (cont.)

23~

~ ~ 10

¢@ 125 EACH FACE C(v)

2650

T 1ooo 10

¢ 0 25o c;c EACH FACE (H)

- . ,. .. . ~ "" .. . .. .

I· 1000

280, .. ,720~ a I= 1000 ,

I· 3750

23~ f-(c) SECTION A-A

IO~ @ T 2so c;c (H) s Nos

' l 2ao

t-t----+-10¢ @ 125 C/C BOTHWAYS

T 750 2650

t 1000

• -.,

721

C/L I

I r

.,

C/l I

l 100

rr

-T T 50 P.C.C.1: 4:8 r- 1000 1000 _, ·I· I

~8~72~ I r- 3750

(d) SECTION B-B . F'ig 22 wattr tank

· ·I 6 Rein forcemeot details in square

t 1 n !IPIH 1 IN IN'. n l l< ll liH ·:

II \ 11 I I " l ll ,,, llltll) ~I \ ,, ,,, '" \\ hll h I, I ll h ,,,,.. I ,, l l lllrnht J'h 11 I tl b !Ill' \'.lp.tltl 11llh lll lh IS !Ul j.,N II •

1 1111

I '' " ' II t t•n 1

l ls M,n

I tth I,' fl, \ 111 ' tl '' ·'

I 1 100 11\lll

l th II p nn h 1111 (l(l() Ill Ill

I r I 01 I r 11l tIl l II th nn • h In\ 1 4000 11\111

tr tl \ ttl 11 hlp I 00 mill

1 I ' ,,II • I th holt om lOO nm1

''I I • p h. I' l 0111 s I 00 mnt

Ill h I rt1 ll ' .111 h h mdo .

h I 000 m1

r h ., ., ... - Ill

t n { 1 th 1 ml 1 h1m n 111 I i • 2. .17.

f I b

It• n mu.; I It 1 1 • uJ ,, Jom

l h1 kn ''' dorm IOOmm

R1 nl Jom • 1 '\00 11\t\1

, R d1u

h {)

Nh

1.4i - t l 1 m -l ...

l l iom II 10 ' ~ 2 4i 1\N /m ~

I I ln. rl k Nfm ( .1 )

I I I I Nlm

m n th II,

I ll (} (l 7 l

o md 11 n o ~ 1

I 2250

I

I

~oo , : 4: a P.c.c. 1 oo

FiJ! . .22. 17 C' . Th~ tnt lr ular wutt•r t mk-partiall lr~ dorn l' i . . I s Ill lllnp comprl ·s ion in c 0 • 11. 0

'" R I t cnsO

'''rd~o 1~ o ~ Ml lllllllll llllllll Sit.: I Ill hMh dir cti,nh

- ( lll<l ld 8

Ill ill h It {.!I 200 IIlii\

und r n•und

~1 °4 '

m

~'''ll 1 ~ 0 Omm mm

"I 1 \ ' 111 '1 I () I olin, 1 1lln,

1 ° 1 00 nun di.un 1 1

(' s· '0 nt•n I h

· lluit,11

t 1 ~ n' \\lth mm ht lfl Ill\ . \ ' f'l i I ) ll 111 r" 1 m nt111l ,,, tltl b)( n1 n

''· · ' . IN

t H,Ull REl \INING SlRlJCTURF

U HTENING --11 CONDUCTOR

60 @ 1so c;c BOTHWAYS

I I

ISA 50X50X6 ALL ROUND

75

ISA 50X50X6

----------4-8¢ RINGS

900 DIA

(a) VENTILATOR

I·

M.S. COVER 650X650X5

600X600

(b) MANHOLE

• I

,__-.~..._ EXPANDED 300 WIRE MESH

l_ TOP DOME

"' 75 r

HINGE

TOP DOME

~ 6(1) M S @ 250 C/C • . rr r-T

I I I I T

I I I I I I I I I I I I I I I

250 C/C

t 250 C/C

j_

(c) STEEL LADDER

Fig. 22.18

sreeJiadder

A steel ladder_ is provided inside the 1

all ing and repairs etc. The ladde ank so that

725

cle r cons 1st a Perso placed at 500 mm apart. 6 mm M.s bars (a) 2~t two ISA 50 n rnay enter the tank to forrt1 step!>. __, mm c/c are Wel~e~Ot " 6 angle sectJ~~r

o the an I s Ring beam g e sectiOns

Radial out ward horizontal thrust from dome

Hoop tension == 485 x 0.843 "'41 kNtrn

- 41 X D/2- 41 - X J2 :: 492 kN

_ 492xtooo Area of hoop steel required ISO "'3280 rnm2

Provide 16 - 16 mm rings.

Let us provide a ring beam of 600 x 600 mm section

Tensile hoop stress in concrete ::: 492x 103

600 x 600 + 13 x 3200 "' 1.20 MPa

Provide 8 mm tie@ 300 mm c/c in the ring be am. Vertical wall

Horizontal pressure due to water at the base

- Ywh= 10x2.25=22.5kNm~

Hoop tension at the base - Th = Yw h x D/2

- 10 x 2.25 x (24- OJ).~= ~67k m

Tensile hoop stress crh -

Where t _ h" -w - t tckness of wall

tw 1000

_ 267 x 1000 = 0_89 MPa 300x 1000

P < 1.2 MPa rovide 0 '">4 o

- ro hoop steel in the wall. Adopt 12 ntm bars.

At h - 2.25 m. t" -' 300 mm. ,

Area of hoop steel - 0.24 x 300 x 1000 "' 720 rnm·

No of 12 mm bars

100

= E.£ "' oJ 7 113

1000

6.J7 !50 nlln c c

OK

OK

0

I

I -

ll l H I J I I I I I I t lJI I

h

II

\ t r I t h

h I --= I I

l m the \\all. hei ht equal to it

19 106

btw2 16

106

Ill I 'lltlll

l\1 ttun

111orn r11 ,, i 11 1 I • l C\ "I)

It HM • t lltll dl '"' ll'llt d

' tth tpprnprtat I Jltlt C IOltttiJ,It}

""' 19 kNm.lm

3002 == 1.27 MPa < I. 7 MPa Of\

Provide 12 mm bars@ 200 mm c/c on the development length (say 1000 mn

1 > l

'II ""

re of eel pro\·ided = 113 1000 = 565 mm2.

200 1- mm • 0 mm de on the other face upto full height of the wall.

eel pro ed = I 13 I 000 = 452 mm2 250

....... ~ near the bottom of the wall

,.. 565 + 452 = I 017 mm2

> 720 mm2 OK

tapered to 100 mm thickness near the top. The minimum steel near the tor is

0.35

100

100 - x 1000 = 175 mm2

2

< 52 mm2 OK

· b m ·md snctUld en to ensure that the centre Jines of top dome, top rmg ea ' lJ con urrent far a po sible. Otherwise, a local bcndi.ng moment .wJII

b

mem meetmg at the joint Suitable reinforcement should be provuJcd r 1 I aJ bendmg moment.

b · diredly e w JJ and r ... .,~ on oil The weight ot the \\' ter 1

1 1 Juc

d • tf b I C S I } l bearin tensl., fi rc i d velope m tc ' • II · d d 1 •ar the ' of a tr m th II Joe I mom nt r I o pro tu;c 1 c

for lh foflowm 1 for

oil reaction, rw·•rd

{tl u . ntal t n!>ion from walls, and hllrlf!)

pr) . , moment from wall . bendrng .

ttill ·mbcr is ubjccted to axial ten . a nlC IS 3 l<>n and be When 1 water face, • : 370 require th·

1 ndmg rno

00 t 1e . . . a th tre l'llen oecurs . condJtton . e rn t rJcuon

~ 0 t abt -+ ~ ,. J I ~

._, t obt

fbase slab near the vertical wall llltckness o . ., 300 rnrn

I . . tal thrust of water on the wall == Honzoll - y hl I 2 W "' 2 ;, JQ X 2 2S}.

= 25.3 kNtm

. . al BM in base slab due to the horizontal tension Addti!Oil

::: ( ~ x 0.3) 25.3 == 3.8 kNm/rn

Total BM in the base near the wall - 19 i" 3.8 :: 22.& k.Nm!m

or.

Gross area of the slab - bD 1::: 1000 x 300 = 30 x 1Q4 nun2

. 3 Gross moment of inertia of slab - boJ/12 = 1000 x 300

12 or Section modulus - bD2/6 = 1000 x 3()()2 6 =15 IOfl mml

or

< I

21t Rhw Weight/m of circumference of dome -nD

-2 x 22.32 3.5x4 = 26L m

24

Vertical load from the dome = 26 kt /m V 0.6 0.6 25"'9k m ertical load from the ring beam =

V crtical load from the wall = (0.10;0.30)

i .., .2~ k m ntcaiJ d • 0.3 OJ • oa trom slab just below the wall •

8 tq,: I ' Total lllad = 4 ·• Th,~ I • f " II end tht ba e au h Oad \viii di per_ c at 60° at the jun II\ n °

& ht•anng At the bottom of the base lah,

0

I

I

0

0 - ---I

h

1 oo I

100 J 17 n

m 1 II , an

11500

I ld 0 mm

...A---------4 25--------

mm H t e t p ur (

I r

Ill

I I )

I '

I

II ' I

t I I I

' hr II

I '

II II Rl I I 'IN,. 11-tllt llJHI·S

1- n m nd ri ,, 0 nun cit.

I H)

1.1 ' MP<1 1.2 MPa

2 1t •· QAJ

l n lo d on top dum 46 kN I. OJ 75 X 0.07;<; X 25 -

- lts.2o kN

d r d du fo t p nng beam 0.050 0.25 X (it X 8.6) ><. 25

_ - 8.4s kN ue<Jof ddueto\ rticah\all = (rc 8.6)xO.IOx3.15x25==212 .75 kN

De d 1 d due to raper in wall = (n x 8.6) x (_!_ x 0.05 x 1) x 25

_

Total load - 400 kN 2 - 16.88 kN

Total load perm - 400

8 6 = 14.8 kN!m

1t X •

l4.80x 103

103 x 150

= 0.099 MPa Compressive stress -

< 5 MPa

Hoop t n ion at any depth x from top= Yw h D/2

x, m Hoop tension, kN/m 0 0

1m 42.5 2m 85 3m 127.5

re of eel required to resist hoop tension

A 127.5xi03 2 r - = 8-<iO mm

ProVJd 12 hoop @ 125 mm clc.

Area provided

150

- J. f 3 X IOOO = 904 mm2 125

01(

> 850 mm2 OK

Ten 1le stress in concrete 127.5x 103

12 s x 1 ooo + r 2 >< 904

• 0.94 MPa .-- 1.2 MPa OK

The hoop te J m Y be curtailed according to hoop tension at different heights along rhe wall s hown in Fig. 22.20 c.

ILLUS'I RA'J IV(· 1 ' -:XAMI'J

1 1inimurn vertical reinforcem

14i ~ o n tnt, I hat . jC 0 • • I ' l\ IOtn !-.... 1111.'~ I' ' \Jrtf

,. Jo'''' , I ~0 rrrr:n 10,~' face of the bottom dome lllak 't ~~ oP sur es an an I

ptte 1

4.05 &e 2 (J at 1 sin a = 0 1 tentr

5. 925 ·6ll, C(Js r1 .. 5 'J2s~ 1 ·-~ semi central angle u "" 42.840 .. s.rns o 73

. 5)04&· will be m hoop compression dorne .

enure . h f b r to self wetg to ottom dome == l-oad due 25 ,.. 0. IO "'2 S '-

dial water column (3 m- 1.6 rn) , · o.Nr01

2 due to ra 1.4 " I 0 ..,

Load dial water pressure varying for z 14 kN!tn2 due to ra ero at cro"

d .. n to tn ·

7JJ

LOa = 1.6 x I 0 - ilXttnum tn 1.6 - 16 kNtrn2 m depth

. al thrust at edges ~eridton

. - wR2 + Ywh R2 + YwR2 2[!+coscx-2cos2] T 1 1 + cos a 2 6 I ~

+coscx

2.5x5.925 14x5.925 10xs.n52[1 Tl - + + HJ.73-2,..07)21 I + 0. 73 2 6 - I · _

+0.73 r, _ 72.54 kN/m

THESE TWO 8 AR S MUST PASS THRU CROWN

8¢ @ 200 C/C

J

8¢ @ 200 C/C

BARS

BOTTOM RING BEAM

(c) PART PLAN OF BOTIOM DOME

Fig. 22.20 . . 0 H Tank t Cl nt.)

Rrin forcemcnt detail in ctrculat . .

t H_l tl lll Ri I \1 IN{, S I Rt ll Ill!' I s

11 cl>St(

0 72 MPa

" IP.t

51 '-.N t Ill

• mmmJUm mh1r ,·m ·nt 111 tht• dllnlt: in e·'ch .• 1• • " u rectwn

0 35 I 00 \ t Ji)'() 2 >< I ooo 175 lllm2;m

n m bMS 1 ~0() mm c t: r.Jdr.Jll)- .llld circumferenthll)· .. • .ts .shown in F.

B TIES 250 C/C

Bilt ' QQDS 125 C/C

4 OS.

100 HOOPS 25 C/C

4 OS.

2¢ HOOPS 125 C/C

5 OS.

(d)

tg. 22.2oct.

C L

8¢ @ 250 C/C BOTHWAY

I I I I I I

8¢ @ 150 C/C VERTICALLy

8¢ @ 250 C/C I LOOPS 5-NOS.

8¢ @ 200 C/C

I C/L

8¢ @ 200 C/C (BOTHWAYS)

p, '· 22.20 Heinforc ment detaa·l . m cir~ular 0. H. Tank (cont.)

ILL\JS'I RA ILV! I· • ·'<At-.1.-, r s

he am ,~

~,,, rl 1

the bottom of wall ~~ · ohl a

vJtlt> . •ht of bottom dome "' wetg

400 kN

"' (7tx86),.-03s

144 2~ Weight of water

Weight of ring beam

Total weight - 18 ): 0.35 "2S 87kN,..1900kN "' 827SkN

- 1900 n 8 6 "' 704 kNim

Weight per unit length ..

. beam be supported on 6 - 300 mm dia

73s

t the nng columns as h ~ ~mF~n~

C/L

350

sona.J RING BEAM

. T.- · C/L so·

'(

\ COLUMN 6 NOS.

PLAN OF RING BEAM

(e) Fig. 22.20 Reinforcement details in circular 0. H. Tan!..

It can be designed as a circular beam in plan supported on i. \.Oiumn di lU d 'n Chapter 12.

Sagging moment at midspan - 2 ,., r2 of..

Hogging moment at support - - 2 " r1 o f..'

Maxrmum torsional moment 2' r2 a 'A" fh ntrc \me 0! th l\1\U ·n

e maxtmum torsional moment occurs at 12.75o trom the

I

II till Rl 1 I I , " IR!Il' ltml S

I)' 70 Ull

~ J n ' 70. J 2726 \ ,... ~

U1 mum Ill -7 ;:6 0 045 1.!27 kNrn

- :.6 0 OSQ - 2.J2 6 1-:Nm

'"'7"6 - ... 0009 24.5 kNm

p t:eru;fOll c u t"d b) the 01 nJronal thrust from the bottom dome

D r1 cos ax ~ 72.54 x o 73 x 4.12s -

- 218.4 1-:N

Th nn be.lm b !U t beltm the wa~l and b~lcony. Thus, part _of vertical Wall anct I 1 rn the force mducc:d m the rmg beam. The st1ffnesses of Wall n. o • J and 1 OO} re quue high. Th ring beam should be destgne

the uppon for a bendmg moment =- 142 6 kNm and direct tension , 218.4 kN

M Eccentncllv ~ - - - I II 0 mm > 350 mm

T

u) rh mrd Jhin for a bendmg moment 122.7 J..:Nm and direct tension - 218.4 kN

M EccentriCit) e - 562 mrn 350 mm

T (Ill) 14 r hear force due to 70.-l kN m load and twisting moment 24.5 kNm at 12.75o

fr m the centre line ofrhc column

I )

Hn r

• d

C un~:d pan c c

urved dear pan

rrx8.6

6 4 50 m

4.50 - 0.30 4 . .:20 m

r fl rce at the face of rh · column due to 70.4 k: N/m load

r f '" t the clwn of ma unum tor ional moment

11) 5

- 70 x (} CJ57 ' 91 kN

f'fu•v I nt hcarforceVe 1 v t- 1.6-h

d 1 nth of 0 r tlfru

CJI + J.6

146 kN

24.5

0.35 203 kN

twn lor b ndin moment using l;qs. 22.21, 22.22 and 22.23, Jn~ Pr Vi 8-l 6 mr11 bar in the rin , beam . A I so, provide I 0 ~ •

u 200 mm <. c .1 •.howJJ in Hg 22.20f.

1 OQI- 2 LEGGED STIRRUPs o

200 I ~ Cft 1 ~~~ COLUMN

Fig. 22.20 (t)

Reinforcement details in . Circular 0 Lr

1 · n. ank 12 JOINTS zz. e categorized as follows : Joints ar

oovement Jomts - There are three types of mo . (I) tv'' vement JOin .

(a) Contraction Joint is a movement jotnt w' h d but no initial gap between the concrete 0~

1 .tha eliberate d1 c ntmu •

f h · · · · et er std of th jom The p~~os~ o t ts JOint ts to accommodate contractton o on A dtstmctton should be made between a complet ''-'"''"

. F. 2., 21 . e Cotltnlctlon JO shown m tg. ~. a.m which both concrete nd reinfcrci

1 ~

interrupted and a parttal contraction JOint as sho"n in Ft .2 1

b which only the concrete is interrupted, where the rein~ rem st~l run through. A water bar should be provtded either centr \h

10 11 or

on the soffit of a floor. •

CONCRETE BUT NO DISCONTINUITY IN INITIAL GAP WATER BAR

·. A

. . 0.· 6

•

OISCONTINUilY IN STEEL

) 1 t contraction joint (a omp .. t Fig. 20.11 ontrac:tion JOln

.... .., • ) 0

.L\

UQUil) Rl IAININO SIIUJ("flJRI S

JOINT SEALING COMPOUND

6

..

. 6 . .

STRIP PAINTING

6

•6.

DISCONTINUITY IN CONCRETE BUT NO INITIAL GAP

CONTINUITY OF STEEL

(b) Partial contraction joint Fig. 20.21 Contraction joints

(hi Erpanston Joint is a movement joint with complete discontinu·ty . d . d I In bolh reinforcement and concrete an mten ed to accommodate . h

_ . en er expansion or c~ntract_wn of the structure. An expanston type water bar should be provtded erther centrally m a wall as shown in Fig. 22.

22 or

on the soffit of a floor. A centre-bulb water bar may be used in Wall In general, such a jomt requires the provi~ion of an initial gap betwee~ the adjoining parts of a structure which by closing or openin accommodates the expansion or contraction of the structure. g

INITIAL GAP

~- ~

JOINT FILLER ;/'

DISCONTINUITY IN BOTH

CONCRETE AND STEEL

ig. 22.22 f.tpansion joint

ILLUS'l RA ltV F., EXAMI't,Hs

7 .

11 , Jmnt is a movement jo•nt '.11\th .

·) .<hd< ~co mont •nd '"""ao " •h•<lt :""•loo •-..,., ~boo, 1

' re•nfo " "'"'" "'"""''"' in tho ,..::"1 ""''"" • mOdo ~ f '" I ot

0

1; on ;, botw «n wall ""d 0""' ; of tl<o ioim A typ""

applica. p· 22 23. n IOtne tylindrltal t-• shown m rg.

. . fl

A

A

.· A .

STRIP PA!NitNG

JOINT SEALING COMPOUNo - -1 ..

·a, • A·

PREPARED SLIDING SURFACE OR RUBBER PAD

Fig. 22.23 Sliding joint

1) Construction Jomt ~ A join~ in the concrete introduced for convenience tn (.. struction at wh1ch spec1al measures are taken to achieve subsequ • con . . &' fu h . en,

tinu itv without provtsion •Or rt er relative movement is called construction con • 1. . . b .

. . t A typical app 1cat10n IS etween successwe lifu in a reservo1r wall as JOin · h . . d

hown in Fig. 22.24. T e pos1t1on an arrang~:ment of all construction joints 5

hould be predetem1ined by the engineer. Consideration should be giwn to 5

1. iting the number of such joints and to keeping them free from possibility of 1m . . . percolations in a similar manner to contraction JOints

. '

PREPARED JOINT SURFACE

CONTINUITY OF STEEL

Fig. 22.24 Construction joint

IIQUIPIU I I I 'to SIRtJCiliRI· S

t .J, I t Jl tempttlllrtl\ I n bet .. , . . .. l'n the 1 tnt ·llJr '' htdt ·•Iter a \II liable 1111 lllncr • rvaJ a d tte

ut rnto u . 1 ttflcd "''h mortar or ~:un~:rctc e 1 n hefur HI hh

. , . litre elh r "11 t rn~· u~wn of uituble · · . 0 1llplet 1 e . fOJOhn" e y

1d c In th• tom1er c:1se the \\ tdth 1. o tnaten•• ilt . _ o the .,. "'' tde tu bt prep.tred before hlhng. •AP should bt

PREPARED JOINT SURFACEs

(a)

INITIAL. GAP LATER FILLED WITH CONCRETE

STRIP ~

PAINTING

INITIAL GAP -JOINT SEALING

COMPOrD P

(b)

(c)

I

JOINT SEALING COMPOUND

STRIP PAINTING

A

- MORTAR FILLING

ig. 22.25 Temporary open joint

•\ ,~tJ /(' '

ll.LUS I RA'IIVI

fJ oE·TAifiNU Of JOIN1S

• '-XAMI'I.I ~

. movement JOint should . 01 J atrn .

,~,, t'lcient tunct10n1ng : iil lKht VC'- [ore '' v,n~ th ~,e> e follow

# . t should accommodate rcpe~t d tng destrabt 1orl1 u e mo

I roe tightness vernent of n I' ~~· ater . tt tructur

d si••n should prov1de for exclu~· e With 1 1~ f

rhe e "' . . ton of - o bl ·ng of the JOmt. gr11 and d b l the ctost . e rt~ "'htch Wo

aterial used tn the constructi uld Pt'event fhe rn . on of Ill

( 1 ·ng properties : overnel'lt . ' folloW I )OlltlS h ould ha\e the

(i) It should not suffer permanent d' d1splaced by flutd pressure •stonton or elltru

ton and h ld (ii) Jt should not slump unduly in hot ou not be

weather. Weather or b ecome brmle m cold

(rii) It should be insoluble and du bl 1- h ra e and h exposure to tg tor by evaporati f s ould not be

on o solvent or pi . . affected b) · 1 h asttctsers

(iv) In spec•a cases, t e materials sh ld . I d b- ou be non t . to chem 1ca an •ological action · • 0Xlc, taintless 0 as may be specifted r reststant

)f4C!NG Of JOINTS

J ints in general and construction joints in partie 1 o h u ar are the deti ·

.,,cture and, therefore, t e number of joints should be k .'c1ent areas in the lu" -d- - bl . ept to the mtnlmu Th' "'realized by provt mg sutta e remforcement for cont 1 f m. IS may l' • . . ro o cracks If it· ~

movement or JOint should be provtded at the design 5 . · ts not teas•ble, . . pacmg. The movem t · ·

~~ould be prov1ded at the followmg spacings: en JOtnts

(a) In reinforced concrete floors, movement J-oints should b d . . e space at not mon: than 15 m apart tn two dtrections at right anoles The wall a d fl · . . _ . . e - n oor JOmts should be m I me except where shdmg JOints occur at the base of the y,a\1 in wh1ch case correspondence is not so important.

(b) For floors with only nominal percentage of remforcement, the concrete tloor should be cast in panels with sides not more than 4.5 m.

(c) In concrete walls, the vertical movement joint~ should normal\) be pia d at a maximum spacing of 15m.

{d) Amongst the movement joints in tloors and wall , e ·pan ion joints hould normally be provided at a spacing of not more than 30m bet\\etn 'UC 1\t

exransion JOints or between the end of the structure and the next e. P n.tOI\

JOint, all other joints being of the contraction ty~.

(e) If the temperature change to be accommodated are abno~a\ ~r occur 70: frequently than usual as in the case of storage of arm rtqutd or m untn u at

root slabs, a spacing smaller than 30m ·hould be adopted.

I Jllllll> RL I' ININtr S'l RlH II)\ hS •

H. hi'E ltt:N< 'ES

0 (I 96 ) 'onct etc .\'trllt Ill/ I' for the Stomgc of l.llfllll/s, l';ut I

E ur 1u oflmh Ill Si.md,u I.. Nl'" I >clh i 10

IV

1 , ; (-00 dr.llll 'oncrctt ' Stmctun· fin· tlze Stora~c of LuJuicls P· Bureau ,,f lrHti n Standards, N~'' lklhi . ' uri

1 1111d II

1 · ·. II''• ;'l •19,~5) ('riteriajor J)e.,ign ofRC( StaginK f'or 0\'<'rheud "' u - ' ' to (1/('r r k.s Burc, u oflndtan ._'tandurds, N~w Delhi. U/1 ·,

EXERCISES

ectton 1

subjc~tcd t? a direct tensio~ of~OO kN/m and a moment of 30 kNm De ign the ~cction on ( 1) uncrackcd basis, (II) cracked bas1s. Use M 25 c · - • oncrete and e 415 grade steel.

- · 200 nun thick section of a wall is subjected to a direct pull of 50 kN and bending moment of 15 kNm in the horizontal plane. It is remforced with IO a

. d h . mm bnr (fL 125 mm clc on each face. Ftn t e max1mum stresses in concrete and steel if concrete is of M 25 grade and steel is H. S. D. bars.

3. 0 ign a rectangular water tank of 250 kL capacity in a space of 15 m x 5 m area. Jt i a covered tank and placed 1m below the ground level.

4. De 1gn a ckar water reservoir of 2000 kL capacity. lt is square in plan and completely under-ground. The depth of the tank should not be more than 6 m. The net b~.:aring capacity of the earth is I 00 kN/m2 and depth of water table is 3 m. As ume necessary data and prepare detailed drawings.

5. Rcde ign the clear water reservoir of exercise 4 if it is circular in plan.

6 Oe 1gn n Circular overhead water tank of 200 kL capacity over a staging of 25m. A ume uitable proportions and prepare detailed structural drawings.

a 0.