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PRESTRESSED CONCRETECIRCULAR STORAGE TANKS

AND SHELL ROOFS

11 .l INTRODUCTION

Prestressed concrete circular tanks are usually the best combination of structural formand material for the storage of liquids and solids. Their performance over the past half-century indicates that, when designed with reasonable skill and care, they can functionfor 50 years or more without significant maintenance problems.

The first effort to introduce circumferential prestressing into circular structures wasthat of W. S. Hewett, who applied the tie rod and turnbuckle principle in the early 1920s(Ref. 11.6). But the reinforcing steel available at that time had very low yield strength,limiting the applied tension to not more than 30,000 to 35,000 psi (206.9 to 241.3 MPa).Indeed, significant long-term losses due to concrete creep, shrinkage, and steel relaxationalmost neutralized the prestressing force. As higher strength steel wires became avail-able, J. M. Crom, Sr., in the 1940s successfully developed the principle of winding high-tensile wires around the circular walls of prestressed tanks. Since that time, over 3,000circular storage structures have been built of various dimensions up to diameters in ex-cess of 300 feet (92 m).

Two 583,000-bbl (92,.500-m3) double-wall prestressed concrete tanks for liquefied natural gas stor-age, Philadelphia. (Courtesy, N.A. Legatos, Preload Technology, Inc., New York.)

11.2 Design Principles and Procedures 643

The major advantage in performance and economy of using circular prestressing inconcrete tanks over regular reinforcement is the requirement that no cracking be al-lovved. The circumferential **hugging” hoop stress in compression provided by externalwinding of the prestressing wires around the tank shell is the natural technique for elimi-nating cracking in the exterior walls due to the internal liquid. solid. or gaseous loads thatthe tank holds. Other techniques of circumferential prestressing using irlcliviciurrl tendonsvvhich are anchored to buttresses have been more widely used in Europe than in NorthAmerica for reasons of local economy and technological status.

Containment vessels utilizing circumferential prestressing. which can be eithersitu-cast or precast in segments. include water storage tanks. wastewater tanks and ef-fluent clarifiers. silos. chemical and oil storage tanks. offshore oil platform structures.cryogenic vessels. and nuclear reactor pressure vessels. All these structures arc consid-ered thin shells because of the exceedinglv small ratio of the container thickness to its di-ameter. Because no cracking at working-load levels is permitted. the shells are expectedto behave elastically under vvorking-load and overload conditions.

11.2 DESIGN PRINCIPLES AND PROCEDURES

11.2.1 Internal Loads

Considering the behavior of circular tanks involves examining both the interior pressuredue to the material contained therein acting on a thin-walled cvlindrical shell cross sec-tion and the exterior radial and sometimes vertical prestressing forces balancing the inte-rior forces. The interior pressure is horizontally radial. but varies vertically depending onthe type of material contained in the tank. If the material is water or a similar liquid. thevertical pressure distribution against the tank walls is rl-iarzglrlrrr, with maximum intensityat the base of the wall. Other liquids which are accompanied by ’ *as would give a constmtbhorizontal pressure throughout the height of the wall. The vertical pressure distributionin tanks used for storage of granular material such as grain or coal would be essentiallysimilar to the gas pressure distribution. with a constant value along most of the depth ofthe material contained. Figure I I. 1 shovvs the pressure distributions for these three casesof loading.

The basic elastic theory of cylindrical shells applies to the analysis and design of thewalls of prestressed tanks. A rin g force causes ring tension in the thin cylindrical walls,assumed unrestrained at the ends at each horizontal section. The magnitude of the forceis proportional to the internally applied pressure. and HO vertical moment is producedalong the height of the walls. If the wall ends are restrained. the magnitude of the ringforce changes and a bending moment is induced in the vertical section of the tank wall.The magnitudes of the ring forces and vertical moments are thus a function of the degreeof restraint of the cy,lindrical shell at its boundaries and are computed from the elasticshell theory and its simplifications and idealizations to be discussed subsequently.

Liquid Load and Freely Sliding Base. From basic mechanics. the ring force is

(1 l.la)

and the ring stress is

where ti = diamctcr of cylinderI’ = radius of cylinder

(ll.lb)

6 4 4 Chapter 11 Prestressed Concrete Circular Storage Tanks and Shell Roofs

0, = ring thrust or shearMO = restraining moment

at base of f ixed wall

(a) (d)

Figure 11.1 Tank internal pressure diagrams. (a) Tank cross section, showingradial shear Q, and restraining moment MO at base for fixed-base walls. (b) Liquidpressure, triangular load. (c) Gaseous pressure, rectangular load. (d) Granularpressure, trapezoidal load.

r = thickness of wall corep = unit internal pressure at wall base = yHy = unit weight of material contained in vessel.

The tensile ring stress nt any point below the surfke of the material contained in the ves-sel becomes

fR = y(H - y,g = y(H - I’) 5 (11.2a)

where H is the height of the liquid contained and y is the distance above the base. Thecorresponding ring force is

F = y(H - y)r (11.2b)

The maximum tensile ring stress at the base of the freely sliding tank wall for y = 0 be-comes, as in Equation 1 l.lb,

(11.2c)

Gaseous Load on Freely Sliding Base. Again from basic principles of mechanics.the constant tensile ring stress is

( 11.3)

Note that while theoretically the centerline diameter dimension is more accurate to use,the ratio r/d is so small that the use of the internal diameter O’ is appropriate.

Liquid and Gaseous Load on a Restrained Wall Base. If the base of the wall isfixed or pinned, the ring tension at the base vanishes. Because of the restraint imposed

Photo 11.1 4.0 Million Gallon Preload Tank, City of Troy, Ohio. (Courtesy, N.A.Legatos, Preload Inc., Garden City, New York.)

on the base, the simple membrane theory of shells is then no longer applicable, due to theimposed deformations of the restraining force at the wall base. Instead, bending modifi-cations to the membrane stresses become necessary (see Refs. 11.2 and 11.6) and the de-viation of the ring tension at intermediate planes along the wall height must beapproximated as in Ref. 11.2 and the discussion in Sec. 11.3.

If the vertical bending moment in the horizontal plane of the wall at any height isMY, the flexural stress in compression or tension in the concrete becomes

f, = fC = $ = T per unit height

The distribution of the flexural stress across the thickness ofFigure 11.2.

(11.4)

the tank wall is shown in

(al (b)

Figure 11.2 Ring tension and flexural stresses. (a) Ring tension internal force Fin the horizontal section. (b) Flexural stress due to bending moment M in the wallthickness of the vertical section.

646 Chapter 11 Prestressed Concrete Circular Storage Tanks and Shell Roofs

11.2.2 Restraining Moment MO and Radial Shear Force 0,at Freely Sliding Wall Base Due to Liquid Pressure

11.2.2.1 Membrane Theory. The study of forces and stresses in a circular untrackedtank wall is an elasticity problem in cylindrical shell analysis. If the shell is free to deformunder the influence of the internal liquid pressure. the basic membrane equations ofequilibrium apply. The longitudinal unit force N,. the “hugging” circumferential unit

force N8, and the central unit shears NH and N,, are shown in the differential element ofFigure 113(b). Note that these follr unknowns all act in the plane of the shell.

The basic three equations of equilibrium for these four unknown unit forces are

dN,H;l\l+pHr==O (I 1%)

(a)

Y

t f

No +

No, +

P = YHliiiiilTH

I

t-4(c)

a Noa e

%I,a s

E

N, + 2 dy N,, + 2 dy

(b)

x

0

z

(d)

Figure 11.3 Membrane forces in cylindrical tank. (a) Tank shell geometry.(b) Shell membrane forces. (c) Liquid-filled tank elevation. (d) Axisymmetrical in-ternal pressure at any horizontal plane.

Photo 11.2 Panel Being Lifted in a Preload Prestressed Tank (Cuurtesy, N.A.Legatos, Preload Inc., Garden City, New York.)

dN. dNfhr-...2+.dY

ae +p,r=o

Nfl- = +pz = 0I

(11Sb)

(11Sc)

where aN,, = JN,, due to loading symmetry. The unknowns are thus reduced to three,representing a statically determinate structure subjected to direct forces only.

For axisymmetrical loading as in Figure 11.3(c), pe = py = 0 and pz = p * f(y), inde-pendent of 0. Hence,

Pz = -m-Y) (11.6)

and the solution to Equation 11.5 is

NY, = NY = 0

and

No = YW - Y)’ (11.7)

11.2.2.2 Bending Theory. The introduction of restraint at the boundary of the vesselinduces radial ring horizontal shear and vertical moments in the shell. Consequently, themembrane force equations presented in the previous section have to be modified by su-perimposing these additional moments and shears. The modified expressions are de-


noted the bending theory of circular shells; the theory accounts for strain compatibility re-quirements in the induced deformations caused by the induced shears and moments.

The bending moments and central shears in the axisymmetrically loaded cylindricalshell are shown by force and moment vectors in Figure 11.4. The infinitesimal elementABCD shows the points of application and sense of the unit moments MJ about thex-axis and M, about the y-axis, the circumferential unit moments M,, and AI,,., the unitnormal shear Q,. acting in the plane of the vertical shell generator and perpendicularly tothe shell axis, and the unit radial shear Qe acting through the shell radius in the plane ofthe shell parallels.

Superposition of the moments and shears in Figure 11.4 on the forces in Figure11.3(b) results in the following equilibrium equations:

aN, aN,.,

aB+~ - Qe + per = 0

aY

dN.Ir+

aNtI,

ay~ + py = 0

a0

aQe at?,tl+$r+N,,+pzr=O

aM, aMbH-pr+pay ay

+ Qvr = 0

i)M, aM VH-+-

a0 ayr - Qer = 0

(11.8a)

( 11 .Sb)

(ll.Sc)

( 11.8d)

(ll.Xe)

Due to symmetry of loading, N,., = N,,. = M., = M,,. = 0, and dQH can be disregarded,reducing the partial differential equations 11.8 to the set of the ordinary differentialequations

dN>pr + p,r = 0riy

(ll.C)a)

A shell

Figure 11.4 Bending moments and normal shears in a cylindrical shell wall.

11.2 Design Principles and Procedures 6 4 9

dQy-r+N,+p,r=Ody

d4--r+Q,r=OdY

(11.9b)

(11.9c)

With the central membrane forces NY constant and taken to be zero (see Refs. 11.1 and11.3) the remaining equations 11.9b and 11.9~ can be written in the following simplifiedform having the three unknowns N,, Q,, and My:

dQ,-++N@=-p;dydM.v--Qy=Ody

(ll.lOa)

(ll.lOb)

In order to solve these equations, displacements have to be considered and equations ofgeometry developed.

Force Equations. If v and w are the displacements in the y and z directions, thenthe unit strains in these directions are, respectively,

dv% = dy

and

Wl o=--r

which give

(lllla)

Photo 11.3 250,000-bbl (39,750-m3) prestressed concrete propane gas storage con-tainer, Winnipeg, Manitoba, Canada. (Courtesy, N.A. Legatos, Preload Technol-ogy, Inc., New York.)


a n d

N, =Et7 (9, + WI =1 - IJ--

(Il.llb)

where k = Poisson’s ratioI = thickness of the wall core.

From Equation Il. 11 a,

(11.12a)

From Equation 11.11 b.

N, = -E$ (1 l.12b)Y

Moment Equations. Due to symmetry. there is no change in curvature in the cir-cumfercntial direction: hence. the curvature in the J’ direction has to be equal to-tf \Yc!\*~. Using the same moment expressions for thin elastic plates results in

M,, = FM,

M, =-D$

where D = E?/lZ( 1 - k-‘) is the shell or plate tlexural ricgidity.Introducing Equations I 1.12 and 1 I. 13 into Equations 11.10 results in

(I l.l.%)

(1 1.1%)

If the wall thickness t is constant. Equation I 1.14 becomes

p’=L= 3(1 - $)4r’D (rt)’

Equation I 1 .lS becomes

(11.14)

(11.15)

(I 1.16)

Equation 11.16 is the same as is obtained for a prismatic bar with flexural rigidity D sup-ported by a continuous elastic foundation and subject to the action of a unit load intcn-sityp-. The general solution to this equation (Ref. 11.1) for the rrrtiirrl displacement in the:-direction is

(11.17)

where ,f(>jJ is the particular solution of Equation 1 I. 16 as a membrane solution @ving dis-placement


11.2.3 General Equations of Forces and Displacements

Solving Equation 11.17 and introducing the notation

the expression for radial deformation in the z direction and its consecutive derivatives atany height y above the wall base can be evaluated from the following simplified expres-sions as a function of the wall base unit moments M,, and unit radial shears Q,,:

Deflection 1%‘ = ~ ______’ [PM,d@~~) + Q,~‘(PJ~)I2p”D(11.18a)

(11.1%)

(I 1.1Xc)

(ll.lXd)

The shell functions @(PJ,). $(PJ‘). O( PJ*). and j(py) are given in the standard influence co-efficients of Table I 1.1 (Ref. I 1.1). for a range 0 5 PI* I 3.9.

The maximum radial displacement or deflection at the restrained wall base. fromEquation 11.18a. is

(W),=(, = - ,k, (PM,, + QJ (I 1.19a)

and the maximum rotation of the wall at the base, from Equation 11.18b. becomes

, -0 = & CW4,, + Q,J (11.19b)

where M,, and Q,, are respectivelv the restraining moment and the ring shear at the baseshown in Figure 11.1.

For tanks with constant wall thickness. the unit forces along the wall height are asfollows:

M, = PM, ( 11.2Oc)

(11.20d)


Table 11 .l Table of Functions @, 9, 0, and i

PY a

0 1 . o o o o 1 .OOOO 1 .OOOO 0

0.1 0.9907 0 .8100 0.9003 0.0903

0.2 0.965 1 0 .6398 0 .8021 0 .1627

0.3 0 .9267 0 .4888 0 .7077 0.21x9

0.3 0 .8784 0 .3563 0 .6174 0 .2610

0.5 0.823 1 0.731 s 05323 0.2YOX

0.6 0 .7628 0.1431 0 .4530 0 .3099

0.7 0 .6997 O.OSYY 0 .3798 0.;199

0.8 0 .6354 -0.0093 0.3131 0 .3223

0.9 0 .5712 -0.0657 0.2527 0 .3185

1.0 0.5083 -0. I 108 O.lY88

1.1 0 .4476 -0.1457 0.1510

1.2 0 .3899 -0.1716 0.1091

1.3 0.3355 -0.1897 0.072Y

1.4 0 .2849 -0.201 1 0.04 19

15 0.2383 -0.2068 O.OlSX

1.6 O.lY5Y -0.2077 -O.OOSY

1.7 0 .1576 -0.x47 -0.0235

1.8 0 .1234 -0.1985 -0.03761.9 0.0932 418YY -0.0484

2.0 0 .0667 -0.1791 -0.0s632.1 0.043Y -0. I675 -0.06 1 X2.2 0.0244 -0.1548 -0.06522.3 0.00x0 -0.1416 -0.066X2.4 -0.0056 -0.1282 -0 .0664 ,

2.5 -0.0166 -0.1 14Y -0.0658

2.6 -0.0254 -0.1019 -0.0636

2.7 -0.0320 -o.oxYs -0.0608

2.8 -0.0369 -0.0777 -0.0573

2.9 -0.0403 -0.0666 -0.0534

3.0 -0.04233.1 -0.043 I3.2 -0.043 1

3.3 -0.04223.4 -0.04081-2.3 -0.0389

3.6 -0.03663.7 -0.034 13.8 -0.03143.Y -0.0286

-0.0563 -0.0393-0.0469 -0.0350-0.0383 -0.0407

-0.0306 -0.0364-0.0237 -0.0323

-0.0177 -0.0283

-0.0124 -0.0235-0.007Y -0.02 IO-0.0040 -0.0177-0.0008 -0.0147

* 0 5


From Equations 11.18~. 11.18d. 11.2Ob, and 11.20d, the expressions for vertical momentsand horizontal radial shears at the base of the wall, where y is zero, become (Ref. 11.1)

(Q,),=c = Qo = - (WH - 1) v-tm)

(11.21a)

(11.21b)

The expression for the vertical moment at any level y above the wall base can be ob-tained from

(11.22)

The o,ffwt ring shear force AQY corresponds to a radial displacement w, of the wallat a height y above the base when the tank is empty and the values of Q, and M, due to afull liquid or full gas load are induced, as shown in Figure 11.5. This force can be ex-pressed as either

o r

AQ, = + h(:3-r!f“[PW~~ + QoWY)I (11.23)

The ring shear Q, at a plane ~3 above the base would be equal to the difference betweenthe ring force for a freely sliding base and AQ?:

Q, = F - AQ, (11.24)

It is important to be consistent in the sign convention used throughout a solution.The easiest approach is to draw the deflected shape of the wall and use a positive (+) no-tation for the following conditions:

1. Moment causing tension on the outside extreme fibers.

2. Ring tension radial forces.3. Thrust inwards toward the vertical axis. Here. the same sense is used as for ring ten-

sion forces in order to draw the diagram for the balancing prestressing forces on thesame side as the ring tension forces for comparison.

4. Lateral wall movement inwards toward the vertical axis.

5. Anticlockwise rotation.

Pinned Wall Base, Liquid Pressure. When the wall base is pinned and carrying aliquid load moment M,, = 0 at the base.

Q,, = +2P-‘yH(rt)’

12(1 - t.P)


R e d u c t i o n

T

Tankdepth

f f

I--Ring tensionfreely sliding

id)

Ring tensionfreely sliding

(el

Figure 11.5 Wall base restraint in empty tank inducing MO and C?,, for full liquidor gas pressure. (a) Deformed walls of empty tank. (b) Moment along vertical sec-tion ( + represents tension on outside). (c) Ring tension force Fin horizontal sec-tion (always positive). (d) Offset AQ, for liquid pressure. (e) Offset IQ,, for gaspressure.

or

(11.15)

The value of the shell constants (3. p’. and p’ for use in the preceding equations can easilybe computed from the expression for p’ as follows:

(I 1.36a)

( 1 1 .?hb)

pJ=E’= 3(1 - p2)3;o (?$

P‘ =[.i(l - /A’):” 4

(r-t): 2

@ = c-31 ~ tw 2(rt)

(1 1.26c)

Photo 11.4 Wire Winding Operation (Courtesy, N.A. Legatos, Preload Inc., Gar-den City, New York.)

@ = [3(1 - F2P(rt)‘j2

(11.26d)

11.2.4 Ring Shear Q, and Moment MO Gas Containment

If the edges of the shell are free at the wall base, the internal pressure produces onlyhoop stress fR = pr/t and the radius of the cylinder increases by the amount

rfR v2W=F=Et

Also, for full restraint at the wall base,

(W)v=” = &PM, + Qdand

= --&PM, + Q,) = 02P2D

Solving for MO and Q, gives

Mo=-2p2Dw=-_4L.=- prt2P2 41211 - /L2)

and

Q,=+4B3Dw== +p= +p(2rt)“’

P [12(1 - $)]“4

(11.27)

(11.28a)

(11.28b)

(11.29a)

(11.29b)


Table 11.2 Equations for Liquid-Retaining Tanks

Parameter Equation Number

Flexural rigidity. DRing s t ress . f,

Ring force. FPressure . P,

Radial deflection. w

Rotation $

Maximum deflection. (M,),=,~

Maximum rotation $c ‘>, /,

Q,, = (Q,L,,

M,

Empty tank offset. lQ,

Q,

Q,, when M,, = 0 (P inned base)

Tank Cons tan ts : p-’

P’

P

Et’4 12( 1 - CL?)]y(H ~ Y)r/tY(H - ?.)rY(H -VI

’ IW’,JJ(PY) + Q,,WPY)2p‘ D

’ PW,,Wv) + Q,P(PJ)I2p’ D

& (PM,, + Q,,)& G’W,, + Qd = 0

yHrt

12(1 ~ I*$

+ (WH - 1) q&

+ +‘l’!B?.i + Q,,i@~)l

+ 6(;3-t;)’ lP~,A(P>,) + (Q,,(Px)Ir+(F--1Q,)

yHVrr,.2+ [12(1 ~ I*‘)]”

[3( 1 - k’)]“l(rtJ”[3( 1 - k2]’ ‘irf[3( 1 - p?)]’ ‘/(rt)’ 2

11.2 a11.2 b

11.2 b

11.18 a

1 I.18 b

11.19 a

I I.19 b

11.21 a

11.21 h

Il.22

11.23

Il.24

I I .2s

I I .26 hI 1.26 c

11.26 d

Pinned Wall Base, Gas Pressure. If the wall base is pinnrti and carrying a gas loadmoment M, = 0 at the base.

Q 0

or

r t ”Q() = [12(1 Y $)]I -I 20

( 11 JO)

Table 11.2 presents a summary of the design equations for liquid-retaining tanks.and Table 11.3 gives a similar summary for gas-retaining tanks.

11.3 MOMENT MO AND RING FORCE C?,, IN LIQUID RETAINING TANK

Example 11.1

A pres t rcssed concre te c i rcu lar tank i s fu l ly res t ra ined a t the wal l base . I t has an in ter ior d i -ameter ri = 125 ft (38.1 m) and retains water having height H = 25 ft (7.62 m). The wall thick-

11.3 Moment M, and Ring Force Q, in Liquid Retaining Tank 657

Table 11.3 Equations for Gas-Retaining Tanks

Parameter Equation Number

& (P.‘lf,, + Q,,) 1 1.28 a

11.28 b& (W% + Q,,) = 0

p,‘tI I.29 a

I 12(1 ~ CL?)

o,, = (42, ),-,I

Q,, \vhcn .Lf,, = 0 (Pinned base)

ncss t = 10 in. (25 cm). Compute (a) the unit \t‘rtical moment M,, and the radial ring force Q,,at the base of the wall. and (b) the unit wrtical moment ‘21, at 74 ft (2.29 m) above the base.CJse Poisson’s ratio TV = 0.2 and unit water weight y = 62.3 lbift: (1 .OOO kg/m”).

Solution:

Ir = 7 x 125 = 62.5 ft (19 Ill)

t = 10 in. = 0.S ft (3 m)

From Equation I I .Xd.1

p = 13(1 ~ I*-);’ ’ = [3( 1 - 0.2 x 0.2)]’ -I

(rt)’ 2 (623 x 0.83)’ 2= 0.181

From Equ;ltion I 1 .2 I a.

= - 18.574 ft-lb ft (7.68 kN-m 111) of circumference

From Equat ion I I .2 1 b .

Q,, = +(3PH - 1)yrt

VlZ(1 - I*‘)

= +7.677 lb ft (112 kN;m) of circumference

Water height = (II ~ !,) = 25 ~ 7.5 = 17.5 ft (5.33 m)

Chapter 11 Prestressed Concrete Circular Storage Tanks and Shell Roofs

Height ratio = 1 - $ = 1 ~ g = 0.7c >

pc = 0.181 x 7.5 = 1.36

From Equation 1 1.22.

I

M, = + (1:x1 (~0.181 x 18.574 x 0.311 + 7.677 x 0.37)

= +-I.012 ft.lb ft of circumfurcnce

11.4 RING FORCE Qy AT INTERMEDIATE HEIGHTS OF WALL

Example 11.2

Compute the radial ring force Q, in Example I I.1 at (a) J‘ = 7) ft (2.79 m) and (h) J‘ = IO t’t(3.05 m) above the wall base.

Solution: The freeI> sliding base rins force F = yHr = 63.3 x 25 x 61.5 = 97.500 Itdft ( I .17.3kN/m). From Equation 11.23. the rinp force offset is

From Example 1 1, I. p = 0. I81 : hence. p.’ = O.OWY.(a) Q, (it 7.5 ft ntm~~ Wdl Bmc

p!, = 0.1x1 x 7.5 = I.33

From Table 11. I for BJ, = 1.36.

t/J = ~O.lYh5

h( 1 - 0.01)-1Q, = +-

0.005Y x 62.5(0.83)’

x ~0.1x1(~1t(.571)(~0.1065) + 7.677(+0.05‘43)]

= 24.431 lb ft (356 kN:m)

From Equation 11.2b. the ring force F = y(lf ~ >,)r = 62.4 x (75 x 7.5) x 62.5 = hS.250Ihift. So QTi = F- IQ, = 68.150 - 23.431 = 43.819 lbift (705 kN/m) of circumference. asshown in Figure 11.6(a): (a) At 75 ft abo~c the base: (b) At IO ft ahwc the baw.

(h) Q, (11 lO.O.fi nhow Wrrll Bmc

p>, = O.ltll x 10 = 1.81

FromTable 11.1 for p, = 1.81.

11.5 Cylindrical Shell Membrane Coefficients 6 5 9

(a) (id

Figure 11.6 Radial ring force profile. (a) At 71 ft above the base. (b) At 10 ftabove the base in Ex. 11 .l.

-1Q, =6( 1 - 0.04)

O.OOSY x 62.X~(O.83)~

x [0.181(-18.s73)(~0.1Y84) + 7.677(-0.0387)j = 8.387 lb:ft

The ring force F = y(H - ~)r = 62.3(25 - IO)625 = 58.500 Ihift. So Q,,, = F - .lQ, =5X.500 ~ 8.387 = 50.113 lbift (73 1 kN/m) of circumference. as shown in Fiyre 11.6(b).Compare how close this value is to Q = SO.1 15 lbift obtained by usins membrane cocf-ficients in Example 11.3.

11.5 CYLINDRICAL SHELL MEMBRANE COEFFICIENTS

The bending moment at any level along the height above the base of a cylindrical tankcan be computed from the bending moment expression for a cantilever beam. This is ac-complished by multiplying the cantilever moment values by coefficients whose magni-tudes are functions of the geometrical dimensions of the tank and which are termedtuctnhrutle co~fficierzrs. The basic moment expressions developed in Section 1 1.2 for thecircular container can be rearranged into a factor H’irlr denoting georurrry and a factoryH: or pH2 denoting crrr~filrver effecr, for liquid and gaseous loading. respectively (Ref.11.2).

The tank constant p in Equation 11.26d is a function of rr or rlt. where rl is the tankdiameter. Using Poisson’s ratio k F 0.2 for concrete. we have

p = [3(1 - & 4 1X) 1 .x4

(rf)’ ’ (rt)’ ’ ((if)’ ’

The factor l/PH used in the basic bending expressions of Section 11.2 can be rewritten interms of (r/r/H’)’ ’ since p = 1.84/(dt)’ ‘. The product pV can also be rewritten in terms ofA( H’/dr)’ ’ using y = AH, where y is the height above the base.

Consequently. the moment M,. of Equation II.22 in a wall section a distance yabove the base can be represented in terms of the form factor H’idr and the cantileverfactor y H-’ or p H’ as follows:

M, = numerical variant X form factor X cantilever factor


Photo 11.5 Two-and-a-half-million-gallon tendon prestressed concrete tank withthe horizontal and vertical tendons utilizing plastic sheathing to protect the pre-stressing steel from seepage through the wall. (ColLrtesy, Jorgenson, Hendricksonand Close, Denver, Colorado.)

or

variant x g x [yH71 o r pH2] (11.31)

The form factor H2/dt is constant for the particular structure being designed. Hence, theproduct of the variant and the form factor produces the membrane coefficient C, so thatEquation 11.31 becomes

M, = CyH’ (11.32a)

for a liquid load and

M, = CpH2 (11.32b)

for a gaseous load.

Tables 11.4 to 11.16 from Ref. 11.5 give the membrane coefficients C for variousform factors H2/dt and most expected boundary and load conditions. They significantlyreduce the computational efforts normally required in the design and analysis of shells,without loss of accuracy in the results. Using the membrane coefficients for the solution

Photo 11.6 Prestressing preload circular tank wall with wire winder. (Courtesy,N.A. Legatos, Preload Technology, Inc., New York.)

of the circular tank forces and moments should give results reasonably close to those ob-tained from the bending solutions presented in Section 11.2 and the sets of equationslisted in Tables 11.2 and 11.3.

11.6 PRESTRESSING EFFECTS ON WALL STRESSES FOR FULLY HINGED,PARTIALLY SLIDING AND HINGED, FULLY FIXED, AND PARTIALLY FIXED BASES

The liquid or gas contained in a cylindrical tank exerts outward radial pressure yh or p onthe tank walls, inducing ring tensions in each horizontal section of wall along its height.This ring tension in turn causes tensile stresses in the concrete at the outside extreme wallfibers, resulting in impermissible cracking. To eliminate this cracking that causes leaksand structural deterioration, external horizontal prestressing is applied which induces in-ward radial thrust that can balance the outward radial tension. Additionally, in order toprevent the development of cracks in the inside walls when the tank is empty, verticalprestressing is induced to reduce the residual tension within the range of the modulus ofrupture of the concrete and with an adequate safety factor.

In order to ensure against the development of cracking at the outside face of thetank wall, it is good practice to apply somewhat larger horizontal prestressing forces than

(text continues on page 676)


Table 11.4 Moment Influence Coefficients, Triangular Load

Moments in Cyl indr ica l Wal lTriangular LoadFixed Base. Free TopMom. = cocf. X yH’ ft. lb. per ft.Positive: siy ind ica t e s t ens ion in the ou t s ide

L iqu id Load

H2dt

0 . 4

0 . 8

1.2

I.62.0

3.04.0

5.06.08.0

IO.0

12.014.016.0

Coefficients at Point

0.1 H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H l.OH

+.002 1+.0063+.0077

+.0075+.006X

+.0047+.002x+.0016

+.000x+.0002

+.ooo 1

+.ooo 1.oooo

-.OOOl

+.0007+.00x0+.0103+.o IO7+.OOYY

+.007 1+.0017+.0029+.OOlr)

+.000x

+.0004+.0002

.OOOO-.0002

-.0042+.0070+.Ol 12+.Ol21

+.0120

+.OOYO+.0066+.0046

+.0032+.0016

+.0007

+.0003+.OOOl-.OOOl

-

-.0150+.0023

+.OOYO+.Ol I I+.Ol IS

+.0097

+.0077+.0059+.0036

+.1)02S

+.1)019+.oo I3+.000x

+.0004-

Notes: I-Tnbles I I .4 to I I, I6 Adapted from Ref. 1 1.5.

-.0302 -.(I520 -.0816 -. 1205-.0068 -.0224 -.0465 -.Ol%+.0022 -.0108 -.03 I I -.0602+.005x -.005 1 -.0232 -.osos+.0075 -JO2 1 -.01x5 -.0436

+.(I077

+.OMY+.0059+.005 I

+.0038

+.0029+.0023+.OOlY

+.OOl3

+.0012+.0023+.0028

+.0029+.002Y

-.Ollc)

-.0(x30-.005x-.004 I

-.0022

-.0012-.0005-.OOOl

+.OOOl

-.0333-.026X-.0222

-.Ol87-.0146

+.0028

+.0026+.0023+.OOlc)

-.0122

-.0104-.OOYO-.0079

2.O.OH 1s the top and 1 .OH is the bottom of the wall. except it’wall is flxrd at top and with shear and momt‘nt at top.3.Shear actins inv,ards is positiLL’: moment applied at an edge is posltlve when outaard rotation results at that edge.

11.6 Prestressing Effects on Wall Stresses 663

Table 11.5 Moment Influence Coefficients, Rectangular Load

Moments in Cylindrical WirllRcctan~~ular LoadFixed Base. Free TopMom. = cocf’. x’ pH’ ft. lb. per ft.Positi\,e sip indicates tension in the outside

H2-cft 0.1 H 0.2H 0.3H 0.4H


0.5H 0.6H 0.7H 0.8H 0.9H l.OH

0.40,sI.’1 .67.0

3.04.0i.06.0s.0

10.017.014.0Ih.0

-.0033

.oooo

+.ooos

+.oo I I+.OOlO

+.0007+.000-t+.0002+.ooo I+.oooo

.oooo.oooo.oooo.oooo

-.OOY3

-.0006

+.0036

+.003h

+.00.36

+.0026

+.oo IS

+.000x

+.001)4

+ . o o o 1

-.ooo I.oooo.oooo.oooo

-.0227-.0025+.0037+.0062+.0066

+.005 1+.0033+.OOlY+.oo I I+.0003

.oooo-.OOOl

.oooo

.oooo

-.043Y-.00x3+.002Y+.0077+.00X8

+.0071+.0052+.0035+.(I022+.000x

+.OOO?.oooo.oooo

~.OOOl

+.OOYl+.0068+.005 1+.0036+.OOlS

-.I018-.(I362-.OOSY+.oo I I+.OOSY

+.0083+.0075+.OOh 1+.004Y+.(X)3 I

+.(X)2 1+.0014+.OOlO+.0006

-.I455

-.OSY4-.0227-.OOY3-.oo I Y

+.0042+.oos3+.0052+.0048+.0038

+.om+.0(123+.oo 1X+.0012

-.2wo-.(I917- .046X-.(I670-.()I67

-.0053-.oo 13+.0007+.OOl7+.0024

+.0026+.0022+.002 I+a020

-.25Y3 -.3310-.I325 -.I835-.08lS -.I 178-.052Y -.0876-.(I389 -.07 I Y

+.0223 -.0483-.()I45 -.0365-.OlOl -.0293-.0073 -.0242-.0040 -.Ol84

-.(I022 -.()I47-.oo I 2 -.(I123-.0007 -.0105-.ooos -.OOY I


Table 11.6 Moment Influence Coefficients, Trapezoidal Load

Moments in Cyl indr ica l Wal lTrapezoidal LoadHinged Base, Free TopMom. = coef. x (yH2 + pH2) ft. lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion i n t he ou t s ide IBl-lIf* Coefficients at Point

2% 0.1 H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H l.OH

0.4 +.0020 +.0072 +.01510.3 +.OOl!, +.0064 +.0133

1 .2 +.0016 +.0058 +.Ol 11

1.6 +.0012 +.0044 +.OOY 12.0 +.0009 +.0033 +.0073

3.0

4.0

5.0

6.0

8.0

10.012.0

14.0

+.0004

+.OOOl

.oooo

.oooo

.OOOO

.OOOO

.oooo

.OOOO

.OOOO

+.0015+.0007

+.OOOl

.oooo

.oooo

.oooo

.oooo

.oooo

.oooo

+.0040

+.0016+.0006

+.0002

-.0002

I 16.0

-.0002

-.OOOl

-.ooo 1

.oooo

+.0230 +.0301

+.0207 +.0271+.0177 +.0237

+.0145 +.OlY5+.0114 +.0158

+.0063 +a092

+a033 +a057

+.0016 +.0034

+.0008 +.0019

.oooo +.0007

-.OOOl +.0002

-.0002 .oooo

-.OOOl -.ooo 1

-.OOOl -.0002

+.0348 +.0357 t.0312 +.0197

+.031Y +.032Y +.02Y2 +.01x7

+.0280 +.0296 +.0263 +.0171+.0236 +.OZS +.0232 +.0155

+.OlYY +.021Y +.0205 +.0145

+.0127 +.0152 +.0153 +.OI I I

+.00x3 +.0109 +.011x +.ow2

+.0057 +.0080 +a094 +.0078

+.0039 +.0062 +.0078 +.006X

+.0020 +.0038 +.0057 +.0054

+.OOl 1 +.0025 +.0043 +.0045

+.0005 +.0017 +.0032 +.0039

.oooo +.0012 +.0026 +.0033

-.0004 +.000x +.0022 +.002Y


Table 11.7 Moment Influence Coefficients, Empty Tank (Shear Applied at Top Base Fixed)

Moments in Cylindrical WallShear Per Ft.. Q. Applied at TopFixed Base. Fret Top

E m p t ytank

Mom. = coel’. X VH ft. lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion i n t he ou t s ide

Ringt e n s i o n c+

-

M o m e n t

H2 Coefficients at Point

2% 0.1 H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H l.OH

0 . 40 . xI.3I.62 . 0

3 . 04 . 05 . 06.0x . 0

10.017.014.0

16.0

+O.OY.J

+o.oxs

+0.0X?

+O.O7Y+0.077

+0.072

+0.06X+0.064+0.062-to.057

to.053+0.049+0.046

i-o.044

+o. 172+o. I35+0.132

+o. 122+O.l I5

+o. 100+o.o<x<x

+0.07x+0.070+0.0.5x

+O.O4Y+0.042+0.036

1-0.03 I

+0.240+o. I85

+0.157+o. 13’)-to. I26

+o. 100

+0.0x 1+0.067+0.056

i-O.04 1

+0.029+0.022+o.o 17

+0.012

+0.300+0.20x+(I.161

+o. 138+o. 1 I Y

+0.0X6

+0.063+(I.04710 .036+0.02 I

-to.012+0.007+0.004

+O.OOl

+0.354+0.220+0.159

to. I25+o. 103

+0.066

+0.043+0.028+0.01x

+a007

+0.0020 .000

-0.001

-0.002

+0.402 +0.44x +0.492

+0.224 +0.223 -to.219+O. 145 +&I27 +o. 106

i-0. I OS +0.08 1 +O.OSh+0.080 +0.056 +0.03 I

-to.044

-to.025+0.013+0.006

0 .000

+0.006-0.001

-0.003-0.003-0.003

-0.002-0.002-0.002

-0.002

+a025+O.OlO

+0.0030 .000

-0.002

-0.002-0.002-0.001-0.001

-0.002-0.00 I-0.001

0.000

+0.535 +0.578

+0.214 +0.208+0.0x4 +0.062

+0.030 +0.004

+0.006 i-o.019

-0.010 -0.024-0.010 -0.01 Y

-0.007 -0.011-0.005 -0.006-0.002 -0.001

-0.00 1 -0.0000.000 0 .0000 .000 0 .0000 .000 0 .000


Table 11.8 Moment Influence Coefficients, Empty Tank (Shear Applied at Top Hinged Base)

Moments in Cyl indr ica l Wal lMoment Per Ft., M. Applied at BaseHinged Base. Free Top

Em@T a n k

Mom. = coef. X !bf ft. lh. per ft.Pos i t i ve s i gn i nd i ca t e s t en s ion i n the ou t s ide

-

H2z

0.4

0.XI.2

1.62.0

3.04.05.06.0

x.0

10.012.0

14.016.0

0.1 H 0.2H

+0.013 +0.05 1

+O.OOY +o.o-to+0.006 +(I.027+0.003 +o.o 1 I-0.002 4.002

-0.007 -0.021-0.00x -0.026-0.007 -0.024-0.005 -0.0 I x-0.001 4OOY

0.000 -0.002

0.000 0 .0000 .000 0 .0000 .000 0 .000


0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H 1 .OH

+o. 109 +o. I96 +0.296 +0.-l I4 +0.547 +0.692 +().X-l3 +I ,000+O.OYO -to. I64 +0.253 to.375 +0.503 tO.659 +o.<s21 + 1 .ooo

+0.063 +0.125 +0.206 +0.3 1 h +0.454 +O.hl h +0.x02 + I .ooo+O.O35 +0.07s +(I.152 +0.253 +0.393 +0.570 +0.775 + I .ooo+0.013 +0.034 +().(I’)6 +o. I Y.; +0.340 +0.5 I Y +0.71x + I .ooo

-0.030 -0.02Y +O.OlO +0.0x7 +(I.227 +0.4x +OhY2 +I ,000-0.04 -0.05 I -0.034 +0.023 O.l50 +0..351 +O.h45 +I .ooo-0.045 -0.06 I -0.057 -0.015 +0.095 +0.X7 0.606 +I ,000-0.040 -0.05x -0.065 -0.037 +0.057 +0.x +0.572 +I .ooo-0.022 -0.044 -0.06s -0.063 +o.w +o. 17X +(I.5 15 +I . o o o

-O.OOY 402x -0.053 4.067 -0.03 I +o. 133 +0.467 +I .ooo-0.003 -0.016 -0.030 -0.06-1 -0.04Y +O.OSl +0.-l’-! +I . o o o

0 .000 -0.008 4.020 -0.05Y -0.060 +0.043 +03x7 +I .ooo+O.OO’ -0.003 -0.02 I -0.05 I -0.066 +0.03 +o..sl +I .ooo


Table 11.9 Shear Q Influence Coefficients

Shear at Base of Cylindrical Wall‘yH’ lb. (triangular)

Q = cod. x \ pH lb. (rectangular)

I M/H lb. (mom. at base)P o s i t i v e sip1 ind ica tes shear ac t ing inward

H* Triangular load, Rectangular load,dt fixed base fixed base

0 . 4 0.336 0.7550 . x 0374 0.5521.2 0.339 0 . 3 6 01 .6 0.;17 0.407

2 . 0 0.2YY 0 . 3 7 0

Triangular orrectangular load,

hinged base

0.2350 . 2 3 40 . 2 2 00 . 2 0 4

0. IX’)

-3.0

4.05.06.0x.0

10 .012 .0

l-4.0

16 .0

0 .310

0.2710.2430.2220.193

0.1720.15X0.147

0 . 1 3 7

0 .1580.137

0.1210 .110O.OYh

0.0x70 . 0 7 9

0.0730 . 0 6 8


Table 11.10 Ring Tension Influence Coefficients, Triangular Load (Fixed Base)

Tension in Circular RingsTriangular LoadFixed base. Free TopF = cod. x yHR lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion

‘Liquid Load’-Fiscd

H2Coefficients at Point

dt O . O H 0.1 H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H

0.4 +0.1490.8 +0.2631.2 +0.2831.6 +0.265

2.0 +0.234

3.0 +o. 1344.0 +0.0675.0 +0.025

6.0 +0.0188.0 +O.Ol 1

10.0 -0.0 1 112.0 -0.00s14.0 -0.002

16.0 0 .000

+0.134

+0.23Y+0.271+0.268

+0.25 I

+0.203+O. 164+o. 137

+0.119+o. 104

+O.OYS+O.OY 7+O.OYX

+O.OYY

+o. 120+0.215

+0.25-I+0.268+0.273

1-0.267+0.256

to.245+0.234

+0.218

+0.20x+(I.202+0.200

+o. 19’1

+O.lOl+o. 1 YO

+0.234+0.266+0.285

1-0.322+0.33Y

to.346+0.344

+0.335

+0.323+o.i 12+0.306+0.304

+0.082+O. 160

+0.209+0.250+0.2x5

1-0.357+0.403

tO.128+0.441

+0.443

+0.437tO.42Y+0.120

+0.4 12

+0.066-to. 130

+o. 180+0.266+0.274

i-0.362+O.-12Y

+a-!77+o.s04

+0.534

+0.5‘42+0.5-E+o 5 3 Y

+0.53 1

+O.O4YtO.096

+o. l-12+o. 185+0.232

+0.330+0.4OY

+0.46’S+0.514

+0.575

+0.60X+0.62S+0.639

+0.64 1

+0.029tO.063+O.OYY+0.134+0.172

+0.x2+0.334

+0.39x+0.147

+0.530

+o.wtO.633+0.666

+0.6X7

+0.014to.034+0.04s+0.075+o. 1 o-1

to. 157+0.210

+0.2S‘)+0.30 I

+0.3s 1

+0.4-K)+0.4Y-l+O.SJ 1

+o.w

+0.004to.01 0+o.o 16+0.02.3+0.0.3 I

+0.052+0.073+0.092+o. I I3

+o. IS 1

+o. 179to.2 I I

+0.2-t 1+0.x5

11.6 Prestressing Effects on Wall Stresses

Table 11.11 Ring Tension Influence Coefficients, Rectangular Load (Fixed Base)

Tension in Circular RingsRectangular LoadFixed Base. Free TopF= wet’. x pR lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion B

--Pi

‘Gas’ Load-Fixed

HZ Coefficients at Point

It O . O H O.lH 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H

0 .4

0 . x

1.2

1.6

7.0

3.0

4.0

s.0

6.0

x.0

10.0

12.014.016.0

+0.5x2+l.OS?+1.21x

+1.257+I .253

+I.160

+ I .0x5+I .037

+I.010+O.YXY

+O.YXY

+(I.904

+O.YY7

+ 1 .ooo

+0.505

+O.Y2l+ 1.078

+1.141+1.144

+I.112

+ 1.073+I.044

+ 1 ,024+ 1.005

+O.YYS

+O.YY7

+O.YYX

+O.YYY

+0.43 1

+0.796

+O.Y46

+1.009+I.041

+I.061+ I .057+I ,047

+1.038+1.022

+l.OlO

+ I .003+ I .ooo+0.9YY

+0.353

+0.669

+0.808

+0.88 I+O.Y29

+O.Y98

+1.02Y+I.042

+1.045+1.036

+1.023

+I.014+1.007+1.003

+0.277

+0.542

+0.665

iO.742

+0.806

+O.Y12

+O.Y97

+l.OlS

+1.034+ I ,044

+1.039

+I.031+I ,022+I.015

+0.206

+0.41s

+0.519

+0.600

+0.667

+0.796

+0.887

+0.94Y

+0.986

+1.026

+1.040

+1.043

+1.040

+1.032

+0.145

+0.289

+0.37x

+0.449

+o.s14

+0.646

+0.746

1-0.825

+0.x79

+0.953

+0.996

+1.022

+I ,035+I .040

+0.092

+0.179

+0.246

+0.294

+0.345

+0.459

+0..553

+0.629

+0.694

+0.788

+0.859

+0.911

+0.949

+0.97s

+0.046

+0.089

+0.127

+0.153

+0.186

+0.258

+0.322

+0.379

+0.430

+O.s19

+o.s91

+0.652

+0.70s

+0.750

+0.013

+0.024

+0.034

+0.045

+0.055

+0.081

+0.105

+0.128

+0.149

+0.189

+0.226

+0.262

+0.294

+0.321


Table 11.12 Ring Tension Influence Coefficients, Triangular Load (Pinned Base)

Tension in Circular RingsTriangular hadHinged Base. FJW TopF = cod. x ?/HR lb. per ft.Posirive s ign ind ica t e s t ens ion

-

H2ii


O.OH 0.1 H 0.2H 0.3H 0.4H-

0.4 to.474 +0.440 +0.395 +0.352 tO.3080.8 +0.423 +0.402 +0.38 1 +0.358 +0.330

1.2 +0.350 +0.355 +0.36 1 +0.362 +0.35x1.6 to .271 +0.303 +0.341 tO.36Y +0.3x52.0 +0.205 +0.260 -to.321 +0.373 +0.411

3.04.0

5.06.08.0

+0.281+0.253

+0.235+0.223+0.208

to.449tO.36YtO.46Y

+().#I3to.443

10.012.014.016.0

+0.074+0.017

-0.008-0.011-0.015

-0.008-0.002

0.000+0.002

+0.179+0.137+0.114

to. 103+O.OY6

+0.095+0.097

+0.098+o. 100

+0.200+0.197

+0.197+o. 198

+(I.375tO.367

+0.356to.343to.324

+0.311to.302

to.299to.299

+0.428+0.317

+0.408+0.403

0.5H 0.6H

tO.264 to.215+0.2’)7 tO.24Y

to.343 tO.3OYtO.385 +o..K!to.434 +0.41 Y

+0.X)6 to.519+0.545 tO.57Y

t0.562 +0.617+O.i66 +I).639tO.M4 t0.661

+(I.552 +0.666+0.541 t0.664

to.53 1 tO.659

+0.52 1 +0.h50

-0.7H 0.8H 0.9H

+o. 165 i-o.1 1 1 +0.057to.202 to.145 to.07h

+0.256 10.1 Sh +O.OYS+0.31-I to.2.33 +o. 12-JtO..3bY +0.x0 +().I51

+0.479to.553

tO.606+o. b-J3+o.w7

+0.730+0.7.s0

tO.76 1+(I.764

co.375+0.-117+0.50.3

to.547+0.62 I

+0.67s+0.7x

+0.753+0.776

+(I.210+o.wl+0.2Y-l

+0.:27+0.356

+0.4x+().A77

to.3 13to.543

11.6 Prestressing Effects on Wall Stresses 6 7 1

Table 11.13 Ring Tension Influence Coefficients, Rectangular Load (Hinged Base)

Tension in Circular RingsRectangular LoadHinged Base. Free TopF= coef. x pR lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion

‘Gas’ Load-Pinned

H2-dt O . O H

0 .4 +I ,4740 . 8 +1.423

1.2 +1.350

1.6 +1.2712.0 +I .205

3.0 +1.0744.0 +1.0175.0 +(I.992

6.0 +0.989X . 0 +O.Y85

10.0 +O.Y92

12.0 +0.99814.0 +l.OOO16.0 +1.002

0.1 H 0.2H 0.3H


0.4H 0.5H 0.6H 0.7H 0.8H 0.9H

-1.340

+1.302+1.255

+I .203+1.160

+1.079+1.037+1.014

+ I .003+0.996

+0.995

+0.997+0.998+l ,000

+1.195

+1.181+1.161

+1.141+1.121

+1.081+1.053

+1.035+1.023+1.008

+l.OOO

to.997to.997tO.998

+1.052

t1.058+1.062

t1.069+1.173

t1.075t1.067

t1.056+I ,043t1.024

+l .Ol 1

t1.002to.999+(I.999

to.903

+0.930

tO.958tO.985t1.011

+I ,049+1.069+1.069

+1.063+1.043

t1.028

+1.017t1.008t1.003

to.764

to.797+0x43

tO.885to.934

+1.006t1.045t1.062

t 1.066+1.064

+l.OS?

+1.041+1.031t1.021

+0.615to.649

to.709+0.756tO.819

to.919to.979t1.017

t1.039-1-1.061

+1.066

+1.064+1.059+I .050

+0.465to.502

to.556to.614to.669

+0.779tO.853+l.Y06

+0.943to.997

+I ,030

+I .050t1.061t1.064

to.3 I Ito.345

to.386+0.433tO.480

to.575to.647to.703

1-0.747+0.82 I

tO.878to.920

+0.952to.976

to.1541-0.166

to. 198+0.224to.25 1

to.310to.356to.394

i-O.427to.486

to.523+0.577

+0.613+0.543

672 Chapter11 Prestressed Concrete Circular Storage Tanks and Shell Roofs

Table 11.14 Empty Tank Ring Tension Influence Coefficients, Fixed Base

Tension in Circular RingsShear per Ft., (2, Applied at TopFixed Base, Free TopF = coef. x VR/H lb. per ft.Pos i t i ve s i gn i nd i ca t e s t ens ion Ring

tension, I

7- Empty TankM o m e n t


Ft O . O H 0.1 H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H

0.4 - 1.57

0.8 -3.091.2 -3.951.6 -4.572.0 -5.12

3.0 -6.324.0 -7.345.0 -8.226.0 -9.02

8.0 -10.42

10.0 -11.6712.0 -12.76

14.0 -13.7716.0 -14.74

-1.32 -1.08 -0.86 -0.65-2.55 -2.04 -1.57 -1.15-3.17 -2.44 -1.79 -1.25-3.54 -2.60 -1.80 -1.17-3.83 -2.68 -1.74 -1.02

-4.37 -2.70 -1.43 -0.58-4.73 -2.60 -1.10 -0.19-4.99 -2.45 -0.79 +O.ll

-5.17 -2.27 -0.50 +0.34-5.36 -1.85 -0.02 +0.63

-5.43 -1.43 +0.36 +0.78-5.41 -1.03 +0.63 +0.83-5.34 -0.68 +o.so +0.81-5.22 -0.33 +0.96 +0.76

-0.47 -0.3 1 -0.18

-0.80 -0.51 -0.28-0.81 -0.48 -0.25-0.69 -0.36 -0.16-0.52 -0.21 -0.05

-0.02+0.26+0.47+0..59+0.66

-0.15

+0.3x+0.50+0.53+0.46

+0.62+0.52+0.42+0.32

+0.33+0.21+0.13+o.os

+O.lY

+0.33+0.37+0.35+0.24

+0.12+0.0-t0.00

-0.04


Table 11.15 Empty Tank Ring Tension Influence Coefficients, Hinged Base

Empty Tank

HZdt O . O H


O.lH 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.6H 0.9H

0 . 40,s1.2I .h2 . 0

3 . 01.05.0h.0s.0

0 . 02 . 04 . 06.0

+1.70+2.02+ 1 .06+().I2-0.6s

-1.7s-I.S7

-1.5-l-I .01

-0.21

to.2 Ito..37

+0.26

+0.27

+‘..-rO+2.06+I.42tO.7Y+0.7’

-0.7 I-I .oo

-1 .03-0.M-0.53

-0.2.;-0.05

+0.0-l

+0.07

t2.30

+2.10+1.7Y+1 .A3+I.10

+0.4.3-0.0s

-0.42-0.59-0.73

-0.64-0.46-0.75-0.0s

+2.12

+?.I4+2.03+1.0-i+7.02

+ I .60+I .0-l+O.IS-0.05

-0.67

-0.Y-l-0.96-0.76-0.64

+I.91+2.10

+2.46t2.72+Z.YO

+2.‘)5t7.17+lM+1.21-0.02

-0.7.?-1.15

-1.X

-1.2s

+I .6Y

+2.02+2.65+3.25+3.6Y

+-I.20+-l.-3 I+3.Y3+334

+2.05

+o.s2d).lS

-0.87-1.30

+I.41+1.95+?.HO+3.56+4.X)

+5.66+6.34+6.60+6.54

i5.87

i-I.70+3Xi2.29+I.12

+1.13 tO.80 +0.44+1.75 +1.3Y +0.80+2.60 i2.22 +1.37+3.SY +3.13 +2.01i4.54 +4.0x +2.75

+6.5S +6.55 +4.73+S.lY +8.82 +6.81+Y.-ll +I 1.u.i +9.02

+ IO.28 +13.0x +11.41

+I 1.32 +I652 +16.06

+I 1.63 +19.4x +20.87+I 1.27 +2 1.80 +25.73+I055 +23.50 +30.34+9.67 i24.53 +34.65


Table 11.16 Supplementary Influence Coefficients for Values of /-/‘/c/t Greater Than 16 for Tables 11.4-I 1 .15

Table 11.4a Table ll.Sa



z .80H .85H .9OH .95H 1 .OOH z .8OH .85H .9OH .95H l.OOH

2 0 +.OOlS +.oo l-1 +.0005 -.001x -.0063 20 +.0015 +.0013 +.0002 -.0024 -.007.:24 +.0012 +.oo 12 +.0007 -.0013 -.0053 24 +.0012 +.0012 +.0004 -.OOIS -.OOh I32 +.0007 +.OOOY +.9007 -.ooos -.0040 32 +.ooos +.OOOY +.0006 -.oo I O -.004640 1.0002 +.0005 +.0006 -.0005 -.0032 40 +.0005 +.0007 +.ow7 -.0005 -.00.~74x .oooo +.ooo I +.0006 -.0003 -.0026 4s +.0004 +.0006 +.OOOh -.0003 -.oos3 I56 .oooo .oooo +.0004 -.ooo I -JO23 56 +.0002 +.0004 +.0005 -.OOOl -.0026

Table 1 lha Table 11.7a



t .75H .8OH .85H .9OH .95H dt .05H .lOH .15H .2OH .25H

2 0 +.000x +.oo I3 +.0020 +.0023 +.002024 +.0005 +.OOlO +.oo 15 +.0020 +.001732 .oooo +.0005 +.OOOY +.0014 +.oo 1340 .oooo +.OOO.~ +.0006 +.OOl I +.oo I 1

3x .oooo +.ooo I +.0004 +.000x +.OOlO56 .oooo .oooo +.0003 +.0007 +.000x

Table 11.8a Table 11.9a

H2z

202432404856


.8OH .85H

-0.0 15 +O.OY5-0.037 +0.057-0.062 +O.OO?-0.067 -0.03 1-0.06‘4 -0.049-0.059 -0.060

.90H .95H 1 .OOH

+0.206 +0.606 + I .ooo

+o.m +0.572 + 1 .ooo+o. 17x +0.515 + 1.000+0.123 +0.467 + I .ooo+0.081 +0.424 + 1.000+o.o‘M +0.387 + 1 .ooo

H2-dt


Tri. Fixed Rect. Fixed T. or R. Hinged

+O.l 1-l +o. 122 +0.062+o. 102 +o. I I I +0,055+O.OSY +0.096 +O.(NS+o.oso +0.0S6 +o.o-!;+0.072 +0.079 +0.039+0.067 +0.07-1 +o.oxl

11.6 Prestressing Effects on Wall Stresses

Table 11.16 Continued

Table ll.lOa Table ll.lla


cv .75H .8OH .85H .9QH .95HH2


dt .75H .8OH .85H .9QH .95H

20 +0.716 +0.654 +0.520 +0.32s +o. 11524 +0.7-K? +0.702 +0.577 iO.372 +o. 13737 +0.7x7 +0.76S +0.663 +0.459 +o. 18240 +o.soo +0.x05 +0.73 1 +0.530 +0.2 17-Is +0.7Y I +0.X28 +0.785 +0.593 iO.25456 +0.763 +0.83X +0.x23 +0.636 +(I.285

20 +0.949 +0.825 +0.629 +(I.379 +o. 12824 +0.986 iO.879 +0.694 +0.3X) +().I4932 +1.026 +0.953 +0.788 +O.SlY +o. 18940 +1.040 +0.996 +0x59 +0.5Y I +022648 +I ,043 +1.022 +0.911 +0652 +0.26256 +I ,040 +I ,035 +O.949 +0.705 +0.2’)4

Tahle 11.12a


dt .75H .80H .85H .90H .95H

Table 11.13a


dt .75H .80H .85H .90H .95H

20 +o.s12 +o.s17 +(I.756 +0.603 +0..34474 +O.Slf1 +O.MY +0.793 +0.647 ioc377-?1 - +o.s14 +O.Sf,l +o.t(37 +0.72 I +0.43640 +o.sw +O.Shh +O.S80 -CO.778 +0.#34s +0.7Yl +O.S64 +O.YOO +0.x20 -co.52731 +0.7s I +o.hw +O.Yl 1 +0.x52 +o.xl.3

20 + 1 ,062 +I.017 +O.Y06 -to.703 +0.39424 +1.066 + 1.039 +O.Y33 +(I.747 +0.42732 + 1 ,064 +1.061 +O.YY7 +0.x2 1 +0.48630 + 1.052 +I ,066 + I ,030 +0.X78 +0.53338 +1.041 +I .063 +I ,050 +O.Y20 io.57756 +I.021 +I .05Y +1.061 +O.Y52 +0.613

Table 11.14a Table 11.15a

H2Coefficients at Paint

cft .OOH .05H .lOH .15H .2QHH2


dt .75H .8QH .85H .90H .95H

20 -16.44 ~ 9.98 -4.YO -1.5Y +0.2724 -1X.04 -10.34 -4.54 - 1 .oo +0.6X.32 -20.s- l -10.72 -3.70 -0.04 +1.2640 -23,; 4 -10.86 -2.X6 +0.72 +1.564s -25.52 - 10.82 -2.06 +0.26 +I.6656 -27.51 -10.6s -1 36 +I.60 +I .62

20 +lS.30 +25.Y +36.9 +43.3 + 35.324 +13.20 i25.9 +30.7 +S1.8 f 45.332 + 8.10 +23.2 i45.9 +65.4 + 63.640 + 3.28 +19.2 i46.5 i77.Y + 83.548 - 0.70 +14.1 +45. I +87.2 +103.056 - 3.40 + 9.2 f42.2 +94.0 +121.0


Photo 11.7 Shotcrete Application Covering the Wire (Courtesy, N.A. Legatos,Preload Inc., Garden City, New York.)

are required to neutralize or balance the outward radial forces caused by the internal liq-uid or gas, thereby producing residual compression in the tank when it is full (Ref. 11.2).Such an increase in circumferential prestressing forces through the use of additional hor-izontal prestressing steel, and sometimes mild vertical steel, also counteracts the effectsof temperature and moisture gradients across the wall thickness in an adverse environ-ment.

11.6.1 Freely Sliding Wall Base

When the boundary condition is such that the wall at its base can freely slide when thetank is internally loaded, there is no moment in the vertical wall due either to liquid loador to prestressing when the tank is totally filled to height H. Only a small nominal mo-ment develops when the tank is partially filled, partially prestressed, or empty, and novertical prestressing is necessary. The deflected shape of the freely sliding tank is shownin Figure 11.7.

While free sliding is an ideal condition that renders the structure statically determi-nate and hence most economical, it is difficult to achieve in practice. Frictional forcesproduced at the wall base after the tank becomes operational and the difficulty of achiev-ing liquid tightness render this alternative essentially unimplementable.

11.6.2 Hinged Wall Base

For walls with a hinged connection to the base, the maximum radial forces due to the liq-uid retained and the prestressing at the critical section a distance y above the base are al-most equal to those in the freely sliding case at height y. But vertical moments areintroduced, and vertical prestressing becomes necessary to reduce the tensile stresses inthe concrete at the outer wall face.


Residual r ingcompression

B

Outside

IBefore

stressingAfter stressing Part ia l ly F u l l

and empty f u l l Maximum ringtension from

(8) liquid

compressionfrom horizontal

prestressing

--IH

-1

Figure 11.7 Freely sliding tank. (a) Deflected shape. (b) Residual ring compression.

The deflected shape of the hinged wall is shown in Figure 11.8. Note that the criticalsection for ring forces is not necessarily at the same height as the moment critical section.

In order to minimize the possibility of cracking, a residual ring compression of aminimum value of 200 psi (1.38 MPa) is necessary for wire-wrapped prestressed tankswithout diaphragms, and 100 psi (0.7 MPa) for tanks with a continuous metal diaphragm.The maximum tension at the inside face of the wall should not exceed 3e at working-load level as given in Table 11.17 in a later section. The deflected shape of the tank wallsand the stress variations in the concrete across the thickness of the section when the tankis empty and when it is full are shown in Figure 11.8. For tanks prestressed with preten-sioned and post-tensioned tendons, the minimum residual compressive stress should beas stipulated in Section 11.10.

11.6.3 Partially Sliding and Hinged Wall Base

A partially sliding and hinged wall-base system is accomplished by providing a slot in thewall-base supporting slab such that the wall can slide within its base during the prestress-ing. After prestressing and all losses due to creep, shrinkage, and relaxation have takenplace, the slot is sealed and the tank wall behaves as hinged under service-load condi-tions. The magnitude of sliding can be controlled such that either full or partial sliding isallowed before hinging is accomplished. A partial slide of about 50 percent of the fullslide with hinging at the end of the wall movement has the structural advantages of bothfull sliding and hinging, and the sealing of the wall-base slab-pinned joint against leakageof liquids or gases is more dependable than if full sliding prior to anchorage is allowed.The deformed shape of the wall during the prestressing procedure, together with the ringforces, vertical moments, and concrete stress variations across the wall thickness, isshown in Figure 11.9. The vertical prestress needed for the partial slide-pinned case canbe considerably smaller than the fully pinned case without sliding.

11.6.4 Fully Fixed Wall Base

Full fixity of the wall at its base means full restraint against rotation at the wall base. Thiscondition can be accomplished if the lower segment of the wall is cast monolithically andis well anchored into a base slab of a similar stiffness. But such an indeterminate system


Residual compression(minimum - 200 psi)

Tension when tank empty

( m a x i m u m - 3fl)

f Vertical

Beforestressing

After stressingand empty

F u l l Ringforcesper ft .

Verticalmoments

per ft .

(al - : Due to prestress

---: Due to l iquid pressure

Outs ide ins ide

Horizontal prestress

t+ .45r, bMaximum 0 --*’

-VLiquid or gas

Vertical prestress

Tank full

(4

Figure 11.6 Hinged-base tank. (a) Deflected shape of tank wall. (b) Horizontalring forces and vertical moments. (c) Concrete stresses across wall thickness.(d) Resultant wall stresses.


DeviationResidual f r o m

comoression full sltde

Outride

Before stressing After stressing,tank empty

Wall base pinned,tank filled

Outside

I

Inside

I

I

MCT -

%

+

Horizontal prestrers

Liquid or gas

Ring forcesper ft. of

circumference

Prestress,fullv Liquid

Vertical momentsper ft. of

c i r c u m f e r e n c e

(b)

Tank empty

r\ + $I,,. 34

Tank empty

Tank full

Vertical prestress

Figure 11.9 Partially sliding and hinged-base tank. (a) Deflected shape. (b) Hor-izontal ring forces and comparative vertical moments. (c) Concrete stressesacross wall thickness. (d) Resultant wall stresses.

is difficult to fully achieve and is not economical as well, since a tank base area is verylarge and partial fixity becomes necessary (see shortly). The radial horizontal forces fromboth prestressing and the contained internal pressure are unchanged from the triangularform for liquid. rectangular for gas, and trapezoidal for granular contained material. Therestraint imposed by the horizontal slab base. however. modifies the ring forces and in-troduccs additional moment in the vertical section of the wall. Because of fixity at thebase. no displacement takes place at either the bottom or the top of the wall, and achange in curvature along the height of the wall above the base takes place when the tankis empty, as is shown in Figure 11.10. Note that the wall should be designed to become es-sentially vertical. with a minimum residual compressive stress due to prestressing of 200


~~s~i~-~~~~~~~“~-----I,

Before stressing After stressing, Tank filled Freely sl iding Pos i t ive 1 Negative momentand empty due to m o m e n t

prestress(a)

Ring forces Vertical momentspar ft. o f per ft of

circumference circumference

fb)

,,1/1 -+-‘I/II,Horizontal prestress

L i q u i d

-4 r-:-l , Tank;

Tank full Tank partial ly full

(f)

Vertical prestress

(4 (4

Figure 11.10 Fully fixed-base tank. (a) Deflected wall shape. (b) Horizontal ringforces and vertical moments. (c) Concrete stresses across wall for full tank. (d)Concrete stresses across wall for partially full tank. (e) Resultant stresses, fulltank. (f) Resultant stresses, partially full tank.

I

psi as in the previous cases. The vertical prestress needed for tanks with fully fixed wallbases is considerably greater than the vertical prestress needed for the other boundaryconditions. This is necessary in order to offset the high tensile stresses in the wall base atthe outside face caused by the large negative movement at the base [see Figure I 1.10(a)and (b)] and the reverse curvature near it. It is sometimes more economical to use mildsteel reinforcement at the lower portion of the wall in addition to prestressing, in order tobe able to use lesser vertical prestressing and assign the excess negative moment to thenonprestressed reinforcement. The tensile stresses in the concrete can also be reduced byusing eccentric vertical prestressing with the appropriate eccentricity achieved by trialand adjustment, as well as by using additional mild steel. Vertical prestressing in tanks isexpensive, however, due to the required anchorages at the top and bottom of the tankwall. Thus, reducing the level of vertical prestress needed in the design adds to the econ-omy of the total design of the system.


11.6.5 Partially Fixed Wall Base

11.6.5.1 Rotational Restraint. As indicated previously. full restraint against rotationat the wall base is difficult to achieve. The reasons are essentially threefold: (1) one has toprovide the necessary stiffness in the tank floor slab at the wall junction for total fixity:(2) subsoil movement under the wall can cause rotation of the wall base: and (3) a con-centration of anchorages is required. for both the vertical prestressing of the wall and thehorizontal circumferential prestressing of the wall-base segment since the wall and baserings are separately prestressed.

Because the floor slab area is large. its restraining or stiffening influence is limitedto the narrow peripheral toe cantilevering from the wall bottom. The choice of the cor-rect width of the toe or base ring determines whether or not the assumed degree of fixityof the wall base gives the correct stiffness values in the design. Figure 11.11 schematicallydemonstrates the effect of the base ring width on the rotation of the wall and the defor-mation of the ring. Part (c) of the figure gives an equilibrium state where the tip of thering is at the same level as the bottom of the wall. whereas the conditions represented inparts (a) and (b) involve deformations below the bottom of the wall and are conse-quently unsatisfactory.

The theoretical formulation of the solution to the critical ring base width can be at-tained through the use of the principle of superposition by combining the case of a freelyrotating wall with that of a totally fixed wall as shown in Figure 11.12. Let

M,, = theoretical fully fixed moment at the wall baseM,, = partial moment at the wall base caused by the loaded cantilever toe

8, = free rotation of wall base when pinned only. corresponding to deflection .I, ofa stiff unloaded toe

8, = wall base rotation due to restraining moment M,,, corresponding to deflectionA, of a straight unloaded toe

8: = rotation of the tip of the stiffening toe as a cantilever under vertical load. cor-responding to deflection A3 of the toe tip due to the vertical load

(b)

Figure 11.11 Base ring effective width. (a) Full base slab. (b) Large cantilever.(c) Equilibrium condition.


(b)

Figure 11.12 Deformation and rotation of wall base. (a) Fully free wall. (b) Fullyfixed wall. (c) Superposition of (a) and (b).

L = width of stiffening toeq = unit load applied to the stiffening toe = yH, where H is the height of a tank

whose diameter is tl. whose wall thickness is t. and whose base slab thicknessis 11.

Then the unit rotation 8 of the wall at its base due to moment AI,,. but without radial dis-placement. can be obtained from Equation 1 1. I Xa by setting ~1‘ = 0 to set Q = -PM. Equa-tion 1 l.lXb for unit rotation then becomes

Mo,=*.

Ml102 = 2pD

Hence. we have

(I 1.3)

( I 1.34)

If the stiffening wall toe is considered a cantilever subjected to a transverse loadyH, the maximum cantilever moment M,, and the corresponding deflection -1; are. re-spectively.

(I 1.35)


The moment at the fixed watt base can be obtained using the membrane coefficient Cfrom Table 11.4 for the applicable form factor H’idr and type of toad. For liquid toad.

M,, = CyHi (11.36)

The deflected form due to full toad. from Figure I 1.12(c). is

A, = 1, + -1;

As a rcasonabte approximation. assume

k = 0.2 and p = 2/fi.

Substituting for I? and A, from Equation 11.34 into Equations 11.35 and 11.36 and rear-ranging terms gives

(11.37)

a n d

M = yHL20 ~2

(11.38)

Now let the term

(11.39)

in Equation I 1.37 be designated a ~~orfi,f~i~zg ,frrcror ,for pnrrial ,fi.ui!\: This factor is nor-mally small and represents the difference between the total fixity moment M,, and thepartial restraint moment M,,. Hence.

M,, = M,,(l - S) ( 1 1.40)

The value of L in the rle,lor~7ilmtor of Eq. 1 I.37 is conservatively assumed = 1 for simplifi-cation in modifying the factor S.

If the value of S is very small. as is the case in large-diameter tanks (diameter largerthan 125 to 150 ft). the expressions for L and M,, become expressions for full fixity.namely.

and

M,, = Cy H’

11.652 Base Radial Deformation. The radial deformation -1, of the base ring sub-jected to radial force in its plane can be obtained from the theory of circular plates withconcentric holes. The expression for the deflection of the plate shown in Figure 11.13(a)i s

(11.41)

where p = Poisson’s ratio - 0.2 for concrete and E is the modulus. The horizontal radialthrust per unit of circumference required to induce unit displacement in a solid circularstab is

6 8 4 Chapter 11 Ppstressed Concrete Circular Storage Tanks and Shell Roofs

2.s E(22 = 7

0

and the corresponding value of the radiant thrust applied to the outer ring is

Qi = g0

(11.42)

( 11.43)

where

and rf = inside diameter of base ring = ((i,, - 2L).The relative stiffness of the vvall to the base is determined in terms of the force re-

quired to produce a rrrrir deformation in the wall and the base slab from the principles ofvirtual work as shown in Figures 11.13(b) and (c). The distribution of the prestressing en-ergy between the wall and base slab ring is a function of their relative radial stiffness:hence. determining the relative stiffness is necessary. In doing so. however, one mustkeep in mind that the stiffness response of the base ring in a prestressed tank to radialcompression in its own plane is considerably larger than the response of the cylindricalwall of the tank under radial internal pressure. Thus, the loss of prestress from the differ-

d o -

- - - - - - -

A,= 1

(a)

Figure 11.13 Deformation of circular wall base ring. (a) Ring plan and crosssection. (b) Deflected wall bottom due to radial force Q’. (c) Deflected ring basedue to radial force 4.


ence in stiffness is insignificant in large-diameter tanks (Ref. 11.2). but should be consid-ered in small-diameter tanks.

The unit deformation .A due to the radial force (2’ per unit of circumference withoutmrrtiorl at the foot base can be obtained from Equation 11.18b using 2BM = -Q for rota-tion A*/c!\* = 0. The unit deflection A in Equation 1 1.1 Xa becomes

or

Q ’A=-

4p-’ D(11.44)

where

D =Et’

12(1 - k?)

LJsing k - 0.2. Equation 11.44 for unit radial displacement of the wall at the wallbase without rotation becomes

where E is the modulus of concrete. From Equation 11.42. the radial force per unit of cir-cumference required to produce unit radial displacement in the solid circular slab is

By superimposing Q’ on Q1. the total force exerted at the wall-slab base junction is dis-tributed to the wall and the slab base in proportion to the relative energy required to pro-duce unit deformation in each.

The proportion of the total force Q’ + Q, to be carried by the wall is

Q’R= Q’+Q2

Sal’

1

1 + s,

Rearranging terms while combining Equations 11.45 and 11.46 results in

s = 2.S(h/d)

’ 2.2(1/d)” 2

assuming that r/ - tl,,, or

If S, is small. the proportion of the horizontal force transferred from the slab base to thewall can be taken, with sufficient accuracy. to be

R = y percentI

(11.48)


photo 11.8 Sis-million-gallon tendon-prcstrosscd circular t a n k scc11 from IllSidC\vith situ-cast LV:IIIS. (Corrr.rcs>~. Jorgensen. Hendrickson and Close. Inc.. Dcn\.cr.Colorado.)

When only the outer ring of the slab is compressed 1~: radial thrusl at the rim. thevalue of Q, has to be modified from that obtained by Equation 11.32. and S, in Equation11.38 becomes

( I 1 .-IV)

where. from before.

in which tl is the inner slab ring diameter = tl,, = ZL and ti,, is the outer diameter.

11.7 RECOMMENDED PRACTICE FOR SITU-CAST AND PRECASTPRESTRESSED CONCRETE CIRCULAR STORAGE TANKS

77.Z7 Stresses

General guidelines for situ-cast and precast prestressed concrete circulnr storage tanksare provided bg the Prestressed Concrete Institute (Ref. 11.6). the American ConcreteInstitute (Refs. 1 I .7-l 1 .Y). and the Post-Tensionin g Institute (Ref. 11.10) for choosingthe applicable allowable stresses. dimensioning. minimum wall thickness. and construc-tion and erection procedure. The allowable stresses in concrete and shotcrete arc given inTable I I. I 7 (Ret: I I. 7). with modifications to accommodate the recommerldrd .strcs.st:s in

Rt%' 1 f. h. A//o w:~bk sfresxs h fhr rc/jlfirccnxw~ are g’wn nil Z~b/e /I. /of

11.7 Recommended Practice for Situ-Cast and Precast Prestressed Concrete Circular Storage Tanks

Table 11.17 Allowable Concrete Stresses in Circular Tanks

6 8 7

Type and limit of stress

Concretesitu-cast and precast

ServiceTemporary= load

stresses stressesfci, psi fc, psi

Shotcretesitu-cast

ServiceTemporarya load

stresses stressesfQi, psi fa, psi

Axial compression. .f;

Axia l t ens ionFlexural compression. ji

Maximum flexural tensionh.,f;

Minimum residual compression. ,f[,

0.45f~,hut not morethan 1 .600 +

40ft ps i

00.45,f ;,

mxf’;

00.3Xf;.

200 psi

“Before creep and shrinkage lossrs.

“Film atres. 5 in preconiposed tension z0ne

Table 11.18 Stresses in Reinforcement

Type of Stress

Tendon jacking forceImmediately after prcstress transfer

Post-tensioning tendons a t anchorage and couple r s .immediately after tendon anchorage

Service load stress. .f;),,Nonpres t ressed mi ld s tee l a t in i t i a l prestressing.,f,,Fina l serv ice load strcss.,f, (ps i ) . po tab le bvater s torage.

60 grade steelcorros ive s toragedry s torage

*I .OOO psi = 6.895 Pa.

Max allowable stress*

0.94f,,, 2 O.X.sf;,)(

o.x2,f;,, < 0.75,/;,,,0.7ot;,,,

0.55.t;,,,j;:l.6

23.00018.000/;/1.x

11.7.2 Required Strength Load Factors

The structure, together with its components and foundations, would have to be designedso that the design strength exceeds the effect of factored load combinations specified byAC1 318. ANSUASCE 7-M. or as justified by the engineer based on rational analysis.with the following exceptions:

Feature Load factor

In i t i a l l i qu id p ressu reInternal lateral pressure from dry materialPrestressirtg ,forccx

Final prest ress af ter lossesStrength reduction factor for both reinforcement and concrete. +

I.3

I.7

1.7

0.9


The nominal moment strength equation M, is similar to the one used for linear prestress-ing, i.e.,

or

when mild vertical steel A, is used and

where A,,, = vertical prestressing steel per unit width of circumference. in-‘.fpx = stress in prestressed reinforcement at nominal strength, psi

fV = yield strength of mild steel, psi

11.7.3 Minimum Wall-Design Requirements

11.7.3.1 Circumferential Forces

Liquid

Initial F, = yr(H - y) T per foot of wallf/‘A

Backfill

Initial Fhr = p(r + t)

where t is the total wall thickness.

( 11 .%)a)

(ll.SOb)

(ll.Sla)

(ll.Slb)

11.7.3.2 Thickness and StressesCore Wall Thickness

It - -- c > - :

(11.52)

but not less than the minimum wall thickness to be set out in subsection 11.7.3.6.

Final Stress Due to Backfill and Initial Prestress

(11.53)

11.7.3.3 Deflections. The unrestrained initial elastic radial deflection of the wall dueto initial prestressing is

A, = F,rt,.,, E,

( 11.54)

where r = tank inner radiust,, = thickness of wall core at top or bottom of wallE,. = 57,000 fi psi for both normal-weight concrete and shotcrete.

The final radial deflection Af may reach 1.5 to 3 times the initial unrestrained deflection.For normal conditions, the final permitted radial deflection can be taken as

Af = 1.7h, (11.55)

11.7 Recommended Practice for Situ-Cast and Precast Prestressed Concrete Circular Storage Tanks 6 8 9

11.7.3.4 Restraint Effects

Maximum Vertical Wall Bending Due to Radial Shear

M, = 0.24Q,, KThis moment occurs at a distance

(1156a)

from the base or top edge.

J’ = 0.68&t,,, (11 S6b)

Radial Shear for Monolithic Base Details Which Ma)! he Assumed to ProvideHinged Connection

EQ,, = 0.38 F,\ ‘;:’ (11.57)

This type of detail should be used only with situ-cast tanks which incorporate a di-aphragm in their wall construction.

11.7.3.5 Mild Steel for Base Anchorage. If a diaphragm is used. extend the full areaof the inside bars in a U-shape a distance

j‘, = 1.4.\/yr,, (1133a)

above the base. If no diaphragm is used. extend to

j’? = 1.86 (11.58b)

above the base. Note that anchorage length has to be added to y, or y?. The minimumarea of nominal vertical steel at the base region is

A, = 0.00Sr,,, (11.59)

and should be extended above the base a distance of 3 ft or

\‘- =. ., 0.75 < (11.60)

whichever is greater.

11.7.3.6 Minimum Wall ThicknessSitu- Cast Walls

Type of tank Minimum wall thickness

Shotcrcte-steel diaphragm tanksTanks wi thout ver t i ca l p res t ress ingTanks wi th ver t ica l p res t ress ing

31 in.8 in.7 in.

Precast Walls

Type of tank Minimum wall thickness

Tanks wi th ver t i ca l p re tens ion ing andexternal circumferential prestress

Tanks wi th ver t i ca l p re tens ion ing andinternal c i rcumferent ial prestress

Tanks wi th ve r t i ca l pos t - t ens ion ing andinternal c i rcumferent ial prestress

5 in.

6 in.

7 in.


It should be noted that for tanks prestressed with tendons. a thickness not less than9 in. is advisable for practical considerations.

11.8 CRACK CONTROL IN WALLS OF CIRCULAR PRESTRESSED CONCRETE TANKS

Vessey and Preston in Ref. 11.14 recommend the following expression based on Nawy’swork in Ref. 11.15 for the maximum crack width at the exterior surface of the prestressedtank wall:

It’,,,;,, = 4.1 x lo-” E,, E,], V% (I 1.61)

where E,, = tensile surface strain in the concrete

I, =grid index

s, = reinforcement spacing in direction “2”3; = reinforcement spacing in perpendicular direction “1” (horizontal)t,, = concrete cover to center of steel

+, = diameter of steel in main direction “1.”

The tensile strain can be computed from

a, .fil,E -<i

E,,,

where cy, = stress parameter %,f,,/A,,,f,, = actual stress in the prestressing steel

.f,,, = initial prestress before losses.

For liquid-retaining tanks. the maximum allowable crack width is 0.003 in.

(11.62)

11.9 TANK ROOF DESIGN

Roofs for storage tanks are constructed in the form of a shell dome or as flat roofs sup-ported internally on columns. The cost of the roof is generally about one-third of theoverall cost of the structure. In the case of flat roofs. whether precast or situ cast. the de-sign follows the normal design principles of floor systems for reinforced or prcstresscdconcrete one-way- or two-way-action floors as stipulated in the AC1 318 Code. If the roofis made out of precast prestressed elements. and the tank diameter is not exceedingI>large. no interior columns are necessary. Otherwise. the added cost of interior columnsand the accompanying footings would increase the cost of the overall structure.

A shell roof in the form of a dome has distinct advantages for tanks not exceeding150 ft. in diameter. namely. that the dome does not need supporting interior columns andcan also be economical in underground storage tanks in withstanding backfill load.Hence, the shell form and the manner of its connection to the tank walls have a signifi-cant effect on cost. Preferably. the roof shell should be supported by tank walls with acompletely ,flexihle joint: otherwise the design of both the tank wall and the roof domewill have to be modified in relation to their degree of interrestraint and relative stiffness.with the concomitant added construction cost.

A spherical shell of low rise-to-diameter ratio h’iri of approximately Q is reasonableto use. Such a flat dome or axisymmetrical shell introduces outward horizontal thrust atthe springing. which has to be resisted by a properly designed prestressed ring beam atthe support level. The type of support of the ring beam determines the extent to whichredundant reactions and moments due to end restraint impose additional direct andbending stresses in the shell near the springing. In other words. the membrane solution

11.9 Tank Roof Design 6 9 1

has to be adequately modified by superimposing on it the bending effects determined bythe strain compatibility requirements of the bending theory.

11.9.1 Membrane Theory of Spherical Domes

11.9.1 .l Shell of Revolution. The basic membrane equations of equilibrium for thedirect forces in a shell of revolution as shown in Figure Il.14 are used for defining theunit meridional forces N,,. unit tangential forces N,. and unit central shears N,,,,, and NH<,,in terms of the gravity loads I),,~. p,,. and ix. These equations are as follows:

Parallels or thnes of latitude

Merldlans orlines of longitude

Gravityload

W

ia)

N,, r, cos 0 dQ dO

Cc)

(b)

W = -2sroN,, sin 0

r, and r2 = radll of curvature

rO = radius of parallel circle

Figure 11.14 Membrane forces in a shell of revolution. (a) Meridian and parallellines. (b) Membrane forces on infinitesimal surface element. (c) Component offorce N,,r,& in the y direction needed to simplify the basic equation 11.63a. (d)Dome cross section with total gravity load W.


qN*r,,) .Meridional: ~ -

WI,

34N,$ + ~ r, + p* r,, r, = 0

dB(11.63a)

Tangential:?JN, C-Jr,,z rl + NHcb 7 +

dN,,

rJ4__ r, + pH r,, r, = 0d4

(11.63b)

(11.63c)

Because of loading symmetry. all terms involving i,B vanish. and those involving iJH canbe rewritten as total differentials ri$ since nothing varies with respect to 8. Also. the cir-cumferential load component pti = 0, as the shear resultants vanish along the meridionaland parallel circles. Hence. Equations 11.63 can be rewritten as

& (N, r,,) - N,, r, cos + + p,r,r,, = 0

( 1 1.64b)

11.9.1.2 Spherical DomeMrmhrrrnr Analysis of‘ the Eqtrilihrilrm Forces. The spherical dome has a uniform

curvature. Consequently, r, = r, = r(,. Assuming that the radius of the sphere = (1. thenr(, = u sin 4 in Figure 11.14(c). and. setting!>- = K’!) for self-weight. the general equilibriumequations 11.64 become

N,, = (1~‘~ ’1 + cos 4 - cos 4

( 1 1.6Sa)

and

where PV~~ is the intensity of self-weight per unit area. It is plain from Equation 11.6Sbthat the meridional force N, is always negative. Therefore. compre.ssiotz develops alongthe meridians and increases as the angle + increases: when b = 0. N, =-i UW~): and when4 = 7712, N, = -awLI.

The tangential force NH is negative. i.e.. compressive. only for limited values of theangle +. Setting N, = 0 in Equation 11.6Sa. l/( 1 + cos +) - cos + = 0 gives 4 = Sl”40’. Thisdetermination indicates that for 4 greater than Sl”49’. tensile stresses develop in the di-rection perpendicular to the meridians. The distribution of the meridional stresses N,,and the tangential stresses NH for both the self-weight M‘~) and the external live load ~3~. isshown in Figure 11.15.

If the external load is uniform. such as snow. giving a projection intensity M’[.. themeridional force N, is obtained from free-body equilibrium by equating the externalload to the internal meridional force. i.e.. - 7i(cU2)‘w, = Z~(LI sin b)N,. Since c//2 = (I sin 6.we obtain

Hence. N, is constant throughout the shell depth. as is plain in Figure 11.15.N, due to the live load u’~ is

N, = - IIW,. cos2 4 + y = (lw, (; - cos’4) = ycos24 (11.66b)


(-1: Compression(+): Tension

-+w,a

A% -WDa

-1jwLa% -I

-;wLa

WD.

a

k

- - - -_

G

a ( 51”49’

dJ2a= -sin 9

-:,,a

+w,a

-{wLa----BN,+

+tw,a

Figure 11.15 Gravity membrane force distribution in a spherical dome. (a) Flatdome segment of rise h’. (b) Membrane stresses due to self-weight w, (N, = 0 forC$ = 51”, 49’). (c) Membrane stresses due to snow load w, (A/, = 0 for I$ = 45“).

For the case of N, = 0, the shell angle $ = 45”. Consequently, shell stresses due to tangen-tial forces N, for + less than 45 degrees are compressive. eliminating cracking. From thedistribution of the tangential forces N,, it can be concluded that roofs of storage tanksshould beflur, i.e., the ratio h’id in Figure 11.1.5(b) should not exceed Q, so that the con-crete will be totally in compression due to both N+ and N,, as angle + is less than 51”49’for meridional forces and 45” for tangential forces.

As discussed at the outset, the support type at the springing level, if restrained, intro-duces indeterminate reactions that result in direct and bending stresses in the shell near thespringing level. Accordingly, the bending theory, a rigorous procedure beyond the scope ofthis text, has to be applied. Refs. 11.1 and 11.3, on the subject of plates and shells, can beused for determining the resulting bending stresses. The following covers the design of the


prestressed ring beam at the springing level to counter the horizontal component of themeridional compressive thrust N& which causes the edge of the dome to move inwards.

From Equations 11.6Sb and 11.66a. the meridional thrust. N*. for self-weight it’,)per unit surface area and uniform live load w[. per unit projected area can be written as

Nd, = - (I ‘2‘n +w’1 + cos c$ 2

(I 1.67)

where (I = ~112 sin + is the radius of the shell.Note that the thrust, Nd,. becomes vertical at the springing (6 = n/2) of a hemispher-

ical dome and is equal to W = 0/2(2w,, + lt’[.) per unit width. At other values of 4, N,. it isinclined and the value of its horizontal component is needed for the design of the pre-stressed ring beam at the springing level. namely. the shell rim. This horizontal compo-nent is p = N,, cos +. If P is the prestressing force per beam height in the ring beam, thenP = ~~02 from Equation I l.la. and

P = + (NC!, cos do) (1 I .6X)

Evidently. if the force P could be applied directly to the dome rim, the stresses in thedome would be those defined by Equation 11.67. This is usually not feasible, since thelarge amount of prestressing steel needed due to P cannot be accommodated in the smallthickness of the shell. and the stress in the concrete in the rim zone would be very high in-deed. Thus, an edge beam has to be provided. transforming the shell into a statically de-terminate structure.

Prestressing the Statically Indeterminate Flat Dome. The simplest boundarycondition is obtained when the edge beam reaction is vertical and without any supportrestraint. as shown in Figure 11.16, where the dome thrust N, passes through the beamcentroid. If an imaginary cut along line A-A is made. the horizontal thrust N, cos C$Icauses the dome edge to move inwards a distance (Ref. 1 1.16)

(I 1.69)

where p = Poisson’s ratio = 0.2 for concretetl = shell span

Displacements dueto N, cos 4

Reaction

(a) (b)

Figure 11.16 Ring beam effects. (a) Simply supported beam with thrust linepassing through ring beam centroid. (b) Shell displacements at rim; rotations dis-regarded.


and the tangential unit force is obtained from Equation 1 1.65a as

W[,d 1N, =

2 sin &I 1 + cos +- cos + - &(cos2$) ( I 1.70)

Conversely. the meridional thrust N,,, causes the ring beam to move OU~I~WU/S a distance

-I,, =N,,,(cos +)d’

4Ehh(11.71)

The prestressing force must therefore be sufficient to move the ring beam i~l~rr& a totaldistance

-1, = -1, + A,,

so that the total force acting on the ring beam cross section is

(11.72)

Lvhcre II is the total ring beam depth. A comparison of Equations 11.72 and 11.68 showsthat the effective prestressin,~7 force needed in the former is greater than that required inthe latter. The magnitude of this increase is about 5 to 10 percent. The same conditionsalso hold true for domes in which the line of thrust from the dome does not pass throughthe centroid of the rins beam and the beam itself is rigidly attached to the wall as in Fig-urc I I. 17(a). The required prestressin,(7 force P can be obtained approximately by in-creasing the value of P in Equation 11.68 by IO percent (Ref. 11.16). In such a case. thestresses in the shell itself at the springing level zone can significantly differ from those ob-tained in the membrane solution. and the bending solution modifications have to bemade as in Ref. 1 1.1 or 1 13.

If the horizontal radial prestressin g force in the ring beam is larger than required.escessive bending deformation develops in the shell rim. as is shown in Figure 1 I .17(b).\vith a significant increase in the \,alue of the tangential force N,, as compared to the in-crcasc in the meridional force N,,. As a result. the bending stresses in the concrete in theaffected zone could exceed the maximum allowable at service load. If the initial prestressbefore losses is P,. the area of the beam cross section is

Meridionalthrust

N*

Prestressingforce

P/centroid

IA VerticalA-/ reaction

II

II

I I D e f o r m e d

I I shell shape

I at the rim

I

(a) lb)

Figure 11 .17 Edge ring beam monolithic with tank wall. (a) Thrust A$, not pass-ing through ring beam centroid-general case. (b) Shell deformed shape due toexcessive prestressing.

Photo 11.9 1.55 Million Gallon Reactor Tank, Bishop Texas. (Courtesy, N.A.Legatos , Preload Inc. , Garden Ci ty , New York.)

A, = 3L

where P, = initial prestressing force P/Tf, = allowable compressive stress in the concrete7 = residual stress percentage.

(11.73)

It is desirable to maintain a low value off,, about 0.2ff and not exceeding 800 to 900 psi,in order to minimize any excessive strain that develops in the edge ring beam, which inturn could produce high stresses in the shell at the springing zone.

The area of the prestressing steel in the dome ring is

Unit A,, = 3fPl

(11.74a)

where fpr is the allowable stress, in psi, in the prestressing reinforcement before losses. Ifaccurate analysis to determine A,, is not required, the steel area can be taken as

wcot+A,, = ~

2l.r fpe(11.74b)

where W = total dead and live load on the dome due to wD + wLfpe = effective steel prestress after losses, psi.

The minimum thickness of the dome required to withstand buckling (Ref. 11.7)may be taken to be

Min hd = aJ

1JPu+Pi Pc Ec

(11.75)

11 .l 1 Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs 697

where (I = radius of dome shellp,, = ultimate uniformly distributed design unit pressure due to dead load and live

load = (1.40 + I .7L)/l44+ = strength reduction factor for material variability = 0.7p, = buckling reduction factor for deviations from true spherical surface due to

imperfectionsp, = (n/r,)‘. where r, I 1.40

p, = buckling reduction factor for creep. material nonlinearity. and cracking =0.44 + 0.003 W, , but not to exceed 0.53

EC = initial modulus of concrete = 57.000 v? psi.

11 .lO PRESTRESSED CONCRETE TANKS WITH CIRCUMFERENTIAL TENDONS

Instead of wrapping the prcstressing wires or strands. as is done in the Preload System,internal or external horizontal tendons are used. These tendons are stressed after theyare placed within or on the wall. Vertical post-tensioning is incorporated in the walls aspart of the vertical reinforcement. The concrete walls are either cast in place or precast,and the core wall is considered to be the portion of the concrete wall that is circumferen-tially prestressed. No steel diaphragms are used in this type of construction as comparedwith wrapped-wire prestressin,.0 where the tank walls can be either with or without steeldiaphragms.

The internal prcstressed reinforcement is protected by the concrete cover as re-quired in AC1 31X. and the ducts or sheathing have to be filled with corrosion-inhibitingmaterials or grouted. The bonded post-tensioned tendon reinforcement has to be pro-tected by portland cement grout as required in the AC1 318 code, and external tendonsshould be protected by a shotcrete cover of 1 -in. (25mm) minimum thickness.

The wall design procedures are similar to those of circular tanks prestressed by wireor strand wrapping. and the same requirements for crack control and water or liquidtightness apply. A minimum residual compressive stress of 200 psi (1.4 MPa) in the con-crete wall after all prestress losses has to be provided in the design when the tank is filledto the design level. If the tank is not covered. a residual compressive stress of 400 psi(2.X MPa) has to be provided at the wall top. reducing linearly to not less than 200 psi at0.66 from the top of the liquid level.

Typical Wall Base and Dome Roof Connections. From the foregoing discus-sions. it is clear that the boundary conditions at the base of the circular prestressed tankand at the ring beam support for the roof dome determine the practicality, economy, andsuccess of the entire design. Consequently. accumulated experience in developing theconnections at these boundary conditions is invaluable. A selection of connection detailstaken from Refs. 1 1.6 to 11 .Y is given in Figures 11.18 through 11.22.

11 .l 1 STEP-BY-STEP PROCEDURE FOR THE DESIGN OF CIRCULARPRESTRESSED CONCRETE TANKS AND DOME ROOFS

The following trial-and-adjustment procedure is recommended for designing a pre-stressed concrete circular tank and its roof shell:

1. Select the prestressing system. the type of prestressing wire, the concrete strength,and the type of restraint that can be accomplished under local conditions.


4Note: This detail commonly used for

small diameter shotcrete tanks

Shotcrete connection

(al

Dome ring

Note: This detail commonly used for-+-I- intermediate diameter shotcrete

tanks

Shotcrete closure stripplaced after wire winding

Dome ring

Elastomeric padz

Figure 11.18 Cast-in-place tanks. (a) Monolithic base joint; monolithic and fullyrestrained against translation before and after wire winding. (b) Monoli thic basejoint; hinged with limited restraint against translation during wire winding, andmonolithic and fully restrained against translation after wire winding. (c) Sepa-rated base joint, allows translation, rotation, or both (d) Monolithic dome-wall con-nection. (e) Separated dome-wall connection.

11 .ll Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs 6 9 9

Greasedsteel shims

Tank wall

/ 1Sealant

/ Wall footing

f:’ cl:l‘Non-shrink’

groutFil l

Sealant

/ ,Wall footing

’ Bearing pads

Figure 11.19 Wall base joints for precast tanks

Low shrinkage mix of concrete,shotcrete or grout

Sandblasted surfaces

Circumferentialprestressing tendons

Vertical pretensioning

hotcrete cover

andblasted surfaces

\P/C wall panel

Figure 11.20 Vertical wall joints for precast tanks.


/Diaphragm

Prestressing wireConcrete spherical dome

. . . . ./.n A.--~

Gun&cover w

Horizontal prestressingwires and gunite cover

0.0173” inner

Tlll-lv steel diaphragm

Elastomeric bearingpad a n d sponge filler

Concrete waterstop encasement

Sponge filler9” PVC waterstop (cont.)

‘\

\6 mil polyethylene filmunder floor and footing

LV a r i e s (2’-6” m i n . ) - /\

Compacted gravel fill

‘For seismic zones or for tankswith unequal backfill only

Figure 11.21 Typical tank section of a domed preload prestressed concretetank with an inner steel diaphragm. (Courtesy, Preload Technology, Inc., NewYork.)


Chord steelDowel a t wal l jo in ts

\ I SloDe I

Galvan ized tubef i l led with foam

insulat ion

T-Allowance for radial movement

‘Neoprene bearing pad, continuous

Neoprene pad

Precast single tee

Mild steel reinforcing

Precast wall panel

Figure 11.22 Connections for precast tank roofs

2. Determine the contained material pressure on the wall: yH for liquid and p for gas.Use the trapezoidal distribution for granular or solid containment.

Find the unit ring force F = y(H - y)r for a completely sliding base, where r is theradius of the tank and v is the distance above the base.

3. Choose, from Tables 11.4 through 11.16, the applicable vertical moment coeffi-cients for the particular load type and wall base restraint condition caused by liquidpressure

M, = +$WWY) + Q,,i(Py)l

and determine the corresponding horizontal radial ring tensions

Q,, = +WH - 1) wm


and Q!. = (F- A&), where the offset

and

p = [3(1 - k41’ ’(r.t)’ z

where k = .20 for concrete.4. Find the applicable membrane coefficients C from Tables I I .-I through 1 1.16. Con-

pute the applicable ring force F = CyHr.5. Compute the critical vertical moments in the \~a11 usin g the applicable membrane

coefficient C. The equation for moment due to liquid load is

AZ, = C(yH‘ + pH’)

or

due to gas load if applicable. Compute the moment at the base. Lvhcrc applicable.and at the critical J* plane above the base.

6. Choose the level of vertical prestressing force.

7. Compute the concrete stresses across the thickness of the wall both for the condi-tion when the tank is empty and for when it is totally full. Allow maximum residualaxial compressive stress .f;,6=3vyj

= 200 psi at service and a masimum tensile strcasas shown in Table 11.17.

8. Design both the horizontal and the vertical prestressing steel limiting stresses tothose given in Table 1 I. 18.

9. Compute the factored moment M,, using the applicable load factors given in subsec-tion 11.7.2. The required M,, = M,,/+. where 6 = O.Y. Compute the available nominalmoment strength M,, = A,), .f,,,(d,, - rri2). or M,, = A,,, .f;,,(d!, - r//2) + A, ,f;(d - u/3). Theavailable M,, has to be greater than or equal to the rcqulred M,,.

10. Design the length L of the annular ring at the base of the wall from the equation

L? = 2CH-‘I + (r/41)”

((if y

where t is the thickness of the wall and /z the thickness of the base slab.

11. Compute the percentage of prestress in the base to be transferred to the wall fromthe formula

1Percentage R = ~

1+s

where S = 1.1 (/z/f) x ((l/r) I’.

When only the outer rim of the slab ring is compressed by radial thrust at the rim.the value of S is modified to


where

in which d,, = outer diameterd = inner slab ring diameter = d(, - 2L.

12. Check the minimum wall thickness requirements. and evaluate the unrestrained ini-tial elastic radial deflection

A, = F,r

where E, =57,000-\/zt,,, E,

t < 0 = thickness of wall core at top or bottom of wallr = Id.

The final radial deflection &= 1.71,.13. Anchor the steel from the base to the wall such that the steel extends into the wall a

distance y2 = 1.X< or 3 ft. whichever is greater. Also. ensure that the minimumnominal vertical steel at the base region is

A, = O.OOSt,,,

14. Verify the maximum crack width M’,,, = 4.1 x IO-“e,.,E,,, fi.

where E,., = tensile surface strain in the concrete = (h&)/(E,,,),(, = actual stress in the steel

.f,,, = initial prestress before losses4 -.&J&,

S, = spacing of reinforcement in direction “1”+, = diameter of steel in direction “1”

s? = spacing of reinforcement in direction “2”th = concrete cover to center of steel. in.

Note that maximum allowable w,,,~~ = 0.004 in. for liquid-retaining tanks.15. Design the roof cover dome after selecting the type of connection at the top of the

tank wall. Limit the ratio of the rise h’ of the dome to its base d such that h’/d doesnot exceed 6.

Compute the required horizontal radial prestressingbeam from the equation

force P for the edge

P = F(& - /.LN”) +d(N, cos 4)

2

where

N, = ~ w,d1

- -2 sin 4 1 + cos +

c o s 4 1 6 (COS 24)

N,x-(~ w D1 + cos 4

+2

andh = total depth of rim beamh = ring beam width


Photo 11.10 Two prest ressed concrete anaerobic digester tanks during construc-t ion. (Courtesy, N.A. Legatos , Preload Technology, Inc . , New York.)

W D = intensity of self-weight of shell per unit area (dead load)wL = intensity of live-load projection.

16. Compute the ring-edge beam cross section

where P, = initial prestressing force = Ph7 = residual stress percentagef, = allowable compressive stress in the concrete, not to exceed 0.2fi, but

not more than 800-900 psi, in the edge beam.

17. Compute the area of the edge beam prestressing tendon

where f,, is the allowable stress in the prestressing steel before losses, or

Wcot glA,, = ~

hfpeif accurate analysis is not performed, In the latter, W is the total dead and live loadon the dome due to wD + wL and fpe is the effective prestress after losses.

11 .l 1 Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs

18. Check the minimum dome thickness required to withstand buckling, i.e.,

705

Min. hd = l&4a4Pi Pc Ec

where N = radius of dome shell

P,, = ultimate uniformly distributed design unit pressure due to dead loadand live load = (1.40 + 1.7L)/144

4 = strength reduction factor for material variability = C.7

p, = buckling reduction factor for deviations from true spherical surfacedue to imperfections

p, = (a/r,)‘, where r, 5 1.4ap, = buckling reduction factor for creep, material nonlinearity, and crack-

ing = 0.44 + O.O03W,. but not to exceed 0.53EC = initial modulus of concrete = 57,OOOe psi.

Figure 11.23 gives a step-by-step flowchart for a recommended sequence of opera-tions to be performed in the design of circular prestressed concrete tanks and theirshell roofs.


START

01 I

Input: d, H. r. h. a, ~0. h’. 70 P. W,, W,, f;. f;;, f,, f,, fcv,

f&w f,,. f,,. f,,, f,.

I

m)Assume wall thickness r and type of wall base joint. ComputeF = y(H - y)r for freely sliding base. Select membrane coefficientC from Tables 10.4-10.16

P=[3(1 - $)I”4

(f-r)‘”Compute max. M, at y above base

M, = cl+/3 + PM)

M,, 0,. AQ, and 0,

0, = +l2PH - 1) J12,yL p2)

AQ, = + $$ [PM,$(fly) + Q,O(Ov)lr

(3, = F - AQ,

03I

Choose vertical prestress P,. Compute concrete fiber stressesat critical base section when tank is empty and when full

f-; fM,c Mpc

-7+-T-

where M, = liquid load vertical unit moment

MP = prestress vertical unit moment

Max f, = 0.45f;

Min r = 7 in. with vertical prestress

Max. allow. residual axial f,, = 200 psi

Max. allow. tensile stress ft = 3fi

Revise wallsection details

Compute factored moment MU using load factors:

Rqd M, 5 available M, = 2

Avail. M, = A,sfPs(dP - z) + A,f, (Ed - :)

Figure 11.23 Flowchart for the design of circular prestressed tanks and their flatdome roofs.

11 ,l 1 Step-by-Step Procedure for the Design of Circular Prestressed Concrete Tanks and Dome Roofs 7 0 7

601 Revise wall Yes

section 1

C7Compute slab base ring length L and thickness h

L2 -2CH2,+(tlh)3

(dt)“2

where t = wall thickness

h = base slab thickness

d = tank interior diameter.

Compute percentage R = r+‘s of moment to be

transferred to wall where

S = l.l(hlt)~

or S’ = i (h/t1 m when only the outer rim of

the slab ring is compressed by radial thrust at the rim,

where K =

06 tCheck if the elastic radial long-term deflection A, = 1.7

,i 1$ is

coacceptable, where F, = initial thrust, r = fd, r,, = thicknessof wall

core at top or bottom of wall, E, = 57,OOOfi

09 1

Anchor steel from base to wall up to minimum distance y2 above

base, where yI = fi but not less than 3 ft. above top of base.

Min. vertrcal steel A, = O.O05t,,.

010 I

Max. allow crack width = 0.004 in. for liquid-retaining tanks

I where I, = f (y)

0, = diameter of wire in main direction

s2 = spacing of wire in perpendicular direction

tb = core thickness

XfEC, = tensile surface strain in the concrete 9

L‘where X, - f,/fp,

fp = actual stress in steel

f,, = initial prestress

b

Figure 11.23 Continued


wrise h’

Design roof shell dome: -d

5 1. Assume ring beam8

section b X h = A,. Select shell thickness t and check formin. t required to resist buckling from step 15. Edge ringbeam prestressing force:

P= $ (N, - gN,) + ; (N,cos#)

where

tangential Nsw,d 1

= - - -2 sin 4 1 t cos d

-clx$ 1 * (cos 291

-S-+2Wmeridional NQ = -a

1+cong 2

1 b = beam width, h = beam depth, w,, = dead load, wL = live load

-1- Compute rqd. A,I

- -= P,lf,. where P, = PI,, 7 = residual stress percentage,

f, = allowable concrete mmpressive stress < O.ZOf; 5 800 to 900 psi

II

t

Compute edge ring beam prestress reinforcement AP, = P, /fs, or

Wcm$.Apr = - If accurate analysis is not performed

2”fP.

W = total dead and live load (w, + W,) on the dome

fpe = effective prestress after losses

@@

Check min. dome thickness t to withstand buckling

where a = radius of dome shell

p,= 1 . 4 0 + 1.7L.r$=0.7,~,~0.50,

p, = 0.44 + o.O03w, 2 0.53,

EC = 57,000~

23END

Figure 11.23 Continued

11.12 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 7 0 9

11.12 DESIGN OF CIRCULAR PRESTRESSED CONCRETE WATER-RETAINING TANKAND ITS DOMED ROOF

Example 11.3

Determine the maximum horizontal ring forces and vertical moments, and design the wallprestressing reinforcement. for a circular prcstressed concrete tank whose diameter d = 125 ft(38.1 m) and which retains a water height H = 25 ft (7.62 m) for the following conditions ofwall base support: (a) hinged. (b) fully fixed. (c) semisliding. and (d) partially fixed. Also, de-sign the prcstresscd concrete ring edge beam for the domed roof shell assuming that the shellrise-span ratio h’id = k. Use a flat shell roof having shell angle do = 36”. and find the area ofprcstressing reinforcement for both wire-wrapped and tendon reinforced conditions. Givendata arc as follows:

f“ = 5.000 psi (33.5 MPa). normal-weight concretecf,‘, = 3.750 psi (25.9 MPa)

f; = 212 psi (0.86 MPa) 5 3e

f; = 0.45f,’ = 2.250 psi (15.5 MPa)

residual ,f;, = 225 psi (1.55 MPa)

f;,,, (wire) = 250.000 psi (I .724 MPa)

/;>ji (strands and tendons) = 250.000 psi (1,724 MPa)

,6,, = 0.7&, = 175.000 psi (1.207 MPa)

f,,$ = 220.000 psi (I 517 MPa)

I,‘, = 15 psf (718 Pa) for snow load on dome

Assume 26 percent total loss in prestress for all long-term effects.

Saluticm: Disregard the weight of the wall and the roof dome effect as insignificant on thestresses as compared to the effect of the vertical prestress forces. Consider the water pres-sure distribution shown in Figure I 1.24 on the tank wall giving

y = 62.4 Ibjft’ (1 .OOO kg/m”)

dr=--

2- F = 62.5 ft (19.1 m)

Assume the wall thickness f = 10 in. = 0.83 ft (25.4 cm). Then the form factor

H’ 25 x 25-= =dt 12s x 0.83

6

and yHr = 62.4 x 25 x 62.5 = Y7.500 lbift of circumference.

Basic Forces and Moments. Tables 11 .I 9 through 11.21 give the basic forces and mo-ments in the tank wall.

Figure 11.24 Liquid ring tension F, wall base freely sliding.


Table 11.19 Maximum Ring Tension F= C(yHr) Ib/ft Circumference, Example 11.3

Freely Sliding Wall Base Fixed Base Hinged Base

Table ll.lOfor !$ =6 Table 11.12 for $ = 6

C=l c = 0.514 C = 0.643

F = 97,500 F = 0.514 x 97,500 = 50,115 F = 0.643 X 97,500 = 62,693

*Compare with 50.113 lbift in the detailed method of Example 11.2.

Table 11.20 Vertical Moments M= C(Y/-/~) ft-lb/ft, Example 11.3. Positive (+) = Tensionin Outside Face

Freely Sl iding Bass

M, = MO = 0

Fixed Wall Bass

Table 11.4

C = +0.0051 for 0.7H = 17.5 ft

C = -0.0187 for l.OH = 25 h

M, = +0.0051 X 62.4(25j3= +4,973

M, = -0.0187 X 62.4(25b3= -18,233

Hinged Bass

Table 11.6

C = +O.O07B for 0.0H = 20 ft

C = 0 for l.OH, or full height

M, = +0.0078 X 62.4(25j3= +7,605

M,=O

*This moment value is very close to the value obtained by using the detailed method and the moment functions ofTable 11.1 and Example 11.1 (M,, = -18.574).

11 .I2 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 7 1 1

Table 11.21 Prestressing Effects Using 225psi Residual Radial Compression, Example 11.3. RingForces Q Ib/ft, Vertical Moments M,, ft-lb/ft

Freely Sliding Base

Ring Forces

Residual

Fixed Wall Base Hinged Base

Ring Forces Ring Forces

Liquid

x = 97,500 + [Res. camp. X t X 1 ft.1= 97,500 + 1225 X 10 X 121

= 124,500

x=50,115+[225X10X121 x = 62,693 + I225 X 10 X 121= 77.115 = Q,, = 89,693 = Q,7,5

Moments

[M,l,=,, = M, =O

Moments Moments

77 115y = 4 ,973 X 50115 = 7 ,652 y = +7,605 X E

= +M, = +10,880 = +M,

M = _ 18 ,233 x 77 ,115 M, = 00

50 ,115

= - 2 8 , 0 5 6


Wall Maximum Concrete Stresses at 20 ft from Top: Hinged Base. By trial and

adjustment, provide vertical concentric prestress P,. = 50,000 Ib/ft (730 kN/m) of circum-ference. Then for a wall thickness t = 10 in. compute the resulting stresses as shown inFigure 11.25.

Outside Inside

f++L 10,880 x 1212(10)2

= ?653 psi

6

M 7,605 x 12f+ = =s 12(10)2

= 2456 psi

6

P, 50,000fu = A, = 12 x 10- = -417 psi

f4=Qf@Max. fi = 236 psi = 3*,000

= 212 psi, O.K.

Max. fc = -1,070 psi < 0.45f:, O.K.

-65&J +6530 Horizontal P/S moment

+466+-466

@ Liquid moment

Max. fc = -614 psi < 0.45f:, 0.K

-417

JJ Vertical P/S

- -417

-614 y -220

05 Tank full

Figure 11.25 Stress at maximum moment, 20 ft from top, psi. Negative (-) =compression, positive (+) = tension.

11 .12 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 7 1 3

Wall Maximum Concrete Stress at I7ft 6 in. from Top: Fully Fixed Base. The max-imum positive moment M, is at 17 ft 6 in. from the top of the wall. By trial and adjustment,use eccentric vertical prestressing P, = 100.000 lbift closer to the outer face [e = 1.05 in.(26.7 m)]. Then compute the resulting stresses in the wall as shown in Figure 11.26.

Outs ide Ins ide

M 7,652 x 12f+ = -s = 12(10)*

= S459 psi

6

f+=!+ 4,973 x 1212(10)*

= k298 psi

6

- = -833 psi

f, = P,ek) 100,ooo x 1.05- =t I 12(10j2= +525 psi

6

f5=@+@+@

Max. J = + 151 psi < 3e = 212, O.K.

Max. fc = -1,817 psi < 0.45fL = -2,250 psi, O.K.

Max. fc = -1,519 psi < 0.45fJ, O.K.

45gL ---;+45g@ Horizontal P/S moment

+2g8 4 -2g8@ Liquid moment

(ij Vert ical P/S

@ Vertical P/S moment

-1,817

“

-

+151

05 Tank empty

06 Tank full

Figure 11.26 Stresses at maximum positive (+) moment, 17 ft, 6 in. from top,psi. Negative (-) = compression, positive (+) = tension.


Wall Maximum Concrete Stress at Base: Fully Fixed Base. Use eccentric vertical

prestress P,. = 100.000 lb closer to the outer face (e = 1.05 in.). Then compute the resultingstresses in the wall as shown in Figure 11.27.

Outs ide Ins ide

f+=!+f+=!+ -28,056-28,056 x 12x 12 -= 1 , 6 8 312( lo)*12( lo)*

= ?1,683 psi?1,683 psi

66

f+=!+f+=!+ -18,233-18,233 x 12x 1212(10J212(10J2

= t1,094 psi= t1,094 psi +1,683+1,683

66

fc = 2 = -l~@o = -833 psi1 Horizontal P/S moment

c 12 x 10

f, = P,e(c) 100,000 x 1.05-=c I 12( lo)*= T525 psi m,,_:_1-

6+

fs=@+@+@+ 1,094

Max. fc = -1,991 psi < 0.45 fi. O.K. @ Liquid moment

Max. fi = +325 psi at base when tank is empty.This stress will rapidly decrease wellbelow 3e within one foot abovebase, hence O.K.

-833 71 -833

f6=Q+@+@+@@ Vertical P/S

Max. fc = -897 psi < 0.45 fi, O.K.

-525 vv +525

@ Vertical P/S moment

+325 4 -l’ggl

0 Tank empty

-769-897

-1 1

06 Tank full

Figure 11.27 Stresses at maximum negative (-) moment at wall base, psi. Neg-ative (-) = compression, positive (+) = tension.

11 .12 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 7 1 5

Wall Maximum Concrete Stress: Semisliding Base. By trial and adjustment, useconcentric vertical prestress P,. = 20,400 Ib/ft (297 kN/m). Then semislide M = a (+10.880)= 5.440 ft-lbift. and compute the resulting stresses in the wall as shown in Figure 11.28.

\waterstop

= ?326 psi

6

f+=$ +7,605 x 1212( 1o)2

= 5456 psi

6

ft - ;“ - -2090 _ - 170 psic 12 x 10

f4=0+@

Max.& = -496 psi < 0.45 f:, O.K.

Max. fi = + 156 psi < 3X@, 0.K

f5 = 0 + @ + @

Max. f;. = -300 psi < 0.45 f:, O.K.

Max.f, = -40 psi < 3*, O.K.

F u l l y h i n g e d /

Outside

-

t10,880-

(b)

Inside

-=‘-ih A +326

@ Horizontal P/S moment

+456 # -456

@ Liquid moment

- 1 7 0

& Vertical P/S -

] - 1 7 0

-4g6 L +I56

04 Tank empty

- 4 0 t-300

05 Tank ful l

(4

Figure 11.28 Stresses at maximum positive (+) moment, psi. (a) Wall base de-tails. (b) Semislide moment, ft-lb/ft. (c) Concrete stresses, psi.

Photo 11.11 Arco Floating LPG Barge: ABAM-designed largest floating pre-stressed hull in the world. (Courtesy, ABAM Engineers, Tacoma, Washington.)

Partial Fixity at the Wall Base. The restraint moment is Mp = IV, (1 - S), wherethe full fixity moment M, = 18,233 ft-lb/ft. The modifying factor for partial fixitys = (tAz)3/(dt)“2.

Figure 11.29 shows the deformed shape of the base slab. If the base slab thicknessh = 10 in., then, from Equations 11.39 and 11.40,

s = uo/w3

(125 x 0.83)“2 = OS”

and

Mp = M,(l - S) = l&233(1 - 0.1) = 16,410 ft-lb/ft.

The moment loss due to partial fixity = 18,233 - 16,410 = 1,823 ft-lb/ft. From Equation11.37 for the base ring width L,

2CH2L2 = -1+s

*Use temperature steel

7in liningt

t

t

“4 in. lining

zzf$ )zg..h

(a) (b)

Figure 11.29 Deformed shape of base slab. (a) Wall base. (b) Deformed sec-

11 .I 2 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 7 1 7

Also, from Table 11.4. the membrane coefficient at the base for form factor (H2)l(dt) = 6is C = -0.0187. Thus, we have

L2 = 2 x 0.0187(25)*

1 + 0.1= 21.25

and it follows that

L = 4.61 ft = 4 ft 74 in.

Accordingly, use a ring slab base width L = 4 ft 9 in. (145 cm). Since for large-diametertanks S has a very small value, the degree of fixity, as the solution shows, is almost thesame for both fully fixed and partially fixed wall bases.

From Equations 11.47 and 11.48, the percent R of prestress in the base that is trans-ferred to wall = 100/S,. where

s, = l.l($)(~~ 7 = l.l($(~~~’ = 13.50%

Consequently,

R= ‘00~ = 7.4%13.50

which means that the required design prestress for the wall can be slightly reduced, assome compression is available from the base ring.

Design of Prestressing Reinforcement

Horizontal Prestressing. Use the same size wire to wrap the circular wall, varyingthe spacing of the wire hoops in S-ft bands along the tank height. In the case of the freelysliding tank wall. the minimum spacing is in the lowest band at the base, as presentedgraphically in Figure 11.30.

In order to determine the variation of wire pitch throughout the height of the wall,additional computations of the horizontal ring thrust QY have to be made at the bottomof each band. Consequently. only one typical calculation of size and wire distribution willbe made for purposes of illustration.

Taking the case of the fixed wall base from Table 11.21, the maximum Q,, = 77,115lbift of circumference per foot height of wall. So trying 0.192-in. dia (4.88 mm) prestress-ing 250-K wire, we obtain A,,, = 0.0289 in.’ per wire and ,fbl = 0.7&,, = 0.7 x 250,ooO =175.000 psi (1,207 MPa).

Now assume 26-percent prestress loss for elastic shortening, seating, creep, shrink-age. and steel relaxation. Then

Figure 11.30 Horizontal-prestress wire distr ibution bands.


f,(, = 0.74 X 175.000 = 129,500 psi (893 MPa)

A,,77.115

= ~ =129,500

0.60 in.’ per 1 ft of wall height

No. of wire loops in 5-ft band =0.60 x 5

0.0289= 104

Hence, use 104 wire loops in the 5-ft wall band whose base is 15 ft below the top of thewater level. Also. use 2-in. shotcrete to cover the wrapped horizontal 0.192-in. dia wires.

If the tank were prestressed with I-in. dia 250-K 7-wire strand tendons. A,,, wouldbe 0.144 in.‘/strand and the required number of strands in a 5-ft-height band would be0.60 x 510.144 E 20 tendons.

Verfical Prestressing. For proportioning the vertical prestressing reinforcement.P,. = 100,000 lbift at e = 1.05 in. (1,459 N/m at e = 26.7 mm) on the outer force side. Hence.try $-in. dia (17.7-mm dia) 7-wire 250-K strands. We obtain

A,, = 0.144

fi,,, = 250.000 psi (1,724 MPa)

h,, = 0.7&,, = 0.7 X 250.000 = 175.000 psi (1.207 MPa)

Assume 26-percent total prestress loss. Then Ii,<, = 0.74 x 175.000 = 129.500 psi (X89 MPa).the required A,,, per foot of circumference = 100,000/129,500 = 0.772 in.’ (4.98 cm?). andthe number of vertical strands per foot of circumference = 0.77210.144 = 5.36. Thus. use$-in. dia 7-wire 250-K strands for vertical prestressing at 2: in. center-to-center spacing =0.769 in.’ E 0.772 in.‘, O.K.

Nominal Moment Strength Check of Tank Wall. The maximum wall vertical mo-ment for a fixed-base wall. from Table 11.21. is M = 28.056 ft-lbift or in.-lb/in. of circum-ference. We thus have:

S.F. = 1.3 (step 5 of flowchart)

M,, = 1.3 x 28.056 = 36.473 in.-lb/in

M,, 36.473W Mu = z = ~ = 40.525 in.-lb/in.o.y

d = F + 1.05 = 6.05 in. (15.37 cm)

0.144A,,, = __ = 0.064 in.‘jin. width

2.25

A,,, .t;,, 0.064 x 220,000“=0.8.5f:,= =0.85 x 5.000 x 1 3.31 in.

Available M,, = A,,, J,, = 0.064 X 220.000

II= 61.882 in.-lb/in. >> Rqd. M,, = 40.525 in.-lb/in.. O.K.

The wall design should include a check of the deflection as described in step 8 of theflowchart. Also. a determination should be made of the anchor steel at the base of thewall as well as the crack width u’,,,:,~ in step 9 of the flowchart. Finally. a check of temper-

11 .I 2 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 719

\ \\ a\\

Ring beamCritical shell

section

3

t

Figure 11.31 Tank dome shell roof. (a) Geometry of dome. (b) Edge ring beam.(c) Equivalent ring beam.

ature and creep effects has to be made to ascertain whether any additional nonpre-stressed mild steel has to be added to the prestressed wall reinforcement.

Design of Roof Dome Prestressed Edge Ring Beam. Use a rise-span ratio h’ld =&. Also, choose a freely supporting reaction at the top of the tank wall, using a neoprenepad under the edge ring beam. The shell would then have the form shown in Figures11.31 and 11.32.

Since ti = 125 ft.. h' = 12518 = 15.63 ft (4.76 m). Also. since C$ = 36” is less than 51”49’.the entire shell would be in compression, and only temperature reinforcement is needed.

Ring

Layer ofp n e u m a t i c - -

mortar

ia I”Circumferential

P/S wiresI 4

n I d

xtical P / Swires

Figure 11.32 Dome prestressed ring beam support detail in Example 11.3.


Photo 11.12 Olympic oval at the University of Calgary, Calgary, Canada. Struc-tural engineers: Simeson, Lester, Goodrich; Calgary, Alberta, Canada. (Courtesy,Prestressed Concrete Institute.)

The shell radius is

42 62.5a = - = - = 106 ft (32.3 m)

sin 4, 0.588

From Equation 11.75, the minimum shell thickness to withstand buckling is

hd = a J 1.5P,

44% Pc Ec

Hence, assuming that t = 3.0 in., we have

P,, = 1.40 + 1.7L = 1.4 z x 150 + 1.7 x 15 = 78 Ib/ft2(’ >

c$ = 0.7

p, = (a,r,)2 = (l.4~lo6y = 0.51

p, = 0.44 + 0.003 X 15 = 0.49 < 0.53, use p, = 0.49

E, = 57,000~5,000 = 4.03 X lo6 psi

1.5 x 78Minh = a

0.7 x 0.51 x 0.49 x 4.03 x lo6

= 1.36 in. (3.5 cm) < 3 in., O.K.

So use a shell t = 3 in. (7.6 cm). Then sin 4 = sin 36” = 0.59, cos 4 = cos 36” = 0.81, and a =sphere radius = 106 ft.

11 .I2 Design of Circular Prestressed Concrete Water-Retaining Tank and Its Domed Roof 721

From Equation 11.70, the tangential force per unit length of circumference is

WDdNo = ~1

- -2 sin + 1 + cos C$ cos + 1 s (cos W)

37.5 x 125 1=2 x 0.59 1 + 0.81

- 0.81 1 - yo*g (0.31)

= - 1,269 lb/ft

From Equation 11.67, the meridional force per unit length of circumference, witha = 106 ft, is

N+=-a w D +T1 + cos + >

=-106(%+:)=-2,99llb/11(43.6kN/m)

From Equation 11.72, the radial prestressing force in the ring beam required to pro-duce compatibility of deformation with the shell rim is

To determine the cross-sectional area bh of the ring beam, use P = (d/2)(N, cos 4) for thefirst trial, since the first term of the equation has less than 10 percent of the total value ofP (see the discussion accompanying Equation 11.62). We obtain

P = f (N+ cos +) = F (-2,991 X 0.81) = -151,149 lb per ft

Given that the total prestress loss is 26 percent, it follows that

7 = 1 - 0.26 = 0.74

and

P, =151 419) = 204,620 Ib/ft0.74

. Use a maximum concrete compressive stress f, = 800 psi (5.52 MPa) in order to min-imize excess strain in the edge beam, which could produce high stresses in the shell rim.The required cross-sectional area of the prestressed ring beam is

204 620A,+,h=t:=~fc 8 0 0

= 256 in.2

Try b = 14 in. and h = 20 in. Then A, = 280 in2 Substituting into Equation 11.72, we get

P = g -& x 1,269 - 0.2(-2,991). [ 1 + F(-2,991 X 0.81)

Use

= -5,217 - 151,419 = -156,636 lb/ft

P, = 156,636~ = 211,671 lb (717 kN)

0.74

7 2 2

s o


Hatch

V e n t D o m eSprl”gl”g

l i n e .

ins ideladder

Temporary /construction

manhole

/Waterstop

I- Half elevation ~__tc_ Half section __r(

Figure 11.33 Typical elevation and section of a domed prestressed concrete cir-cular tank.

From before,

fpl = 0.7f,, = 175,000 psi

Pi 2 1 1 , 6 7 1-=Ap,y = g = 175,000 1.21 in2 (7.56 cm2)

Trying $-in. dia (12.7-mm) 7-wire 250-K strands, we obtain

APT/strand = 0.144 in.2

and

No. of strands = &$4 = 8.4

If the prestress loss is slightly more than 26 percent, the number of strands should be ap-proximately 9. Hence, use nine+-in. dia 7-wire strands to prestress the edge ring beam.

Check the Concrete Stress in the Critical Section t = 3 in. of the Shell Rim. Themeridional compression N+ = -2,991 lb/ft of circumference, and the compressive stressf, = 2,991/(12 x 3) = 83 psi only, which is satisfactory. The support details of the edge ringbeam and the roof are shown in Figure 11.32. Note that the ring beam is supported verti-cally on a neoprene pad, which enables sliding. A typical elevation and section of adomed prestressed circular tank is shown in Figure 11.33.

REFERENCES

11.1 Timoshenko, S., and Woinowsky-Krieger, S. Theory of Plates and Shells. 2d ed. McGraw Hill, NewYork, 19.59.

11.2 Creasy. L. R. Prestressed Concrete Cylindrical Tanks. John Wiley & Sons, New York, 1961.

Problems 723

11.3 Billington, D. P. Thin Shell Concrete Structures. 2d ed. McGraw Hill, New York, 1982.11.4 Ghali, A. Circular Storage Tanks and Silos. E. & F. N. Spon Ltd., London, 1979.11.5 PCA, “Circular Concrete Tanks without Prestressing,” Concrete Information Series ST-57, Port-

land Cement Association, Skokie, Ill., 19.57,32 pp.11.6 PC1 Committee on Precast Prestressed Concrete Storage Tanks. “Recommended Practice for Pre-

cast Prestressed Concrete Circular Storage Tanks.” Prestressed Concrete Institute, Chicago, 1987.11.7 AC1 Committee 344. Design and Construction of Circular Prestressed Concrete Structures, ACI

344R. American Concrete Institute, Farmington Hills, MI, 1970.11.8 AC1 Committee 344. Design and Construction of Circular Wire and Strand Wrapped Prestressed

Concrete Structures, ACZ 344-R, American Concrete Institute, Farmington Hills, MI, 1989.11.9 AC1 Committee 344. Design and Construction of Circular Prestressed Concrete Structures with Cir-

cumferential Tendons, ACI 344.2R, American Concrete Institute, Farmington Hills, MI, 1989.11.10 Post-Tensioning Institute. Post-Tensioning Manual. 6th ed. Post-Tensioning Institute, Phoenix,

2000.11.11 Prestressed Concrete Institute. PCI Design Handbook. 5th ed. Prestressed Concrete Institute,

Chicago, 1999.11.12 Tadros, M. K. “Expedient Service Load Analysis of Cracked Prestressed Concrete Sections.” Jour-

nal of the Prestressed Concrete Institute, Vol. 27, No. 6, Nov-Dee, Chicago, 1983, 137-1.58.11.13 Brondum-Nielsen, T. “Prestressed Tanks.” Journal of the American Concrete Institute, Detroit,

July-August 1985, pp. 500-509.11.14 Vessey J.V., and Preston, R. L. A Critical Review of Code Requirements for Circular Prestressed

Concrete Reservoirs. F.I.P., Paris, 1978.11.15 Nawy, E. G., and Blair, H., Further Studies of Flexural Crack Control in Structural Slab Systems.

American Concrete Institute, Farmington Hills, MI, SP-30,1971.11.16 Abeles, P. W., and Bardhan-Roy, B. K. Prestressed Concrete Designer’s Handbook. 3d ed. View-

point Publications, London, 1981.

PROBLEMS

11.1 Solve Example 11.3 if the tank diameter is 120 ft (36.6 m) and the water height is 30 ft (9.1 m). As-sume that the total prestress loss is 20 percent, and use a rise-span ratio h’/d = & for the roof dome,assuming that half the shell angle is + = 45”.

11.2 A circular prestressed concrete tank has an internal diameter d = 85 ft (26 m) and retains water to aheight H= 22 ft (6.7 m). Determine the maximum horizontal ring forces and vertical moment, anddesign the prestressing reinforcement using both horizontal and vertical prestressing. Also, designa roof dome shell for the tank assuming a rise-span ratio h’/d = B and half shell angle 4 = 30”. Solvefor (a) hinged, (b) partially fixed, and (c) sliding wall base fixity, and design the prestressing rein-forcement for both wire-wrapped and tendon prestressing conditions. Given data are:

f: = 6,000 psi (41.4 MPa), normal weight

f :, = 4.250 psi (29.3 MPa)

ft I 3a = 230 psi (1.59 MPa)

A. = 0.45fi = 2,700 psi (18.6 MPa)

fey = 250 psi (1.72 MPa)-residual compressive stress

fp,, for both wire and strand or tendon = 250,000 psi (1,724 Pa)

hj = 0.7fp,i = 175,000 psi (1,207 MPa)

Snow load intensity wI. = 20 1b/ft2 (985 Pa)

Assume 20-percent total loss in prestress.

prestressed concrete circular storage …imcyc.com/biblioteca/archivospdf/tanques de concreto/4...

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