the islamic university of gaza department of civil...

Design of Rectangular Concrete Tanks

The Islamic University of Gaza

Department of Civil Engineering

RECTANGULAR TANK DESIGN

The cylindrical shape is structurally best suited for tank construction, but rectangular tanks are frequently preferred for specific purposes

Easy formwork and construction process

Rectangular tanks are used where partitions or tanks with more than one cell are needed.


The behavior of rectangular tanks is different from the behavior of circular tanks

The behavior of circular tanks is axi-symmetric. That is the reason for the analysis to use only unit width of the tank

The ring tension in circular tanks was uniform around the circumference


The design of rectangular tanks is very similar in concept to the design of circular tanks

The loading combinations are the same. The modifications for the liquid pressure loading factor and the sanitary coefficient are the same.

The major differences are the calculated moments, shears, and tensions in the rectangular tank walls.


The requirements for durability are the same for rectangular and circular tanks.

The requirements for reinforcement (minimum or otherwise) are very similar to those for circular tanks.

The loading conditions that must be considered for the design are similar to those for circular tanks.


The restraint condition at the base is needed to determine deflection, shears and bending moments for loading conditions.

Base restraint conditions considered in the publication include both hinged and fixed edges.

However, in reality, neither of these two extremes actually exist.

It is important that the designer understand the degree of restraint provided by the reinforcing bars that extends into the footing from the tank wall.

If the designer is unsure, both extremes should be investigated.


Buoyancy forces must be considered in the design process

The lifting force of the water pressure is resisted by the weight of the tank and the weight of soil on top of the slab

Plate Analysis Results

This chapter gives the coefficients of deflections Cd, Shear Cs and moments (Mx, My, Mxy) for plates with different end conditions. Results are provided from FEM analysis of two dimensional plates subjected to our-of-plane loads.

The Slabs was assumed to act as a thin plate.

For square tanks the moment coefficient can be taken directly from the tables in chapter 2.

For rectangular tank, adjustments must be made to account for redistribution for bending moments to adjacent walls.

The design coefficient for rectangular tanks are given in chapter3

Tank Analysis Results

This chapter gives the coefficients of deflections Cd and moments (Mx, My, Mxy). The design are based on FEM analysis of tanks.

The shear coefficient Cs given in chapter 2 may be used for design of rectangular tanks.

The effect of tension force, if significant should be recognized.

RECTANGULAR TANK BEHAVIOR

x

y

y

z

Mx = moment per unit width about the x-axis

stretching the fibers in the y direction when the plate is in the x-y plane. This moment determines the steel in the y (vertical direction).

My = moment per unit width about the y-axis

stretching the fibers in the x direction when the plate is in the x-y plane. This moment determines the steel in the x or z (horizontal direction).

Mz = moment per unit width about the z-axis

stretching the fibers in the y direction when the plate is in the y-z plane. This moment determines the steel in the y (vertical direction).


Mxy or Myz = torsion or twisting moments for plate or wall in the x-y and y-z planes, respectively.

All these moments can be computed using the equations

Mx=(Mx Coeff.) x q a2/1000

My=(My Coeff.) x q a2/1000

Mz=(Mz Coeff.) x q a2/1000

Mxy=(Mxy Coeff.) x q a2/1000

Myz=(Myz Coeff.) x q a2/1000

These coefficients are presented in Tables of Chapter 2 and 3 for rectangular tanks

The shear in one wall becomes axial tension in the adjacent wall. Follow force equilibrium.


The twisting moment effects such as Mxy may be used to add to the effects of orthogonal moments Mx and My for the purpose of determining the steel reinforcement

The Principal of Minimum Resistance may be used for determining the equivalent orthogonal moments for design

Where positive moments produce tension:

Mtx = Mx + |Mxy|

Mty = My + |Mxy|

However, if the calculated Mtx < 0,

then Mtx=0 and Mty=My + |Mxy2/Mx| > 0

If the calculated Mty < 0

Then Mty = 0 and Mtx = Mx + |Mxy2/My| > 0

Similar equations for where negative moments produce tension

Moment coefficient for Slabs with various edge Conditions

MultiCell Tank

Moment coefficients from chapter 3, designated as L

coefficients, apply to outer or L shaped corners of

multi-cell tanks.

Corner of Multicell Tank:

MultiCell Tank

Three wall forming T-Shape:

If the continuous wall, or top of the T, is part of the long sides

of two adjacent rectangular cells, the moment in the continuous

wall at the intersection is maximum when both cells are filled.

The intersection is then fixed and moment coefficients,

designated as F coefficients, can be taken from Tables of

chapter 2.

MultiCell Tank


If the continuous wall is part of the short sides of two adjacent

rectangular cells, moment at one side of the intersection is

maximum, when the cell on that side is filled while the other

cell is empty.

For this loading condition the magnitude of moment will be

somewhere between the L coefficients and the F coefficients.

MultiCell Tank


If the unloaded third wall of the unit is disregarded, or its

stiffness considered negligible, moments in the loaded walls

would be the same L coefficients.

If the third wall is assumed to have infinite stiffness, the

corner is fixed and the F coefficients apply.

The intermediate value representing more nearly the true

condition can be obtained by the formula.

where n: number of adjacent unloaded walls

2

nEnd Moments L L F

n

MultiCell Tank

MultiCell Tank

Intersecting Walls:

If intersecting walls are the walls of square cells,

moments at the intersection are maximum when any

two cells are filled and the F coefficients in Tables 1,

2, or 3 apply because there is no rotation of the joint.

If the cells are rectangular, moments in the longer of

the intersecting walls will be maximum when two

cells on the same side of the wall under consideration

are filled, and again the F coefficients apply.

MultiCell Tank

Intersecting Walls:

Maximum moments in the

shorter walls adjacent to

the intersection occur

when diagonally opposite

cells are filled, and for this

condition the L

Coefficients apply.

Example 1

A

C

E

B F D

The tank shown has a clear height of a = 3m. horizontal

inside dimensions are b = 9.0 m and c = 6.0 m.

Height of the soil against wall is 1.5m.

Design of Single-Cell Rectangular Tank

2 2300 / and =4200 /c yf kg cm f kgAssume cm

The tank will consider fixed at the base and free at

the top in this example.

Design of Wall for Loading Condition 1 (Leakage Test)

Design for Shear Forces (Top Free anbd bottom Fixed)

According to Case 3 for : b/a = 3.0 and c/a = 2.0 (Page 2-17)

Example 1 (Design of Rectangular Tank)

Assume the wall thickness is 30 cm

Check for shear at bottom of the wall


0.5 1 3 3 4.5

1.4

1.4 4.5 6.3

s

u

V C q a

ton

V V

ton

`0.75 ( )( )

0.53 0.75 300(100)(24.3) /1000

16.7

30 5 1.4 / 2 24.3

c c

u

V f b d

ton V

d cm

Check for shear at side edge of the long wall

This wall is subjected to tensile forces due to shear in the short

wall

Shear in the short wall


0.37 1 3 3 3.33

1.4 1.4 3.33 4.67

s

u

V C q a ton

V V ton

`1 ( )( )35

3.4 10000.53 0.75 1 300(100)(24.3) /1000

35 35 100

16.3

c c

g

u

NV f b d

A

ton V

0.27 1 3 3 2.43

1.4 1.4 2.43 3.4

s

u

V C q a ton

V V ton

Note when design of Wall for Loading Condition 3 (cover

in place) (Top hinged and bottom fixed)

Case 4 page 2-23 for the shear coefficient is smaller than

previous case.


Design of Wall for Loading Condition 1 (Leakage Test)

Design for Vertical Reinforcement (Mx)

Moments are in ton.m if coefficients are multiplied by

qa2/1000= 3*9/1000=0.027

Moment coefficients taken from Table 5-1 for b/a = 3 and c/a = 2

For Sanitary Structures


Required Strength = factored load=

factored load1.0 :

unfactored load

0.9 420165 from diagram 1.6

1.4 165

1.6 1.4 0.027 . 0.0605 .

d d

y

d

s

s d

ux x x

S S U

fS where

f

f S

M M Coef M Coef

38

Vertical Bending Reinforcement:

Inside Reinforcement (Mu=-7.8 t.m)

5

min2

2

0.85(300) 2.61(10) (7.8)1 1 0.0036

4200 100(24.3) (300)

0.0036 100 24.3 8.75 /sA cm m

This maximum positive moment is very small and will controlled by

minimum reinforcement.

The required reinforcing of the interior face of the wall is

0.0605 129 7.8 .uxM ton m

Outside Reinforcement (Mu=-7.8 t.m)

0.0605 10 0.605 .uxM ton m

Use 812 mm/m on the inside of the wall.


39

Design for Horizontal Reinforcement (My) Horizontal Bending Reinforcement:

Inside Reinforcement

5

min2

2

0.85(300) 2.61(10) (4.7)1 1 0.0021

4200 100(24.3) (300)

0.0033 100 24.3 8.0 /sA cm m

This maximum positive moment is very small and will controlled by

minimum reinforcement.

0.0605 78 4.7 .uxM ton m

Outside Reinforcement

0.0605 24 1.45 .uxM ton m

Use 812 mm/m on the inside of the wall.


Note when design of Wall for Loading Condition 3 (cover

in place) (Top hinged and bottom fixed)

Case 4 page 3-39 for the moment coefficient is smaller than

previous case.


41


Slab Reinforcement Details

812/m

3m

812/m

30 cm

7.5cm

10 cm

Walls Reinforcement Details

42

Design for Uplift force under Loading Condition 3

Weight of the Tank

The weight of the slab and walls as well as the soil resting on the footing

projection must be capable of resisting the upward force of water.

Walls = height × length × thickness × 2.5 t/m3

=3 ×(9+9+6+6) ×0.3 ×2.5=67.5 ton

Bottom slab = length × width × thickness × 2.5 t/m3

=(9+0.6)×(6+0.6) ×0.3 ×2.5=47.5 ton

Top slab = length × width × thickness × 2.5 t/m3

=(9)×(6) ×0.3 ×2.5=40.5 ton

Soil on footing overhang =soil area ×soil height × 1.2 t/m3

=[(9.6 ×6.6)-(9 ×6)] ×1 ×1.2=11.2 ton

Total Resisting Load =67.5+47.5+40.5+11.2 =166.7 ton


43

Design for Uplift force under Loading Condition 3

Buoyancy Force

Buoyancy Force=Bottom slab area ×water pressure

=(9.6 ×6.6) ×1 ×1.3=82.4 ton

Assume the soil is 1m above the base slab.

Factor of Safety = Total resisting Load/Buoyancy Force

=166.7 /82.4 2.0


44

Example 1 (Design of Roof Slab)

Coef. Mtx = Coef. Mx + Coef. |Mxy| for +ve B.M. along short span

Design of Roof Slab

It is assumed that the tank has a simply supported roof

The slab is designed using plate analysis result of case 10

chapter 2 with a/b =9/6=1.5 page 2-62 For Positive Moment along short span

45


Coef. Mty = Coef. My + Coef. |Mxy| for +ve B.M. along long span

For Positive Moment along long span

46


Coef. Mtx = Coef. Mx - Coef. |Mxy| for -ve B.M. along short span

if Mtx>0 then Mtx=0

For Negative Moment along short span

47


Coef. Mty= Coef. My - Coef. |Mxy| for -ve B.M. along long span

if Mtx>0 then Mtx=0

For Negative Moment along long span

48

Steel in short direction Positive moment at center


DL factors of 1.2 for slab own weight

LL assumed to be 100 kg/m2

2

2

1.2 1.6

1.6 1.

., . 78

1000

0.3 1 2.5 0.1 1.7 /

78 1.7 (6) /1000 7

2 1.6

.1 6 . /.6

tx utx tx

u d

u

M coef q aM Maximun M coef

q S DL LL

q t m

M t m m

5

min2

2

0.85(300) 2.61(10) (7.6)1 1 0.0034

4200 100(24.3) (300)

0.0034 100 24.3 8.26 /sA cm m

Use 812 mm/m for bottom Reinforcement

49

Steel in long direction Positive moment at center


2

2

., . 51

1000

51 1.7 (6) /1000 5.0 .1.6 /

tx utx tx


M t m m

5

min2

2

30 5 1.2 0.6 23.2

0.85(300) 2.61(10) (5.0)1 1 0.0025

4200 100(23.2) (300)

0.0033 100 23.2 7.7 /s

d

A cm m


50

Moment near corners Maximum Mtx and Mty Coef. =49


2

2

., . 49

1000

51 1.7 (6) /1000 4.8 .1.6 /

tx utx tx


M t m m

5

min2

2

30 5 1.2 0.6 23.2

0.85(300) 2.61(10) (4.8)1 1 0.0024

4200 100(23.2) (300)

0.0033 100 23.2 7.7 /s

d

A cm m


51

812/m 812/m

812/m 812/m

25cm

1.5m


Slab Reinforcement Details

Two-Cell Tank, Long Center Wall

The tank in Figure consists of two adjacent cells, each with

the same inside dimensions as the single cell tank (a clear

height of a =3m. Horizontal inside dimensions are b = 9.0

m and c = 3.0 m). The top is considered free.

The tank consists of four L-shaped and two T-shaped units.

The Bending moments in the walls of multicell tanks are

approximately the same as in single tank, except at locations

of where more than two walls intersect.

The same coefficients of single-cell tank can be directly used

except at the T-shaped wall intersections.

L-(L-F)/3 coefficient are applicable for the three intersecting

walls of the two T-intersections

The coefficient are determined as follow:

Determine the BM Coef. In two-cell as if it were two

independent tanks.

Determine L and F factors to be used in adjustment of BM coef.

at T-shaped

Adjust bending moment coef. At T-shaped wall locations.


Determine the BM Coef. as if it were two independent Tanks

The BM coef. Are determined using table on page 3-30. For

b/a=3 and c/a=1 are given as follow:


BM coef. (Mx)for single-Cell-Tank –Long outer Wall


BM coef. (My) for single-Cell-Tank –Long outer Wall

BM coef. (Mx) for single-Cell-Tank –short outer Wall


BM coef. (My) for single-Cell-Tank –short outer Wall

BM coef. (Mx) for single-Cell-Tank –Center Wall


BM coef. (My) for single-Cell-Tank –Center Wall

Determine L & F factor to adjust BM for at T-shape wall location

The L and F factors are required to determine the bending

moment coefficient taking into account that the tank is multi-

cell.

L-factors for short wall for b/a=3 & c/a=1are taken from page 3-

30 and F factors for b/a=1are taken from page 2-21 of chapter 2.

L-factors for center wall b/a=3 & c/a=1are taken from page 3-30.

and F factors for b/a=3are taken from page 2-18 of chapter 2.

Note that coef is not needed for long outer wall since it not have

intersection with more than one wall.



L and F factors for short outer Wall

L and F factors for center Wall



Adjust bending moment coef at T-shaped intersections

Coef.=L-(L-F)/3

Two-Cell Tank, Short Center Wall

The tank in Figure consists of two cells with the same

inside dimensions as the cells in the two-cell tank with

the short center wall. (a clear height of a =3m.

Horizontal inside dimensions are b = 4.5 m and c = 6.0

m).

Determine the BM Coef. As if it were two independent Tanks

The BM coef. Are determined using table on page 3-31. For

b/a=2 and c/a=1.5 are given as follow:


BM coef. (Mx)for single-Cell-Tank – 6m Long outer Wall


BM coef. (My) for single-Cell-Tank – 6 m Long outer Wall

BM coef. (Mx) for single-Cell-Tank –4.5 Long Wall


BM coef. (My) for single-Cell-Tank –4.5 Long Wall

BM coef. (Mx) for single-Cell-Tank – Center Wall


BM coef. (Mx) for single-Cell-Tank – Center Wall

Determine L & F factor to adjust BM for at T-shape wall location

The L and F factors are required to determine the bending

moment coefficient taking into account that the tank is multi-

cell.

L-factors for short wall for are taken from page 3-31 and F

factors for b/a=2 and b/a=1.5 are taken from page 2-19 and 2-20

respectively.


L and F factors for 4.5m Wall

L and F factors for center 6m Wall



Adjust bending moment coef at T-shaped intersections

Coef = F for Col. 1 and Col 2

Coef.=L-(L-F)/3 for Col. 3and 4

Two-Cell Tank, Short Center Wall

6m 8m

Details at Bottom Edge

All tables except one are based on the assumption that the bottom

edge is hinged. It is believed that this assumption in general is

closer to the actual condition than that of a fixed edge.

Consider first the detail in Fig. 9, which shows the wall

supported on a relatively narrow continuous wall footing,


In Fig. 9 the condition of restraint at the bottom of the footing

is somewhere between hinged and fixed but much closer to

hinged than to fixed.

The base slab in Fig. 9 is placed on top of the wall footing and

the bearing surface is brushed with a heavy coat of asphalt to

break the adhesion and reduce friction between slab and

footing.

The vertical joint between slab and wall should be made

watertight. A joint width of 2.5 cm at the bottom is considered

adequate.

A waterstop may not be needed in the construction joints when

the vertical joint is made watertight


In Fig. 10 a continuous concrete base slab is provided either

for transmitting the load coming down through the wall or for

upward hydrostatic pressure.

In either case, the slab deflects upward in the middle and tends

to rotate the wall base in Fig. 10 in a counterclockwrse

direction.


The wall therefore is not fixed at the bottom edge and it is

difficult to predict the degree of restraint

The waterstop must then be placed off center as indicated.

Provision for transmitting shear through direct bearing can be

made by inserting a key as in Fig. 9 or by a shear ledge as in

Fig. 10.

At top of wall the detail in Fig. 10 may be applied except that

the waterstop and the shear key are not essential. The main

thing is to prevent moments from being transmitted from the

top of the slab into the wall because the wall is not designed

for such moments.

Tanks Directly Built on Ground

Tanks on Fill or Soft Weak Soil

The stress on the soil due to weight of the tank and water is

generally low (~0.6 kg/cm2 for a depth of water of 5m)

But it is not recommended to construct a tank directly on

unconsolidated soil of fill due to serious differential

settlement.

Soft weak clayey layers and similar soils may consolidate to

big values even under small stresses.

It is recommended to support the tank on columns and isolated

or strip footings if the stiff soil layers are at a reasonable depth

from the ground surface (see Figure 1).



It is recommended to support the tank on columns and isolated

or strip footings if the stiff soil layers are at a reasonable depth

from the ground surface (see Figure 1).

Figure 1



In case of medium soils at foundation level, raft foundation

may be used (see Figure 2).

Figure 2



If the incompressible layers are deep or the ground water level

is high one may support the tank on piles. The piles cap may

acts as column capitals (see Figure 3).

Figure 3


Tanks on Rigid Foundation.

If the tank supported by a rigid foundation then it the vertical

reaction of the wall will be resisted by area beneath it.

The distance L beyond which no deformation or bending

moment can be calculated approximately as follow:

Figure 4

3

0 224 6

wL ML ML

EI EI w


Tanks on Compressible Soils

Floors of tanks resisting on medium clayey or sandy soils may

be calculated in the following manner:

The internal forces transmitted from the wall to the floor may

be assumed to be distributed on the soil by the distance L=0.4

to 0.6H.

The length L is chosen such that the maximum stress 1 is

smaller than the allowed soil bearing pressure, 2 > 1/2 on

clayey soils and 2 > 0 on sandy soils.

This limitations are recommended in order to prevent

relatively big rotations of the floor at b.


Tanks on Compressible Soils

Figure 5

G1 = weight of the wall and roof

G2 = weight of the floor cb

W= weight of water on cb

Approximate Analysis

Design of Rectangular Concrete Tanks Approximate Analysis

Deep Tanks

Where H/L>2 and H/B >2

The effect of fixation of the wall will be limited to a small part at the base

The rest of the wall will resist water pressure horizontally by closed frame action

H

L

B

(3/4H)

H

Deep Tanks: Square sections

It is assumed that the maximum internal pressure take place at ¾ H from the top or 1m from the bottom whichever greater

2

2

at support12

at center24

C

m

PLM

PLM

Direct Tensio n :2

PLT

Mc

Mm

It is assumed that the maximum internal pressure take place at ¾ H from the top

2 2

2

1

2 2

at support12

8

2 224

C

m c

PM L LB B

PLM M

PL LB B

Mc

M1m

M2m

L

B

Deep Tanks: Rectangular sections

2

2 2

2 2 28 24

m c

PB PM M B LB L

Direct Tension in long Wall

Direct Tension in short Wall

2

2

PBT

PLT

Deep Tanks: Rectangular sections

B) Shallow Tanks

Where H/L and H/B <1/2

The water pressure is resisted by vertical action as follows:

a) Cantilever walls

Wall fixed to the floor and free at top may act as simple cantilever walls (suitable for H<3 m)

Tension in the floor = Reaction at the base

R=H/2

M=H3/6

H

B) Shallow Tanks

b) Wall simply supported at top and fixed at Bottom

Wall act as one way slab and resist water pressure in vertical direction (suitable for H<4.5 m)

R=0.4H

M=H3/15

H

R=0.1H

H3/15

H3/33.5 +

B) Shallow Tanks

c) Wall fixed at top and fixed at Bottom

R=0.35H

M=H3/20

H

R=0.15H

M=H3/20

M=H3/20

M=H3/20

H3/46.6

+

-

C) Medium Moderate Tanks

In moderate or medium tanks where

The water pressure is resisted by vertical and horizontal action

Different approximate methods is used to determine the internal distribution Some of them:

a) Approach 1: According to L/B ratio (Deep tank action)

b) Approach 2: Strip method (coefficient method)

0.5 & 2H H

L B


Approach 1: According to L/B ratio

For rectangular tank in which L/B<2 the tanks are designed as continuous frame subjected to max. pressure at H/4 from the bottom

The bottom H/4 is designed as a cantilever

Mc

M1m

M2m

L

B (3/4H)

H


Approach 1: According to L/B ratio

For rectangular tank in which L/B>2

The long wall are designed as a cantilever

The short walls as a slab fixed supported on the long walls

The bottom H/4 portion of the short wall is designed as a cantilever

R=H/2

M=H3/6

H


Approach 1: According to L/B ratio > 2

3

For Long Wall

Direct Tens

6

3

4 2

ion

base

HM

BT H

R=H/2

M=H3/6

H


Approach 1: According to L/B ratio >2

2

sup

2

3

3

4 12

For Short Wall

a) Horizontal Moment

a) Vertical Mom

3

4 24

1 1

2

ent

4 3 4 96

port

center

H BM

H BM

H H HM H

(3/4H)

H

wH2/12

+

-

wH2/24


Approach 1: According to L/B ratio > 2

Direct Tension

It is assumed that the end one meter width of the long wall contribute to direct tension on the short wall

Direct Tension Short Wall

1T H


Approach 2: The Strip Method

This method gives approximate solution for rectangular flat plates of constant thickness, supported in four sides and subjected to uniform hydrostatic pressure

Walls and floors supported on four sides and having L/B<2 are treated as two-way slabs.

Grashof, Marcus, or Egyptian code coefficient can be used to evaluate loads transferred in each direction



Load distribution of two-way slabs subjected to triangular loading is approximately the same as uniform load.

P=Pv + Ph

Where:

P: hydrostatic pressure at specific depth

Pv: Pressure resisted in the vertical direction

Ph: Pressure resisted in the horizontal direction

Pv Ph

H/4

3H/4



The fixed Moment at bottom due to pressure resisted vertically

The shear at a

The shear at b is evaluated from equilibrium

The moments due to horizontal pressure are evaluated as discussed before at (3H/4)

2 2

15 117f V h

H HM P P

10 540v h

H HRa P P

Pv Ph

H/4

3H/4

a

b

Ra

Design of section subjected to eccentric tension or compression

If the resultant stress on the liquid side is compression the section is to be designed as ordinary RC cracked section

If the resultant stress on the liquid side is tension the section must have

Adequate resistance of cracking

Adequate strength

+ve for tension

-ve for compression

'

2

62

r

c

My Nf

I bt

M Nf

bt bt

Design of section subjected to eccentric tension or compression

Reinforcement for direct tension can be added to reinforcement required to resist bending using strength design method.

'

u u uM M P e

Pu

Mu

Pu

Mu’

e

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