ExRectangularTanks-25-3-2015.pdf

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 + |Mxy
2/Mx| > 0
 If the calculated Mty < 0
 Then Mty = 0 and Mtx = Mx + |Mxy
2/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
n
End Moments L L F
n
  


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
 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 2
300 / and =4200 /
c y
f kg cm f kg
Assume 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 1000
0.53 0.75 1 300(100)(24.3) /1000
35 35 100
16.3
c c
g
u
N
V 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 = factoredload=
factoredload
1.0 :
unfactoredload
0.9 420
165 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
f
S where
f
f S
M M Coef M Coef



 
  

  

     

38
 Vertical Bending Reinforcement:
 Inside Reinforcement (Mu=-7.8 t.m)
5
min
2
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 /
s
A 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 .
ux
M ton m
    
 Outside Reinforcement (Mu=-7.8 t.m)
 
0.0605 10 0.605 .
ux
M ton m
  
Use 812 mm/m on the inside of the wall.

39
 Design for Horizontal Reinforcement (My)
 Horizontal Bending Reinforcement:
 Inside Reinforcement
5
min
2
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 /
s
A cm m
 
 
    
 
 
 
   
This maximum positive moment is very small and will controlled by
minimum reinforcement.
 
0.0605 78 4.7 .
ux
M ton m
    
 Outside Reinforcement
 
0.0605 24 1.45 .
ux
M ton m
  
Use 812 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
812/m
3m
812/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 u
tx tx
u d
u
M coef q a
M Maximun M coef
q S DL LL
q t m
M t m m
 
 
    
       
    
5
min
2
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 /
s
A cm m
 
 
    
 
 
 
   
Use 812 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 u
tx tx
M coef q a
M Maximun M coef
M t m m
 
 
    
5
min
2
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 u
tx tx
M coef q a
M Maximun M coef
M t m m
 
 
    
5
min
2
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
812/m 812/m
812/m
812/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 2
24 6
wL ML M
L
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
atsupport
12
at center
24
C
m
PL
M
PL
M
 

Direct Tension :
2
PL
T 
Mc
Mm

It is assumed that the maximum internal pressure take
place at ¾ H from the top
 
 
2 2
2
1
2 2
atsupport
12
8
2 2
24
C
m c
P
M L LB B
PL
M M
P
L LB B
   
 
  
Mc
M1m
M2m
L
B
Deep Tanks: Rectangular sections

 
2
2 2
2 2 2
8 24
m c
PB P
M M B LB L
    
Direct Tension in long Wall
Direct Tension in short Wall
2
2
PB
T
PL
T


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=0.4H
M=H3/15
H
R=0.1H
H3/15
H3/33.5
+

B) Shallow Tanks
 c) Wall fixed at top and fixed at Bottom
R=0.35H
M=H3/20
H
R=0.15H
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 & 2
H 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
H
M
B
T 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 B
M
H B
M
H H H
M 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
1
T 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
15 117
f V h
H H
M P P
 
  
 
 
10 540
v h
H H
Ra 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
6
2
r
c
My N
f
I bt
M N
f
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 u
M M P e
 
Pu
Mu
Pu
Mu’
e

ExRectangularTanks-25-3-2015.pdf

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