The Islamic University of Gaza Department of Civil Engineering
Design of Rectangular Concrete Tanks
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.
RECTANGULAR TANK DESIGN
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
RECTANGULAR TANK DESIGN
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.
RECTANGULAR TANK DESIGN
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.
RECTANGULAR TANK DESIGN
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.
RECTANGULAR TANK DESIGN
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-ofplane 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 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). y
Mz = moment per unit width about the z-axis
z
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). y x
RECTANGULAR TANK BEHAVIOR
M xy or M yz = 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
2 Myz=(Myz Coeff.) x q a /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.
RECTANGULAR TANK BEHAVIOR
The twisting moment effects such as Mxy may be used to add to the effects of orthogonal moments M x 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 M tx < 0,
If the calculated Mty < 0
then Mtx=0 and Mty=My + |Mxy2 /Mx| > 0 Then Mty = 0 and Mtx = Mx + |Mxy2 /My| > 0
Similar equations for where negative moments produce tension
RECTANGULAR TANK BEHAVIOR
Where negative moments produce tension:
Mtx = Mx-|Mxy| Mty = My - |Mxy| However, if the calculated M tx > 0,
then Mtx=0 and Mty=My - |Mxy2 /Mx| < 0
If the calculated Mty > 0
Then Mty = 0 and Mtx = Mx - |Mxy2 /My| < 0
Moment coefficient for Slabs with various edge Conditions
MultiCell Tank Corner of Multicell Tank:
Moment coefficients from chapter 3, designated as L coefficients, apply to outer or L shaped corners of multi-cell tanks.
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 Three wall forming T-Shape:
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 Three wall forming T-Shape:
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.
End Moments L
n n 2
L F
where n: number of adjacent unloaded walls
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 Design of Single-Cell Rectangular Tank
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. 2 2 Assume f c 300kg / cm and f y =4200 kg / cm
The tank will consider fixed at the base and free at the top in this example. A
C
E
Example 1 (Design of Rectangular Tank)
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
V C s q a
0.5 1 3 3 4.5ton V u 1.4 V
1.4 4.5 6.3ton ` V c 0.75 f c (b )(d )
0.53 0.75 300(100)(24.3) / 1000 16.7 ton V u d 30 5 1.4 / 2 24.3cm
Example 1 (Design of Rectangular Tank)
Check for shear at side edge of the long wall V C s q a 0.37 1 3 3 3.33ton
V u 1.4 V 1.4 3.33 4.67ton This wall is subjected to tensile forces due to shear in the short wall
Shear in the short wall
V C s q a 0.27 1 3 3 2.43ton V u 1.4 V 1.4 2.43 3.4ton N ` f c (b )(d ) V c 1 35 A g 3.4 1000 0.53 0.75 1 300(100)(24.3) / 1000 35 35 100 16.3ton V
Example 1 (Design of Rectangular Tank)
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.
Example 1 (Design of Rectangular Tank)
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 = S d factored load=S d U S d
s
f y f s
1.0
where :
165 from diagram
M ux
S d
factored load unfactored load 0.9 420
1.6
1.4 165 1.6 1.4 0.027 M x Coef . 0.0605 M x Coef .
Example 1 (Design of Rectangular Tank)
Example 1 (Design of Rectangular Tank) Vertical Bending Reinforcement: Inside Reinforcement (Mu=-7.8 t.m)
The required reinforcing of the interior face of the wall is M ux 0.0605 129 7.8 ton.m 0.85(300) 2.61(10)5 (7.8) 1 1 0.0036 min 2 4200 100(24.3) (300)
A s
0.0036 100 24.3 8.75 cm 2 / m
Use 812 mm/m on the inside of the wall. Outside Reinforcement (Mu=-7.8 t.m)
M ux 0.0605 10 0.605 ton.m
This maximum positive moment is very small and will controlled by minimum reinforcement. 38
Example 1 (Design of Rectangular Tank)
Design for Horizontal Reinforcement (My)
Horizontal Bending Reinforcement: Inside Reinforcement
M ux 0.0605 78 4.7 ton.m
0.85(300) 2.61(10)5 (4.7) 1 1 0.0021 min 2 4200 100(24.3) (300)
A s
0.0033 100 24.3 8.0 cm 2 / m
Use 812 mm/m on the inside of the wall. Outside Reinforcement
M ux 0.0605 24 1.45 ton.m
This maximum positive moment is very small and will controlled by minimum reinforcement. 39
Example 1 (Design of Rectangular Tank)
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.
Example 1 (Design of Rectangular Tank)
30 cm
10 cm 812/m 3m 812/m 7.5cm
Slab Reinforcement Details Walls Reinforcement Details
41
Example 1 (Design of Rectangular Tank)
Design for Uplift force under Loading Condition 3 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. Weight of the Tank
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
42
Example 1 (Design of Rectangular Tank)
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
43
Example 1 (Design of Roof Slab)
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 Coef. M tx = Coef. M x + Coef. |M xy | for +ve B.M. along short span
44
Example 1 (Design of Rectangular Tank) For Positive Moment along long span Coef. M ty = Coef. M y + Coef. |M xy | for +ve B.M. along long span
45
Example 1 (Design of Rectangular Tank) For Negative Moment along short span
Coef. M tx = Coef. M x - Coef. |M xy | for -ve B.M. along short span if M tx >0 then M tx =0
46
Example 1 (Design of Rectangular Tank) For Negative Moment along long span Coef. M ty = Coef. M y - Coef. |M xy | for -ve -ve B.M. along long span if M tx >0 then M tx =0
47
Example 1 (Design of Rectangular Tank)
Steel in short direction Positive moment at center
M tx
M tx coef . q u a 2 1000
,
Maximun Maxim un M tx coef . 78
qu S d 1.2 DL 1.6 L L qu 1.6 1.2 0.3 1 2.5 1.6 0.1 1.7t / m M 1.6 78 1.7 (6) 2 / 1000 7.6 t .m / m
0.85(300) 2.61(10)5 (7.6) 1 1 0.0034 min 2 4200 100(24.3) (300)
A s
DL factors factors of 1.2 1 .2 for slab own weight LL assumed assumed to be 100 kg/m2 k g/m2
0.0034 100 24.3 8.26 cm 2 / m
Use 812 mm/m for bottom bo ttom Reinforcement
48
Example 1 (Design of Rectangular Tank)
Steel in long direction Positive moment at center
M tx
M tx coef . q u a 2 1000
,
Maximun M tx coef . 51
M 1.6 51 1.7 (6) 2 /1000 5.0t .m / m d
30 5 1.2 0.6 23.2
0.85(300) 2.61(10)5 (5.0) 1 1 0.0025 min 2 4200 100(23.2) (300)
A s
0.0033 100 23.2 7.7 cm 2 / m
Use 812 mm/m for bottom Reinforcement
49
Example 1 (Design of Rectangular Tank)
Moment near corners Maximum Mtx and Mty Coef. =49
M tx
M tx coef . q u a 2 1000
,
Maximun M tx coef . 49
M 1.6 51 1.7 (6) 2 /1000 4.8t .m / m d
30 5 1.2 0.6 23.2
0.85(300) 2.61(10)5 (4.8) 1 1 0.0024 min 2 4200 100(23.2) (300)
A s
0.0033 100 23.2 7.7 cm 2 / m
Use 812 mm/m for bottom Reinforcement
50
Example 1 (Design of Rectangular Tank)
812/m
812/m
812/m
812/m
1.5m
25cm
Slab Reinforcement Details
51
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.
Two-Cell Tank, Long Center Wall
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.
Two-Cell Tank, Long Center Wall
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
Two-Cell Tank, Long Center Wall
BM coef. (My) for single-Cell-Tank – Long outer Wall
BM coef. (Mx) for single-Cell-Tank short outer Wall
Two-Cell Tank, Long Center Wall
BM coef. (My) for single-Cell-Tank – short outer Wall
BM coef. (Mx) for single-Cell-Tank – Center Wall
Two-Cell Tank, Long Center Wall
BM coef. (My) for single-Cell-Tank – Center Wall
Two-Cell Tank, Long 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 multicell. L-factors for short wall for b/a=3 & c/a=1are taken from page 330 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.
Two-Cell Tank, Long Center Wall
L and F factors for short outer Wall
L and F factors for center Wall
Two-Cell Tank, Long Center Wall
Adjust bending moment coef at T-shaped intersections
Coef.=L-(L-F)/3
L and F factors for center Wall
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).
Two-Cell Tank, Long Center Wall
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
Two-Cell Tank, Long Center Wall
BM coef. (My) for single-Cell-Tank – 6 m Long outer Wall
BM coef. (Mx) for single-Cell-Tank – 4.5 Long Wall
Two-Cell Tank, Long Center Wall
BM coef. (My) for single-Cell-Tank – 4.5 Long Wall
BM coef. (Mx) for single-Cell-Tank – Center Wall
Two-Cell Tank, Long 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 multicell. 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.
Two-Cell Tank, Long Center Wall
L and F factors for 4.5m Wall
L and F factors for center 6m Wall
Two-Cell Tank, Long Center 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
L and F factors for center Wall
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,
Details at Bottom Edge
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
Details at Bottom Edge
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.
Details at Bottom Edge
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).
Tanks Directly Built on Ground Tanks on Fill or Soft Weak Soil
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
Tanks Directly Built on Ground Tanks on Fill or Soft Weak Soil
In case of medium soils at foundation level, raft foundation may be used (see Figure 2).
Figure 2
Tanks Directly Built on Ground Tanks on Fill or Soft Weak Soil
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 Directly Built on Ground 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: wL3 24 EI
ML 6EI
0
L 2
Figure 4
M w
Tanks Directly Built on Ground 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 Directly Built on Ground Tanks on Compressible Soils
G1 = weight of the wall and roof G2 = weight of the floor cb W= weight of water on cb
Figure 5
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
(3/4H)
B
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
M C M m
PL2 12
PL
at support
Mc
2
24
Direct Tension :
Mm
at center
T
PL 2
Deep Tanks: Rectangular sections It is assumed that the maximum internal pressure take place at ¾ H from the top
M C M 1m
P
P
L 12
PL2
L 24
8 2
2
LB B
2
at support M1m
M c
Mc
2LB 2B
2
B
M2m L
Deep Tanks: Rectangular sections
M 2 m
PB 8
2
Mc
P
B 24
2
2LB 2L
Direct Tension in long Wall T
PB
Direct Tension in short Wall T
2 PL 2
2
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
H
R=H/2 M=H3 /6
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=0.1H H3 /33.5
+
H H3 /15
R=0.4H M=H3 /15
B) Shallow Tanks
c) Wall fixed at top and fixed at Bottom
M=H3 /20 M=H3 /20
R=0.15H
+ H3 /46.6
M=H3 /20
H
R=0.35H M=H3 /20
C) Medium Moderate Tanks In moderate or medium tanks where 0.5
H L
&
H B
2
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)
C) Medium Moderate Tanks 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 M1m Mc (3/4H)
B
M2m
H
L
C) Medium Moderate Tanks 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
H
R=H/2 M=H3 /6
C) Medium Moderate Tanks Approach 1: According to L/B ratio > 2
For Long Wall M base
H
3
6 Direct Tension
3 B T H 4 2
H
R=H/2 M=H3 /6
C) Medium Moderate Tanks Approach 1: According to L/B ratio >2
For Short Wall a) Horizontal Moment 2 3 H B
M sup port
4 12
(3/4H)
2 3 H B
M center
4 24 a) Vertical Moment
H H 1 H M H 2 96 4 3 4 1
H -
3
+
wH2 /24
wH2 /12
C) Medium Moderate Tanks 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 T 1H
C) Medium Moderate Tanks 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
C) Medium Moderate Tanks Approach 2: The Strip Method Load distribution of two-way slabs subjected to triangular loading is approximately the same as uniform load.
P=Pv + Ph 3H/4
H/4
Where: Pv Ph P: hydrostatic pressure at specific depth Pv: Pressure resisted in the vertical direction Ph: Pressure resisted in the horizontal direction
C) Medium Moderate Tanks Approach 2: The Strip Method The fixed Moment at bottom due to pressure resisted Ra a 2 2 vertically H H P h M f PV 15 117 3H/4
The shear at a Ra R a Pv
H 10
P h
H 540
H/4 Pv
Ph
b
The shear at b is evaluated from equilibrium The moments due to horizontal pressure are evaluated as discussed before at (3H/4) ( 3H/4)
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 My I 6 M bt 2
N bt N
f r
bt
2 f c'
+ve for tension -ve for compression
