ENVIRONMENTAL ENGINEERING CONCRETE
STRUCTURES
CE 498 – Design Project
November 16, 21, 2006
OUTLINE
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INTRODUCTION
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LOADING CONDITIONS
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DESIGN METHOD
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WALL THICKNESS
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REINFORCEMENT
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CRACK CONTROL
INTRODUCTION
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Conventionally reinforced circular concrete tanks have been
used extensively. They will be the focus of our lecture
today
Structural design must focus on both the strength and
serviceability. The tank must withstand applied loads
without cracks that would permit leakage.
This is achieved by:
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Providing proper reinforcement and distribution
Proper spacing and detailing of construction joints
Use of quality concrete placed using proper construction
procedures
A thorough review of the latest report by ACI 350 is
important for understanding the design of tanks.
LOADING CONDITIONS
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The tank must be designed to withstand the loads that it
will be subjected to during many years of use. Additionally,
the loads during construction must also be considered.
Loading conditions for partially buried tank.
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The tank must be designed and detailed to withstand the
forces from each of these loading conditions
LOADING CONDITIONS
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The tank may also be subjected to uplift forces from
hydrostatic pressure at the bottom when empty.
It is important to consider all possible loading conditions
on the structure.
Full effects of the soil loads and water pressure must be
designed for without using them to minimize the effects of
each other.
The effects of water table must be considered for the
design loading conditions.
DESIGN METHODS
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Two approaches exist for the design of RC members
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The use of strength design was considered inappropriate
due to the lack of reliable assessment of crack widths at
service loads.
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Strength design, and allowable stress design.
Strength design is the most commonly adopted procedure for
conventional buildings
Advances in this area of knowledge in the last two decades
has led to the acceptance of strength design methods
The recommendations for strength design suggest inflated
load factors to control service load crack widths in the
range of 0.004 – 0.008 in.
Design Methods
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Service state analyses of RC structures should include
computations of crack widths and their long term effects
on the structure durability and functional performance.
The current approach for RC design include computations
done by a modified form of elastic analysis for composite
reinforced steel/concrete systems.
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The effects of creep, shrinkage, volume changes, and
temperature are well known at service level
The computed stresses serve as the indices of performance
of the structure.
DESIGN METHODS

The load combinations to determine the required strength
(U) are given in ACI 318. ACI 350 requires two
modifications
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Modification 1 – the load factor for lateral liquid pressure is
taken as 1.7 rather than 1.4. This may be over conservative
due to the fact that tanks are filled to the top only during
leak testing or accidental overflow
Modification 2 – The members must be designed to meet the
required strength. The ACI required strength U must be
increased by multiplying with a sanitary coefficient

The increased design loads provide more conservative design
with less cracking.
Required strength = Sanitary coefficient X U
Where, sanitary coefficient = 1.3 for flexure, 1.65 for direct
tension, and 1.3 for shear beyond the capacity provided by the
concrete.
WALL THICKNESS
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The walls of circular tanks are subjected to ring or hoop
tension due to the internal pressure and restraint to
concrete shrinkage.
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Any significant cracking in the tank is unacceptable.
The tensile stress in the concrete (due to ring tension from
pressure and shrinkage) has to kept at a minimum to prevent
excessive cracking.
The concrete tension strength will be assumed 10% f’c in this
document.
RC walls 10 ft. or higher shall have a minimum thickness of
12 in.
The concrete wall thickness will be calculated as follows:
WALL THICKNESS
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Effects of shrinkage
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Figure 2(a) shows a block of concrete
with a re-bar. The block height is 1 ft, t
corresponds to the wall thickness, the
steel area is As, and the steel percentage
is r.
Figure 2(b) shows the behavior of the
block assuming that the re-bar is absent.
The block will shorten due to shrinkage.
C is the shrinkage per unit length.
Figure 2(c) shows the behavior of the
block when the re-bar is present. The rebar restrains some shortening.
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The difference in length between Fig.
2(b) and 2(c) is xC, an unknown quantity.
WALL THICKNESS
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The re-bar restrains shrinkage of the concrete. As a result,
the concrete is subjected to tension, the re-bar to
compression, but the section is in force equilibrium
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Concrete tensile stress is fcs = xCEc
Steel compressive stress is fss= (1-x)CEs
Section force equilibrium. So, rfss=fcs
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The resulting stresses are:
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fss=CEs[1/(1+nr)]
and
fcs=CEs[r/(1+nr)]
The concrete stress due to an applied ring or hoop tension
of T will be equal to:
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Solve for x from above equation for force equilibrium
T * Ec/(EcAc+EsAs) = T * 1/[Ac+nAs] = T/[Ac(1+nr)]
The total concrete tension stress = [CEsAs + T]/[Ac+nAs]
WALL THICKNESS
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The usual procedure in tank design is to provide horizontal
steel As for all the ring tension at an allowable stress fs as
though designing for a cracked section.
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Assume As=T/fs and realize Ac=12t
Substitute in equation on previous slide to calculate tension
stress in the concrete.
Limit the max. concrete tension stress to fc = 0.1 f’c
Then, the wall thickness can be calculated as
t = [CEs+fs–nfc]/[12fcfs]* T
This formula can be used to estimate the wall thickness
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The values of C, coefficient of shrinkage for RC is in the
range of 0.0002 to 0.0004.
Use the value of C=0.0003
Assume fs= allowable steel tension =18000 psi
Therefore, wall thickness t=0.0003 T
WALL THICKNESS
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The allowable steel stress fs should not be made too small.
Low fs will actually tend to increase the concrete stress and
potential cracking.
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For example, the concrete stress = fc = [CEs+fs]/[Acfs+nT]*T
For the case of T=24,000 lb, n=8, Es=29*106 psi, C=0.0003
and Ac=12 x 10 = 120 in3
If the allowable steel stress is reduced from 20,000 psi to
10,000 psi, the resulting concrete stress is increased from
266 psi to 322 psi.
Desirable to use a higher allowable steel stress.
REINFORCEMENT
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The amount size and spacing of
reinforcement has a great effect
on the extent of cracking.
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The amount must be sufficient
for strength and serviceability
including temperature and
shrinkage effects
The amount of temperature and
shrinkage reinforcement is
dependent on the length
between construction joints
REINFORCEMENT
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The size of re-bars should be chosen recognizing that
cracking can be better controlled by using larger number of
small diameter bars rather than fewer large diameter bars
The size of reinforcing bars should not exceed #11.
Spacing of re-bars should be limited to a maximum of 12
in. Concrete cover should be at least 2 in.
In circular tanks the locations of horizontal splices should
be staggered by not less than one lap length or 3 ft.
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Reinforcement splices should confirm to ACI 318
Chapter 12 of ACI 318 for determining splice lengths.
The length depends on the class of splice, clear cover, clear
distance between adjacent bars, and the size of the bar,
concrete used, bar coating etc.
CRACK CONTROL
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Crack widths must be minimized in tank walls to prevent
leakage and corrosion of reinforcement
A criterion for flexural crack width is provided in ACI 318.
This is based on the Gergely-Lutz equation z=fs(dcA)1/3
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Where z = quantity limiting distribution of flexural re-bar
dc = concrete cover measured from extreme tension fiber to
center of bar located closest.
A = effective tension area of concrete surrounding the
flexural tension reinforcement having the same centroid as
the reinforcement, divided by the number of bars.
CRACK CONTROL
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In ACI 350, the cover is taken equal to 2.0 in. for any cover
greater than 2.0 in.
Rearranging the equation and solving for the maximum bar
spacing give: max spacing = z3/(2 dc2 fs3)
Using the limiting value of z given by ACI 350, the maximum
bar spacing can be computed
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For ACI 350, z has a limiting value of 115 k/in.
For severe environmental exposures, z = 95 k/in.
ANALYSIS OF VARIOUS TANKS
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Wall with fixed base and free top; triangular load
Wall with hinged base and free top; triangular load and
trapezoidal load
Wall with shear applied at top
Wall with shear applied at base
Wall with moment applied at top
Wall with moment applied at base
CIRCULAR TANK ANALYSIS
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In practice, it would be rare that a base would be fixed
against rotation and such an assumption would lead to an
improperly designed wall.
For the tank structure, assume
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Height = H = 20 ft.
Diameter of inside = D = 54 ft.
Weight of liquid = w = 62.5 lb/ft3
Shrinkage coefficient = C = 0.0003
Elasticity of steel = Es = 29 x 106 psi
Ratio of Es/Ec = n = 8
Concrete compressive strength = f’c = 4000 psi
Yield strength of reinforcement = fy = 60,000 psi
CIRCULAR TANK ANALYSIS
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It is difficult to predict the behavior of the subgrade and its
effect upon restraint at the base. But, it is more reasonable
to assume that the base is hinged rather than fixed, which
results in more conservative design.
For a wall with a hinged base and free top, the coefficients
to determine the ring tension, moments, and shears in the
tank wall are shown in Tables A-5, A-7, and A-12 of the
Appendix
Each of these tables, presents the results as functions of
H2/Dt, which is a parameter.
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The values of thickness t cannot be calculated till the ring
tension T is calculated.
Assume, thickness = t = 10 in.
Therefore, H2/Dt = (202)/(54 x 10/12) = 8.89 (approx. 9 in.)
Table A-5 showing the ring tension values
Table A-7, A-12 showing the moment and shear
CIRCULAR TANK ANALYSIS
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In these tables, 0.0 H corresponds to the top of the tank,
and 1.0 H corresponds to the bottom of the tank.
The ring tension per foot of height is computed by
multiplying wu HR by the coefficients in Table A-5 for the
values of H2/Dt=9.0
wu for the case of ring tension is computed as:
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wu = sanitary coefficient x (1.7 x Lateral Forces)
wu = 1.65 x (1.7 x 62.5) = 175.3 lb/ft3
Therefore, wu HR = 175.3 x 20 x 54/2 = 94, 662 lb/ft3
The value of wu HR corresponds to the behavior where the
base is free to slide. Since, it cannot do that, the value of
wu HR must be multiplied by coefficients from Table A-5
CIRCULAR TANK ANALYSIS
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A plus sign indicates tension, so there is a slight
compression at the top, but it is very small.
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The ring tension is zero at the base since it is assumed that
the base has no radial displacement
Figure compares the ring tension for tanks with free sliding
base, fixed base, and hinged base.
CIRCULAR TANK ANALYSIS
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Which case is conservative? (Fixed or hinged base)
The amount of ring steel required is given by:
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As = maximum ring tension / (0.9 Fy)
As = 67494/(0.9 * 60000) = 1.25 in2/ft.
Therefore at 0.7H use #6bars spaced at 8 in. on center in
two curtains.
Resulting As = 1.32in2/ft.
The reinforcement along the height of the wall can be
determined similarly, but it is better to have the same bar
and spacing.
Concrete cracking check
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The maximum tensile stress in the concrete under service
loads including the effects of shrinkage is
fc = [CEsAs + Tmax, unfactored]/[Ac+nAs] = 272 psi < 400 psi
Therefore, adequate
CIRCULAR TANK ANALYSIS
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The moments in vertical wall strips
that are considered 1 ft. wide are
computed by multiplying wuH3 by
the coefficients from table A-7.
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The value of wu for flexure =
sanitary coefficient x (1.7 x lateral
forces)
Therefore, wu = 1.3 x 1.7 x 62.5 =
138.1 lb/ft3
Therefore wuH3 = 138.1 x 203 =
1,104,800 ft-lb/ft
The computed moments along the
height are shown in the Table.
The figure includes the moment for
both the hinged and fix conditions
CIRCULAR TANK ANALYSIS
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The actual restraint is somewhere in between fixed and
hinged, but probably closer to hinged.
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For the exterior face, the hinged condition provides a
conservative although not wasteful design
Depending on the fixity of the base, reinforcing may be
required to resist moment on the interior face at the lower
portion of the wall.
The required reinforcement for the outside face of the wall
for a maximum moment of 5,524 ft-lb/ft. is:
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Mu/(f f’c bd2) = 0.0273 ………(where d = t – cover – dbar/2)
From the standard design aid of Appendix A, take the value
of 0.0273 and obtain a value for w from the Table.
Obtain w=0.0278
Required As = w bdf’c/fy = 0.167 in2
CIRCULAR TANK ANALYSIS
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r=0.167/(12 x 7.5) = 0.00189
rmin = 200/Fy = 0.0033 > 0.00189
Use #5 bars at the maximum allowable spacing of 12 in.
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The shear capacity of a 10 in. wall with f’c=4000 psi is
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As = 0.31 in2 and r = 0.0035
Vc = 2 (f’c)0.5 bwd = 11,384 kips
Therefore, f Vc = 0.85 x 11,284 = 9676 kips
The applied shear is given by multiplying wu H2 with the
coefficient from Table A-12
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The value of wu is determined with sanitary coefficient = 1.0
(assuming that no steel rft. will be needed)
wuH2 = 1.0 x 1.7 x 62.5 x 202 = 42,520 kips
Applied shear = Vu = 0.092 x wuH2 = 3912 kips < fVc
RECTANGULAR TANK DESIGN
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The cylindrical shape is structurally best suited for tank
construction, but rectangular tanks are frequently preferred
for specific purposes
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Rectangular tanks can be used instead of circular tanks when
the footprint needs to be reduced
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
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The behavior of circular tanks is axisymmetric. That is the
reason for our analysis of only unit width of the tank
The ring tension in circular tanks was uniform around the
circumference
RECTANGULAR TANK DESIGN
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The design of rectangular tanks is very similar in concept
to the design of circular tanks
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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. This is related to crack width control,
which is achieved using the Gergely Lutz parameter z.
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
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The restraint condition at the base is needed to determine
deflection, shears and bending moments for loading
conditions.
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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 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
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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
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 (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
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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
2
 Mx=(Mx Coeff.) x q a /1000
2
 My=(My Coeff.) x q a /1000
2
 Mz=(Mz Coeff.) x q a /1000
2
 Mxy=(Mxy Coeff.) x q a /1000
2
 Myz=(Myz Coeff.) x q a /1000
These coefficients are presented in Tables 2 and 3 for rectangular
tanks
The shear in one wall becomes axial tension in the adjacent wall.
Follow force equilibrium - explain in class.
RECTANGULAR TANK BEHAVIOR
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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
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Where positive moments produce tension:
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Mtx = Mx + |Mxy|
Mty = My + |Mxy|
However, if the calculated Mtx < 0,
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If the calculated Mty < 0
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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