ENVIRONMENTAL ENGINEERING CONCRETE
STRUCTURES
CE 498 – Design Project
September 26, 2006
OUTLINE

INTRODUCTION

PERFORMANCE CRITERIA

DESIGN LOADS AND CONDITIONS

STRUCTURAL DESIGN

CONCRETE MIX DESIGN

ADDITIONAL CRITERIA
INTRODUCTION
Why concrete?
 Concrete is particularly suited for this application because
it will not warp or undergo change in dimensions
 When properly designed and placed it is nearly
impermeable and extremely resistant to corrosion
 Has good resistance to natural and processing chemicals
 Economical but requires significant quality control
What type of structure?
 Our focus will be conventionally reinforced cast-in-place or
precast concrete structures
 Basically rectangular and/or circular tanks
 No prestressed tanks
INTRODUCTION
How should we calculate loads?
 Design loads determined from the depth and unit weight of
retained material (liquid or solid), the external soil
pressure, and the equipment to be installed
 Compared to these loads, the actual live loads are small
 Impact and dynamical loads from some equipments
What type of analysis should be done?
 The analysis must be accurate to obtain a ‘reasonable’
picture of the stress distribution in the structure,
particularly the tension stresses
 Complicated 3D FEM analysis are not required. Simple
analysis using tabulated results in handbooks etc.
PERFORMANCE CRITERIA
What are the objective of the design?
 The structure must be designed such that it is watertight,
with minimum leakage or loss of contained volume.
 The structure must be durable – it must last for several
years without undergoing deterioration
How do you get a watertight structure?
 Concrete mix design is well-proportioned and it is well
consolidated without segregation
 Crack width is minimized
 Adequate reinforcing steel is used
 Impervious protective coating or barriers can also be used

This is not as economical and dependable as the approach of
mix design, stress & crack control, and adequate reinforcem.
PERFORMANCE CRITERIA
How to design the concrete mix?
 The concrete mix can be designed to have low permeability
by using low water-cement ratio and extended periods of
moist curing
 Use water reducing agents and pozzolans to reduce
permeability.
How to reduce cracking?
 Cracking can be minimized by proper design, distribution of
reinforcement, and joint spacing.
 Shrinkage cracking can be minimized by using joint design
and shrinkage reinforcement distributed uniformly
PERFORMANCE CRITERIA
How to increase durability?
 Concrete should be resistant to the actions of chemicals,
alternate wetting and drying, and freeze-thaw cycles
 Air-entrainment in the concrete mix helps improve
durability. Add air-entrainment agents
 Reinforcement must have adequate cover to prevent
corrosion
 Add good quality fly-ash or pozzolans
 Use moderately sulphate-resistant cement
DESIGN LOADS AND CONDITIONS


All the loads for the structure design can be obtained from
ASCE 7 (2006), which is the standard for minimum design
loads for building structures – endorsed by IBC
Content loads

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Raw Sewage ………………… 63 lb/ft3
Grit from grit chamber ….. 110 lb/ft3
Digested sludge aerobic…. 65 lb/ft3
Digested sludge anerobic… 70 lb/ft3
For other numbers see ACI 350.
Live loads


Catwalks etc 100 lb/ft2
Heavy equipment room 300 lb/ft2
DESIGN LOADS AND CONDITIONS

When using the LRFD (strength or limit states design
approach), the load factors and combinations from ACI 318
can be used directly with one major adjustment


The load factors for both the lateral earth pressure H and the
lateral liquid pressure F should be taken as 1.7
The factored load combination U as prescribed in ACI 318
must be increased by durability coefficients developed from
crack width calculation methods:




In calculations for reinforcement in flexure, the required
strength should be 1.3 U
In calculations for reinforcement in direct tension, including
hoop tension, the required strength should be 1.65 U
The required design strength for reinforcement in shear
should be calculated as fVs> 1.3 (Vu-fVc)
For compression use 1.0 U
STRUCTURAL DESIGN




Large reinforced concrete reservoirs on compressible soil
may be considered as beams on elastic foundations.
Sidewalls of rectangular tanks and reservoirs can be
designed as either: (a) cantilever walls fixed at the bottom,
or (b) walls supported at two or more edges.
Circular tanks normally resist the pressure from contents
by ring tension
Walls supporting both interior water loads and exterior soil
pressure must be designed to support the full effects of
each load individually

Cannot use one load to minimize the other, because
sometimes the tank is empty.
STRUCTURAL DESIGN

Large diameter tanks expand and contract appreciably as
they are filled and drained.


The connection between wall and footing should either
permit these movements or be strong enough to resist them
without cracking
The analysis of rectangular wall panels supported at three
or four sides is explained in detail in the PCA publication
that is available in the library and on hold for the course


It contains tabulated coefficients for calculating stress
distributions etc. for different boundary conditions and can
be used directly for design
It also includes some calculation and design examples
STRUCTURAL DESIGN

Reinforced concrete walls at least 10 ft. high that are in
contact with liquids should have a minimum thickness of
12 in.


For crack control, it is preferable to use a large number of
small diameter bars for main reinforcement rather than an
equal are of larger bars


The minimum thickness of any minor member is 6 in., and
when 2 in. cover is required then it is at least 8 in.
Maximum bar spacing should not exceed 12 in.
The amount of shrinkage and temperature reinforcement is
a function of the distance between joints in the direction


Shrinkage and temperature reinforcement should not be less
thank the ratios given in Figure 2.5 or ACI 350
The reinforcement should not be spaced more than 12 in.
and should be divided equally between the two surfaces
STRUCTURAL DESIGN

Figure showing minimum shrinkage reinforcement and
table showing minimum cover for reinforcement required
STRUCTURAL DESIGN

In order to prevent leakage, the strain in the tension
reinforcement has to be limited




The strain in the reinforcing bars is transferred to the
surrounding concrete, which cracks.
Hence, minimizing the stress and strain in the reinforcing bar
will minimize cracking in the concrete.
Additionally, distributing the tension reinforcement will
engage a greater area of the concrete in carrying the strain,
which will reduce cracking even more.
The strength design requires the use of loads, load
combinations and durability coefficients presented earlier
STRUCTURAL DESIGN



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
Serviceability for normal exposures
For flexural reinforcement located in one layer, the quantity
Z (crack control factor of ACI) should not exceed 115
kips/in.
The designer can use the basic Gergley-Lutz equation for
crack width for one way flexural members.
The reinforcement for two-way flexural member may be
proportioned in each direction using the above
recommendation too.
Alternate design by the working stress method with
allowable stress values given and tabulated in ACI 350. Do
not recommend this method for us.
STRUCTURAL DESIGN



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Impact, vibration, and torque issues
When heavy machines are involved, an appropriate impact
factor of 1.25 can be used in the design
Most of the mechanical equipment such as scrapers,
clarifiers, flocculators, etc. are slow moving and will not
cause structural vibrations
Machines that cause vibration problems are forced-draft
fans and centrifuges for dewatering clarifier sludge or
digester sludge
The key to successful dynamic design is to make sure that
the natural frequency of the support structure is
significantly different from frequency of disturbing force
STRUCTURAL DESIGN




To minimize resonant vibrations, ratio of the natural
frequency of the structure to the frequency of the
disturbing force must not be in the range of 0.5 to 1.5.
It should preferably be greater than 1.5
Methods for computing the structure frequency are
presented in ACI 350 (please review if needed)
Torque is produced in most clarifiers where the entire
mechanism is supported on a central column

This column must be designed to resist the torque shear
without undergoing failure
MATERIAL DESIGN

The cement should conform to:





Portland cement ASTM C150, Types I, IA, II, IIA, ….
Blended hydraulic cement ASTM C595
Expansive hydraulic cement ASTM C845
They cannot be used interchangeably in the same structure
Sulfate-resistant cement must have C3A content not
exceeding 8%. This is required for concrete exposed to
moderate sulfate acctak (150 to 1000 ppm)



Portland blast furnace slab cement (C595 may be used)
Portland pozzolan cement (C595 IP) can also be used
But, pozzolan content not exceed 25% by weight of
cementitous materials
MATERIAL DESIGN

The air entraining admixture should conform to ASTM C260




Improves resistant to freeze-thaw cycles
Improves workability and less shrinkage
If chemical admixtures are used, they should meet ASTM
C494. The use of water reducing admixtures is
recommended
The maximum water-soluble chloride ion content,
expressed as a % of cement, contributed by all ingredients
of the concrete mix should not exceed 0.10%
MATERIAL DESIGN

Mix proportioning – all material should be proportioned to
produce a well-graded mix of high density and workability




Type of cement – as mentioned earlier
Maximum water-cement ratio = 0.45


28 day compressive strength of 3500 psi where the concrete
is not exposed to severe weather and freeze-thaw
28 day compressive strength of 4000 psi where the concrete
is exposed to severe weather and freeze-thaw
If pozzolan is used, the maximum water-cement + pozzolan
ratio should be 0.45
Minimum cementitious material content



1.5 in. aggregate max – 517 lb/yd3
1 in. aggregate max – 536 lb/yd3
0.75 in. aggregate max – 564 lb/yd3
MATERIAL DESIGN

Air entrainment requirements



Slump requirements



5.5 ± 1 % for 1.5 in. aggregate
6.0 ± 1 % for 1.0 or 0.75 in. aggregate
1 in. minimum and 4 in. maximum
Concrete placement according to ACI 350 (read when you
get a chance)
Curing using sprinkling, ponding, using moisture retaining
covers, or applying a liquid membrane-forming compound
seal coat

Moist or membrane curing should commence immediately
after form removal
ADDITIONAL CRITERIA



Concrete made with proper material design will be dense,
watertight, and resistant to most chemical attack. Under
ordinary service conditions, it does not require additional
protection against chemical deterioration or corrosion
Reinforcement embedded in quality concrete is well
protected against corrosive chemicals
There are only special cases where additional protective
coatings or barriers are required


The steel bars must be epoxy coated (ASTM A775)
In special cases, where H2S evolves in a stagnant
unventilated environment that is difficult or uneconomical to
correct or clean regularly, a coating may be required
REFERENCES



ACI 350 (1989)
Books on reserve in the library
Emails from Jeffrey Ballard, structural engineer, HNTB. He
will visit to talk with us soon.
ENVIRONMENTAL ENGINEERING CONCRETE
STRUCTURES
CE 498 – Design Project
November 16, 21, 2006
OUTLINE

INTRODUCTION

LOADING CONDITIONS

DESIGN METHOD

WALL THICKNESS

REINFORCEMENT

CRACK CONTROL
INTRODUCTION



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:




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


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.

The tank must be designed and detailed to withstand the
forces from each of these loading conditions
LOADING CONDITIONS




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

Two approaches exist for the design of RC members



The use of strength design was considered inappropriate
due to the lack of reliable assessment of crack widths at
service loads.


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


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.


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


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

The walls of circular tanks are subjected to ring or hoop
tension due to the internal pressure and restraint to
concrete shrinkage.





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

Effects of shrinkage



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.

The difference in length between Fig.
2(b) and 2(c) is xC, an unknown quantity.
WALL THICKNESS

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



Concrete tensile stress is fcs = xCEc
Steel compressive stress is fss= (1-x)CEs
Section force equilibrium. So, rfss=fcs


The resulting stresses are:


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:


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

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.





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




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

The allowable steel stress fs should not be made too small.
Low fs will actually tend to increase the concrete stress and
potential cracking.




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

The amount size and spacing of
reinforcement has a great effect
on the extent of cracking.


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



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.



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


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



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



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


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


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



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.



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



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:



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

A plus sign indicates tension, so there is a slight
compression at the top, but it is very small.


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


Which case is conservative? (Fixed or hinged base)
The amount of ring steel required is given by:






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



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

The moments in vertical wall strips
that are considered 1 ft. wide are
computed by multiplying wuH3 by
the coefficients from table A-7.





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

The actual restraint is somewhere in between fixed and
hinged, but probably closer to hinged.



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:




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



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.


The shear capacity of a 10 in. wall with f’c=4000 psi is



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



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

The cylindrical shape is structurally best suited for tank
construction, but rectangular tanks are frequently preferred
for specific purposes



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


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