Upload
anonymous-bdx1gkmmy
View
220
Download
0
Embed Size (px)
Citation preview
7/28/2019 CE498 Lecture Sept 26
1/57
ENVIRONMENTAL ENGINEERING CONCRETESTRUCTURES
CE 498 Design Project
September 26, 2006
7/28/2019 CE498 Lecture Sept 26
2/57
OUTLINE
INTRODUCTION
PERFORMANCE CRITERIA
DESIGN LOADS AND CONDITIONS
STRUCTURAL DESIGN
CONCRETE MIX DESIGN
ADDITIONAL CRITERIA
7/28/2019 CE498 Lecture Sept 26
3/57
INTRODUCTION
Why concrete?
Concrete is particularly suited for this application becauseit will not warp or undergo change in dimensions
When properly designed and placed it is nearlyimpermeable 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 orprecast concrete structures
Basically rectangular and/or circular tanks
No prestressed tanks
7/28/2019 CE498 Lecture Sept 26
4/57
INTRODUCTION
How should we calculate loads?
Design loads determined from the depth and unit weight ofretained material (liquid or solid), the external soilpressure, 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. Simpleanalysis using tabulated results in handbooks etc.
7/28/2019 CE498 Lecture Sept 26
5/57
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 severalyears without undergoing deterioration
How do you get a watertight structure?
Concrete mix design is well-proportioned and it is wellconsolidated 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.
7/28/2019 CE498 Lecture Sept 26
6/57
PERFORMANCE CRITERIA
How to design the concrete mix?
The concrete mix can be designed to have lowpermeability by using low water-cement ratio andextended 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
7/28/2019 CE498 Lecture Sept 26
7/57
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 improvedurability. Add air-entrainment agents
Reinforcement must have adequate cover to preventcorrosion
Add good quality fly-ash or pozzolans
Use moderately sulphate-resistant cement
7/28/2019 CE498 Lecture Sept 26
8/57
DESIGN LOADS AND CONDITIONS
All the loads for the structure design can be obtained from
ASCE 7 (2006), which is the standard for minimum designloads for building structures endorsed by IBC
Content loads
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
7/28/2019 CE498 Lecture Sept 26
9/57
DESIGN LOADS AND CONDITIONS
When using the LRFD (strength or limit states design
approach), the load factors and combinations from ACI 318can be used directly with one major adjustment
The load factors for both the lateral earth pressure H and thelateral liquid pressure F should be taken as 1.7
The factored load combination U as prescribed in ACI 318must be increased by durability coefficients developed fromcrack 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 shearshould be calculated as fVs> 1.3 (Vu-fVc)
For compression use 1.0 U
7/28/2019 CE498 Lecture Sept 26
10/57
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 contentsby ring tension
Walls supporting both interior water loads and exterior soilpressure 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.
7/28/2019 CE498 Lecture Sept 26
11/57
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 themwithout cracking
The analysis of rectangular wall panels supported at threeor four sides is explained in detail in the PCA publicationthat is available in the library and on hold for the course
It contains tabulated coefficients for calculating stressdistributions etc. for different boundary conditions and canbe used directly for design
It also includes some calculation and design examples
7/28/2019 CE498 Lecture Sept 26
12/57
STRUCTURAL DESIGN
Reinforced concrete walls at least 10 ft. high that are in
contact with liquids should have a minimum thickness of12 in.
The minimum thickness of any minor member is 6 in., andwhen 2 in. cover is required then it is at least 8 in.
For crack control, it is preferable to use a large number ofsmall diameter bars for main reinforcement rather than anequal are of larger bars
Maximum bar spacing should not exceed 12 in.
The amount of shrinkage and temperature reinforcement isa function of the distance between joints in the direction
Shrinkage and temperature reinforcement should not be lessthank 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
7/28/2019 CE498 Lecture Sept 26
13/57
STRUCTURAL DESIGN
Figure showing minimum shrinkage reinforcement andtable showing minimum cover for reinforcement required
7/28/2019 CE498 Lecture Sept 26
14/57
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, loadcombinations and durability coefficients presented earlier
7/28/2019 CE498 Lecture Sept 26
15/57
STRUCTURAL DESIGN
Serviceability for normal exposures
For flexural reinforcement located in one layer, thequantity Z (crack control factor of ACI) should not exceed115 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 aboverecommendation too.
Alternate design by the working stress method withallowable stress values given and tabulated in ACI 350. Donot recommend this method for us.
7/28/2019 CE498 Lecture Sept 26
16/57
STRUCTURAL DESIGN
Impact, vibration, and torque issues
When heavy machines are involved, an appropriate impactfactor 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 ordigester sludge
The key to successful dynamic design is to make sure thatthe natural frequency of the support structure issignificantly different from frequency of disturbing force
7/28/2019 CE498 Lecture Sept 26
17/57
STRUCTURAL DESIGN
To minimize resonant vibrations, ratio of the natural
frequency of the structure to the frequency of thedisturbing 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 entiremechanism is supported on a central column
This column must be designed to resist the torque shearwithout undergoing failure
7/28/2019 CE498 Lecture Sept 26
18/57
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 notexceeding 8%. This is required for concrete exposed tomoderate 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 ofcementitous materials
7/28/2019 CE498 Lecture Sept 26
19/57
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 ASTMC494. The use of water reducing admixtures is
recommended The maximum water-soluble chloride ion content,
expressed as a % of cement, contributed by all ingredientsof the concrete mix should not exceed 0.10%
7/28/2019 CE498 Lecture Sept 26
20/57
MATERIAL DESIGN
Mix proportioning all material should be proportioned to
produce a well-graded mix of high density and workability 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 Type of cement as mentioned earlier
Maximum water-cement ratio = 0.45
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
7/28/2019 CE498 Lecture Sept 26
21/57
MATERIAL DESIGN
Air entrainment requirements
5.5 1 % for 1.5 in. aggregate
6.0 1 % for 1.0 or 0.75 in. aggregate
Slump requirements
1 in. minimum and 4 in. maximum
Concrete placement according to ACI 350 (read when youget a chance)
Curing using sprinkling, ponding, using moisture retainingcovers, or applying a liquid membrane-forming compoundseal coat
Moist or membrane curing should commence immediately
after form removal
7/28/2019 CE498 Lecture Sept 26
22/57
ADDITIONAL CRITERIA
Concrete made with proper material design will be dense,
watertight, and resistant to most chemical attack. Underordinary service conditions, it does not require additionalprotection 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 stagnantunventilated environment that is difficult or uneconomical tocorrect or clean regularly, a coating may be required
7/28/2019 CE498 Lecture Sept 26
23/57
REFERENCES
ACI 350 (1989)
Books on reserve in the library
Emails from Jeffrey Ballard, structural engineer, HNTB. Hewill visit to talk with us soon.
7/28/2019 CE498 Lecture Sept 26
24/57
ENVIRONMENTAL ENGINEERING CONCRETESTRUCTURES
CE 498 Design Project
November 16, 21, 2006
7/28/2019 CE498 Lecture Sept 26
25/57
OUTLINE
INTRODUCTION
LOADING CONDITIONS
DESIGN METHOD
WALL THICKNESS
REINFORCEMENT
CRACK CONTROL
7/28/2019 CE498 Lecture Sept 26
26/57
INTRODUCTION
Conventionally reinforced circular concrete tanks have
been used extensively. They will be the focus of ourlecture today
Structural design must focus on both the strength andserviceability. 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 constructionprocedures
A thorough review of the latest report by ACI 350 isimportant for understanding the design of tanks.
7/28/2019 CE498 Lecture Sept 26
27/57
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
7/28/2019 CE498 Lecture Sept 26
28/57
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 ofeach other.
The effects of water table must be considered for thedesign loading conditions.
7/28/2019 CE498 Lecture Sept 26
29/57
DESIGN METHODS
Two approaches exist for the design of RC members
Strength design, and allowable stress design.
Strength design is the most commonly adopted procedure forconventional buildings
The use of strength design was considered inappropriate
due to the lack of reliable assessment of crack widths atservice loads.
Advances in this area of knowledge in the last two decadeshas led to the acceptance of strength design methods
The recommendations for strength design suggest inflatedload factors to control service load crack widths in therange of 0.004 0.008 in.
7/28/2019 CE498 Lecture Sept 26
30/57
Design Methods
Service state analyses of RC structures should include
computations of crack widths and their long term effectson the structure durability and functional performance.
The current approach for RC design include computationsdone 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 performanceof the structure.
7/28/2019 CE498 Lecture Sept 26
31/57
DESIGN METHODS
The load combinations to determine the required strength
(U) are given in ACI 318. ACI 350 requires twomodifications
Modification 1 the load factor for lateral liquid pressure istaken as 1.7 rather than 1.4. This may be over conservative
due to the fact that tanks are filled to the top only duringleak testing or accidental overflow
Modification 2 The members must be designed to meet therequired strength. The ACI required strength U must beincreased by multiplying with a sanitary coefficient
The increased design loads provide more conservative designwith less cracking.
Required strength = Sanitary coefficient X U
Where, sanitary coefficient = 1.3 for flexure, 1.65 for directtension, and 1.3 for shear beyond the capacity provided by the
concrete.
7/28/2019 CE498 Lecture Sept 26
32/57
WALL THICKNESS
The walls of circular tanks are subjected to ring or hoop
tension due to the internal pressure and restraint toconcrete 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 preventexcessive cracking.
The concrete tension strength will be assumed 10% fc in thisdocument.
RC walls 10 ft. or higher shall have a minimum thickness of12 in.
The concrete wall thickness will be calculated as follows:
7/28/2019 CE498 Lecture Sept 26
33/57
WALL THICKNESS
Effects of shrinkage
Figure 2(a) shows a block of concretewith a re-bar. The block height is 1 ft, tcorresponds to the wall thickness, thesteel area is As, and the steel percentageis r.
Figure 2(b) shows the behavior of theblock assuming that the re-bar is absent.The block will shorten due to shrinkage.Cis the shrinkage per unit length.
Figure 2(c) shows the behavior of theblock when the re-bar is present. The re-bar restrains some shortening.
The difference in length between Fig.2(b) and 2(c) is xC, an unknown quantity.
7/28/2019 CE498 Lecture Sept 26
34/57
WALL THICKNESS
The re-bar restrains shrinkage of the concrete. As a result,
the concrete is subjected to tension, the re-bar tocompression, 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 Solve for x from above equation for force equilibrium
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 tensionof T will be equal to:
T * Ec/(EcAc+EsAs) = T * 1/[Ac+nAs] = T/[Ac(1+nr)]
The total concrete tension stress = [CEsAs + T]/[Ac+nAs]
7/28/2019 CE498 Lecture Sept 26
35/57
WALL THICKNESS
The usual procedure in tank design is to provide horizontal
steel As for all the ring tension at an allowable stress fs asthough designing for a cracked section.
Assume As=T/fs and realize Ac=12t
Substitute in equation on previous slide to calculate tensionstress in the concrete.
Limit the max. concrete tension stress to fc= 0.1 fc
Then, the wall thickness can be calculated as
t = [CEs+fsnfc]/[12fcfs]* T
This formula can be used to estimate the wall thickness
The values of C, coefficient of shrinkage for RC is in therange 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
7/28/2019 CE498 Lecture Sept 26
36/57
WALL THICKNESS
The allowable steel stress fs should not be made too small.
Low fs will actually tend to increase the concrete stress andpotential 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 from266 psi to 322 psi.
Desirable to use a higher allowable steel stress.
7/28/2019 CE498 Lecture Sept 26
37/57
REINFORCEMENT
The amount size and spacing of
reinforcement has a great effecton the extent of cracking.
The amount must be sufficientfor strength and serviceability
including temperature andshrinkage effects
The amount of temperature andshrinkage reinforcement isdependent on the length
between construction joints
7/28/2019 CE498 Lecture Sept 26
38/57
REINFORCEMENT
The size of re-bars should be chosen recognizing that
cracking can be better controlled by using larger number ofsmall 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.
7/28/2019 CE498 Lecture Sept 26
39/57
CRACK CONTROL
Crack widths must be minimized in tank walls to preventleakage 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 tocenter of bar located closest.
A = effective tension area of concrete surrounding theflexural tension reinforcement having the same centroid asthe reinforcement, divided by the number of bars.
7/28/2019 CE498 Lecture Sept 26
40/57
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 fs
3)
Using the limiting value of z given by ACI 350, the maximumbar spacing can be computed
For ACI 350, z has a limiting value of 115 k/in.
For severe environmental exposures, z = 95 k/in.
7/28/2019 CE498 Lecture Sept 26
41/57
ANALYSIS OF VARIOUS TANKS
Wall with fixed base and free top; triangular load
Wall with hinged base and free top; triangular load andtrapezoidal 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
7/28/2019 CE498 Lecture Sept 26
42/57
CIRCULAR TANK ANALYSIS
In practice, it would be rare that a base would be fixed
against rotation and such an assumption would lead to animproperly designed wall.
For the tank structure, assume
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 = fc = 4000 psi
Yield strength of reinforcement = fy = 60,000 psi
7/28/2019 CE498 Lecture Sept 26
43/57
It is difficult to predict the behavior of the subgrade and its
effect upon restraint at the base. But, it is more reasonableto assume that the base is hinged rather than fixed, whichresults 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 thetank wall are shown in Tables A-5, A-7, and A-12 of theAppendix
Each of these tables, presents the results as functions ofH2/Dt, which is a parameter.
The values of thickness t cannot be calculated till the ringtension T is calculated.
Assume, thickness = t = 10 in.
Therefore, H2/Dt = (202)/(54 x 10/12) = 8.89 (approx. 9 in.)
CIRCULAR TANK ANALYSIS
7/28/2019 CE498 Lecture Sept 26
44/57
Table A-5 showing the ring tension values
7/28/2019 CE498 Lecture Sept 26
45/57
Table A-7, A-12 showing the moment and shear
7/28/2019 CE498 Lecture Sept 26
46/57
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 thevalues 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 thebase is free to slide. Since, it cannot do that, the value ofwu HR must be multiplied by coefficients from Table A-5
7/28/2019 CE498 Lecture Sept 26
47/57
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 thatthe base has no radial displacement
Figure compares the ring tension for tanks with free slidingbase, fixed base, and hinged base.
7/28/2019 CE498 Lecture Sept 26
48/57
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 intwo curtains.
Resulting As = 1.32in2/ft.
The reinforcement along the height of the wall can bedetermined similarly, but it is better to have the same barand 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
7/28/2019 CE498 Lecture Sept 26
49/57
CIRCULAR TANK ANALYSIS
The moments in vertical wall strips
that are considered 1 ft. wide arecomputed by multiplying wuH
3 bythe coefficients from table A-7.
The value of wu for flexure =
sanitary coefficient x (1.7 x lateralforces)
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 theheight are shown in the Table.
The figure includes the moment for
both the hinged and fix conditions
7/28/2019 CE498 Lecture Sept 26
50/57
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 lowerportion of the wall.
The required reinforcement for the outside face of the wallfor a maximum moment of 5,524 ft-lb/ft. is:
Mu/(ffc bd2) = 0.0273 (where d = t cover dbar/2)
From the standard design aid of Appendix A, take the valueof 0.0273 and obtain a value for w from the Table.
Obtain w=0.0278
Required As = wbdfc/fy = 0.167 in2
7/28/2019 CE498 Lecture Sept 26
51/57
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.
As = 0.31 in2 and r = 0.0035
The shear capacity of a 10 in. wall with fc=4000 psi is Vc= 2 (fc)
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
7/28/2019 CE498 Lecture Sept 26
52/57
RECTANGULAR TANK DESIGN
The cylindrical shape is structurally best suited for tank
construction, but rectangular tanks are frequentlypreferred for specific purposes
Rectangular tanks can be used instead of circular tanks whenthe footprint needs to be reduced
Rectangular tanks are used where partitions or tanks withmore than one cell are needed.
The behavior of rectangular tanks is different from thebehavior of circular tanks
The behavior of circular tanks is axisymmetric. That is thereason for our analysis of only unit width of the tank
The ring tension in circular tanks was uniform around thecircumference
7/28/2019 CE498 Lecture Sept 26
53/57
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 sanitarycoefficient 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 rectangularand 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 thedesign are similar to those for circular tanks.
7/28/2019 CE498 Lecture Sept 26
54/57
RECTANGULAR TANK DESIGN
The restraint condition at the base is needed to determine
deflection, shears and bending moments for loadingconditions.
Base restraint conditions considered in the publicationinclude both hinged and fixed edges.
However, in reality, neither of these two extremes actuallyexist.
It is important that the designer understand the degree ofrestraint provided by the reinforcing that extends into thefooting from the tank wall.
If the designer is unsure, both extremes should beinvestigated.
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
7/28/2019 CE498 Lecture Sept 26
55/57
RECTANGULAR TANK BEHAVIOR
x
y
y
z
Mx= moment per unit width about the x-axisstretching the fibers in the y direction when theplate is in the x-y plane. This momentdetermines the steel in the y (vertical direction).
My= moment per unit width about the y-axisstretching the fibers in the x direction when the
plate is in the x-y plane. This momentdetermines the steel in the x (horizontaldirection).
Mz= moment per unit width about the z-axisstretching the fibers in the y direction when the
plate is in the y-z plane. This moment determinesthe steel in the y (vertical direction).
RECTANGULAR TANK BEHAVIOR
7/28/2019 CE498 Lecture Sept 26
56/57
RECTANGULAR TANK BEHAVIOR
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 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
7/28/2019 CE498 Lecture Sept 26
57/57
RECTANGULAR TANK BEHAVIOR
The twisting moment effects such as Mxy may be used to
add to the effects of orthogonal moments Mx and My forthe purpose of determining the steel reinforcement
The Principal of Minimum Resistance may be used fordetermining the equivalent orthogonal moments for design
Where positive moments produce tension: Mtx = Mx + |Mxy|
Mty = My + |Mxy|
However, if the calculated Mtx < 0,
then Mtx=0 and Mty=My + |Mxy2/Mx| > 0
If the calculated Mty < 0
Then Mty = 0 and Mtx = Mx + |Mxy2/My| > 0
Similar equations for where negative moments producetension