Post on 14-Apr-2018
transcript
7/27/2019 CE498 Lecture Nov 16
1/34
ENVIRONMENTAL ENGINEERING CONCRETESTRUCTURES
CE 498 Design Project
November 16, 21, 2006
7/27/2019 CE498 Lecture Nov 16
2/34
OUTLINE
INTRODUCTION
LOADING CONDITIONS
DESIGN METHOD
WALL THICKNESS
REINFORCEMENT
CRACK CONTROL
7/27/2019 CE498 Lecture Nov 16
3/34
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.
7/27/2019 CE498 Lecture Nov 16
4/34
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
7/27/2019 CE498 Lecture Nov 16
5/34
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 onthe 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.
7/27/2019 CE498 Lecture Nov 16
6/34
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.
7/27/2019 CE498 Lecture Nov 16
7/34
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 performance
of the structure.
7/27/2019 CE498 Lecture Nov 16
8/34
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 theconcrete.
7/27/2019 CE498 Lecture Nov 16
9/34
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:
7/27/2019 CE498 Lecture Nov 16
10/34
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.
7/27/2019 CE498 Lecture Nov 16
11/34
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]
7/27/2019 CE498 Lecture Nov 16
12/34
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 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
7/27/2019 CE498 Lecture Nov 16
13/34
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.
7/27/2019 CE498 Lecture Nov 16
14/34
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
7/27/2019 CE498 Lecture Nov 16
15/34
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.
7/27/2019 CE498 Lecture Nov 16
16/34
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 to
center 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.
7/27/2019 CE498 Lecture Nov 16
17/34
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 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.
7/27/2019 CE498 Lecture Nov 16
18/34
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
7/27/2019 CE498 Lecture Nov 16
19/34
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
7/27/2019 CE498 Lecture Nov 16
20/34
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
7/27/2019 CE498 Lecture Nov 16
21/34
Table A-5 showing the ring tension values
7/27/2019 CE498 Lecture Nov 16
22/34
Table A-7, A-12 showing the moment and shear
7/27/2019 CE498 Lecture Nov 16
23/34
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
7/27/2019 CE498 Lecture Nov 16
24/34
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.
7/27/2019 CE498 Lecture Nov 16
25/34
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 serviceloads including the effects of shrinkage is
fc = [CEsAs + Tmax, unfactored]/[Ac+nAs] = 272 psi < 400 psi
Therefore, adequate
7/27/2019 CE498 Lecture Nov 16
26/34
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
7/27/2019 CE498 Lecture Nov 16
27/34
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/(ff
cbd2) = 0.0273 (where d = t cover d
bar/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
7/27/2019 CE498 Lecture Nov 16
28/34
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
7/27/2019 CE498 Lecture Nov 16
29/34
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
7/27/2019 CE498 Lecture Nov 16
30/34
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.
7/27/2019 CE498 Lecture Nov 16
31/34
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
7/27/2019 CE498 Lecture Nov 16
32/34
RECTANGULAR TANK BEHAVIOR
x
y
y
z
Mx= moment per unit width about the x-axis
stretching 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).
7/27/2019 CE498 Lecture Nov 16
33/34
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=(M
yCoeff.) 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.
7/27/2019 CE498 Lecture Nov 16
34/34
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
+ |Mxy
2/Mx
| > 0
If the calculated Mty < 0
Then Mty = 0 and Mtx = Mx + |Mxy2/My| > 0
Similar equations for where negative moments producetension