CE498 Lecture Nov 16

7/27/2019 CE498 Lecture Nov 16

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ENVIRONMENTAL ENGINEERING CONCRETESTRUCTURES

CE 498 Design Project

November 16, 21, 2006


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OUTLINE

INTRODUCTION

LOADING CONDITIONS

DESIGN METHOD

WALL THICKNESS

REINFORCEMENT

CRACK CONTROL


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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.


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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


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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.


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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.


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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.


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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.


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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:


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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.


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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]


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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


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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.


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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


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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.


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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)

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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 theflexural tension reinforcement having the same centroid asthe reinforcement, divided by the number of bars.


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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.


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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


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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


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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 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.)



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Table A-5 showing the ring tension values


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Table A-7, A-12 showing the moment and shear


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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 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


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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.


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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


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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


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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


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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.

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


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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


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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.


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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


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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).


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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

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.


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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 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

CE498 Lecture Nov 16

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