Advanced Reinforced Concrete Analysis and Design

7/30/2019 Advanced Reinforced Concrete Analysis and Design

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Advanced Reinforced

Concrete Analysis and

DesignAccording to ACI-318 2005This book presents some example of using ACI Codes in the design of various structura

elements

2008

Eng. Mohammed Osama Yousef


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Ad d R i f d C t A l i & D i

Water Tanks:

Resting on ground

Elevated Circular

Under Ground


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Rectangular Spherical

Intz Conical Bottom


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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

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 is not required. Simple analysis using tabulated results in handbooks etc.


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What are the objectives 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 reinforcement.

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

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


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Design Load 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

Raw Sewage 10 kN/m3

Grit from grit chamber .. 17.5 kN/m3

Digested sludge aerobic. 10 kN/m3

Digested sludge anaerobic 11 kN/m3

For other numbers see ACI 350.

Live loads

Catwalks etc 5 kN/m2

Heavy equipment room 14.5 kN/m2

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 1.7The 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 Vs> 1.3

(Vu - Vc)

For compression use 1.0 U


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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

(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 supportthe full effects of each load individually

Cannot use one load to minimize the other, because sometimes the tank is empty.

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.

It contains tabulated coefficients for calculating stress distributions etc. for different

boundary conditions and can be used directly for designIt also includes some calculation and design examples

Reinforced concrete walls at least 3 m high that are in contact with liquids should have a

minimum thickness of 300 mm.

The minimum thickness of any minor member is 150 mm, and when 50 mm. cover is

required then it is at least 200 mm.

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

Maximum bar spacing should not exceed 300 mm.

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 300 mm and should be divided

equally between the two surfaces


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Figure showing minimum shrinkage reinforcement and table showing minimum cover for

reinforcement required


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In order to prevent leakage, the strain in the tension reinforcement has to be limited; the strain inthe 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

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 flexuralmembers.

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.

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, flocculates, 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

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 than1.5

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.


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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 concreteexposed to moderate sulfate attack (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

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%

Mix proportioning all material should be proportioned to produce a well-graded mix of high

density and workability

28 day compressive strength of 24 MPa where the concrete is not exposed to severe

weather and freeze-thaw

28 day compressive strength of 28 MPa 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

40 mm aggregate max 517 lb/yd3


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Air entrainment requirements

5.5 1 % for 40 mm aggregate

6.0 1 % for 1.0 or 15 mm aggregate

Slump requirements

25 mm minimum and 100 mm maximum

Concrete placement according to ACI 350

Curing using sprinkling, bonding, 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 isdifficult or uneconomical to correct or clean regularly, a coating may be required

Permissible Concrete Stresses:

Permissible concrete stresses in calculation relating to resistance the cracking.

Grade of

Concrete

Permissible Stress in tension

Direct (direct

tension)

Due to Bending

(rupture)Shear Stress

fc' = 30MPa 1.3 MPa 1.7 MPa 1.9 MPa

fc' = 35 MPa 1.4 MPa 1.8 MPa 2.0 MPa

fc' = 40 MPa 1.5 MPa 2.0 MPa 2.2 MPa

fc' = 45 MPa 1.6 MPa 2.2 MPa 2.5 MPa

fc' = 50 MPa 1.7 MPa 2.4 MPa 2.7 MPa


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Permissible Stresses in Steel:Tensile stress in member in direct tension, Serviceability Limit State:

fs = 100 MPa ; this is for the crack limit of 0.1 mm.

fs = 130 MPa ; this for the crack limit of 0.2 mm.

1) Circular tanks resting in ground:When the joints at base are flexible, hydrostatic pressure induces maximum increase in diameter

at base and no increase in diameter at top. This is due to fact that hydrostatic pressure varies

linearly from zero at top and maximum at base. Deflected shape of the tank is shown in Figure

When the joint at base is rigid, the base does not move. The vertical wall deflects as shown in

figure

Tank with flexible base

Tank with rigid base


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Conventionally reinforced circular concrete tanks have been used extensively. They will be the

focus of our lecture today

Structural design must focus on both the strength and serviceability. The tank must withstandapplied 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

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.


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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 for conventional buildings

The use of strength design was considered inappropriate due to the lack of reliable assessment of

crack widths at service loads.

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 0.1 02 mm.

Service state analyses of RC structures should include computations of crack widths and theirlong term effects on the structure durability and functional performance.

The current approach for RC design includes 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.

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 forshear 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.


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The tensile stress in the concrete (due to ring tension from pressure and shrinkage) has to

keep at a minimum to prevent excessive cracking.

The concrete tension strength will be assumed 10% fc in this document.

RC walls 33 m or higher shall have a minimum thickness of 300 mm.

The concrete wall thickness will be calculated as follows:

Effects of shrinkage

Figure 2(a) shows a block ofconcrete with a re-bar. The blockheight is 300 mm 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 shortendue to shrinkage. C is the

shrinkage per unit length.

Figure 2(c) shows the behavior ofthe block when the re-bar is

present. The re-bar restrains some

shortening.

The difference in length betweenFig.

2(b) and 2(c) is xC, an unknown

quantity.

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

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 tension of T will be equal to:

T * Ec/(EcAc+EsAs) = T * 1/[Ac+nAs] = T/[Ac(1+nr)]


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The total concrete tension stress = [CEsAs + T]/[Ac+nAs]

The usual procedure in tank design is to provide horizontal steel A s 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 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 =125 MPa

Therefore, wall thickness t=0.0003 T

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=105 kN

n=8

Es=200000 MPa

C=0.0003 and

Ac=1000 x 250=250000 mm2

If the allowable steel stress is reduced from 140 MPa to 70 MPa, the resulting concretestress is increased from 1.8 MPa to 2.2 MPa.

Desirable to use a higher allowable steel stress.


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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

The size of re-bars should be chosen recognizing that cracking can be better controlled by usinglarger number of small diameter bars rather than fewer large diameter bars

The size of reinforcing bars should not exceed 32 mm bar. Spacing of re-bars should be limited toa maximum of 300 mm. Concrete cover should be at least 50 mm.

In circular tanks the locations of horizontal splices should be staggered by not less than one laplength or 1 m.

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 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.

In ACI 350, the cover is taken equal to 50 mm for any cover greater than 50 mm

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 20562 kN/m

For severe environmental exposures z = 16986 kN/m

Analysis of Various Tanks: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


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Circular Tanks 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

Height = H = 6 m

Diameter of inside = D = 16.5

Weight of liquid = w = 10 kN/m3

Shrinkage coefficient = C = 0.0003

Elasticity of steel = Es = 200000 MPa

Ratio of Es/Ec = n = 8

Concrete compressive strength = fc = 28 MPa

Yield strength of reinforcement = fy = 420 MPa

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 = 250 mm

Therefore, H2/Dt = (62)/(16.5 x 0.25) = 8.89 (approx. 9)


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In these tables, 0.0 H corresponds to the top of the tank, and 1.0 H corresponds to the bottom of

the tank.

The ring tension per foot of height is computed by multiplying wu HR by the coefficients in Table

A-5 for the values of H2/Dt=9.0

wu for the case of ring tension is computed as:

wu = sanitary coefficient x (1.7 x Lateral Forces)

wu = 1.65 x (1.7 x 10) = 28 kN/m3

Therefore, wu HR = 28 x 6 x 16.5/2 = 1386 kN/m3

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

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.


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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 = (1386 x 0.713 x 1000) / (0.9 * 420) = 2615 mm2/m

Therefore at 0.7H use 20mm bars spaced at 200 mm on center in two curtains.

Resulting As = 3140 mm2/m

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] = 1.96 MPa < 2.8 MPa

Therefore, adequate

The moments in vertical wall strips that

are considered 1 m 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 10 = 22.1

kN/m3

Therefore wuH3

= 22.1 x 63

= 4988.7

kN-m/m

Mu = 0.005 x 4988.7 = 24.9 kN.m

The figure includes the moment forboth the hinged and fixes conditions


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The actual restraint is somewhere in between fixed and hinged, but probably closer to hinge.

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 25 kN-m/m is:

Mu/( bd 2) = 0.77

where d = t cover dbar/2 = 250 50 20/2 = 190 mm

1 21 (1 )

mR

m fy

0.00187

Required As = bd = 355 mm2

min = 1.4/Fy = 0.0033 > 0.00189

Use 16 mm bars at the maximum allowable spacing of 250 mm

The shear capacity of a 250 mm wall with fc=28 MPa is

Vc = 1/6 (fc)0.5 bwd = 220 kN

Therefore, Vc = 0.75 x 220 = 165 kN

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 will be

needed)

wuH2 = 1.0 x 1.7 x 10 x 62 = 612 kN

Applied shear = Vu = 0.092 x wuH2 = 57 kN < Vc


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T

T

Design of Circular Tanks resting on ground with flexible base:

Maximum hoop tension in the wall is developed at the base. This tensile force T is computed by

considering the tank as thin cylinder

2

DHT ; Quantity of reinforcement required in form

of hoop steel is computed asstst

st

2/HDTA or

0.3 % (minimum)

In order to provide tensile stress in concrete to be less to

be less than permissible stress, the stress in concrete is

computed using equation

ststc

cA)1m(t1000

2/HD

A)1m(A

TIf c cat

where: cat= from table

then the section is from cracking, otherwise the thickness has to be increased so that c is less than cat.

While designing, the thickness of concrete wall can be estimated as t = 30xH + 50 mm, where H is in

meters. Distribution steel in the form of vertical bars are provided such that minimum steel area

requirement is satisfied. As base slab is resting on ground and no bending stresses are induced hence

minimum steel distributed at bottom and the top are provided

Example:

Design a circular water tank with flexible connection at base for a capacity of 4,00,000 liters. The tank

rests on a firm level ground. The height of tank including a free board of 200 mm should not exceed

3.5m. The tank is open at top. Use fc= 30 MPa concrete and fy= 420 MPa Steel. Draw to a suitable scale:

i) Plan at baseii) Cross section through centre of tank.

Dimension of tank:

Depth of water H = 3.5 -0.2 = 3.3 m

Volume V = 4,00,000/1000 = 400 m3

Area of tank A = 400/3.3 = 121.2 m2

Diameter of tank m42.12A4

D 13 m

The thickness is assumed as t = 30H+50 = 149 160 mm


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Design of Vertical wall:

Max hoop tension at bottom kN5.2142

133.310

2

D

HT

Area of steel 23

mm1650130

105.214

stst

st

TTA

Minimum steel to be provided

Ast min=0.3% of area of concrete = 0.003x1000x160 = 480 mm2

The steel required is more than the minimum required

Spacing of 16 mm diameter bar = 100/(1650/201) = 123 mm c/c

Provide #16 @ 100 c/c as hoop tension steel

Check for tensile stress:

Area of steel provided Ast provided= 201x1000/100 = 2010 mm2

Modular ratio m= 33.1373

280

3

280

cbc

Stress in concrete 23

N/mm16.12010)133.13(1601000

105.214

)1(1000 stc

Amt

T

Permissible stress cat = 1.3 MPa

Actual stress is equal to permissible stress, hence safe.

Curtailment of hoop steel:

Quantity of steel required at 1m, 2m, and at top is tabulated. In this table the maximum spacing is

taken an 3 x 160 = 480 mm

Height from top Hoop tension

T = HD/2 (kN)

Ast= T/ st Spacing of #16

mm c/c

2.3 m 149.5 996 200

1.3 m 84.5 563.33 350

Top 0 Min steel (480 mm2) 400


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Vertical reinforcement:

For temperature and shrinkage distribution steel in the vertical reinforcement is provided @ 0.3 %

Spacing of 10 mm diameter bar = 79x1000/480 =164.5 mm c/c 150 mm c/c

Tank floor:

As the slab rests on firm ground, minimum steel @ 0.3 % is provided. Thickness of slab is

assumed as 150 mm 8 mm diameter bars 200 c/c is provided both directions at top and botom


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2) Rectangular Tanks:

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 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

Rectangle 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 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 isrelated to crack width control, which is achieved using the Gergely Lutz parameter z.

The requirements for reinforcement 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.

The restraint condition at the base is needed to determine deflection, shears and bending moments

for loading conditions.

Base restraint considered in the publication includes both hinged and fixed edges.

However, in reality, neither of these two extremes actually exists.

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

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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Ad d R i f d C t A l

Rectangle Tank Behavior:

Mx = moment per unit width

plate is in the x-y plane. This

My = moment per unit width

plate is in the x-y plane. This

Mz = moment per unit width

plate is in the y-z plane. This

Mxy orMyz = torsion or twisting

All these moments can be co

Mx=(Mx Coeff.) x q a2

My=(My Coeff.) x q a2

Mz=(Mz Coeff.) x q a2/ Mxy=(Mxy Coeff.) x q

Myz=(Myz Coeff.) x q a

These coefficients are presente

The shear in one wall beco

explain in class.

i & D i

about the x-axis stretching the fibers in the y dir

oment determines the steel in the y (vertical direct

about the y-axis stretching the fibers in the x dir

oment determines the steel in the x (horizontal dir

about the z-axis stretching the fibers in the y dir

oment determines the steel in the y (vertical direct

moments for plate or wall in the x-y and y-z plane

puted using the equations

1000

1000

1000

2/1000

2/1000

d in Tables 2 and 3 for rectangular tanks

es axial tension in the adjacent wall. Follow for

30

ection when the

ion).

ection when the

ection).

ection when the

ion).

, respectively.

e equilibrium -


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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 orthogonalmoments 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 produce tension

For rectangular tanks in whichL/B tanks walls are designed as continuous frame subjected to

pressure varying from zero at top to max at H/4 or 1m.

Mcantilever=

give us vertical reinforcements inside and for outside use min. reinforcementHorizontal Reinforcements:

P = H-h) take 1m strip

Now we have frame 1m with force from the inside.

Use moment distribution where stiffness = , 1/L and 1/B

For more detailed calculation see next material from PCA.

Rectangular tank with fixed base resting on ground:

Rectangular tanks are used when the storage capacity is small and circular tanks prove uneconomical for

small capacity. Rectangular tanks should be preferably square in plan from point of view of economy. It

is also desirable that longer side should not be greater than twice the smaller side.

Moments are caused in two directions of the wall ie., both in horizontal as well as in vertical direction.

Exact analysis is difficult and such tanks are designed by approximate methods. When the length of the

wall is more in comparison to its height, the moments will be mainly in the vertical direction, ie., the

panel bends as vertical cantilever. When the height is large in comparison to its length, the moments will

be in the horizontal direction and panel bends as a thin slab supported on edges. For intermediate

condition bending takes place both in horizontal and vertical direction.


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32


In addition to the moments, the walls are also subjected to direct pull exerted by water pressure on some

portion of walls. The walls are designed both for direct tension and bending moment.

`

Maximum vertical moment = Mx wa3 ( for x/a = 1, y = 0)

Maximum horizontal moment = My wa3 (for x/a = 0, y = b/2)

Tension in short wall is computed as Ts = pL/2

Tension in long wall TL = pB/2

Horizontal steel is provided for net bending moment and direct tensile force

Ast=Ast1+Ast2;jd

'MA

st

1st

Ast2=T/ st

M=Maximum horizontal bending moment T x

x= d-D/2

a-

+

TTCB

PLAN @ BASE

DA

p= H

TT

T

FBD OF AB

FBD OF AD

y

L

B

x

D/2d T

Bending moment diagram

0.5b 0.5b

x


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33


Example:

Design a rectangular water tank 5m x 4m with depth of storage 3m, resting on ground and whose walls

are rigidly joined at vertical and horizontal edges. Assume fc=30 MPa concrete and fy= 420 MPa grade

steel. Sketch the details of reinforcement in the tank

Analysis for moment and tensile force:

Long wall:

L/a = 1.67 1.75

at y = 0 and x/a =1 Mx = -0.074

at y = b/2 and x/a =1/4 My= -0.052

Max vertical moment = Mx wa3 = -19.98

Max horizontal moment = My wa3 = -14.04

Tlong = w ab/2 = 60 kN

Short wall:

B/a = 1.33 1.5

at y = 0 and x/a = 1 Mx= -0.06

at y = b/2 and x/a = 1/4 My= -0.044

Max vertical moment = Mx wa3 = -16.2

Max horizontal moment = My wa3 = -11.88

Tshort = w aL/2 = 75 kN

E

B

A

F

D

C

Free

a=H=3m

b=4m

L=5m

Fixed


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Design constants:

cbc= 7 MPa st = 130 MPa m = 13.33

41.0stcbc

cbc

m

mk

j =1- (k/3) = 0.87

Q = cbcj k = 1.15

Design for vertical moment:

For vertical moment, the maximum bending moment from long and short wall

(Mmax)x = -19.98 kN-m

mm8.1311000x15.1

10x98.19

Qb

Md

6

Assuming effective cover as 33mm, the thickness of wall is

t = 131.88+33 = 164.8 mm 170 mm

dprovided= 170-33 = 137mm

26

5.128913787.0130

1098.19mm

xx

x

jd

MA

st

st

Spacing of 14 mm diameter bar = cmmcx

/1195.1289

1000154(Max spacing 3d=411mm)

Provide #14 @ 100 mm c/c

Distribution steel:

Minimum area of steel is 0.3% of concrete area

Ast=(0.3/100) x1000 x 170 = 510 mm2


/100510

100024.50

Provide #8 @ 100 c/c as distribution steel.

Provide #8 @ 100 c/c as vertical and horizontal distribution on the outer face.


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35


CA

11.88

14.4

B

Design for Horizontal moment:

Horizontal moments at the corner in long and short wall produce unbalanced moment at the joint. This

unbalanced moment has to be distributed to get balanced moment using moment distribution method.

56.020/9

4/1

44.020/9

5/1

20

9K;

5

1;

5

1

AB

AC

ACAC

DF

DF

KK

Moment distribution Table

Joint A

Member AC AB

DF 0.44 0.56

FEM -14 11.88

Distribution 0.9328 1.1872

Final Moment -13.0672 13.0672

The tension in the wall is computed by considering the section at height H 1 from the base. Where, H1 is

greater of

i) H/4ii) 1mEx:

i) 3/4=0.75ii) 1m

H1= 1m

Depth of water h = H-H1 = 3-1-2m; p = wh = 10 x 2 = 20 kN/m2

Tension in short wall Ts = pL/2 = 50 kN

Tension in long wall TL = pB/2 = 40 kN


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Net bending moment M = M-Tx, where: x= d-D/2 = 137-(170/2) = 52mm

M =13.0672-50 x 0.052 = 10.4672 kN-m

26

1 5.67513787.0130

104672.10mm

xx

xAst

23

2 385130

1050mm

xAst

Ast= Ast1 + Ast2 = 1061 mm2


/5.1061061

1000113(Max spacing 3d=411mm)

Provide #12@100 mm c/c at corners

Base Slab:

The slab is resting on firm ground. Hence nominal thickness and reinforcement is provided. The thickness

of slab is assumed to be 200 mm and 0.3% reinforcement is provided in the form of #8 @ 150 c/c. at top

and bottom

A haunch of 150 x 150 x 150 mm size is provided at all corners

Note: More tables and examples included at appendix A from PCA notes for rectangle water tanks.


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Rectangular Concrete Tanks

While cylindrical shapes may be structurally best fortank construction, rectangular tanks frequently are pre-ferred for specific purposes. Special processes or oper-ations may make circular tanks inconvenient to use.When several separate cells are required, rectangulartanks can be arranged in less space than circular tanksof the same capacity. Tanks or vats needed inside a

building are therefore often made in rectangular orsquare shapes. For these and other reasons, breweries,tanneries, and paper mills generally use rectangulartanks.

Data presented here are for design of rectangulartanks where the walls are subject to hydrostatic pres-sure of zero at the top and maximum at the bottom.Some of the data can be used for design of counter-forted retaining walls subject to earth pressure for whicha hydrostatic type of loading may be substituted in thedesign calculations. Data also can be applied to designof circular reservoirs of large diameter where lateralstability depends on the action of counterforts built inte-grally with the wall.

Another article on tank construction, Circular Con-crete Tanks Withouf Prestressing, has been publishedby the Portland Cement Association.

Moment Coefficients

Moment coefficients were calculated for individualpanels considered fixed along vertical edges, and coef-ficients were subsequently adjusted to allow for a cer-tain rotation about the vertical edges. First, three sets ofedge conditions were investigated, in all of which verti-cal edges were assumed fixed while the other edgeswere as follows:

1. Top hinged-bottom hinged2. Top free-bottom hinged3. Top free-bottom fixed*

Moment coefficients for these edge conditions aregiven in Tables 1, 2, and 3, respectively. In all tables, adenotes height and b width of the wall. In Tables 1, 2,and 3, coefficients are given for nine ratios of b/a, thelimits being b/a = 3.0 and 0.5. The origin of the coordi-nate system is at midpoint of the top edge; the Y axis ishorizontal; the X axis is vertical and its positive directiondownward. The sign convention for bending moments isbased on the coordinate fiber that is being stressed. Forexample, A$ stresses fibers parallel to the X axis, Thesign convention used here is not compatible with two

other conventions-namely, that (1) the subscript is theaxis of the moment, and (2) that the moment is in a par-

titular principal plane. Coefficients are given-exceptwhere they are known to be zero-at edges, quarterpoints, and midpoints both in X and Y directions.

The slab was assumed to act as a thin plate, for whichequations are available in textbooks such as Theory otPlates and Shells by S. Timoshenko, but since only asmall portion of the necessary calculations for moment

coefficients for specific cases is available in the engi-neering literature, they have been made especially forthis text.

Table 4 contains moment coefficients for uniformload on a rectangular plate considered hinged on allfour sides. The table is for designing cover slabs andbottom slabs for rectangular tanks with one cell. If thecover slab is made continuous over intermediate sup-ports, the design can follow procedures for the design ofslabs supported on four sides.

Coefficients for individual panels with fixed sideedges apply without modification to continuous wallsprovided there is no rotation about vertical edges. In asquare tank, therefore, moment coefficients can be

taken directly from Tables 1, 2, or 3. In a rectangulartank, however, an adjustment must be made, as wasdone in Tables 5 and 6, similar to the modification offixed-end moments in a frame analyzed by momentdistribution.

In this procedure the common-side edge of two ad-jacent panels is first considered artificially restrained sothat no rotation can take place about the edge. Fixed-edge moments taken from Tables 1,2, or 3 are usuallydissimilar in adjacent panels and the differences, whichcorrespond to unbalanced moments, tend to rotate theedge. When the artificial restraint is removed these un-balanced moments will induce additional moments inthe panels, Adding induced and fixed-end moments at

the edge gives final end moments, which must be iden-tical on both sides of the common edge.Moment distribution cannot be applied as simply to

continuous tank walls as it can to framed structures,because moments must be distributed simultaneouslyalong the entire length of the side edge so that momentsbecome equal at both sides at any point of the edge. Theproblem was simplified and approximated to some ex-tent by distributing moments at four points only: quarterpoints, midpoint, and top. The end moments in the twointersecting slabs were made identical at these fourpoints and moments at interior points adjusted accord-ingly.

Applicable tn cases where wa l l s lab , counter fo r t , and base s lab area l l built I n tegra l lyPublIshed by McGraw-HI11 Book Co, New York, 1959


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Procedure for using these coefficients to determinemoments and design of the wall is similar to that illus-trated for the open-top single-cell tank shown in Fig. 3.

Details at Bottom Edge

Note that all tables except one are based on the as-sumption that the bottom edge is hinged. It is believedthat this assumption in general is closer to the actualcondition than that of a fixed edge. Consider first thedetail in Fig. 9, which shows the wall supported on arelatively narrow continuous wall footing, and then Fig.10 in which the wall rests on a bottom slab.

Fig. 9.

In Fig. 9 the condition of restraint at the bottom of thefooting is somewhere between hinged and fixed butmuch closer to hinged than to fixed. Resultant of pres-sure on the subsoil lies well within the edge of the foot-ing, and the product of resultant and its eccentricity isusually much smaller than the moment at the bottom ofthe wall when it is assumed fixed. Furthermore, thefoot-ing must rotate about a horizontal axis in order to pro-duce eccentric loading on the subsoil and rotation itselfrepresents a relaxation of restraint.

When the wall footing is not capable of furnishingmuch restraint, it is not necessary to provide for hingeaction at the construction joint in Fig. 9. The dowels areclose to the surface, leaving the center of the joint freefor insertion of a shear key. Area of steel in the dowelsalong each face may be taken as not less thanO.O025bd,and extension of the dowels above the construction

joint may be made not less than say 3 ft.

The base slab in Fig. 9 is placed on top of the wall foot-ing and the bearing surface is brushed with a heavy coatof asphalt to break the adhesion and reduce friction

between slab and footing. The vertical joint betweenslab and wall should be made watertight. A joint width of1 in. at the bottom and 1% in. at the top is consideredadequate. As indicated in Fig. 9, the bottom of the jointmay be filled with oakum, the middle with volcanic clayof a type that expands greatly when moistened, and theupper part sealed with mastic. Any leakage will makethe clay penetrate into fissures and expand, pluggingthe leak. Mortar mixed with iron powder has been usedextensively for joints such as in Fig. 9, and so has lead

joint filler, but both iron powder and lead are not alwaysreadily available. A waterstop may not be needed in theconstruction joints when the vertical joint in Fig. 9 ismade watertight.

In Fig. 10 a continuous concrete base slab is providedeither for transmitting the load coming down through thewall or for upward hydrostatic pressure. In either case,the slab deflects upward in the middle and tends to ro-tate the wall base in Fig. 10 in a counterclockwrse direc-tion. The wall therefore is not fixed at the bottom edge.It is difficult to predict the degree of restraint. The rota-

tion may be great enough to make the bottom edgehinged or may be even greater. Under the circum-stances it is advisable to avoid placing moment rein-forcement across the joint and to cross the dowels atthe center. The waterstop must then be placed off cen-ter as indicated. Provision for transmitting shear throughdirect bearing can be made by inserting a key as in Fig. 9or by a shear ledge as in Fig. 10.

The waterstop in Fig. 10 may be galvanized steel,copper, preformed rubber, or extruded plastic.

At top of wall the detail in Fig. 10 may be applied ex-cept that the waterstop and the shear key are not essen-tial. The main thing is to prevent moments from beingtransmitted from the top of the slab into the wall because

the wall is not designed for such moments.

Fig. 10.

14


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Metric Conversion Factors

To convert from To Multiply by

inch (in.) meter (m) 0.0254

feet (ft) meter (m) 0.3048

square feet (sq ft) square meter (m2) 0.0929pound (lb) kilogram (kg) 0.4536

kip (1000 lb) kilogram (kg) 453.6

Ib/lin ft kg/m 1.488kip/lin ft kg/m 1488.Ib/sq ft kg/m2 4.88Ib/cu ft kg/m3 16.02ft-kips newton-meter (Nm) 1356.

ft-kips kilogram-meter (kgm) 138.2

The prefixes and symbols listed are commonly usedto form names and symbols of the decimal multiplesand submultiples of the SI units.

Multiplication Factor Pr efix Symbol

1 000 000 000 = 109 giga G1 oooooo= 106 mega M1000=10~ kilo kl= l -

0.001 = 10-3 milli m0.000 001 = 10-6 micro I-10.000 000 001 = 1 o-9 nano n

Documents

Advanced Reinforced Concrete Analysis and Design