Advanced Reinforced Concrete Analysis and DesignDesign According to ACI-318 2005 This book presents...

Advanced Reinforced Concrete Analysis and Design According to ACI-318 2005 This book presents some example of using ACI Codes in the design of various structural elements

2008

Eng. Mohammed Osama Yousef 8/14/2008

Chapter One 2

Advanced Reinforced Concrete Analysis & Design Eng. Mohammed Osama yousef

Water Tanks:

Resting on ground

Elevated

Circular

Under Ground

WATER TANK

BASED ON PLACEMENT OF

TANK

BASED ON SHAPE OF TANK

1. RESTING ON GROUND

2. UNDER GROUND

3. ELEVATED

1. CIRCULAR

2. RECTANGULAR

3. SPHERICAL

4. INTZ

5. CONICAL BOTTOM

Chapter One 3


Rectangular Spherical

Intz Conical Bottom

Chapter One 4


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 hand books etc.

Chapter One 5


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

Chapter One 6


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 1.7

The factored load combination U as prescribed in ACI 318 must be increased by durability coefficients developed from crack width calculation methods:

· In calculations for reinforcement in flexure, the required strength should be 1.3 U

· In calculations for reinforcement in direct tension, including hoop tension, the required strength should be 1.65 U

· The required design strength for reinforcement in shear should be calculated as fVs> 1.3 (Vu - fVc)

· For compression use 1.0 U

Chapter One 7


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 support the 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 design

· It 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

Chapter One 8


Figure showing minimum shrinkage reinforcement and table showing minimum cover for reinforcement required

Chapter One 9


In order to prevent leakage, the strain in the tension reinforcement has to be limited; the strain in the reinforcing bars is transferred to the surrounding concrete, which cracks, hence, minimizing the stress and strain in the reinforcing bar will minimize cracking in the concrete.

Additionally, distributing the tension reinforcement will engage a greater area of the concrete in carrying the strain, which will reduce cracking even more.

The strength design requires the use of loads, load combinations and durability coefficients presented earlier

Serviceability for normal exposures

For flexural reinforcement located in one layer, the quantity Z (crack control factor of ACI) should not exceed 115 kips/in.

The designer can use the basic Gergley-Lutz equation for crack width for one way flexural members.

The reinforcement for two-way flexural member may be proportioned in each direction using the above recommendation too.

Alternate design by the working stress method with allowable stress values given and tabulated in ACI 350. Do not recommend this method.

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 than 1.5

Methods for computing the structure frequency are presented in ACI 350 (please review if needed)

Torque is produced in most clarifiers where the entire mechanism is supported on a central column, this column must be designed to resist the torque shear without undergoing failure.

Chapter One 10


Material Design:

The cement should conform to:

· Portland cement ASTM C150, Types I, IA, II, IIA…

· Blended hydraulic cement ASTM C595

· Expansive hydraulic cement ASTM C845

· They cannot be used interchangeably in the same structure

Sulfate-resistant cement must have C3A content not exceeding 8%. This is required for concrete exposed to moderate sulfate 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

Chapter One 11




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 is difficult 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

Chapter One 12


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

Chapter One 13


Conventionally reinforced circular concrete tanks have been used extensively. They will be the focus of our lecture today

Structural design must focus on both the strength and serviceability. The tank must withstand applied loads without cracks that would permit leakage.

This is achieved by:

· Providing proper reinforcement and distribution

· Proper spacing and detailing of construction joints

· Use of quality concrete placed using proper construction procedures

A thorough review of the latest report by ACI 350 is important for understanding the design of tanks.

Loading Conditions:

The tank must be designed to withstand the loads that it will be subjected to during many years of use. Additionally, the loads during construction must also be considered.

Loading conditions for partially buried tank.

· The tank must be designed and detailed to withstand the forces from each of these loading conditions

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.

Chapter One 14


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 their long 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 for shear beyond the capacity provided by the concrete.

Wall Thickness:

The walls of circular tanks are subjected to ring or hoop tension due to the internal pressure and restraint to concrete shrinkage.

· Any significant cracking in the tank is unacceptable.

Chapter One 15


· 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% f’c 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 of concrete with a re-bar. The block height 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 shorten due to shrinkage. C is the shrinkage per unit length.

· Figure 2(c) shows the behavior of the block when the re-bar is present. The re-bar restrains some shortening.

· The difference in length between Fig. 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)]

Chapter One 16


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

The usual procedure in tank design is to provide horizontal steel As for all the ring tension at an allowable stress fs as though designing for a cracked section.

Assume As=T/fs and realize Ac=12t

Substitute in equation on previous slide to calculate tension stress in the concrete.

Limit the max Concrete tension stress to fc = 0.1 f’c

Then, the wall thickness can be calculated as

t = [CEs+fs–nfc]/[12fcfs]* T

This formula can be used to estimate the wall thickness

· The values of C, coefficient of shrinkage for RC is in the range of 0.0002 to 0.0004.

· Use the value of C=0.0003

· Assume fs= allowable steel tension =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 concrete stress is increased from 1.8 MPa to 2.2 MPa.

· Desirable to use a higher allowable steel stress.

Chapter One 17


Reinforcement:

Ø 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

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

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

Ø In circular tanks the locations of horizontal splices should be staggered by not less than one lap length 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.

Chapter One 18


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

Chapter One 19


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 = f’c = 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)

Advanced Reinforced Concrete Analysis

Chapter One


Chapter One 20

Eng. Mohammed Osama yousef


Chapter One


Chapter One 21


Chapter One 22


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.

Chapter One 23


Chapter One 24


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 Ø 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 wuH

3 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 for both the hinged and fixes conditions

Chapter One 25


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/(f bd2) = 0.77

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

1 21 (1 )

mRm fy

ræ ö

= ´ - -ç ÷ç ÷è ø

r = 0.00187

Required As = r bd = 355 mm2

rmin = 1.4/Fy = 0.0033 > 0.00189

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

The shear capacity of a 250 mm wall with f’c=28 MPa is

Vc = 1/6 (f’c)0.5 bwd = 220 kN

Therefore, f 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 < fVc

Chapter One 26


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

2D

HT g= ; Quantity of reinforcement required in form

of hoop steel is computed as stst

st

2/HDTA

sg

=s

= 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

ststcc A)1m(t1000

2/HDA)1m(A

T-+

g=

-+=s If sc £ scat

where: scat= from table

then the section is from cracking, otherwise the thickness has to be increased so that sc is less than scat. 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 base ii) 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 =p

= »13 m

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

Chapter One 27


Design of Vertical wall:

Max hoop tension at bottom kN5.2142

133.3102D

HT =´´

=g=

Area of steel 23

mm 1650130

105.214=

´===

ststst

TTA

ss

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

2803280

cbc

=´

=s

Stress in concrete 23

N/mm 16.12010)133.13(1601000

105.214)1(1000

=-+´

´=

-+=

stc Amt

Ts

Permissible stress scat = 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 = gHD/2 (kN)

Ast= T/sst 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

Chapter One 28


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

Chapter One 29


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 is related 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


Rectangle Tank Behavior:

Mx = moment per unit width about the xplate is in the x-y plane. This moment determines the steel in the y (vertical direction).

My = moment per unit width about the yplate is in the x-y plane. This moment determines the steel in the x (horizontal direction).

Mz = moment per unit width about the zplate is in the y-z plane. This moment determi

Mxy or Myz = torsion or twisting moments for plate or wall in the x

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 a

· Myz=(Myz Coeff.) x q a

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.

Chapter One


= moment per unit width about the x-axis stretching the fibers in the y direction when the y plane. This moment determines the steel in the y (vertical direction).

= moment per unit width about the y-axis stretching the fibers in the x direction when the y plane. This moment determines the steel in the x (horizontal direction).

= moment per unit width about the z-axis stretching the fibers in the y direction when the z plane. This moment determines the steel in the y (vertical direction).

= 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

/1000

/1000

/1000

Coeff.) x q a2/1000

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

Chapter One 30


axis stretching the fibers in the y direction when the y plane. This moment determines the steel in the y (vertical direction).

x direction when the y plane. This moment determines the steel in the x (horizontal direction).

axis stretching the fibers in the y direction when the nes the steel in the y (vertical direction).

z planes, respectively.

The shear in one wall becomes axial tension in the adjacent wall. Follow force equilibrium -

Chapter One 31


The twisting moment effects such as Mxy may be used to add to the effects of orthogonal moments Mx and My for the purpose of determining the steel reinforcement

The Principal of Minimum Resistance may be used for determining the equivalent orthogonal moments for design

· 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 which L/B ≤ 2; tanks walls are designed as continuous frame subjected to pressure varying from zero at top to max at H/4 or 1m.

Mcantilever = b挠黄寡 ℎ 萍脑

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

· P = γ (H-h) take 1m strip · Now we have frame 1m with force from the inside. · Use moment distribution where stiffness = 恼琵疲痞 , 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.

Chapter One 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 = Mxgwa3 ( for x/a = 1, y = 0)

Maximum horizontal moment = Mygwa3 (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'M

Ast

1st s=

Ast2=T/sst

M’=Maximum horizontal bending moment – T x

x= d-D/2

a -

+

T T C B

PLAN @ BASE

D A

p=gH

T T

T

FBD OF AB

IN PLAN

FBD OF AD

IN PLAN

y

L

B

x

D/2 d T

Bending moment diagram

0.5b 0.5b

x

Chapter One 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 gwa3 = -19.98

Max horizontal moment = My gwa3 = -14.04

Tlong = gw 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 gwa3 = -16.2

Max horizontal moment = My gwa3 = -11.88

Tshort = gw aL/2 = 75 kN

E

B

A

F

D

C

Free

a=H=3m

b=4m

L=5m

Fixed

Chapter One 34


Design constants:

scbc= 7 MPa sst = 130 MPa m = 13.33

41.0=+

=stcbc

cbc

mm

kss

s

j =1- (k/3) = 0.87

Q = ½ scbc j 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

mm8.1311000x15.110x98.19

QbM

d6

===

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

xxx

jdM

Ast

st ===s

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.

Chapter One 35


C A

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.

56.020/94/1

44.020/95/1

209

K ; 51

; 51

==

==

=== å

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 H1 from the base. Where, H1 is greater of

i) H/4 ii) 1m

Ex:

i) 3/4=0.75 ii) 1m

\H1= 1m

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

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

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

Chapter One 36


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 5.67513787.0130104672.10

mmxxx

Ast ==

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.

1

CO~r@tC

INFORMATION

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

Q Portland Cement Association 1969 Revised 1961

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 momentcoefficients 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 betaken 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 atthe edge gives final end moments, which must be iden-tical on both sides of the common edge.

Moment distribution cannot be applied as simply tocontinuous 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 a rea l l built I n teg ra l l y

“PublIshed by McGraw-HI11 Book Co, New York, 1959

Tables 1, 2, 3, and 4. Moment Coefficients for Slabs with Various Edge Conditions

Table 1 Table 2

Moment = Coef. x wa’ /j f-lmi_ Moment=Coef.xwa3 d mlr

-

b l a

-

3.00

2.50

2.00

1.75

1.50

1.25

1.00

0.75

0.50

x l ay - o

MX MV

+0.035 +0.010+0.057 +0.016+0.051 +0.013

to031 +0.011+0.052 +0.017+0047 +0.015

+0.025 +0.013+0.042 +0.020+0.041 +0.016

to.020 +0013+0.036 +0.020+0.036 +0.017

+0.015 +0.013+0.028 +0.021+0.030 +0.017

+o 009 +0.012+0.019 +0.019+0.023 +0.017

+0.005 +0.009+0011 +0.016+0.016 +0014

+0.001 +0.006to.005 +0.011+0.009 +0.011

0 +0.003+0.001 + 0 . 0 0 5+0.004 +0.007

_wa Xl

y = b/4 y = b/2

vx 4 Y MV

+0026 +0.011 -0.008 - 0 . 0 3 9+0.044 +0.017 -0.013 -0.063+0.041 +0.014 -0.011 -0055

+0.021 +0010 -0.008 -0.038+0.036 +0.017 -0.012 -0.062+0.036 +0.014 -0.011 -0.055

+0.015 +0.009 -0.007 -0.037+0.028 +0.015 -0.012 -0.059+0.029 +0.013 -0.011 -0.053

+0.012 +0.008 -0.007 -0.035+0.023 +0.013 -0.011 -0.057+0025 +0.012 -0.010 -0051

+0.008 +0007 -0.006 -0.032+0.016 +0.011 -0.010 -0.052+0.020 +0.011 -0.010 -0.048

+0.005 +0.005 -0006 -0.028+0.011 +o 009 -0 009 -0.045+0.014 +0.009 -0.009 -0.043

+0.002 +0.003 -0.004 -0020+0006 +0.006 -0.007 -0.035+0.009 +0.007 -0.007 -0.035

0 +0.002 -0.002 -0.012+0.002 +0.003 -0.004 -0022+0.005 to.005 -0.005 -0.025

0 +0.001 -0.001 -0.005+0.001 +0.001 -0.002 -0.010+0.002 +0.002 -0.003 -0.014

Minus s,gn lndlcates ienslon on the loaded side I” all tables

b l a

-

3.00

2.50

2.00

1.75

150

1.25

1.00

075

0.50

y - o y = b / 4

4 MY Y MY

0 to.070 0 +0.027to.028 +0.061 +0.015 +0.028kO.049 +0.049 +0032 +0026bO.046 +0.030 +0.034 +0.018

0 +0.061 0 +0.019bO.024 +0053 +0.010 +0.022bO.042 +0.044 +0.025 +0.022bO.041 10027 +0.030 +0.016

0 +0.045 0 +0.011bO.016 +0042 +0.006 +0.014kO.033 +0.036 +0.020 +0.016bO.035 r0.024 +0.025 +0.014

0 +0.036 0 +0008IO.013 +0.035 +0005 +0.011bO.028 +0.032 10017 co.01410.031 +0.022 r0.021 10012

0 +0027 0 +0.005moo9 +0.028 +0.003 +0008bO.022 +0.027 +0012 +0.01110.027 +0020 +0.017 +0.011

0 co.017 0 +0.00310.005 +0020 +0.002 +0.005bO.017 +0023 +0.009 +0.00910.021 +0.017 10013 +0.009

0 +O.OlO 0 +0.002to.002 +0013 +o 000 io.003'0 010 +0.017 +0005 +0006too15 +0.015 +o 009 +0.007

0 +0005 0 +0001~0.001 +0008 +o.ooo +0.002to.005 +0011 +0002 +0004~0.010 +0.012 +0.006 +0.004

0 +0.002 0 0~0.000 +0004 +o.ooo +0001bO.002 +0.006 +0001 +0.002a007 +0008 +0.002 +0.002

! y = b/2

MX 4

0 -0.196-0.034 -0.170-0.027 -0.137-0.017 -0.087

0-0.026-0023-0.018

0-0.019-0.018-0.013

0-0015-0015-0.012

0-0012-0013-0.010

0-0008-0010-0.009

-0.138-0132-0.115-0.078

-0.091-0.094 ~-0.089

/ ,"

-0.071-0.076-0076-0.059

-0052-0.059-0.063-0.052

-0034-0.042-0.049-0.044

0-0005-0.007-0007

0-0.003-0004-0.005

0-0.001-0002-0.003

-0.019-0.025-0.036-0036

-0.008-0013-0022-0026

-0.003-0.005-0.010-0014

:-0.065

Table 3 Table 4

Moment = Coef. x wa3 ID;;

,w” 4 X l

bla xla

1.25

1.00

0.75

0.50

y = o

MX MY

+0025

y = b/4

Mx 4

y = b/2

M* MY0

+0.010+0.005-0.033-0.126

+0.019+0.010-0.004-0.025

0 to.014 0 -0.082+0.007 +0.013 -0014 -0.071+0.008 +0.010 -0.011 -0055-0.018 -0 000 -0.006 -0.028-0.092 -0.018 0 0

0 +0.027 0 +0.013 0 -0.074

+0.012 +0.022 +0.007 +0.013 -0.013 -0.066+0.011 +0.014 +0.008 +0010 -0.011 -0.053

-0.021 -0.001 -0.010 +0.001 -0.005 -0.027-0.108 -0.022 -0.077 -0.015 0 0

0 +0.027 0 +0.009 0 -0.060+0.013 +0.023 +0.006 +0.010 -0.012 -0.059+0.015 +0.016 +0.010 +0.010 -0.010 -0 049

-0.008 +0.003 -0.002 +0.003 -0.005 -0.027-0.086 -0.017 -0.059 -0.012 0 0

0 +0.025 0 +0.007 0 -0.050+0.012 +0.022 +0.005 +0.008 -0010 -0.052

+0.016 +0016 +0.010 +0.009 -0 009 -0.046

-0.002 +0.005 +0.001 +0.004 -0.005 -0.027-0.074 -0.015 -0.050 -0.010 0 0

0 +0.021 0 +0.005 0 -0.040to.008 +0.020 +0.004 +0.007 -0.009 -0.044

to.016 +0.016 +0.010 +0.008 -0.008 -0042to.003 +0006 +0.003 +0.004 -0005 -0.026-0.060 -0.012 -0.041 -0.008 0 0

0to.005+0.014+0.006-0.047

0to.002+0.009+0.008-0.035

+0.015 0 +0.003+0.015 +0.002 +0.005+0.015 +0.008 +0.007+0.007 +0.005 +0.005-0.009 -0031 -0.006

0-0.007-0.007-0.0050

0-0.005-0.006-0.0040

0-0.002-0.003-0.0030

0-0.001-0.002-0.0010

-0.029-0.034-0.037-00240

+0.009 0+0.011 0+0.013 +0.005+0.008 +0.005-0.007 -0.022

0 +0.004 0+0.001 +0.008 0+0.005 +0.010 +0.002+0.007 +0.007 +0.003-0024 -0.005 -0.015

0 +0.001 00 +0.005 0

+0.002 +0.006 +0.001+0.004 +0.006 +0.001-0.015 -0.003 -0.008

+0.002+0003to.005+0.004-0.005

+0.001+0.002+0.003+0.003-0.003

0+0.001+0.001+0.001-0.002

-0018-0.023-0.029-0.0200

-0007-0.011-0.017-0.0130

-0002-0.004-0.009-0.0070

- yi~hngea 1Moment q Coef. x bvaz

‘_II m

1 P”@

MX

bla

3.00

2.50

2.00

1.75

1.50

125

1.00

0.75

0.50

y = o

Mx MY

IO.089to.118

to022+o 029

+0077 to.025+0101 to.034

bO.085 +0.024 +0.070 +0.027to112 +0.032 +o 092 +0.037

0076 +0.027 +0.061 +0.028+0.100 +0.037 +0.078 +0038

+0.070 +0.029 +0.054 +o 029+0091 +0040 +0.070 +0.039

to.061 +0.031 +0047 to.029t0.076 +0.043 +0.059 +0.040

to 049 +0.033 +0038 +0.029+0063 +0.044 +0.047 +0.039

to.036 +0.033 +0.027 +002710.044 +0044 +0033 +0036

to.022 +0.029 +0.016 +0023to.025 +0.038 +0.018 +0.030

.O.OlO to.020 +0007 bO.015-0 009 +0025 +0.007 +0.019

y = b/4

MX M”

3

I

Table 5. Moment Coefficients for Tanks with Walls Free at Top and Hinged at Bottom

b/a = 3.0

$y = o

Mx M”

y = b / 4

M, 4

y = b / 2

4 4

z = cl4

4 MZ

0 +0027+0.015 +0026+0.032 +0.026+0.034 +0.018

0 to013+0.009 +0.014+0.023 +0.017+0.029 +0014

0 -0.005+0.002 -0.002+o.o1f3 +0.005+0.022 +0008

0 -0.018-0.003 -0.012+0011 -0.003+0.018 +0.004

0 -0033-0.007 -0.024+o 005 -0.012+0013 0

0 -0.052-0.011 -0 0390 -0.022

+0.008 -0006

0 -0074-0.015 -0.056-0005 -0.034to003 -0.014

0 -0 098-0.020 -0 079-0.011 -0.051-0.002 -0.025

0 -0.126-0.024 -0105-0.016 -0.073-0.007 -0.040

I=0

M, M,c / a

-

3.00

2.50R

Moment q Coef. x wa3

1.25

0 +0.070 0 +0.027+0.028 +0.061 +0.015 +0.028+0.049 +0.049 +0.032 +0026+0.046 +0.030 +0.034 +0.018

0 -0196-0.034 -0170-0.027 -0137-0.017 -0087

0 +0.073 0 +0033 0 -0.169+0.028 +0.063 +0.016 +0033 -0030 -0.151+0.049 +0.050 +0.033 +o 029 -0.025 -0.126+0.046 +0.030 +0.037 +0.020 -0.017 -0.084

0 +0.075 0 +o 039 0 -0.146+0.029 +0.065 +0.017 +0.036 -0.027 -0133+0.050 f0.051 +0.035 +0.032 -0.023 -0.113+0.046 +0.031 +0.037 +0.021 -0016 -0078

0 +0.076 0 +0041 0 -0.137+0.029 +0065 +0.018 +0038 -0025 -0.125+0.050 +0.052 +0036 +0033 -0.021 -0.106+0.046 +0.031 +0037 +0.021 -0.015 -0074

0 +0.077 0 +0.043 0 -0.129+0.030 +0.066 +0.018 +0.039 -0.024 -0.118+0050 +0.053 +0.037 +0.034 -0.020 -0.100+0.046 +0031 +0.038 +0.022 -0.014 -0.070

0 +0.078 0 +0.045 0 -0.122+0.030 +0.067 +0.019 +0.041 -0022 -0111+0.050 +0.054 +0.038 +0.035 -0.019 -0 095+0.047 +0.032 +0.038 +0.023 -0.014 -0.068

0 +0.079 0 +0.047 0 -0118+0.030 +0067 +0.020 +0.043 -0.021 -0105+0.051 +0054 +0.038 +0.036 -0.018 -0.090+0.047 +0032 +0.038 +0.023 -0.013 -0.065

0 +o 079 0 +0.047 0+0.029 +0.066 +0.020 +0.042 -0.021+0.051 +0.053 +0.037 +0.036 -0.018+0.047 +0.031 +0.037 +0.022 -0.013

0 +0.078 0 +0.047 0+0.029 +0.065 +0.019 +0.042 -0.023+0.050 +0.053 +0.035 +0.035 -0.019+0.046 +0.031 +0.036 +0.021 -0.014

-0.120-0107-0 090-0066

-0.130-0.115-0.095-0.068

0 +0.070+0028 +0061+o 049 to.049+0046 +0030

0 +0057+0022 +0050+0041 +0043+0040 +0027

0 +0031+0.013 +0.03210030 +o 029+0034 +0020

0 +0014+0.007 +0.018+0.023 +0.020+0.027 +0015

0 -0.006+0.002 +0.004+0.015 +0.010+0021 +0.010

0 -0.031-0.004 -0.018+0.008 -0.005+0.016 +0.001

0 -0.060-0.010 -0.042+0001 -0.022+0.009 -0.009

0 -0.092-0.016 -0.070-0.006 -0045+0.003 -0024

0-0.022-0.013-0.004

-0123- 0 1 0 1

b/a = 2.5

c/ay = b/4 y = b/2 z=oz = cl4

4 M*

0 +0.019+0010 +0.022+0025 +0.022+0.030 +0.016

0 +0.003+0005 +0.006+0.018 +0.011+0023 +0.011

0 -0.006+0.001 -0002+0.013 +0004+0019 +0.008

0 -0018-0.003 -0.012+0.008 -0.002too15 +0004

0 -0.030-0.006 -0.024+0.003 -0012+0011 -0002

0 -0.045-0010 -0.036-0.003 -0.021+0.006 -0.008

0 -0062-0014 -0053-0.008 -0.035+0.002 -0016

0 -0.081-0.019 -0.072-0.014 -0.056-0.003 -0030

MX MY 4 MY M” MZ0 +0.0.!31 0 co.019

+0.024 +0.053 +0010 +0.022+0042 +0.044 +0025 +0.022+0.041 to.027 +0.030 +0016

0 -3138-0.026 -0132-0023 -0.115-0.016 -0.078

0 +0065 0 +0.026 0 -0.118+0.025 +0.055 +0.012 to.027 -0.023 -0.113+0.043 +0.046 iO.028 +0.025 -0.020 -0.102+0042 +0.028 +0.031 +0.018 -0014 -0.070

0 .-.,7 0 +0.030 0 -0.108+0.025 +0.057 +0.013 +0.030 -0.021 -0.104+0.044 +0.047 +0.029 +0.027 -0.019 -0.096+0043 +0.028 +0.033 +0.019 -0013 -0.066

0 to.068 0 +0033 0 -0100+0.026 +0.058 +0.014 +0032 -0.019 -0.097+0.045 +0047 +0.030 +0.029 -0.018 -0.089+0043 +0.029 +0.034 +0.019 -0.013 -0.063

0 +0 069 0 +0035 0 -0.092+0.026 +0.059 +0.015 +0034 -0.018 -0.089+0.045 +0.048 +0.031 +0031 -0.016 -0.082+0.044 +0.029 +0.034 +0.020 -0012 -0 059

0 +0.070 0 +0037 0 -0087+0.026 +0.060 +0.015 +0036 -0.017 -0.083+0.046 +0.048 +0.031 +0.032 -0.015 -0077+0.044 +0.029 +0.033 +0021 -0.011 -0056

0 +0.070 0 +0.038 0 -0082+0.025 +0.060 to.015 +0.037 -0.016 -0.078+0.045 +0.047 +0.030 +0.032 -0.014 -0071+0043 +0.029 +0.033 +0.020 -0.011 -0.054

0 +0.069 0 +0.039 0 -0080+0.025 +0.059 +0.014 +0.038 -0.015 -0075+0.044 +0.046 +0028 +0.032 -0.014 -0.068+0.042 +0.028 +0.032 +0.019 -0.010 -0.052

0 +0.061+0.024 +0053+0042 +0.044+0.041 +0027

0 +0.038+0.015 +0.037+0.032 +0.033+0034 +0.022

0 +0025+0010 +0026to.025 +0025+0.029 +0.019

0 +0.008+0004 +0013+0017 +0.017+0024 +0015

0 -0010-0.002 -0.003+0.008 +0.007+0.018 +0008

0 -0032-0.008 -0021-0.001 -0008+0011 0

0 -0.055-0.014 -0.042-0.009 -0.025+o 005 -0.011

0 -0.080-0019 -0.068-0017 -0.048-0002 -0.026

2.50

1.25

4

\-0071-0042

t

y = o y = b/4

M, MY M, M,

0 +0.045 0 +0.011+0016 +0.042 +0.006 +0.014+0033 +0.036 +o.ozo +0.016+0.036 +0.024 +0.025 +0.014

0 +0048 0 +0.015+0.017 +0.044 +0.007 +0.017+0.034 +0.038 +0.021 +0019+0036 +0.024 +0.025 +0.015

0 +0.050 0 +0019+0018 +0.046 +o.ooa +0.021+0.035 +0.039 +0.022 +0.021+0.036 +0.025 +0.026 +0.016

0 +0.052 0 +0023+0019 +0.048 a009 +0.024+0036 +0.041 +0.023 +0.023+0037 +0.025 +0.026 +0.017

0 +0054 0 +0.027+0.019 +0.050 +0.010 +0.027+0.037 +0.042 +0.024 +0.025to.037 +0.026 +0.027 +0.018

0 +0.055 0 +0.030+0.018 +0.051 +0.011 +0.029+0.038 +0.043 +0.025 +0026+0.037 +0.026 +0.027 +0.018

0 +0054 0 +0.030+0.018 +0.052 +0.011 +0.029+0.038 +0.044 +0.025 +0.025+0037 +0.026 +0.026 +0.017

= 2.0

c a

2.00

1.75

1.50

1.25

100

075

0.50

z = cl4y = b/2

M, MY

-00910-0.019-0.018-0.013

-0.094-0.089-0.065

z=o

4 4to0110

+0.006+0020+0025

+0.014+0016to014

M, MZ0 co.045

0-0.017-0.017-0.012

0-0015-0.015-0.012

0-0.014-0.014-0.011

0-0.012-0.013-0.010

0-0.012-0.012-0.010

0-0.014-0.013-0.010

-0.081 0 -0001-0.085 +0.003 +0006-0.083 +0.015 +0011-0061 +0.020 +0012

-0072 0 -0.010-0077 0 -0002-0.076 +o 009 +0004-0.058 +0.016 +0.008

-0064 0 -0021-0.068 -0.002 -0.013-0.069 +0.005 -0004-0054 +0.011 +0.002

-0.058 0 -0037-0.062 -0.005 -0025-0.064 0 -0.015-0051 +0006 -0006

-0.058-0.062-0.062-0.049

-0 049-0040-0 029-0.015

-0.065-0068-0064-0.050

0-0.009-0.005+o.ooz

0-0012-0010-0.003

-0.064-0056-0.045-0026

+0.016 +0.042+0.033 +0.036+0.036 +0024

0 ~0032+0.012 co.032+0.027 +0.02910.031 +0021

0 +0.018to.007 '0020+0.020 co.022+0.025 10017

0 0+0.001 '0005+0011 +0012+0016 +0.011

0 -0.023-0.005 -0.013+0.001 0+0.008 to.004

0 -0.044-0010 -0031-0.007 -0.015+0.001 -0.004

0 -0061-0014 -0051-0.012 -0034-0.004 -0.018

I I = 1.5

y = b/4 y = b/2

4 M Y 4 MY

0 +0.005 0 -0.052+0.003 +0.008 -0.012 -0.059+0.012 +0.011 -0.013 -0.063+0.017 +0011 -0.010 -0.052

0 +0.008 0 -0.045+0.005 +0.012 -0010 -0.050+0.014 +0.014 -0.011 -0.056+0.018 +0.012 -0.010 -0.048

0 +0013 0 -0.038+0.006 +0.016 -0008 -0.042+0.015 +0.017 -0.010 -0.049to.019 +0.014 -0.009 -0.045

0 +0.016 0 -0.034+0.007 +0.018 -0.008 -0.038+0.016 +0.019 -0.008 -0.042+0.019 +0.015 -0.008 -0.041

0 +0.017 0 -0.036+0.007 +0.019 -0.008 -0040+0.017 +0.020 -0.009 -0.044+0.018 +0.016 -0.008 -0.040

”1.50

w/a* = c/4 z=o

M" 4 MX MZ

0 +0005 0 co.027+0003 +0.008 +0.009 +0028+0.012 +0.011 +0.022 to.027+0.017 +0.011 +0.027 to.020

0 -0005 0 coo11+0.001 -0.001 +0.004 +0015+0007 +0006 +0.014 +0.02060.013 +0006 +0.018 10016

0 -0.016 0 -0.006-0.002 -0010 -0001 +0.001+0.002 -0.003 +0006 +0010+0.008 +0.002 +o 009 +0.010

0 -0024 0 -0.019-0.005 -0.020 -0.004 -0013-0.002 -0.014 -0.001 -0.004+0.003 -0.007 +0.002 +0001

0 -0030 0 -0.028-0.008 -0031 -0.007 -0.027-0.006 -0.027 -0.006 -0020-0.002 -0018 -0.004 -0010

y - o

M. M”I

0 +0.027+0.009 +0.028+0.022 +0.027+0.027 +0.020

0 +0.031+0.010 +0.031+0.024 +0.030+0.027 +0.021

0 +0.035+0.011 +0.034+0.025 +0.032+0.028 +0.022

0 +0.038+0.011 +0.036+0.025 +0.033+0028 +0.022

0 +0.040+0.010 +0.037+0.024 +0.034+0.028 +0.022

125

b/a = 1.0

c/a x/ay - o Y = b / 4 Y = b / 2 z = c/4 z=o

M” Mb M” M” M” M,

0 +0.010 0 +0.002 0 -0.019 0 +0.002+0.002 +0.013 0 +0.003 -0.005 -0.025 0 +0.003+0.010 +0.017 +0.005 +0006 -0.007 -0036 +0.005 +0006+0015 +0.015 +0.009 +0.007 -0.007 -0036 +0.009 +0.007

MX MZ

0 +0.010+0002 f0.013+0.010 +0017+0.015 +0.015

0 +0.016 0 +o.oc)7 0 -0.013 0 -0.004 0 +0.003+0.003 +0.017 +0.001 +0.008 -0.004 -0.020 -0.001 -0.005 -0.001 +0.003+0.011 +0.020 +0.006 +0.009 -0.007 -0.033 +0.002 -0.001 +0.005 +0.007+0.016 +0.014 +0.009 +0.009 -0.006 -0.032 +0.004 +0.002 +0.009 fO008

0 +0.020 0 +0.011 0 -0.011 0 -0.007 0 -0.005+0.003 +0.018 +0.001 +0.010 -0.004 -0.018 -0002 -0.012 -0.003 -0.007+0.012 +0.021 +0.008 +0.010 -0.006 -0.032 +0.001 -0.009 +0.002 -0.005+0.017 +0.013 +0.010 +0.009 -0.006 -0.031 +0002 -0.005 +0.006 +0.001

5

Table 6. Moment Coefficients for Tanks with Walls Hinged at Top and Bottom

Moment = Coef. x wa3

b/a = 3.0

C/a

3.00

2.50

2.00

1.75

1.50

1.25

1.00

0.75

0.50

f

w/ay = o y = b/4 y = b/2 z = c/4 .?=O

MX MY Mx MY MX MI+0.035 +0.010 +0.026 +0011+0.057 +0.016 +0.044 +0.017+0.051 +0.013 +0.041 +0014

+0.035 +0.010 +0.026 +0.011+0057 +0.016 +0.044 +0.017+0.051 +0.013 +0.041 +0.014

+0.035 +0.010 +0.026 +0.011+0.057 +0.016 +0045 +0.017to.051 +0013 +0.042 +0.014

+0.035 +0010 +0027 +0.011+0.057 +0.015 +0.045 +0017+0051 +0013 +0.042 +0.014

+0.035 +0.010 +0.027 +0.011+0.057 +0.015 +0.045 +0.017+0.051 +0.013 +0.042 +0.014

+0.035 +0010 +0027 +0.011+0.057 +0.015 +0.046 +0.017+0.051 +0.013 +0.042 +0.014

+0.035 +0.010 +0.027 +0.011+0.057 +0.015 +0.046 +0.017+0.051 +0013 +0.043 +0.014

+0.035 +0.010 +o.o2a +0.011+0.057 +0.015 +0.046 +0.017+0.052 +0.013 +0043 +0.014

+0036 +0.010 +0.028 +0.011+0.057 +0.015 +0.047 +0.017+0.052 +0.013 +0.043 +0.014

-0 008-0.013-0011

-0.008-0.012-0.011

-0.008-0.012-0011

-0.007-0.012-0.011

-0.007-0.011-0.010

-0.006-0.011-0.010

-0.006-0.010-0.009

-0.005-0.008-0.008

-0.004-0007-0.007

-0 039-0.063-0.055

-0.039-0.062-0055

-0.038-0.062-0.054

-0037-0.060-0053

-0.035-0.057-0.051

-0.032-0.053-0.048

-0 029-0.048-0.044

-0.025-0.042-0.039

-0.021-0035-0.033

+0026 +0.01110044 +0.017+0041 +0.014

+0.021 +0010+0036 +0.017+0036 to014

+0.015 to010+0.028 +0015+0.029 +0.013

+0011 +0008+0021 +0.013+0024 '0012

+0007 to.006to.015 +0.010+0019 +0011

+0.003 +0.003+0.008 to.006+0013 +0008

-0001 0to.002 +0002+0.007 +0.004

-0.003 -0005-0.003 -0005+0.002 -0002

-0004 -0.011-0007 -0.016-0.004 -0.010

+0035 +0.010+0057 +0016+0051 +0.013

+0031 +0011+0.052 +0017+0047 +0.014

+0025 to.013+0043 to020+0041 +0016

+0.020 +0013+0036 +0020+0.036 +0016

to.014 +0013+0027 +0020to 029 to.017

+0008 +001110017 a017+0.021 +0.016

to002 +0.008+0.007 +oc14+0.013 +0.013

-0002 +0.001-0.001 +0.007+0.006 +0.007

-0.005 -0.008-0.006 -0010-0001 -0.004

b/a = 2.5

c/a

2.50

2.00

1.75

1.50

1.25

100

0.75

0.50

y = b/2

4 M”-0.008 -0.038-0.012 -0.062-0.011 -0.055

-0.008 -0.038-0.012 -0.061-0.011 -0.054

-0.007 -0.037-0.012 -0.059-0.011 -0.053

-0.007 -0.035-0.011 -0057-0.010 -0.051

-0.006 -0.032-0.011 -0.053-0.010 -0.048

-0.006 -0028-0.010 -0048-0.009 -0044

-0005 -0.024-0.008 -0041-0.008 -0.039

-0004 -0.021-0.007 -0035-0.007 -0034

z = c/4

M, MZ

+0.021 +0010+0036 +0017co.036 +0.014

+0.015 +o 009+0.028 +0015+0.029 +0.013

+0011 +0.008+0.022 +0.013+0.024 +0.012

+0.007 +0.006+0.015 +0.010+0019 +0.010

+0003 +0.004+0.008 +0.007+0.014 +0.008

-0.001 0+0.002 +0002to.007 +a004

-0003 -0.005-0.003 -00050 -0002

-0004 -0.011-0007 -0016-0.004 -0.010

z=o

4 MZ+0.031 +0.011 +0.021 +0.010+0.052 +0.017 +0.036 to.017+0.047 +0.015 +0.036 +0.014

+0.031 +0.011 +0.021 +0.010+0.052 +0.017 +0.036 to.017+0.047 +0.015 +0.036 +0.014

+0.032 +0.011 +0.021 +0.010+0.052 +0.018 +0.036 +0.017+0.047 +0.015 +0.036 +0.014

+0.032 +0.011 +0.022 +0.010to.052 +0.018 +0.037 +0.017+0.047 +0.015 +0036 +0.014

+0.032 +0.011 +0.022 +0010+0.052 +0.018 +0.038 +0.017+0.048 +0.015 +0.037 +0.014

+0.032 +0.011 +0.023 +0.011+0.053 +0.018 +0038 +0.017+0.048 +0.015 +0.038 +0.015

+0.033 +0.011 +0024 co.011+0.054 +0.018 +0.039 +0.017+0.049 +0.015 +0.038 +0.015

+0.033 +0.012 +0.024 +0.011+0.054 +0.018 +0040 +0.017+0.049 +0.015 +o 039 to015

+0031 +0.011+0.052 +0.017+0.047 +0.015

+0025 +0.012+0042 +0.020+0.041 +0.016

+0.020 +0.012+0035 +0021+0.035 +0.017

'0.014 +0013+0.027 +0.021+0.029 +0.017

+0.007 +0.012co.018 +0019+0022 to.018

+0.002 +0.008to007 +0014+0013 +0013

-0.002 -0.0020 +000510006 +0006

-0.005 -0.008-0.006 -0010-0.001 -0004

6

b/a = 2.0l-

y - o y = b/4 y = b/2

Mx 4 Mx 4 4 MY+0.025 +0.013 +0.015 +0.009 -0007 -0.037+0.042 +0.020 +0.028 +0.015 -0012 -0.059+0.040 +0.016 +0.029 +0.013 -0.011 -0.053

+0.025 +0.013 +0.015 +0.009 -0.007 -0.036+0.042 +o.ozo +0.028 +0.015 -0.012 -0.058+0.040 +0.016 +0.029 +0.013 -0.010 -0052

+0.025 +0.013 +0.016 +0.009 -0.007 -0034+0.043 +0.020 +0.028 +0.015 -0.011 -0.056+0.041 +0.016 +0.029 +0.013 -0.010 -0.050

+0.0?6 +0.013 +0.016 +0.010 -0.006 -0.032+0.043 +0.020 +0.029 +0.015 -0.010 -0052+0.041 +0.016 +0.030 +0.013 -0.010 -0.048

+0.026 +0.013 +0.017 +0.010 -0.006 -0.028+0.044 +0.020 +0.030 +0.016 -0.009 -0.046+0.041 +0.016 +0.031 +0.014 -0.009 -0.044

+0.027 +0.013 +0.018 +0.010 -0.005 -0.024+0.045 +0.020 +0031 +0.016 -0.008 -0.040+0.042 +0.016 +0.032 +0.014 -0.008 -0.041

+0.027 +0.013 +0.019 +0.010 -0.004 -0.021+0.046 +0.020 +0.033 +0.017 -0.007 -0.034+0.042 +0.016 +0.032 +0.015 -0.007 -0.037

- -

c/az=o

M, Mz+0.025 +0.013+0.042 +0020+0.040 +0.016

+0.020 +0.013+0.035 +0.021+0.035 +0.017

+0.014 +0.013+0027 +0.021+0.029 +0.017

+0.007 +0.011+0.018 +0.019+0.021 +0.016

+0.002 +0.008+0.007 +0.014+0.013 +0.013

-0.001 +o.m20 +0.005

to.005 to.008

-0.004 -0.007-0.006 -0.009-0.002 -0.003

z = cl4

4 4

to015 +0.009+0028 +0.015+0.029 +0013

+0.011 +0.008+0.022 +0.013+0.024 +0012

+0.007 +0.006+0.015 +0.011+0.019 +0010

+0.003 +0003+0.008 +0007+0.013 +0008

-0.001 0+0.002 +0.002+0.007 +0.004

-0.003 -0.004-0002 -0004+0.002 -0002

-0.004 -0010-0.006 -0015-0.003 -0010

2.00

1.75

1.50

125

1.00

0.75

0.50

1b/a = 1.5

c/az=o

4 4

+0.015 +0.013+0.028 +0.021+0.030 +0.017

+0.009 +0.012+0.018 +0.019+0023 +0.016

+0.003 +0008+0.008 +0014+0.014 +0.014

-0.001 to.002+0.001 +0.005+0.006 +0.008

-0004 -0006-0.005 -0007-0.001 -0001

z = c/4

4 MZ

+0008 +0.007+0.016 +0.011+0.020 +0.011

+0.004 +0.004+0.009 +0008+0.014 +o 009

0 +0.001to.003 +0.003+0008 +0.005

-0002 -0.003-0.002 -0.004+0.002 0

-0.003 -0 009-0.006 -0.014-0003 -0.008

y = b/4 y = b/2

4 MY 4 MY

+0.008 +0.007 -0.006 -0.032+0.016 +0.011 -0.010 -0.052+0.020 +0.011 -0.010 -0.048

+0.009 +0.008 -0.006 -0.029+0.017 to.012 -0.010 -0.049+0.020 +0.012 -0.009 -0.045

+0.010 +0.009 -0005 -0.025+0.019 +0.012 -0.009 -0.043+0.021 +0.013 -0.008 -0.041

+0.011 +0.010 -0.004 -0.021+0.021 +0.014 -0.007 -0036+0.022 +0.014 -0.007 -0.036

+0.013 +0.012 -0.003 -0.017+0.023 +0.018 -0.006 -0.031+0.024 +0.016 -0.007 -0.033

y - o

4 MY

+0015 +0013+0.028 +0.021+0.030 +0.017

+0.016 +0.013+0.029 +0.021+0.030 +0.017

+0.016 +0.013+0.030 +0.021to.031 to.017

+0.018 +0.014+0.032 +0.022+0.032 +0.018

+0.020 +0.016+0035 +0.024+0034 +0.020

150

125

b/a = 1.0

y = b/2 I = c/4 z=oc/a

M.x 4+0.005 +0.009+0.011 +0.016+0.016 +0.015

+0.001 +0.005+0.005 +o 009+0.008 +0010

-0.003 -0.002-0.003 -0.0020 +0.001

y = o y = b/4

4 M” 4 MY

+0.005 +0.009 +0.002 +0.003+0.011 +0.016 +0.006 +0.006+0016 +0.015 +0.009 +0.007

+0.006 +0010 +0.003 +0.004+0.013 +0.017 +0.008 +0.008+0.017 +0016 +0.010 +0.008

to.007 +0.011 +0.005 +0.006+0.015 +0018 +0.010 +0.010+0.018 +0.016 +0.012 +0.010

M. M” M” M,

+0.002 +0.003+0.006 +0.006+0.009 +0.007

-0.004 -0.020-0.007 -0.035-0.007 -0.035

-0003 -0.016-0.006 -0.029-0.006 -0.031

-0.002 -0.010-0.004 -0.021-0.005 -0.026

0 0+0.001 +0.001+0.004 +0.003

-0.002 -0.005-0.003 -0.007-0.001 -0.004

7

In this manner, moment coefficients were computedand are tabulated in Tables 5 and 6 for top and bottomedge conditions as shown for single-cell tanks with alarge number of ratios of b/a and c/a, b being the largerand c the smaller of.the horizontal tank dimensions. Mo-ments in vertical and horizontal directions equal thecoefficients times wa3, in which w is the weight of theliquid. Note that the loading term is wa3 for all wall slabssubject to hydrostatic pressure but is wa2 for the floorslab in Table 4, which has uniformly distributed load. Inthe first case, w is weight per cubic foot, but in the latterit is weight per square foot.

There is a peculiarity about the horizontal end mo-ments in the slabs at the free top edge. Calculations ofsuch moments by means of the trigonometric seriesused result in a value of zero, whereas these momentsactually have finite values and may even be compara-tively large. Horizontal end moments at the free edgewere therefore established by extrapolation. The con-sistency of extrapolated moment coefficients waschecked by plotting and studying curves. This gavereasonably good results, although coefficients thusdetermined are probably not quite as accurate as thecoefficients that were computed. A condition prevails atthe quarter point of the free edge, similar to that at theend point but to a lesser degree. At the midpoint of thefree edge the coefficients were computed, extrapolationbeing used only for checking purposes.

When a tank is built underground, the walls must beinvestigated for both internal and external pressure. Thelatter may be due to earth pressure or to a combinationof earth and groundwater pressure. Tables and otherdata presented can be applied in ‘the case of pressurefrom either side but the signs are opposite. In the caseof external pressure, actual load distribution may notnecessarily be triangular as assumed in the tables.Consider for illustration a tank built below ground withearth covering the roof slab and causing a trapezoidaldistribution of lateral earth pressure on the walls. In thiscase it gives a fairly good approximation to substitute a

triangle with the same area as the trapezoid represent-ing the actual load distribution. The intensity of load isthe same at middepth in both cases and when the wallis supported at both top and bottom edges, the discrep-ancy between triangle and trapezoid has relatively littleeffect at and near the supported edges

Shear Coefficients

Shear values along the edges of a tank wall are neededfor investigation of shear and development stresses.Along vertical edges, shear in one wall is also used asaxial tension in the adjacent wall and must be combinedwith bending moment to determine tensile reinforce-ment.

Various data for shear were computed and are givenin Table 7. The wall is considered fixed at the two verti-cal edges while top and bottom edges are assumed tobe hinged. The wall panel with width b and height a issubject to hydrostatic pressure due to a liquid weighingw lb per cubic foot.

The first five lines in Table 7 are shears per linear footin terms of wa*. The remaining four lines are total shearsin kips or pounds depending on how w is given. Shearsper linear foot are for ratios of b/a = %, 1,2, and infinity.The difference between the shear for b/a = 2 and infinityis so small that there is no necessity for computing co-efficients for intermediate values.

When b/a is large, a vertical strip of the slab near mid-point of the b dimension will behave essentially as a

0.50waz, of which two-thirds or 0.33wa2is the reaction at the bottom support and one-third or0.17wa2 is the reaction at the top. Note in Table 7 thatshear at midpoint of the bottom edge is 0.3290waz forb/a = 2.0, the coefficient being very close to that of one-third for infinity. In other words, maximum bottom shearis practically constant for all values of b/a greater than

Table 7. Shear at Edges of Slabs Hinged at Top and Bottom

b l a ‘h 1 2 5 10 lnfmtty

Midpoint of bottom edge +o 1407wa’ +o 2419weCorner at bottom edge -0 2575wa” -0 4397wa’

M,dpo,“t of flxed side edge +o 1260wa’ +O 2562wa’Lower third-pant of side edge *o 173&v@ +o 3113wa’Lower quarter-pant of side edge ‘0 1919wP +o 3153w.e

Total at top edge 0 OOOOwa’b 0 0052wa’bTotal at bottom edge 0 0460wa-‘b 0 0960wa’bTotal at one foxed side edge 0 2260wazb 0 1 9 9 4 w a ’ bTotal at all four edges 0 5000wa’b 0 5000waJb

‘Negatwe s,gn lndlcates reaction acts I” darectlon of loadtEsbmated

+o 3290w.a’- 0 5633w.F

+0.3604waz‘0 4023wa’‘0 3904w.3’

0 0536w.+b 0 ,203~~b0 1616wa’b 0 2715wa’b0 1322wa’b 0 0541wa.b0 5000wa’b 0 5 0 0 0 w a . b

+o 3333waz-0 6000wa’

‘0 3912w.a’+0 4116wa’‘0 39t30wa.

0 1435wa’b0 3023wa’b0 0271 wa’b0 5000wa’b

0 1667wa’b0 3 3 3 3 w a . b0 275wav0 5000wa’b

8

\asimply supported one-way slab. Total pressure on a Jstrip 1 ft wide is

2. As will be shown, this is correct only when the topedge is supported, not when it is free.

At the corner, shear at the bottom edge is negativeand numerically greater than shear at midpoint. Thechange from positive to negative shear occurs approxi-mately at the outer tenth points of the bottom edge.These high negative values at the corners arise be-cause deformations in the planes of the supportingslabs are neglected in the basic equations and aretherefore of only theoretical significance. These shearscan be disregarded in checking shear and developmentstresses.

Unit shears at the fixed edge in Table 7 were used forplotting the curves in Fig. 1. There is practically nochange in shear curves beyond b/a = 2.0. Maximumvalue occurs at a depth below the top somewhere be-tween 0.6a and 0.8a. Fig. 1 is useful for determination ofshear or axial tension for any ratio of b/a and at anypoint of a fixed side edge.

Total shear from top to bottom of one fixed edge inTable 7 must equal the area within the correspondingcurve in Fig. 1, and this relationship was used for check-ing the curves ‘Total shears computed and tabulated fora hanged top were also used in making certain adjust-ments to determine approximate values of shear forwalls with free too-recorded in Table 8.

For b/a = % in Table 7, total shear at the top edge is sosmall as to be practically zero, and for b/a = 1 .O totalshear, 0.0052, is only 1% of total hydrostatic pressure,0.5000.’ Therefore, it is reasonable to assume that re-moving the top support will not materially change totalshears at any of the other three edges when b/a = Y2 and1. At b/a = 2.0, there is a substantial shear at the top

‘Loading term is omitted here.

9" 0.3

gij a4

fg 0.5

% a6p

B a7

1.00

Fig. 1.

0.1 0.2 0.3

Shear per lin. ft. = coef x wa2

Table 8. Shear at Edges of Slabs Free at Topand Hinged at Bottom*

tJa 1 2 3

Mldpolnt o f b o t t o m e d g e ‘ 0 141wa: *o 242w.F ‘ 0 3awa7 ‘ 0 45wetCorner at bottom edge -0 258wa-$ -0 440-a. - 0 583~9 - 0 sowa

Top of flxed side edge 0 ooowa. ‘0 olowa’ *o 100wa -0 165wa-

Mldpomf of flxed side edge +O 128wa‘ *O 258wa’ .o 375wa: ‘ 0 406WWLower third-wont of side edae *o 174ws *0311wa- *o 406W% *o 416w.F

Lower quarter-WI”, of side edge ‘0 192-a- *o 315w.T *o 390wa ,O 398waTotal at bottom edge 0 0 4 8 w a b 0 0 9 6 w a . b 0 2 0 4 w a . b 0 286wa.bTotal at one faxed s,de edge 0 226wa b 0 2 0 2 w a . b 0 148wa:b 0 107wa.bTotal al all four edges 0 500w.s. b 0 500wa-b 0 500wa. b 0 500~4 b

‘Data dewed by modlfymg values compufed for waifs hanged fop and boflom

tThls value could not be esflmated accurately beyond two decimal places

Wegat~ve s!gn lndlcates react~o” acts I” d,recfwn of load

edge when hinged, 0.0538, so that the sum of totalshears on the other three sides is only 0.4462. If the topsupport is removed, the other three sides must carry atotal of 0.5000. A reasonable adjustment is to multiplyeach of the three remaining total shears by 0.5000/0.4462 = 1 .12, an increase of 12%. This was done in pre-paring Table 8 for b/a = 2.0. A similar adjustment wasmade for b/a = 3.0, where the increase is 22%.

Total shears recorded in Table 8 were used to deter-mine unit shears also recorded in that table. Considerfor illustration the shear curves in Fig. 1 and imagine thetop is changed from hinged to free. As already stated, forb/a = % and 1 it makes little difference in total shear-the area within shear curves-whether the top is sup-ported or not. Consequently, curves for b/a = % and 1remain practically unchanged. They were transferredalmost without modification to Fig. 2, which covers thecase with top free. For b/a = 2 an adjustment was made.A change in the support at the top has little effect uponshear at the bottom of the fixed edge. Consequently, thecurves in Figs. 1 and 2 are nearly identical at the bottom.Gradually, as the top is approached curves for the freetop deviate more and more from those for the hingedtop, as in Fig. 2. By trial, curve for b/a = 2 was so ad-justed that its area equals the total shear for one fixededge for b/a = 2.0 in Table 8. A similar adjustment wasmade for b/a = 3.0, which is the limit of moment coeffi-cients given.

One point of interest stands out In a comparison ofFigs. 1 and 2. Whereas for b/a = 2.0 and 3.0 total shearis increased 12% and 22%, respectively, when top isfree instead of hinged, maximum shear is increased butslightly, 2% at most. The reason is that most of the in-crease in shear is near the top where shears are rela-tively small.

The same general procedure was applied, but notillustrated, for adjustment of unit shear at midpoint ofbottom, but in this case the greatest change resultingfrom making the top free is at midpoint where shear is

9

0

Fig. 2.

0.1 @.2 0.3

Shear per lin ft. - coef Y wa2

04

large for the hinged-top condition. For illustration, forb/a = 3.0, unit shear at midpoint of the bottom is0.33wa2with hinged top but 0.45wa2 with free top-an increaseof approximately one-third.

Shear data were computed for wall panels with fixedvertical edges. They can be appli$d with satisfactoryresults to any ordinary tank wall even if vertical edgesare not fully fixed.

Open-Top Single-Cell Tank

The tank in Fig. 3 has a clear height of a = 16 ft. Hori-zontal inside dimensions are b = 40 ft and c = 20 ft. Thetops of the walls are considered free and the bottomhinged. The tank contains water weighing 62.5 lb percubic foot.

Coefficients for moment and shear are selected fromtables or diagrams for b/a = 40/16 = 2.50 and c/a =20/l 6 = 1.25. Moments are in foot-kips if coefficientsare multiplied by wa3/1000 = 62.5 x 163/1000 = 256;and shears are in kips if coefficients are multiplied bywaz/lOOO = 62.5 x 16z/1000 = 16.

Moment coefficients taken from Table 5 for b/a = 2.50and c/a = 1 .25 are tabulated below. Coefficients for x =a (bottom edge), being equal to zero, are omitted.

0 0 +0069 0 +0035 0 - 0 . 0 9 2 0 - 0 . 0 3 0'A +0.026 GO59 +0015 +0034 - 0 . 0 1 6 - 0 . 0 8 9 -0006 - 0 . 0 2 4 -:002 I:::;;H +0.045 +0046 +0.031 +0.031 - 0 . 0 1 6 - 0 . 0 6 2 ' 0 . 0 0 3 - 0 . 0 1 2 +O.@X3 +OOO,% +0.044 to.029 ' 0 . 0 3 4 +0020 - 0 . 0 1 2 -0059 +001, - 0 . 0 0 2 +o,o,i3 +0008

The largest moment occurs in the horizontal direc-tion at the top of the corner common to both walls andequals -0.092wa3 = -0.092 x 256 = -23.6 ft kips. Thenegative sign simply indicates that tension is on the in-

10

Fig. 3.

side and need not be considered In subsequent calcu-lations.

Maximum horizontal moment at midpoint of the longerwall is +0.069wa3 = +0.069 x 256 = +17.7 ft kips. Thepositive sign shows that tension is in the outside of thewall. There is also some axial tenslon on this sectionthat can be taken equal to end shear at the top of theshorter wall. For use in connection with Fig. 2, ratio ofb/a for the shorter wall is 20/ 16 = 1.25. Shear is 0.03wa2= 0.03 x 16 =0.48 kips. The effect of axial tension is neg-ligible in this case and the steel area can be determinedas for simple bending.

Horizontally at x = a/2, axial tension taken from Fig. 2for b/a = 1.25 is equal to N = -0.30wa2 = -0.30 x 16 =-4.80 kips per linear foot, which is not negligible. Mo-ment is M = 0.048wa3 = 0.048 x 256 = 12.3 ft kips.

In the shorter wall, positive moments are all relativelysmall. Maximum positive moment is vertical: 0.01 8wa3

x 256 = 4.6 ft kips.Maximum Mx in the vertical strip at midpoint ot longer

0.045wa3 = 0.045 x 256 = 11.5 ft klps.Maximum shear at the bottom taken from Table 8 is

V = 0.42wa2 = 0.42 x 16 = 6.72 kips.

Closed Single-Cell Tank

The tank in this section differs from the preceding oneonly in that the tops of the walls are considered hingedrather than free. This condition exists when the tank iscovered by a concrete slab with dowels extending fromthe wall into the slab without moment reinforcementacross the bearing surface.

Moment coefficients taken from Table 6 are givenbelow. All coefficients for x = 0 (top edge) and x = a (bot-tom edge), being equal to zero, are omitted.

With a free top, maximum M, = +0.045wa3 and maxi-mum My = -0.092wa3. With a hinged top, maximum n/l,= +0.052wa3 and maximum My = -0.053wa3. It is to beexpected that a wall with hinged top will carry more loadvertically and less horizontally, but it is worth noting thatmaximum coefficient for vertical moment is only 13%

A= 0.018

,’panel is

less for wall with free top than with hinged top.Another noteworthy point is that maximum M, coeffi-

cient at y = 0 is +0.069 for a free top but +0.018 for ahinged top. Adding top support causes considerablereduction in horizontal moments, especially at y = 0.

Maximum moment is -0.053~~1~ = -0.053 x 256 =-13.6 ft kips.

Maximum moment in a vertical strip is M = 0.052~~1~ =0.052 x 256 = 13.3 ft kips. Axial compression (N) on thesection subject to this moment, and loads per linear footcan be taken as follows:

8-ft-high wall: 8 x 1 .08 x 0.150 = 1.3 kips12-in. top concrete slab: 0.150 x20/2 = 1.5 kips’

3-ft fill on top of slab: 0.300 x 20/2 =3.0 kips’Live load on top of fill: 0.100 x 20/2 = 1

forN =6.8 kips and to design tensile steel for N = 1 .3 + 1.5 = 2.8kips, in which fill and live load are disregarded.

Top and Base Slabs

The closed single-cell tank is covered with a concreteslab. Assume the slab is simply supported along all foursides and has a live load of 100 psf and an earthfillweighing 300 psf.

Estimating slab thickness as 12 in. gives a total designload of 100 + 300 + 150 = 550 psf. From Table 4, for aratio of 40/20 = 2, select maximum coefficient of 0.100,which gives maximum M = 0.1 00wa2 F 0.100 x 0.550 x20.02 = 22.0 ft kips.

At the corners, a two-way slab tends to lift off thesupports; and if this tendency is prevented by dowelingslab to support, cracks may develop in the top of theslab across its corners. Nominal top reinforcementshould therefore be supplied at the corners, say0.005bdsq in. per foot in each direction. Length of these bars canbe taken as %a = l/4 x 20 = 5 ft.

Assume the closed single-cell tank has a base slab ofreinforced concrete. Weight of base slab and liquid doesnot create any bending or shearing stresses in concreteprovided the subsoil is uniformly well compacted. Weighttransferred to the base through the bottom of the wall is

Top slab: 0.550 x 22 x 42 = 510 kipsWalls: 16x0.162(2x41.1 +2~21,1)=320kips

830 kipsIf the base slab extends 9 in. outside the walls, its area

is 43.7 x 23.7 = 1035 sq ft. The average load of w =830,000/ 1035 = 800 psf is used for design of the baseslab just as w = 550 psf was used for design of the topslab.

Total average load on the subsoil is 16 x 62.5 + 800 +weight of base slab, say 1000 + 800 + 200 = 2000 psf,which the subsoil must be able to carry.

If there is an appreciable upward hydrostatic pressureon the base slab, the slab should also be investigated forthis pressure when the tank is considered empty.

\ -‘ProportIons of tank being deslgned are such that for determlning

axial compression In sde walls, all the top load may be consideredcarned the short way

Multicell Tank

Multicell tanks do not lend themselves readily to mathe-matically accurate stress analysis It is possible, how-ever, with the tables presented here for single-cell tanksand for individual wall panels with fixed vertical edges toestimate moment coefficients for symmetrical multicelltanks with sufficient accuracy for design purposes. Whileresults obtained by the following procedure are approxi-mate and should therefore be considered as a guide toengineering judgment, the procedure does give a con-servative design.

Because a rotation of one corner has comparativelylittle effect on moments at adjacent corners in atankwithwall panels supported on three or four sides, moments inthe walls of a multicell tank are essentially the same asin single-cell tanks-except at corners where more thantwo walls intersect. Moment coefficients from Tables 5and 6, designated as L coefficients, apply to outer or L-shaped corners of multicell tanks (see Fig. 4a) as well asto interior sections in all walls, that is, sections desig-nated as y = b/4, y = 0, z = c/4, and z = 0. Moment coeffi-cients for design sections at corners where more thantwo panels intersect depend on the loading conditionproducing maximum moment and on the number of inter-secting walls.

In Fig. 4b, three walls form a T-shaped unit. If the con-tinuous wall, or top of the T, is part of the long sides oftwo adjacent rectangular cells, the moment in the con-tinous wall at the intersection is maximum when bothcells are filled. The intersection is then fixed and mo-ment coefficients, designated as F coefficients, can betaken from Tables 1, 2, or 3, depending on edge condi-tions at top and bottom. These three tables cover panelswith fixed side edges. If the continuous wall is part of theshort sides of two adjacent rectangular cells, momentat one side of the intersection is maximum, when the cellon that side is filled while the other cell is empty. Like-wise the end moment in the center wall is maximumwhen only one cell is filled. For this loading condition themagnitude of moment will be somewhere between theLcoefficients and the F coefficients. If the unloaded thirdwall of the unit is disregarded, or its stiffness considerednegligible, moments in the loaded walls would be thesame as in Fig. 4a, that is, the L coefficients apply. If thethird wall is assumed to have infinite stiffness, the corneris fixed and the f coefficients apply. The intermediatevalue representing more nearly the true condition canbe obtained by the formula:

End moments = L -nG2(i -F)

(4 b)

Fig. 4.

411

.O kips’

6.8 kips

tIt is conservative to check compressive stress

in which n denotes number of adjacent unloaded walls.This formula checks for n equal to zero and infinity. In anL-shaped unit n equals 0 and the end moments equalL - O(L - F) = L. Inserting n equal to infinity will givenl(n + 2) = 1 and the end moments equal L - 1 (L - F) = f,which also checks.

In Fig. 4c, two continuous walls form a cross. If inter-secting walls are the walls of square cells, moments atthe intersection are maximum when any two cells arefilled and the F coefficients in Tables 1,2, or 3 apply be-cause there is no rotation of the joint. If the cells are rec-tangular, moments in the longer of the intersecting wallswill be maximum when two cells on the same sideof thewall under consideration are filled, and again the F coef-ficients apply. Maximum moments in the shorter wallsadjacent to the intersection occur when diagonallyopposite cells are filled, and for this condition the L coef-ficients apply.

Fig. 5 shows moment coefficients at wall intersec-itions in two- and four-cell tanks. Where coefficients arenot shown, L coefficients of Tables 5 and 6 apply.

Two-Cell Tank, Long Center Wall

The tank in Fig. 6 consists of two adjacent cells, eachwith the same inside dimensions as the open-top single-cell tank and the closed single-cell tank. The top is con-sidered free.

In accordance with the types of units in Fig. 4, the tankconsists of four L-shaped and two T-shaped units. L co-eff icients from Table 5 for b/a = 2.50 and c/a = 1.25, andF coefficients from Table 2, for b/a = 2.50 and 1.25, aretabulated as follows:

Long outer walls

L = coeffuents from Table 5 forbla = 2 . 5 0 a n d cla = 1 . 2 5

x / ay = b / 2

Iy = b / 4

Iy = o

M” Mv M. Mv Mx Mv

0 I 0 -0.092 I 0 +0035 I 0 to.069% I - 0 0 1 8 - 0 . 0 8 9 to.015 to.034 +0.026 +0.059‘h - 0 0 1 6 - 0 0 8 2 to031 to.031 +0.045 +o.o4a% -0.012 -0 059 +0034 +0.020 I +0.044 +0.029

Short outer walls

L = coefficients from Table 5 for I F =Tca~fetTfz Ib l a = 2 . 5 0 a n d c/a = 1 2 5 b/a = 1 25

L .L&.? = c/2 L = Cl4 z=o 3

x/a .? =c/2Mx Ml 4 Mz MN Mz M.Y M, 4 MZ

0 0 -0.092 0 -0.030 0 -0.010 0 - 0 0 3 4 0 -0.073% -0.018 -0.089 -0.006 - 0 0 2 4

I

- 0 0 0 2 -0.003 -0.008 -0.042 -0.015 -0.073‘h - 0 0 1 6 -0.082 +0.003 -0.012 ‘0.008 +0.007 -0.010 -0.049 -0.014 -0.071% -0.012 -0.059 +0.011 -0.002 +0.018 +0.008 -0.009 - 0 0 4 4 -0.011 -0.054

Center wall

F = coef f r o mL = coefficients from Table 5 for Table 2 for

bla = 2.50 and c/a = 1.25 bla = 2.50 L L-F

XIBy = o y = b / 4 y = b/2 y = b / 2 3

4 MY 4 MY MI MY 4 MY 4 MY0 0 +0.069 0 +0.035 0 -0.092 0 -0.138 0 -0.107

‘74 +0.026 to.059 to.015 to.034 -0.018 -0.089 -0.026 -0.132 -0.021 -0.103‘h +0.045 +0.048 io.031 +0.031 -0.016 -0.082 -0.023 -0.115 -0.018 -0.093% +0.044 +0.029 +0034 +0.020 -0.012 -0.059 - 0 0 1 6 -0.078 -0.013 -0.065

‘VLLgFig. 5.

._r

12

Fig. 6.

Note that f coefficients in this tabulation are used

only for calculation of coefficients L-L-that are to be3

used for design at the intersection of the center andouter walls as shown in Fig. 5a.

Coefficients for the center wall are for one cell filled,the negative sign indicating tension on the loaded side.All signs must be reversed when the other cell is filled.

Shear coefficients in Tables 7 and 8 as well as in Figs.1 and 2 can be applied both to center and outer walls.

Two-Cell Tank, Short Center Wall

M, coefficient thatoccurs at the center wall of -0.138 instead of -0.092 atthe corner in the tank in Fig. 6. Maximum moment is M,, =-0.1 38wa3 = -0.138 x 256 = -35.3 ft kips.

Short outer walls

L = coefflcents from Table 5 forb/a = 2 50 and c/a = 1 25

Long outer walls

F = coel fromL = coeffxwXs from Table 5 for Table 2 for

b a = 2 50 and cia = 1 25 b/a = 2 50y = b/2 y = b / 4 y=o y = b/2

0 lo -0 138

I

-0092 I 0 +0035 I 0 10069 1 0‘1 -0016 - 0 069

I

+ o 015 +o 034 *O 026 to 059 -0 026 -0 1 3 2I -0 016 -0 062 +o 0 3 1 *o 0 3 1 ‘ 0 045 ‘ 0 046 -0 023 - 0 115

J/r -0 012 -0 059 +o 034 -0 020 ‘ 0 044 *o 029 - 0 0 1 6 -0 078

Center wall

F = cod fromL = coefflclents from Table 5 for Table 2 for

b/a = 2.50 and c a = 1 25 ba= 1 2 5L-F

z=o I = Cl4 z = Cl2 .? = Cl2 L-T

maMx Mz Mx Mz Mx Mz 4 MZ Mx Mz

0 0 -0.010 0 -0030 0 - 0 0 9 2 0 -0.034 0 - 0 0 7 3% - 0 0 0 2 - 0 0 0 3 -0.006 -0 024 - 0 016 -0 089 - 0 0 0 6 -0.042 -0.015 -0 073% +0 OQ6 10007 to.003 - 0 . 0 1 2 -0.016 -0 062 - 0 0 1 0 -0 049 -0.014 -0.071y. +0018 +ooo6 +0.011 -0002 - 0 0 1 2 - 0 0 5 9 -0.009 - 0 0 4 4 - 0 0 1 1 -0.054

I t I t I

Counterforted Tank Walls

In a tank or reservoir with large horizontal dimensions,say three or four times the height, and without a rein-forced concrete cover slab, it becomes necessary todesign walls as cantilevers or, when they are quite high,as counterforted walls. The slab in Fig. 8 is free at the topand may be considered fixed at the bottom. If counter-forts are spaced equidistantly, the slab may also betaken as fixed at counterforts. For this type of construc-tion, coefficients in Table 3 apply.

Fig. a.Consider for illustration a wall panel of a counterfort-

ed wall in which spacing of counterforts is b = 40 ft andheight is a = 20 ft. From Table 3, for b/a = 40/20 = 2,select the following coefficients.

r I ty = o y = b/4 I y = b/2

Mx MY 4 MV f"% MV

Fig. 7.

1 3

‘.

The tank in Fig. 7 consists of two cells with the sameinside dimensions as the cells in the two-cell tank withthe long center wall. The difference is that the centerwall is 40 ft wide in the previously discussed tank, but20 ft wide in this example.

Design procedure is identical for both two-cell tanks,but the schedule of coefficients is different because thelonger side of the cell in Fig. 7 is continuous instead ofthe shorter side as in Fig. 6.

Note from the following tabulation that the long wallmust be designed for a maximum

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 constructionjoint 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 frictionbetween 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 becausethe wall is not designed for such moments.

Fig. 10.

14

Metric Conversion Factors

To convert from To Multiply byinch (in.) meter (m) 0.0254

feet (ft) meter (m) 0.3048

square feet (sq ft) square meter (m2) 0.0929

pound (lb) kilogram (kg) 0.4536

kip (1000 lb) kilogram (kg) 453.6

Ib/lin ft kg/m 1.488

kip/lin ft kg/m 1488.

Ib/sq ft kg/m2 4.88

Ib/cu ft kg/m3 16.02

ft-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 Prefix Symbol

1 000 000 000 = 109 giga G1 oooooo= 106 mega M

1000=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

Advanced Reinforced Concrete Analysis and DesignDesign According to ACI-318 2005 This book presents...

Documents

Transcript of Advanced Reinforced Concrete Analysis and DesignDesign According to ACI-318 2005 This book presents...