Advanced Reinforced Concrete Analysis and Design

2008

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

Eng. Mohammed Osama Yousef 8/14/2008

Chapter One

2

Water Tanks:

WATER TANK

BASED ON PLACEMENT OF TANK

BASED ON SHAPE OF TANK 1. CIRCULAR

1. RESTING ON GROUND

2. RECTANGULAR

2. UNDER GROUND

3. SPHERICAL

3. ELEVATED

4. INTZ 5. CONICAL BOTTOM

Resting on ground

Elevated

Under Ground

Circular

Advanced Reinforced Concrete Analysis & Design

Eng. Mohammed Osama yousef

Chapter One

Rectangular

3

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 freezethaw 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 fc' = 30MPa fc' = 35 MPa fc' = 40 MPa fc' = 45 MPa fc' = 50 MPa

Permissible Stress in tension Direct (direct Due to Bending tension) (rupture) 1.3 MPa 1.7 MPa 1.4 MPa 1.8 MPa 1.5 MPa 2.0 MPa 1.6 MPa 2.2 MPa 1.7 MPa 2.4 MPa


Shear Stress 1.9 MPa 2.0 MPa 2.2 MPa 2.5 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)] Advanced Reinforced Concrete Analysis & Design


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 fs3) 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)



Chapter One


20


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 wuH3 by the coefficients from table A-7. The value of wu for flexure = sanitary coefficient x (1.7 x lateral forces) Therefore, wu = 1.3 x 1.7 x 10 = 22.1 kN/m3 Therefore wuH3 = 22.1 x 63 = 4988.7 kN-m/m Mu = 0.005 x 4988.7 = 24.9 kN.m The figure includes the moment 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

r=

1 æ 2mR ö ´ çç 1 - (1 )÷ m è fy ÷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

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

D ; Quantity of reinforcement required in form 2 T gHD / 2 of hoop steel is computed as A st = or = s st s st T = gH

T

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

T

sc =

0.3 % (minimum)

T gHD / 2 If sc £ scat = A c + ( m - 1) A st 1000 t + ( m - 1) A st

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) ii)

Plan at base 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 D =

4A = 12.42m »13 m p

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



Chapter One

27

Design of Vertical wall: Max hoop tension at bottom T = gH

Area of steel Ast =

D 10 ´ 3.3 ´ 13 = = 214.5kN 2 2

T T 214.5 ´ 103 = = = 1650 mm 2 s st s st 130

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=

280 280 = = 13.33 3s cbc 3 ´ 7

Stress in concrete s c =

T 214.5 ´ 103 = = 1.16 N/mm 2 1000t + (m - 1) Ast 1000 ´ 160 + (13.33 - 1)2010

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

Ast= T/sst

T = gHD/2 (kN)

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



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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



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Rectangle Tank Behavior: Mx = moment per unit width about the xx-axis axis stretching the fibers in the y direction when the plate is in the x-yy plane. This moment determines the steel in the y (vertical direction). My = moment per unit width about the yy-axis stretching the fibers in the x direction when the plate is in the x-yy plane. This moment determines the steel in the x (horizontal direction). Mz = moment per unit width about the zz-axis axis stretching the fibers in the y direction when the plate is in the y-zz plane. This moment determi determines nes the steel in the y (vertical direction).

Mxy or Myz = torsion or twisting moments for plate or wall in the xx-y and y-zz planes, respectively. All these moments can be computed using the equations ·

Mx=(Mx Coeff.) x q a2/1000

·

My=(My Coeff.) x q a2/1000

·

Mz=(Mz Coeff.) x q a2/1000

·

Mxy=(Mxy Coeff.) x q a2/1000

·

Myz=(Myz Coeff.) x q a2/1000

These coefficients are presented in Tables 2 and 3 for rectangular tanks The shear in one wall becomes axial tension in the adjacent wall. Follow force equilibrium explain in class.



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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

ℎ

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 = , 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. TT C

B p=gH

A

T B

D

L PLAN @ BASE

T

FBD OF AD IN PLAN T FBD OF AB IN PLAN 0.5b

+

0.5b

y a

-

Bending moment diagram ` Maximum vertical moment = Mxgwa3

x ( 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; A st1 =

M' s st jd d

Ast2=T/sst

D/2

T x

M’=Maximum horizontal bending moment – T x x= d-D/2



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:

E C

A

Free a=H=3m F Fixed D

b=4m

B

L=5m 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



Chapter One

34

Design constants: scbc= 7 MPa

k=

sst = 130 MPa

m = 13.33

ms cbc = 0.41 ms cbc + s st

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

M 19.98x10 6 d= = = 131.8mm Qb 1.15x1000 Assuming effective cover as 33mm, the thickness of wall is t = 131.88+33 = 164.8 mm » 170 mm dprovided = 170-33 = 137mm

M 19.98 x10 6 Ast = = = 1289.5mm 2 s st jd 130 x0.87 x137 Spacing of 14 mm diameter bar =

154 x1000 = 119 mmc / c (Max spacing 3d=411mm) 1289.5

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 Spacing of 8 mm diameter bar =

50.24 x1000 = 100 mmc / c 510

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

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. A

14.4

C

1 1 ; K AC = ; 5 5 1/ 5 = = 0.44 9 / 20 1/ 4 = = 0.56 9 / 20

K AC =

11.88

DFAC B

DFAB

9

å K = 20

Moment distribution Table Joint

A

Member

AC

AB

DF

0.44

0.56

FEM

-14

11.88

0.9328

1.1872

-13.0672

13.0672

Distribution Final Moment

The tension in the wall is computed by considering the section at height H1 from the base. Where, H1 is greater of i) ii)

H/4 1m

Ex: i) ii)

3/4=0.75 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 Net bending moment M’ = M-Tx,

36

where: x= d-D/2 = 137-(170/2) = 52mm

M’ =13.0672-50 x 0.052 = 10.4672 kN-m

Ast1 =

10.4672 x10 6 = 675.5mm 2 130 x0.87 x137

Ast 2 =

50 x103 = 385mm 2 130

Ast= Ast1 + Ast2 = 1061 mm2 Spacing of 12 mm diameter bar =

113 x1000 = 106 .5 mmc / c (Max spacing 3d=411mm) 1061

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.



CO~r@tC 1

INFORMATION

Rectangular Concrete Tanks While cylindrical shapes may be structurally best for tank construction, rectangular tanks frequently are preferred for specific purposes. Special processes or operations may make circular tanks inconvenient to use. When several separate cells are required, rectangular tanks can be arranged in less space than circular tanks of the same capacity. Tanks or vats needed inside a building are therefore often made in rectangular or square shapes. For these and other reasons, breweries, tanneries, and paper mills generally use rectangular tanks. Data presented here are for design of rectangular tanks where the walls are subject to hydrostatic pressure of zero at the top and maximum at the bottom. Some of the data can be used for design of counterforted retaining walls subject to earth pressure for which a hydrostatic type of loading may be substituted in the design calculations. Data also can be applied to design of circular reservoirs of large diameter where lateral stability depends on the action of counterforts built integrally with the wall. Another article on tank construction, Circular Concrete Tanks Withouf Prestressing, has been published by the Portland Cement Association.

Moment

Coefficients

Moment coefficients were calculated for individual panels considered fixed along vertical edges, and coefficients were subsequently adjusted to allow for a certain rotation about the vertical edges. First, three sets of edge conditions were investigated, in all of which vertical edges were assumed fixed while the other edges were as follows: 1. Top hinged-bottom hinged 2. Top free-bottom hinged 3. Top free-bottom fixed* Moment coefficients for these edge conditions are given in Tables 1, 2, and 3, respectively. In all tables, a denotes height and b width of the wall. In Tables 1, 2, and 3, coefficients are given for nine ratios of b/a, the limits being b/a = 3.0 and 0.5. The origin of the coordinate system is at midpoint of the top edge; the Y axis is horizontal; the X axis is vertical and its positive direction downward. The sign convention for bending moments is based on the coordinate fiber that is being stressed. For example, A$ stresses fibers parallel to the X axis, The sign convention used here is not compatible with two other conventions-namely, that (1) the subscript is the axis 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-except where they are known to be zero-at edges, quarter points, and midpoints both in X and Y directions. The slab was assumed to act as a thin plate, for which equations are available in textbooks such as Theory ot Plates and Shells by S. Timoshenko,” but since only a small portion of the necessary calculations for moment coefficients for specific cases is available in the engineering literature, they have been made especially for this text. Table 4 contains moment coefficients for uniform load on a rectangular plate considered hinged on all four sides. The table is for designing cover slabs and bottom slabs for rectangular tanks with one cell. If the cover slab is made continuous over intermediate supports, the design can follow procedures for the design of slabs supported on four sides. Coefficients for individual panels with fixed side edges apply without modification to continuous walls provided there is no rotation about vertical edges. In a square tank, therefore, moment coefficients can be taken directly from Tables 1, 2, or 3. In a rectangular tank, however, an adjustment must be made, as was done in Tables 5 and 6, similar to the modification of fixed-end moments in a frame analyzed by moment distribution. In this procedure the common-side edge of two adjacent panels is first considered artificially restrained so that no rotation can take place about the edge. Fixededge moments taken from Tables 1,2, or 3 are usually dissimilar in adjacent panels and the differences, which correspond to unbalanced moments, tend to rotate the edge. When the artificial restraint is removed these unbalanced moments will induce additional moments in the panels, Adding induced and fixed-end moments at the edge gives final end moments, which must be identical on both sides of the common edge. Moment distribution cannot be applied as simply to continuous tank walls as it can to framed structures, because moments must be distributed simultaneously along the entire length of the side edge so that moments become equal at both sides at any point of the edge. The problem was simplified and approximated to some extent by distributing moments at four points only: quarter points, midpoint, and top. The end moments in the two intersecting slabs were made identical at these four points and moments at interior points adjusted accordingly. ‘Applicable tn cases where wall slab, counterfort, and base slab are a l l built Integrally “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

Xl

_wa

bla -

y - o xla

y = b/4

MX

MV

3.00

+0.035 +0.057 +0.051

+0.010 +0.016 +0.013

2.50

to031 +0.052 +0047

2.00

y = b/2 4

Y

MV

+0026 +0.044 +0.041

+0.011 +0.017 +0.014

-0.008 -0.013 -0.011

-0.039 -0.063 -0055

+0.011 +0.017 +0.015

+0.021 +0.036 +0.036

+0010 +0.017 +0.014

-0.008 -0.012 -0.011

-0.038 -0.062 -0.055

+0.025 +0.042 +0.041

+0.013 +0.020 +0.016

+0.015 +0.028 +0.029

+0.009 +0.015 +0.013

-0.007 -0.012 -0.011

-0.037 -0.059 -0.053

1.75

to.020 +0.036 +0.036

+0013 +0.020 +0.017

+0.012 +0.023 +0025

+0.008 +0.013 +0.012

-0.007 -0.011 -0.010

-0.035 -0.057 -0051

1.50

+0.015 +0.028 +0.030

+0.013 +0.021 +0.017

+0.008 +0.016 +0.020

+0007 +0.011 +0.011

-0.006 -0.010 -0.010

-0.032 -0.052 -0.048

1.25

+o 009 +0.019 +0.023

+0.012 +0.019 +0.017

+0.005 +0.011 +0.014

+0.005 +o 009 +0.009

-0006 -0 009 -0.009

-0.028 -0.045 -0.043

1.00

+0.005 +0011 +0.016

+0.009 +0.016 +0014

+0.002 +0006 +0.009

+0.003 +0.006 +0.007

-0.004 -0.007 -0.007

-0020 -0.035 -0.035

0.75

+0.001 to.005 +0.009

+0.006 +0.011 +0.011

0 +0.002 +0.005

+0.002 +0.003 to.005

-0.002 -0.004 -0.005

-0.012 -0022 -0.025

0.50

0 +0.001 +0.004

+0.003 + 0.005 +0.007

0 +0.001 +0.002

+0.001 +0.001 +0.002

-0.001 -0.002 -0.003

-0.005 -0.010 -0.014

Minus s,gn lndlcates

vx

bla -

y - o 4 0

y = b/4 MY

Y 0

MY

to.028 kO.049 bO.046

to.070 +0.061 +0.049 +0.030

+0.015 +0032 +0.034

+0.027 +0.028 +0026 +0.018

2.50

0 bO.024 bO.042 bO.041

+0.061 +0053 +0.044 10027

0 +0.010 +0.025 +0.030

2.00

0 bO.016 kO.033 bO.035

+0.045 +0042 +0.036 r0.024

1.75

0 IO.013 bO.028 10.031

150

!

y = b/2 MX

4

-0.034 -0.027 -0.017

-0.196 -0.170 -0.137 -0.087

+0.019 +0.022 +0.022 +0.016

0 -0.026 -0023 -0.018

-0.138 -0132 -0.115 -0.078

0 +0.006 +0.020 +0.025

+0.011 +0.014 +0.016 +0.014

0 -0.019 -0.018 -0.013

-0.091 -0.094 -0.089 -0.065

+0.036 +0.035 +0.032 +0.022

0 +0005 10017 r0.021

+0008 +0.011 co.014 10012

0 -0015 -0015 -0.012

-0.071 -0.076 -0076 -0.059

0 moo9 bO.022 10.027

+0027 +0.028 +0.027 +0020

0 +0.003 +0012 +0.017

+0.005 +0008 +0.011 +0.011

0 -0012 -0013 -0.010

-0052 -0.059 -0.063 -0.052

1.25

0 10.005 bO.017 10.021

co.017 +0020 +0023 +0.017

0 +0.002 +0.009 10013

+0.003 +0.005 +0.009 +0.009

0 -0008 -0010 -0.009

-0034 -0.042 -0.049 -0.044

1.00

0 to.002 '0 010 too15

+O.OlO +0013 +0.017 +0.015

0 +o 000 +0005 +o 009

+0.002 io.003 +0006 +0.007

0 -0005 -0.007 -0007

-0.019 -0.025 -0.036 -0036

075

0 ~0.001 to.005 ~0.010

+0005 +0008 +0011 +0.012

0 +o.ooo +0002 +0.006

+0001 +0.002 +0004 +0.004

0 -0.003 -0004 -0.005

-0.008 -0013 -0022 -0026

0.50

0 ~0.000 bO.002 a007

+0.002 +0004 +0.006 +0008

0 +o.ooo +0001 +0.002

0 +0001 +0.002 +0.002

0 -0.001 -0002 -0.003

-0.003 -0.005 -0.010 -0014

3.00

ienslon on the loaded side I” all tables

0

~ / ,"

:

Table 4

Table 3

Moment = Coef. x

wa3

ID;;

,w” 4 xla

q

Coef. x

bvaz

Xl

i~hngea 1 - y ‘_I I

M

y = b/4

y=o bla

Moment

X

m

1P”@

y=o

y = b/2

y = b/4

bla

MX

MY

Mx

4

M*

MY

0 +0.010 +0.005 -0.033 -0.126

+0025 +0.019 +0.010 -0.004 -0.025

0 +0.007 +0.008 -0.018 -0.092

to.014 +0.013 +0.010 -0 000 -0.018

0 -0014 -0.011 -0.006 0

-0.082 -0.071 -0055 -0.028 0

0 +0.012 +0.011 -0.021 -0.108

+0.027 +0.022 +0.014 -0.001 -0.022

0 +0.007 +0.008 -0.010 -0.077

+0.013 +0.013 +0010 +0.001 -0.015

0 -0.013 -0.011 -0.005 0

-0.074 -0.066 -0.053 -0.027 0

0 +0.013 +0.015 -0.008 -0.086

+0.027 +0.023 +0.016 +0.003 -0.017

0 +0.006 +0.010 -0.002 -0.059

+0.009 +0.010 +0.010 +0.003 -0.012

0 -0.012 -0.010 -0.005 0

-0.060 -0.059 -0 049 -0.027 0

0 +0.012 +0.016 -0.002 -0.074

+0.025 +0.022 +0016 +0.005 -0.015

0 +0.005 +0.010 +0.001 -0.050

+0.007 +0.008 +0.009 +0.004 -0.010

0 -0010 -0 009 -0.005 0

-0.050 -0.052 -0.046 -0.027 0

0 to.008 to.016 to.003 -0.060

+0.021 +0.020 +0.016 +0006 -0.012

0 +0.004 +0.010 +0.003 -0.041

+0.005 +0.007 +0.008 +0.004 -0.008

0 -0.009 -0.008 -0005 0

-0.040 -0.044 -0042 -0.026 0

1.25

0 to.005 +0.014 +0.006 -0.047

+0.015 +0.015 +0.015 +0.007 -0.009

0 +0.002 +0.008 +0.005 -0031

+0.003 +0.005 +0.007 +0.005 -0.006

0 -0.007 -0.007 -0.005 0

-0.029 -0.034 -0.037 -0024 0

1.00

0 to.002 +0.009 +0.008 -0.035

+0.009 +0.011 +0.013 +0.008 -0.007

0 0 +0.005 +0.005 -0.022

+0.002 +0003 to.005 +0.004 -0.005

0 -0.005 -0.006 -0.004 0

-0018 -0.023 -0.029 -0.020 0

0.75

0 +0.001 +0.005 +0.007 -0024

+0.004 +0.008 +0.010 +0.007 -0.005

0 0 +0.002 +0.003 -0.015

+0.001 +0.002 +0.003 +0.003 -0.003

0 -0.002 -0.003 -0.003 0

-0007 -0.011 -0.017 -0.013 0

0.50

0 0 +0.002 +0.004 -0.015

+0.001 +0.005 +0.006 +0.006 -0.003

0 0 +0.001 +0.001 -0.008

0 +0.001 +0.001 +0.001 -0.002

0 -0.001 -0.002 -0.001 0

-0002 -0.004 -0.009 -0.007 0

Mx

MY

MX

3.00

IO.089 to.118

to022 +o 029

+0077 +0101

to.025 to.034

2.50

bO.085 to112

+0.024 +0.032

+0.070 +o 092

+0.027 +0.037

2.00

0076 +0.100

+0.027 +0.037

+0.061 +0.078

+0.028 +0038

1.75

+0.070 +0091

+0.029 +0040

+0.054 +0.070

+o 029 +0.039

1.50

to.061 t0.076

+0.031 +0.043

+0047 +0.059

to.029 +0.040

125

to 049 +0063

+0.033 +0.044

+0038 +0.047

+0.029 +0.039

1.00

to.036 10.044

+0.033 +0044

+0.027 +0033

+0027 +0036

0.75

to.022 to.025

+0.029 +0.038

+0.016 +0.018

+0023 +0.030

to.020 +0025

+0007 +0.007

bO.015 +0.019

0.50

.O.OlO -0 009

M”

3

I

Table 5. Moment Coefficients for Tanks with Walls Free at Top and Hinged at Bottom b/a = 3.0 y = o c/a Mx

M”

M,

4

4

4

4

MZ

3.00

0 +0.028 +0.049 +0.046

+0.070 +0.061 +0.049 +0.030

0 +0.015 +0.032 +0.034

+0.027 +0.028 +0026 +0.018

0 -0.034 -0.027 -0.017

-0196 -0170 -0137 -0087

0 +0.015 +0.032 +0.034

+0027 +0026 +0.026 +0.018

2.50

0 +0.028 +0.049 +0.046

+0.073 +0.063 +0.050 +0.030

0 +0.016 +0.033 +0.037

+0033 +0033 +o 029 +0.020

0 -0030 -0.025 -0.017

-0.169 -0.151 -0.126 -0.084

0 +0.009 +0.023 +0.029

0 +0.029 +0.050 +0.046

+0.075 +0.065 f0.051 +0.031

0 +0.017 +0.035 +0.037

+o 039 +0.036 +0.032 +0.021

0 -0.027 -0.023 -0016

-0.146 -0133 -0.113 -0078

0 +0.029 +0.050 +0.046

+0.076 +0065 +0.052 +0.031

0 +0.018 +0036 +0037

+0041 +0038 +0033 +0.021

0 -0025 -0.021 -0.015

0 +0.030 +0050 +0.046

+0.077 +0.066 +0.053 +0031

0 +0.018 +0.037 +0.038

+0.043 +0.039 +0.034 +0.022

0 +0.030 +0.050 +0.047

+0.078 +0.067 +0.054 +0.032

0 +0.019 +0.038 +0.038

0 +0.030 +0.051 +0.047

+0.079 +0067 +0054 +0032

0 +0.029 +0.051 +0.047 0 +0.029 +0.050 +0.046

-

R

Moment

q

z = cl4

y =b/2

y = b/4

Coef. x wa3

1.25

$

I=0 M,

M,

0 +0028 +o 049 +0046

+0.070 +0061 to.049 +0030

to013 +0.014 +0.017 +0014

0 +0022 +0041 +0040

+0057 +0050 +0043 +0027

0 +0.002 +o.o1f3 +0.022

-0.005 -0.002 +0.005 +0008

0 +0.013 10030 +0034

+0031 +0.032 +o 029 +0020

-0.137 -0.125 -0.106 -0074

0 -0.003 +0011 +0.018

-0.018 -0.012 -0.003 +0.004

0 +0.007 +0.023 +0.027

+0014 +0.018 +0.020 +0015

0 -0.024 -0.020 -0.014

-0.129 -0.118 -0.100 -0.070

0 -0.007 +o 005 +0013

-0033 -0.024 -0.012 0

0 +0.002 +0.015 +0021

-0.006 +0.004 +0.010 +0.010

+0.045 +0.041 +0.035 +0.023

0 -0022 -0.019 -0.014

-0.122 -0111 -0 095 -0.068

0 -0.011 0 +0.008

-0.052 -0 039 -0.022 -0006

0 -0.004 +0.008 +0.016

-0.031 -0.018 -0.005 +0.001

0 +0.020 +0.038 +0.038

+0.047 +0.043 +0.036 +0.023

0 -0.021 -0.018 -0.013

-0118 -0105 -0.090 -0.065

0 -0.015 -0005 to003

-0074 -0.056 -0.034 -0.014

0 -0.010 +0001 +0.009

-0.060 -0.042 -0.022 -0.009

+o 079 +0.066 +0.053 +0.031

0 +0.020 +0.037 +0.037

+0.047 +0.042 +0.036 +0.022

0 -0.021 -0.018 -0.013

-0.120 -0107 -0 090 -0066

0 -0.020 -0.011 -0.002

-0 098 -0 079 -0.051 -0.025

0 -0.016 -0.006 +0.003

-0.092 -0.070 -0045 -0024

+0.078 +0.065 +0.053 +0.031

0 +0.019 +0.035 +0.036

+0.047 +0.042 +0.035 +0.021

0 -0.023 -0.019 -0.014

-0.130 -0.115 -0.095 -0.068

0 -0.024 -0.016 -0.007

-0.126 -0105 -0.073 -0.040

0 -0.022 -0.013 -0.004

-0123 -0101 -0071 -0042

b/a = 2.5

2.50

1.25

4

y = b/2

y = b/4 c/a

z=o

z = cl4

MX

MY

4

MY

4

M*

M”

MZ

0 +0.024 +0042 +0.041

+0.0.!31 +0.053 +0.044 to.027

0 +0010 +0025 +0.030

co.019 +0.022 +0.022 +0016

0 -0.026 -0023 -0.016

-3138 -0132 -0.115 -0.078

0 +0010 +0025 +0.030

+0.019 +0.022 +0.022 +0.016

0 +0.024 +0042 +0.041

+0.061 +0053 +0.044 +0027

0 +0.025 +0.043 +0042

+0065 +0.055 +0.046 +0.028

0 +0.012 iO.028 +0.031

+0.026 to.027 +0.025 +0.018

0 -0.023 -0.020 -0014

-0.118 -0.113 -0.102 -0.070

0 +0005 +0.018 +0023

+0.003 +0.006 +0.011 +0.011

0 +0.015 +0.032 +0034

+0.038 +0.037 +0.033 +0.022

0 +0.025 +0.044 +0043

.-.,7 +0.057 +0.047 +0.028

0 +0.013 +0.029 +0.033

+0.030 +0.030 +0.027 +0.019

0 -0.021 -0.019 -0013

-0.108 -0.104 -0.096 -0.066

0 +0.001 +0.013 +0019

-0.006 -0002 +0004 +0.008

0 +0010 to.025 +0.029

+0025 +0026 +0025 +0.019

0 +0.026 +0.045 +0043

to.068 +0.058 +0047 +0.029

0 +0.014 +0.030 +0.034

+0033 +0032 +0.029 +0.019

0 -0.019 -0.018 -0.013

-0100 -0.097 -0.089 -0.063

0 -0.003 +0.008 too15

-0018 -0.012 -0.002 +0004

0 +0004 +0017 +0024

+0.008 +0013 +0.017 +0015

0 +0.026 +0.045 +0.044

+0 069 +0.059 +0.048 +0.029

0 +0.015 +0.031 +0.034

+0035 +0034 +0031 +0.020

0 -0.018 -0.016 -0012

-0.092 -0.089 -0.082 -0 059

0 -0.006 +0.003 +0011

-0.030 -0.024 -0012 -0002

0 -0.002 +0.008 +0.018

-0010 -0.003 +0.007 +0008

0 +0.026 +0.046 +0.044

+0.070 +0.060 +0.048 +0.029

0 +0.015 +0.031 +0.033

+0037 +0036 +0.032 +0021

0 -0.017 -0.015 -0.011

-0087 -0.083 -0077 -0056

0 -0010 -0.003 +0.006

-0.045 -0.036 -0.021 -0.008

0 -0.008 -0.001 +0011

-0032 -0021 -0008 0

0 +0.025 +0.045 +0043

+0.070 +0.060 +0.047 +0.029

0 to.015 +0.030 +0.033

+0.038 +0.037 +0.032 +0.020

0 -0.016 -0.014 -0.011

-0082 -0.078 -0071 -0.054

0 -0014 -0.008 +0.002

-0062 -0053 -0.035 -0016

0 -0.014 -0.009 +o 005

-0.055 -0.042 -0.025 -0.011

0 +0.025 +0.044 +0.042

+0.069 +0.059 +0.046 +0.028

0 +0.014 +0028 +0.032

+0.039 +0.038 +0.032 +0.019

0 -0.015 -0.014 -0.010

-0080 -0075 -0.068 -0.052

0 -0.019 -0.014 -0.003

-0.081 -0.072 -0.056 -0030

0 -0019 -0017 -0002

-0.080 -0.068 -0.048 -0.026

\

t y = o

ca

2.0

=

y = b/2

y = b/4

M,

MY

2.00

0 +0016 +0033 +0.036

+0.045 +0.042 +0.036 +0.024

1.75

0 +0.017 +0.034 +0036

1.50

M, 0

M,

M, 0

z=o

z = cl4

MY

4

4

M,

0 +0.006 +0020 +0025

to011 +0.014 +0016 to014

+0.016 +0.033 +0.036

co.045 +0.042 +0.036 +0024

MZ

0

+0.006 +o.ozo +0.025

+0.011 +0.014 +0.016 +0.014

-0.019 -0.018 -0.013

-0091 -0.094 -0.089 -0.065

+0048 +0.044 +0.038 +0.024

0 +0.007 +0.021 +0.025

+0.015 +0.017 +0019 +0.015

0 -0.017 -0.017 -0.012

-0.081 -0.085 -0.083 -0061

0 +0.003 +0.015 +0.020

-0001 +0006 +0011 +0012

0 +0.012 +0.027 10.031

~0032 co.032 +0.029 +0021

0 +0018 +0.035 +0.036

+0.050 +0.046 +0.039 +0.025

0 +o.ooa +0.022 +0.026

+0019 +0.021 +0.021 +0.016

0 -0015 -0.015 -0.012

-0072 -0077 -0.076 -0.058

0 0 +o 009 +0.016

-0.010 -0002 +0004 +0.008

0 to.007 +0.020 +0.025

+0.018 '0020 co.022 10017

1.25

0 +0019 +0036 +0037

+0.052 +0.048 +0.041 +0.025

0 a009 +0.023 +0.026

+0023 +0.024 +0.023 +0.017

0 -0.014 -0.014 -0.011

-0064 -0.068 -0.069 -0054

0 -0.002 +0.005 +0.011

-0021 -0.013 -0004 +0.002

0 +0.001 +0011 +0016

0 '0005 +0012 +0.011

100

0 +0.019 +0.037 to.037

+0054 +0.050 +0.042 +0.026

0 +0.010 +0.024 +0.027

+0.027 +0.027 +0.025 +0.018

0 -0.012 -0.013 -0.010

-0.058 -0.062 -0.064 -0051

0 -0.005 0 +0006

-0037 -0025 -0.015 -0006

0 -0.005 +0.001 +0.008

-0.023 -0.013 0 to.004

075

0 +0.018 +0.038 +0.037

+0.055 +0.051 +0.043 +0.026

0 +0.011 +0.025 +0.027

+0.030 +0.029 +0026 +0.018

0 -0.012 -0.012 -0.010

-0.058 -0.062 -0.062 -0.049

0 -0.009 -0.005 +o.ooz

-0 049 -0040 -0 029 -0.015

0 -0010 -0.007 +0.001

-0.044 -0031 -0.015 -0.004

0.50

0 +0.018 +0.038 +0037

+0054 +0.052 +0.044 +0.026

0 +0.011 +0.025 +0.026

+0.030 +0.029 +0.025 +0.017

0 -0.014 -0.013 -0.010

-0.065 -0068 -0064 -0.050

0 -0012 -0010 -0.003

-0.064 -0056 -0.045 -0026

0 -0014 -0.012 -0.004

-0061 -0051 -0034 -0.018

I

”

w/a

1.5

I=

y = b/4

y-o

y = b/2

z=o

* = c/4

MY

M"

4

MX

MZ

-0.012 -0.013 -0.010

-0.052 -0.059 -0.063 -0.052

0 +0003 +0.012 +0.017

+0005 +0.008 +0.011 +0.011

0 +0.009 +0.022 +0.027

co.027 +0028 to.027 to.020

+0.008 +0.012 +0.014 +0.012

0 -0010 -0.011 -0.010

-0.045 -0.050 -0.056 -0.048

0 +0.001 +0007 60.013

-0005 -0.001 +0006 +0006

0 +0.004 +0.014 +0.018

coo11 +0015 +0.020 10016

0 +0.006 +0.015 to.019

+0013 +0.016 +0.017 +0.014

0 -0008 -0.010 -0.009

-0.038 -0.042 -0.049 -0.045

0 -0.002 +0.002 +0.008

-0.016 -0010 -0.003 +0.002

0 -0001 +0006 +o 009

-0.006 +0.001 +0010 +0.010

+0.038 +0.036 +0.033 +0.022

0 +0.007 +0.016 +0.019

+0.016 +0.018 +0.019 +0.015

0 -0.008 -0.008 -0.008

-0.034 -0.038 -0.042 -0.041

0 -0.005 -0.002 +0.003

-0024 -0.020 -0.014 -0.007

0 -0.004 -0.001 +0.002

-0.019 -0013 -0.004 +0001

+0.040 +0.037 +0.034 +0.022

0 +0.007 +0.017 +0.018

+0.017 +0.019 +0.020 +0.016

0 -0.008 -0.009 -0.008

-0.036 -0040 -0.044 -0.040

0 -0.008 -0.006 -0.002

-0030 -0031 -0.027 -0018

0 -0.007 -0.006 -0.004

-0.028 -0.027 -0020 -0010

M.

M ”I

1.50

0 +0.009 +0.022 +0.027

125

4

4

M

+0.027 +0.028 +0.027 +0.020

+0.003 +0.012 +0.017

+0.005 +0.008 +0.011 +0011

0 +0.010 +0.024 +0.027

+0.031 +0.031 +0.030 +0.021

0 +0.005 +0.014 +0.018

0 +0.011 +0.025 +0.028

+0.035 +0.034 +0.032 +0.022

0 +0.011 +0.025 +0028 0 +0.010 +0.024 +0.028

0

Y

0

b/a = 1.0 c/a

Y = b/4

y-o

x/a

M”

0 +0.002 +0.010 +0015

+0.010 +0.013 +0.017 +0.015

0 +0.003 +0.011 +0.016 0 +0.003 +0.012 +0.017

0

Y = b/2 Mb

0 +0.005 +0.009

+0.002 +0.003 +0006 +0.007

+0.016 +0.017 +0.020 +0.014

0 +0.001 +0.006 +0.009

+0.020 +0.018 +0.021 +0.013

0 +0.001 +0.008 +0.010

M”

0

z=o

z = c/4 M”

-0.005 -0.007 -0.007

-0.019 -0.025 -0036 -0036

+o.oc)7 +0.008 +0.009 +0.009

0 -0.004 -0.007 -0.006

+0.011 +0.010 +0.010 +0.009

0 -0.004 -0.006 -0.006

M”

0

M,

MX 0

MZ

0 +0.005 +0.009

+0.002 +0.003 +0006 +0.007

+0002 +0.010 +0.015

+0.010 f0.013 +0017 +0.015

-0.013 -0.020 -0.033 -0.032

0 -0.001 +0.002 +0.004

-0.004 -0.005 -0.001 +0.002

0 -0.001 +0.005 +0.009

+0.003 +0.003 +0.007 fO008

-0.011 -0.018 -0.032 -0.031

0 -0002 +0.001 +0002

-0.007 -0.012 -0.009 -0.005

0 -0.003 +0.002 +0.006

-0.005 -0.007 -0.005 +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

y=o

w/a

y = b/4

.?=O

z = c/4

y = b/2

MX

MY

Mx

MY

-0 008

-0 039

+0.011 +0.017 +0.014

f

MX

MI

+0035 +0057 +0051

+0.010 +0016 +0.013

3.00

+0.035 +0.057 +0.051

+0.010 +0.016 +0.013

+0.026 +0.044 +0.041

+0011 +0.017 +0014

-0.013 -0011

-0.063 -0.055

+0026 10044 +0041

2.50

+0.035 +0057 +0.051

+0.010 +0.016 +0.013

+0.026 +0.044 +0.041

+0.011 +0.017 +0.014

-0.008 -0.012 -0.011

-0.039 -0.062 -0055

+0.021 +0036 +0036

+0010 +0.017 to014

+0031 +0.052 +0047

+0011 +0017 +0.014

2.00

+0.035 +0.057 to.051

+0.010 +0.016 +0013

+0.026 +0045 +0.042

+0.011 +0.017 +0.014

-0.008 -0.012 -0011

-0.038 -0.062 -0.054

+0.015 +0.028 +0.029

to010 +0015 +0.013

+0025 +0043 +0041

to.013 to020 +0016

1.75

+0.035 +0.057 +0051

+0010 +0.015 +0013

+0027 +0.045 +0.042

+0.011 +0017 +0.014

-0.007 -0.012 -0.011

-0037 -0.060 -0053

+0011 +0021 +0024

+0008 +0.013 '0012

+0.020 +0036 +0.036

+0013 +0020 +0016

1.50

+0.035 +0.057 +0.051

+0.010 +0.015 +0.013

+0.027 +0.045 +0.042

+0.011 +0.017 +0.014

-0.007 -0.011 -0.010

-0.035 -0.057 -0.051

+0007 to.015 +0019

to.006 +0.010 +0011

to.014 +0027 to 029

+0013 +0020 to.017

1.25

+0.035 +0.057 +0.051

+0010 +0.015 +0.013

+0027 +0.046 +0.042

+0.011 +0.017 +0.014

-0.006 -0.011 -0.010

-0.032 -0.053 -0.048

+0.003 +0.008 +0013

+0.003 to.006 +0008

+0008 10017 +0.021

+0011 a017 +0.016

1.00

+0.035 +0.057 +0.051

+0.010 +0.015 +0013

+0.027 +0.046 +0.043

+0.011 +0.017 +0.014

-0.006 -0.010 -0.009

-0 029 -0.048 -0.044

-0001 to.002 +0.007

0 +0002 +0.004

to002 +0.007 +0.013

+0.008 +oc14 +0.013

0.75

+0.035 +0.057 +0.052

+0.010 +0.015 +0.013

+o.o2a +0.046 +0043

+0.011 +0.017 +0.014

-0.005 -0.008 -0.008

-0.025 -0.042 -0.039

-0.003 -0.003 +0.002

-0005 -0005 -0002

-0002 -0.001 +0.006

+0.001 +0.007 +0.007

0.50

+0036 +0.057 +0.052

+0.010 +0.015 +0.013

+0.028 +0.047 +0.043

+0.011 +0.017 +0.014

-0.004 -0007 -0.007

-0.021 -0035 -0.033

-0004 -0007 -0.004

-0.011 -0.016 -0.010

-0.005 -0.006 -0001

-0.008 -0010 -0.004

b/a = 2.5 y = b/2

c/a

6

z=o

z = c/4

4

M”

M,

MZ

4

MZ

2.50

+0.031 +0.052 +0.047

+0.011 +0.017 +0.015

+0.021 +0.036 +0.036

+0.010 to.017 +0.014

-0.008 -0.012 -0.011

-0.038 -0.062 -0.055

+0.021 +0036 co.036

+0010 +0017 +0.014

+0031 +0.052 +0.047

+0.011 +0.017 +0.015

2.00

+0.031 +0.052 +0.047

+0.011 +0.017 +0.015

+0.021 +0.036 +0.036

+0.010 to.017 +0.014

-0.008 -0.012 -0.011

-0.038 -0.061 -0.054

+0.015 +0.028 +0.029

+o 009 +0015 +0.013

+0025 +0042 +0.041

+0.012 +0.020 +0.016

1.75

+0.032 +0.052 +0.047

+0.011 +0.018 +0.015

+0.021 +0.036 +0.036

+0.010 +0.017 +0.014

-0.007 -0.012 -0.011

-0.037 -0.059 -0.053

+0011 +0.022 +0.024

+0.008 +0.013 +0.012

+0.020 +0035 +0.035

+0.012 +0021 +0.017

1.50

+0.032 to.052 +0.047

+0.011 +0.018 +0.015

+0.022 +0.037 +0036

+0.010 +0.017 +0.014

-0.007 -0.011 -0.010

-0.035 -0057 -0.051

+0.007 +0.015 +0019

+0.006 +0.010 +0.010

'0.014 +0.027 +0.029

+0013 +0.021 +0.017

1.25

+0.032 +0.052 +0.048

+0.011 +0.018 +0.015

+0.022 +0.038 +0.037

+0010 +0.017 +0.014

-0.006 -0.011 -0.010

-0.032 -0.053 -0.048

+0003 +0.008 +0.014

+0.004 +0.007 +0.008

+0.007 co.018 +0022

+0.012 +0019 to.018

100

+0.032 +0.053 +0.048

+0.011 +0.018 +0.015

+0.023 +0038 +0.038

+0.011 +0.017 +0.015

-0.006 -0.010 -0.009

-0028 -0048 -0044

-0.001 +0.002 to.007

0 +0002 +a004

+0.002 to007 +0013

+0.008 +0014 +0013

0.75

+0.033 +0.054 +0.049

+0.011 +0.018 +0.015

+0024 +0.039 +0.038

co.011 +0.017 +0.015

-0005 -0.008 -0.008

-0.024 -0041 -0.039

-0003 -0.003 0

-0.005 -0005 -0002

-0.002 0 10006

-0.002 +0005 +0006

0.50

+0.033 +0.054 +0.049

+0.012 +0.018 +0.015

+0.024 +0040 +o 039

+0.011 +0.017 to015

-0004 -0.007 -0.007

-0.021 -0035 -0034

-0004 -0007 -0.004

-0.011 -0016 -0.010

-0.005 -0.006 -0.001

-0.008 -0010 -0004

b/a = 2.0 c/a

y - o

l-

y = b/4

Mx

4

y = b/2

z=o

z = cl4

Mx

4

4

MY

4

4

M,

Mz

2.00

+0.025 +0.042 +0.040

+0.013 +0.020 +0.016

+0.015 +0.028 +0.029

+0.009 +0.015 +0.013

-0007 -0012 -0.011

-0.037 -0.059 -0.053

to015 +0028 +0.029

+0.009 +0.015 +0013

+0.025 +0.042 +0.040

+0.013 +0020 +0.016

1.75

+0.025 +0.042 +0.040

+0.013 +o.ozo +0.016

+0.015 +0.028 +0.029

+0.009 +0.015 +0.013

-0.007 -0.012 -0.010

-0.036 -0.058 -0052

+0.011 +0.022 +0.024

+0.008 +0.013 +0012

+0.020 +0.035 +0.035

+0.013 +0.021 +0.017

1.50

+0.025 +0.043 +0.041

+0.013 +0.020 +0.016

+0.016 +0.028 +0.029

+0.009 +0.015 +0.013

-0.007 -0.011 -0.010

-0034 -0.056 -0.050

+0.007 +0.015 +0.019

+0.006 +0.011 +0010

+0.014 +0027 +0.029

+0.013 +0.021 +0.017

125

+0.0?6 +0.043 +0.041

+0.013 +0.020 +0.016

+0.016 +0.029 +0.030

+0.010 +0.015 +0.013

-0.006 -0.010 -0.010

-0.032 -0052 -0.048

+0.003 +0.008 +0.013

+0003 +0007 +0008

+0.007 +0.018 +0.021

+0.011 +0.019 +0.016

1.00

+0.026 +0.044 +0.041

+0.013 +0.020 +0.016

+0.017 +0.030 +0.031

+0.010 +0.016 +0.014

-0.006 -0.009 -0.009

-0.028 -0.046 -0.044

-0.001 +0.002 +0.007

0 +0.002 +0.004

+0.002 +0.007 +0.013

+0.008 +0.014 +0.013

0.75

+0.027 +0.045 +0.042

+0.013 +0.020 +0.016

+0.018 +0031 +0.032

+0.010 +0.016 +0.014

-0.005 -0.008 -0.008

-0.024 -0.040 -0.041

-0.003 -0002 +0.002

-0.004 -0004 -0002

-0.001 0 to.005

+o.m2 +0.005 to.008

0.50

+0.027 +0.046 +0.042

+0.013 +0.020 +0.016

+0.019 +0.033 +0.032

+0.010 +0.017 +0.015

-0.004 -0.007 -0.007

-0.021 -0.034 -0.037

-0.004 -0.006 -0.003

-0010 -0015 -0010

-0.004 -0.006 -0.002

-0.007 -0.009 -0.003

-

-

b/a = 1.5 c/a

1

y - o

y = b/4

y = b/2

z = c/4

z=o

4

MY

4

MY

4

MY

4

MZ

4

4

150

+0015 +0.028 +0.030

+0013 +0.021 +0.017

+0.008 +0.016 +0.020

+0.007 +0.011 +0.011

-0.006 -0.010 -0.010

-0.032 -0.052 -0.048

+0008 +0.016 +0.020

+0.007 +0.011 +0.011

+0.015 +0.028 +0.030

+0.013 +0.021 +0.017

125

+0.016 +0.029 +0.030

+0.013 +0.021 +0.017

+0.009 +0.017 +0.020

+0.008 to.012 +0.012

-0.006 -0.010 -0.009

-0.029 -0.049 -0.045

+0.004 +0.009 +0.014

+0.004 +0008 +o 009

+0.009 +0.018 +0023

+0.012 +0.019 +0.016

+0.016 +0.030 to.031

+0.013 +0.021 to.017

+0.010 +0.019 +0.021

+0.009 +0.012 +0.013

-0005 -0.009 -0.008

-0.025 -0.043 -0.041

0 to.003 +0008

+0.001 +0.003 +0.005

+0.003 +0.008 +0.014

+0008 +0014 +0.014

+0.018 +0.032 +0.032

+0.014 +0.022 +0.018

+0.011 +0.021 +0.022

+0.010 +0.014 +0.014

-0.004 -0.007 -0.007

-0.021 -0036 -0.036

-0002 -0.002 +0.002

-0.003 -0.004 0

-0.001 +0.001 +0.006

to.002 +0.005 +0.008

+0.020 +0035 +0034

+0.016 +0.024 +0.020

+0.013 +0.023 +0.024

+0.012 +0.018 +0.016

-0.003 -0.006 -0.007

-0.017 -0.031 -0.033

-0.003 -0.006 -0003

-0 009 -0.014 -0.008

-0004 -0.005 -0.001

-0006 -0007 -0001

b/a = 1.0 c/a

y=o

y = b/4 4

z=o

I = c/4

y = b/2

4

M”

MY

M.

M”

M,

M.x

4

+0.005 +0.011 +0016

+0.009 +0.016 +0.015

+0.002 +0.006 +0.009

+0.003 +0.006 +0.007

-0.004 -0.007 -0.007

-0.020 -0.035 -0.035

+0.002 +0.006 +0.009

+0.003 +0.006 +0.007

+0.005 +0.011 +0.016

+0.009 +0.016 +0.015

+0.006 +0.013 +0.017

+0010 +0.017 +0016

+0.003 +0.008 +0.010

+0.004 +0.008 +0.008

-0003 -0.006 -0.006

-0.016 -0.029 -0.031

0 +0.001 +0.004

0 +0.001 +0.003

+0.001 +0.005 +0.008

+0.005 +o 009 +0010

to.007 +0.015 +0.018

+0.011 +0018 +0.016

+0.005 +0.010 +0.012

+0.006 +0.010 +0.010

-0.002 -0.004 -0.005

-0.010 -0.021 -0.026

-0.002 -0.003 -0.001

-0.005 -0.007 -0.004

-0.003 -0.003 0

-0.002 -0.002 +0.001

M”

7

triangle with the same area as the trapezoid representing the actual load distribution. The intensity of load is the same at middepth in both cases and when the wall is supported at both top and bottom edges, the discrepancy between triangle and trapezoid has relatively little effect at and near the supported edges

In this manner, moment coefficients were computed and are tabulated in Tables 5 and 6 for top and bottom edge conditions as shown for single-cell tanks with a large number of ratios of b/a and c/a, b being the larger and c the smaller of.the horizontal tank dimensions. Moments in vertical and horizontal directions equal the coefficients times wa3, in which w is the weight of the liquid. Note that the loading term is wa3 for all wall slabs subject to hydrostatic pressure but is wa2 for the floor slab in Table 4, which has uniformly distributed load. In the first case, w is weight per cubic foot, but in the latter it is weight per square foot. There is a peculiarity about the horizontal end moments in the slabs at the free top edge. Calculations of such moments by means of the trigonometric series used result in a value of zero, whereas these moments actually have finite values and may even be comparatively large. Horizontal end moments at the free edge were therefore established by extrapolation. The consistency of extrapolated moment coefficients was checked by plotting and studying curves. This gave reasonably good results, although coefficients thus determined are probably not quite as accurate as the coefficients that were computed. A condition prevails at the quarter point of the free edge, similar to that at the end point but to a lesser degree. At the midpoint of the free edge the coefficients were computed, extrapolation being used only for checking purposes. When a tank is built underground, the walls must be investigated for both internal and external pressure. The latter may be due to earth pressure or to a combination of earth and groundwater pressure. Tables and other data presented can be applied in ‘the case of pressure from either side but the signs are opposite. In the case of external pressure, actual load distribution may not necessarily be triangular as assumed in the tables. Consider for illustration a tank built below ground with earth covering the roof slab and causing a trapezoidal distribution of lateral earth pressure on the walls. In this case it gives a fairly good approximation to substitute a

Shear Coefficients Shear values along the edges of a tank wall are needed for investigation of shear and development stresses. Along vertical edges, shear in one wall is also used as axial tension in the adjacent wall and must be combined with bending moment to determine tensile reinforcement. Various data for shear were computed and are given in Table 7. The wall is considered fixed at the two vertical edges while top and bottom edges are assumed to be hinged. The wall panel with width b and height a is subject to hydrostatic pressure due to a liquid weighing w lb per cubic foot. The first five lines in Table 7 are shears per linear foot in terms of wa*. The remaining four lines are total shears in kips or pounds depending on how w is given. Shears per linear foot are for ratios of b/a = %, 1,2, and infinity. The difference between the shear for b/a = 2 and infinity is so small that there is no necessity for computing coefficients for intermediate values. When b/a is large, a vertical strip of the slab near midpoint of the b dimension will behave essentially as a simply supported one-way slab. Total pressure on a strip 1 ft wide is 0.50waz, of which two-thirds or 0.33wa2 is the reaction at the bottom support and one-third or 0.17wa2 is the reaction at the top. Note in Table 7 that shear at midpoint of the bottom edge is 0.3290waz for b/a = 2.0, the coefficient being very close to that of onethird for infinity. In other words, maximum bottom shear is practically constant for all values of b/a greater than

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

‘h

bla

1

2

5

10

lnfmtty

Midpoint of bottom edge Corner at bottom edge

+o 1407wa’ - 0 2575wa”

+o 2419we -0 4397wa’

+o 3290w.a’ - 0 5633w.F

+o 3333waz -0 6000wa’

M,dpo,“t of flxed side edge Lower third-pant of side edge Lower quarter-pant of side edge

+o 1260wa’ *o 173&v@ ‘0 1919wP

+O 2562wa’ +o 3113wa’ +o 3153w.e

+0.3604waz ‘0 4023wa’ ‘0 3904w.3’

‘0 3912w.a’ +0 4116wa’ ‘0 39t30wa.

Total at top edge Total at bottom edge Total at one foxed side edge Total at all four edges ‘Negatwe s,gn lndlcates tEsbmated

8

reaction

0 0 0 0

OOOOwa’b 0460wa-‘b 2260wazb 5000wa’b

acts I ” darectlon

0 0 0 0

0052wa’b 0960wa’b 1994wa’b 5000waJb

of load

0 0 0 0

0536w.+b 1616wa’b 1322wa’b 5000wa’b

0 0 0 0

,203~~b 2715wa’b 0541wa.b 5000wa.b

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

0 0 0 0

1667wa’b 3333wa.b 275wav 5000wa’b

\a J

As will be shown, this is correct only when the top edge is supported, not when it is free. At the corner, shear at the bottom edge is negative and numerically greater than shear at midpoint. The change from positive to negative shear occurs approximately at the outer tenth points of the bottom edge. These high negative values at the corners arise because deformations in the planes of the supporting slabs are neglected in the basic equations and are therefore of only theoretical significance. These shears can be disregarded in checking shear and development stresses. Unit shears at the fixed edge in Table 7 were used for plotting the curves in Fig. 1. There is practically no change in shear curves beyond b/a = 2.0. Maximum value occurs at a depth below the top somewhere between 0.6a and 0.8a. Fig. 1 is useful for determination of shear or axial tension for any ratio of b/a and at any point of a fixed side edge. Total shear from top to bottom of one fixed edge in Table 7 must equal the area within the corresponding curve in Fig. 1, and this relationship was used for checking the curves ‘Total shears computed and tabulated for a hanged top were also used in making certain adjustments to determine approximate values of shear for walls with free too-recorded in Table 8. For b/a = % in Table 7, total shear at the top edge is so small as to be practically zero, and for b/a = 1 .O total shear, 0.0052, is only 1% of total hydrostatic pressure, 0.5000.’ Therefore, it is reasonable to assume that removing the top support will not materially change total shears at any of the other three edges when b/a = Y2 and 1. At b/a = 2.0, there is a substantial shear at the top 2.

‘Loading

term is omitted here.

9" 0.3 g ij a4 f g 0.5 % a6 p B a7

1.0

0

0.1

0.2

0.3

Shear per lin. ft. = coef x wa2

Fig. 1.

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

tJa

Mldpolnt

1

2

3

‘ 0 141wa: -0 258wa-$

*o 242w.F -0 440-a.

‘ 0 3awa7 - 0 583~9

‘ 0 45wet

Top of flxed side edge Mldpomf of flxed side edge Lower third-wont of side edae

0 ooowa. +O 128wa‘

‘0 olowa’

*o 100wa

*O 258wa’ *0311wa-

.o 375wa: *o 406W%

-0 165wa-

*o 174ws

Lower quarter-WI”, of side edge Total at bottom edge Total at one faxed s,de edge Total al all four edges

‘0

*o 315w.T

*o 390wa

Corner

of bottom edge at

bottom

edge

192-a-

0 048wa b 0 226wa b

0 096wa.b 0 202wa.b

0 500w.s.

0 500wa-b

b

0 204wa.b 0 148wa:b 0 500wa. b

- 0 sowa ‘ 0 406WW *o 416w.F ,O 398wa

0 286wa.b 0 107wa.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 total shears on the other three sides is only 0.4462. If the top support is removed, the other three sides must carry a total of 0.5000. A reasonable adjustment is to multiply each of the three remaining total shears by 0.5000/ 0.4462 = 1 .12, an increase of 12%. This was done in preparing Table 8 for b/a = 2.0. A similar adjustment was made for b/a = 3.0, where the increase is 22%. Total shears recorded in Table 8 were used to determine unit shears also recorded in that table. Consider for illustration the shear curves in Fig. 1 and imagine the top is changed from hinged to free. As already stated, for b/a = % and 1 it makes little difference in total shearthe area within shear curves-whether the top is supported or not. Consequently, curves for b/a = % and 1 remain practically unchanged. They were transferred almost without modification to Fig. 2, which covers the case with top free. For b/a = 2 an adjustment was made. A change in the support at the top has little effect upon shear at the bottom of the fixed edge. Consequently, the curves in Figs. 1 and 2 are nearly identical at the bottom. Gradually, as the top is approached curves for the free top deviate more and more from those for the hinged top, as in Fig. 2. By trial, curve for b/a = 2 was so adjusted that its area equals the total shear for one fixed edge for b/a = 2.0 in Table 8. A similar adjustment was made for b/a = 3.0, which is the limit of moment coefficients given. One point of interest stands out In a comparison of Figs. 1 and 2. Whereas for b/a = 2.0 and 3.0 total shear is increased 12% and 22%, respectively, when top is free instead of hinged, maximum shear is increased but slightly, 2% at most. The reason is that most of the increase in shear is near the top where shears are relatively small. The same general procedure was applied, but not illustrated, for adjustment of unit shear at midpoint of bottom, but in this case the greatest change resulting from making the top free is at midpoint where shear is

9

Fig. 3.

0

0.1

@.2

0.3

04

Shear per lin ft. - coef Y wa2

Fig. 2. large for the hinged-top condition. For illustration, for b/a = 3.0, unit shear at midpoint of the bottom is0.33wa2 with hinged top but 0.45wa2 with free top-an increase of approximately one-third. Shear data were computed for wall panels with fixed vertical edges. They can be appli$d with satisfactory results to any ordinary tank wall even if vertical edges are not fully fixed.

Open-Top Single-Cell Tank The tank in Fig. 3 has a clear height of a = 16 ft. Horizontal inside dimensions are b = 40 ft and c = 20 ft. The tops of the walls are considered free and the bottom hinged. The tank contains water weighing 62.5 lb per cubic foot. Coefficients for moment and shear are selected from tables or diagrams for b/a = 40/16 = 2.50 and c/a = 20/l 6 = 1.25. Moments are in foot-kips if coefficients are multiplied by wa3/1000 = 62.5 x 163/1000 = 256; and shears are in kips if coefficients are multiplied by waz/lOOO = 62.5 x 16z/1000 = 16. Moment coefficients taken from Table 5 for b/a = 2.50 and c/a = 1 .25 are tabulated below. Coefficients for x = a (bottom edge), being equal to zero, are omitted.

0 'A H %

0 +0.026 +0.045 +0.044

+0069 GO59 +0046 to.029

0 +0015 +0.031 '0.034

+0035 +0034 +0.031 +0020

0 -0.016 -0.016 -0.012

-0.092 -0.089 -0.062 -0059

0 -0006 '0.003 +001,

-0.030 -0.024 -0.012 -0.002

-:002 +O.@X3 +o,o,i3

I:::;; +OOO, +0008

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

10

side and need not be considered In subsequent calculations. Maximum horizontal moment at midpoint of the longer wall is +0.069wa3 = +0.069 x 256 = +17.7 ft kips. The positive sign shows that tension is in the outside of the wall. There is also some axial tenslon on this section that can be taken equal to end shear at the top of the shorter wall. For use in connection with Fig. 2, ratio of b/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 negligible in this case and the steel area can be determined as for simple bending. Horizontally at x = a/2, axial tension taken from Fig. 2 for b/a = 1.25 is equal to N = -0.30wa2 = -0.30 x 16 = -4.80 kips per linear foot, which is not negligible. Moment is M = 0.048wa3 = 0.048 x 256 = 12.3 ft kips. In the shorter wall, positive moments are all relatively small. Maximum positive moment is vertical: 0.01 8wa3 = 0.018 x 256 = 4.6 ft kips. Maximum Mx in the vertical strip at midpoint ot longer panel is 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 one only in that the tops of the walls are considered hinged rather than free. This condition exists when the tank is covered by a concrete slab with dowels extending from the wall into the slab without moment reinforcement across the bearing surface. Moment coefficients taken from Table 6 are given below. All coefficients for x = 0 (top edge) and x = a (bottom edge), being equal to zero, are omitted.

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

A

,’

less for wall with free top than with hinged top. Another noteworthy point is that maximum M, coefficient at y = 0 is +0.069 for a free top but +0.018 for a hinged top. Adding top support causes considerable reduction 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 the section subject to this moment, and loads per linear foot can be taken as follows: 8-ft-high wall: 8 x 1 .08 x 0.150 = 1.3 kips 12-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 .O kips’ 6.8 kips It is conservative to check compressive stress forN = 6.8 kips and to design tensile steel for N = 1 .3 + 1.5 = 2.8 kips, in which fill and live load are disregarded.

t

Top and Base Slabs The closed single-cell tank is covered with a concrete slab. Assume the slab is simply supported along all four sides and has a live load of 100 psf and an earthfill weighing 300 psf. Estimating slab thickness as 12 in. gives a total design load of 100 + 300 + 150 = 550 psf. From Table 4, for a ratio of 40/20 = 2, select maximum coefficient of 0.100, which gives maximum M = 0.1 00wa2 F 0.100 x 0.550 x 20.02 = 22.0 ft kips. At the corners, a two-way slab tends to lift off the supports; and if this tendency is prevented by doweling slab to support, cracks may develop in the top of the slab across its corners. Nominal top reinforcement should therefore be supplied at the corners, say0.005bd sq in. per foot in each direction. Length of these bars can be taken as %a = l/4 x 20 = 5 ft. Assume the closed single-cell tank has a base slab of reinforced concrete. Weight of base slab and liquid does not create any bending or shearing stresses in concrete provided the subsoil is uniformly well compacted. Weight transferred to the base through the bottom of the wall is Top slab: 0.550 x 22 x 42 = 510 kips Walls: 16x0.162(2x41.1 +2~21,1)=320kips 830 kips If 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 base slab just as w = 550 psf was used for design of the top slab. 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 pressure on the base slab, the slab should also be investigated for this pressure when the tank is considered empty. \

Multicell Tank Multicell tanks do not lend themselves readily to mathematically accurate stress analysis It is possible, however, with the tables presented here for single-cell tanks and for individual wall panels with fixed vertical edges to estimate moment coefficients for symmetrical multicell tanks with sufficient accuracy for design purposes. While results obtained by the following procedure are approximate and should therefore be considered as a guide to engineering judgment, the procedure does give a conservative design. Because a rotation of one corner has comparatively little effect on moments at adjacent corners in atankwith wall panels supported on three or four sides, moments in the walls of a multicell tank are essentially the same as in single-cell tanks-except at corners where more than two walls intersect. Moment coefficients from Tables 5 and 6, designated as L coefficients, apply to outer or Lshaped corners of multicell tanks (see Fig. 4a) as well as to interior sections in all walls, that is, sections designated as y = b/4, y = 0, z = c/4, and z = 0. Moment coefficients for design sections at corners where more than two panels intersect depend on the loading condition producing maximum moment and on the number of intersecting walls. In Fig. 4b, three walls form a T-shaped unit. If the continuous wall, or top of the T, is part of the long sides of two adjacent rectangular cells, the moment in the continous wall at the intersection is maximum when both cells are filled. The intersection is then fixed and moment coefficients, designated as F coefficients, can be taken from Tables 1, 2, or 3, depending on edge conditions at top and bottom. These three tables cover panels with fixed side edges. If the continuous wall is part of the short sides of two adjacent rectangular cells, moment at one side of the intersection is maximum, when the cell on that side is filled while the other cell is empty. Likewise the end moment in the center wall is maximum when only one cell is filled. For this loading condition the magnitude of moment will be somewhere between theL coefficients and the F coefficients. If the unloaded third wall of the unit is disregarded, or its stiffness considered negligible, moments in the loaded walls would be the same as in Fig. 4a, that is, the L coefficients apply. If the third wall is assumed to have infinite stiffness, the corner is fixed and the f coefficients apply. The intermediate value representing more nearly the true condition can be obtained by the formula: End moments = L -nG2(i -F)

-

‘ProportIons of tank being deslgned are such that for determlning axial compression In sde walls, all the top load may be considered carned the short way

(4

b)

4

Fig. 4.

11

in which n denotes number of adjacent unloaded walls. This formula checks for n equal to zero and infinity. In an L-shaped unit n equals 0 and the end moments equal L - O(L - F) = L. Inserting n equal to infinity will give nl(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 intersecting walls are the walls of square cells, moments at the intersection are maximum when any two cells are filled and the F coefficients in Tables 1,2, or 3 apply because there is no rotation of the joint. If the cells are rectangular, moments in the longer of the intersecting walls will be maximum when two cells on the same sideof the wall under consideration are filled, and again the F coefficients apply. Maximum moments in the shorter walls adjacent to the intersection occur when diagonally opposite cells are filled, and for this condition the L coefficients apply. Fig. 5 shows moment coefficients at wall intersecitions in two- and four-cell tanks. Where coefficients are not 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, each with the same inside dimensions as the open-top singlecell tank and the closed single-cell tank. The top is considered free. In accordance with the types of units in Fig. 4, the tank consists of four L-shaped and two T-shaped units. L coeff icients from Table 5 for b/a = 2.50 and c/a = 1.25, and F coefficients from Table 2, for b/a = 2.50 and 1.25, are tabulated as follows: Long outer walls L = coeffuents from Table 5 for bla = 2 . 5 0 a n d cla = 1 . 2 5 y = b/2 x/a

%

Mv

-0018 -0016 -0.012

-0.092 -0.089 -0082 -0 059

0 I 0 ‘h %

I

y =b/4

I

M”

I

y = o

I

M.

Mv

0 to.015 to031 +0034

+0035 to.034 to.031 +0.020

Mx

Mv

I

0 +0.026 +0.045 I +0.044

to.069 +0.059 +o.o4a +0.029

._r

Short outer walls L = coefficients from Table 5 for b l a = 2 . 5 0 a n d c/a = 1 2 5 x/a 0

%

‘h %

.? = c/2 Mx

L = Cl4 Ml

0 -0.018 -0016 -0.012

I

-0.092 -0.089 -0.082 -0.059

F

=Tca~fetTfz b/a = 1 25

z=o

.? =c/2

4

Mz

MN

Mz

M.Y

0 -0.006 +0.003 +0.011

-0.030 -0024 -0.012 -0.002

0 -0002 ‘0.008 +0.018

-0.010 -0.003 +0.007 +0.008

0 -0.008 -0.010 -0.009

I

M, -0034 -0.042 -0.049 -0044

I L .L& 3 4

MZ

0 -0.015 -0.014 -0.011

-0.073 -0.073 -0.071 -0.054

Fig. 5.

Center wall F = coef f r o m Table 2 for bla = 2.50

L = coefficients from Table 5 for bla = 2.50 and c/a = 1.25 XIB 0 ‘74 ‘h %

12

y=o 4 0 +0.026 +0.045 +0.044

MY +0.069 to.059 +0.048 +0.029

y = b/4 4 0 to.015 io.031 +0034

y = b/2 MY

+0.035 to.034 +0.031 +0.020

MI 0 -0.018 -0.016 -0.012

MY -0.092 -0.089 -0.082 -0.059

y =b/2 4 0 -0.026 -0.023 -0016

MY -0.138 -0.132 -0.115 -0.078

L L-F 3 4 0 -0.021 -0.018 -0.013

MY -0.107 -0.103 -0.093 -0.065

‘VLLg

Short outer walls L = coefflcents from Table 5 for b/a = 2 50 and c/a = 1 25

Long outer walls

Fig. 6. y = b/2

Note that f coefficients in this tabulation are used only for calculation of coefficients L-L-that are to be 3 used for design at the intersection of the center and outer 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

‘.

The tank in Fig. 7 consists of two cells with the same inside dimensions as the cells in the two-cell tank with the long center wall. The difference is that the center wall is 40 ft wide in the previously discussed tank, but 20 ft wide in this example. Design procedure is identical for both two-cell tanks, but the schedule of coefficients is different because the longer side of the cell in Fig. 7 is continuous instead of the shorter side as in Fig. 6. Note from the following tabulation that the long wall must be designed for a maximum M, coefficient that occurs at the center wall of -0.138 instead of -0.092 at the corner in the tank in Fig. 6. Maximum moment is M,, = -0.1 38wa3 = -0.138 x 256 = -35.3 ft kips.

F = coel from Table 2 for b/a = 2 50

L = coeffxwXs from Table 5 for b a = 2 50 and cia = 1 25

0

-0092 I 0 069 + o 015 -0 062 +o 0 3 1 -0 059 I +o 034

lo

‘I1

I

J/r

y = b/4

-0016 -0 016

+0035 +o 034 *o 0 3 1 -0 020

-0

-0 012

y=o

I

0 *O 026 ‘ 0 045 ‘ 0 044

y = b/2

10069 to 059 ‘ 0 046 *o 029

1 0 -0 026 -0 023 -0016

-0 -0 -0 -0

138 132 115 078

Center wall F = cod from Table 2 for ba= 1 2 5

I = Cl4

.? = Cl2

z=o

ma 0 % % y.

L = coefflclents from Table 5 for b/a = 2.50 and c a = 1 25

I

Mx

Mz

0 -0002 +0 OQ6 +0018

-0.010 -0003 10007 +ooo6

Mx

t

z = Cl2

Mz

0 -0030 -0.006 -0 024 to.003 - 0 . 0 1 2 +0.011 -0002

I

L-F L-T

Mx

Mz

4

MZ

Mx

0 - 0 016 -0.016 -0012

-0092 -0 089 -0 062 -0059

0 -0006 -0010 -0.009

-0.034 -0.042 -0 049 -0044

0 -0.015 -0.014 -0011

t

I

Mz -0073 -0 073 -0.071 -0.054

Counterforted Tank Walls In a tank or reservoir with large horizontal dimensions, say three or four times the height, and without a reinforced concrete cover slab, it becomes necessary to design walls as cantilevers or, when they are quite high, as counterforted walls. The slab in Fig. 8 is free at the top and may be considered fixed at the bottom. If counterforts are spaced equidistantly, the slab may also be taken as fixed at counterforts. For this type of construction, coefficients in Table 3 apply.

Fig.

a.

Consider for illustration a wall panel of a counterforted wall in which spacing of counterforts is b = 40 ft and height is a = 20 ft. From Table 3, for b/a = 40/20 = 2, select the following coefficients. r

I

y=o Mx

MY

t

y = b/4 4

y = b/2

I MV

f"%

MV

Fig. 7.

13

Procedure for using these coefficients to determine moments and design of the wall is similar to that illustrated 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 assumption that the bottom edge is hinged. It is believed that this assumption in general is closer to the actual condition than that of a fixed edge. Consider first the detail in Fig. 9, which shows the wall supported on a relatively narrow continuous wall footing, and then Fig. 10 in which the wall rests on a bottom slab.

joint filler, but both iron powder and lead are not always readily available. A waterstop may not be needed in the construction joints when the vertical joint in Fig. 9 is made watertight. In Fig. 10 a continuous concrete base slab is provided either for transmitting the load coming down through the wall or for upward hydrostatic pressure. In either case, the slab deflects upward in the middle and tends to rotate the wall base in Fig. 10 in a counterclockwrse direction. The wall therefore is not fixed at the bottom edge. It is difficult to predict the degree of restraint. The rotation may be great enough to make the bottom edge hinged or may be even greater. Under the circumstances it is advisable to avoid placing moment reinforcement across the joint and to cross the dowels at the center. The waterstop must then be placed off center as indicated. Provision for transmitting shear through direct bearing can be made by inserting a key as in Fig. 9 or 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 except that the waterstop and the shear key are not essential. The main thing is to prevent moments from being transmitted from the top of the slab into the wall because the wall is not designed for such moments.

Fig. 9.

In Fig. 9 the condition of restraint at the bottom of the footing is somewhere between hinged and fixed but much closer to hinged than to fixed. Resultant of pressure on the subsoil lies well within the edge of the footing, and the product of resultant and its eccentricity is usually much smaller than the moment at the bottom of the wall when it is assumed fixed. Furthermore, thefooting must rotate about a horizontal axis in order to produce eccentric loading on the subsoil and rotation itself represents a relaxation of restraint. When the wall footing is not capable of furnishing much restraint, it is not necessary to provide for hinge action at the construction joint in Fig. 9. The dowels are close to the surface, leaving the center of the joint free for insertion of a shear key. Area of steel in the dowels along each face may be taken as not less thanO.O025bd, and extension of the dowels above the construction joint may be made not less than say 3 ft. The base slab in Fig. 9 is placed on top of the wall footing and the bearing surface is brushed with a heavy coat of asphalt to break the adhesion and reduce friction between slab and footing. The vertical joint between slab and wall should be made watertight. A joint width of 1 in. at the bottom and 1% in. at the top is considered adequate. As indicated in Fig. 9, the bottom of the joint may be filled with oakum, the middle with volcanic clay of a type that expands greatly when moistened, and the upper part sealed with mastic. Any leakage will make the clay penetrate into fissures and expand, plugging the leak. Mortar mixed with iron powder has been used extensively for joints such as in Fig. 9, and so has lead

14

Fig. 10.

Metric Conversion Factors To convert from inch (in.) feet (ft) square feet (sq ft) pound (lb) kip (1000 lb) Ib/lin ft kip/lin ft Ib/sq ft Ib/cu ft ft-kips ft-kips

To meter (m) meter (m) square meter (m2) kilogram (kg) kilogram (kg) kg/m kg/m kg/m2 kg/m3 newton-meter (Nm) kilogram-meter (kgm)

Multiply by 0.0254 0.3048 0.0929 0.4536 453.6 1.488 1488. 4.88 16.02 1356. 138.2

The prefixes and symbols listed are commonly used to form names and symbols of the decimal multiples and submultiples of the SI units. Multiplication Factor

Prefix

Symbol

1 000 000 000 = 109 1 oooooo= 106 1000=10~ l=l = 10-3 0.001 = 10-6 0.000 001 0.000 000 001 = 1 o-9

giga mega kilo

G M k m I-1 n

milli micro nano

Advanced Reinforced Concrete Analysis and Design

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