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Presentation on theme: "ENVIRONMENTAL ENGINEERING CONCRETE STRUCTURES"— Presentation transcript:

CE 498 – Design Project November 16, 21, 2006


3 INTRODUCTION 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.

4 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

5 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.

6 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 – in.

7 Design Methods 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 include 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.

8 DESIGN METHODS 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.

9 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. The tensile stress in the concrete (due to ring tension from pressure and shrinkage) has to kept at a minimum to prevent excessive cracking. The concrete tension strength will be assumed 10% f’c in this document. RC walls 10 ft. or higher shall have a minimum thickness of 12 in. The concrete wall thickness will be calculated as follows:

10 WALL THICKNESS Effects of shrinkage
Figure 2(a) shows a block of concrete with a re-bar. The block height is 1 ft, 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.

11 WALL THICKNESS 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)] The total concrete tension stress = [CEsAs + T]/[Ac+nAs]

12 WALL THICKNESS 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 to Use the value of C=0.0003 Assume fs= allowable steel tension =18000 psi Therefore, wall thickness t= T

13 WALL THICKNESS 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=24,000 lb, n=8, Es=29*106 psi, C= and Ac=12 x 10 = 120 in3 If the allowable steel stress is reduced from 20,000 psi to 10,000 psi, the resulting concrete stress is increased from 266 psi to 322 psi. Desirable to use a higher allowable steel stress.

14 REINFORCEMENT The amount size and spacing of reinforcement has a great effect on the extent of cracking. The amount must be sufficient for strength and serviceability including temperature and shrinkage effects The amount of temperature and shrinkage reinforcement is dependent on the length between construction joints

15 REINFORCEMENT 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 The size of reinforcing bars should not exceed #11. Spacing of re-bars should be limited to a maximum of 12 in. Concrete cover should be at least 2 in. In circular tanks the locations of horizontal splices should be staggered by not less than one lap length or 3 ft. Reinforcement splices should confirm to ACI 318 Chapter 12 of ACI 318 for determining splice lengths. The length depends on the class of splice, clear cover, clear distance between adjacent bars, and the size of the bar, concrete used, bar coating etc.

16 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.

17 CRACK CONTROL In ACI 350, the cover is taken equal to 2.0 in. for any cover greater than 2.0 in. Rearranging the equation and solving for the maximum bar spacing give: max spacing = z3/(2 dc2 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 115 k/in. For severe environmental exposures, z = 95 k/in.

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

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 = 20 ft. Diameter of inside = D = 54 ft. Weight of liquid = w = 62.5 lb/ft3 Shrinkage coefficient = C = Elasticity of steel = Es = 29 x 106 psi Ratio of Es/Ec = n = 8 Concrete compressive strength = f’c = 4000 psi Yield strength of reinforcement = fy = 60,000 psi

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 = 10 in. Therefore, H2/Dt = (202)/(54 x 10/12) = 8.89 (approx. 9 in.)

21 Table A-5 showing the ring tension values

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

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 62.5) = lb/ft3 Therefore, wu HR = x 20 x 54/2 = 94, 662 lb/ft3 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.

Which case is conservative? (Fixed or hinged base) The amount of ring steel required is given by: As = maximum ring tension / (0.9 Fy) As = 67494/(0.9 * 60000) = 1.25 in2/ft. Therefore at 0.7H use #6bars spaced at 8 in. on center in two curtains. Resulting As = 1.32in2/ft. 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] = 272 psi < 400 psi Therefore, adequate

The moments in vertical wall strips that are considered 1 ft. 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 62.5 = lb/ft3 Therefore wuH3 = x 203 = 1,104,800 ft-lb/ft The computed moments along the height are shown in the Table. The figure includes the moment for both the hinged and fix conditions

The actual restraint is somewhere in between fixed and hinged, but probably closer to hinged. For the exterior face, the hinged condition provides a conservative although not wasteful design Depending on the fixity of the base, reinforcing may be required to resist moment on the interior face at the lower portion of the wall. The required reinforcement for the outside face of the wall for a maximum moment of 5,524 ft-lb/ft. is: Mu/(f f’c bd2) = ………(where d = t – cover – dbar/2) From the standard design aid of Appendix A, take the value of and obtain a value for w from the Table. Obtain w=0.0278 Required As = w bdf’c/fy = in2

r=0.167/(12 x 7.5) = rmin = 200/Fy = > Use #5 bars at the maximum allowable spacing of 12 in. As = 0.31 in2 and r = The shear capacity of a 10 in. wall with f’c=4000 psi is Vc = 2 (f’c)0.5 bwd = 11,384 kips Therefore, f Vc = 0.85 x 11,284 = 9676 kips The applied shear is given by multiplying wu H2 with the coefficient from Table A-12 The value of wu is determined with sanitary coefficient = 1.0 (assuming that no steel rft. will be needed) wuH2 = 1.0 x 1.7 x 62.5 x 202 = 42,520 kips Applied shear = Vu = x wuH2 = 3912 kips < fVc

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 tanks 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

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 (minimum or otherwise) 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 conditions considered in the publication include both hinged and fixed edges. However, in reality, neither of these two extremes actually exist. 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

Mx = moment per unit width about the x-axis stretching the fibers in the y direction when the plate is in the x-y plane. This moment determines the steel in the y (vertical direction). My = moment per unit width about the y-axis stretching the fibers in the x direction when the plate is in the x-y plane. This moment determines the steel in the x (horizontal direction). x y z Mz = moment per unit width about the z-axis stretching the fibers in the y direction when the plate is in the y-z plane. This moment determines the steel in the y (vertical direction).

Mxy or Myz = torsion or twisting moments for plate or wall in the x-y and y-z planes, respectively. All these moments can be computed using the equations Mx=(Mx Coeff.) x q a2/1000 My=(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.

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


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