211577041 Design of Reinforced Concrete Water Tanks

Design of Reinforced Concrete Tanks

TYPES OF TANKS BASED ON PLACEMENT OF TANK RESTING ON GROUNDUNDER GROUNDELEVATEDBASED ON SHAPE OF TANKCIRCULARRECTANGULARSPHERICALINTZCONICAL BOTTOM

RESTING ON GROUND

ELEVATED

CIRCULAR

RECTANGULAR

SPHERICAL

INTZ

CONICAL BOTTOM

Design ConsiderationsLoading Density of Retained Liquids

Under Ground

Design ConsiderationsWall Thickness - Extra thickness will cause higher thermal stress when the concrete is hardening - Minimum wall thickness tw = 1/10 of the span for a simple cantilever - Rectangular tanks tw = Ls / 16 not less than 250 mm tb = Ls /12 not less than 400 mm

Design Considerations - Circular tanks tw =1.73 H2 / r n2 not less than 200 mm

fct = 1.7 1.8 N/mm2 gw =10-5 N/mm2 H = Height of tank r = Radius of tank

Design Considerations tb = r / 10 12 or tw + 100 which is larger not less than 250 mm for elevated tank tb = r / 6 8 or tw + 100 which is larger not less than 400 mm for rested tank

Design ConsiderationsConcrete mix Design C35 mix to BS 8007 - minimum cement content: 325 kg/m3 - maximum cement content: 400 kg/m3 - maximum water/cement ratio : 0.55 - minimum cover : 40 mm

Design ConsiderationsReinforcement Details - High yield bars (460 N/mm2) = 0.0035 - mil steel bars (250 N/mm2) = 0.0064 Walls and Suspended Slab - If h 500 mm, the required reinforcement is calculated for the whole area of concrete and half the reinforcement is provided on each face.

Design Considerations - If h > 500 mm, the required reinforcement is calculated for the outer 250 mm depth of concrete and half the reinforcement is provided on each face. Ground Slabs: h < 300mm, the required reinforcement is calculated on the basis of top half of the slab only. Provide this area of reinforcement in the top half of the slab.

Design Considerations300< h 500 mm, Provide reinforcement for the upper half of the slab, In addition calculate the reinforcement for the 100 mm depth of the slab in contact with ground and provide the sameH > 500 mm, Calculate and provide reinforcement as for (ii) above, except that the depth of the upper half is limited to 250 mm only

Figure A.1 Surface zones: walls and suspended slabs

Figure A.2 Surface zones: ground slabs

Design ConsiderationsMaximum spacing of reinforcement is 300 mm or wall thickness whichever is lesser.Minimum reinforcement should not be less than the required thermal and shrinkage reinforcement.Minimum reinforcement should not be less than the reinforcement required to control the crack limited width

Design ConsiderationsFloatation

Limit State DesignThe principal steps are Ultimate limit state design calculationsServiceability limit state design calculations with either a) Calculations of crack width in mature concrete (due to flexural and direct tension) b) restrained thermal and shrinkage movement in immature concrete (due to direct tension)

Shrinkage and Thermal Reinforcement

T2 Depend on the changes in environmental temperature between casting and subsequent used

Crack WidthFlexural tension in mature concrete The design surface crack with may be calculated from equation

Where em = e1 - e2

for limiting crack width of 0.2 mm

Crack Width

For limiting crack width of 0.1 mm

Crack WidthDirect tension in mature concrete The maximum likely surface crack width due to direct tension may be calculated from equation

Where em = e1 - e2 for 0.2 mm limiting crack width

for 0.1 mm limiting crack width

Combine flexural and direct tension in mature concrete Where flexural and direct tension are combined, the strains due to each must be added together in calculating crack width in the mature concrete An alternative to such calculations of crack widths, Table 3.1 of the code offers maximum service for the reinforcement and if these values can be shown to be satisfied it may be assumed the maximum crack width will be below the limiting value

Where

and (x/d) determine from figure 4.29 in Mosley (text book) modular ratio Es = 200 KN/mm2

Analysis For calculating moments in the walls of the tank, ready made tables of moment coefficients are available. These coefficients have been obtained from elastic analysis of thin plates using analytical methods or using the finite element method. Table 17.9 in Macginley (Text book) and 2.53 in Reynolds (Hand book) or tables in pages 173-175 in Anchor (ref. Book)are determine the moment coefficients of elastic analysis for triangular distributed loads on panels.

Analysis An additional surcharge pressure with rectangular distribution can be determine by reference to Table 3.14 in BS 8110

Table 17.9 (Macginley)

Anchor Pages 170-175

DESIGN OF A RECTANGULAR COVERED TOP UNDER GROUND WATER TANKSpecification: Design a rectangular water tank with two equal compartments as shown in fig. Soil: Unit weight g=18 kN/m3 Soil: Submerged unit weight g=(18gw)=8 kN/m3 Coefficient of friction f = 30 Surcharge: 12 kN/m2. Unit weight of water gw=10 kN/m3

Consider the possibility of water logging up to 1 m below the ground level. Design for severe exposure, design crack width=0.2 mm. Use C35A concrete and 460 grade steel. Assume walls and slabs are 400 mm thick. The roof is not integrally connected to the walls and is simply supported on the external walls but continuous over the central dividing wall with 250 mm thick.

Check uplift: Total weight W of the tank when empty: W={5108(820.4)(1030.4)(5-0.25-0.4)}24 =2985 kN Uplift Pressure of water under the floor due to 4 m head of water Uplift pressure=104=40 kN/m2 Uplift force=81040=3200 kN Design Uplift force = 3200 1.1 = 3520 KN Additional weight required to have a factor of safety against floatation of 1.1 Additional weight= 3520 2985= 535 kN

This can be provided by extending the base as shown in Fig.

The submerged unit weight of the soil =1810 = 8 kN/m3 Pressure due to 1 m high dry soil plus 3.6 m of submerged soil =118+3.68 = 46.8 kN/m2 Submerged weight of additional slab =(24 0.410x0.4)=5.6 kN/m2 Total additional weight of soil and additional slab = 46.8 +5.6 = 52.4 Kn/m2If b= 0.55 m, the additional weight is {(11.1 9.1) (108)} 52.4 = 1100 KNWeight of tank = 2985+1100= 4085KN> 3520 KN

(b) Pressure calculation on the walls:Case 1: Tank empty: Coefficient of active earth pressure:

Pressure due to surcharge =ka x 12=4 kN/m2 The wall is 5000 400 250=4350 mm high. For the top (1000250)=750 mm, unit weight of soil=18 kN/m3 Below this level submerged unit weight of soil=8 kN/m3

In addition to the soil pressure there is also the pressure due to ground water. The pressures at different levels are:At 250 mm below ground: p = 4 kN/m2 due to surcharge+ka180.25 = 5.5 kN/m2(ii) At 1000 mm below ground: p = 4 kN/m2 due to surcharge+ka181.0 =10.0 kN/m2

(iii) At 4600 mm below ground: p = 4+ka 83.6+ 103.6 due to ground water = 49.5 KN/m2Case 2: Tank full: Ignore any passive pressure due to soil and assume that the ground is dry.At 4600 mm below ground p= 104.35=43.5 kN/m2

(c) Check shear capacity: Effective depth:d = 40040 mm cover12 mm bar /2= 354 mmCase 1: Tank empty:Total shear force at base is approximatelyV = 0.55.50.25+0.510.00.75 +0.549.53.6 = 93.5 kN/m v = 93.5103/ (1000354)=0.26 N/mm2

Assuming minimum area of steel As=0.35%

Section thickness is adequate.Case 2: Tank full. Total shear force at base is approximately V = 0.543.54.35=94.6 kN/m v = 94.6103/(1000354)=0.27 N/mm2 v < vc Section thickness is adequate.

(d) Minimum steel: From Table A1 of 8007, crit= 0.0035 for 460 grade steel. Minimum steel As area required = 0.00351000400=1400 mm2/m Wmax = 0.2 mm =12106 from Table 3.2 of BS 8110, Part 2 T1=25C (Table A.2 of BS 8007) From Table A1 of 8007, fct/fb= 0.67 for deformed bars of type 2. Choose bar diameter f = 12 mm

r = 0.003 < 0.0035 Using continuous construction for full restraint (Table 5.1 of BS 8007), minimum steel required is As = 0.00351000400=1400 mm2/m Provide T12150 mm c/c=755 mm2/m on each face. Total steel are=1510 mm2/m.

(e) Design of walls for bending at serviceability limit state: Typical results are shown in Table 17.9 for the case of side and bottom edges being clamped and the top edge being free (i) Transverse walls: The wall is designed as a 7.2 m4.35 m slab clamped on three sides and free at top and subjected to a hydrostatic loading giving base pressures of 49.5 kN/m2 for case 1 (Tank empty) and 43.5 kN/m2 for case 2 (Tank full). Since the pressure difference is not large, design for Case 1 and use the same steel area for case 2.

Vertical bending moment at base From Table 17.9, interpolating between b/a of 1.5 and 2.0, bending moment coefficient for triangle distributed load =(0.084+0.058)/2=0.071 Vertical bending moment M at SL: M= 0.07149.54.352= 66.5 kN.m/m (SLS) Vertical bending moment at base (ULS) M=1.466.5=93.1 kN.m/m (ULS)

Or bending moment coefficient for triangle distributed load for soil and water loads and rectangular distribution load for surcharge - Triangular load Coefficient = 0.071, Load = 45.5 Kn/m2 - Rectangular load Coefficient = 0. 5 from Table 3.14 BS8110, Load = 4.0 Kn/m2Vertical bending moment M at SL: M= 0.07145.54.352+ 0.06344.352 = 65.9 kN.m/m (SLS)

Vertical bending moment at base (ULS) M=1.465.9=92.3 kN.m/m (ULS)

therefore lever-arm factor , la = o.95 hence

This could provide on both faces T12-150 with area of 754 mm2.

(2) Horizontal bending moment at fixed vertical edges From data in Table 17.9, interpolating between b/a of 1.5 and 2.0,bending moment coefficient =(0.064+0.039)/2=0.052 M at SLS=0.05249.54.352=48.71 kNm/mOR =0.05245.54.352+0.03744.352 =47.6 kNm/m

(3) Horizontal bending moment at mid-span From data in Table 17.9, interpolating between b/a of 1.5 and 2.0,bending moment coefficient =(0.027+0.021)/2=0.024 M at SLS=0.02449.54.352=22.5 kNm/m This could provide on both faces T12-150 with area of 754 mm2.

(4) Direct tension in walls In case 2 (Tank full) there is also direct tension in the horizontal direction in the wall due to water pressure on the 10 m long walls. Average pressure p is approximately p=0.543.5=21.8 kN/m2 Ignoring the resistance provided by the base, tensile force N per meter is N=0.55.021.8=54.5 kN/m. OR from Table 2.53 Reynolds N=0.26543.54.35=50.15 kN/m.

Check Crack width Using deemed to satisfy conditions, check the service stress in reinforcement

Therefore

From figure 4.29

Check width of vertical cracks (Horizontal reinforcement) using the following data Service moment and tension force M = 48.71 KN.m/m, N = 54.5 KN/m

This is greater than 130 N/mm2 allowable stress and the steel area must be increased if deemed to satisfy requirement are to be met

In addition, reinforcement for tension force should be added

Total minimum area of steel required to satisfy crack width of 0.2 mm is = 1146+419/2 = 1355 mm2 provide T16-125 , Area provided 1610mm2

Check width of Horizontal cracks (Vertical reinforcement) using the following data Service moment and tension force M= 0.07143.54.352= 58.4 kN.m/m (SLS)

This is greater than 130 N/mm2 allowable stress and the steel area must be increased if deemed to satisfy requirement are to be met

provide T16-125 , Area provided 1610mm2

(ii) longitudinal walls: The wall is designed as a 4.4 m4.35 m slab clamped on three sides and free at top and subjected to a hydrostatic loading giving base pressures of 49.5 kN/m2 for case 1 (Tank empty) and 43.5 kN/m2 for case 2 (Tank full). Since the pressure difference is not large, design for Case 1 and use the same steel area for case 2.

Vertical bending moment at base From Table 17.9, b/a of 1.0, bending moment coefficient for triangle distributed load = 0.032 Vertical bending moment M at SL: M= 0.03249.54.352= 30 kN.m/m (SLS) Vertical bending moment at base (ULS) M=1.430=42 kN.m/m (ULS) Use minimum steel, Provide T12150 mm c/c=755 mm2/m on each face.

(2) Horizontal bending moment at fixed vertical edges From data in Table 17.9, b/a of 1.0, bending moment coefficient = 0.028 M at SLS=0.02849.54.352=26.2 kNm/m Use minimum steel, Provide T12150 mm c/c=755 mm2/m on each face.(3) Horizontal bending moment at mid-span bending moment coefficient = 0.013 M at SLS=0.01349.54.352=12.2 kNm/m This could provide on both faces T12-150 with area of 754 mm2.

(4) Direct tension in walls In case 2 (Tank full) there is also direct tension in the horizontal direction in the wall due to water pressure on the 8 m long walls. Average pressure p is approximately p=0.543.5=21.8 kN/m2 Ignoring the resistance provided by the base, tensile force N per meter is N=0.58.021.8=87.2 kN/m. The tensile stress due to tensile force is =87.2103/(2754)=57.8 N/mm2

Check width of vertical cracks (Horizontal reinforcement) using the following data Service moment and tension force M = 26.2 KN.m/m, N = 57.8 KN/m

This is less than 130 N/mm2 allowable stress for deemed to satisfy requirement no additional reinforcement required This could provide on both faces T12-150 with area of 754 mm2.