Design of Reinforced Concrete Tanks

TYPES OF TANKS

BASED ON PLACEMENT OF TANKI. RESTING ON GROUNDII. UNDER GROUNDIII. ELEVATED

BASED ON SHAPE OF TANKI. CIRCULARII. RECTANGULARIII. SPHERICALIV. INTZV. CONICAL BOTTOM

RESTING ON GROUND

ELEVATED

CIRCULAR

RECTANGULAR

SPHERICAL

INTZ

CONICAL BOTTOM

Design Considerations

Loading Density of Retained Liquids

LiquidWeight (KN/m3)

Water10

Raw Sewage11

Digested Sludge Aerobic11.4

Digested Sludge Anaerobic11.3

Sludge from Vacuum Filters12

Under Ground

Design Considerations

Wall 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

w =10-5 N/mm2

H = Height of tank r = Radius of tank

25.05.02

w

ct

r

Hfn

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 Considerations

Concrete 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 Considerations

Reinforcement 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:I. 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 Considerations

II. 300< 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 same

III. H > 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 Considerations

Maximum 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 Design

The principal steps are 1.Ultimate limit state design calculations2.Serviceability 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

21max 22

TTwf

f

b

ct

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

from equation

Where m

for limiting crack width of 0.2 mm

xhca

aw mcr

minmax

21

3

xdAE

xaxhb

ss

t

3

'2

Crack Width

For limiting crack width of 0.1 mm

xdAE

xaxhb

ss

t

3

'5.12

Crack WidthDirect tension in mature concrete The maximum likely surface crack width due

to direct tension may be calculated from equation

Wherem

for 0.2 mm limiting crack width

for 0.1 mm limiting crack width

mcraw 3max

ss

t

AE

hb

3

22

ss

t

AE

hb2

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

3x

dA

Mf

s

s

2/c

se E

E

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 =18 kN/m3

Soil: Submerged unit weight =(18−w)=8 kN/m3

Coefficient of friction °

Surcharge: 12 kN/m2. Unit weight of water w=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.

(a) Check uplift: Total weight W of the tank when empty: W={5×10×8−(8–2×0.4)(10–3×0.4)(5-0.25-

0.4)}×24 =2985 kN Uplift Pressure of water under the floor due to

4 m head of water Uplift pressure=10×4=40 kN/m2

Uplift force=8×10×40=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 =18−10 = 8 kN/m3

Pressure due to 1 m high dry soil plus 3.6 m of submerged soil

=1×18+3.6×8 = 46.8 kN/m2

Submerged weight of additional slab =(24× 0.4–10x0.4)=5.6 kN/m2

Total additional weight of soil and additional slab = 46.8 +5.6 = 52.4 Kn/m2

If b= 0.55 m, the additional weight is {(11.1× 9.1) – (10×8)} × 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 (1000–250)=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:(i)At 250 mm below ground: p = 4 kN/m2 due to surcharge+ka×18×0.25

= 5.5 kN/m2

(ii) At 1000 mm below ground: p = 4 kN/m2 due to surcharge+ka×18×1.0

=10.0 kN/m2

(iii) At 4600 mm below ground: p = 4+ka× 8×3.6+ 10×3.6 due to ground water

= 49.5 KN/m2

Case 2: Tank full: Ignore any passive pressure due to soil and assume

that the ground is dry.(i) At 4600 mm below ground p= 10×4.35=43.5 kN/m2

(c) Check shear capacity: Effective depth:d = 400–40 mm cover—12 mm bar /2= 354 mmCase 1: Tank empty:Total shear force at base is approximatelyV = 0.5×5.5×0.25+0.5×10.0×0.75 +0.5×49.5×3.6 = 93.5 kN/m v = 93.5×103/ (1000×354)=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 = 0.5×43.5×4.35=94.6 kN/m v = 94.6×103/(1000×354)=0.27 N/mm2

v < vc

Section thickness is adequate.

(d) Minimum steel: From Table A1 of 8007, ρcrit= 0.0035 for 460

grade steel. Minimum steel As area required

= 0.0035×1000×400=1400 mm2/m Wmax = 0.2 mm

α=12×10−6 from Table 3.2 of BS 8110, Part 2 T1=25°C (Table A.2 of BS 8007)

From Table A1 of 8007, fct/fb= 0.67 for deformed bars of type 2.

Choose bar diameter = 12 mm

= 0.003 < 0.0035 Using continuous construction for full

restraint (Table 5.1 of BS 8007), minimum steel required is

As = 0.0035×1000×400=1400 mm2/m

Provide T12–150 mm c/c=755 mm2/m on each face.

Total steel areα=1510 mm2/m.

)(22 21max TT

f

fw

b

ct

(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 m×4.35 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.

(1) 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= 0.071×49.5×4.352= 66.5 kN.m/m (SLS) Vertical bending moment at base (ULS) M=1.4×66.5=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/m2

Vertical bending moment M at SL: M= 0.071×45.5×4.352+ 0.063×4×4.352

= 65.9 kN.m/m (SLS)

Vertical bending moment at base (ULS) M=1.4×65.9=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.

021.0353541000

101.932

6

2

cufbd

M

mmmzf

MA

ys /5.633

35495.046095.0

101.93

95.02

6

(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=0.052×49.5×4.352=48.71 kNm/mOR =0.052×45.5×4.352+0.037×4×4.352

=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=0.024×49.5×4.352=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=0.5×43.5=21.8 kN/m2

Ignoring the resistance provided by the base, tensile force N per meter is

N=0.5×5.0×21.8=54.5 kN/m. OR from Table 2.53 Reynolds N=0.265×43.5×4.35=50.15 kN/m.

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

reinforcement

Therefore

From figure 4.29

2/27 mmKNEc

8.142/27

200

2/

c

sc E

E

03.03541000

7548.14

bd

Asc

mmdx 4.8135423.023.0

- 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

26

/6.197

34.81

354754

1071.48

3

mmNx

dA

Mf

s

s

21146754130

6.197mmAs

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

23

419130

105.54mm

f

NA

ss

- Check width of Horizontal cracks (Vertical reinforcement) using the following data

Service moment and tension force M= 0.071×43.5×4.352= 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

26

/237

34.81

354754

104.58

3

mmNx

dA

Mf

s

s

21375754130

237mmAs

(ii) longitudinal walls: The wall is designed as a 4.4 m×4.35 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.

(1) 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= 0.032×49.5×4.352= 30 kN.m/m (SLS) Vertical bending moment at base (ULS) M=1.4×30=42 kN.m/m (ULS) Use minimum steel, Provide T12–150 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=0.028×49.5×4.352=26.2 kNm/m Use minimum steel, Provide T12–150 mm

c/c=755 mm2/m on each face.(3) Horizontal bending moment at mid-span bending moment coefficient = 0.013 M at SLS=0.013×49.5×4.352=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=0.5×43.5=21.8 kN/m2

Ignoring the resistance provided by the base, tensile force N per meter is

N=0.5×8.0×21.8=87.2 kN/m. The tensile stress due to tensile force is =87.2×103/(2×754)=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.

26

/3.106

34.81

354754

102.26

3

mmNx

dA

Mf

s

s