DESIGN OF UNDERGROUND TANK
DESIGN INFORMATION
CLIENT : PROPOSED UNIVERSITY OF ILORIN
DESIGNED BY
SATELLITE TOWN
EFUNTOYE.
O.T
UNDERGROUND WATER TANK
Intended use
of structure
1. BS8007: Design of concrete structures for retaining aqueous liquids.
Relevant
2. BS8110: The structural use of concrete.
codes of
3. Design of Liquid retaining concrete structures by Anchor
Practice and
design
manuals
Allowable Bearing pressure = 250kN/m2
Subsoil
Ground water level = 3.75m below GL
conditions
Angle of repose, φ = 300
Density of Soil = 18kN/m3
Self-weight of concrete=24kN/m3
General
Surcharge on roof =5.0kN/m2
Loading
conditions
Surcharge on soil due to vehicles=10kN/m2
Unit weight of water, γw =10kN/m3
Partial factor of safety for loads=1.4
Partial factor at serviceability=1.0
1
Characteristic strength of steel, Grade of ribbed high-yield bars, fy
Material
(main bars and links) =460N/mm2
strength
Characteristic strength of concrete, fcu=35N/mm2
Concrete grade = Grade 35A
Severe
Exposure
Concrete cover, c = 40mm
condition
maximum allowable crack width, wmax = 0.2mm
Variation temperature, T2=20oc
Fall in temperature from hydration peak to ambient, T1=300c
Co-efficient of linear expansion of concrete αc=12
Restraint factor, R=0.5
Population growth rate, r=2.47%
2
REFERENCE
CALCULATIONS
OUTPUT
Assumed
Average Number of people in building = 8 people
Population
Population of
Number of Houses in the Satellite town=500
=4000 people
the
Total Number of people in the Satellite town
satellite town
= 8×500 =4000 people
Population growth rate in Nigeria=2.47%
C/A World fact
Design year = 20years
book
Pn = Population at 20years
(2014 est,)
P0 = Initial Population
n = Number of Years
r = Population growth rate
Punmia et Al
1995
Using the Arithmetic growth rate formula
= 5976 people say
Pn = P0 (1 +nr)
Pn = 4000(1 + (20 × 0.0247)) = 5976 people say 6000
W.H.O Sule et Al
2010
Design population
WATER DEMAND
Average Per capita water demand per day = 0.05−0.1m3
Average Per Capita water demand per day of Ilorin residents
6000 people
Per Capital water
demand for
densely
= 0.046 − 0.115m3
populated area
Average Water demand for densely populated area say
say 0.120m3
0.120m3
water demand = Population × per capita water Demand=
6000 × 0.120 = 720m3
720m3 design
volume
3
REFERENCE
CALCULATION
OUTPUT
DIMENSIONING OF TANK
Designing for a tank of 720m3 The tank is divided into
two compartments. Each compartment is 360m3 The
longer side is twice the shorter side for economic reason
Length=7.5m
Breath=7.5m
Volume=L x B x H
Height=6.4m
Volume = (6.4 × 7.5 × 7.5)2 =720m3
LOCATING OF TANK
The tank is sited or located far from sewage and septic
Tanks to prevent contamination. The tank location is also
of relatively high altitude the tank is also located on a soil
with optimum bearing capacity
STRUCTURAL DESIGN OF THE UNDERGROUND TANK
SOIL PROPERTIES
Soil Density = 18kN/ m3
Angle of repose = 30o
Bearing capacity = kN/m2
4
Volume=720m3
REFERENCE
CALCULATION
OUTPUT
All dimensions are in millimetres
LOADING
-Surcharge due to loaded vehicles on retained soil
=10kN/m2
Weight of stone aggregate on roof =28.7kN/m3×0.1m
= 2.87kN/m2
soil=2.87kN/m2
LAYOUT OF STRUCTURE
The size of the structure is less than the length which
would require and expansion joint (say 70)
DESIGN ASSUMPTIONS
(a)Loads
Soil Pressure= Kaγ
Active Earth pressure coefficient 𝐾
𝑎
=
1−sin ∅
1+sin ∅
Soil density γ = 18kN/m3
φ = 30o
Ka =
1−sin 30
1+sin 30
Surcharge on
1
= 3 = 0.33
1
Soil pressure = 18 ∗ = 6kN/m2
3
Design should be based on ground water head of 0.75m of
the depth i . e 1.35m
5
REFERENCE
CALCULATIONS
OUTPUT
Assume density of water =10kN/m3
USING BS8007 AND BS8110
(b) DESIGN FOR SEVERE EXPOSURE.
Design crack width=0.2mm
(c) MATERIALS
Concrete grade 35A with a minimum cement content of
325Kg/m3 of finished concrete. Reinforcement steelribbed high yield bars grade 460
(d) Cover to outer layer of steel = 40mm
(e) DESIGN
All floor, wall and roof slab panel are designed as
continuous two way spanning panels.
(f) JOINTS
In view of the size of the structure, no improvement
joints are desirable as they are potential sources of
leakage
The structure will be designed as a monolithic structure.
LOADING CASES
(a) WELLS
External soil, ground water and surcharge pressures.
Soil=38.01kN/m3
Ground water
=25.01kN/m3
Surcharge
=3.33kN/m3
6
REFERENCE
CALCULATION
OUTPUT
COMBINED DIAGRAM OF PRESSURE
Where ground water is present the effective density of
the soil is reduced due to the buoyancy effect. The
effective extra load due to the presence of ground water
1
is 10 × (1 − 3)H = 6.67H
(b)WET WELL FULL
Pressure due to water in well
Pressure due to
water in
well=64kN/m2
7
REFERENCE
BS80072.3
CALCULATION
OUTPUT
No allowance will be made for the pressure resistance
of the external soil
BS8110
THICKNESS OF SECTIONS
For ease of construction of 6.4m high wall, the
minimum thickness should be 300mm.
The allowable ultimate shear strength of 35 grade
concrete with an assumed 0.5% of reinforcement is
Vc =
BS8110 Table 3.9
100As
bd
0.79 ×
1
1
1
Fcu 3
( 25 )
100As 3
( bd )
400 4
(d)
×
×
γm
Is assumed to be 0.5% for a minimum thickness of
300mm effective depth, d.
1
Vc =
0.79 ×
35 3
(25)
1
3
× (0.5) ×
1
400 4
(300)
1.25
=0.60N/mm2
Maximum ultimate force at the foot of the walls due to
maximum.
EXTERNAL LOADING
1
Total Force From
Soil=2 ×B×H,
The Pressure
1
Water=2×B×H
Diagram =
Surcharge = L × B
5
1
1
X 1.4 X [[ X 38.01 X 6.4] + [ X 25.01 X 3.75]
8
2
2
+ [3.33 X 6.4]] = 166.12kN/m
8
166.12kn/M
REFERENCE
CALCULATION
THE MINIMUM EFFECTIVE DEPTH OF WALL
d=
OUTPUT
D=276.85mm
V
166.2 X 103
=
= 276.85mm
Vc X b
0.6 X 1000
Ø
Over all thickness = d + 2 + Ødistribution + C
=276.85+12.5+16+40
= 345.35mm
Use wall 350mm Thickness and floor 400mm thickness
TOP COVER SLAB
Only one needed to be designed because both sides are
similar
The joints between the roof slap and the walls are
assumed partially free. The roof slab is a continuous twoway spanning over its supports.
9
Walls=350mm
Floor=400mm
REFERENCE
CALCULATION
OUTPUT
Ly = Lx = 7.85m
𝑙𝑦
= 1.0𝑚
𝑙𝑥
SLAB SIZING
Roof slab, d
Using a slab 250mm thick
=200mm
SLAB LOADING
Weight of stone aggregates to reduce Thermal cracking
28.7kN/m3 × 0.1m = 2.87kN/m2
Self weight of roof slab= 0.25 × 24 = 6.0kN/m2
Gk =dead load
Total dead load = 2.87 + 6.0 = 8.87kN/m2
=8.87kN/m2
Live load due to partial access on roof (Vehicles are
Qk =Live load
excluded) and light construction traffic = 5.0kN/m2
=5.0kN/m2
Design load; n= 1.4Gk + 1.6Qk = 1.4 (8.87) + 1.6 (5.0)
n= 20.42kN/m2
= 20.42kN/m2
BENDING MOMENT COEFFICIENTS,
Maximum span moment, Msx =BSXnl2x
Msy = BSynl2x
Support moment, Mx = -Bxnlx2
My = -Bynlx2
LONG SPAN:
Mid-span, Msy = 0.044 X 20.42 X 7.852=55.37kNm
Continous edge My = -0.058 X 20.42 X 7.852
= -68.05kNm
Msy =55.37kNm
My =−68.05kNm
SHORT SPAN:
Msx =52.85kNm
Mid-span, Msx = 0.042 X 20.42 X 7.852 = 52.85kNm
Mx =0kNm
Continuous edges, mx = 0kN m
10
REFERENCE
CALCULATION
OUTPUT
MAIN REINFORCEMENT
Assuming the use of 16mm diameter steel in direction of
short-span.
∅
d=h–c-2
= 250 − 40 −
16
2
= 202mm
SHORT SPAN:
Midspan moment Msx = 52.85kNm
k=
M
52.68 X 10^6
= 35 X 1000 X 2022 = 0.037
Fcu bd2
K=0.037
K<0.156 : no need of compression reinforcement
Z=lad
la = 0.5 + √0.25 −
la = 0.5 + √0.25 −
k
0.9
0.037
0.9
la = 0.957
z = 0.957 X 202 = 193.32
m
As = 0.87 ×f
y lad
52.85 X 106
As = 0.87 X 460 X 0.957 X 202
Z=193.32
As = 683.13mm2
Provide Y16@250mmc/c B(804mm 2)
Y16@250mm c/c
This is the minimum reinforcement that satisfied the
B(804mm2)
strength
CALCULATION OF DESIGN CRACK WIDTH
Thickness of wall = 350mm
Applied service moment= 350m
Reinforced area = 804mm2, cover= 40
11
REFERENCE
CALCULATION
OUTPUT
∅
Depth (d) = h – c - 2
d = 350 – 40 – ½ x 16 = 302mm
ᵨ=
As required
bd
=
804
100 ×302
= 2.66 × 10−3
modular ratio = 𝜀 s, = 200KN/mm2
𝜀 c, = ½ x 27 = 13.5KN/mm2
єs
αe = єc =14.8
αeᵨ = 14.8 x 0.00266 = 0.394
DEPTH OF NEUTRAL AXIS IS GIVEN BY
x
𝟐
= αeᵨ (√𝟏 + αeᵨ − 𝟏)
d
x
𝟐
= 0.0394 (√𝟏 + 0.0394 − 𝟏)
d
x
= 0.244
d
x = 0.244 × 302 = 73.7mm
Lever arm Z = d – x/3 = 302 – 73.7/3
Z= 277.4mm
ms
Steel tensile stress, Fs = Z
As
52.85 X 106
FS = 277.4 X 804 = 236.96 N/mm2
CONCRETE COMPRESSIVE STRESS, FCB
2ms
Fcb = Z
bx
=
2x52.85 × 106
277.4 × 1000 × 73.7
Fcb = 5.19N/mm2
CHECK STRESS LEVEL
Fs =236.96 < 368
0.45 Fcu = 0.45 × 35 = 15.7
Fcb=5.19<15.7
Fcb = 5.19<15.7 ok
stress level
OK
0.8Fy = 0.8 × 460 = 368
Fs = 236.96 < 368 Ok
12
REFERENCE
CALCULATION
ϵ1 =
h−x
d−x
Fs
×
Es
350−73.7
= 302−73.7 ×
OUTPUT
236.96
200 ×103
= 1.43 × 10−3
2
2
s
∅
∅
∝cr = √(2) + (c + 2) - 2
200 2
= √(
2
) + (40 +
16 2 16
2
) -
2
= 102.9mm
FOR LIMITING DESIGN SURFACE CRACK
b(h − x)(a − x)
3Es As (d − x)
ϵ2 =
ϵ2 =
1000(350 − 73.7)(350 − 73.7)
3 × 200 × 103 × 804(302 − 73.7)
ϵ2 = 6.9318 × 10−4
Em = ϵ1 − ϵ2
Em = 1.43 × 10−3 − 6.9318 × 10−4
Em = 7.368 × 10−4
DESIGN CRACK WIDTH
w=
w=
3acr Em
1+2 (
acr − Emin
n−x
3 × 102.9 × 7.368 × 10−4
102.9− 40
1 + 2 ( 350−73.7 )
W = 0.156mm
VERTICAL REINFORCEMENT
Table A2.6
)
Design Cack width=0.2m
h=350mm
Limiting moment 60.8kNm
Service steel stress 190N/mm2
Ultimate shear capacity 167KN/m
13
Crack width =
0.156mm less than
allowable
0.2mm
REFERENCE
CALCULATION
OUTPUT
Moment =52.85KNm
Fs =
52.85
60.8
× 190 = 165.16N/mm2
165.16 < 190 satisfactory
m
As = zf
s
52.85 × 106
As = 190 × 193.32
As = 1438.8mm2
Provide Y16 @ 125mm c/c
Y16@ 125mm c/c
(As provided 1610)
DEFLECTED CHECK FOR ROOF SLAB
Basic span depth ratio of 2-way slab=26
Fs =
2
As required
× 460 ×
3
As provided
mf = 0.55 +
Fs =
477 − fs
120 (0.9 +
m
bd2
)
2
683.13
× 460 ×
= 166.26
3
1260
m
52.85 × 106
=
= 1.29
bd2
1000 × 2022
477 − 166.26
mf = 0.55 +
= 1.73
120 (0.9 + 1.29)
limiting span
= basic ratio × mf
effective depth
=26 × 1.73 = 44.98
actual span
7850
=
= 38.86
effective depth
202
Limiting span ratio
44.98 > Actual span
Limiting span ratio > Actual span ratio,
ratio= 38.86
Deflection
deflection Ok
ok
14
REFERENCE
CALCULATION
CALCULATION OF MINIMUM OF CONTINUOUS
CONSTRUCTION
= (7.5 × 2) + (0.35 × 3) = 16.05m
Fall in temperature Hydration peak to ambient, T1 = 300c
Variation in temperature, T2 = 200c
Design crack width=0.2mm
αc = coefficient of linear expansion of concrete= 12
Restraint factor R=0.5
NET EFFECTIVE CONTRACTION STRAIN
= R × αc × T1 = 0.5 × 12 × 30
= 180microstrain
Maximum allowable crack width = 0.2mm
w
0.2
Smax = R × α
c ×T
= 180 ×10−6
= 1111mm
Also Smax =
fct
fb
×
Ø
2ρ
fct 2
=
fb 3
2
Smax = 3 ×
Smax =
ρ=
Table 5.1 Anchor
16
2ρ
5.33
ρ
5.33
× 100 = 0.48%
1111
Also Pcrit =
fct
fy
=
1.6
460
× 100 = 0.35%
FOR THE ROOF SLAB
Minimum area of reinforcement
=0.48 × area of reinforcement
=0.48% × 1000 × 250 =1200mm2
15
OUTPUT
REFERENCE
CALCULATION
Provide Y12@75mm c/c(1510mm2) each face
OUTPUT
Provide Y12@
75mmc/c for
distribution
FOR WALL
Minimum area of reinforcement= 0.48% × Area
=
0.48
100
× 1000 × 350= 1680mm2
Provide Y16@100mm c/c for both faces (2010mm2)
Provide Y16 @
100mm c/c both
faces (2010mm2)
FORCES ON WALLS
Effective height of walls,
center to center of roof slab to floor slab.
= 6400 +
350
2
+
250
2
assuming a base Thickness of 350mm
= 6700mm
The walls panels are loaded triangularly due to water and
soil pressure and rectangularly due to surcharge pressure.
It is convenient to replace the surcharge pressure by the
equivalent soil pressures.
16
REFERENCE
CALCULATION
By using the similar triangles approach
6.7
x
=
40.2
43.53
0.7 × 43.53
x=
= 7.255
40.2
Say 7.260
h = equivalent height = 7.26m
LOADING
CASE1
External soil pressure (including surcharge) =43.53kN/m2
Ground water = 6.67 × H = 6.67 × 1.5 =10.00kN/m2
Total = 53.53kN/m2
CASE 2
Internal water pressure =
γH = 10 × 6.7 = 67kN/m2
WALL A
height of wall = 6.7m
length of wall = 7.85m
ratio
Lx
Lz
7.85
= 6.70 = 1.17
Mv = Vertical span
MH = Horizontal span
+M = Tension on unloaded face
-M = Tension on loaded face
CASE 1
SOIL + WATER (TANK EMPTY)
−Mv = 0.046 × 53.53 × 7.26 × 6.7 = 119.8
+Mv = 0.017 × 53.53 × 7.26 × 6.7 = 44.26
−MH = 0.02 × 53.53 × 7.852 = 65.97
17
OUTPUT
REFERENCE
CALCULATION
+MH = 0.008 × 53.53 × 7.8522 = 26.39
CASE 2
INTERNAL WATER (TANK FULL)
−Mv = 0.046 × 67 × 6.722 = 138.4
+Mv = 0.017 × 67 × 6.722 = 51.13
−MH = 0.02 × 67 × 7.8522 = 82.57
+MH = 0.008 × 67 × 7.8522 = 33.03
WALL B
Length of wall = Lx = 7.850
Height of wall = Lz = 6.70m
l
Ratio = lx =
z
7.850
6.70
= 1.17
CASE 1: one compartment full, the other empty
As case 2 wall A, on each face
Wall C just as wall A
WALL D
Length of wall = Lx = 7.850
Height of wall = Lz = 6.70m
l
Ratio = lx =
z
7.850
6.70
= 1.17
CASE 1, SOIL AND WATER (TANK EMPTY)
Same as case 1 wall A
CASE 1 INTERNAL WATER (TANK FULL)
Same as case 2 wall A walls E,F,G same as Wall A
DIRECT FORCES
All External loads cause compressive horizontal forces in
the walls which are resisted by the concrete in
compression.
The value of the compressive stress is low and
18
OUTPUT
REFERENCE
CALCULATION
may be ignored.
The load due to water in both wells cause tension in the
walls which is evaluated below
TENSION IN WALL D
Tension in wall D due to pressure on walls A and B.
Maximum water pressure = 10 × 6.7 = 67kN/m2
Average water pressure over the lowest 1m height of wall
= 10 ×
(6.7+5.7)
= 62kN/m2
2
Total force = 62 × 7.85 = 486.7kN/m height
Force per meter height on wall
D and F =
486.7
2
=243.4kN
The effect of the floor is neglected but it is conservative
Assuming a service steel stress of Fs = 240N/mm2
T
243.4 × 103
As =
=
= 507mm2
2fs
2 × 240
FLOOR SLAB
The soil pressure under the floor slab is due to the
imposed weight of the structure.
TAKING A COMPART AND ANALYZING THE LOADS:
(1) Roof Self weight = 24 × 0.25 × 16.05 × 8.2 = 789.66kN
(2) Imposed load on roof Stone aggregate and live load
= (2.87 + 5) × 18.05 × 8.2 = 1035.77kN
(3)Walls DEFG = 2(16.05×0.35×6.4×24) = 1725.6kN
Walls ABC = 3(7.5 × 0.35 × 6.4 × 24) = 1209.6kN
(4)Floor slab = 24 × 0.40 × 16.05 × 8.2 = 1263.5kN
Total Load = 5234.43kN
Assuming uniform distribution over the floor area the soil
pressures
19
OUTPUT
REFERENCE
CALCULATION
FOS load
Bearing pressure = Base Area =
1.5 ×5234.43
16.05 ×8.2
OUTPUT
= 59.66kN/m2
Designing for the base of one compartment and
providing for the other.
Before designing the base, structure has to be checked
for flotation
Weight of Empty Tank
Roof self weight = 789.66kN
Walls DEFG = 1725.6kN
Walls ABC = 1209.6kN
Floor slab = 1263.5kN
Total weight = 4988.36kN
UPLIFT DUE TO GROUND WATER
weight of water displaced = 10 × 16.05 × 18.2 × 1.9 =
2566.39kN
Weight of empty tank
4988.36
=
= 3990.7kN
Factor of safety
1.25
weight of empty tank > uplift due to ground water
20
REFERENCE
CALCULATION
OUTPUT
Floatation satisfied
L
For the floor slab Ly =
x
Floatation check
7850
7850
satisfied.
=1
Floor slab is two way slab
Assuming simply supported and allowing for fixing
moments later
+Mx = 0.055 × 65.66 × 7.852 = 222.5kNm
+My = 0.056 × 65.66 × 7.852 = 226.59kNm
UDL FROM WALL E
24 × 0.35 × 6.4 × 7.85
= 6.43kN
8.2 × 16.02
UDL FROM WALL C
=
24 × 2 × 0.35 × 6.4 × 7.85
= 6.43kN
8.2 × 16.02
Moment from wall E =
wl2
8
=
6.43 × 7.852
8
= 49.53kNm
WALL B (INTERNAL)
max + mx = 222.5 −
49.53
= 197.7kNm
2
max + mx = 226.5 − 49.58 = 177.7kNm
21
REFERENCE
CALCULATION
OUTPUT
FLOOR REINFORCEMENT
Floor Thickness=400mm
Cover = 40mm
Effective depth of inner layer= 400-40-20-10=300mm
+Mx = 197.7kNm
K=
m
fcu bd2
=
197.7 × 106
35 × 1000 × 3002
K = 0.0627 < 0.156
Z = lad
la = 0.5 + √0.25 −
0.0627
0.9
Z = 0.924d
Z = 277.4
197.7 × 106
As =
0.87 × 460 × 277.4
As = 1780.85mm2
Provide Y20@ 150mm c/c TF (2090mm2)
+My = 177KNm
K=
m
177 × 106
=
fcu bd2
35 × 1000 × 3002
22
Floor Y20@ 150mm
REFERENCE
CALCULATION
OUTPUT
K = 0.0562 < 0.156
Z = lad
la = 0.5 + √0.25 −
0.0562
0.9
Z = 0.93d
Z = 279.9
177 × 106
As =
0.87 × 460 × 279.9
As = 1580.13mm2
Provide Y20@175mm TF(As Provided 1800mm2)
MINIMUM REINFORCEMENT IN FLOOR
Fall in temperature from hydration peak to ambient,
T1 = 300c
Variation in temperature,
T2 = 200c
Design Crack width = 0.2mm
αc = Coefficient of linear
expansion of concrete = 12
Restraint factor, R=0.5
Net effective contraction strain
R × αc × T1
0.5 × 12 × 30 = 180 micro strain
Maximum allowable crack width = 0.2mm
w
: Smax = crack space = R × ∝
c
Ø = 20mm
Also Smax =
𝑓𝑐𝑡
𝑓𝑏
×
∅
2𝜌
𝑓𝑐𝑡
2
=
𝑓𝑏
3
23
×T
=
0.2
180 × 10−6
= 1111mm
Y20@ 175mm
REFERENCE
CALCULATION
Smax =
Smax =
3
3
OUTPUT
20
×
2𝜌
20
3𝜌
Smax = 1111mm
:ρ=
20
1111 ×3
Also ρcrit =
= 0.60%
fct
fy
1.6
= 460 × 100 = 0.35%
Asmin = %bh
Asmin = 0.60% × 1000 × 400
Asmin = 2400mm2
Provide Y20@ 125mm c/c As provided (2510mm2)
ARRANGEMENT OF REINFORCEMENT FOR FLOOR SLAB
TOP FACE
Provide Y20@ 150mm c/c(TF) As provided(2090mm2)
Provide Y20@ 175mm c/c(TF) As provided (1800mm2)
BOTTOM FACE
Provide Y20@125mm c/c As provided (1800mm2)
WALL REINFORCEMENTS
Table A2.6 Anchor
Wall Thickness = 350mm
Cover=40mm
Effective depth to inner layer of reinforcement
=350-40-10-12=286mm
The minimum reinforcement calculated earlier is
Y16@100mm c/c for each face (2010mm2)
For Y16@100mm c/c
moment=98.1KNm
24
Y20@ 125mm EW
REFERENCE
CALCULATION
OUTPUT
WALL A
HORIZONTAL REINFORCEMENT
All moment are less than 98.1kNm/m
Therefore for bending use
Y16@100mm c/c As=2010mm2
Adding reinforcement area to cater for direct Tension
=2010mm2 + 507mm2 = 2517mm2
Y20@100mm EF
Provide Y20@100mm c/c EF(3140mm2)
for walls
VERTICAL REINFORCEMENT
Maximum vertical moment=138.4kNm/m
Y20@100EF for all
Allowable moment = 173kNm/m
walls
Wall B, C, D, E, F, G are provided with the same
reinforcement.
25