Intze Overhead Water Tank Design by Working Stress - IS Method.pdf

DESIGN OF AN INTZE TANK
COURSE CODE: CE - 4102
COURSE TITLE: STRUCTURAL ANALYSIS AND DESIGN -III SESSIONAL
Designed By:
SUMAN JYOTI
Student Id: 191125
Department of Civil Engineering
Dedicated To:
TRIBHUVAN UNIVERSITY
Kritipur, Nepal
KATHMANDU UNIVERSITY
Banepa, Nepal
DEPARTMENT OF CIVIL ENGINEERING
DHAKA UNIVERSITY OF ENGINEERING & TECHNOLOGY, GAZIPUR

Design an Intze tank from following data:
Future population = 3175
Per capita demand = 130 litre/day
Height of the cylindrical wall = 11 ft.
Concrete strength, f´c = 5030 psi
Steel strength, fy = 60 ksi = 415
Mpa
Solution:
1. GEOMETRIC DESIGN:
Total quantity of water = 3175 × 130 = 412750 lit/day = 412.75 m3
/ day
= 14564.94 ft3
/ day
Let, the diameter of the tank = D in ft.
Now, the volume of the tank = capacity of the tank
 14564.94 =
𝜋𝐷2
4
× 11ft
 D = 41.56 ft ≈ 42 ft
1.1 Dimension and Illustration of Water Tank:
Figure: Dimension Proportioning of Intz Water Tank

2. STRUCTURAL DESIGN:
2.1 Materials Property:
2.1.1 Permissible concrete stresses to resistance to cracking:
Direct tension, fc = 5√𝑓′𝑐 psi = 5√5030 psi psi = 354.61 psi = 2.445 MPa
Tension due to bending, fc = 12√𝑓′𝑐 psi = 12√5030= 851.06 psi = 5.87 MPa
Shear = √f′c psi = 2√5030 psi =141.85 psi = 0.978 MPa
Direct compression, fc = 0.25 f ′c psi = 1257.5 psi = 8.67 MPa
Compression due to bending, fc = 0.45 fc’ psi = 2263.5 psi = 15.06 MPa
Modular ratio, n =
29∗106
57000√f′c
=
29∗106
57000√5030
= 7.17
2.1.2 Allowable tensile strength of reinforcement (According to BNBC):
fs= 0.4 fy = 0.4 × 60 = 24 ksi =165.5 Mpa
2.2 Design of Top Spherical Dome:
2.2.1 Allowable tensile strength of reinforcement (According to BNBC):
Radius of top spherical dome = R
Let, r be the radius of the dome
So, h (2r-h) = R2
or, 5.25 × (2r – 5.25) = 212
or, r = 44.625 ft = 13.6 m
Now, sin θ = 21/ 44.625
θ = 28.07 º < 51.8 º (so, compression dome)
cos θ = 0.882
2.2.2 Load calculation:
Let, the thickness of dome slab = 100 mm
Self-weight = 0.1 x 25 = 2.5 kN/m2
Live load = 1.5 kN/m2
So, Total Distributed Load, w = 4 kN/m2
r
r-h
R = 21ft
h = 5.25 ft
R

2.2.3 Check for stresses:
Let, the thickness of dome
Hoop stress at any angle θ, =
𝑤𝑟
𝑡
(
cos2 θ+cosθ−1
1+𝑐𝑜𝑠θ
)
Maximum hoop stress will occur at θ = 0,
So, Hoop stress =
𝑤𝑟
𝑡
×
1
2
=
4∗1000∗13.6
0.1∗ 2
= 0.272× 106
N/m2
= 0.272 N/mm2
< 5 kN/mm2
( Ok )
Meridional stress at any angle θ, =
𝑤𝑟
𝑡
(
1
1+𝑐𝑜𝑠θ
)
Maximum meridional stress will occur at θ = 28.07 º = Ø,
So, Meridional stress =
𝑤𝑟
𝑡
×
1
1+𝑐𝑜𝑠𝜃
=
4∗1000∗13.6
0.1
×
1
1+0.882
= 0.29× 106
N/m2
= 0.29 N/mm2
< 5 kN/mm2
( Ok )
2.2.4 Area of reinforcement calculation:
Both stresses are within safe limit, hence a minimum reinforcement may be
provided.
Ast = 0.002bt
= 0.002×1000×100
= 200 mm2
per meter (both direction)
Use 8 mm dia. bar @ 200 mm c/c (both direction)
2.2.5 Detailing of reinforcements:
8 mm @ 200 mm c/c
at mid of the slab thickness
100 mm thick dome slab

2.3 Design of Top Ring Beam:
2.3.1 Load calculation:
The horizontal component of Meridional thrust, P = T Cos φ
P = 0.29 × 100 ×1000 ×(0.882) = 25578 N
Total tension tending to rupture the ring beam per meter length of its
circumference,
= P × D/2
= 25578 × (12.8/2)
= 163700 N
Since steel in contact with water then fs = 165.5 N/mm2
[fs = 0.4fy ]
So, Area of reinforcement required,
Ast =
Tensile force
Allowable stress of steel
=
163700
165.5
= 990.12 mm2
Using 16 mm dia bar (as = 200.1 mm2
), required bar=
990.12
200.1
= 4.94 ≈ 5 nos.
2.3.3 Size of beam calculation:
Equivalent area of the composite section = A + ( n-1 ) Ast
= A + ( 7.17-1 ) × 5 × 200.1
= A + 6174
Take the allowable stress to the composite section is 1.2 kN/mm2
.
or, 1.2 =
163700
A + 6174
or, A = 130242 mm2
Let, beam width is, b = 500 mm and the depth, d = 260.48 ≈ 260 mm
2.3.4 Minimum reinforcement calculation:
Using (#3) 10 mm dia bar, then spacing, Sv =
2.5 As∗fy
b
As = 2 × 0.11 in2
( for two legs )
So, Spacing, Sv =
2.5∗78.54∗413.7
260
= 312.42 mm ≈ 300 mm c/c

2.4 Design of Cylindrical Wall:
2.4.1 Load Calculation:
Let, joint condition is simple joint, i.e., cylindrical wall is free to move at top
and bottom.
Maximum hoop tension at the base of the wall is,
T = γh
𝐷
2
= 10000 × 3.534 ×
12.8
2
[Using, γw = 10000 N/m3
]
T = 226176 N
Area of the required reinforcement, Ast =
226176
165.5
= 1366.6 mm2
Area of reinforcement on each face = 1366.6/2 = 683.3 mm2
≈ 684 mm2
Using 16 mm dia. bar, the spacing =
1000 ∗ 201
684
= 293.86 mm ≈ 280 mm c/c
Cut 50% bar at a depth of 1.8 m below the top cylinder than base,
So, spacing is = 560 mm c/c
2.4.3 Thickness of cylindrical wall calculation:
Equivalent area of composite section = t×103
+ (7.17–1) × (
1000
280
×2×201)
= 1000t + 8858.35
Since the tension is direct then permissible direct tensile stress = 1.2 N/mm2
or, 1.2 =
226176
1000𝑡+8858.35
or, t = 179.62 mm ≈ 180 mm
Since hoop tension is smaller at the top of cylinder than base,
So, use thickness 150 mm at the top of cylindrical wall.

2.4.4 Distribution of reinforcement calculation:
Average thickness of the wall slab =
180+150
2
= 165 mm
Area of steel required is = 0.0022 × 1000 × 165 = 330 mm2
Using 12 mm bar, the spacing =
113∗1000
330
= 342.4 mm c/c ≈ 300 mm c/c
2.4.5 Detailing of reinforcement:

2.5 Design of bottom ring beam:
Load due to top dome = Area of slab x Meridional stress x sin φ
= 100 x 1000 x 0.29 x sin28.07º
= 13646 N /m
Load due to top ring beam = 0.26 x (0.5 - 0.15) x 25000
= 2275 N/m
Load due to cylindrical wall = 0.165 x 3.35 x 25000
= 13819 N /m
Assume, beam size, 900 mm x 500 mm
Self-weight of beam = 0.5 x (0.9 – 0.5) x 25000 = 5000 N /m
Total w1= 34740 N/ m
Horizontal thrust due to vertical load = w1 tan β
= 34740 x tan 45º = 34740 N/ m
Hoop tension due to vertical load, H1 =
34740∗12.8
2
= 222336 N
Hoop tension due to water, H2 = 10000 x 3.35 x 0.6 x 12.8/2 = 128640 N
Total hoop tension, H = 222336 + 128640 = 350976 N
2.5.2 Area of Reinforcement Calculation:
350976
165.5
= 2120.7 mm2
Using 16 mm dia bar, Nos. of bar =
2120.7
201
= 10.55 ≈ 11 Nos.
2.5.3 Check for Tensile stresses:
Maximum tensile stress =
350976
900∗500+(7.17−1)∗11∗201
= 0.0006 N/mm2
<1.2 N/mm2
( Ok )
2.5.4 Shear reinforcement calculation:
Using 10mm dia bar Spacing, Sv =
2.5 𝐴𝑠𝑣 𝑓𝑦
𝑏
=
2.5∗ 78.54∗4∗165.5
500
= 260 mm c/c ≈ 250 mm c/c

Fig. Detailing of bottom ring beam
2.6 Design of conical dome:
Average diameter of Conical dome = 10.4 m
Average depth of water = 3.35 + 2.4/2 = 4.55m
Assume, Thickness of slab 400 mm
Self-weight of slab = (π x 10.8 x 3.39) x 0.4 x 25000 = 1150203 N
Weight of water = (π x 10.8 x 4.55 x 2.4) x 10000 = 370508 N
Load from Top dome, cylindrical wall, bottom ring beam, = 34740× π×10.4
= 370508 N
Total load on Conical dome =1150203+ 370508 + 370508
= 1891219 N
Load at per meter of conical dome base =
1891219
3.1416∗4.54
= 132598 N/m
Meridional thrust, T =
132598
𝑐𝑜𝑠45
= 187522 N
Meridional stress =
187522
1000∗400
= 0.5 N/mm2 < 5.0 N/mm2
, So Safe.
2.6.2 Hoop tension calculation:
General expression for hoop tension

Diameter of conical dome at h m height from base, Dh = 8 + 2 × 2.4/2.4 × h
= 8 + 2h
Intensity of water pressure at height h from base, p
or, p = (5.76 – h) x 10000 x 1 = (57600 – 10000h) N/m
Weight of Conical dome, w1 = 0.4 x 1 x 25000 = 10000 N/m
Hoop tension at any height h, = (𝑝𝑠𝑖𝑛β+ w1 tanβ) ×
8 + 2h
2
= {(57600 – 10000h) sin45 + 10000 tan45)} ×
8 + 2h
2
Maximum Hoop tension (at h = 2.44) = 216056 N
Area of Reinforcement Calculation, Ast =
216056
165.5
= 1306mm2
Area of steel on each face =
1306
2
= 653 mm2
Using 16 mm dia. bar, Spacing =
201∗1000
653
= 307.8 mm C/C ≈ 300 mm C/C
Check for maximum tensile stress-
=
216056
400∗ 1000 +(12.8 − 1)∗(
1000
100
∗201∗2)
= 0.00001 N/mm2
< 1.2 N/mm2
(Ok)
2.6.4 Distribution Reinforcement Calculation:
Assuming distribution bar as 0.18 %
Ast = 0.0018× 400 × 1000 = 720 mm2
Area of steel on each face = 720 /2 = 360 mm2
Using 12 mm bar, spacing =
113.1∗1000
360
= 314.2 mm ≈ 300 m c/c

2.6.5 Area of reinforcement for B.M.
The conical dome also acts as a slab,
Load on slab, W =
1150203 + 370508
3.1414 ∗8
= 60507.16 N/m
Load on 1m wide and 3m spanned slab, W = 60507.16 N/m
So, Maximum Bending moment =
60507.16 ∗3
12
= 15126.8 N-m
Let, main bar Ø16mm, then effective depth = 400 – 40 -16/2 = 352 mm
Ast =
15126.8∗1000
115 ∗ 352 ∗ 0.866
≈ 430 mm2
Use #5 (16mm) bar, Spacing =
201.1∗1000
430
= 467.67 mm C/C ≈ 400 mm C/C
Since flexural reinforcing bar spacing = 400 mm C/C,
Which will also act as distribution bar as well as main bar.
Use #4 (12mm) bar @ 400 mm C/C as distribution bar at the top of the slab.
2.6.6 Reinforcement Detailing
Fig. Detailing of conical dome slab

2.7 Design of bottom spherical dome:
Assume, slab thickness = 400 mm
Let, Radius of dome rb,
rb2 = (rb- 1.6)2 + 4.0 2
∴ rb = 5.8 m
or, sin φ =
4
5.8
= 0.6897
∴ φ = 43.60 < 51.8º So Compression dome
∴ cos φ = 0.724
Self-weight of dome = (2π rb × h) × t × γc = 2π x 5.8 x 1.6 x 0.4 x 25000
= 583080 N
Volume of water above conical dome, V = πR2
hwater – 1
3
π x h2
x (3 x rb – h)
= π × 42
× 4.55 – 1
3
π × 1.62
× (3 × 5.8 – 1.6) = 186.35 m3
Weight of water = 186.35 x 10000 = 1863508 N
Total weight = 583080 + 1863508 = 2446588 N
2.7.2 Check for stress:
Load per unit area w =
2446588
2π x 5.8 x 1.6
= 41960 N/m2
∴ Maximum Hoop stress =
wr
𝑡
×
1
2
=
41960 x 5.8
0.4
×
1
2
= 0.304 x 106
N/m2
= 0.304 N/mm2
< 5 N/mm2
∴ Maximum Meridional stress =
wr
𝑡
×
1
(1+𝑐𝑜𝑠 φ)
=
41960 x 5.8
0.4
×
1
(1+0.724)
= 0.353 x 106
N/m2
= 0.353 N/mm2
< 5 N/mm2
ρ = 0.18 %
0.18
100
x 400 x 1000 = 720 mm2
Using #4 (12mm) bar, Spacing =
113.1∗1000
720
= 157.08 mm c/c ≈ 150 mm c/c
in both directions.
R = 4 m
h = 1.6 m
r-h
rb

2.7.4 Details of reinforcement:
Fig: Detailing of Bottom Spherical Dome.
2.8 Design of circular bottom beam:
Meridional thrust from conical dome, T = 187522 N/m
Inward thrust from conical dome = 187522× sin 45º = 132598.07 N/m
Meridional thrust from bottom spherical dome = 0.353 × 400 × 1000
= 141200 N/ m
Outward thrust from bottom spherical dome = 141200 × Cos 43.6°
= 102253 N/m
So, Net inward thrust, P = 132598.07 – 102253
= 30345 N/m (Compressive)
Hoop compressive force in Beam =
𝑃𝐷
2
=
30345∗
26.25
3.28
2
= 121380 N
Assuming, Beam size = 450 mm × 800 mm
∴ Hoop stress =
121380
450∗800
= 0.3372 N/mm2
< 5 N/mm2
Vertical load from conical dome = 132598 N/m (Computed earlier)
12 mm @ 150 mm c/c
at mid of the slab thickness
00 mm thick dome slab

Vertical load from bottom spherical dome,
= 2446588/ (π x 8) = 97346.6 N/m
Total Vertical load on beam = 132598 + 97346.6 = 229945 N/m
Self-weight of beam = 0.45 × 0.8 × 25000 = 9000 N/ m
∴ The design load for the beam, 𝜔 = 229945 + 9000 = 238945 N/m
2.8.2 Moment Calculation:
Load on per meter beam, w = 238945 N/m
It is proposed to support the beam by 6 Columns.
Maximum (-ve) Bending moment at support = k1wR2
×
1
3
π
M1 = 0.089 × 238945 × 42
×
1
3
π = 356317 N- m
Maximum (+ve) Bending moment at mid span = k2wR2
×
1
3
π
= 0.045 × 238945 × 42
×
1
3
π = 180160 N- m
Maximum (-ve) Twisting moment (T) at 12.75º from support = k3wR2
×
1
3
π
= 0.009 × 238945 × 42
×
1
3
π = 36032 N- m
2.8.3 Shear Calculation:
Shear force at support section, V1= w ×
2𝜋𝑅
𝑁𝑜.𝑜𝑓 𝐶𝑜𝑙𝑢𝑚𝑛
×
1
2
= 238945 ×
2𝜋∗4
6
×
1
2
= 500445 N
Shear force at the point of maximum torsion,
V2 = 500445 – 500445 ×
12.75
30
= 287755.88 N ( rough, 360/6 column × 2 )

2.8.4 Design of support section (For Moment):
M = 356317 N- m, V1= 500445 N
Effective depth, d = √
Moment 𝑖𝑛 𝑁−𝑚𝑚2
0.5∗𝑓𝑐∗𝑗∗𝑘∗𝑏
deff = √
356317 ∗1000
0.5∗15.06∗0.866∗0.403∗450
= 548.9 mm
Let clear cover = 40 mm, and 2 layers of main bar
Ø of main bar = 25 mm & stirrup = 10 mm
So, Actual effective depth, d = 800 – 40 – 10 – (25 + 25/2)
= 712.5 mm > 548.9 mm, Safe (ok).
Area of Reinforcement =
356317 × 1000
165.5 × 0.866 × 712.5
= 3490 mm2
Use 20 mm Ø bar. So, Nos. of bars = 3490/314.2 = 11.10 ≈ 12 Nos.
2.8.5 Design of support section (For Shear):
Developed shear stress, 𝜏v =
𝑉
𝑏𝑑
=
500445
450 ∗712.5
∴ 𝜏v = 1.56 N/mm2
Maximum allowable shear stress, 𝜏allow = 3.4 √𝑓𝑐
′
∴ 𝜏allow = 3.4 √5030 = 241.13 psi = 1.66 N/mm2
Since 𝜏v < 𝜏allow, this section is safe for shear.,
Here, Percentage of steel ratio ρ =
12.5
30
= 0.417 %
So, allowable concrete shear stress, 𝜏c = 1.1 √𝑓𝑐
′
𝜏c = 1.1 √5030 = 78.01 psi = 0.538 N/mm2
Since 𝜏v > 𝜏c Shear reinforcement is required.

Here, Allowable concrete shear force, Vc = 𝜏c × bd
= 0.538 × 450 × 712.5 = 172496.25 N
Use #3 (10mm) bar 6-legged bar as stirrup, Spacing,
S =
𝐴𝑠𝑣∗𝑓𝑠𝑣∗𝑑
𝑉−𝑉𝑐
=
(6∗78.54) × 210 × 712.5
500445 −172496.25
= 215 mm C/C ≈ 200 mm C/C
2.8.6 Design for torsion:
The maximum torsion occurs at the point of contraflexure
Equivalent moment, Me = M + MT
At point of contraflexure M = 0
And, we know, MT =
𝑇
1.7
(1 +
𝐷
𝐵
)
=
36032∗1000
1.7
(1 +
800
450
) = 175.54 × 106
N-mm
So, the equivalent moment, Me = 58.87 × 106
N-mm
58.87 × 106
165.5 x 0.866 x 712.5
= 576.55 mm2
Use 16 mm Ø bar. So, Nos. of bars = 576.55/201.1 = 2.87 ≈ 4
The number of bars at the point of contraflexure is more than the 3, So no
additional bars are required.
Equivalent shear, Ve = V +1.6
𝑇
𝑏
= 500445 + 1.6 ×
36032
450
= 500573 N
Equivalent nominal shear stress, 𝜏ve =
500573
450∗712.5
= 1.56 N/mm2
< 1.66 N/mm2
Let, 10mm Ø bar with 6 legged will be used as stirrup,
so, Asv = 6 × 78.54 = 471.2 mm2
Allowable concrete shear force, Vc = 0.538 × 450 × 712.5 = 172496.25 N

(i) Spacing considering torque and shear force at point of contraflexure
Asv =
𝑇∗𝑆𝑣
𝑏1 ∗ 𝑑1 ∗ 𝑓𝑠𝑣
+
𝑉2∗𝑆𝑣
2.5 𝑑1 ∗𝑓𝑠𝑣
471.2 =
36032∗1000∗𝑆𝑣
384 ∗ 719 ∗210
+
287755.88 ∗𝑆𝑣
2.5 ∗719 ∗ 210
∴ 𝑆v = 340 ≈ 300 mm c/c
b1 = 450 – 2×25 – 2×8
= 384 mm
d1 = 800 – 40 – 25 – 2 × 8
= 719 mm
(ii) Spacing considering equivalent shear force
𝐴vs =
(𝑉𝑒−𝑉𝑐)𝑆𝑣
𝑓𝑠𝑣∗𝑑
or, Sv =
𝐴𝑠𝑣∗𝑓𝑠𝑣∗𝑑
𝑉𝑒−𝑉𝑐
=
471.2 ∗ 210∗712.5
(500573 − 172496.25 )
= 214.9 mm c/c ≈ 200 mm c/c
Hence provide 10 mm-6-legged stirrup at 200 mm c/c from column face to
the point of contraflexure.
2.8.7 Design of mid-section:
Effective depth, d = 712.5 mm
Bending Moment = 180160 N- m
180160 × 1000
165.5 × 0.866 × 712.5
= 1764.24 mm2
Use 20 mm Ø bar. So, Nos. of bars = 1764.24/314.2 = 5.62 ≈ 6
2.8.8 Detail Drawing:
Fig. Detailing of Circular Bottom Beam

Fig:-Section showing the details reinforcement in bottom circular beam, column and bracing.

Fig:- Details reinforcement of Intze Tank

Intze Overhead Water Tank Design by Working Stress - IS Method.pdf

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Intze Overhead Water Tank Design by Working Stress - IS Method.pdf

Similar to Intze Overhead Water Tank Design by Working Stress - IS Method.pdf (20)

More from Dhaka University of Engineering & Technology, Gazipur

More from Dhaka University of Engineering & Technology, Gazipur (20)

Recently uploaded

Recently uploaded (20)

Intze Overhead Water Tank Design by Working Stress - IS Method.pdf