Content; 1. Top spherical dome. 2. Top ring beam. 3. Cylindrical wall. 4. Bottom ring beam. 5. Conical dome. 6. Circular ring beam. The basics of enticing water tank design and the related components are broadly calculated in this document. The next few documents will demonstrate the design of Intze tank members like column, bracing and foundation. Keep following the updates.....

1. 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

2. 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

3. 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

4. 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

5. 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
2.3.2 Area of reinforcement calculation:
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

6. 2.3.5 Detailing of reinforcements:
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
2.4.2 Area of reinforcement calculation:
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.

7. 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:

8. 2.5 Design of bottom ring beam:
2.5.1 Load Calculation:
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:
Area of the required reinforcement, Ast =
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

9. 2.5.5 Detailing of reinforcements:
Fig. Detailing of bottom ring beam
2.6 Design of conical dome:
2.6.1 Load Calculation:
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

10. 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
2.6.3 Area of reinforcement calculation:
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

11. 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

12. 2.7 Design of bottom spherical dome:
2.7.1 Load Calculation:
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
2.7.3 Area of reinforcement calculation:
ρ = 0.18 %
Area of the required reinforcement, Ast =
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

13. 2.7.4 Details of reinforcement:
Fig: Detailing of Bottom Spherical Dome.
2.8 Design of circular bottom beam:
2.8.1 Load Calculation:
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

14. 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 )

15. 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.

16. 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
Area of Reinforcement =
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

17. (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
Area of Reinforcement =
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

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

19. Fig:- Details reinforcement of Intze Tank