Reinforced Concrete Elevated tank Analysis and Design (Wind & Seismic) According to ECP-2013

Analysis & Design of Inze Tank
for Wind and Earthquake Forces
Produced by : Eng. Alaa Metwally
R.C Structures

Analysis and Design of Inze TankReinforced Concrete Structures
© All Copyright reserved for Alaa Metwally
1
Analysis and Design of Inze Tank
1.1 Introduction.
This paper contains a complete design for Inze tank or accurately redesign the tank designed
at the textbook “Analysis and Design of R.C tanks” Dr. Mohamed Hilal 1988. Thanks to
him as he is one of the pioneers in design in Egypt after a complete control from the foreign
design institutions for all projects in this country. Inze tank is simple easy and considered as
an old-fashion tank, but from academic point of view and for theoretical approach it is
considered as an important one to enhance and build a deep understanding for tanks structural
elements, load flow and design methods. The last mentioned textbook adopted elastic design
method and lateral load analysis is only for wind load. Hence, we adopted ultimate limit states
design method which adopted by ECP-203 and lateral analysis contains wind and seismic
loads.
Capacities.
Cylinder = π x 62
x 6.5 = 735.13 m3
Cone (part of cone) = π
2
3
(62
+42
+6x4) = 159.17 m3
Shaft = π x 0.752
x 7.4 = 13.07 m3
Dome = π
1.1
6
(3x42
+42
+1.12
) = 28.34 m3
Capacity = 735.13 + 159.17 – (13.07+28.34) = 853 m3
Note that the volume of shaft and lower dome are subtracted because no water filled this
parts. The water fills only the cylinder at its height so that the upper dome not taken into
calculation of water capacity.
lantern Upper dome
Lower Dome
1.1 m
Cylindrical Walls
Shaft
1.2 m
6 m
6.5 m
2 m
Fig. (1) Tank Elements
1.5 m
Cylindrical Beam
Ring Beam

2
It’s now the time to analyze the tank member one by one and with a short caption or hints to
keeps it as simple as we can for students and engineers.
1.2 Roof Dome.
a=
R2+Y2
2Y
=
62+1.42
2x1.4
=13.6 m
x=√13.62
-62
=12
Sin ϕ = 6/13.6
Cos ϕ = 12/13.6
A = 2πay = 2 π x13.6 x1.4 = 119.6
Assume thickness of dome is 8 cm.
g=γC ts+ Superimposed & live=2.5x0.08+0.15=0.35 ton/m2
W=g x A= 0.35x119.6=41.86 t
P1=
W
2 π r
=
41.86
2 πx6
=1.11 t m-⁄
H1=
P1
tan ϕ
=
1.11
(
6
12
)
= 2.22 t/m-
T1=H1R=2.22x6=2.22 t m-⁄
Ring Beam will be designed to resist tension force only.
t.b = 60 T1
Assume t = 15 cm = thickness of dome @ footing
b=
60x13.32
15
=50 cm
As=
T1
(fy/γs)
=
13.32x103
(3600/1.15)
=4.225 cm2
Asm=
As
βcr
=
44.225
0.85
= 4.97 cm2
use 6 ϕ 12
Stresses of Roof Dome
All forces developed in dome are Meridian force (Compression) and Ring tension which
provide Compressive Stresses in the dome. Hence, a light reinforcing steel mesh used as
concrete are sufficient to resist that kind of stresses.
@ Crown. ϕ = 0 → cos ϕ = 0
Pa
2
=
0.35x13.6
2
=2.53 t m-⁄=Nθ=Nϕ
ϕ
H
P1
R=6m
a
x
a
12m
1.4m
Fig. (1) Forces in Roof Dome

3
@ Footing.
Nϕ=
ga
1+cosϕ
→ ϕ=26°
33’
54’’
→ cos ϕ =
12
13.6
Nϕ=
0.35x13.6
1+(
12
13.6
)
=2.53 t m’⁄
Nθ=ga
(cosϕ-
1
1+cosϕ
) =.35x13.6(
12
13.6
-
1
1+(
12
13.6
)
=1.67 t/m’
𝜎c =
( 2.53x103)
100x15
= 1.69
kg
cm2
< 60
kg
cm2
ok safe
Use minimum area of steel → 5 ϕ 10/m’
Part of dome considered to resist moment at a distance X which calculated roughly as in the
following equation. This is attributed to considering the connection between the Ring beam
and dome as a partially fixed connection.
X=0.6√at = 0.6√13.06x.15 = 0.86 m
M=P
X2
2
= 0.35x
.862
2
= 0.129 t.m/m'
1.3 Cylindrical Wall
Ring tension, field moment and shear forces calculated for wall with hinged base. However,
bending moment calculated for fixed base from Portland cement association tables after
determining the preliminary thickness of wall and the tank shallowness (shallow, medium or
deep).
Tmax =0.75x1x6.5x6= 29.25 t
𝐷
H
=
6
6.5
= 1.85 → medium tank
X
Fig. (2) Moment resisted by dome
5 ϕ 10/m’
6 ϕ 12

4
t=0.8 tmax = 23.4 cm → take t=25 cm
H2
D t
=
6.52
12x.25
=14 → from P.C.A tables
Tmax= 0.761xWHR=.761x1x6x6.5=29.68 t @ 4.55 m
M =0.0033 W H3
=0.0033x1x6.53
= 0.91 t.m @ 5.85m
Qmax
=0.075 W H3
=0.075x1x6.53
=3.1 t @ base
Mmax -ve =-0.009 WH3
=-0.009x1x6.53
=-2.47 t.m @ base
N1=P1=1.1
t
m’
N2=P1+γ c
twhw=1.11+2.5x.25x6.5=5.17
t
m’
Design of sections
 Ring tension Tmax = 29.68 ton
As =
Tu
(
fy
γc
)
=
1.4x29.68
(
3600
1.15
)
=13.27cm2
For one sideAs one branch =
13.27
2
=6.635 cm2
6 ϕ 12
σt=
29.68x103
100x25+10x6.635
=11.57
kg
cm2
<18
kg
cm2
ok safe
= 0.91 t.m N = 4.764 tmaxM1-Sec 1-
t=√
M
3b
-3 =14 cm
29.68 t
1.55 m
3 m
1.55 m
0.91 t 2.47 t5.17 t
1.11 t
1 1
2 2

5
take 25 cm
= -1.19+8.736=5.826
kg
cm2
fct = -
N
A
+
6M
bt2
=
-4.764x103
100x25
+
6x.91x105
100x252
choose min 5 ϕ 12 /m’
t= 2.74 t.m N = 5.17maxM2-2Sec-
t=√
M
3b
-3 cm =√2.47x105
3x100
-3 = 25 cm
= -25.78
kg
cm2
< fct = 95
kg
cm2
ok safefct =
-5.17x103
100x25
-
6x2.47x105
100x252
e=
M
N
=
2.47
5.17
=0.48 m
e
t
=
0.48
0.25
=1.92 big eccentricity
= 0.48+
.25
2
-0.03=0.575 mes= e+
t
2
-cover
Mus=Nu x es =1.4x5.17x0.575= 4.16 t.m
d=c1√
Mu
fcux b
→22
=c1√4.16x105
250 x 100
→C1=5.39→ J=0.826
As=
4.16x105
0.826x3600x22
-
1.4x5.17x103
(
3600
1.15
)
= 4cm2
Choose 5 ϕ 12 m’
Asm=
As
βcr
=
4
0.85
= 4.76 cm2
- Ring Beam 50x80 cm2
Ring Beam B2 resist an outward thrust force plus the shear force at the cylindrical wall.
H=N2+Qmax
= 5.17+3.1= 8.27
t
m’
Tmax=HR=8.27x6=49.62 t
Choose 14 ϕ 16As=
1.4x49.62x103
(
3600
1.15
)
= 2.22 cm2
ft=
1.4x49.62x103
50x80+10x27.75
=16.24< ft=18
kg
cm2
Ok safe
Φ=45
N2
N 2
Fig. (2) Moment resisted by dome

6
1.4 Conical floor.
Conical floor subjected to Meridian force Ns and Ring tension Nθ. If we find the vertical
component of total loads at section we can find the meridian force at that section with
triangular functions (Sin, Cos...).
@ Section 1
Ns=5.17√2=7.3 t/m’
Nϕ=R2 Pr= R2 [water pressure + g cos(ϕ)]
g = Dead load of cone= γc x tc
=2.5x.4=1 t/m2
Nθ=6√2 (1x6.5+1x
1
√2
) =61.15 t/m’
@ Section 2
Ns = [
wt of wall + wt of cone + water
π D
] √2
= [
5.17xπx12+ {
π(12+10)
2
} x1√2x1+π(62
-52
)(
6.5+7.5
2
)
πx10
] √2
= 15.46√2= 21.86 t/m
Nθ= R2Pr=5√2(1x7.5+1x (
1
√2
) =58.03t/m’
@ Section 3
NS= [
5.17xπx12+πx (
12+8
2
) x2√2x1+π(62
-42
) (
6.5+8.5
2
) x1
πx8
] √2
= 30.04√2 ≅ 42.5
t
m’
Nθ=R2Pr =4√2 [1x8.5+1x (
1
√2
)] =40.77 t/m’
Hcone=
30.04
tan(45)
=30.04 t/m’
→ Reaction of Dome at Circular Beams
5 √2
4 √2
Φ=45
6 √2
Fig. (4) Sections in Cone
1
1
2
3
3

7
How to Design the Tank?
Moment distribution method used to calculate moments at rigidly connected members
(Cone, Dome and Circular Beam B3). First we calculate the Fixed End Moment (F.E.M) by
the equations mentioned in Prof. Dr. Mohamed Hilal Handbook and calculate the
stiffness of each member. Then get the final moment at each member. We should note that
stiffness of circular beam taken into account in order to calculate relative stiffness but its
moment is zero. Hence we get the balanced moment at point b in order to have equilibrium
at that point.
h=12.5
S = 6√2x2√2=5.66
l =6√2=8.5
ϕ = 45
sin ϕ = cos ϕ =
1
√2
t=40 cm
p=5.17 t/m' concentrated load
g = γc
tc =25x.4=1 t/m2
K3=1.3068 √(tan(45) )/0.4 = 2.07
E ∆r° =
1x5.662
0.4
[1-
0.167
2x0.5
(1-
8.52
5.662)] x0.5x1+
5.17x8.5
0.4
x1
+
1x5.66
0.4
x[ 5.66 (
12.5
0.71
-5.66) +
0.167
2x5.66
+(
12.5
0.71
{8.52
-5.662
}
-
2
3
{(8.52
-5.662
)}] x 0.5= 48.4 + (1x5.662
)/0.4) (0.5 (1- (
8.562
5.662) ) x 0.5
= 48.4+109.9+3446.25= 3604.5
Eθ°=
1x5.662
0.4
[ 0.5 {1- (
8.52
5.662) +0.167-(2+0.167)x0.5}] x
1
0.71
- [
5.17x8.5
0.4x5.66
x (
0.71
0.5
)] + [
1
0.4
(3x5.662
x0.71-2x12.5)] x 1
- [
1
0.4
{
12.5
2x5.66
(8.52
-5.662
)+
0.71
3x5.66
(8.53
-5.663
)}] x1 = -174.2-27.6+42.3= -159.4
h = 12.5 m
6.5 m
6 m
D=12 m
ϕ = 45

8
M=
0.4x0.71
2x5.66x2.073
x0.5
x (3604-
.71
2.07
x1-159.41) = 20.7 m.t
S=2x0.43
x2.07= 0.265
1.5 Floor Dome
Ignoring shaft and consider a dome R=4.
a=
R2+Y2
2Y
=
4+1.12
2x1.1
=7.8 m
A=2πaY=2xπx7.8x1.1 ≅ 54 m2
X=√7.82
-42
=6.7 m
Assume that the thickness of dome at footing 30 cm and at
crown 15 cm. The Dome resisting its own weight and water
pressure
o.w @ crown = γc
tD=2.5x.15=0.375
t
m2
o.w @ footing = γc
xtD=2.5x0.3= 0.75 t/m2
average weight= [
(0.375+0.75)
2
] x 54 =30.4 t
Total water pressure = volume of water x γw
=(πx42
x8.5-28)x1= 398.6 t
Total weight=30.4+398.6= 429 t
However loads and different from that at roof dome. But, stress in roof dome are
compressive stresses due to meridian & Ring tension.
@ Crown.
Nϕ= Nθ=
Pa
2
.Nϕ=Nθ=
[(8.5−1.1)+0.375] x 7.8
2
= 30.3
t
m’
@ Footing
Nϕ=
[
429
𝜋𝐷
]
cos(ϕ)
=
429
πx8x
4
7.8
=33.3 t/m’
Nϕ=a Pr - Nϕ = 7.8 (8.5x1x0.75x
6.7
7.8
) -33.3= 38.025
t
m’
ϕ
H
P
R=4 m
a
x
12m
1.1 m
Fig. (6) Forces in Floor Dome
ϕ
DomeH
429
𝜋𝐷
= 17 t/m’
Fig. (7) Reactions of Dome

9
a=7.8 m h=15.2 m
v = 0.167= 1/6 (Poisson’s ratio)
t0 =0.15 m
t=0.3 m
g°
=0.375
t
m2
g=0.75 t/m2
w=1t/m2
ϕ=30°
50’
16’’
= 0.538 →radians
sin ϕ= 0.513
cos ϕ = 0.859
sin2
ϕ =0.263
cos2
ϕ =0.737
K2=1.3068√
7.8
0.3
=6.66
=
0.375x7.8
0.15x0.3
[
1+0.167
0.263
(0.15-0.3x0.859-
0.3-0.15
0.583
) -0.3x0.859] 𝑥 0.513 +0rE∆
1x7.83
0.4
[
(1-0.167)x15.2
2x7.8
-0.859+
1+0.167
3
(0.859+
1
1.859
)] x 0.513
= -694.7+301.9= -392.
Eθ°=
0.375x7.8
0.15x0.3
[ {
0.3-0.15
0.583
x0.859-2.167x0.3x0.513 +
1.167x0.15
(0.3x0.583x0.263)
}
+ {0.15x0.3x0.513-
0.3
1.167
x0.859x0.263}
1𝑥7.83
0.15𝑥0.3
{0.513+
0.15
0.3x0.538
(1.167) (
15.2
7.8x2
-
1
3
x0.859+
1
1+0.859
-
0.15
0.3x0.538
(
15.2
7.8
-0.859))} ]
=260.5+10.5=271
M=-
0.3
2x6.662 (-
392.8
0.513
-
7.8
6.66
x271) =3.66mt
S=2x0.33
x
6.66
7.8
=0.046
Fig. (8) Reactions of Dome
h = 15.2 m
8.5 m
6.7 m
h = 15.2 m
ϕ

10
1.6 Circular Beam
b =50cm t = 120cm R=4m
S=0.972 x
bt3
R2 =0.972x
0.5x1.23
42 =0.05
DomeConeMember
0.0460.265Stiffness
.1270.73D.F
+3.66-20.7F.E.M.
-17.04
+2.16+12.44D.M
+5.82-8.26Final Moment
Torsional moment of circular beam due to unbalanced moment
= -17.04 x (
0.052
0.265+0.043+0.052
) = -2.46 m.t
Notes:
 From the fig the positive moment is always with the tension side at top.
 When using moment distribution method Assume a positive sign (clockwise or
counter clock wise ) and determine whether moment In the diagram +or –
 Stiffness of circular beam taken into account for calculating D.F. because the
moment Beam (Mtorsion) needed.
 No F.E.M for circular beam.
4
1
1
2
2 3
3
4
Positive Direction
Fig. (9) Straining Action

11
1.7 Design of Sections
1.7.1 Sec 1-1
Nϕ=61.15
t
m’
t=40 cm
As=
Tu
(
fy
γs
)
=
1.4x61.15x103
=
3600
1.15
=227.34 cm2
Asoneside=
27.34
2
=13.67 cm2
Asm=
13.67
0.75
= 18.23cm2
choose 8 ϕ 18/m’
σt=
1.4x61.15x103
100x40+10x18.23
=20.5 kg/cm3
Not safe
Increase Dimensions
t=45 σt=18.2 safe
1.7.2 Sec 2-2
Mϕ= -8.25 t.m/m’
Water section Nϕ= -42.5 t/m’
ft=
-42.5x103
100x45
-
6x8.25x105
100x452 = -33.9 kg/cm2
e= (
Mϕ
Nϕ
) =
8.25
24.5
=0.194 m
e
t
=
0.194
0.45
=0.43<0.5 small eccentricity.
from table 5-1a page 260 from Elbehairy R.C Handbook → ftm =95> ft=33.9 safe
1.7.3 Sec 3-3
Mϕ=+5.82 t.
m
m’ Nϕ=-3.33
t
m’ Nθ=39.025
t
m’
t=30 cm
ft=
-3.33x103
100x30
+
6x5.82x105
100x302 =27.7 kg/cm2
e=
Mϕ
Mθ
=
5.82
33.3
=0.175

12
e
t
=
0.175
0.3
=0.583>0.5 big eccentricity.
From table
ftcu=95
kg
cm2
>ft ok safe
σc=
Nθ
Ac
=
38.025x103
100x30
=12.675 <Fct
As=As’
=0.4 % Ac=
0.4
100
x(100x30)=12 cm2
use 6 ϕ 16 / m'
5 ϕ 12/m’
longitudinal
1.7.4 Sec 4-4
Nϕ=Nθ=30.3 Safe
Use the same as at the foot ring
σc=σt=
30x103
100x15
=20.2 kg/cm2
Safe
σc all=95 kg/cm2
1.7.5 Design of circular Beam
Beam supported on columns
V=VDome+Vcone=17+30.04=47.04 t/m’
H=Hdome+ Hcone= -30.04 +28.6= - 1.44 t/m’
(Compression)
C =-1.44 x 4 = - 5.76t
From Prof. Dr. Shaker Elbehairy R.C Design Handbook page 132
P=2πr.g
g=o.w +V= γc
b t + V=2.5x.5x1.2x47.04 = 48.54 t/m’
P=2πx4x48.54= 1220 t total vertical columns
From tables
Qmax
=
P
16
=
1220
16
= 76.25 t
Mmax +ve =0.0042 pv
= 0.0042x1220x4=20.5 t.m

13
Mmax -ve = -0.0083 p.r = -0.0083x1220x4= -40.5 t.m
Mt {
due to un balanced moment = -2.46 m. t
0.0006 pv =2.928 m.t
1.7.5.1 Sec 1-1
M=-40.5 ton
C=-5.76 ton
B=150cm
A=50x120+2x30x50=9000 cm3
Y’
=
50x120+2x30x50x15
50x120x2x30x15
=45 cm
Ix=
50x1203
12
+50x120x(60-45)2
+(2x
50x303
12
+2x30x50 x(45-15)2
=11.475x106
cm4
fct=
N
A
±
Mx
Ix
y =
-5.76x103
9000
-
40.5x105
11.475x106 x45= -16.53 < 17.6 safe
e=
M
N
=
40.5
5.76
=7.03
e
t
=
7.03
1.2
=5.86>0.5 big ecc
es=e+d-z=7.03+1.15-0.45=7.73m
z=y’
=45 cm
d=115 cm
Mu=Nu.es=1.4x5.76x7.73= 62.33
d=c1√
Mu
fcB
→ 115= c1√62.33x105
250x150
C1=8.92 →J=.826
As=
62.33x105
.826x3600x115
-
1.4x5.76x103
3600
1.15
= 15.65
ASm=
15.65
0.75
=20.87 cm2
choose 9 ϕ 18
1.7.5.2 Sec 2-2
M=+20.5
N=+5.76
ft=
5.76x103
9000
+
20.5x105
11.475x106 x45=
8.68kg
cm2
<
18kg
cm2
safe section
e=
M
M
=
20.5
5.76
=3.55-1.15+0.45=2.85
Mus=Nues=1.4x5.76x2.85=22.98 mt
2
2
1
1

14
d=c1√
Mu
fcB
→115√22.98x105
250x150
C1=14.69 → J=0.826
As=
2290x105
0.826x3600x115
+
1.4x5.76x103
3600
1.15
=9.3 cm2
Asm=
9.3
0.8
= 11.82
choose 6 ϕ 16
1.7.5.3 Shear Design
qu
=
Qu
bd
Qu
=1.4x76.25=106.75 t
qu
=
106.75x103
50x115
=18.57
kg
cm2
qall
=19
kg
cm2
ok
use 4 braches at the breadth of the beam more than 40 cm
use 5ϕ10’m 4 branches
Torsion Design
qt
=
3m
b2
t
=
3x2.982x105
502
x120
=2.982<16kg /cm2
ok
No web resistance due torsion needed
900 mm
500 mm
6 ϕ 16
9 ϕ 18

15
1.8 Lateral Load Analysis and Design
1.8.1 Wind loads
Calculating the applied vertical loads
Wt of full container = 1220 ton
Wt of fall container per column = 1220/8 = 152.5 t
Wt of water = volume x γw
=853x1= 853 ton
Wt of water per column =
853
8
= 106.625 ton
Length of column or start has weight of 0.75 t/m
ℓ = 0.383 D
Level Column
Weight
Strut Weight total Cum total
0-I 4.5 2.298 6.798 6.798
I-II 4.5 2.585 7.085 13.883
II-III 4.5 2.873 7.373 21.256
III-IV 4.5 3.159 7.659 28.915
IV-V 6 3.447 9.447 38.362
∑ - - - 38.362
O.w of water tank per column @ 0-0
=container – water = 152.5-106.625=45.875t
Total weight per column @ V-V =106.625+45.875+38.362=190.862 t
Assume pressure =𝟏𝟎𝟎 𝐤𝐠/𝐦 𝟐
Element Area Y from 0-0 A.y
Lantern 1x1.5 11.5 17.25
Upper dome 12.5x1.5 10 187.5
Cylinder 12.5x6.5 8.5 690.625
Cone (12.8+8.6)/2 x2 2 42.8
Ring Beam 8.6x1 .5 4.3
∑ 131.5 - 942.475

16
Total wind force = ∑Area x pressure = 131.5x100= 13150 kg = 13.15 ton
y=
942.475
131.5
=7.17 m
Y → center of gravity of container (the point at which wind force concentrated )
1.8.1.1 Wind force on staging
wind force per meter=widthxpressure0.6x0.15=0.06 t
Horizontal
Level Hz force of wind Result
0-0 8x3x0.06 1.44
I-I (8x6+9x2)x0.06 3.96
II-II (8x6+10x2)x0.06 4.08
III-III (8x6+11x2)x0.06 4.2
IV-IV (8x6+12x2)x0.06 4.32
V-V No Hz force at foundation -
Vertical
Level Ht Mw (m.t) Resist
Mw
Cum
Mw
r Vmax
0-0 1.44 13.15
I-I 3.96 14.59 13.15(7.17+3)+1.44x3 138.06 138.06 8.5 4.06
II-II 4.01 18.55 14.59x6+3.96x3 99.42 237.48 9.5 6.25
III-III 4.2 22.63 18.55x6+4.08x3 123.54 361.02 10.5 8.6
IV-IV 4.32 26.83 22.63x6+4.2x3 148.38 509.4 11.5 11.1
V-V - 31.13 26.83x6+4.32x4 173.94 683.34 12.65 13.5
Total vertical at columns
Level Vertical force Result Wind vertical Total vertical
0-I 152.5+6.798 159.298 ±4.06 163.36
I-II 159.298+7.085 166.383 ±6.25 172.63
II-III 166.383+7.373 173.756 ±8.60 182.37
III-IV 173.756+7.659 181.415 ±11.1 192.52
IV-V 181.425+9.477 190.862 ±13.5 204.36

17
Horizontal and max B.M at columns
Level Hz per column Max Bending moment
0-I 14.59/8 1.824 1. 824x3 5.472
I-II 18.55/8 2.319 2.319x3 6.957
II-III 22.63/8 2.829 2.829x3 8.487
III-IV 26.83/8 3.354 3.354x3 10.062
IV-V 31.13/8 3.891 3.891x4 154.564
Straining action at struts
For plan the plane frame of octagonal staging all forces are multiplied by 0.146. For more
information review Dr. Mohamed Hilal Handbook
13.15 t
10 m
1.44 t
3.96 t
4.08 t
4.2 t
4.32 t
V5 t
V4 t
V3 t
V2 t
V1 t
6 m
6 m
6 m
6 m
6 m
40 m
Fig. (10) Wind Analysis of Inze tank

18
Level Wind in struts Result
0-0 - -
I-I 0.146x3.96 0.578
II-II 0.146x4.08 0.596
III-III 0.146x4.2 0.613
IV-IV 0.146x4.3 0.628
Note that at level 0-0 there is a ring beam not struts.
V1=
13.15+1.44
2
x0.146=1.065
V2=
14.59+3.96
2
x0.146=1.354
V3=
18.55+4.08
2
x0.146=1.652
V4=
22.63+4.2
2
x0.146 =1.959
V5=
26.83+4.32
2
x0.146=2.274
Moment in struts
Level Moment Result
I-I 1.065x3+1.354x3 7.257
II-II 1.354x3+1.652x3 9.018
III-III 1.652x3+1.959x3 10.018
IV-IV 1.959x3+2.274x3 14.976
1.8.2 Seismic Loads.
1.8.2.1 Calculating weights of tank elements.
W1 = A t γc = 119.6x0.15x2.5 = 44.85 t
W2 = π D b t γc = πx12x0.5x0.15x2.5 = 7.07 t
W3 = π D bw tw γc = πx12x0.25x6.5x2.5 = 153.15 t
W4 = π
𝐷1+𝐷2
2
ℓcone t γc = π
12+8
2
x2√2 x0.45x2.5 = 99.96 t
W5 = 54x0.3x2.5 = 40.5 t
W6 = πx8x1.2x0.5x2.5 = 37.70 t
W7 = πx12x0.5x0.8x2.5 = 37.7 t
W8 = 4.5+4.5+4.5+4.5+6 = 24 t

19
W9 = 8x0.383x[8 + 9 + 10 + 11] = 116.432 t
Weight of empty container = W1+W2+W3+W4+W5+W6+w7
= 420.93 t
Weight of Staging = W8+W9 = 140.432 t
ms = Weight of Structure (Mass Source) = Wc+
1
3
Ws = 420.93+
1
3
x 140.432 = 467.74 t
1.8.2.2 Center of Gravity of tank & parameters.
Y = [44.85x11.05+7.07x9.625+153.15x6.45+99.96x2.2+37.7(2.25x0.6)+40.5x2.12]/420.93= 4.46 m
Weight of Water = 853 t
D = 12 m
From ECP-203 clause 10-6-1-3 tank with irregular shape converted to an equivalent circular
tank with the same Diameter.
853 =
𝜋 𝐷2
4
h =
𝜋 122
4
x h → h = 7.542 m
h/D = 7.542/12 = 0.63
From shape in page 189 spring parameters can be determined as follows.
𝑚 𝑖
𝑚
= 0.65 → 𝑚𝑖 = 0.65x853 = 554.45 t
𝑚 𝑐
𝑚
= 0.38 → 𝑚 𝑐 = 0.38x853 = 554.45 t
ℎ 𝑖
ℎ
= 0.4 → ℎ𝑖 = 0.4x7.542 = 3.02 m
ℎ 𝑐
ℎ
=
ℎ∗ 𝑖
ℎ
= 0.68 → ℎ 𝑐 = ℎ ∗ 𝑐 = 0.68x7.542 = 5.13 m
Assume that Lateral Stiffness for the tank equals 8-10 ms
Ks =
8+10
2
x467.74 = 4206.99 t/m2/m
T𝑖 = 2π√
𝑚𝑖+𝑚 𝑠
𝑘 𝑠
= 2π√
554.45+467.74
4209.66
= 3.1 sec.
T𝑐 = Cc√
𝐷
𝑔
→ for h/D=0.63 → Cc = 3.25 sec.
1.8.2.3 Response Spectrum Design.
Subsoil Class = C @ Second Region.
R = 2.5 γ = 1 ag = 1.227 S = 1.5

20
TB = 0.1 Tc = 0.25 TD = 1.2
Sd(Ti) = 1.227x1x1.5x
2.5
2.5
x[
0.25𝑥1.2
3.12
]= 0.0575
Sd(Tc) = 1.227x1x1.5x
2.5
2.5
x[
0.25𝑥1.2
3.62
]= 0.0426
Sv(Tc) = 1.227x1x1.5x
2.5
2.5
= 1.841
Where T = 0.3 for all Types of tanks.
Basic Shear.
Vi = Sd(Ti) (mi+ms) = 0.0575(554.45+467.74) = 58.78 t
Vc = Sd(Tc) mc = 0.0426x324.14 = 13.8 t
V = √Vi
2
+Vc
2
= √58.782
+13.82
=60.38 t
Basic Moment.
M*i = Sd(Ti) {mi(h*i+hs)+msxhc.g} = 0.0575 {554.45(5.13+32)+467.74x4.46} = 1303.67
m.t
M*c = Sd(Tc) mc (h*c + hs) = 0.0426x324.14(5.66+32) = 520.02 m.t
M* = √M*i2
+M*c2
= √520.022
+1307.72
= 1403.6 m.t
For Staging the stiffness’s of columns are the same. Therefore Earthquake forces distributed
by the same value or equally.
M*/each Column =
1403.6
8
= 175.45 m.t
V*/each Column =
6038
8
= 7.5475 t
Hydrodynamic Pressure.
Maximum pressure occurs at the base of the tank wall.
@ y=0 → ϕ=0
Piw = Qiw(y) Sd(Ti) ρ h Cosϕ
@ Base
𝑦
ℎ
=
0
7.542
= 0 &
ℎ
𝐷
= 0.63 → Qiw = 0.45
Piw = 0.45x0.0575x1x7.542x1 = 0.195 t
Pcw = Qcw(y) Sd(Tc) ρ D (1-
1
3
Cos2
ϕ )Cosϕ

21
@ Base y=0 → Qcw=0.11
Pcw = 0.11x0.0426x1x12 [1 −
1
3
𝑥1]x1 = 0.0375 t
Pww = Sd(Ti) t ρ m = 0.0575x0.25x2.5 = 0.0359 t
Pv = Sv(T) ρ h (1-
𝑦
ℎ
) / R
@ Base y=0 → Sv(T) = 1.841
Pv =
1.841𝑥1𝑥7.542(1−0)
2.5
= 5.554 t
Pmax = √(Piw+Pcw)2
+Pww
2
+Pv
2
= √(0.195+0.0375)2
+0.03592
+5.5542
= 5.559 t
Max Hydrodynamic pressure at the wall of the tank is approximately equal to the maximum
pressure due to gravity loads and. Hence, considered safe.
1.9 Design of Staging Elements.
The Staging may be designed as a Shaft with a concrete slab at level from 0-V or as columns
as a preliminary design. However, tank is mainly supported by columns but shaft considered
as a logical solution for the seismic loads at the base are very large.
1.9.1 Design of staging for Wind.
M=15.564 m.t N=204.36 t
Mu=21.79 m.t Nu=286.1 t
e=
M
M
=
15.564
204.36
=0.0762
use I.D
Pu
fcub t
→→ use b x t =500x500 mm2
Pu
fu b t
=
286.1x104
25x500x500
=0.46
Mu
fcub t2
=
21.79x107
25x500x5002 =0.07
ζ=
t-2d’
t
=
500-2x25
500
=0.9
∝=1
uniform steel
∴ ρ = 8
μ=8x25x10-4
=0.02
Astotal=μ x b x t=0.02x500x500=5000
use 16 ϕ 20 (5026m2)
500 mm
500 mm
16 ϕ 20

22
Asmin=0.8% Ac=
0.8
100
x500x500=2000mm2
OK Safe
1.9.2 Design of Staging for Earthquake.
Shaft Designed as a Hollow Core Circular Column with I.D or With SAP2000 as follows.
M=1403.6 m.t N=204.361634.88 t
Dinner = 8 m
Using the thickness of wall as 50 cm → Douter = 7.5 m
Pu
fcu 𝐴 𝑐
=
1634.88𝑥103
25x6083750
= 0.011
Mu
fcu 𝐴 𝑐 𝑟
=
1403.6𝑥106
25x6083750x4250
= 0.03
ζ = 0.95 &
𝑟 𝑖
𝑟
= 0.94 → ρ = 1
Then use minimum area of steel.
@ Outer diameter use 126 ϕ 12
@ Inner diameter use 118 ϕ 12
r'
As total =244 ϕ 12
8 m

23
References
ECCS 203-2007, “The Egyptian Code the Design and Construction of Reinforced Concrete Structures”,
Housing and Building Research Center, Giza, Egypt.
ECCS 201-2013, “The Egyptian for Calculating loads and forces for Building and Structures”, Housing
and Building Research Center, Giza, Egypt.
Mohamed Hilal, “Design of Reinforced Concrete Tanks” Marcou & Co, 1988, Cairo University ,
Giza, Egypt.
Mashhour Ghoneim & Mahmoud El-Mihilmy, “Design of Reinforced Concrete Structures”,
Volume 1, Cairo University, 2014, Giza, Egypt.
Mashhour Ghoneim & Mahmoud El-Mihilmy, “Design of Reinforced Concrete Structures”,
Volume 2, Cairo University, 2014, Giza, Egypt.
Awad Elmansy, “Design of Reinforced Concrete Tanks”, Nile Academy for Sciences &
Technology, 2016, Mansoura, Egypt.

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