1130030 - Extra Materials

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Reliability Analysis on Shell Design of Large Oil Storage Tanks
Wang He1, a, Li-chuan Liu1, b, Jin-lin Yang1, c
1
Dept. of petroleum supplying engineering, Logistical Engineering University, Chongqing, 401311,China
(ahewang326@163.com, blicliu@sina.com, cyangjinlin9075@163.com)
Abstract - The paper studies the procedures of the shell
design and reliability analysis of large oil tanks and presents
the results of calculation and analysis on the strength-stress
model of tanks’ wall reliability by checking point’s method.
Taking a 100000m3 tank for example, shown is the analysis
and calculation on the reliability of the wall design with
different numbers of shell courses, and also resulted is the
numbers of shell courses based on the reliability of the
holistic tank wall.
Keywords - Large oil tank, Tank-wall design, Shell
thickness, Shell course, Structural reliability.
I. INTRODUCTION
Usually, the amount of steel used in a tank wall
would be 50% of the whole used. So the design of a tank
wall doesn’t only affect its function and reliability, but
also does its cost. According to the requirement for a
tank’s strength, its wall thickness should increase linearly
along with static hydraulic pressure from top to bottom.
However, in engineering practice, the wall of tanks is
made of the steel-plates with different thickness. On the
premise of optimal radius and height, here more important
is what steel-plate width of the shell courses could keep
higher reliability of the holistic tank-wall. Therefore,
based on the design of tank-wall, analyzed is reliability of
a tank-wall structure.
II. METHODOLOGY
A. Wall Design
On the basis of design capacity and least amount of
material used, the formula of calculating the
tank-wall-height with different thickness of shell plates
can be described as Eq.1
H  
(1)
Where H is tank-wall-height (m); is a coefficient,
related to tank design parameter;  is thickness
summation of tank-bottom and tank-roof. With the above
equation, the tank diameter and height could be defined as
the basic structure parameters of a designed tank.
The wall design of oil steel tanks [1][2] can be divided
into two groups, with the same and different shell wall
thickness. The capacity below 1000m3 is more economic
with the former, while the tanks with a larger capacity
should be designed of the variable shell thickness. So the
thickness of every course decreases along with the wall
from bottom to top. A floating-roof tank should keep the
all courses with the same diameter, inside for the floating
tray available to move up and down. The maximum stress
of every course is at the point a little higher than the
bottom of shell course end because of the below course’s
hoop stress. The one-foot method [3][4], usually, is applied
in the tanks with a diameter less than 60 meter. However,
it is not irrational for the tanks with diameter more than
60m, so the variable point method [3][4][5] is applied to
reduce the required shell thickness with the same
allowable stresses. In order to determine the position of
the maximal stress of every course-plate, the impact
between the shell wall’s courses is considered in the
wall-thickness calculation called as the variable point
method. The variable point calculation of the
wall-thickness of oil tanks is shown as followed.
The bottom shell course thickness of tanks can be
obtained from the following expression [6]:
t01 
 ( H  0.3) D
2[ ]
t01  [1.06 
or
0.0696 D
H 9.8HD
]
H
[ ] 2[ ]
(2)
Where t01 is the design thickness of the wall bottom
course, mm; H is the uprightness height from bottom of
shell to top angle or to bottom of overflow, m; D is the
nominal tank diameter, m;  is the per- weight of the
liquid, t/m3;  is joint efficiency; [ ] is the steels
allowable design stress for calculating plate thickness,
and taken by minimum between
0.75 s and
3
b
7
(MPa).
The thickness of the second shell course of tank wall
is calculated by the following procedures. When the
bottom shell course of the wall-thickness t01 has been
calculated, the width of the course-plate should be judged
first by the following three governing conditions and then
the thickness of the second shell course given by the
respective expressions [6]:
hi 1
 1.375,t0i  t0(i 1) .
Rt0(i 1)
hi 1
 2.625,t0i  tai .
Rt0( i 1)
1.375 
(3)
hi 1
 2.625 ,
Rt0(i 1)
hi 1
t  tai  (t
 t )(2.1 
)
0i
0(i 1) ai
1.25 Rt0(i 1)
Where
hi 1 is the height of bottom shell course, m;
R is the tank radius, m; t0i is the final thickness of the
ith shell course, mm; tai is the thickness of the ith shell
course by the variable point method, mm.
tai can be
[6]
obtained by resolving the following equations .
 (H i  X i )D
2[ ]
X i1  0.61 Rt ai  0.32Ci H i
t ai 
(4)
X i 2  Ci H i
(5)
X i 3  1.22 Rt ai
H i is the hydraulic height of the ith shell
course’s bottom, m; K i is the distance above the lower
( K i  1) K i
1  Ki Ki
;
K i is
thickness lower course at joint/thickness upper course at
t0( i 1)
tai
(7)
r s
Where R(t) , P ,
f r (r ) , f s ( s) are reliability degree,
probability, density function about strength, density
function about stress.
In engineering practice, it is complex even almost
impossible to calculate the above integral with an explicit
result, even if the simplest functional function [7].
According to the engineering statistical investigation,
random variables about stress and strength would follow
the normal distribution or logarithm normal distribution
model. When random variables are assumed as normal
distribution and independent one another, the reliability
could be expressed as [5]
R(t )  1  (
and
Where
joint, equated with
R(t)=P(y>0) =  f r (r ) f s ( s)drds .
r  s
 r2   s2
)  (
r  s
 r2   s2
)
(8)
Where  r ,  s are the average of strength and stress
X i  min( X i1,X i 2,X i 3 )
end of the ith shell course; Ci 
physical model in the reliability design of static
structures.
Then limit state equation [7] can be expressed by:
y (r , s )  r  s
(6)
Where r is the strength; s is the hoop stress.
When the distributing for the random variables is
known, the reliability can be calculated by the following
equation [4]:
; others are the same as the
meaning shown above.
The thickness of the upper shell courses is related
with the thickness of lower shell courses. The thickness of
each course is determined by a common stress, and the
theoretical location of the design point is at a variable
distance above the bottom of the course. The distance is
lowest value obtained from the above expressions (4) (5).
B. Reliability Analysis of Tank Wall
Every engineering design should be provided with
certain of its reliability, and so does the wall design of oil
tanks. The reliability of structures is defined as the
probability that the structure performances its function
during a certain period of time and in the designed
condition [7].
In the cases of static loading, stress-strength
interference model is usually followed as the invalidation
 r ,  s are
stress.
the standard deviation of strength and
r  s
Let  
 r2   s2
,
And then, R (t ) =  (  )
(9)
(10)
If can be shown that the parameter  indicate the
reliability of tank shell.
The reliability model about tank wall is a weakest-chain
or series model as the following [8].
n
n
i 1
i 1
R(t )   Ri (t )   ( i )
(11)
There are several methods to calculate  , such as
central point, checking points, mapped transform,
practicality analysis, etc. Here, checking points is applied
to calculate the reliability index of tank wall and analyze
its effect on shell courses on tank wall to be analyzed.
According to membrane theory, the stress in the wall of
vertical cylindrical tank [6] is given by:
PD
(12)
2t
Where s is the hoop stress; P is the liquid pressure;
t is the uniform thickness of shell, equated with t0i .
s
So, the limit state equation becomes
Where,  r ,  D ,
PD
2t0i
y (r , s)  g ( X 1 , X 2 ,..., X n )  r  s  r 
(13)
 g[20  (i  1)hi ]D
D
r
r
0
2t0i
t0 i
r *   r   r cos 1
D*   D   D cos  2
(14)
t0i  t0i   t0i cos3
*
Where cos  X i is sensitivity coefficient, which can
be given by:
cos  X i 
X
i
2
 D* 
  
   D *     t0 i t *2 
0i
 t  


D *
t
2
 y


i 1  X i
 r
n
2
2
cos  2 
  r 
p*

* X 
p
i

cos 1 
  r 
y
X i
2
cos 3 
(15)
* 2
D 
  
   D *     t0i t *2 
0i
 t  

*
D
 t0 i t *2
2
  r 
 D* 
  
   D *     t0 i t *2 
0i
 t  

2
2
equation balanced.
A 2  B  C  0
A   r t 0 i cos 1 cos 3
B   r  cos 1  r t0 i cos 3   D cos  2
C  r    D
toi
t0i ;  r ,  D ,  t0i are the standard
(16)
t0i . Cyclic iterative
C. Example Case
Taken consideration of is the case: the volume
capacity of the tank is: 100000cbm, steel material:
thickness range: 10–34mm, strength: 610–730MPa, yield
stress r: 490MPa, allowable stress [ ] :327MPa, shell
width:
1000–3000mm,
store
liquid
density  :1000 kg/cm3, variation coefficient of strength
limitation : 0.05–0.1, variation coefficient of steel-size :
0.03-0.05.
First, according to (1), H is 21.8m, set it into 22m,
For the 100000cbm of volume capacity, the diameter is
about 80m.
Then, suppose tank made of the same width
shell-course, number of shell-course is from 7 to 14 for a
tank-wall. The wall thickness is calculated by compiling
the program of Microsoft Visual Basic 2005. If the steel
width hi 1 is known, the wall thickness can be obtained.
As shown in figure 1 is the wall thickness with different
course-number.
And then, in order to calculate the reliability index,
the average and standard deviation of three random
variables must be known.
The variation coefficient of strength limitation gets
0.05–0.1 and that average gets the geometry size and
criterion deviation gets 1/3 deviation for its statistical
meaning[8]. 0.1 is used as the variation coefficient of
strength, namely   490,    49 MPa. Literature [9]
namely
Put the average of every random variable as test
checking point P0 ( r ,  D , t0 i ) , and make the following
t0 i
are the average of strength,
method can be used to calculate the checking points and
 for different courses and different thickness.
points
0i
2
toi
deviation of strength, diameter and
Where  is a coefficient and constant to every
course with the designed shell width. By the above limit
state equation, there are three random variables. In order
to decide the design-check points for the reliability, the
values of these random variables at the design-check
point need to be calculated as the following.

diameter and

out
that
D  76m,  D  0.021
D  80,  D  0.021 m.
,
Given the courses,
thickness can get a fixed value, which can act as average
of wall thickness. 0.04 is used as the variation coefficient
of the thickness of tank-wall.
In order to calculate reliability indices, a program is
compiled and calculated checking points and reliability
indices  for different shell-course number and different
thickness. The results of the calculation are shown in the
following [10].
26
5
24
4.8
4.6
reliability index
thickness
22
20
18
16
14
4.4
4.2
4
3.8
3.6
3.4
12
3.2
10
1
2
3
4
5
6
7
8
3
9 10 11 12 13 14
1
2
3
each shell-course
4
5
6
7
8
each shell-course
9courses
10courses
11courses
12courses
13courses
14courses
Fig. 1 Thickness of every shell-course with different
numbers of shell-course
9courses
12courses
10courses
13courses
11courses
14courses
Fig. 3 Reliability index of every shell-course with
different numbers of shell-course
0.97
77.6
0.965
0.96
0.955
77.2
reliability
steel consumption amount
77.4
77
76.8
0.95
0.945
0.94
0.935
76.6
0.93
0.925
76.4
0.92
1
76.2
9
10
11
12
13
2
Fig. 2 Steel consumption amount in volume about tank
wall with different numbers of shell-course
4
5
6
7
8
each shell-course
14
numbers of shell-course
3
9courses
10courses
11courses
12courses
13courses
14courses
Fig. 4 Reliability of every shell-course with different
numbers of shell-course
of shell-course, the more obvious is the influence from
the restraint of the tank bottom in the shell-course design.
The heights of minimum thickness and numbers of
shell-course have several impacts on tank-wall’s whole
reliability and steel consumption amount for large tanks
in fig. 5.and fig. 6. There is one shell-course number to
make whole reliability of tank-wall maximum when
taking no account of influence on reliability of weld
number in progress of reliability calculation. But if weld
is considered in reliability, the optimal number of
shell-course will change and number of shell-course will
become fewer. The upper part of the tank-wall can be
taken place by other steel material that strength is lower
because it requires lower respectively.
0.9994
all tank-wall reliability
0.9992
0.999
0.9988
0.9986
0.9984
0.9982
0.998
9
10
11
12
13
REFERENCES
14
[1]
numbers of shell-course
Fig. 5 Reliability of the whole tank-wall with different
numbers of shell-course
[2]
[3]
height of mininum thickness
11.5
[4]
11
[5]
10.5
10
[6]
9.5
[7]
9
[8]
8.5
9
10
11
12
13
14
[9]
numbers of shell-course
[10]
Fig. 6 height of minimum thickness with different
numbers of shell-course
III. RESULTS AND DISCUSSIONS
Wall thickness is not variable with different numbers
of shell-course at bottom course in fig. 1. The steel
consumption amount in volume about the whole
tank-wall becomes fewer with numbers’ increasing as a
whole in fig.2. Reliability is relatively higher when the
thickness calculated less than the maximum that
design-code requires in fig. 3 and fig. 4. Reliability of the
second shell-course is increasing with shell-course
number increasing, that is to say, the lower are numbers
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