 Open Access
 Total Downloads : 806
 Authors : Dr. Mohammad Tariq, Er. Mohammed Naeem Mohammed
 Paper ID : IJERTV2IS3096
 Volume & Issue : Volume 02, Issue 03 (March 2013)
 Published (First Online): 08032013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Parametric Study Of The Material On Mechanical Behavior Of Pressure Vessel
Dr. Mohammad Tariq
Assistant Professor Mechanical Engineering Department
SSET, SHIATSDeemed University, Naini, Allahabad,
U.P., India211016
Er. Mohammed Naeem Mohammed Midland Refineries Company, Ministry of Oil, Republic of Iraq
Abstract
The parametric study for pressure vessel which is considered as one of the most significant applications in the daily life under action of the internal pressure that resulting from its operation was implemented. The parametric study of the considered pressure vessel is viewed from two main aspects, material problem and study the effect of structural thickness on the structural behavior of the vessel. A176 carbon steel alloy is used as modeling material. The study of thickness problem is viewed topologically for pressure vessel by varying the vessel thickness gradually to predict the mechanical behavior of the vessel versus thickness variation. The technique was included a finite element modeling of the vessel using highorder isoperimetric plate elements and used to create thick wall cylindrical. The vessel model has been drawn using Ansys Parametric Design Language, in order to implement the parametric study of the problem together with the Ansys package capability.
Keywords pressure vessels, stress, strain, ansys, FEA.

Introduction
Two types of analysis are commonly applied to pressure vessels. The most common method is based on a simple mechanics approach and is applicable to thin wall pressure vessels which by definition have a ratio of inner radius (r) to wall thickness (t) of
( 10). The second method is based on elasticity
solution and is always applicable regardless of the
(r/t) ratio and can be referred to as the solution for
thick wall pressure vessels. Both types of analysis are discussed here, although for most engineering applications, the thin wall pressure vessel can be used [22].
Since the vessel is under static equilibrium, it must satisfy Newton's first law of motion. In other words, the stress around the wall must have a net resultant to balance the internal pressure across the crosssection [18].
The internal pressure generates three principal stresses, i.e., a circumferential stress (t), an axial stress (a) and a radial stress (r). As in the case of the cylinder, it has to be examined the tri axial status of stresses, and determine an ideal stress through one of the theories of failure. By setting the ideal stress almost equal to the allowable stress, an equation will be obtained to calculate the minimum required thickness [1].

Parametric study variables
The pressure vessels performance depends on many parameters. These parameters can be classified as geometrical parameters, material properties and operating conditions. For each configuration, certain properties of the wall can be calculated. Ranges for each parameter are determined, taking A176 as modeling material reference.
Geometrical parameters are selected to be the wall thickness of the vessel, modeling material of the vessel depending on the application field. It is easily noticeable, that these parameters define the cylindrical vessel with open ends. Hence, different combinations of these variables can be picked to examine the system and obtain relations with the cylindrical wall performance [14].
Solution of the problem is expressed in terms of two parametric functions. The relationship between them suggests that Lames elastic solution and solution for
perfectly elastic material depend on special choices for these parameters. Both solutions use linear material behavior [4].

Thick cylinder
Thick walled cylinders subjected to high internal pressure are widely used in various industries. In general, vessels under high pressure require a strict analysis for an optimum design for reliable and secure operational performance. Solutions have been obtained either in analytical form or with numerical implementations.

Stress analysis in thick cylinder
In general, pressure vessels designed in accordance with the ASME code, section VIII, division 1, are designed by rules and do not require a detailed evaluation of all stresses. It is recognized that high localized and secondary bending stresses may exist but are allowed for by use of a higher safety factor and design rules for details. It is required, however, that all loadings (the forces applied to a vessel or its structural attachments) must be considered.
While the code gives formulas for thickness and stress of basic components, it is up to the designer to select appropriate analytical procedures for determining stress due to other loadings. The designer must also select the most probable combination of simultaneous loads for an economical and safe design. The code establishes allowable stresses and states that the maximum general primary membrane stress must be less than allowable stresses outlined in material sections. Further, it states that the maximum primary membrane stress plus primary bending stress may not exceed 1.5 times the allowable stress of the material sections. These higher allowable stresses clearly indicate that different stress levels for different stress categories are acceptable [8].


Finite element model (FEM)
The proposed finite element model and their mesh along global X and Ycoordinate, location and numbers of the generated nodes and element for the two cases have shown in figs 1.1 and 1.2.
For the case 1, Solid42 plane 4node elements are used to build this model. The drawn rectangular surface is discretized in to 155 plane elements at size smart 0.1mm. For case 2, Solid82 plane 8node
elements are used to build this model. The drawn rectangular surface is discretized into 429 plane elements at size smart 0.05mm.
ANSYS finite element package can be created to build and analyze of this model, according to the following steps:
The domain of the model is assumed to be rectangular drawn through the plane of global X and Y axis; their dimensions are R1 and R2 along global Xaxis and depth (height) along global Yaxis for all the four cases taken in this work.
Fig 1.1 Finite element modeling of case 1
Fig 1.2 Finite element modeling of case 2

Geometrical shape of case 1 and case 2
Case 1 and 2 having the same geometrical shape and this can be explained as shown in fig (1.3).
Fig 1.3 geometrical shape of the cylinder
Elements shown in figs. 1.4 and 1.5 have been taken in present work. The stress t constant along the circumference, is exercised on sides AB and CD.
Fig. 1.4 Illustrate the forces in a cylinder
Fig. 1.5 Illustrate the element is subjected to stresses
As far as the congruence of deformations are concerned, by assuming the circular ring of thickness dr shown in fig 1.6.
Fig 1.6 the circular ring
Because of the circumferential r the radius of circle (), has an elongation (t). Deformation given by,
r = t = r (6)
The radius of circle () in turn has an elongation
r = t + dt (r + dr) (7)
To impose congruence, the difference between these
two elongations must correspond to the thickness increment of the ring i.e.
t = r dr (8)
or
r r = t (9)
From (6) and (8)
r t
r t
+ r dt = 0 (10)
dr
Equation (10) is the equation of congruence of the
cylinder and,
E
E
t = 1 t ta (11)
E
E
r = 1 r ta (12)
E
E
a = 1 a tr (13)
E is normal modulus of elasticity and is the Poissons ratio.
r1 + (14)
r1 + (14)
p
p
Principal stresses are calculated by using Lames equations as follows,
Forces exerted on sides AC and BD are given by
FAC = r rd (1)
FBD = r + dr r + dr d (2)
t =
r =
2
e
a2 1 r2
r1 (15)
r1 (15)
p
p
2
e
a2 1 r2
Finally, on the sides AB and CD
FAB = FCD = t dr (3)
The resultant based on such two forces in the radial
direction is
p
a =
a =
a21

Displacement calculation
(16)
FBD = t drd (4)
According to the previous equations, with d having a nonzero value, will obtain
The displacement in the thick wall cylinder is given by,
dr
u = 1 pi r po r r + 1+ pi po r r 1
2 2 2 2
2 2 2 2
(17)
o
r = 0 (5)
i o i o
i
i
r 2 2
t r dr
E r2o ri
E ro r2 r
This is the equilibrium equation of the cylinder.

Coordinates systems
There are three types of coordinates systems used in above finite element models, these are:

The global coordinates system.
N1 , n = 1 1 (1 ) N , n = 1 1 + (1 )
4
4
2
2
4
N3 , n = 1 1 + (1 + )
(20)

The nodal coordinates system. 4

The element coordinates system.


StressStrain Relationship
The axisymmetric stressstrain relations in cylindrical coordinates aligned with principle material directions is given by,
N4 , n = 1 1 (1 + )
4
4
where and are local coordinate for elements. For this element the matrix [B] is given by,
J
J
B = 1 B1 B2B3B4 (21)
Bi is given by,
a Ni b Ni 0
r (1 ) 0 r
= E(1)
(1 ) 0
(18)
B =
0 c Ni d Ni
(22)
(1+)(12)
(12)
i
r 0 0
2 r
c Ni d Ni
a Ni b Ni
where r is the radial stress, is the tangential stress, r is the shear stress, is the Poissons ratio
Where;
and E is the modules of elasticity.

Element Parameters

a = 1 y1 1 + y2 1 + y3 1 + + y4 1
4
4
4
4
b = 1 y1 1 + y2 1 + y3 1 + + y4 (1 ) c = 1 x1 1 + x2 1 + x3 1 + + x4 (1 )
4
All above finite element models have been created using linear fournode quadrilateral plane and eight node elements. This type of element is used for idealization of pressure vessel (Thick cylinder) in 2D. In this section, the parameters that are concerned with the selected element are discussed. These parameters are basically included the element property parameters and the material properties at each node
d = 1 x1 1 + x2 1 + x3 1 + + x4 (1 )
4
4
(23)
Since the shape functions N are functions of the local coordinates rather than Cartesian coordinates, a relationship needs to be established between the derivatives in the two coordinates systems. By using the chain rule, the partial differential relation can be expressed in matrix form as,
of the structure pressure vessel. The material
Ni
x y
Ni
properties for whole vessel are specified as isotropic
=
x (24)
material. The element degrees of freedom are
Ni
x y
Ni
y
assigned at each node along the element coordinate system. The displacement at each node is given by,
where [J] is the Jacobian matrix and the elements of this matrix can be obtained by differentiating the
k
k
u(, n)
=
N (, n)
ui
(19)
following equations;
v(, n)
i=1 i
vi
x , = 4
N , x
k is 4 for plane 42 four node and 8 for plane 82 eight
i=1 i
i (25)
node, Ni is the shape function, ui and vi are global
y , = 4 1 Ni , yi
i=
i=
nodal displacements, and are local coordinate for
4 Ni xi 4
Ni yi
J =
i=1
i=1
(26)
x
x
elements. It is obvious that each node has two degree of freedom, and then the element is of eight degrees
4 Ni i=1 i
4 Ni i=1 i
y
y
of freedom in plane 42 and sixteen degrees of freedom in plane 82. Not all but some of the element
Then, the derivatives of the shape function with respect to Cartesian coordinates can be given as:
degrees of freedom are considered at each of the
Ni
Ni
finite element models, depending upon the function
x = J 1 (27)
(boundary conditions) of that model.
Ni
y
Ni
Where [J]1is the inverse of Jacobian matrix given by,
2.5.1 Linear four node quadrilateral plane42
The shape function for plane 42 four node is represented as,
J 1 = x
y
x (28)
y
The determinant J is given by,
J = 1 x1 x2 x3 x4 =
8
0 1 1
The geometrical nonlinearity is not considered in the
1
0
1
1 +
0
1 + (29)
present work hence, the engineering components of strain can be expressed in terms of first partial
1 + 1 0
For the case of plane stress, matrix D is given by,
1 0
derivatives of the displacement components.
Therefore, the linear straindisplacement relation at any point on element and for two degrees of freedom
2
2
D = E
1
0 (30)
1
per node can be written as;
r = u , = u and z = w .
0 0
2
For the case of plane train, matrix D is given by,
r r z
r = u + w
(38)
s
1 0
z r
D = E
1+ (12)
1 0
0 0 12
2
(31)
2.7 Boundary conditions
The specify boundary conditions include the
2.5.2 Quadratic eightnode quadrilateral plane 82
The shape functions are;
4
4
N1 = 1 1 (1 ) 1
2
2
N = 1 1 + (1 ) 1
4
N3 = 1 + (1 + ) + 1
4
4
N = 1 1 1 + + 1
4
constraint and free state of the degrees of freedom at each node in the finite element models. The state of the degree of freedom is constrained or free, depending upon the model itself. From the all model to the eight model, the cylindrical structural model is fixed from the bottom joining nodes thus all these nodes are constraint in all their degrees of freedom. Thus in the corresponding finite element model, the
5
5
N = 1 1 1 + 1
4
6
6
N = 1 1 + 1 + 1
4
7
7
N = 1 1 + 1 + 1
4
(32)
nodes that are located on the end of cylinder are of constrained degrees of freedom, while all other nodes above the bottom are free of degrees of freedom as shown in fig 2.1.
4
4
N8 = 1 1 1 + 1
The Jacobian Matrix for this element is given by,
x
x
x
J =
y
y
y
(33)
Where x and y are given by:
N1x1 + N2x2 + N3x3 + N4x4+ N5x5 + N6x6 + N7x7 + N8x8 N1y1 + N2y2 + N3y3 + N4y4 + N5y5 + N6y6 + N7y7 + N8y8
(34)
The matrix [B] for this element is given as follows:
Figure 2.1 boundary condition of the cylinder
B = D N (35)
2.8 Static analysis
0
0
y () y ()
Static analysis is achieved on each of the finite
D = 1
0 x () x ()
(36)
element models for each function with their
J
corresponding boundary conditions and load sets.
x () x ()
y () y ()
Static analysis solution has been included the
N =
calculation of the effects of the applied static
N1 0 + N2 0 + N3 0 + N4 0 + N5 0 + N6 0 + N7 0 + N8 0
0 N1 + 0 N2 + 0 N3 + 0 N4 + 0 N5 + 0 N6 + 0 N7 + 0 N8
(37)
2.6 Straindisplacement relationship
distributed loads on each model for each function with the corresponding boundary conditions. These effects included displacements, strains, and stresses that are induced in the structure due to the applied
loads. The static analysis is governed by the following equilibrium equations (in matrix notation): [K].{u} = {F} (39)
where [K] the assembled stiffness matrix is given by,

Case study
As parametric study, the pressure vessel problem is based on geometrical investigation. The
=1
=1
=
(40)
implementation of comparative study between two
ANSYS solve the above equilibrium equations to obtain the following results:

Displacements of each node along their free degrees of freedom.

Strains and stresses at each element along element coordinate axis.

Principle stresses and strain values and directions with respect to the element coordinates axis at each element.

VonMises stresses and the maximum shear stresses.


Material modeling
The modeling material for pressure vessel is selected from carbon steel alloys scheduled in table (1).
Table (1) Properties of modeling material
Material
E (Gpa)
yield (Mpa)
ultimate (Mpa)
(kg/m3)
Carbon steel (A176)
206.83
205
415
7800
0.3

Specification of cases studies
Table (2) State of cases
Case
R1(m)
R2(m)
L
(m)
Material
Type of element
Pressure (Mpa)
Thickness (m)
1
1
1.4
3
A176
Solid 42
5
0.4
2
1
1.4
3
A176
Solid 82
5
0.4

Assumptions
1 Internal pressure applied to inner area. 2 Fixed base of cylinder (Ux, Uy) = 0.

Modeling element restricted to solid element only.

Using Vonmises criterion in the estimation of stress level of elements.

Modeling material restricted to isotropic material only.

No bending, no torsion, no external pressure and no thermal effect.

Axisymmetric pressure vessel is considered in present work.

Open end thick cylinder.


Results
element types solid42 and solid82, by selecting the most applicable materials used for manufacturing of pressure vessels (A176 austenitic stainless steel type 410S UNS S41008).

Detail of case studies
The present study is restricted to take 2case studies to illustrate the difference among them by showing the contour plots that captured immediately from ANSYS program.

Comparison with maximum stress
Table (3) Maximum stress of cases
Cases
Stress(Mpa)
r
von mises
Case 1
5.0271
15.541
18.518
Case 2
4.9558
15.372
18.359

Comparison with maximum displacement
Table (4) Maximum displacement of cases
Cases
Displacement (m)
Ur (m)
Uabsolute (m)
Case 1
82407E9
93831E9
Case 2
81790E9
93510E9

Comparison with maximum strain
Table (5) Maximum strain of cases
Cases
Strain (max)
r (m)
(m)
z (m)
von mises
Case 1
47316E10
81962E10
1604E10
11639E9
Case 2
4625E10
81512E10
15109E10
11539E9

Comparison at A176 and element at thickness 0.4m
Table (6) Comparison between case 1 and case 2
Mechanical behavior
Case1 solid42
Case 2solid 82
r max (Mpa)
5.0271
4.9558
max (Mpa)
15.541
15.372
von mises max
(Mpa)
18.518
18.359
Ur max (mm)
82407E9
81790E9
Uabsolute max
(mm)
93831E9
93510E9
r max
47316E9
46250E9
max
81962E9
81512E9
z max
16040E9
15109E9
von mises max
116390E9
115390E9
Mechanical behavior
Case1 solid42
Case 2solid 82
r max (Mpa)
5.0271
4.9558
max (Mpa)
15.541
15.372
von mises max
(Mpa)
18.518
18.359
Ur max (mm)
82407E9
81790E9
Uabsolute max
(mm)
93831E9
93510E9
r max
47316E9
46250E9
max
81962E9
81512E9
z max
16040E9
15109E9
von mises max
116390E9
115390E9

VonMises stress distribution in (pa) case 1

VonMises stress distribution in (pa) case 2

Fig. 3.1 Illustrative contours for nodal VonMises stress distribution
3.2 Parametric study: Case 1 and Case 2
3.2.1 Stress investigation
Through the investigation of stress for case 1, it is noted that the stress VonMises having the maximum value at node 36 which is equal to 18.518 Mpaas compared to case 2 as mentioned in table (3). It is noted that von mises max for case 2 having value

Mpa at node 1 which is less than that obtained
for case 1 due to different in type of element used as shown in figures (3.1 a, b).

Absolute displacement distribution (case 1)

Absolute displacement distribution (case 2)
Figure 3.2 Illustrative contours for nodal absolute displacement distribution
3.3 Analytical calculation
For a thick wall cylinder with the parameters as shown in table (2) the stresses, strains and displacement calculated by the model using equations
(11) through (17) are given for all the cases and the results were compared with the numerically analyzed by finite element method.

VonMises strain distribution case (1)
The results obtained have been plotted in Origin 6.1 for given cases. Two different methods have been applied for the calculation of stresses and strains at internal pressure (5MPa).
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Theoretical FEA
CASE 1
Theoretical FEA
CASE 1
5 5
4 4
Radial Stress
Radial Stress
3 3
2 2
1 1
0 0

VonMises strain distribution case (2)


Fig 3.2 Illustrative contours for nodal VonMises strain distribution
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.1) Radial stress vs. thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40

Displacement investigation
Through the investigation of displacement for case 1, it is noted that the displacement having the maximum value at node 36 which is equal to 93831E9 (m) as compared to case 2 as listed in table (4). It is noted that Uabsolute max for case 2 having value 93510E9
(m) at node 70 which is less than that obtained for
case1 due to different in type of element used as shown in figure (3.2a, b).

Strain investigation
Through the investigation of strain for case 1, it is noted that the strain having the maximum value at node 36 which is equal to 116390E9 as compared to case 2 as shown in table (5). It is noted that
von mises max for case 2 having value 115390E9 at node 1 which is less than that obtained for case1 due to difference in type of element used as shown in figure (3.3 a, b).
16 16
Theoretical FEA 

CASE 1 

Theoretical FEA 

CASE 1 

15 15
Tangential Stress
Tangential Stress
14 14
13 13
12 12
11 11
10 10
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.2) Tangential stress vs. thickness of cylinder
CASE 1 
Theoretical FEA 

CASE 1 
Theoretical FEA 

1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
0.000082
0.000080
Displacement
Displacement
0.000078
0.000076
Theoretical FEA
CASE 1
0.000082
0.000080
0.000078
0.000076
1.51E005
1.51E005
1.51E005
1.51E005
Axial Strain
Axial Strain
1.51E005
1.51E005
1.51E005
1.51E005 1.51E005
1.51E005
1.51E005
1.51E005
0.000074
0.000072
0.000070
0.000074
0.000072
0.000070
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.51E005
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.3) Displacement vs. thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
0.00005 .000050
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.6) Axial strain vs thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
0 0
0.000045
0.000040
Radial Strain
Radial Strain
0.000035
0.000030
Theoretical FEA
CASE 1
0.000045 5
0.000040 4
Radial Stress
Radial Stress
0.000035
3
0.000030
CASE 2
5
Theoretical FEA
4
3
0.000025
2 2
0.000025
0.000020
0.000020 1 1
0.000015
0.000015
0 0
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.4) Radial strain vs. thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.7) Radial stress vs. thickness of cylinder
0.000085
0.000085
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
0.000080
0.000075
Tangential Strain
Tangential Strain
0.000070
Theoretical FEA
CASE 1
16
0.000080
15
0.000075
Tangential Stress
Tangential Stress
0.000070 14
16
CASE 2
15
Theoretical
FEA
14
0.000065
0.000065 13 13
0.000060
0.000060
12 12
0.000055
0.000050
0.000055
11 11
0.000050
10 10
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.5) Tangential strain vs. thickness of cylinder Fig (3.8) Tangential stress vs. thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Theoretical FEA 

CASE 2 

Theoretical FEA 

CASE 2 

0.000082 
0.000082 

0.000080 
0.000080 

0.000078 
0.000078 

0.000076 
0.000076 

0.000074 
0.000074 

0.000072 
0.000072 

0.000070 
0.000070 
0.000082 
0.000082 

0.000080 
0.000080 

0.000078 
0.000078 

0.000076 
0.000076 

0.000074 
0.000074 

0.000072 
0.000072 

0.000070 
0.000070 
1.51E005
1.51E005
1.51E005
CASE 2
Theoretical FEA
1.51E005
1.51E005
1.51E005
1.51E005
Displacement
Displacement
Axial Strain
Axial Strain
1.51E005
1.50E005
1.50E005

E005

E005
1.51E005
1.50E005
1.50E005
1.50E005
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.9) Displacement vs. thickness of cylinder
1.0 1.1 1.2 1.3 1.4
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.12) Axial strain vs. thickness of cylinder
3.4 Comparison of the analytical and
0.000050
0.000045
0.000040
Radial Strain
Radial Strain
0.000035
0.000030
0.000025
0.000020
0.000015
0.000010
CASE 2
Theoretical FEA
0.000050
0.000045
0.000040
0.000035
0.000030
0.000025
0.000020
0.000015
0.000010
numerical results
The results of the stress, strain and displacement distribution obtained from analytical (thickwalled cylinder theory, Lames equations) and numerical techniques (FEA) were compared with respect to radial stress versus radius at internal pressure. The stress obtained from the two methods change linearly and gradual decrease from inner radius to outer radius with the applied internal pressure along the wall thickness of the cylinder is shown in figs (3.1) and (3.7). The highest radial stress is found at inner
1.0 1.1 1.2 1.3 1.4
Radius
Fig (3.10) Radial strain vs. thickness of cylinder
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
radius i.e. at the inner wall of the cylinder. It can be seen that the results obtained from the two techniques are in good agreement.
The tangential stress obtained from the analytical and FEA methods change linearly with the applied
CASE 2 Theoretical FEA 

CASE 2 Theoretical FEA 

0.000085
0.00000
0.000075
Tangential Strain
Tangential Strain
0.000070
0.000065
0.000060
0.000055
0.000050
0.000085
0.000080
0.000075
0.000070
0.000065
0.000060
0.000055
0.000050
internal pressure. In figs (3.2) and (3.8), gradual decrease in the tangential stress from inner to outer radius has been shown. The highest tangential (hoop) stress is found at the inner radius i.e. at the inner wall of the cylinder. It can be seen that the results obtained from the two techniques are in good agreement.
The displacement distribution obtained from two methods shows a gradual decrease. The highest displacement is found at inner radius as shown in figures (3.3) and (3.9). It can be seen that the results obtained from the two techniques are in good
1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40
Radius
Fig (3.11) Tangential strain vs. thickness of cylinder
agreement.
The radial strain obtained from two methods change linearly with the applied internal pressure. The highest radial strain is found at inner radius and decreases gradually to outer radius as shown in figures (3.4) and (3.10). It can be seen that the results
obtained from the two techniques are in good agreement.
The tangential strain obtained from two methods change linearly with the applied internal pressure as shown in figures (3.5) and (3.11). It can be seen that the results obtained from the two techniques are in good agreement.
The axial strain obtained from two methods is remained constant through wall thickness as shown in figures (3.6) and (3.12). The axial strain in FEA change through wall thickness is highest at inner radius and decreases to outer radius and in theoretical method the axial strain remain constant through wall thickness. It can be seen that the results obtained from the two techniques are in good agreement.
List of symbols
F – Force
P – Pressure

– Diameter
t – Thickness
r – Radial Stress
– Tangential Stress
a – Axial Stress
– Radians of circle
r – Radial Elongation
/t – Tangential Elongation
z – Axial Elongation

– Modulus of elasticity
– Poissons ratio
ur – Radial displacement
Ux, Uy – Nodal displacement
Sx, Sy – Nodal stress
x, y – Global coordinates
R1 – Inner radius of model
R2 – outer radius of model
r – Shear Strain
r – Shear Stress
Ni – Shape function
, – Local Coordinate
[B] – Strain matrixJ – Jacobian
[D] – Constitutive matrix– Potential energies
SE – Strain energy
WF – Work done
WF – External work
[K] – Stiffness matrix [F] – External applied force matrixJ – Determinate at the jacobian matrix
L – Length
t – Hoop stress
a – Longitudinal stress
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