Parametric Study Of The Material On Mechanical Behavior Of Pressure Vessel

Parametric Study Of The Material On Mechanical Behavior Of Pressure Vessel

Dr. Mohammad Tariq

Assistant Professor Mechanical Engineering Department

SSET, SHIATS-Deemed University, Naini, Allahabad,

U.P., India-211016

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 high-order 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.

1. 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 cross-section [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].

   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].

   2. 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.

   3. 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].

2. Finite element model (FEM)

   The proposed finite element model and their mesh along global X- and Y-coordinate, 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 4-node 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 8-node

   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 X-axis and depth (height) along global Y-axis 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

   1. 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 A-B and C-D.

      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 A-C and B-D 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 A-B and C-D

      FAB = FCD = t dr (3)

      The resultant based on such two forces in the radial

      direction is

      p

      a =

      a =

      a21

   2. 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.

   3. Coordinates systems

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

      1. 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)

      2. The nodal coordinates system. 4

      3. The element coordinates system.

   4. Stress-Strain Relationship

      The axi-symmetric stress-strain 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.

   5. 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 four-node 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 eight-node 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,

1. 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:

   1. Displacements of each node along their free degrees of freedom.

   2. Strains and stresses at each element along element coordinate axis.

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

   4. Von-Mises stresses and the maximum shear stresses.

1. 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

2. 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

3. Assumptions

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

   1. Modeling element restricted to solid element only.

   2. Using Von-mises criterion in the estimation of stress level of elements.

   3. Modeling material restricted to isotropic material only.

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

   5. Axisymmetric pressure vessel is considered in present work.

   6. Open end thick cylinder.

1. 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).

1. Detail of case studies

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

2. 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

3. Comparison with maximum displacement

   Table (4) Maximum displacement of cases

   Cases	Displacement (m)
   	Ur (m)	Uabsolute (m)
   Case 1	82407E-9	93831E-9
   Case 2	81790E-9	93510E-9

4. Comparison with maximum strain

   Table (5) Maximum strain of cases

   Cases	Strain (max)
   	r (m)	(m)	z (m)	von mises
   Case 1	-47316E-10	81962E-10	-1604E-10	11639E-9
   Case 2	-4625E-10	81512E-10	-15109E-10	11539E-9

5. Comparison at A176 and element at thickness 0.4m

   Table (6) Comparison between case 1 and case 2

   Mechanical behavior	Case1 solid42	Case 2-solid 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)	82407E-9	81790E-9
   Uabsolute max (mm)	93831E-9	93510E-9
   r max	-47316E-9	-46250E-9
   max	81962E-9	81512E-9
   z max	-16040E-9	-15109E-9
   von mises max	116390E-9	115390E-9

   Mechanical behavior	Case1 solid42	Case 2-solid 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)	82407E-9	81790E-9
   Uabsolute max (mm)	93831E-9	93510E-9
   r max	-47316E-9	-46250E-9
   max	81962E-9	81512E-9
   z max	-16040E-9	-15109E-9
   von mises max	116390E-9	115390E-9

   1. Von-Mises stress distribution in (pa) case 1

   2. Von-Mises stress distribution in (pa) case 2

Fig. 3.1 Illustrative contours for nodal Von-Mises 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 Von-Mises 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

1. 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).

   1. Absolute displacement distribution (case 1)

   2. 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.

      1. Von-Mises 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

      2. Von-Mises strain distribution case (2)

Fig 3.2 Illustrative contours for nodal Von-Mises 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

1. 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 93831E-9 (m) as compared to case 2 as listed in table (4). It is noted that Uabsolute max for case 2 having value 93510E-9

   (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).

2. 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 116390E-9 as compared to case 2 as shown in table (5). It is noted that

von mises max for case 2 having value 115390E-9 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

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

Axial Strain

Axial Strain

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005 -1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

0.000074

0.000072

0.000070

0.000074

0.000072

0.000070

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

-1.51E-005

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

1.51E-005

1.51E-005

1.51E-005

CASE 2

Theoretical FEA

1.51E-005

1.51E-005

1.51E-005

1.51E-005

Displacement

Displacement

Axial Strain

Axial Strain

1.51E-005

1.50E-005

1.50E-005

1. E-005

2. E-005

1.51E-005

1.50E-005

1.50E-005

1.50E-005

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 (thick-walled 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

1. – Diameter

   t – Thickness

   r – Radial Stress

   – Tangential Stress

   a – Axial Stress

   – Radians of circle

   r – Radial Elongation

   /t – Tangential Elongation

   z – Axial Elongation

2. – 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 matrix

J – Jacobian

[D] – Constitutive matrix

– Potential energies

SE – Strain energy

WF – Work done

WF – External work

[K] – Stiffness matrix

[F] – External applied force matrix

J – Determinate at the jacobian matrix

L – Length

t – Hoop stress

a – Longitudinal stress

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