Home >Documents >Chapter 9site.iugaza.edu.ps/marafa/files/FEM-Chapter-9-2017-18.pdf · Finite Element Method Chapter...

Chapter 9site.iugaza.edu.ps/marafa/files/FEM-Chapter-9-2017-18.pdf · Finite Element Method Chapter...

Date post:21-Jul-2020
Category:
View:24 times
Download:2 times
Share this document with a friend
Transcript:
  • Finite Element Method

    Chapter 9

    Axisymmetric Elements

  • Definition of an axisymmetric solid

    An axisymmetric solid (or a thick-walled body) of revolution is defined as a 3-D body that is generated by rotating a plane and is most easily described in cylindrical coordinates. Where z is called the axis of symmetry.

    If the geometry, support conditions, loads, and material properties are all axially symmetric (all are independent of ), then the problem can be idealized as a two-dimensional one.

  • Problems such as soil masses subjected to circular footing loads, thick-walled pressure vessels, and a rocket nozzle subjected to thermal and pressure loading can often be analyzed using axisymmetric elements.

    Examples of an axisymmetric solid

  • Examples of an axisymmetric solid

    Enclosed pressure vesselEngine valve stem

  • axisymmetric problems can be analyzed by a finite element of revolution, called axisymmetric elements. Each element consists of a solid ring, the cross-section of which is the shape of the particular element chosen (triangular, rectangular, or quadrilateral elements). An axisymmetric element has nodal circles rather than nodal points

    FE axisymmetric elements

  • Equations of Equilibrium:

    The three-dimensional elasticity equations in cylindrical coordinates can be summarized as follows

    011

    021

    01

    bzrzzzr

    rzr

    rrzrrr

    Zrzrr

    Yrzrr

    Xrzrr

  • Equations of Equilibrium:

    The three-dimensional strain-displacement relationships of elasticity in cylindrical coordinates were u, v, w are the displacements in the r, ,dz , respectively, are:

    w

    rz

    v

    z

    w

    z

    u

    r

    w

    r

    uv

    r

    r

    v

    r

    vu

    rr

    u

    zz

    zr

    rr

    1,

    ,1

    1,

  • Equations of Equilibrium:

    The three-dimensional stress-strain relationships for isotropic elasticity are:

    z

    zr

    r

    z

    r

    z

    zr

    r

    z

    r

    E

    2

    210000

    02

    21000

    002

    21000

    0001

    0001

    0001

    )21()1(

  • Axisymmetric Stresses and Strains

    In axisymmetric problems, because of the symmetry about the z -axis, the stresses are independent of the coordinate. Therefore, all derivatives with respect to vanish and the circumferential (tangent to direction) displacement component is zero; therefore,

    0 0

    , , ,

    r z r z

    r z r z

    and

    u u w u w

    r r z z r

  • Axisymmetric Stresses and Strains

    ,

    ,

    r

    z r z

    u u

    r r

    w u w

    z z r

  • The stress-strain relationship for isotropic axisymmetric problems

    zr

    z

    r

    zr

    z

    r

    E

    2

    21000

    01

    01

    01

    )21()1(

    }{][}{ D

  • Derivation of the Stiffness Matrix and Equations

    Typical slice through an axisymmetric solid Discretized into triangular elements

    Step 1: Discretize and Select Element Type

  • Step 1: Discretize and Select Element Type

    m

    m

    j

    j

    i

    i

    m

    j

    i

    w

    u

    w

    u

    w

    u

    d

    d

    d

    d}{

    (ui , wi ) displacement components of node i in the r and z directions,respectively.

  • Step 2: Select Displacement Functions

    zaraazrw

    zaraazru

    654

    321

    ),(

    ),(

    6

    5

    4

    3

    2

    1

    654

    321

    1000

    0001

    ),(

    ),(}{

    a

    a

    a

    a

    a

    a

    zr

    zr

    zaraa

    zaraa

    zrw

    zru

  • Step 2: Select Displacement Functions

    mmm

    jjj

    iii

    mmm

    jjj

    iii

    zaraaw

    zaraaw

    zaraaw

    zaraau

    zaraau

    zaraau

    654

    654

    654

    321

    321

    321

    In Matrix Form

    m

    j

    i

    mm

    jj

    ii

    u

    u

    u

    zr

    zr

    zr

    a

    a

    a1

    3

    2

    1

    1

    1

    1

    m

    j

    i

    mm

    jj

    ii

    w

    w

    w

    zr

    zr

    zr

    a

    a

    a1

    6

    5

    4

    1

    1

    1

  • mm

    jj

    ii

    yx

    yx

    yx

    A

    1

    1

    1

    2

    mji

    jimimjmji yyxyyxyyxA

    )()()(2

    Solving for the a’s

    A is the area of the triangle

    m

    j

    i

    mji

    mji

    mji

    u

    u

    u

    A

    a

    a

    a

    2

    1

    3

    2

    1

    m

    j

    i

    mji

    mji

    mji

    w

    w

    w

    A

    a

    a

    a

    2

    1

    6

    5

    4

  • ijmmijjmi

    jimimjmji

    jijimimimjmjmji

    rrrrrr

    zzzzzz

    rzzrrzzrrzzr

    3

    2

    1

    1}{

    a

    a

    a

    zru

    m

    j

    i

    mji

    mji

    mji

    u

    u

    u

    zrA

    u

    12

    1}{

  • 1{ } 1

    2

    1( , ) ( ) ( ) ( )

    2

    1( , ) ( ) ( ) ( )

    2

    ( , ){ }

    ( , )

    i i j j m m

    i i j j m m

    i i j j m m

    i i i i j j j j m m m m

    i i i i j j j j m m m m

    i i j

    u u u

    u r z u u uA

    u u u

    u r z r z u r z u r z uA

    w r z r z w r z w r z wA

    N u Nu r z

    w r z

    j m m

    i i j j m m

    u N u

    N v N v N v

  • ( , ){ }

    ( , )

    1

    2

    1

    2

    1

    2

    i i j j m m

    i i j j m m

    i i i i

    j j j j

    m m m m

    N u N u N uu r z

    N v N v N vw r z

    N r zA

    N r zA

    N r zA

  • [ ]

    0 0 0{ }

    0 0 0

    { } { }

    0 0 0

    0 0 0

    i

    i

    i j m j

    i j m j

    m

    m

    i j m

    i j m

    u

    w

    N N N u

    N N N w

    u

    w

    d

    N N N

    N

    N

    N NN

  • Step 3: Define the Strain/Displacement and Stress/Strain Relationships

    2

    6

    312

    3 5

    { }

    { }

    r

    z

    rz

    r

    z

    rz

    u

    r

    w

    z

    u

    r

    u w

    z r

    a

    a

    a zaa

    r r

    a a

  • }{][}{ dB

    0 0 0

    0 0 0

    0 0 0

    0

    0

    0

    i

    i j m

    i

    i j m

    j

    j ji i m mji j m

    mi i j j m m

    m

    i

    ii

    ij

    i j m i i ij i

    mi i

    m

    u

    w

    uzz z

    wr r r r r r

    u

    w

    u

    w

    uB B B B z

    wr r

    u

    w

    B is a function of r and z

  • Stress Strain Relationship

    }{][][}{ dBD

    zr

    z

    r

    zr

    z

    r

    E

    2

    21000

    01

    01

    01

    )21()1(

  • Step 4 :Derive the Element Stiffness Matrix and Equations

    [ ] [ ] [ ][ ]

    [ ] 2 [ ] [ ][ ]

    T

    V

    T

    A

    k B D B dV

    k B D B r dr dz

    1) Numerical integration (Gaussian quadrature)2) Explicit multiplication and term-by-term integration.3) Evaluate [B] at a centroidal point of the element

    3 3

    [ ( , )] [ ]

    [ ] 2 [ ] [ ][ ]

    i j m i j m

    T

    r r r z z zr r z z

    B r z B

    k r A B D B

  • }{}{][}{][}{ PdSTNdVXNf

    S

    T

    V

    T

    Concentrated nodal forcesBody forces

    Surface Tractions

    Step 4 :Derive the Element Stiffness Matrix and Equations

  • { } [ ] { } 2 [ ]bT T

    b

    bV A

    Rf N X dV N r dr dz

    Z

    for a machine part rotating with:a constant angular velocity about the z axis (zero angular acceleration); : material mass density (kg/m3) and r is the radial coordinate.Zb= body force per unit volume due to the force of gravity

    (weight density, kN/m3).

    Distributed Body ForcesBody forces can either : Gravity force in the direction of z axis, or Centrifugal forces in rotating machine parts in the direction of the r axisThe nodal equivalent body forces can found using

    2

    bR r

  • 0{ } 2

    0

    2{ }

    3

    i b

    bi

    i bA

    bir b

    bi

    biz b

    N Rf r dr dz

    N Z

    f RArf

    f Z

    b

    b

    b

    b

    b

    b

    bmz

    bmr

    bjz

    bjr

    biz

    bir

    b

    Z

    R

    Z

    R

    Z

    R

    Ar

    f

    f

    f

    f

    f

    f

    f3

    2}{

    rRb 2

    Distributed Body Forces

  • { } [ ] { } [ ]rT T

    s

    zS S

    pf N T dS N dS

    p

    dzrp

    p

    zr

    zr

    Af j

    z

    rz

    zjjj

    jjj

    jsm

    j

    2

    0

    0

    2

    1}{

    For example: along the vertical face jm of an element, let uniform loads pr and pz be applied, as shown along r=rj.

    Surface Forces

  • dzrp

    p

    zr

    zr

    Af j

    z

    rz

    zjjj

    jjj

    jsm

    j

    2

    0

    0

    2

    1}{

    z

    r

    z

    rjmj

    smz

    smr

    sjz

    sjr

    siz

    sir

    s

    p

    p

    p

    pzzr

    f

    f

    f

    f

    f

    f

    f

    0

    0

    2

    )(2}{

    Surface Forces

    evaluated at r = rj ; z = z

  • Step 5: Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions

    N

    e

    ekK1

    )( ][][

    }{][}{ dKF

    N

    e

    efF1

    )( ][][

    Step 6: Solve for the Nodal Displacements

    Step 7: Solve for the Element Stresses

  • Steps 5 through 7

    Steps 5 through 7, are analogous to those of Chapter 6 for the CST element, except the stresses are not constant in each element.

    They are usually determined by: Determine the centroidal element stresses, Determine the nodal stresses for the element

    and then average them. The latter method has been shown to be more

    accurate in some cases

  • Example 1

    For the element of an axisymmetric body rotating with a constant angular velocity = 100 rev/min.Evaluate the approximate body force matrix, include the weight of the material, where the weight density w is 0.283 lb/in

    3.The coordinates of the element (in inches) are shown in the figure.

    The body forces per unit volume evaluated at the centroid of the element are

    Zb =0.283 lb/in3

  • All r-directed and z-directed nodal

    body forces are equal

    Example 1 -Continue

  • Example 2

    For the long, thick-walled cylinder under internal pressure p equal to 1 psi shown, determine the displacements and stresses.

  • Example 2

    To illustrate the finite element solution for the cylinder, we first discretize the cylinderinto four triangular elements, as shown. A horizontal slice of the cylinderrepresents the total cylinder behavior.

  • Example 2

    Assemblage of the Stiffness Matrix

    0.5, 0, 1.0, 0, 0.75, 0.25

    ( 1; 2,

    For element

    5 )

    1

    i i j j m mr z r z r and z

    i j and m

  • Example 2 – Stiffness Matrix for ELEMENT 1

  • Example 2 – Stiffness Matrix for ELEMENT 1

  • Example 2 – Stiffness Matrix for ELEMENT 1

  • 1.0, 0, 1.0,

    For element 2

    0.5, 0.75, 0.25

    ( 2; 3, 5 )

    i i j j m mr z r z r and z

    i j and m

    Example 2 – Stiffness Matrix for ELEMENT 1

  • Example 2 – Stiffness Matrix for ELEMENT 2 & 3

  • Example 2 – Stiffness Matrix for ELEMENT 1

  • Example 2 – Total Stiffness Matrix

  • HW:

    xxxx

    ../../../../../Documents and Settings/Suvranu De/Local Settings/Temp/Logan/Logan-START.ppt#2. Table of Contents../../../../../Documents and Settings/Suvranu De/Local Settings/Temp/Logan/Logan-START.ppt#2. Table of Contents
of 44/44
Finite Element Method Chapter 9 Axisymmetric Elements
Embed Size (px)
Recommended