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					M. Vable                Notes for finite element method: Axi-symmetric, plates and shells, 3-D



                                             Polar Coordinates
   The small strain-displacement equations in polar coordinates are:
                                ∂u r                 u r 1 ∂v θ                        ∂w
                      ε rr =                  ε θθ = --- + --
                                                       - -                    ε zz =
                                ∂r                    r r ∂θ                           ∂z
                             1 ∂u r ∂v θ v θ                               ∂w ∂u r                  1 ∂w ∂v θ
                      γ rθ = --
                              -    +    – ----
                                             -                  γ rz =       +               γ zθ = --
                                                                                                     -   +
                             r ∂θ ∂r        r                              ∂r ∂z                     r ∂θ ∂z
   The Generalized Hooke’s Law can be written as:
                            2G                                                           τ rθ
               σ rr = ------------------- [ ( 1 – ν )ε rr + νε θθ + νε zz ]
                                        -                                        γ rθ = ------
                      ( 1 – 2ν )                                                          G
                            2G                                                            τ rz                       E
               σ θθ = ------------------- [ ( 1 – ν )ε θθ + νε rr + νε zz ]
                                        -                                         γ rz = -----
                                                                                             -                                -
                                                                                                        G = -------------------
                      ( 1 – 2ν )                                                           G                2( 1 + ν)
                            2G                                                            τ zθ
               σ zz = ------------------- [ ( 1 – ν )ε zz + νε θθ + νε rr ]
                                        -                                        γ zθ = -------
                      ( 1 – 2ν )                                                           G
   Axi-symmetric problems
   For a problem to be axi-symmetric the following requirements must be met:
       1. The geometry must be symmetric about an axis of revolution.
       2. The material properties must be symmetric about the axis of revolution.
       3. The loading and boundary conditions must be symmetric about the axis of
          revolution.
   Implications: Displacements and stresses must be independent of angular location
   (θ) and there can be no twist (vθ must be zero).
               ∂u r                ur                        ∂w                                       ∂w ∂u r
      ε rr =                ε θθ = ---
                                     -             ε zz =                  γ rθ = 0          γ rz =     +                  γ zθ = 0
               ∂r                   r                        ∂z                                       ∂r ∂z

   • Note radial displacement causes tangential normal strain.
                                                                     σzz
                                                                     τrz
                                            σθθ                                        σrr


                                 z


                                        r         θ




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M. Vable              Notes for finite element method: Axi-symmetric, plates and shells, 3-D




                                    z
                                        r
                                                       Smooth
                     Pressurized                       Surface



   3-node Triangular Element
   • Displacements are linear in r and z directions
                                        ur = a0 + a1 r + a2 z            w = b0 + b1 r + b2 z
                                                                                                       (e)
      z            (e)
                                                                          3               ⎧ u r1 ⎫
                  w3                    (e)                                      (e)      ⎪        ⎪
                              (e)
                            u r3
                                    w2            u r ( r, z ) = ∑ N i ( r, z )u i        ⎪ (e) ⎪
                 3                                               i=1                      ⎪ w1 ⎪
        (e)                         2             (e)
                                                                                          ⎪        ⎪
       w1                                       u r2                                      ⎪u  (e) ⎪
                                                                                     (e)  ⎪ r2 ⎪
            1         (e)                                         3                {d } = ⎨        ⎬
                     u r1                   r
                                                                               (e)        ⎪ w( e ) ⎪
                                                 w ( r, z ) = ∑ N i ( r, z )v i           ⎪ 2 ⎪
                                                                                          ⎪ (e) ⎪
                                                                i=1                       ⎪ u r3 ⎪
                                                                                          ⎪        ⎪
                                                                                          ⎪ (e) ⎪
                                                                                          ⎩ w3 ⎭

             ∂u r                        ∂w                              ∂u r ∂w
     ε rr =       = a1          ε zz =      = b2                γ rz =       +   = a2 + b1      Same as CST
             ∂x                          ∂z                              ∂z ∂r
              ur    a0           a2 z
     ε θθ   = --- = ---- + a 1 + -------
                -      -               -                        Tangential normal strain is not constant
               r      r             r

   • You can use any 2-D element, but will need to post-process the results of
      displacements and strains to get ε θθ , σ rr , σ θθ , σ zz .




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M. Vable           Notes for finite element method: Axi-symmetric, plates and shells, 3-D



                                                  Thin Plate
   A thin two-dimensional structural element that is subjected to bending loads.
   • Plane stress in z-direction
                                              z

                                      t               y
                                                                         pz(x,y)
                                          x                                                              Mid-surface
                                                                                                         is neutral surface

   • Mid-plane is initially flat
   • Plane sections before deformation remain plane after deformation. (displacements
      u and v are linear in z, i.e., through the thickness.)
   Kirchhoff Plate Theory
   • Plane sections initially perpendicular to the mid-surface remains perpendicular
      after deformation ( γ xz ≈ 0                   γ yz ≈ 0 ) --Shearing action is small)

                                                          ∂w                       ∂w
                                              u = –z                  v = –z
                                                          ∂x                       ∂y
   w is the displacement in the z-direction and is only a function of x and y. u and v are
   displacements in x and y direction.
   For small strain:
                                2                                      2                                            2
                    ∂u     ∂w                           ∂v     ∂w                             ∂u ∂v      ∂w
           ε xx   =    = –z 2                 ε yy    =    = –z 2                    γ xy   =   +   = –z
                    ∂x     ∂x                           ∂y     ∂y                             ∂y ∂x      ∂ x∂y

   Stresses in plane stress:
                                                                                    2           2
                                        [ ε xx + νε yy ]                     – Ez ⎛ ∂ w           ∂ w⎞
                             σ xx   = E ----------------------------- = ------------------ ⎜ 2 + ν 2 ⎟
                                                                                         -
                                                                        (1 – ν )⎝∂x               ∂y ⎠
                                                            2                          2
                                             (1 – ν )
                                                                                    2           2
                                        [ ε yy + νε xx ]                     – Ez ⎛ ∂ w           ∂ w⎞
                             σ yy   = E ----------------------------- = ------------------ ⎜ 2 + ν 2 ⎟
                                                                                         -
                                                                        (1 – ν )⎝∂y               ∂x ⎠
                                                            2                          2
                                             (1 – ν )
                                                                               2
                                               E                    –E z ∂ w
                             τ xy   = ------------------- xy = -------------------
                                                        -γ                       -
                                      2(1 + ν)                 2 ( 1 + ν ) ∂ x∂y




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M. Vable                               Notes for finite element method: Axi-symmetric, plates and shells, 3-D



   Internal Forces and Moments:
                     t⁄2                                                        t⁄2                                            t⁄2

    Mx =              ∫       zσ xx dz                               My =         ∫      zσ yy dz                  M xy =        ∫      zτ xy dz
                    –t ⁄ 2                                                      –t ⁄ 2                                         –t ⁄ 2
                   t⁄2                                                 t⁄2

    qx =             ∫      τ xz dz                             qy =     ∫      τ yz dz
                   –t ⁄ 2                                              –t ⁄ 2
   The moments and shear forces have units of moments and forces per unit length.
   Moment Curvature Formulas:
                                   2                    2                                         2                2                                      2
                                  ∂w                ∂w                                        ∂w               ∂w                                         ∂w
    Mx = –D                                +ν                            My = –D                          +ν                     M xy      = –D ( 1 – ν )
                                  ∂x
                                       2
                                                    ∂y
                                                            2
                                                                                              ∂y
                                                                                                      2
                                                                                                               ∂x
                                                                                                                       2                                  ∂ x∂y
                                                    3
                       Et
   where, D = ------------------------- is called the plate rigidity.
                                   2
                                       12 ( 1 – ν )
   Differential Equation: Bi-harmonic Equation
       4              4                         4
    ∂w             ∂w                       ∂w                                            4               2    2
           4
               +          2        2
                                       +            4
                                                            = p z ( x, y ) or ∇ w = ∇ ∇ w = p z ( x, y )
    ∂x             ∂ x ∂y                   ∂y
                                                2                2
                              2            ∂                ∂
   where, ∇ =                                   2
                                                    +            2
                                                                     is the harmonic operator.
                                           ∂x           ∂y

                                                                                                                                                         ∂w ∂w
   • A kinematically admissible deflection w requires continuity of w,                                                                                     ,
                                                                                                                                                         ∂x ∂y
                                                                                                                                                               at all
       points.
                                                                                              2                2                                                  2
   •   At a corner the requirement that ∂ w = ∂ w results in the condition that ∂ w
                                        ∂ x∂y ∂ y∂x                             ∂ x∂y
       be continuous at the corner.
   • Rectangular element: Each node has four degrees of freedom (dof) per node:
                                                    2
          ∂w ∂w ∂ w
       w,    ,   ,      . Can be used only with rectangular sides parallel to x and y
          ∂ x ∂ y ∂ x∂y
       axis. Hermite polynomials are used for interpolation functions.
               y


                                                                                                                           21 dof
                     16 dof
                                                            x




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M. Vable            Notes for finite element method: Axi-symmetric, plates and shells, 3-D


                     2          2
   •   To ensure ∂ w = ∂ w at any orientation, requires all second derivatives to be
                 ∂ x∂y ∂ y∂x
       continuous at nodes.
   • Triangular element: Each corner node has six degrees of freedom per node
                           2        2     2
            ∂w ∂w    ∂w ∂w ∂w
       w,     ,    ,      ,      ,       and the middle node on each side has one degree
            ∂x ∂y    ∂x
                        2
                            ∂y
                               2   ∂ x∂y

       of freedom ∂w where the n direction is the normal direction to the side.
                  ∂n
   • The continuity of second derivatives implies that moments must be continuous. If
       there is a line load of moment then this will leads to problems.
   • Non-conforming elements do not satisfy all continuity requirements. Non-
       conforming elements are used in plate analysis.
   Mindlin Plate Theory
   • Mindlin plate theory differs from Kirchhoff plate theory in the same way as
       Timoshenko’s beam theory differs from classical beam theory.
   • The assumption of plane sections initially perpendicular to the mid-surface
       remains perpendicular after deformation is dropped and transverse shear is
       accounted.
   Displacements:
                                              u = zθ y        v = – zθ x
   where, θ x and θ y are the rotation about x and y axis, respectively, of a line that was
   initially perpendicular to the mid surface.
   Strains
             ∂u     ∂θ y                ∂v      ∂θ x                   ∂u ∂v      ∂θ y ∂θ x⎞
    ε xx =      = z            ε yy =      = –z               γ xy =     +   = z⎛
             ∂x     ∂x                  ∂y      ∂y                     ∂y ∂x    ⎝∂y – ∂x ⎠
             ∂u ∂w     ∂w                              ∂u ∂w     ∂w
    γ xz =     +   = ⎛    + θ y⎞              γ yz =     +   = ⎛    – θ x⎞
             ∂z ∂x   ⎝∂x       ⎠                       ∂z ∂x   ⎝∂y       ⎠

                       ∂w           ∂w
   • Note θy      = – ⎛ ⎞ and θ x =    reduces Mindlin’s theory to Kirchhoff’s theory.
                      ⎝∂x ⎠         ∂y
   • Kinematically admissibility requires that w,                  θ x, θ y must be continuous. Can use
       Lagrange polynomial for interpolation functions.




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M. Vable          Notes for finite element method: Axi-symmetric, plates and shells, 3-D



                                Thin Shell Elements
   • Curved plate: Combination of membrane (2-D in-plane) and plate bending.
   • The elements are similar to plate elements but requires definition of curved
      geometry.
   • FEM codes usually have shallow thin shell elements which can be used to also
      simulate plate elements.




                                                                                           6
M. Vable            Notes for finite element method: Axi-symmetric, plates and shells, 3-D



                       Three Dimensional Elements
   Tetrahedron
        • Displacements are linear in x and y, resulting in constant strains.
                                               Constant Strain
                                             u = a0 + a1 x + a2 y + a3 z
                                             v = b0 + b1 x + b2 y + b3 z
                                             w = c0 + c1 x + c2 y + c3 z


                                                   1

                                                                   z
                                                                               z2
                                        x                     y
                                                    xz               yz                z3
                                  2                xy                      2                 z3
                              x                                        y
                                                                                                  z3

                      x3              x2 y                  xy2                y3


               x4           x3y                x2y2                xy3          y4

                        x4y           x3y2                             xy4            y5
           5
           x                                           x2y3




                                                                                                       7
M. Vable                  Notes for finite element method: Axi-symmetric, plates and shells, 3-D



   Hexahedron (Brick) Element
                                                                                     Trilinear
                                     ζ
                         5                               7                            1
                                                                                N 1 = -- ( 1 – ξ ) ( 1 – η ) ( 1 – ζ )
                                                                                       -
                                                                                      8
      6                                                                               1
                                            8                                   N 4 = -- ( 1 + ξ ) ( 1 + η ) ( 1 – ζ )
                                                                                       -
                                                                 η                    8


                                                           3
                 ξ        1
                                           4
         2


                                                                                     20 node quadratic
                                     ζ



                                                                 η

                 ξ


   Iso-parametric:
                                         (e)
    u =        ∑ Ni ( ξ, η, ζ )ui                                         x =   ∑ Ni ( ξ, η, ζ )xi
               i=1                                                              i=1
                n                                                                n
                                         (e)
    v =        ∑ Ni ( ξ, η, ζ )vi                                         y =   ∑ Ni ( ξ, η, ζ )yi
               i=1                                                              i=1
                n                                                                n
                                          (e)
    w =        ∑ Ni ( ξ, η, ζ )wi                                         z =   ∑ Ni ( ξ, η, ζ )zi
                                                                          1 1 1
         (e)                                                                            T
                     ∫ ∫ ∫ [B]                                            ∫ ∫ ∫ [B]
                                 T
    [K         ] =                   [ E ] [ B ] ( dx ) ( dy ) ( dz ) =          ˜          [ E ] [ B ] J ( dξ ) ( dη ) ( dζ )
                                                                                                    ˜
                                                                          –1 –1 –1




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posted:10/13/2011
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