Axisymetric

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    7/31/13 Axisymmetric Analysis - A problem in which the geometry, loadings, boundary conditions and materials are symmetric with respect to an axis is one that can

    mae.uta.edu/~lawrence/ansys/thick_cylinder/thick_cylinder.htm 1

    Axisymmetric Analysis - A problem in which the geometry, loadings, boundary conditions and materials are

    symmetric with respect to an axis is one that can be solved as an axisymmetric problem instead of as a three

    dimensional problem.

    To illustrate this type of analysis consider the problem of finding the stresses in a thick open-ended cylinder with

    an internal pressure (such as a pipe discharging to the atmosphere). The steel cylinder below has an inner radius

    of 5 inches and an outer radius of 11 inches.

    A cross section is shown below

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    In both drawings the length of the object is arbitrary and represents a segment of a long, open-ended cylinder.

    The Y axis is the axis of symmetry. The cylinder can be generated by revolving a rectangle 6 inches wide and of

    arbitrary height 360 degrees about the Y axis.

    Since the height of the segment considered is arbitrary, we will use a segment 1 inch in height for the finite

    element model. The geometry is shown below.

    The ANSYS solution to this problem is given next for an internal pressure of 1000 psi.

    1. Start ANSYS

    Use a quadrilateral element with axisymmetric behavior.

    2. Preprocessor -> Element Type -> Add/Edit/Delete -> Add -> select Quad 4 Node 42 -> OK ->

    Options (Element Behavior)-> Axisymmetric -> OK -> Close

    Enter material property data for steel.

    3. Preprocessor -> Material Props -> Material Models . . . -> Structural -> Linear -> Elastic ->

    Isotropic ->Enter Ex = 3.e7 and PRXY = 0.3 -> OK -> Close.

    Create geometry for rectangle 1 inch by 6 inches starting 5 inches from Y axis. Note: In ANSYS the Y axis is

    always the axis of symmetry for axisymmetric problems..

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    4. Preprocessor -> Create -> Areas -> Rectangle ->By 2 Corners

    Mesh the area and apply loads and boundary conditions.

    5. Preprocessor -> Mesh -> Areas -> Free (Pick rectangular area)

    6. Preprocessor -> Loads ->Apply ->Displacement -> On Lines (Pick the bottom line of the rectangle) uy

    = 0 along this line. This simply prevents rigid body motion in the Y direction. No other displacement boundaryconditions are required. The radial movement is prevented by the 'hoop' tension in the cylinder.

    Preprocessor -> Loads ->Apply ->Pressure -> On Lines (Pick the left hand line of the rectangle). Enter a

    pressure of1000

    7. Solution -> Solve -> Current LS ->OK

    Check the deformed shape to see if it's reasonable. (The dotted line is the undeformed shape.)

    8. General Post Processor -> Plot Results -> Deformed Shape . . . ->Def +undeformed -> OK

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    Examine the stresses ----------

    9. General Postprocessor -> Plot Results -> Element Solu . . . (PickSx then Sy then Sz and examineeach).

    SX --------------------------------------

    The SX stress is the radial stress that is equal to the pressure (-1000 psi) on the interior of the cylinder and is

    zero on the exterior. The stress legend varies from -1001 to -110. We will need to examine the computed values

    more closely evaluate this further.

    SY --------------------------------------

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    The SY stress is the axial stress in the cylinder and should be zero since the ends are open. This is nearly so inthe plot above.

    SZ --------------------------------------

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    SZ is the 'hoop' stress perpendicular to the plane of this rectangle and varies from 1523 to 520 in the legendabove.

    We can calculate SX and SZ from solid mechanics formulas for thick walled cylinders. These equations give

    At the inside of the cylinderSX = -1000 psi and SZ = 15210 psi

    At the outside of the cylinderSX = 0 psi and SZ = 520 psi.

    Thus the ANSYS calculated results agree pretty well with the theory.

    Examine the stresses more closely at the boundaries.

    Number the nodes and elements.

    10. PlotCntls -> Numbering . . .-> Turn on Node & Element Numbers Zoom in on the inside and outside

    walls of the cylinder.

    INSIDE ------------------------------------

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    List the stress results for Elements 1, 16, and 31.

    11. List -> Results ->Element Solution . . . -> Stress -> Components -> OK.

    PRINT S ELEMENT SOLUTION PER ELEMENT

    ***** POST1 ELEMENT NODAL STRESS LISTING *****

    LOAD STEP= 2 SUBSTEP= 1

    TIME= 2.0000 LOAD CASE= 0

    THE FOLLOWING X,Y,Z VALUES ARE IN GLOBAL COORDINATES

    ELEMENT= 1 PLANE42

    NODE SX SY SZ SXY SYZ SXZ

    1 -995.80 7.6070 1518.9 -46.247 0.0000 0.0000

    3 -816.22 2.8291 1337.8 -46.213 0.0000 0.0000

    37 -816.36 2.8243 1337.8 45.349 0.0000 0.0000

    36 -995.93 7.6119 1518.9 45.315 0.0000 0.0000

    ELEMENT= 16 PLANE42

    NODE SX SY SZ SXY SYZ SXZ

    36 -995.47 11.132 1520.2 -47.531 0.0000 0.0000

    37 -815.90 1.2212 1337.4 -47.403 0.0000 0.0000

    38 -815.62 1.2215 1337.6 44.196 0.0000 0.0000

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    35 -995.19 11.132 1520.5 44.067 0.0000 0.0000

    ELEMENT= 31 PLANE42

    NODE SX SY SZ SXY SYZ SXZ

    35 -994.73 18.160 1522.9 -48.075 0.0000 0.0000

    38 -814.76 -3.0243 1336.5 -47.906 0.0000 0.0000

    34 -821.59 -3.2188 1334.7 43.696 0.0000 0.0000

    20 -1001.2 18.360 1521.9 43.527 0.0000 0.0000

    The computed SX and SZ stresses are within 0.4% of the theoretical values.

    OUTSIDE ------------------------------------

    List the stress results for Element 15.

    12. List -> Results ->Element Solution . . . -> Stress -> Components -> OK.

    ELEMENT= 15 PLANE42

    NODE SX SY SZ SXY SYZ SXZ

    16 -19.922 -0.64520 540.57 -8.2588 0.0000 0.0000

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    2 0.90585E-01 -1.4459 520.29 -8.2562 0.0000 0.0000

    18 0.11328 -1.4457 520.31 8.1154 0.0000 0.0000

    63 -19.900 -0.64542 540.58 8.1127 0.0000 0.0000

    Notice at nodes 2 and 18 the SX stress is virtually zero and the SZ stress matches the theoretical value.

    Next compare the computed and theoretical solutions for the radial displacement at the inside and outside of the

    cylinder.

    Theoretical Values:

    Displacement at Inside Surface = 0.305E-4 inches

    Displacement at Outside Surface = 0191E-3 inches

    List the ANSYS results

    13. General Postprocessor -> List -> Results -> Nodal Solution . . . DOF ->All DOFs -> OK (A partial

    listing is shown below)

    NODE UX UY

    1 0.30353E-03 0.0000

    2 0.19107E-03 0.0000

    3 0.28572E-03 0.0000

    4 0.27072E-03 0.0000

    5 0.25798E-03 0.0000

    6 0.24708E-03 0.0000

    7 0.23770E-03 0.0000

    8 0.22958E-03 0.0000

    9 0.22255E-03 0.0000

    10 0.21644E-03 0.0000

    11 0.21113E-03 0.0000

    12 0.20650E-03 0.0000

    13 0.20248E-03 0.0000

    14 0.19897E-03 0.0000

    15 0.19594E-03 0.0000

    16 0.19332E-03 0.0000

    17 0.19110E-03-0.52763E-05

    18 0.19107E-03-0.17508E-05

    19 0.19107E-03-0.35069E-05

    20 0.30376E-03-0.48288E-05

    Nodes 1 and 20 are on the inside surface. Nodes 2 and 17 are on the outside surface. Note that the computed

    displacements are in agreement with the theoretical values.

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    t d / l / /thi k li d /thi k li d ht 10

    2001 Kent L. Lawrence