Скачать презентацию

Идет загрузка презентации. Пожалуйста, подождите

Презентация была опубликована 2 года назад пользователемВладислав Целовальников

1 S3-1NAS105, Section 3, May 2005 SECTION 3 TRICKS

2 S3-2NAS105, Section 3, May 2005

3 S3-3NAS105, Section 3, May 2005 SectionPage TRICKS……………………………………………………………………………………………..3-5 ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS…………………………….3-6 DISPLACEMENT COORDINATES SYSTEMS…………………………………………………3-9 GRIDS POINTS…………………………………………………………………………………….3-15 OFFSETS ON BARS, BEAMS AND PLATES………………………………………………… HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS…………………………….3-18 SHEAR CENTER – BEAMS AND BARS……………………………………………………… BEAM VS. BAR…………………………………………………………………………………….3-34 HOW ARE OFFSETS IMPLEMENTED ON PLATES………………………………………….3-35 HANDY HINTS…………………………………………………………………………………… MORE HANDY HINTS…………………………………………………………………………….3-40 COMPATIBILITIES ……………………………………………………………………………… MODELING CORNERS WITH PLATES AND BEAMS……………………………………… EXAMPLE OF POOR PRACTICE……………………………………………………………… EQUIVALENT ROD……………………………………………………………………………… EQUIVALENT ROD EXAMPLE RESULTS…………………………………………………… TABLE OF CONTENTS

4 S3-4NAS105, Section 3, May 2005 TABLE OF CONTENTS SectionPage PLATES AND SHELLS IN-PLANE ROTATIONAL STIFFNESS…………………………… MESH TRANSITIONS…………………………………………………………………………….3-54 BEAM-TO-PLATE ELEMENTS………………………………………………………………… MORE MESH TRANSITIONS……………………………………………………………………3-62 SOME POSSIBLE PLATE-TO-SOLID TRANSITION………………………………………….3-64 STRESS SORTING……………………………………………………………………………….3-66 DTI, INDTA – BULK DATA ENTRY…………………………………………………………… STRESS SORTING – SAMPLE…………………………………………………………………..3-72

5 S3-5NAS105, Section 3, May 2005 TRICKS n For the sake of discussion, let us define a trick as something you would not expect unless someone told you or you had already experienced it.

6 S3-6NAS105, Section 3, May 2005 ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS n Pressure loadings on elements are replaced by forces at the corners. n For a simple element, such as the QUAD4 or TRIA3, the force is simply the total force divided by the number of connecting GRID points. n For higher order elements, the forces are found based on the element formulation, and may not agree with users physical intuition or common sense. n Example u Grid point loads equivalent to a uniform pressure

7 S3-7NAS105, Section 3, May 2005 ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS n Different element types have different order polynomials used for their displacement fields. n As a result, it is possible that local discontinuities will occur when you mix different element types.

8 S3-8NAS105, Section 3, May 2005 ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS (Cont.) n Method of Testing u Put SPCs on boundary, to induce uniform stress. u Measure SPCF output.

9 S3-9NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS n When you define a GRID entry in MSC.NASTRAN, there are two fields used to select coordinate systems. n CP is the position coordinate system, or the system in which X1, X2, and X3 are measured. n CD is the displacement coordinate system. This defines the coordinate system which is used to measure the displacements at the GRID point. All constraints, MPCs, SPCs, and the BAR/BEAM offsets and orientation vector use this coordinate system. n Coordinate systems may be rectangular (X,Y,Z), cylindrical (R,, Z), or spherical (R,, ).

10 S3-10NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS (Cont.) n In MSC.NASTRAN the following coordinate systems may be used. u Basic Coordinate System – Implicitly defined reference rectangular coordinate system (Coordinate System 0). Orientation of this system is defined by the user through specifying the components of grid point locations. u Alternate (local) Coordinate Systems – Alternate systems can be defined to facilitate geometric input. Each local system must be related directly or indirectly to the basic coordinate system. The six possible alternate coordinate systems are: n All angular coordinates are input in degrees. Output associated for these coordinates is in radians. Rectangular Cylindrical Spherical CORD1R CORD2R CORD1C CORD2C CORD1S CORD2S

11 S3-11NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS (Cont.) U 1 = x direction U 2 = y direction U 3 = z direction Note: A, B, and C are points used to define the local coordinate system. P is a grid point defined in the local system. Rectangular Coordinate System (x, y,z)

12 S3-12NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS (Cont.) Point A = local system origin Point P = grid point defined in local cylindrical system Point C = reference point in the r-z plane at = 0 Point B = reference point for z axis direction (U r,U,U z ) = displacement components of P in local system Cylindrical Local Coordinate System (R,,Z)

13 S3-13NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS (Cont.) Point A = local system origin Point P = grid point defined in local spherical system Point C = reference point in the x-z plane at ( = 0) Point B = reference point for z axis direction ( = 0) (U r,U,U ) = displacement components of P in local system Note: cylindrical spherical Spherical Local Coordinate System (R,, )

14 S3-14NAS105, Section 3, May 2005 DISPLACEMENT COORDINATE SYSTEMS (Cont.) Sample – define a cylindrical system CORD2C,1,,0., 0., 0., 0., 0., 1., 1.,0.,0. Define Grid Points 10 and 20 on a circle. GRID, 10, 1, 10., 45., 0., 0 GRID, 20, 1, 10., 135., 0., 0 where field 3 (CP) references cylindrical coordinate system 1 (defined elsewhere), and field 7 (CD) references basic coordinate system 0. With this definition, all grid point output associated with Grids 10 and 20 will be oriented as shown.

15 S3-15NAS105, Section 3, May 2005 GRIDS POINTS Now, in a separate model, define Grid Points 10 and 20 as : GRID, 10, 1, 10., 45., 0., 1 GRID, 20, 1, 10., 135., 0., 1 With both CP and CD referencing cylindrical coordinate system 1, all output grid information at Grids 10 and 20 will also be in terms of radial and tangential directions. n Remember – your output is in the displacement coordinate system

16 S3-16NAS105, Section 3, May 2005 OFFSETS ON BARS, BEAMS AND PLATES

17 S3-17NAS105, Section 3, May 2005 OFFSETS ON BARS, BEAMS AND PLATES n GRID points connecting BAR/BEAM elements may NOT lie on the beam neutral axis. n GRID points connecting plate elements may NOT lie on the plate mid-surface. n Shear center of Beam sections may NOT lie on the beam neutral axis.

18 S3-18NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? n Bar and Beam elements may be offset from the connecting GRID points. n The offsets are entered on the continuation of the CBAR or CBEAM entry and are in the displacement coordinate system of the GRID points Offset Bar Sample – Axial Load n Let us look at this problem two different ways u Case 1: GRID points using the basic coordinate system as the displacement coordinate system. u Case 2: Define, and Use a coordinate system 100 as the displacement coordinate system on the GRID points

19 S3-19NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.) n First problem – using the basic coordinate system ID OFFSET, BAR MODEL SOL 101 CEND TITLE = TEST OF OFFSET BAR MODEL – FILE offbar.dat SUBT = SHOW WHERE ELEMENT FORCE OUTPUT IS LOAD = 1 DISP = ALL ELFORCE = ALL GPFORCE = ALL set 999 – 7 oload = 999 BEGIN BULK $ $ DEFINE MODEL $ GRID,1,,0.,0.,0.,, GRID,2,,2.,0.,0. =,*(1),=,*(2.),== =(4) CBAR,1,1,1,2,0.,1.,0.,,+CB1 =,*(1),=,*(1),*(1),=,=,=,=,*(1) =(4) $ $ OFFSET 1 UNIT IN THE Y DIRECTION – IN GRID OUTPUT COORD $ +CB1,,,0.,1.,0.,0.,1.,0. *(1),== =(4) PBAR,1,1,1.,12.,12.,24., MAT1,1,30.+6,,.3 pload1,1,6,FX,FR,1.,–1. ENDDATA

20 S3-20NAS105, Section 3, May 2005 SAMPLE OF OFFSET BAR AXIAL LOAD

21 S3-21NAS105, Section 3, May 2005 SAMPLE OF OFFSET BAR AXIAL LOAD (Cont.)

22 S3-22NAS105, Section 3, May 2005 SAMPLE OF OFFSET BAR AXIAL LOAD (Cont.)

23 S3-23NAS105, Section 3, May 2005 SAMPLE OF OFFSET BAR AXIAL LOAD (Cont.)

24 S3-24NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.) n The following summarizes the preceding output n OLOAD RESULTANT – this is the summation of all applied loads about PARAM, GRDPNT (about the BASIC origin if GRDPNT is not specified) in the basic coordinate system. u In this case, it verifies that the applied load was –1.0 units in the X–direction and is offset by 1.0 units (M z = 1.0 = the load multiplied by the offset) n ALWAYS VERIFY THAT THE OLOAD RESULTANT IS CORRECT!!! n SPCFORCE RESULTANT – Similar output for the reaction forces. This should be equal and opposite to the OLOAD RESULTANT. If not, Use GROUNDCHECK Case Control Command. n DISPLACEMENT VECTOR – displacement at the GRID point in the displacement coordinates. u These displacements are in the BASIC coordinate system and verify that the loading is an axial load (there is only axial displacement). Notice that the OLOAD RESULTANT correctly shows a resultant moment about the origin, but the element is offset by 1.0 units.

25 S3-25NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.) n LOAD VECTOR – is the loading at the GRID points in the displacement coordinate system. u In this case, we can now see both the axial component (–1.0 in the X– direction) and the moment due to the offset. n FORCE DISTRIBUTION IN BAR ELEMENTS – forces in the elements (in the element coordinate system) u Once again, this verifies that the loading is an axial loading only, with the element output showing only axial force. u NOTE – the PLOAD1 is a loading applied on the element. In MSC.NASTRAN, the element is considered to begin and end at the offset locations. Therefore, the applied load is along the axis of the element. n GRID POINT FORCE BALANCE – Where is the moment? u Remember that this output is in the displacement coordinate system of the grids. Each element is offset from the GRID points by 1.0 units (internally a rigid offset), and the elements transfer the axial force and the resulting moment to the GRID points.

26 S3-26NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.) u Case 2: GRID points using the coordinate system 100 as the displacement coordinate system. ID OFFSET, BAR MODEL SOL 101 CEND TITLE = TEST OF OFFSET BAR MODEL – FILE offbar100. dat SUBT = USE DISPLACEMENT COORDINATE SYSTEM LOAD = 1 DISP = ALL ELFORCE = ALL GPFORCE = ALL set 999 – 7 oload = 999 BEGIN BULK $ CORD2R,100,,0.,0.,0.,0.,0.,1.,,0.,1.,0. GRID,1,,0.,0.,0.,100, GRID,2,,2.,0.,0.,100 =,*(1),=,*(2.),== =(4) CBAR,1,1,1,2,1.,0.,0.,,+CB1 =,*(1),=,*(1),*(1),=,=,=,=,*(1) =(4) $ OFFSET 1 UNIT IN THE BASIC Y DIRECTION – $ THIS IS IN THE +X IN GRID OUTPUT COORD +CB1,,,1.,0.,0.,1.,0.,0. *(1),== =(4) PBAR,1,1,1.,12.,12.,24., MAT1,1,30.+6,,.3 pload1,1,6,FX,FR,1.,–1. ENDDATA

27 S3-27NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.) n Since coordinate system 100 is the displacement coordinate system, the element offsets will be in the x-direction. n NOTE – For this example, orientation vectors of are used for the BAR elements. Although this appears to be parallel to the elements, it is not, because the orientation vector of BAR and BEAM elements is defined using the displacement coordinate system of the first GRID it connects to.

28 S3-28NAS105, Section 3, May 2005 OFFSET BAR USING DISPLACEMENT COORDINATES

29 S3-29NAS105, Section 3, May 2005 OFFSET BAR USING DISPLACEMENT COORDINATES (Cont.)

30 S3-30NAS105, Section 3, May 2005 OFFSET BAR USING DISPLACEMENT COORDINATES (Cont.)

31 S3-31NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? n The preceding output is similar to that obtained in the original run. n The OLOAD RESULTANT, SPCFORCE RESULTANT, and ELEMENT FORCES are identical. u The RESULTANTs are in BASIC, and the ELEMENT FORCES are in the element coordinate system. n The DISPLACEMENTs are identical, however, they are transformed into the displacement coordinate system (100), so they now show up as Y–translations. n The LOAD VECTOR is also in the displacement coordinate system u We still see the 1.0 unit load and the moment due to the offset, but now they are in system 100. n GRID POINT FORCE BALANCE – now we see the identical results as before, but they are in the displacement coordinate system (system 100).

32 S3-32NAS105, Section 3, May 2005 SHEAR CENTER – BEAMS AND BARS n One of the most common modeling errors is to ignore the offset between the shear center and the neutral axis on a BAR or BEAM element when the cross- section is not doubly-symmetric. n Most users use the BAR element. The BAR element assumes that the shear center and the neutral axis are coincident.

33 S3-33NAS105, Section 3, May 2005 SHEAR CENTER – BEAMS AND BARS (Cont.) n The BEAM element allows the shear center and the neutral axis to be offset from each other. n The offset on the CBEAM entry is from the GRID points to the shear center. The PBEAM entries (N1A, N2A, N1B,N2B) are the offset from the shear center to the neutral axis. n NOTE – the CI, DI, EI, and FI (stress recovery location) are relative to the shear center –they are NOT relative to the neutral axis. n Element loads located at the shear center; not at the neutral axis. n Once again, if the section is doubly-symmetric, or if there is no load effecting the offset, BAR will work fine in linear analysis. Otherwise use BEAM.

34 S3-34NAS105, Section 3, May 2005 BEAM Vs. BAR FEATURES CBEAMCBAR Variable X-Section YES NO Warping X-Section YES NO Shear Relief YES NO Shear Center Offset YES NO Mass Moment of Inertia YES NO Geometric Nonlinear YES NO Plastic Hinges YES NO

35 S3-35NAS105, Section 3, May 2005 HOW ARE OFFSETS IMPLEMENTED ON PLATES n GRID points connecting plate elements may NOT lie on the plate mid-surface. n The plate offsets (ZOFFS) are entered on the CQUADi, and CTRIAi cards. A positive value of ZOFFS implies that the element reference plane is offset a distance of ZOFFS along the positive z-axis of the the element coordinate system. Material matrices and stress fiber locations are relative to the reference plane.

36 S3-36NAS105, Section 3, May 2005 HANDY HINTS n Element strain energy (ESE) is a good tool for determining where to make changes to obtain maximum benefits. n Example – two springs in series Which is the best one to stiffen to reduce the tip deflection? where K 1 = 10 K 2 = 1 P = 1

37 S3-37NAS105, Section 3, May 2005 HANDY HINTS (Cont.) Most of the ESE is in the smaller spring. Therefore, stiffening it is the most efficient way to reduce the tip deflection. n Normal modes analysis is similar in that changing the stiffness of the elements with the most ESE in a mode is often the most efficient way to shift the frequency of that mode.

38 S3-38NAS105, Section 3, May 2005 HANDY HINTS (Cont.) n Determine mesh size based on behavior – in areas of high stress variation, place extra elements u from stiffener on a stiffened cylinder – normally 3 + elem Plot of Moment versus Distance for a Pressurized Stiffened Cylindrical Shell

39 S3-39NAS105, Section 3, May 2005 HANDY HINTS (Cont.) n Curved Shells u Normally QUAD4 can cover a 5–10 angle on a cylindrical surface. u QUAD8 can cover a 10–25 angle. u QUAD8 without midside points uses linear interpolation. This is much less accurate than QUAD4. u A simple rule – for buckling and normal modes, there should be at least 5 GRID points per half sine wave of the deformed shape..

40 S3-40NAS105, Section 3, May 2005 MORE HANDY HINTS n Plates should have same orientation for stress output. n Plate output is usually in the element coordinate system. n Pressure loads on plate are applied as point loads 1/4 at each corner on a QUAD4. u The direction is based on plate orientation. A positive pressure acts in the positive element Z direction. n Plates at a corner are much softer than the actual structure n Each one does not have in-plane rotational stiffness. The model has stability, but does not properly transfer loads.

41 S3-41NAS105, Section 3, May 2005 COMPATABILITIES Plate Beam Membrane – Beam (To improve the compatibility use MPC on z. )

42 S3-42NAS105, Section 3, May 2005 COMPATABILITIES (Cont.) Corners

43 S3-43NAS105, Section 3, May 2005 MODELING CORNERS WITH PLATES AND BEAMS Problem – A corner is much stiffer than is represented by two plate elements coming together Poor Solution

44 S3-44NAS105, Section 3, May 2005 MODELING CORNERS WITH PLATES AND BEAMS (Cont.) Good Solutions (recommended properties: calculated based on ½ of each attached plate elements – use only I1 ad I2 on the PBAR or PBEAM)

45 S3-45NAS105, Section 3, May 2005 EXAMPLE OF POOR PRACTICE Preferred Practice

46 S3-46NAS105, Section 3, May 2005 EQUIVALENT ROD Assume Length of Rod (L) = w Area of Rod (A) = ?

47 S3-47NAS105, Section 3, May 2005 To match this stiffness by a rod of length w, and area A: EA/(w ) = Et/8 A = wt /8 EQUIVALENT ROD (Cont.) Energy Stiffness

48 S3-48NAS105, Section 3, May 2005 EQUIVALENT ROD (Cont.) Compare with constant strain results Factor of 2 difference!

49 S3-49NAS105, Section 3, May 2005 EQUIVALENT ROD EXAMPLE RESULTS E = 6.9 X 10 4 K EQ = N / mn A = 2.21 Results

50 S3-50NAS105, Section 3, May 2005 PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS Q:Why are QUAD plates represented with no in-plane rotational stiffness when, in fact, the real structure does have this stiffness? A:A real plate does have in-plane rotational stiffness, but, this is implicitly represented by the in-plane translational degrees of freedom (T1, T2) in a plate finite element. Consequently, MSC.Nastran, like most general purpose finite element programs, uses either a zero (0.0), or a relatively small value to represent the stiffness with respect to the redundant in-plane rotational DOF (R3) at a plate node. It is left up to the user to either SPC this DOF, at nodes where the connected elements have parallel, or nearly parallel normals.

51 S3-51NAS105, Section 3, May 2005 PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS (Cont.) AUTOSPC, K6ROT, SNORM: Beside the AUTOSPC (Case Control Command in V2004, PARAM in earlier versions), Nastran also has K6ROT (PARAM), for giving a relatively small stiffness to this DOF so the problem can run, without encountering singularity, or round-off problems. However, the user has to be careful against using too large a value for K6ROT, and against errors hidden by the auto- stiffness feature (see example on the following page). PARAMETERS DEFAULT VALUES K6ROT SNORM 20.0

52 S3-52NAS105, Section 3, May 2005 PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS (Cont.)

53 S3-53NAS105, Section 3, May 2005 PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS (Cont.) n The model is to predict first two natural frequencies that are compared to test results. The first mode (in and out of paper) is satisfactory; the second mode (rocking back and forth) does not match the test (too low a frequency). n Problem u The plates had small in-plane rotational stiffness added using K6ROT. This unknowingly resulted in a pin-pin end condition for the beams supporting the vibrating mass and, therefore, the low frequency. n Possible Solutions u Extend the beam one grid into the plates at both ends of the beam. This takes the beam moment out as a shear couple. u Use an MPC to connect the rotational DOF to adjacent translations n Example:

54 S3-54NAS105, Section 3, May 2005 MESH TRANSITIONS n In general, mesh transitions are handled by modern pre–processors (e.g. MSC.Patran), and are not as much of a concern as they were in the past. n General rules for mesh transitions u Keep transitions away from areas of interest. u Try to use compatible elements. u If compatible elements cannot be used, use R-type elements to approximate the dominant behavior. Coarse and Fine MeshElements of Different Types

55 S3-55NAS105, Section 3, May 2005 MESH TRANSITIONS (Cont.) Non-Conforming Element Types Mismatched Shapes

56 S3-56NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS n Situation: Your pre–processors may not handle beam–to–plate, beam–to–solid, or plate–to–solid connections automatically. n If you have any of these connections in your model, they require special modeling efforts. n Example: u Have existing Grid Points 1 through 12 and QUADs Q1 through Q6. It is desired to attach beam 10 to QUAD mesh using Grid Point 20. Several solutions are discussed here. (Note: DOF 6 of Grid Points 1 through 12 might be constrained since QUAD plates d not have stiffness in this direction.)

57 S3-57NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS (Cont.) Option 1 n Grid 20 is not added. Use offsets for BAR. n Beam 10 goes from Grid 30 to Grid 4 with offset from Grid 4 to Beam 10 center line. n Problems u Unrealistic moment in plates is due to the beam offset. u The in-plane rotation must be handled. Otherwise, it is a pinned connection for that DOF.

58 S3-58NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS (Cont.) Option 2 n Add a grid and two beams n Beam properties approximated by section of a QUAD half width and its thickness n Problems u May have added extra stiffness at edge due to beams u May lose some local effects where the beam attaches to the plates Note: DOFs 1 through 6 refer to XYZ coordinate system as defined here. In applying these solutions to another problem, note which DOFs are the out-of-plane and in-plane stiffness.

59 S3-59NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS (Cont.) Option 3 n Add a grid and three triangles n Problems u Need to add RBE3 DOF 6 from Grid Point 20 to Grid Points 4, 5, 7, and 8 in DOFs 1 and 2 u Must add two more elastic elements and one rigid element u Be careful not to constrain DOF 6 at GRID point 20

60 S3-60NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS (Cont.) Option 4 n Add a grid and an RBE3. n RBE3 DOF 1 through 6 at Grid Point 20 to DOFs 1, 2, 3, and 5 of Grid Points 4, 5, 7, 8 n Problems u Lose some local effects near the beam connection

61 S3-61NAS105, Section 3, May 2005 BEAM TO PLATE ELEMENTS (Cont.) n Situation where the beam attaches to an existing grid n Beam 10 extends from Grid Point 30 to Grid Point 7 n RBE3 Grid Point 7 DOF 6 to DOF 1, 2, and 3 of Points 4, 8, and 10 n Do not SPC DOF 6 at Grid Point 7. n Problems u Handling the in-plane rotation

62 S3-62NAS105, Section 3, May 2005 MORE MESH TRANSITIONS Solid Plate* RBE2 = enforce plate theory at transition RSSCON = easy way to make the connection, especially if your preprocessor supports it. Use MPCs or RBEs

63 S3-63NAS105, Section 3, May 2005 MORE MESH TRANSITIONS (Cont.) Higher Order – Lower Order If you do this, always do it away from areas of interest.

64 S3-64NAS105, Section 3, May 2005 SOME POSSIBLE PLATE-TO-SOLID TRANSITION Split Plate Extra Element

65 S3-65NAS105, Section 3, May 2005 SOME POSSIBLE PLATE-TO-SOLID TRANSITION (Cont.) Split Solid Note: The plate may be the same thickness as the solid.

66 S3-66NAS105, Section 3, May 2005 STRESS SORTING n A number of solutions (including 109 and 112) in MSC.NASTRAN offer an option to sort and filter the resultant stresses based on user-defined criteria. This option requires PARAM, S1,0 along with the following optional entries.

67 S3-67NAS105, Section 3, May 2005 STRESS SORTING (Cont.) Remarks 1. The user may override the existing default quantity on which the sort is to be preformed or he may specify the quantity on which the sort is to be performed for elements that have no default. This option is exercised through the following procedure: a. Include DTI,INDTA Bulk Data entry in the Bulk Data Section (see the MSC.NASTRAN Quick Reference Guide). 2. Sorting large amounts of data can be expensive. 3. The parameters SRTELTYP refers to element type numbers that are provided in the table given.

68 S3-68NAS105, Section 3, May 2005 DTI, INDTA – BULK DATA ENTRY Specifies or overrides default item codes for the sorting and filtering of element stresses, strains, and forces.

69 S3-69NAS105, Section 3, May 2005 DTI, INDTA – BULK DATA ENTRY (Cont.) Remarks: 1. This table is recognized only in SOLs 1, 3, 5, 14, 15, 16, 101, 103, 105, 106, 108, 109, 111, 112, 114, 115, 144, 153, and for stress quantities only. One or more of the user parameters S1, S1G, or S1M must be specified with a value greater then or equal to zero in order to request sorting and/or filtering. See also parameters S1AG, and S1AM in the MSC.NASTRAN Quick Reference Guide. 2. If the Ci value is 1, the element type will be suppressed on the output file. An example of this feature could be as follows: If an element type is to be sorted on two different values and output twice, this can be accomplished by two calls to the STRSORT module with two unique DTI tables. However, other element types will be printed twice. This additional print can be suppressed by setting their sort codes to 1.

70 S3-70NAS105, Section 3, May 2005 DTI, INDTA – BULK DATA ENTRY (Cont.) Remarks: (Cont.) 3. Table 1 lists the elements currently that are sort able. In addition, the element type identification number, the default stress output quantity, and the associated stress code identification numbers are provided. If this entry is not specified, then the stresses are sorted based on the default quantity given in Table 1. The Following Should Be Noted A. The element type identification number is used internally by the program to differentiate element types. B. The stress code identification number is merely the word number in the standard printed output for the stress quantity of interest. For example, the thirteenth word of stress output for the CHEXA element is the octahedral shear stress. For this element type, the element identification number and the grid point ID each count as a separate word. Stress codes for the elements are tabulated in the MSC.NASTRAN Quick Reference Guide, Appendix A. C. By default, stress sorting for the membrane and plate elements will be performed on the Hencky-von Mises stress. For maximum shear stress, the STRESS (MAXS) Case Control command should be specified.

71 S3-71NAS105, Section 3, May 2005 DTI, INDTA – BULK DATA ENTRY (Cont.)

72 S3-72NAS105, Section 3, May 2005 STRESS SORTING – SAMPLE C A S E C O N T R O L D E C K E C H O CARD COUNT 1 TITLE = S.E. SAMPLE PROBLEM 1 2 SUBTITLE = S.E. STATICS - RUN 1 - MULTIPLE LOADS 3 DISP = ALL 4 SEALL = ALL 5 PARAM,GRDPNT,1 6 SUBCASE SUPER = 1,1 $ SUPERELEMENT 1 - RESIDUAL LOAD 1 8 LABEL = S.E. 1 PRESSURE LOAD 9 LOAD = STRESS = ALL 11 PARAM,S1,1 12 PARAM,NUMOUT,10 13 $ 20 SUBCASE SUPER = 1,2 $ SUPERELEMENT 1 - RESIDUAL LOAD 2 22 LABEL = S.E = NORMAL LOAD 23 LOAD = $ 30 SUBCASE LABEL = S.E. 1 - OPPOSING LOADS 32 SUPER = 1,3 $ SUPERELEMENT 1 - RESIDUAL LOAD 3 33 LOAD = $ 40 UBCASE LABEL = RES STR LOAD 1 - PRESSURE 42 SPCFORCES = ALL

73 S3-73NAS105, Section 3, May 2005 STRESS SORTING – SAMPLE (Cont.)

74 S3-74NAS105, Section 3, May 2005

Еще похожие презентации в нашем архиве:

© 2017 MyShared Inc.

All rights reserved.