SUPPORT VECTOR MACHINE FOR STRUCTURAL RELIABILITY ANALYSIS

Structural Mechanics 13.0Key ThemesBrake Assembly•Advance nonlinear analysis support in WB •Cyclic symmetry (in Workbench)•Consistent pre-stress modal analysis ***•Beam & shell modeling in WBBeams and shells •Support external loads mapping •GPGPU support for solvers•New advanced materials models •Composites failure analysis -VCCT •3D rezoning•Advances in contact technologyAdvanced Nonlinear AnalysisRestart,Stabilization,Materials,Gasket, CreepRestarts•Ability to generate, manage, and reviewrestart points automatically and manually•Restart analysis :The image part with relationship ID rId3 was not found in the file.–After analysis settings changes,–Command snippet modifications•Static & transient structural analyses•Add load steps(existing loads only) to a completed analysis•Evaluate “Post” command snippets without performing a full solveRestartsWhen convergence failsVisual indication of stateReport –where the next solve will restart from Restart point creation controlled via analysis settings object.Timeline shows other available restart pointsEx. Restart with Post SnippetsSetup before solveSolve modelAdd Post command12Solve model, note that only post commands are issued, (see input file)34Stabilization•Added stabilization as a tool to help achieve convergence in unstable problems.•Support for static and transient structural analyses•Can be turned on/off during each load step•Can be turned on/off within a load step by restart•Result support for stabilization energyNew Material Models in WB•Response function model *new in R13•Chaboche kinematic hardening •Anand plasticity model •Gasket material modelCurve fitting support for Gasket•Temperature dependency for –BISO , BKIN ,–All hyperelastic models including Gent, Blatz-Ko, Arruda Boyce •Enhanced -Anisotropic elasticity UI to check for input errortemperature dependentGasket Analysis•New ‘Stiffness Behavior ’ control (gasket)•Setting above behavior to ‘Gasket’automatically adds a new object ‘Gasket Mesh Control” in the tree under the partInvalid if Material doesn’t haveGasket model•3D solids only•Results support for gasket bodies including pressure and closureCreep•Creep Material Model now supported •New Analysis Settings group –‘Creep Controls’•Load step dependent properties(Need at least 1 creep material added)•Added two new result types:•Equivalent creep strain•Equivalent total strainMisc. Post Enhancements• Results can now be scoped to Named Selections • RMB to automatically create a path from result plots • Result Probes – “Snap to Mesh” • ‘Max over Frequency/Phase’ for harmonic results • Contact area – Result Tracker • RMB to create Contour Results from Response plots(Beta) • Response PSD for remote points(Beta)© 2010 ANSYS, Inc. All rights reserved.11ANSYS, Inc. ProprietaryLinear DynamicsCyclic Symmetry Harmonic Analysis PrePre-Stressed Modal© 2010 ANSYS, Inc. All rights reserved.12 12ANSYS, Inc. ProprietaryCyclic Symmetry Support• Static and modal analysis support • Thermal-stress support • Smart boundary constraints manager • Expansion of the sector • Traveling wave animation • Support pre-stress modal cyclic symmetry • Peak holds results© 2010 ANSYS, Inc. All rights reserved.13ANSYS, Inc. ProprietaryModal Cyclic Symmetry: BCsConstraint manager prevents choice of cyclic sector from affecting distributions of constraints: • Constraint not aligned with cyclic CS • Different numbers of constraints per sector(full model) •Supports: 4 •Faces: 4 •Corners: 4(per sector) •Supports: 2 •Faces: 2 •Corners: 1 •BCs not aligned with Cyclic CS(per sector) •Supports: 1 •Faces: 1 •Corners: 2 •BCs not aligned with Cyclic CS© 2010 ANSYS, Inc. All rights reserved.14ANSYS, Inc. ProprietaryModal Cyclic SymmetryCyclic ExpansionHarmonic Index ControlModes listed by Harmonic IndicesInteractive Frequency-Spectrum Display© 2010 ANSYS, Inc. All rights reserved.15ANSYS, Inc. ProprietaryHarmonic Analysis • Mode reuse for MSUP harmonics – Reuse prior Modal results for a downstream Harmonic analysis • Supports are specified in a modal analysis • Loads can be applied in a harmonic analysisModal reuse© 2010 ANSYS, Inc. All rights reserved.16ANSYS, Inc. ProprietaryConsistent PrePre-Stressed Modal© 2010 ANSYS, Inc. All rights reserved.17 17ANSYS, Inc. ProprietaryPre-Stress Modal AnalysisOld - PSOLVEOnly recreate the last point of the nonlinear analysis Only treat nonlinear materials as linear during expansion pass Only consider contact status as frozen (status remains fixed) Requires user to know ahead of time that they want to do (PSTRESS, EMATWRTE etc.) Requires use of the SOLVE in nonlinear step and PSOLVE in modal step (messy)© 2010 ANSYS, Inc. All rights reserved.New - PERTURBRecreate snapshot at any time point of the nonlinear analysis Treat nonlinear materials as linear or nonlinear during expansion pass Consider contact as frozen (status remains fixed) or as sticking - per contact pair basis Does not require any additional commands ahead of time New procedure removes the need for the PSTRES, EMATWRITE, and PSOLVE18ANSYS, Inc. ProprietaryPre-Stress Modal AnalysisSupported Options• • • • • • Large deflection Cyclic symmetry MODCONTROL,ON (multiple load steps) ANTYPE,MODAL,RESTART (load regeneration) RESVEC,ON (residual vector method) Most newer technology element types (17x contact, 18x structural, etc..) • • •Unsupported @ R13Most older/legacy element types QRdamp eigensolver (MODOPT,QRDAMP) EXPASS,ON (frequencies must be expanded when computed)© 2010 ANSYS, Inc. All rights reserved.19ANSYS, Inc. ProprietaryPre-Stress Modal Analysis • Mechanical now employs the new MAPDL linear perturbation technique for all pre-stress eigenvalue analysis– Support for large deflection, nonlinear static pre-stress modal analysis – Handling true contact status from the static analysis into the eigen-value analysis – Since based upon multi-frame restart files, multiple eigen-value analyses(perturbations) can be run – Pre-stress modal cyclic is supported! – Consistent commands for all variations© 2010 ANSYS, Inc. All rights reserved.20ANSYS, Inc. ProprietaryExample: Pre-Stress Modal in WB1Modes Near 2SlidingModesBlade-Brake Assembly Nonlinearities • Large deformation • Material • Contact status© 2010 ANSYS, Inc. All rights reserved.3Modes Sticking21ANSYS, Inc. ProprietaryEx. Pre-Stress Brake Squeal AnalysisStatic Analysis: Large deflection, frictional contact Pre-stress Modal Solve: QRDamp and UNSYM eigen-solvers QRDamp solution accuracy is close to UNSYMM eigen-solver© 2010 ANSYS, Inc. All rights reserved.22ANSYS, Inc. ProprietaryBeams & Shells© 2010 ANSYS, Inc. All rights reserved.23 23ANSYS, Inc. ProprietaryMesh-Based Shell Connections Connect shell bodies at mesh level:• Shared Nodes at the boundary • No need to topologically join the bodies in the CAD system • Can handle geometric imperfections (gaps and overlaps) • Support for shell edge-face and shell edge-edge only • Need parts to be in a multi-body part3 4 1 2© 2010 ANSYS, Inc. All rights reserved.24ANSYS, Inc. ProprietaryShell Thickness by Face• Define a thickness on a shell body per face • Body thickness is overridden • Import custom shell thickness for a given face from DM • Constant, Tabular or Function input for the thickness • Tabular and Function data is based on either x, y or z locationResults Mesh Geometry© 2010 ANSYS, Inc. All rights reserved.25ANSYS, Inc. ProprietaryBeam Boundary Conditions Beam End Releases• Release some degrees of freedoms at edge vertices of a beam • Under the Connections folder • Coupling or Joint (MPC184)Bolt Pretension on Beams• Bolt pretension loads can now be applied to line bodies - scoping by ‘body’ or ‘edge’© 2010 ANSYS, Inc. All rights reserved.26ANSYS, Inc. ProprietaryExpanded Beam ResultsNew resultsStep changes captured© 2010 ANSYS, Inc. All rights reserved.27ANSYS, Inc. ProprietaryConnection Group • Auto generate contacts or mesh connections for a group of bodies/subassemblies • Use its own tolerance value • Better organization when many connection objects are present • Support joints as well© 2010 ANSYS, Inc. All rights reserved.28ANSYS, Inc. ProprietaryEnhanced - Named SelectionsNew option to define selection by worksheet Worksheet allows set of actions to build up a named selection based on a criteria(size, location, type, coordinate system, convert etc.Able to quickly define criteria when creating from Graphics WindowAKA – Select Logic© 2010 ANSYS, Inc. All rights reserved.29ANSYS, Inc. ProprietaryMisc. Enhancements• Edge Visualization: The edges can be colored to show connectivity - both on mesh and geometry • Option to show vertices of the geometry • Option to split edge virtually Select : Virtual split edge and then press ‘s’ key to dynamically move the split location (with LMB)© 2010 ANSYS, Inc. All rights reserved.30ANSYS, Inc. ProprietaryLoad MappingExternal Data System Dissimilar Mesh for Thermal Stress Expanded Transfer Connections© 2010 ANSYS, Inc. All rights reserved.31 31ANSYS, Inc. ProprietaryExternal Data - Features “External Data” *new system• Import point cloud data to– Static/Transient Structural – Static/Transient Thermal – Thermal-ElectricRMB• Supported loads (Mechanical)– Pressure, temperature, convection – 2D edges, 2D/3D faces, volumetric – Mapping from 2D-2D, 3D-3D & 2D-3D• Supports coordinate systems, units, preview data • Weighting options, advanced control© 2010 ANSYS, Inc. All rights reserved.32ANSYS, Inc. ProprietaryImported Data Inside MechanicalMapped Data is displayed Imported Load Folder createdOption such as scoping and mapping controlsSupport for scale, offset, multiple load steps via worksheetIf source data is 2D, Projection option is available© 2010 ANSYS, Inc. All rights reserved.33ANSYS, Inc. ProprietaryE.g. Thermal-Stress AnalysisAutomatically handle ‘scoping’ to prevent bleed across body boundaries.© 2010 ANSYS, Inc. All rights reserved.34ANSYS, Inc. ProprietaryCFD to Mechanical Volumetric Temperature Transfer2© 2010 ANSYS, Inc. All rights reserved.35ANSYS, Inc. ProprietaryE.g. Mechanical – Maxwell/HFSS• From Maxwell/HFSS: import heat generation/heat flux to Mechanical • From Mechanical: export thermal results to Maxwell/HFSS • From Maxwell: import surface/body force densities to Mechanical© 2010 ANSYS, Inc. All rights reserved.36ANSYS, Inc. ProprietaryExternal Data – Debugging Tools• Graphics Control – Visualize source points on target geometry – For 2D to 3D, option to display the projection plane – Hide/show source points falling inside the target model • Diagnostic Information – Diagnostic information is created to give details about the mapping© 2010 ANSYS, Inc. All rights reserved.37ANSYS, Inc. ProprietaryCore MechanicsSolver Linear Dynamics Materials Elements© 2010 ANSYS, Inc. All rights reserved.38 38ANSYS, Inc. ProprietaryGPU Accelerator Feature• SMP SPARSE and PCG solvers only • Activated with –acc @ command line • Supported on Windows/Linux 64-bit systems • Currently only available for NVIDIA Tesla 20-series (1U) cards•Intel Xeon 5560 (2.8 GHz, 8 cores total) • 32 GB of RAM • Windows XP SP2 (64-bit) • Tesla C2050Overall Simulation Speedups for R12 Benchmark Set© 2010 ANSYS, Inc. All rights reserved.39ANSYS, Inc. ProprietaryEigen-Solvers: UNSYM, DAMPUNSYM, DAMP solves can now be distributed Performances improved with auto-shift logicModal extraction for a 740,000 dof model- Fluid/Structure Interaction - 741K DOF - Calculate 20/40/60 values - Linux64 machine, using 4 processors40ANSYS, Inc. Proprietary© 2010 ANSYS, Inc. All rights reserved.DANSYS Enhancements DANSYS now supports more solution procedures• Features previously blocked in DANSYS will now work, in a sequential fashion on the master process – PSD analyses – MSUP harmonic and transient analyses – Modal reuse (load vector regeneration) – Residual vector method (RESVEC,ON) – Substructure generation and expansion passes • Enhanced scalability – Many enhancements to improve performance of DPCG and DSPARSE solvers© 2010 ANSYS, Inc. All rights reserved.41ANSYS, Inc. ProprietaryLinear DynamicsVT Enforced Motion PML for Acoustics APDL Math© 2010 ANSYS, Inc. All rights reserved.42 42ANSYS, Inc. ProprietaryVT-Based Enhancements• Frequency sweep (VT) for harmonic analyses– All problems which are not frequency dependant automatically use VT method (default) – Mostly a single factorization is needed to cover the entire frequency range• TRNOPT,VT option is now applicable to linear transient analyses– Can reduce the time for this kind of analyses up to a factor 10X© 2010 ANSYS, Inc. All rights reserved.43ANSYS, Inc. ProprietaryEnforced Motion Enforced motion allows a user to apply non-zero displacement/acceleration boundary conditions in a MSUP analysis Full TransientMSUP TransientEnforced motion Analysis of a PCB Structure© 2010 ANSYS, Inc. All rights reserved.44ANSYS, Inc. ProprietaryPML for Acoustic Elements • Sound/pressure radiation from two waveguides – Wave number k = 8π, acoustic speed c = 340 m/s – Element size = 0.05 With PML – FLUID30w/o Boundary ConditionWith PML – FLUID220© 2010 ANSYS, Inc. All rights reserved.With PML – FLUID221ANSYS, Inc. Proprietary45APDL Math Commands • What is it? – A new feature that extends the APDL scripting environment of Mechanical APDL– Access to the powerful matrix manipulation routines in MAPDL product including solvers – You can read them, manipulate them, and write them back out or solve them directly. – This augments the vector and matrix operations in the standard APDL scripting environment. – Import, export features to internal/external data format:• ANSYS files, Nastran files, public formats© 2010 ANSYS, Inc. All rights reserved.46ANSYS, Inc. ProprietaryAPDL Math: ExampleCompute the critical speeds of a rotating object• • •Traditional Method Loop over several rotational • speeds, many eigensolves Build Campbell diagrams • Compute the intersections of the critical speed vs. • curve of Campbell diagram47APDL Math Critical speeds are the eigenvalues of:[K - ω2.(M+jαG0) + jωC]Ф=0Save time (only ONE eigensolve), Gives exact results, very robustANSYS, Inc. Proprietary© 2010 ANSYS, Inc. All rights reserved.Reuse of Modal Extraction Results• After modal analysis, user can:– Apply load for downstream MSUP analysis – Stress expansion – Calculate new residual vectors for downstream MSUP analysis – Apply enforce motion vectors for MSUP transient/harmonic• Supported in QR damp solve, too– Can be reused at the 2nd and subsequent load steps (Useful feature for Campbell plots in rotordynamics)– Can be saved for a later QR damp solution (Useful for efficient Brake Squeal Analysis)© 2010 ANSYS, Inc. All rights reserved.48ANSYS, Inc. ProprietaryPOST1 – RSTMAC Based on Mapping and InterpolationThe RSTMAC command compares the nodal solutions from two results files. The first step is either: – Default - match the nodes of model1 and model2. (For identical or very similar meshes)New in 13.0Or– Map the nodes of model2 into the elements of model1. The solutions of model1 are interpolated. (General procedure, time consuming)Note: the new procedure applies to results obtained from cyclic symmetric models.© 2010 ANSYS, Inc. All rights reserved.49ANSYS, Inc. ProprietaryElement Nodal Forces Summation in Spectrum Analysis• Supports: – Power Spectral Density (PSD) – Single Point Response Spectrum (SPRS) – Multi-Point Response Spectrum (MPRS) • The reaction force calculation in Spectrum Analysis is properly performed accounting for the reaction force sign at support points. – NFORCE – FSUM – PRNLD© 2010 ANSYS, Inc. All rights reserved.Total forces/moments at an interface50ANSYS, Inc. Proprietary。

BOARD TESTOVector vs. Vectorless ICT Test Techniquesby Alan Albee and Michael J. Smith, TeradyneOver the last decade, there has been a move away from powered-up digital in-circuit vector testing to unpowered analog-based (vectorless) device-pin opens testing for large and sometimes small digital devices. Two of the driving factors behind this move are the increasingly complex and custom design of digital devices that are being used and the limited digital in-circuit test (ICT) capabilities of most ICT systems. Most ICT systems on the market today simply do not have the timing and voltage accuracies required to reliably and safely test today’s generation of low-voltage, high-speed digital components.At the same time, analog vectorless measurement techniques that can be used to detect open pins have become more acceptable and widespread. The introduction of active capacitive probes and advanced software algorithms has made the open-pin defect detection provided by these techniques acceptable to most manufacturers.ICT program developers also lack the digital device models required to perform digital vector testing, and they do not have the time required to write digital vector models for complex devices. As a result, many advanced ICT systems also are using analog opens techniques. This has almost leveledFigure 1. Fixture With Probes Fitted, Amplifier, and Multiplier Cardthe defect coverage of low-cost ICT and manufac-turing defect analyzers with the high-performance ICT systems, although the advanced analog open techniques found on the high-performance ICT systems do give superior defect detection on small-geometry devices.Some manufacturers no longer feel that they need to verify that the correct device has been placed and that it is functioning correctly. They are willing to settle for analog opens vectorless testing because it can detect structural defects.All of these factors have combined to make analog opens techniques the preferred solution as manufacturers settle for less test and fast imple-mentation, although there may be potential hidden cost. Costs can arise because analog open vector-less measurement techniques have a number of limitations compared with traditional digital vector testing that have been ignored or forgotten.Additional Fixture CostsBoth digital vectors and analog opens techniques require direct, or in some cases indirect, access to each pin of the DUT. As a result, limited access is not really an issue between the two techniques. Analog opens testing requires that additional hardware be built into the test fixture. The hardware consists of a probe/plate that must be accurately placed over each device that will be tested and a signal amplifier and a central multiplexer/amplifier that interfaces directly with the ICT measurement subsystem (Figure 1).Some test systems also need additional hardware to deal with the signals generated from the fixture hardware, which adds cost to the test system. The additional hardware typically increases fixture costs by $100 to $150 per device, and that doesn’t include the fixture top gate needed to position the probes on the board.Fixture complexity and costs further increase if capacitive probes are required for both the topand bottom sides of the board. In thiscase, the probes must be designed intothe probe plates with CAD systemsto get the position and height settingcorrect. This also can be a problemfor bottom-side analog opens probesin a single-sided fixture. These costsquickly increase with 20 opens tests,on average adding about $2,500 to thecost of a fixture. With vectorless tests,these costs are incurred with each testfixture that is built.In comparison, powered-up vectortests do not require any fixture hard-ware. For that reason, manufacturershave the benefit of lower fixture costsand the possible elimination of expen-sive hold-down gates. Additional Fixture Reliability Problems and Maintenance Including additional hardware in thetest fixture decreases its reliability. Inthe case of analog opens probes, theycan be easily moved and damagedduring the rigors of production testing(Figure 1). They are not expensive toreplace at $30, but any alignment errorsin the probe can produce misleadingresults, false failures, or false passes aswell as potential damage to the UUT.This adds a higher level of fixturemaintenance and test uncertainty. Anddealing with probes mounted into theprobe plates prompts a much higherlevel of support complexity.With powered-up vector testing,there is no special fixture hardwarerequired to execute the tests. Reliabletest results can be more easily achievedbecause there are fewer variables thatmanufacturers must control and lessvariance across different test systemsand test fixtures.ThroughputOne issue not normally consideredis the difference in test time associatedwith using analog opens techniques vs.digital vector tests. Analog opens typi-cally take about 2 ms per pin to test.The amount of time required toexecute a digital test depends on thenumber of vectors and the vector rate.An ICT system applying digital testvectors at a 5-MHz data rate (200 nsper test vector) can execute 10,000 test vectors in the same amount of time ittakes to test one pin using the analogopens technique. Digital test vectorsalso can accommodate multiple pinsin parallel rather than one pin at atime. Even accounting for a typical10-ms vector load time, it is obviousthat digital vectors have a significanttest throughput advantage compared toanalog opens type measurements.A digital test usually takes less than50 ms to execute, and the test time isalmost independent of the number ofpins being tested. A device with 1,000pins uses about 2 s for the analog openstechniques to complete vs. 50 ms for adigital vector test. As a result, there is a40x to 50x faster throughput advantageif digital vectors are used.High-performance ICT systemsalso can test similar digital devicesin parallel; that is, memory devices,again speeding up vector-based test.As manufacturing beat rates increase,the amount of time allowed for ICTdecreases. Moving tests from analogopens techniques to digital tests willdecrease the test time significantlyand either match the line’s beat raterequirement or allow additional testsand device programming on the testsystem.Defect CoverageTypical defect coverage for an openstest is around 85% and can be as highas 99% or as low as 20% dependingon the package geometry and construc-tion. Power pins cannot be tested withanalog opens techniques and are notalways included in defect coveragereports.Digital test defect coverage normallyis very high, especiallywhen boundary scan isused, and we would ex-pect higher than 95%defect coverage. Un-like analog vectorlesstest techniques, the faultcoverage of digital vec-tor testing is not limitedby the device packagegeometry or construc-tion.Analog vectorlesstests utilize a learnedtechnique, and the fault coverage re-ports are just estimates of what defectsthe software thinks the test will detect.Until the defect is actually presenton the board and it gets detected, themanufacturer cannot be confident thatdefects are really being detected. Theycould have false passes on open pinsbecause of board coupling defects orunguarded nets that do not have physi-cal test access.With digital vectors, an advancedICT system can use automated faultinjection techniques to determinewhether or not open pins are detected.This can give manufacturers confi-dence that open pins will be detectedand diagnosed accurately even beforethe program and fixture are sent to themanufacturer.False Fails and Missed FaultsAs with any learned technique, theanalog opens technique can lead tomarginal measurements that can beclose to the limits. Advanced soft-ware algorithms in high-performanceICT systems can help eliminate thisproblem, but many systems use oldersoftware that can cause false failuresand false passes.Digital tests, on the other hand, usesimple logic ones and zeros, makingthem less susceptible to false failures.Digital tests also provide additionalconfidence that the device is workingand has power connected to it.Digital Pin ElectronicsICT always has been an ideal placeto carry out board customization andfunctional test. Simple programmingsuch as board serial numbers completeFigure 2. Low-Voltage Signals vs. Traditional 5-V and 3.3-V LogicBOARD TESTin-system programming,and Flash can be completed at board test but requires the appropriate digital test capability.Boundary scan tests are ideally suited to be run on an ICT system, but again the correct digital resources are required. Even complex functional test can be performed if the test system has a synchronized analog and digital subsystem. For that reason, if you are taking advantage of programming or using boundary scan, you don’t need to use analog opens techniques on those devices.Why Not Use Digital Testing? The two main problems that have stopped the continued use of digital in-circuit techniques when access is available have been the lack of digital vector models and the inadequate capa-bilities of some digital ICT systems. Digital ModelsMany companies and test program suppliers still provide test vector de-vice model programming services. Some are on the back of model devel-opment for component test, but most are related to the use of JTAG within the device design.Most large devices and program-mable devices have boundary scan to either test or program the device. With tools such as Teradyne’s Ba-sicScan, it is easy to automatically generate a digital model quickly from the BSDL, and it can even deal with configurations and tied pins. Other simple devices such as buffers, latches, and memory devices can be modeled, but manufacturers sometimes still use analog opens techniques because it is an easy development option.What Is Required to Provide Digital Vectors?To test the latest generation of digital components, the tester pin electronics must have certain characteristics. To-day’s lower voltage devices are based around 1-V logic which is very differ-ent from the 5-V and 3.3-V logic of theGenerally, measurement and test systems need to be 10 times more accurate than the UUT; in this case, better than 24-mV drive/sense ac-curacy. Dual-level logic thresholds for input and output also are required to guarantee accurate high- and low-logic levels (Figure 3). And to perform whatever function is required for test, the digital pins must be able to drive, sense, tristate, or act as pull-up/down resources for the net in case these are not on the UUT but on another board or backplane. All these features need to be programmable on a pin-by-pin basis; otherwise it becomes very dif-ficult to build fixtures to accommodate the different requirements of each logic family.The complexity of modern devices means that a number of vectors need to be applied. Timing errors between signals must be <5 ns across the pin fields, which could be thousands of channels. When the signals are applied, they need consistent and repeatable timing that only a dedicated digital controller can ensure.The reason for lower voltages is the reduction in power needed in modern electronics and the small die sizes used in device manufacturing. This means that during digital testing the amount of energy needs to be controlled to avoid damaging any of the devices. Not only do you need to limit current, but also the time spent forcing logic values must be controlled.Some devices require 600 mA to force a low logic level to a high, and others would be damaged if 100 mA were used. Some devices can take several milliseconds of backdriv-ing while others should be limited to microseconds. Consequently, the test system needs to be able to measure and control backdrive current and time characteristics on the fly in real time (Figure 3).If the test system does not have these features, then it may be easier and more prudent to use analog opens techniques than risk damaging the UUT.ConclusionAnalog opens has a place within theICT system suite of test techniques. Itis ideal for testing connectors, sockets,and devices that cannot be tested withdigital vector models. However, manu-facturers can use high-performanceICT systems to perform digital vectortesting and gain the benefits of fastertest throughput, lower cost fixtures,more reliable test measurements, andhigher fault coverage.About the AuthorsAlan Albee, the in-circuit test prod-uct manager working in Teradyne’sSystem Test Group, has worked atGenRad and Teradyne for 28 yearsin various engineering, applications,product support, and marketing po-sitions. He has authored numeroustechnical articles on topics relatedto board test and has been awardedtwo patents. Mr. Albee has a B.S. inindustrial science from Fitchburg StateCollege. Teradyne, 700 Riverpark Dr.,MS-NR-7001-1, North Reading, MA01864, 978-370-6238, alan.albee@Michael J. Smith has more than 30years experience in the automatic testand inspection equipment industrywith Marconi, GenRad, and Teradyne.The author of numerous papers andarticles also has chaired iNEMI’sTest and Inspection RoadmapingGroup for many years. Mr. Smith has aBSc(Hons) in control engineering fromLeicester University and is a memberof the Institution of Engineeringand Technology (MIET). smithmj@BOARD TEST。

结构力学答案（Structural mechanics answer）"The test of structural mechanics" chapter 01 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)The object of study, 1 structural mechanics BThe structure of A, B, single pole rodC, shell D, entity structure2, the structure strength calculation purpose is to ensure that the structure of AA, B, economical and safe without excessive deformationC D, beautiful and practical and non rigid motion3, the structure stiffness calculation, is to ensure that the structure of CA, no rigid motion B, beautiful and practicalC, without excessive deformation of D, economical and safe4, there are several fixed hinge support constrained force component? BA, a B, twoC, three D, four5, there are several movable hinge support constrained force component AA, a B, twoC, three D, fourSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, the stability of structure is DEA, structure of resistance failureThe ability of B and non rigid motionC, the structure of the ability to resist deformationD, the resistance ability of instabilityThe structure of E, the ability to keep the original balance formWhat kind of situation is not 2, the following BCDE plane structureA, all of the axis of the rod are positioned in the same plane, the load also function in the planeB, all of the axis of the rod are positioned in the same plane, and the plane vertical loadC, all of the axis of the rod are positioned in the same plane, and the plane parallel loadD, all of the axis of the rod are not located in the same planeEffect of E and load is not in the structure planeWhich of the following conditions should be 3, according to the spatial structure of ABDEA, all of the axis of the rod are positioned in the same plane, and the plane vertical loadB, all of the axis of the rod are not located in the same planeC, all of the axis of the rod are positioned in the same plane, the load also function in the planeD, all of the axis of the rod are positioned in the same plane, and the plane parallel loadEffect of E and load is not in the structure plane4, in order to ensure that the structure is economical and safe, to calculate the structure of BA, strengthB, stiffnessC, stabilityD, forceE, displacement5, the constraints of rigid node is ABA, the rod end cannot move relative to constraintB, binding the rod end can not rotateThe rod end C, constraints can be relatively moved along one directionD, the relative rotation rod end constraintsE, the relatively movable rod end constraintsThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, shell thickness is far less than the other two dimensions.Correct2, the entity structure and thickness of the other two dimensions are of the same order of magnitude.Correct3, in order to ensure that the structure is economical and safe, to carry out stiffness calculation of structure.error4, structural mechanics is a subject of bar structure strength, stiffness and stability.Correct5, the hinge node constraint cannot move relative to the rod end, but the relative rotation.Correct"The test of structural mechanics" chapter 02 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the three rigid geometric composition without redundant constraint invariant system, the number of constraints is necessary for DA, 3 B, 4C, 5 D, 62, what is the hinge to connect four rigid plates node? CA, B, single hinge node incomplete hinge nodeC, D, combined hinge node node3, connecting two rigid plates hinge has several constraints? AA, 2 B, 3C, 4 D, 54, there are several fixed hinge support constraints? BA, a B, twoC, three D, four5, a link system can reduce several degrees of freedom? AA, 1 B, 2C, 3 D, 4Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)Calculate the degrees of freedom ABC 1 geometry systemA, may be greater than zeroB, may be equal to zeroC, may be less than zeroD, must be greater than zeroE, must be equal to zero2, if the system is not a geometric system, then the BCD A, calculate the degrees of freedom may be greater than zero B, calculate the degrees of freedom may be equal to zero C, calculate the degrees of freedom may be less than zero D, degree of freedom must be equal to zeroE, calculate the degrees of freedom must be equal to zero3, from two yuan to remove a non redundant constraint geometric invariant system after the new system ACEA, is free of unnecessary restrictions of geometric invariant systemB, geometry variable systemC, the same degree of freedomD, is a redundant constraint geometric invariant systemE, is a geometric transient system4, the plane system in a link BCEA, can reduce the system of two degrees of freedomB, can reduce the system of one degree of freedomC, there is a constraintD, which has two degrees of freedomE, which has three degrees of freedom5, the following discussion is correctA, transient system in small load will produce great force ABCEB, the only geometric invariant system can be used as a building structureC, building structure degree of freedom is equal to zeroD, building structure calculation of degrees of freedom is equal to zeroE, degree of freedom equal to zero system that can be used as building structureThird questions to determine the problem (1 points for eachquestion 5 questions, a total of 5 points)1, the connection of two restriction effect two link rigid body is equivalent to a single hinge.Correct2 degrees of freedom, equal to zero system that can be used as building structure.Correct3, a link is equivalent to a constraint.Correct4, a single hinge can reduce the system of two degrees of freedom.Correct5, a single hinge is equivalent to two constraints.Correct"The test of structural mechanics" chapter 03 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, if a beam axis is parallel to the shear diagram and bendingmoment diagram for CA, B, parallel to the axis of zeroC, D, two parabolic oblique line2, if the shear diagram of a beam is zero line, then the moment diagram for BA, B, zero line parallel axisC, D, two parabolic oblique line3, if a beam axis is parallel to the shear diagram, the load on the beam is AA, B, no load uniform loadC, D, a couple of concentrated forceIn 4, the mutation at the moment of beam is what force? CA, axial force B, transverse concentrated forceC, D, a couple of forcesThrust 5, three hinge arch has nothing to do with the following factors? DA, high B, load vectorC, D, the span of arch axis shapeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, change the following factors that will not change the internal force of statically determinate beam? BCEA, loadB, section sizeC, material propertiesD, the span of the beamE, the stiffness of the beam2, which of the following factors on static beam internal force does not produce BCDEA, loadB, temperature changeC, mobile supportD, manufacturing errorE, material shrinkage3, if the shear diagram of a simply supported beam is a parallel axis, then the load on the beam may be ABCA, a left support action of a coupleB, have the right support action of a coupleC, a cross between a concentrated momentD, a cross between the uniform load.E, a cross between a concentrated forceThe characteristics of internal force diagram 4, beam without loading section is BA, the shear diagram parallel axisB, linear oblique shear diagramC, two parabolic shear diagramD, the bending moment diagram of parallel axisE, the bending moment diagram of oblique line5, static beam section size change, which of the following factors do not change? The displacement of ABCDA, axial forceB, shearC, momentD, bearing forceE, displacementThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, the external force in the basic part, internal force, deformation and displacement of accessory parts are zero.error2, static structure only by the equilibrium conditions we can find all the internal force and reacting force.Correct3, static structure in the movement of the bearing under the action of internal force, do not produce.Correct4, static structure only by the equilibrium conditions we can find all the internal force and reacting force.Correct5, zero rod force, so it is not needed in the truss rod can be removed.error"The test of structural mechanics" chapter 04 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the parallel chord truss beam (upper and lower internodes, when the upper bearing alignment) and lower load influence line is different is that? DInfluence of A, B, axial force of upper chord line oblique axial force lineThe influence line of effect of C and the bottom chord axial force, axial force vertical line D2, overhanging beam a line shape feature is AA B, a straight line, two lines lineC, the two parallel lines D, parabola3, support the extended beam shear influence line shape feature is CA B, a straight line, two lines lineC, the two parallel lines D, parabolaSection 4, overhanging beam bending moment influence line between supports is BA B, a straight line, two lines lineC, the two parallel lines D, parabola5, by the master-slave structure stress characteristics show that the internal force influence line affiliated part in the basic part of AA, B, all is zeroC andD are all negative, positive or negativeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)Shear effect of 1 section C, beam line, vertical scale C is left a, right C vertical scale is B, the following discussion is correctABA, a P = C 1 C in the left section shear generatedB, B P = C 1 C shear in the right when theC, a P = C C 1 in the left section shear point generatedD, B for P = 1 at C C during right shearE, a P = C 1 C shear in the right when the2, the bearing section overhanging beam bending moment influence line shape is ABCDIn A, the bearing vertical scale is zeroExcept B, the bearing is a straight line.C, the bearing is in the zero lineD, the other at the bearing vertical scale is zeroE, a straight line3, with the static part of the statically indeterminate beam, statically indeterminate part of the internal force influence line is characteristic of ABA, in the static part is on the lineB, in the indeterminate part is curveC, in the static part is curveD, in the indeterminate part is linearIn E, the entire structure is curve4, drawing method of influence line ABA, static methodB, mobile methodC, force methodD, displacement methodE, moment distribution methodEffect of line 5, which of the following is the dimensionless value? ABEA, bearing forceB, shearC, momentD, the restriction momentE, axial forceThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)K effect of line 1, C beam section bending moment in Shubiao said the moment P = K section 1 in the K point of the.error2, if there is a concentrated force on the influence line of vertex, will load the critical position.errorThe maximum vertical bending moment influence line 3 and the fixed end of the cantilever beam section of standard section at the free end.Correct4, beam bending moment influence line is broken.Correct5, the static truss influence line in a straight line between adjacent nodes will.Correct"The test of structural mechanics" chapter 05 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, static frame displacement in the movement of the bearing under the action by what is produced? DA B, the bending deformation and axial deformationC, D, shear deformation and rigid body motion2, using the principle of virtual work, the state should meet what conditions? DA, B, continuous constraintsC, D, physical balance3, the plane frame displacement under load deformation is mainly caused by what? BA B, the bending deformation and axial deformationC D, shear deformation and torsion deformation4, support for mobile indeterminate structure will produce CA B, the internal force and support reaction forceC, D displacement and deformationIn 5, the reciprocal theorem in delta = delta 12, 21, AA, 12 B, 11 sigma deltaC, 22 D, 31 sigma deltaSecond questions, multiple choice questions (2 points for eachquestion 5 questions, a total of 10 points)1, when the force acting on the part ofA, a subsidiary of the internal force is not zero.B, a subsidiary of the displacement is not zeroC, a subsidiary of the deformation is not zero.D, part of the basic force is zeroE, the basic part of the deformation is zero2, the static properties will change the structure of the material, which of the following factors do not change?A, displacementB, deformationC, axial force and shear forceD, momentE, bearing force3, two static frame geometry and size, load, the same material but different section size, the two factors are the sameA, forceB, forceC, stressD, displacementE, deformation4, two static two static frame geometry and the same size, load and section size but with different materials, the two factors are the sameA, forceB, forceC, stressD, displacementE, deformation5, the master-slave structure, only part of temperature rise, the following discussion is correct ACDEA, the internal force of structureB, the whole structure deformationC, part of the deformationD, part of the zero displacementE, the basic part of the deformation is zeroThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, the structure deformation, will cause displacement, in turn, will have deformed structure displacement.error2, static structure temperature changes caused by the displacement calculation formula is only applicable to the static set structure, is not suitable for statically indeterminate structure.Correct3, the virtual work is a fictional work, is actually not possible.error4, static structure in non load under the action of external factors, but does not produce force, displacement.Correct5, can not use the displacement diagram multiplication forthree hinge arch.Correct"The test of structural mechanics" chapter 06 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, the typical force equations, pay coefficient DA, B, constant constant greater than zero is less than zeroC, D, equal to zero can be positive or negative to zero2, get rid of a movable hinge support, equivalent to remove several constraints? AA, 1 B, 2C, 3 D, 43, get rid of a directional support, equivalent to remove several constraints? BA, 1 B, 2C, 3 D, 44, get rid of a fixed hinge bearing, equivalent to remove several constraints? AA, 2 B, 3C, 4 D, 55, the essence of force equation is DA, B, physical condition of equilibrium conditionsC, theorem D and displacement conditionsSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, single span symmetric structure in antisymmetric loads, section ACD on the symmetry axisA, displacement along a symmetry axis direction is zeroThe displacement of B, vertical symmetry axis direction is zeroC, the bending moment is zeroD, the axial force is zeroE, the angular displacement is zero2, two span symmetric structure under symmetrical load, the following discussion is correctRigid node line displacement A, the symmetric axis is zeroB, on the axis of symmetry of the rigid joint angular displacement is zeroC, in the column no momentD, no shear columnE, in the column without axial force3, in the typical force equations in coefficient is ABCDA, can be positiveB, may be negativeC, may be zeroD, there is reciprocal relationE, and the external cause4, the typical force equations can be positive or negative zero isA, 11.B, 12.C, 21.D, 22.E, Delta 1PThe basic system of computing 5, force method can be ABDA, static structureB, no extra bound geometric invariant systemC, geometric transient systemD, a statically indeterminate structureE, geometric constant variable systemThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, open a single hinge, equivalent to removing two constraints.Correct2, get rid of a fixed hinge bearing, equivalent to removing two constraints.CorrectThe basic unknown quantity 3, force method is redundant force.Correct4, the main factor in the typical force equations of constant greater than zero.Correct5, the internal force calculation of statically indeterminate structure should consider the equilibrium condition and deformation continuity condition.Correct"The test of structural mechanics" chapter 07 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, fixed at both ends with statically indeterminate beam angle, the known A terminal of the 0 A bar is internal force error is CA MB, A B, MA 2I = theta theta A = 4IC QAB, 3I A/l D, theta = - QBA = - 6I A/l.In 2, the typical displacement method equation R11 * Z1+r12 * Z2+R1P * Z1+r22 * R21 = 0, Z2+R2P = 0, the following formula is correct BA, R11 B, R12 = R21 = R22C, R12 * Z2 = R21 * Z1 D, R1P = R2P3, one end is fixed with one end hinged with the statically indeterminate beam angle, known fixed end A theta is A, shear force of bar terminal for AA, 3I A/l B, 6I - theta theta A/lC, 12I A/l D, 0.4, A end fixed B end hinged with statically indeterminate beam, fixed end known angle theta A, the following conclusion is wrong CA, MB = 0, MA = B 3I A.C, QBA = 0, QAB = D 3I 9 A/l5, the beam stiffness of single span single infinite non hinged rectangular frame, the following discussion about the basic unknown displacement method, the right is CA, three B, two basic unknown basic unknown quantityC, just two node angle is zero, only one unknown quantity D, just two node angle is not zero, only an unknown quantity.Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, if the additional constraint in the basic system ofdisplacement force (torque) is equal to zero, then AEA, the basic system is consistent with the original structural stress, deformation.B, the basic system and the original structure of the force is not consistent, inconsistent deformation.C, the basic system is consistent with the original structure of the force, but the deformation is not consistent.D, the basic system and the original structure of the force is not the same, but the same deformation.E, the basic system of the original structure and equivalent.2, single span symmetric structure in antisymmetric loads, section ADE on the symmetry axisA, displacement along a symmetry axis direction is zeroThe displacement of B, vertical symmetry axis direction is zeroC, the angular displacement is zeroD, the bending moment is zeroE, the axial force is zero3, two cross symmetric structure under antisymmetric loads, the following discussion is correct AEThe vertical displacement of rigid node A, the symmetric axis is zeroB, on the axis of symmetry of the rigid joint angular displacement is zeroC, in the column no momentD, no shear columnE, in the column without axial force4, using the displacement method to calculate the loads of statically indeterminate structure, the rod relative stiffness is calculated by ABThe real displacement A, node displacement is not structureB, the internal force is correctC, node displacement is the real displacement of structureD, the internal force is not correctE, node displacement and internal force are correct5, the following discussion about the basic system of displacement method is correct ACThe basic system of A, the displacement method is a group ofsingle span statically indeterminate beamThe basic system of B, the displacement method is static structureThe basic system of displacement method C, a structure is uniqueD, a method of structure displacement basic system has many choicesThe basic system of E, the displacement method is geometry variable systemThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as B, angle hinged = - 0 A (CCW).error2, the displacement method can calculate the statically indeterminate structure can also calculate the static structure.Correct3, hyperstatic structure calculation of total principle is the desire of statically indeterminate structure to take a basic system, and then make the basic system in terms of stress anddeformation of the original structure and the same.Correct4, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as MA, bending moment of the fixed end theta = 3I A.Correct5, one end is fixed with one end hinged with the statically indeterminate beam, fixed end angle theta A is known as the QAB, bar shear -3i theta = A/l.Correct"The test of structural mechanics" chapter 08 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, what kind of structure can not be calculated by the method of moment distribution? DThe structure of A, B, non continuous beam node linear displacementC, frame D, node displacement of structure2, member AB A end is a fixed end, no load span, known B end moment is m, then A = AB rod end bending momentC, m/23, the multi node structure, obtained the moment distribution method is CA B, the approximate solution and exact solutionC, D, analytic solution asymptotic solution4, A rod end is fixed and the B end of the sliding line stiffness I, rotational stiffness of C A terminalA, 4I, B, 3IC, I, D, 05, when the distal end is a fixed end, the transfer coefficient is equal to BA, 1 B, 0.5C, -1, D, 0Second questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, only what moment to the distal transfer? BDA, fixed end momentB, the distribution of bending momentFinally, the bending moment CThe proximal bending moment D, the rotation of the rod endThe distal end of the rotating bending moment E, the2, A rod end is fixed and the B end of the sliding line stiffness is I, the following formula is correctA, SBA = IB, SAB = 3IC, SAB = ID, SBA = 3IE, CAB = 13, the moment distribution method can be used to calculate the structure of what kind of? ABCA, continuous beamB, a non sway frameThe structure of C, no node displacementThe structure of D, no node displacementE, the beam stiffness is infinite structure4, continuous beam in a bearing of two adjacent span live load, to both sides of every span live load is the most unfavorable load on what the layoutA, the support section of negative momentB, the force of supportC, the positive moment spanD, the seat on the right side of the maximum shear forceE, the left side of the smallest section shear bearing5, the following discussion about the moment distribution method is correct?A, moment distribution method to obtain exact solutionsB, get the asymptotic solution of moment distribution methodC, from the first node unbalanced torque larger absolute value of node to startD, you want to change the unbalanced joint torque distributionAt the same time, E can relax the adjacent nodesThird questions to determine the problem (1 points for each question 5 questions, a total of 5 points)1, AB A said the moment distribution coefficient of node function unit couple, AB lever A is sharing the rod end moment.Correct2, the moment distribution method are derived based on the displacement method, so the structure can be calculated using displacement method can be calculated by the method of moment distribution.error3, in the moment distribution method, non adjacent nodes can also relax.CorrectScope 4, moment distribution method is continuous beam and non sway frame.CorrectThe end of 5, at a bar of the nodal moment distribution coefficient equals 1.Correct"The test of structural mechanics" chapter 09 OnlineThe first question, multiple-choice questions (1 points for each question 5 questions, a total of 5 points)1, when the harmonic loads for the single degree of freedom system with the particle damping, if the load frequency is far greater than the natural frequency of the system, and then the dynamic load balance is mainlyD, inertial force2, when the harmonic loads for the single degree of freedom system without damping on the particle, if the load frequency is far less than the natural frequency of the system, and then the dynamic load balance is mainly AA, the elastic restoring force3, a single degree of freedom vibration system, the damping ratio zeta, dynamic coefficient, resonance when the following results are correctA, f = 0.05, P = 104, a single degree of freedom system, the initial displacement of 0.685cm, the initial velocity is zero free vibration, vibration of a cycle after the maximum displacement is 0.50cm, the damping ratio for the systemA, 0.05In 5, low damping system can not be ignored the influence of damping on what?C, the free vibration amplitudeSecond questions, multiple choice questions (2 points for each question 5 questions, a total of 10 points)1, a simple point motor, in order to improve the vibration frequency of the beam, the following measures are correctA, shorten the spanB, increasing sectionC, the end of the beam to the fixed endD, reduced qualityE, the motor speed increases2, the free vibration of multi degree of freedom system of the main calculation ABCA, frequencyB, cycleC, modeD, dynamic force。