FEA by nitin gokhale

SpecialElementsandSpecia1Techniques Gaoor mint topoint contact: a. Recommendedwhen number of contact points are less b. Matching mesh (similarmesh) on two surfaces is required c. Specifying value of friction helps in convergence of solution. Observe displacement plot and animation for the partsin contact. Rigid body motion of contact part meansit is not constraint properly. Motion of flying part could be arrestedby constraining necessaryminimum dofs (suchthat it would not affect the results). 9.5 Mass Element Mass element is a zero dimensionor Point element. lt is applied a t C.G. of the component. m:Mass, Co-ordinatesystem and Moment of lnertia To connect the mass with structure, rigid or multi point constraint elements are used. Mass element is frequently used for dynarnic and NVH applications. It is a very good option for restricting total dofs of the model. u+ Morselement For structuralanaiysis of big assembliescommonpracticeis to representnoncriticalcomponents via concentratedmass at C.G. Consider following example, the bracket is mounted on chassis and it is supporting fuel tank. Aim ofanaiysis is body design and fuel tank is not important for the analysis.Weight of fuel tank and fuel could be taken in to account via a single rnass elementat C.G.

How to solve same problem in static analysis without defining mass.lnstead of defining mas and gravity, equivalent force could also be applied directly (say totalmass of fuel tank and fuel is 10 kg, Equivalent Force = mxa= 1019.81 = 98.1 N).Representationof components via mass (insteadof Force)is strongly recommended for dynamicanalysis.Natural frequency calculations wili be affectedif mass is not considered.

Speciol Elements andSpecin1Techniques 9.6 Spring and Damper Elements Spring element: Input data :Translational & rotational stiffness, Co-ordinate system. Applications: Any component could be replaced by a mass, spring and damping element if we know the respective values. 1) Shockabsorber spring 2) Mountings 3) Bearing stiffness 4) Special methods to simulate joints like bolted or welded Consider typical application of shock absorber. Vertical load coming from road would be absorbed by uniaxial spring, rotation of beam element connected to spring is kept free along z axis (using beam end release). uamper element:

- Practical Finite Eiement Anaiysis 1Damping is not applicable for static analysis, but it plays very important role in dynamicanalysis.Input data:Damping coefficient, Co-ordinate system.In the above figure shock absorber at front and rear wheels were modelled using appropriatestiffness (spring element) and damping values (damperelement).Driver's mass represented bymass elements. Mass elementconnected to chair frame via RBE2elements.9.7 Rigid and Constraint ElementsRigid element (node to node connection) has infinite stiffness and transfers al1 the forces andmoments or in other words dofs from one node to other as i t is. Rbar, Rigid, Rigid link areexamplesof rigid elements.RBEZ, RBE3 are multi point constraints (single node connected to many nodes) and should bepreferred over rigid elements.Rigid and RBE2elements add stiffnessto the original structure, while RBE3 does not.

Speciai Elernentsand SpeciaiTechniques RBEZ element distributes the force and moment equally among al1 the connected nodes irrespective of positionof force or moment application. Consider 100 N force is applied at RBEZ element as shown below. It will get equally distributed among the 2 nodes i.e. 50 N eachRBE3 element is a constraintequation to distribute force and moment as per the distance (leastsquare weighted function). In the above example if RBE3 is used instead of RBE2 then it willdistributeit as 75 Nat pt. Band 25 Nat Pt. A.25N '7\" q --- 75 .- RBE B 0 A -9.8 Simple Linear Static AnalysisTechniquesto SimulateContactHow to simulate contact between surfaces, our Company does not have a non linearanalysis solver?Defining contact or gap elements could give better rewlts. But it makes the analysis non-linearand time consuming. Many a times organizations do not have non-linear softwares. For suchsituation a simple linear analysis technique could be to connect two parts via rod element ofnegligible stiffness (rod dia. in the range of 0.1 to 1 mm).Reddotsindicatespots,modeledviu bmm demenaidio. of beam=spot db.)

- Practical FiniteElement AnalvsisStiffener is spot welded to plate. Spot modelled via beam elements. Except spot beams nocontact is definedbetween plate and stiffener. Radelementused for mn (otollthe contactnoder c..... thonspot lacoflons)Could Leakage be found out with beam, rod element contact approach?Yes, it is possible by checking elemental forces in rod / beam elements. Consider followingassembly, its subjected internal pressure. On both the flanges exactly same mesh pattern hasbeen maintained. Rod elements (only tension compression is supported) were usedto connectthe fianges except in bolt areas. Bolts were modelled using beam elements.Apply bolt pre torque to bolt beam elements (as discussed in next chapter). Due to bolt torqueal1the rod elements will be subjected to compression. When internal pressure is applied, twohalves will try to open out and induce tensile stress in the rod. After combining both the loadcases, review results for only rads elements (bydisplaylng only rods on screen). Rods in tensionrepresent leakage area while in compression shows no leakage or good sealing.Our organization does not have non linear solver, i s it possible t o find out equivalentnonlinear stressfrom linear static analysis results ?Linear static analysis shows unrealistic andvery high magnitudeof stress after crossingthe yield

Special Elernents and Speclal Techniques point (well above ultimate stress). But in actual practice component survive for considerable duration. When Non linear anaiysis of the same part is carried out it confirms the practical observation and shows stress higher than yield but lower than ultimate. Is there any way t o find (approximate)non-linear equivalent stresses from available linear static results. Yes, the special trick is to use Neuber's Ruleas shown in the graph. KtZ=Ke*K, K, =Theoretical (based on geometry shape, linear elastic behaviour) stress concentration. Ke = True strain concentration factor = Notch (plastic) strain / Nominal strain Ko = True stress concentration factor = Notch (plastic) stress / Nominal stress

wela, troir, trearing ana snrini<tit bimulation I - Welded Joint Spot Welding Arc Welding(point connection) (lineconnection)Most ofthe reallifefailuresare reported a t either stress concentrationpoints orat bolted, weldedjoints. Failure of bolted welded joints is due to fatigue (dynamic loading). In Our house or office,there are many components which are bolted or welded but since the nature of loadingis static,very rarely we noticefailureof thesecomponents. For examplepleaseconsiderfollowingbracketat Our office.Ithas been there for last 12 yearsand probablyit would remain as it is for next 100 years, becausethe nature of loading is static fwall remains stationaryas well as applied load). Now if the samebracket is mounted on gear box or clutch housing or chassis of a moving vehicle, it will fail inshort time at either bolted or welded joint.Some importantpoints to be remembered while carrying out the weld joint analysis9 Static strength of weld 2 Static strengthof parent materlal9 Fatigue strength of weld <<<Fatigue strengthof parent matetial

Weid, Bolt, BeoringondShrinkFit Simulation Many a times static or dynamic analysis predicts a location of failure not matching with actual field or test failure. In particular this problem is quite common when failure is at weld locations. For example consider wheel disc and rim analysis, static and dynamic analysis will always show failure at either bolt locationsor vent holes [elliptical holes)but neverat the weld a t thejunction of disc and rim. But in real life failures are usually observed a t welding location and we keep on thinking what is wrong with analysis. Whether restraints are wrong or loading or material properties. But when fatigue analysis using weld module is carried out based on same static or dynamic results as input data, it clearly shows failure a t welding. So now the questionis, if static strengthof welding is greater than or equal to the parent material and static analysis can not predict the failure location then why there are so many special techniques to model bolted and welded joints in static analysis. Special techniques are required to 1) Simulateappropriate stiffnessof joint 2)Transfer forces and moments at appropriate locations There are no universal standard guidelines for modeling bolted or welded joints. Different organizations use different methods for the same. Non availability of standards gives CAE engineer freedom to simulate the things as per his perception &engineering skill and it makes field of CAE even more colourful, live and democratic. 10.2 How to Model Spot Weld Spot welding is possible for only sheet metalparts (thickness 2 mm) 1) Classical approach Neglect the welding and assume the two parts as single entity (single part with thickness = combined thickness of two). But this approach results in high stiffness and transfer of forces and moment a t muchmore contact area than the actual one.This leadto safer results i.e. less stress and displacement than the actual one and is not recommended. 2) Beam element Spot weld is modelledusing beam elementat the locationsas specified in drawing. Beam element diameter= Actual spot diameter Beam element should be exactly perpendicular to surfaces (matching mesh pattern on two surfaces).144

3) Rigid elementsEvery thing same as beam except rigid element i s used instead of beam.4) Coupled degree of freedom 1RBE2 elementsSome organizations prefer Coupled dof or RBEZ / RBE3 elements with specific dofs transferredto dependent nodes.5) Spring elementsSpring elements with specific translational and rotational stiffness are used t o represent spotweld.6) ACM (AreaQntact Methodusing brick elements)This is also popular method among CAE engineers. Software automatically generates brickelements with RBE2and RBE3connections at the connector locations defined by user. User hasto specify detaiis ilke spot diameter, thickness ofjoining surfaces and location of spot weld.Comparison of difierent techniques:Beam, rigid element or ACM are commonly accepted methods across the globe and arerecommended.

Weld, Boit, Bearingand ShrinkFit Simulation 10.3 How to Model Arc Weld 1) Shellelement: Weld is represented by shell element (weld element thickness = weld (throat) thickness). Gray colour elementreprerenf arc welding felement thlckners= weidthroat thicknersI 2) Rigidelements: Arcweldingrepresented by rigid elements. connected in the area of welding.

10.4 Practical Considerationsfor WeldedJoints1) Spots are stronger in shear (than the normal load). Most of spot weld failures are caused bytension/compressionor normal stress. Just by changingspot weld orientation (changingnatureof stress to shear) could also avoid the failures.Normalstrerron spot Shearstresronspot notrecommended recornmended2) For very thin sheet metal parts. (below 0.8 mm thickness) spot welding should be preferredover arc welding. Arc welding burns the parent metal itself and causes weakness in the vicinityof the joint.What should be the material of weld elements?Weld properties are usually not available to CAE engineer. Real life experience is even gettingbasic parent materialpropertiesfromdesignengineeritself isa difficult task,and then expectingit forwelding is probably too much!-How t o carry out analysis without knowing weld properties1)Static analysis: For linear static analysis any way materialdo not cornesin the picture (stressis independent of material) and material sameas parent could be assigned to weld.2) Dynamic analysir:Natural frequency calculations(&modal superposition solvers)are basedon density. Generally there is not much difference in the densities of weld and parent materialas well as the welding area and volume of weld material is negligible in comparisonto parent.Hence even for dynamicanalysis samepropertiesas parent materialcould be assumed3)Fatigueanalysis:Commercialsoftwares providesspecial procedureand guidelines to take into account different weld joints (with a materiallibrary) and is recommended.For simulating welding, some organizationscarry out transient thermalanalysis (special materialproperties assigned to weld) first to find out residual stresses due to high temperature and

Weld, Boit, Beafing andShrinkFit Simulation subsequent cooling and then transfer it for structural analysis. This is a very time consurning process and requireslot of computationalefforts. lnsteadfatigueanalysis after performing static or dynamic is muchmore simple,faster and reliable too. Welding static analysislimitations: 1) For staticanalysisprocesslikegas welding,electricarcwelding or thermal welding or pressure welding etc. cannot be takenin to account. 2) For arc welding further details IikeV-joint,U joint, doubleU ordoublev or H -joint etc. cannot be considered with 2-d shell element. Fatigue analysis softwaresconsider abovefactors and is recommended. 10.5 Bolted Joint For modeling bolt joint: Minimum two layers around bolt hole (1.5 to 2.0 times core diameter) representing washer 1bolt head are recommended. 12 to 16 elementson circular hole. Usually aim i s to analyse parent structure and not bolts. Special techniques for bolt simulation are based on representation of bolts by equivalent stiffness modeling via 1-d or RBE2, RBE3 elements. Commonly used methods are as follows 1) Beam elements: Centre of bolt is connected to inner and outer layers via beam elements of diameter equal toW and 2 times'd'(d = core diameter) as follows.

Outerloyers &am dio. =d tcenternodem washeredgeJ Ba/tshankporffonmodeledbyaslnglebeom dio. = d2) Rigid elements:Method 1:Every thing same as beam method except RBE2or RBE3(one to multiple nodes)forwasher area connections (both inner as well outer layers) while bolt shank portion representedvia beam elements (beamdia. equal to core dia.).Method 2:RBE2 and RBE3 elements are created as shown in the figure, shank modelled usingbeam element.Centrr node -1ndependenr(do1 123456) Center node. dependent Id01 1234561 O~terloyernodes- Dependent (do131 -lnner loyer noder Independent (do1 123)In some organizationsit is a standardpractice to neglect stress at outer layer elements (washerarea and one more layer surmunding beamfrigtd connection) due to high stresses observedat beamhigid and shell/solid connections. In this case two more circular layers around thewasher iayer are recommended. Resultsare viewedby neglectingelements in washer area andconnected elements.How t o model bolt threadsShankportion of beam (blueelement) is split in to multiplebeam elements, (very small dia. =0.1to 1 mm) or rigid elements (RBE2spider, greenelements) connected to shelllayers a t an angle =threadhelix angle.

Md,Bdt, BearlngmdShrin4FitSlmoIatEonHow to apply PreloadI bolttorque1) Special commands in commercial software: Most of the commercial softwares havespecialcommands for appiying bolt torque either by directly specifying the torque or otherwiseequivalent axial force produced due to torque. To calculate equivalent compressive axial force,standard formulasavailable in design data books could be used. A simple one (when al1detailsare not available) is as suggested in the book Mechanical Engineering Design by Joseph E.Shigley, McGraw-HillBookCompany, FirstMetric Edition, page number 309 - T=K*F*D (K= 0.15 to 0.2, unitsForce= kN, dia. =mm,T = N.m)2)TemperaturemethodTo achieve the axial compression. beam element (representing shank) is subjected to thermalloading (negative temperature). Negative temperature causes compression of the beam.Temperaturerequiredto produceequivalent compressive force is calculated as followsHookslaw:o=E~Thermal strain: =Al 1L = a At (a = Coeff. of linear thermal expansion)€ = O I E =F/(AE)= aAtAt=F/(AEa)After solution, beam element forces should be checked (can be displayed in post processingvia element force option). If it is not matching with the required axial compressive force then

temperature should be adjusted proportionately. Usually by carrying out 2 iterations desiredresultsare achieved.3) Direct force applicationBolt clamping force is directly applied on washer area as shown below4) Aim is t o design boltsFEA is not recommended for designing standard components like Nuts, Bolts, Gears etc.Thereare dedicated softwaresand standard thumb rulesavailablefor the same (whichalso takesin toaccount manufacturability).In some organizationsbolts are simulatedvia 3-d elements ((asper actual geometry i.e. threadsetc.) and then analysis is carried out by defining contact between threads of mating parts andbolt head and resting surface etc.10.6 Bearing simulation1) Beam 1Rod elements:This method is notas accurate as gap or contact simulation. But gapor contact simulation meansnonlinear analysis and more solution time.When the criticalarea isnot bearing, rodlbeammethodworks well.Shaft is modelled as beam (dia. = shaft dia.) and connected via rod elements (ball bearing) orbeam elements (taper roller bearing) on the circumference (3minimum layers, width = bearingcontact width). Diameterof connection rod/beamelements= 0.1 to 1 mm.

Weld, Bolt, Bearing andShrinkFitSimulotion Shaft lrepresented by begnelemenfa) J iWith the aboveapproachin the first run contact between shaft and casing is 360°.After solution,beam element forcesshouldbe checkedand rods / beamsin tensionshouldbe deleted. Usuallyin 2 iterations we get appropriate contact area.For taper roller bearings beam elements (spider) are connected to shaft at an angle equal totaper angle. Beam Elements Shaft (represented by beam elements)2) Gap elements :Method is same as above except connection beamdrods replaced by gapelements. If shaft and bearing is modelledvia 3-d solid elements then node to nodematchingpatternbetweenthe two contact surfacesi s recommended.

3) Contact simulation:Contact is definedbetween shaft outer and bearing surfaces.4) Direct force application :Shaft is not modeliedand equivalent force in 90\"contact patch isapplied as follows:5) Force application via equation:Some commercial softwares support force application viaequations, Paraboiicequation on 90\" patch with appropriateforce magnitude could be appliedas shown.

Weld, Bolt, BearingondshrinkFitSimulation Comparison of different methods: Last two methods are simple and accurate but it could be used only when contact area (force application area) is known. For practical applications shaft is supported by 2 or more bearings (likegearbox) andsubjectedto multiple loads a t various angleS.This meansthe contactbetween shaft and bearing would be a t arbitrary position, rodfbeam or gap / contact approach handle this situation satisfactorily (finds contact areas on i t s own). Gap-contact would lead to better results and is recommended (type of analysis - non-linear).If only linear solver is availabie then rodlbeamapproach i s recommended.10.7 Shrink Fit SimulationProblem definition: Outer cylinder (OD= 240 mm, ID =180 mm) fitted over inner cylinder (OD:180 mm, ID- 120 mm), shrinkagepressureat the commonjunctionafter compounding= 7.848N /mm2. Find initial stresses developedin both the cylinders.Above problem could be solved using lame's equations, analytical results are displayed ingraphicalformat below inner cylinder--R ~ d l o l d i m b ~ t Ci ~ommpoondCyIInderOuter Cylinder duetorhrinkfit Hqrtrerrdlrrribuhm In Cmpovnd Cylrnderduetorhnnkfit 2805 Nlmm' 20.1 1 Nlmm' Y O N/mml--2825 N/mm> 7848 Nlmm* lnnercylinder OuterKylinder 6rDU,-= 7.90E-03 +6r 6r = 0.0211 mmDisplacement: 6rhm,= 1.32E-02 / Greenelements: lnnercylindei Y Yellowelementr:Outercylinder

Method 1:Contact analysisMethod 2:Gap elementGap seperation Beom elemenfr crerrtedforestrict as -0.021 mm the riqid bodvmuder ofinner& Gap elsrnant ovtercylinder. Detoilsobout DOF: Ilampedrestrainson bwmouter nodes.RototionaiDOf fixed behveen beom&solidelement mnnecfian.

Method 3:Beam element (Temperature method) Boundory Conditions (&am TemperatureAnalysis)

Method 4: Constrainimethod(individualpart analysis)a. Boundary Conditions on lnner Cylinder for Displacementconstraint Displacementrenraln on outer / diameter= -O.Ni79 m mb. BoundaryConditions on Outer Cylinder for Displacement Constraint

Weld, Boit, Bearing andshrink Fit Simulation lnner Cylinder Outer CylinderAnalysis Method

iwareriai rropenies ana tsounaary conaitions11.1 E,G& U'E'Modulusof elasticityis slopeofnormalstress-straincurvein linear elasticdomain.ltis definedas normal stress / normal strain. Units: Nlmm2'G' Modulus of rigidityi s siopeof shearstressandstraincurvein linearelasticdomain. itisdefinedas shear stresslshear strain. Units: N/mm2'u' Poisson's ratio is defined as ratio of lateral strain (A w / W) to longitudinal strain (A I/ L). Itisunit lessentity.Physical interpretation of u : Consider a cube of l x l x l mm dimension. Poisson's ratio 0.30meansif the cube is elongated by 1 mm, lateral directioncontraction would be 0.3 mm.For metals poisson's ratio lies in the range of 0.25 to 0.35.Max. possible value of poisson's ratiois 0.5(for rubber).E, G and u are inter-related by equationOnly two independent materialconstantsarerequiredfor linear static analysis (Le.any two out ofE, u and G).Additional data requiredfor gravity, centrifuga1load anddynamicanalysis is materialdensity and for temperature induced stresses 'a: coefficient of linear thermal expansion.GeneralizedHooke's law & total 36 constantsi n the equationHooke's law familiar to us is 0 = E E. This equation holds good for isotropic material that too inthe linear elastic domain.General equation of Hooks law for an anisotropic materialis

11.2 MaterialClassification Difierentrtreoglh alongfhetheoxir Orthotropic

&f w & W p ~ & ~ t ~ u r e s + # aas ~ ~ c t u ~ r n o t e ~ h I c ~ o ~ r e ~ m ~ m m e n t i e d .1 N = 1 k g s l m/sec2 IF=mXa) IN=lhg*lOOOmm/sec~1 N = IOOOkg 1 mm/sec2nce when forceisspec =1 1 N= 1 tonne * 1 mm/se11.4 Boundary ConditionsBoundary condition isapplication of force and constraint.Different ways to apply forceand moment -1. Concentratedload (at a point or single node)

MaternilProperfles ondBoundary Conditionr 2. Force on line or edge3. Traction'Traction'is force acting on an area in any direction other than normal. Force acting normal toarea is knownas'pressure:4. Dlstributed load (Force varying as equation)5. Pressureand VacuumPmure: Internolpressure>atmosphff~cpressure Vacuum: Internolpressure<otmosphericprersure

6. Hydrostaliic pressureCivil engineeringapplications: Dam design. Mechanicalengineering applicationrVessels/tankscontaining liquid.Hydrostaticpressurei s zero a t the top surface of liquid and ismax. (= pf g * h)a t bottom surface.It varies linearly as shown in thefollowing figure/.Bending momentsConvention for representingforce = singlearrow (dpoin)ted tow,ards force direction.Moment - double arrow,1-( direction of moment is decided by right hand thumb rule

MaterialPropertiesand BoundaryConditions 8. Torque: What is torque?IsTorqueand Bending moment different? Torque is Bendingmoment applied parallelto axis of Shaft (MA. Torque or M, causes shear stresses and angular deformation while effect of other two moments (My,MJ i s normal stress and longitudinal deformation. How t o decidedirection of torque (clockwiseor anticlockwise) It is based on right hand thumb rule. Point thumb of your right hand (Right and right only even for left-handersor left party supporters too!) towards arrow direction. Direction of fingers indicate directionof torque. Howt o apply torque for solid elements (brickltetra): Solid elements have only 3 translationaldofs and no rotational dofs. Hence torque or moments can not be applied directly on solid element nodes. Commonly used techniques for torque applications are as follows:

a. Geometrybased boundary condition (for geometry basedmeshing):Some commercial software5 provlde facility for geometry based (edge) load. Shear force asshown could be applied.Other alternative to apply torque on the edge nodes (when meshing is not automatic) iscalculate x and y components of load (draw a line from node on edge to the centre of circle,forceperpendicular to this llne could be resolvedalong x and y axis as F cos9& F sine. On al1theindividual nodeson circumference respective x and y components (asper angle of node)couldbe applied.-b. Rigldelement connectienCentre no& is connectedto outer edge noder via rigidelement(one to multitl)e, RB~Z).TM~U@isappliedat centre node.

MoteriolPropertiesand BoundaryConditions c. Beam element connection: Same as rigid element except rigid replacedby beam element (diameter 0.1 to 1 mm). Moment is applied at the centre node. d. Shellelement coating: On the brickhetra element outer face additional quadltria (2-d) element coating is created. Thickness of these shell element should be negligible (so that it would not affect the results). Moment could now be applied on al1 the face nodes (momentper node= total moment / no. of nodeson the face). Could you notice rigid, beam or shell element techniques are special methods to impose 3 rotationaldofs on solid element nodes. 9.Temperature loading Suppose metallic scale is lying on ground freely as shown in the figure. If temperature of the room i s increased to 50 degrees, would there be any stress in the scale due to temperature?

There will be no stress in the scale. Itwill just expand (thermalstrain) due to higher temperature.Stress is caused only when there is hindrance or resistance to deformation. Consider anothercase this time one end of metallic object is fixed in rigid wall (non conductive material).Now iftemperature is increased then it will producethermal stress (at fixed end) as shown below.For thermal stress calculations input data is temperature value on nodes, ambient temperature,thermal conductivity & coefficient of linear thermal expansion..10. Gravity loading:Specify direction of gravity and materialdensity11. Centrifuga1loadUser has to specify angular velocity, axis of rotationand materialdensity as input data.12.'g'values (Thumb rules)for full vehicleanalysis:P Verticalacceleration(Impact due to wheel passingover speed braker or pot holes): 39 Lateralacceleration (Cornering force, acts when vehicle takes a turn on curvatures): 0.5to 1gP Axial acceleration (Brakingor suddenacceteration): 0.5 to 1 g

Moierioi Propertiei oiid Boiindory Coriditio~ii 12. One wheel i n ditch FE model should include al1 the components (non critical components could be represented by concentrated mass).Mass of vehicle and FE model mass as well as actual wheel vehicle reactions and femodel wheel reactions should match. While applying constraint, vertical dof of the wheel which is in considered in ditch should be released. Appropriate constraints should be applied on other wheels so as to avoid rigid body motion. Specify gravity direction as downward and magnitude = 3*9810 mm/sec2 Another simple but approximate approach (since most of the time either we do not have al1 the CAD data for entire vehicle or sufficient time for detailed FE niodeling) is to apply 3 times reaction force on the wheel which is in ditch. Le. suppose wheel reaction (as per test data) is 1000 N, apply 3 \"1000 i.e. 3000 N force in vertically upword direction and constraints other three wheels just sufficient to avoid rigid body motion. This approach works well for relative design (for comparison of two designs) e.g. to find quick solution of failure by comparlng results of original and modified design (say10 or 20 % stress reduction achieved at the failure location by increasing thickness or adding the rib etc.). 13.Two wheels i n ditch: Same as discussed aboveexcept now instead of one wheel two wheels are in ditch. One wheel in ditch causes twisting while two wheels in ditch produces bending load. 14. Braking : Linear acceleration (or gravity) along axial direction (opposite to vehicle motion) = 0.5 to 1 g.168

1 Practical Finite Element Analysir15. Cornering:Linear accelerationalong lateral direction = 0.5 to 1 g.11.5 How to Apply ConstraintsA beginner find it difficultto apply boundaryconditions and in particular constraints. Every onewho starts career in CAE faces following two basic question:i) For single component analysis whether forces and constraints should be applied on the individualcomponent (asper free body diagram) or surroundingcomponents should also be considered.ii) At what location & how many dof should be constrained.1) Clutchhousing analysis

MoterioiProperties ondBoundory Conditions Aim is to analyse (only) clutchhousing. Clutchhousing is connected to engine and transmission case via bolts.Thereare 2 possibilities for analysis: Awwroach 1) Only clutch housing consideredfor analysis, apply forces and moments as per free body diagram and constraintbolt holes on both the side faces (al1dof =fix). Aw~roach2 ) Model at least some portion of engine and transmission case at the interface (otherwise model both these components completely with coarse mesh by neglecting small features), represent other components like front axle, rear axle via beam elements tapproximate ch). Apply constraintat the wheels (not al1the dof fix but only minimum dofs required to avoid rigid motion or otherwise inertia relief / kinematic dof approach). Please note clutch housing being the critical area shouldbe meshedfine. Second option is recommended as stiffness representation as well as constraints are more realistic. First approach i.e. fixing both sides of clutch housing results in over constraining and leadto safer results (lesstress and displacement).Also it is not possibleto consider specialload cases like one wheel in ditch, two wheels in ditch etc. 2) Bracket analysis Problem definition:Bracket fix in rigid wall subjectedtovertically downward load (180kg).

.:1BoltholesedgefixedIf;ibove problem is given to CAE engineers working in Idifferent companies then you will finddifferentCAE engineers applying constraintsdifferentlya1s follows. i. Fix the bolt hoieedge directly. ii. Model the bolt via rigid 1beam elements and clamp bolt end. iii. Model the bolt, clamp bolt end, bracket bottom edge translations perpendicular to surfacefixed. Stress Displacement mm NlmmzBolt- modeledvia beam elenentrApplying direct constraint on hole edge causes very high stress. Second method showsdeformation of bracket bottom edge, which is not realistic. Method 3 is recommended. Pleaseobserve difference in magnitude of stress & displacement.

MoteriolPropertiesand Boundary Conditions In some organizationsit is a standard practiceto neglect stress a t washer layer elements(washer area and one more layer surrounding beamlrigid connection) due to high stresses observed a t bearnlrigidand shelllsolidconnections. 3. Pressure vessel lying freely on the ground and plate subjected t o tensile load on both the sides Some times situation demands for unconstrained structure analysis like a pressure vessel lying freely on the ground ( just placed, no constraintlfixing) or plate subjected to tensile load on opposite edges without any constraint. Static analysis problem can not be solved for unconstrained structures. Itshould be fix atleast atone nodeor alternatively at few nodes so as to restrict rigid body motion. if the problem of pressure vessel (subjected to interna1pressure) or plate with tensile load is solved without specifying any constraintsthen either software will comeout of solution giving singularity error message or otherwise report very high stress at unrealistic location if auto singularity option is switchedon. To solve unconstraint structure problemsthere are 2 ways 1) Approximateapproach:Createspring/beamelements(negligiblestiffnessvalue) on entirecircumference(outeredges or surface nodes)andapplyconstraint a t the free end of spring or beam. (as discussedin Chapter 10, shrink fit simulation) 2) Recommended:lnertia relief method or defining kinematic dofs in the model

11.6 Symmetry Symmetry 1 l Plane C symmetry Anti symmrtryCondition for using any type of symmetrySymmetricconditions could be used only when both the following conditions are fulfilled 1) Geometry is symmetric + 21 Roundary conditions (forces and constraints) are symmetrk.Advantages: Half, quarter or a portion of model could be used for analysis, resulting in lesserdofs and computational costLimitations: Symmetric boundary conditions should not be used for dynamic analysis (naturalfrequency and modal superposition solver). Symmetric model (half or portion of part) wouldmiss some of the modes Le.anti nodesor out of phase modes as shown below

MaterialProperties ond BoundaryConditions Natural Frequencycomparisonfor Full and Halfsymmetric model Question: We need to do a simulation of a casting component to decide better option out of two casting materials. What would be the differentiating parameters since the modulus of elasticity and poisson's ratio for both the materials are same. Secondly we do not have stress strain data for both the materials. 1)If natureofloading is static (mostofthe timestaticand rarelydynamic) andthecompany has only linear static solver: Different grades of cast iron have different ultimate strength (or proportionalyield strength).Linear staticstressis independentof materialandboth the materials would report same stress. Decisioncould be made based on ultimate strength (and endurance strength).Say mm.stressas per FEA = 300 Nlmm'. Material1, ultimatestress= 350 and Material 2 = 500 Nlmm2. In this situation second material is clearly the choice. Consider another case suppose max. stress reported by FEA = 90, via basic thumb rulesone can calculateapproximate value of endurance strength like for grey cast iron endurance strength= 0.3 ultimate strength, i.e. for material1, endurance strength = 105 N/mm2and should be preferred over the second one. 2) If nature of loading is dynamic and subjected to sever load:Fatigue analysis i s strongly recommended.Commercialfatigueanalysissoftwares usuallyprovidemateriallibrary. Specifying appropriate materialgrade, surfacefinish etc. andcalculationsfor lifelendurancefactorof safety would leadto realistic and optimum selection out of the two options.

Linear Static Analysis72.1 , ~ e l i n ~ i mIt is the simplest andmost commonly used type of analysis. Software follows thh path fur linearstatic calculatian1 /stress Acnial stress-stiaincurve StrainLinear:Linear means straight line. a =E E is equation of straight line (y = m x) passing through origin.\"E\" Elastic Modulus is slope of the curve & is a constant. In real life after crossing yield pointmaterial follows non liner curve but software follows same straight line. Component brake intotwo separatepiecesafter crossing ultimate stress but software based analysis nevershow failurein this fashion. Itshows single unbroken part only with red colour zone at the location offailure.Analyst has to conclude whether the component is safe or fail by comparing the max. stressvalue with yield or ultimate stress.Static:There are two conditions for static analysis1) No variationof force with respect to time (deadweight)2. Equilibrium condition- 2Force = 0.1 Moments= O

.......12.2 Whlle Startingany FiniteElement Analysis Project This 1s the f m e of o 'Discover-MSI\" motorcycle fmrn Bolal Auto. HyperWorks1s usedas the technology enable for volidating the frarne design. The FE Mode1has been creotedusingHyprMesh.Thestructrualanolysishasbeen dane using Opt15truct. The contour plot shows the post processedresultsfor the fiorne osmgHyperView. SWcturalAnolysisoffrome (Irnoge Sourn?:AltoirC~icndar2W5, Courtesy:Bo]a]Aiito 1td.JIf you are working for a service providing company then following details are not applicable.Overseas client specify al1the details (step by step instructions for meshing, materialproperties,application offorceandconstraints)in aneatlypreparedpower point presentation. CAE engineerhas to follow the instruction and submit report in prescribed format. But if you are working fora manufacturing company, not having standard instructions for analysis then following stepscould be helpful: Study the problem, components, functioning details etc.: Spend sufficient tirne in studying the problem. If the option is available please visit shop floor and observe manufacturing and assembly process. Clear understanding of the functionality is essential for applying appropriateboundaryconditions. Consult and gather further information from service, test, warranty yard, design, experiencedCAE engineer and customers of the product:This is very importantpoint, you might be holdingpost graduateor doctoratedegree from a renowned university of lndia or abroad with excellent academic record but never make a mistake ofnot consultingpeople having years of experiencerelatedto the product, component. Many a times best design solution advice for carrying out further iterations cornes from shop floor people or illiterate workers. Defineobjective and type ofanalysis Availablet h e : a. Appropriate time allotted for the project- Results in good quality (accuracy) of the results. -b. Time is very short Results ln compromisewith result quality. Based on prevlous experience of analysing similar component and relative design concept CAE engineer could still contribute in a major way. To cut short analysis time preference should be given ro i. Auto/batch meshing over manualor semiautomatic techniques ii. Tetra elements over hex/ brick meshing iii. Relax mesh quality criteria



Linear Static Analysis and model shouldbe checked with same interestand patiencetill the very end as like checking Our salary slip, bank pass bookorjourney reservationticket1Even a smallmistake (likedofor properties of connectionelementsnot definedproperly) could leadto substantial differencein analysis result.12.3 How to Check Mesh Model Submitted by a Vendor or Colleague1)Mesh Quality checks1 4 eiements - free Id's, rlgid loops, dependency2 4 elements -warpage < 154aspect<S,skew <459lnterI0rangles(45~<quad <1354200<tria .<12001, Jacobian>0.6, tria <5%3 4 elements -Teira mesh - tet coilapse>O.l,jacoblan >OS,stretch >0.2 - -- - He>:-warpage < 304aspect <7, skew < 459 Interiorface angles (450 <quad 4 3 5 4 200 <tria < 12001,jacobian >0.6, penta < 5 %2) Duplicateelements 3) Dupikate nodeslequivalence4) Deietefreel temp nodesali 1 5) Element normais6) Free edges 1Free faces 7)Min. element length 1t h e step- for crashanaiysis18) Delete empty 1 unused - Cornps, props, assems, 9) Flow pattern - Representing appraprbatepattern ofloadcols, mats, systcolsetc. (Tool- count ) stress waves10)Meshpenetration, deviation from geometry, kinkr 11) Asslgn appropriatedofs for RBE2, RBE3elements[view geometry & mesh both ln rolld view - visoptions)1I I )Renumber- Nodes,elements,mats, propsetc. (tools 13)Elementsummary-Checkelementtype&family,no1-renumber-ail) plot elernentsetc.114) Assign appropriatepropertycards,materiais, beam 15) Compare mass - Actual mass V r Finite Element1ds,thicknessetc. (Port-summary-components) Modelmass161Free- Free run for assemblyof components-first 6 17) Llnear static analysis with dummy boundarymodes ngid, 7'onwards positive deformablemodes. conditions for single componentmesh job12.4 Design Modifications Based on Linear Static Analysis: A Case Study,Bracket Analysis :To analyse given steel bracket for 180 kg load and suggest modificationsif stress is morethan acceptablelimit (200N/mm2). Bracket thickness= 3 mm

* Though theoretically it is possible to mount the bracket on single bolt but practically it willrotate when torque or shear force is applied (side ways) and hence minimum two bolts arealways preferred.qMaterial DroDerties:E = 2.1 E05 N/mm2,v = 0.30. p = 7.89 E-09 tonne /mm: = 250 Nlmm2Factor of safety= 1.25,Acceptablelimit of stress = 25011.25 = 200 N/mm2Constraints:i. Bolt ends clamped (al1dofs fixed)ii. Nodeson bottom edge, only translationx = fix, other dofs are freeResult interpretation and design modifications:Stress plot clearly indicates only upper bolt i s taking al1the load and bottom bolt is just notcoming in to picture. A betteroption could be to arrangethe bolts horizontally.

Linear Static Analysis lteration 1: Horizontal arrangementof bolts Just rearrangement of bolt hole (without changing mass and cost of bracket) resulted in 22 % stress reduction Still stressmagnitude is higher than acceptablelimit (200 Nlmm2)andfurtheriterationscould be carried out by increasing bracket thickness. lteration 2:Increase thickness (horizontal holedesign) At 6 mm thickness stress c 260 N/mm2but mass of the bracket has doubled.Further iterations couldbe carried out toreduce mas ofthe bracket by adding stiffeners.

lteration 3: Singlecentral stiffener (bracket as well stiffener thickness = 3 mm)lteration 4:Two side stiffeners (thlckness =3 mm)Single central stiffener is more effective from stress as well cost (mas) point of view. There isfurther scope t o reduce the cost by reducing the thickness (such that stress magnitude will beaprox. 200 N/mm2)lteration 5: Reduction of thickness

LineorStatic Anolysk Detailed table:Last 4 iterations satisfy required condition i.e. stress c: 200 N/rnm2 but the one with minimumrnassand hence cost would be the autornaticchoice (i.e. single central stiffener with stiffenerasweil as bracketthickness= 2.3 mm).12.5 Unear Static Solvers Linear Static Solvers I Bared on mathematical solutionafthe equation. Baredan assumingthe variable value tilt remainderof equation= O[FI= M[Dl [FI [KI[Dl= 0 Solution guaranteed Solution not guaranteeded Shell, beam. rolid,combinatlun of 1Preferredfor tetrahedronelementr difierent etementr J. diskspace, Time Considerequation Sx-6=ODirect solverwill mlvei t as Conriderrame equatlan i.e Sx-6=0 5x=6 x=1.2 iteratiie5olverwill 10Ive it ar ie.solvhg the actualequations. Ila%umex=O rohwarer remainder'R'=-6 (S*06 ) 2)assumex= 1- Node numbermgIr important. R=-l- [KIcontainslot of zero terms. 3) assume x =2- Word rpatre meanr lesr denrlty le. in the R=+4. rtiffmir matrw there are ierr poritivetermr and At exact 50iut10n R = O le. exact ansver lier between1and 2.Continue more zemr the procesr till acceptable llmlt of Zeror are rhlfted aimard (awayffom diagonal) convergenceis achleved. and matrix along degonal band e conridered. This rearrangementreducer order of matrtxand

12.6 Solution Restait Method* lLoad Case3 IF1= [KIID1 Load case 1Solution restart is a special method to use data generated from previous run. During solutionprocess, maximum time is consumedin stiffness matrixformation.Stiffnessmatrix data is storedin a temporary file which is deleted automatically after completion of the solution. Solutionrestart is a very effectivemethodto savecomputationaltime&cost. Itcouldbeusedforfollowingapplications:1) Additionalloadcasesfsameconstraintandrneshmodell: Multipleruns are required for multiple load cases and for each run software will generate and delete same stiffness matrix data again & again.This could be avoided by storing the data on hard disc duringfirst run and then usingit for subsequentsolutions.2) Additionalout~utreauirementS: uppose there is only one loadcase to be analysed. Aftersubmissionof results,client demandforadditionaloutput (likestrainorstrain energy etc.). With conventionalapproach, analysis will have to be perform again consuming same amount of time as first run. Solution restart, if used (provided stiffness matrix data stored during first run) would save lot of time as weli as computational cost.After performing Linear Qatic Analysis, designengineer decidest o change the material, i sit now necessary to run the analysis againwith modified material? I Alummium Bras rameforceandafea o f d mStress is Force / Area. Itis independent of material. Say 3 geometricallyidentical rods (sameareaof C/S & length)are prepared from Steel, Aluminium &Brass. For samemagnitude of force, stressgenerated in al1the components will be same. Hence if stress is the main criteria then there isno need to rerun the analysis. But displacement depends upon material property'E'(u = FUAE)&it will be different for different materials(Exception to above is body forces which depends ondensitylike gravity, centrifuga1force etc.).

LinearStaticAnaiysis 11.7 h-element us. p-elementElernent length Is reduced for achiwing convergence Eiernentiength: no changeEiement order: no change Eiernent order isincreased for achievingconvergenceApplicablefor ail elernents Rpplicablefor limited typeof elernentsRequiresmore time to converge LesserconvergencetimeRe-rneshing with srnalier elernentlength in criticai areas -Need not rnesh main and ao-ain.. initial coarse rneshand rerunningthe analyrisIs required rnodel ieadstoconvergenceby increaringthecaiculationApplicableforall typesof analysis points lpolynornialorder) internally Lirnited to Linear static and Normal mode dynamics in nostof the commercial software5Who should usep-element method:P-element method is recommended for design engineers (for first hand calculations) i.e. say inthe initial design phase they are not interestedin detailed analysis but just approximate results.Since CAE is not their regular activity, tetrahedron p-elements meshing by picking the volumeand simple boundary conditions based on geometry (by picking surfaces etc.) can lead toreasonable first hand calculations(convergencebeing taken careby the p-methodit self).After finalizing the design, design engineer can transfer CAD data to dedicated CAE groupfor detailed analysis. CAE groups mainly use h-element method. H-element method helps inmodeling al1 the necessary details and use of special elements / techniques which are moreaccurateand not supportedby p-elements.When critical areasare not known and assembly or component is very big, use of would ensuregood accuracy at lesser computational efforts.Fine meshing of entire assembly or component leads to very high dofs. With sub-modelingtechnique, initial run is carried out with coarse mesh. Locations of high stress are separatedout.Separated localized areas are remeshed with smaller element length. Displacement boundaryconditions (these are results based on coarse mesh model) are applied on the boundary oflocalized region. Steps for are as follows:

1 ) Coarse mesh the component(or assemblyl2 ) Analysis with appropriateboundary conditions as usual. ~C-Cy=.~t-t-. dsmallareas /volumearoundcriticaIIocations.Remesh separated localizedareaswith small element length./ Dirplocement results from coarse mesh opplled as constrotnt on the ride faces, no externol force in this case.4)Apply displacementdof(translationsand rotations, Le.displacementresults fromcoarse model)todense mesh mode1boundary (noforce, only constraintsin the form of displacement).