Demonstrative Exam Le Solutions In Static Analysis

Topic 12DemonstrativeExam le Solutionsin Static AnalysisContents: Analysis of various problems to demonstrate, study, andevaluate solution methods in statics Example analysis: Snap-through of an arch Example analysis: Collapse analysis of an elastic-plasticcylinder Example analysis: Large displacement response of a shell Example analysis: Large displacements of a cantileversubjected to deformation-independent and deformation dependent loading Example analysis: Large displacement response of adiamond-shaped frame Computer-plotted animation: Diamond-shaped frame Example analysis: Failure and repair of a beam/cablestructureTextbook:Sections 6.1, 6.5.2, 8.6, 8.6.1, 8.6.2, 8.6.3

Topic Twelve 12-3INTHIS L e:CTU1 .E:) WE'" ANT TOSTU))'! XAt-1PLESOUATlO,aJSOMEI KS":I IAt-fONb-S"UA7 !;j)ISNAi"- TH oLA GHFAILlARE AND'KE'PAIR OF A&EM-1/CA lES'T1t lAC.TWi . EOF A " \A sS A'R(\4E)(.2.FRA.-1ESE)C./c,E) .,ANAL 'ISIS DFC OL.lA1'"SE ANAL'1SIS01 ' AN ELASTO - "?LASTICC'IL'NbER&x.! LARe-E 1lIS'rL.A(E.-MENT OL\IT'ON r:J; A WP''R\C LE'K S \o\E lLCANTIL'\JER'PR SSU E:lAN ERLOA'! INlQMarkerboard12-1

12-4 Example Solutions in Static AnalysisExample: Snap-through of a truss archTransparency12-1L 10.0k A 2.1 X 105 Perform post-buckling analysis usingautomatic load step incrementation. Perform linearized buckling analysis.Postbuckling analysis:Transparency12-2The analytical solution is30,00020,000R10,0000-10,000!:J.

Topic 'l\velve 12-5The automatic load step incrementationprocedure previously described may beemployed.Transparency12-3Using 1 1U -0.1, we obtainanalyticalsolution .s-;;30,000Rfinite elementsolution20,00010,000j lC0 ---- - --- ----l'L- --6-10,000 Solution details for load step 7: The spherical constant are-lengthalgorithm is employed. The initial stiffness matrix is employedfor all iterations, tu .8111, tR 01A RH.1t (i)16,69015,22014,51014,34014,31014,310u(i)A(i) 731Transparency12-4

12-6 Example Solutions in Static AnalysisTransparency12-5Pictorially, for load step 7,i 117,000R16,000A(1)RA15,00014,000i 2i 3 load step 713,000U000.70.80.91.01.1Solution details for load step 8:Transparency12-6 The constant increment of externalwork algorithm is employed. Modified Newton iterations areused, tu .9701 , tR 14,310.it 4t U(i)121.12271.1227A Rt 4t (i)14,74014,500U(i)A(i) R.1526.1526440200!l

Thpic Twelve 12-7Pictorially, for load step 8,Transparencyci ical12-7i 115,000load step 7B/load step 8anasol ution0R10,000Wprescribed area ofquadrilateralABCE5,0000.9tw area ofquadri1ateralABO EU U(1)A1.0E1.11.2We now employ a linearized bucklinganalysis to estimate the collapse loadfor the truss arch.Load25,000 l-.6'R 1,000, TR 25,60020,000 1---------- 'R 10,000, TR 21,1006'R 14,000, TR 16,800115,0006'R 14,500, TR 15,0006l10,0005,000000.51.0Rcr 14,5041.5displacement2.0/to3digitsTransparency12-8

12-8 Example Solutions in Static AnalysisTransparency12-9There are cases for which linearizedbuckling analysis gives buckling loadsfor stable structures. Consider thetruss arch reinforced with a spring asshown:RLoad100,00050,000k* 16,000246displacementThis structure is always stable.Transparency12-10We perform a linearized bucklinganalysis. When the load level is closeto the inflection point, the computedcollapse load is also close to theinflection point.load100,00080,00060,000llR 10,000, TR 60,70040,000 j--------: -- ::::.--l'R 40,000, TR 44,10C20,0000displacement0246

Topic Twelve 12-9Example: Elastic-plastic cylinder underinternal pressure12-11E 8667v 0.3ET OVerylong-Transparency17.32Inner radius 1Outer radius 2cry Goal: Determine the limit load.Finite element mesh: Four 8-nodeaxisymmetric elementsz1.01.0Transparency12-121.0We note that, due to the boundaryconditions and loading used, allstresses are constant in the z direc tion. Hence, 6-node elements couldalso have been used.

12-10 Example SolutioDS in Static AnalysisTransparency12-13Since the displacements are small, weuse the M.N.O. formulation. We employ the following loadfunction:15(the elastic limitload isP 1085P 7.42)0 -1---4---- ----1-- 01Transparency12-145time1015Now we compare the effectiveness ofvarious solution procedures:-Full Newton method with linesearchesFull Newton method without linesearchesBFGS methodModified Newton method withline searchesModified Newton method withoutline searchesInitial stress method

Topic Twelve 12-11The following convergence tolerances areemployed:VVhen any of these procedures are used,the following force-deflection curve isobtained. For P 14, no convergedsolution is found.15P105P 14 (no solution obtained)---.s-- p 13.5 computedsolution a smooth curve fittingthe computed solutions.0-jL-- - -- - o2x10- 4 4x10- 4 6x10- 4displacement of inner surfaceTransparency12-15Transparency12-16

12-12 Example Solutions in Static AnalysisTransparency12-17We now compare the solution times forthese procedures. For the comparison,we end the analysis when the solution forP 13.5 is obtained.MethodFull Newton method with line searchesFull Newton methodBFGS methodModified Newton method with linesearchesModified Newton methodInitial stress .12.2Now we employ automatic load stepincrementation. No longer need to specify a loadfunction Softening in force-deflection curveis automatically taken into account.Here we useETOL 10- 5RTOL O.01RNORM 1.0

Topic Twelve 12-13Result: Here we selected the displacementof the inner surface for the firstload step to be 10- 4 .15computed limit load --.:.----:m:: .-E!ll------e---&-- p Transparency12·1913.810P508)----- --- -- ----- --- -----1f--- o2x10- 4 4x10- 4 6x10- 4 8x10- 4 10x10- 4 12x10- 4displacement of inner surfaceExample: Spherical Shell thinaxisymmetricshell,clamped edges Concentrated forcE;) PTransparency12·20

12-14 Example Solutions in Static AnalysisTransparency12-21r .,tp,U/O.0859 in 5 hine 10.9 E 107 psivTransparency12-22 0.01576R 4.76 in 0.3Finite element mesh: Ten 2-Daxisymmetric elements I: I : I I I:l Deformed configuration for P 100 Ib:1: :r :r :I: :r %-:r :r :J:::::: t:::::::::X: c::::;:::::::rc::::i-T':::;a -

Topic Twelve 12-15Force-deflection curve obtained using10 element 12-23linearanalysis50O-f'---- ---- ----- -- I o0.050.100.150.20displacement of apex (in)Comparison of solution procedures:1) Apply full load (100 Ib) in 10equal steps:Solution procedureFull Newton with line searchesFull Newton without line searchesBFGS methodModified New10n with line searchesModified New10n without line searchesNormalized solution time1.41.0did not convergedid not convergedid not convergeTransparency12-24

12-16 Example Solutions in Static AnalysisTransparency12-25Transparency12-262) Apply full load in 50 equal steps:Solution procedureNormalized solution timeFull Newton with line searchFull Newton without line searchBFGS methodModified Newton with line searchModified Newton without line search1.31.01.61.9did not convergeConvergence criterion employed:Maximum number of iterations permitted 99

Topic Twelve 12-17We may also employ automatic loadstep incrementation:Transparency12-27Here we useETOL 10- 5andIIt atR - t at F(i-1)1121.0R M - RTOlas convergence tolerances.Results: Using different choices of initialprescribed displacements,we obtain(!)100-:1u: 1uTransparency12-28 0.01 in 0.001 inappliedload50(Ib)O --- -----t--- ---- o.05.10.15displacement of apex (in).20

12-18 Example Solutions in Static AnalysisTransparency12-29Example: Cantilever under pressureloading-.Iuniform pressure load p :.:: -- -- -- ojp.2 m10,mE 207000 MPav 0.3Plane strain, width 1.0 m Determine the deformed shape ofthe cantilever for p 1 MPa.Transparency12-30-Since the cantilever undergoeslarge displacements, the pressureloading (primarily the direction ofloading) depends on the config uration of the cantilever: IPdeformation-independentdeformation-dependent

'Ibpic Twelve 12·19Transparency12-31The purpose of this example is tocontrast the assumption of deforma tion-independent loading with theassumption of deformation-dependentloading.Finite element model: Twenty-five two dimensional a-node elements(1 layer, evenly spaced)Solution details: Full Newton method without linesearches is used. Convergence tolerances areETOL 10- 3RTOL 10- 2 ,RNORM 1.0 MNTransparency12-32

12-20 Example Solutions in Static AnalysisTransparency12-33Results: Force-deflection curve For small deflections, there arenegligible differences betweenthe two assumptions.(J,al1.2deformation-independentL loading,"?: .i.6CH4.2c!p Ideformation-dependentloadingl!» eC!)O.----tf----t-----t--- t.002.004.006.008.00vertical displacement of tip (m)Transparency12-34Pictorially, for p 1.0 MPa,undeformed ntloading

Topic 1\velve 12-21Example: Diamond-shaped frameTransparency12-35 frictionless hinge/beam cross-section: 1 in x 1 inE 30x 106 psiv 0.315 inTwenty3-nodeisobeam elements15 inForce-deflection curve, obtained usingthe T.L formulation: A constant load increment of 250Ibs is used.80,000P(Ibs)60,00040,00020,000O- ---- ----- ---l----o102030displacement of top hinge (inches)Transparency12-36

12-22 Example SolutioDS in Static AnalysisTIME ILOAD.HPAComputerAnimationDiamond shapedframetTIMELOADIS/ilI32Sfilfil HPAI",,,,"""""""",,." ,,"",,,, '" ",,yTIMELOAD39975lillillil HPA.A """"'",,,,","" , ,,,, , '" " '"y,,'",,'"'" ",'"

Topic 1\velve 12-23Example: Failure and repair of abeam/cable structureTransparency12-37cable: E 207000 MPaA 10- 4 mno pretension5m 910 mbeam: E 207000 MPav 0.3O"y 200 MPaET 20700 MPap 7800 kg/m 3cross-section:0.1 m x 0.1 mIn this analysis, we simulate thefailure and repair of the cable.Steps in analysis:Load step1EventBeam sags under its weight, butis supported by cable.1 to 2Cable snaps, plastic flow occurs atbuilt-in end of beam.2 to 4A new cable is installed, and istensioned until the tip of the beamreturns to its location in loadstep 1.Transparency12-38

12-24 Example Solutions in Static AnalysisTransparency12-39Finite element model:Two truss elements:/Truss #2 is tensionedby imposing a fictitiousthermal strain.LoadstepActivetruss1234#1none#2#2Five 2-node Hermitian beam elements5 Newton-Cotes integration points in r direction3 Newton-Cotes integration points in s directionTransparency12-40Solution details: The U.L. formulationis employed for the truss elementsand the beam elements.Convergence tolerances:ETOL 10- 3RTOL 10- 2RNORM 7.6x10- 3 MNRMNORM 3.8 x 10- 2 MN-m

Topic 1\velve 12·25Comparison of solution algorithms:Transparency12-41ResultsMethodFull Newton with line searchesAll load steps successful,normalized CPU time 1.0.Full NewtonStiffness matrix not positive definitein load step 2.BFGSAll load steps successful,normalized CPU time 2.5.Modified Newton with or withoutline searchesNo convergence in load step 2.Results:Transparency12-42LoadstepDisp.of tipStressin cableMoment atbuilt-in end1-.008 m64 MPa9.7 KN-m2-.63 m3-.31 m37 MPa22 KN-m4-.008 m72 MPa6.2 KN-m-38 KN-mNote: The elastic limit moment at the built-inend of the beam is 33 KN-m.

12-26 Example Solutions in Static AnalysisPictorially,'Transparency12-43Load step 1:Load step 2:(Displacements are magnifiedby a factor of 10)Load step 3:Load step 4:(Displacements are magnifiedby a factor of 10)

MIT OpenCourseWarehttp://ocw.mit.eduResource: Finite Element Procedures for Solids and StructuresKlaus-Jürgen BatheThe following may not correspond to a particular course on MIT OpenCourseWare, but has beenprovided by the author as an individual learning resource.For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

buckling analysis gives buckling loads for stable structures. Consider the truss arch reinforced with a spring as shown: R Load 100,000 k* 16,000 50,000 246 displacement This structure is always stable. Transparency 12-10 We perform a linearized buckling analysis. When the load level is close to the inflection point, the computed collapse load .