{"id":651,"date":"2020-09-25T13:51:23","date_gmt":"2020-09-25T20:51:23","guid":{"rendered":"http:\/\/gantovnik.com\/bio-tips\/?p=651"},"modified":"2021-06-22T01:25:28","modified_gmt":"2021-06-22T08:25:28","slug":"106-nonlinear-buckling-sol106","status":"publish","type":"post","link":"https:\/\/gantovnik.com\/bio-tips\/2020\/09\/106-nonlinear-buckling-sol106\/","title":{"rendered":"#106: Nonlinear Buckling in MSC Nastran (SOL106)"},"content":{"rendered":"

#106: Nonlinear Buckling (SOL106)<\/p>\n

1) Linear buckling of Euler column. For clamped-free boundary conditions the critical load is:<\/p>\n

Pcrit = (pi**2)*E*I\/(4*(L**2)),
\nwhere,
\nE = 10.5E6, I = 8.333-5, L=10 means Pcrit = 21.59<\/p>\n

2) Nonlinear buckling with PARAM,BUCKLE,2
\nIn f06 result file search for following message (right after eigenvalue table).<\/p>\n

*** USER INFORMATION MESSAGE 9040 (SUBDMAP NLSTATIC)
\nCRITICAL BUCKLING FACTOR (ALPHA)= 3.510785E+00<\/p>\n

In following example Subcase 102 gives ALPHA = 3.510785E+00
\nThis is how one can calculate buckling Load.<\/p>\n

General Formula : Pn + Alpha * (Delta P)<\/p>\n

where
\nPn = Total Load at Subcase Id where Param,buckle,2 is applied
\nDelta P = is the (Delta Load or (P(n) – P(n-1)) \/ No_of_increment.<\/p>\n

Let us consider Subcase 102 (Alpha = 3.510785)
\nTotal Load Pn = 20.00 (subcase 102)<\/p>\n

Delta Load = (20.00 – 19.00) = (Load @ (Subcase 2 – Subcase 1) = 1.00<\/p>\n

Pcr = Pn + (Delat_Load)\/No_of_incr) * Alpha = 20.0 + (1.0\/2) * 3.510785E+00 = 21.755<\/p>\n

Or just print out OLOAD that will give buckling load: Oload = 10.8777 * 2 (at two nodes) = 21.755<\/p>\n

b106-buck.dat<\/strong>
\n
\nTIME 100
\nSOL 106 $ Nonlinear Statics
\nCEND
\n$
\nTITLE = CANTILEVERED BEAM MADE OF PLATES
\n$
\noload = all
\nSUBCASE 101
\nLABEL = Total Load = 19.0 Lbs (0 to 19 lbs)
\nLOAD = 91
\nNLPARM = 1
\nSUBCASE 102
\nLABEL = Total Load = 20.0 Lbs (19 to 20 lbs)
\nLOAD = 92
\nNLPARM = 2
\nparam, buckle,2
\nmethod = 10
\nforce = all
\nspcforce = all
\ndisp = all
\noload = all
\nBEGIN BULK
\n$
\n$ Cantilevered Beam Made of Plates Model
\n$
\n$ \\201 202 203 204 205 206 207 208 209 210 211
\n$ Y \\*----*----*----*----*----*----*----*----*----*----*
\n$ ^ \\| | | | | | | | | | |
\n$ | \\| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
\n$ | \\| | | | | | | | | | |
\n$ +--->X \\*----*----*----*----*----*----*----*----*----*----*
\n$ \\101 102 103 104 105 106 107 108 109 110 111
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nPARAM LGDISP 1
\nparam,post,-1
\n$
\n$
\n$
\nNLPARM 1 5 ITER 1
\nNLPARM 2 2 ITER 1
\n$1234567$1234567$1234567$1234567$1234567$1234567$1234567$1234567$1234567
\nEIGRL 10 3 MASS
\nGRID 101 0. 0. 0. 123456
\nGRID 102 1. 0. 0.
\nGRID 201 0. 1. 0. 123456
\nGRID 202 1. 1. 0.
\n$
\nCQUAD4 1 1 101 102 202 201
\nMAT1 1 10.5E6 .3 2.588-4 1.E-6 0.
\nPSHELL 1 1 .1 1
\n$
\nFORCE 91 111 9.500 -1.0 0. 0.
\nFORCE 91 211 9.500 -1.0 0. 0.
\n$
\nFORCE 92 111 10.00 -1.0 0. 0.
\nFORCE 92 211 10.00 -1.0 0. 0.
\n$
\nFORCE 93 111 10.50 -1.0 0. 0.
\nFORCE 93 211 10.50 -1.0 0. 0.
\n$
\nFORCE 94 111 11.00 -1.0 0. 0.
\nFORCE 94 211 11.00 -1.0 0. 0.
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nENDDATA
\n<\/code><\/p>\n

b106-buck.f06<\/strong>
\n
\nR E A L E I G E N V A L U E S
\nMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZED
\nNO. ORDER MASS STIFFNESS
\n1 1 3.510785E+00 1.873709E+00 2.982100E-01 6.169645E-02 2.166030E-01
\n2 2 3.593726E+02 1.895713E+01 3.017120E+00 1.379434E-01 4.957308E+01
\n3 3 1.113806E+03 3.337373E+01 5.311595E+00 3.651141E-01 4.066663E+02
\n*** USER INFORMATION MESSAGE 9040 (SUBDMAP NLSTATIC)
\nCRITICAL BUCKLING FACTOR (ALPHA)= 3.510785E+00
\n<\/code><\/p>\n

Command line to run Nastran. Note option IFPSTAR=NO (select the MSC Nastran IFP for bulk data processing)
\n
\nnastran.exe old=no scr=yes news=no IFPSTAR=NO mem=estimate b106-buck.dat
\n<\/code><\/p>\n

Run 1: Applied load is 19.
\nTo decide the approximate value of buckling load in the 1st run we will try to find range of the critical load.
\nLOAD FACTOR 0.4000000
\nLOAD FACTOR 0.6000000
\nThis means the critical load is between 0.4*P and 0.6*P, i.e., from 16 to 24. For example, we can take 20.<\/p>\n


\nSOL 106 $ Nonlinear Statics
\nCEND
\n$
\nTITLE = CANTILEVERED BEAM MADE OF PLATES MODEL \/ Linear PCRIT = 21.59
\n$
\n$
\noload = all
\nSUBCASE 101
\nLABEL = Total Load = 19.0 Lbs (0 to 19 lbs)
\nLOAD = 91
\nNLPARM = 1
\n$SUBCASE 102
\n$ LABEL = Total Load = 20.0 Lbs (19 to 20 lbs)
\n$ LOAD = 92
\n$ NLPARM = 2
\nparam, buckle,2
\nmethod = 10
\nforce = all
\nspcforce = all
\ndisp = all
\noload = all
\nBEGIN BULK
\n$
\n$ Cantilevered Beam Made of Plates Model
\n$
\n$ \\201 202 203 204 205 206 207 208 209 210 211
\n$ Y \\*----*----*----*----*----*----*----*----*----*----*
\n$ ^ \\| | | | | | | | | | |
\n$ | \\| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
\n$ | \\| | | | | | | | | | |
\n$ +--->X \\*----*----*----*----*----*----*----*----*----*----*
\n$ \\101 102 103 104 105 106 107 108 109 110 111
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nPARAM LGDISP 1
\n$
\n$
\nNLPARM 1 5 ITER 1
\nNLPARM 2 2 ITER 1
\nEIGRL 10 3
\nGRID 101 0. 0. 0. 123456
\nGRID 102 1. 0. 0.
\nGRID 201 0. 1. 0. 123456
\nGRID 202 1. 1. 0.
\n$
\nCQUAD4 1 1 101 102 202 201
\nMAT1 1 10.5E6 .3 2.588-4 1.E-6 0.
\nPSHELL 1 1 .1 1
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......
\n$FORCE SID G CID F N1 N2 N3
\nFORCE 91 111 20.00 -1.0 0. 0.
\nFORCE 91 211 20.00 -1.0 0. 0.
\n$
\nFORCE 92 111 10.00 -1.0 0. 0.
\nFORCE 92 211 10.00 -1.0 0. 0.
\n$
\nFORCE 93 111 10.50 -1.0 0. 0.
\nFORCE 93 211 10.50 -1.0 0. 0.
\n$
\nFORCE 94 111 11.00 -1.0 0. 0.
\nFORCE 94 211 11.00 -1.0 0. 0.
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nENDDATA
\n<\/code><\/p>\n

Run 2: Applied load is 20 lb. This load is applied in two cases with magnitude as 95% + 5%. In this run we get the exact critical load.
\n
\nSOL 106 $ Nonlinear Statics
\nCEND
\n$
\nTITLE = CANTILEVERED BEAM MADE OF PLATES MODEL \/ Linear PCRIT = 21.59
\n$
\n$
\noload = all
\nSUBCASE 101
\nLABEL = Total Load = 19.0 Lbs (0 to 19 lbs)
\nLOAD = 91
\nNLPARM = 1
\nSUBCASE 102
\nLABEL = Total Load = 20.0 Lbs (19 to 20 lbs)
\nLOAD = 92
\nNLPARM = 2
\nparam, buckle,2
\nmethod = 10
\nforce = all
\nspcforce = all
\ndisp = all
\noload = all
\nBEGIN BULK
\n$
\n$ Cantilevered Beam Made of Plates Model
\n$
\n$ \\201 202 203 204 205 206 207 208 209 210 211
\n$ Y \\*----*----*----*----*----*----*----*----*----*----*
\n$ ^ \\| | | | | | | | | | |
\n$ | \\| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
\n$ | \\| | | | | | | | | | |
\n$ +--->X \\*----*----*----*----*----*----*----*----*----*----*
\n$ \\101 102 103 104 105 106 107 108 109 110 111
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nPARAM LGDISP 1
\n$
\n$
\nNLPARM 1 5 ITER 1
\nNLPARM 2 2 ITER 1
\nEIGRL 10 3
\nGRID 101 0. 0. 0. 123456
\nGRID 102 1. 0. 0.
\nGRID 201 0. 1. 0. 123456
\nGRID 202 1. 1. 0.
\nCQUAD4 1 1 101 102 202 201
\nMAT1 1 10.5E6 .3 2.588-4 1.E-6 0.
\nPSHELL 1 1 .1 1
\n$
\nFORCE 91 111 9.500 -1.0 0. 0.
\nFORCE 91 211 9.500 -1.0 0. 0.
\n$
\nFORCE 92 111 10.00 -1.0 0. 0.
\nFORCE 92 211 10.00 -1.0 0. 0.
\n$
\nFORCE 93 111 10.50 -1.0 0. 0.
\nFORCE 93 211 10.50 -1.0 0. 0.
\n$
\nFORCE 94 111 11.00 -1.0 0. 0.
\nFORCE 94 211 11.00 -1.0 0. 0.
\n$
\n$.......2.......3.......4.......5.......6.......7.......8.......9.......0
\nENDDATA
\n<\/code><\/p>\n","protected":false},"excerpt":{"rendered":"

#106: Nonlinear Buckling (SOL106) 1) Linear buckling of Euler column. For clamped-free boundary conditions the critical load is: Pcrit = (pi**2)*E*I\/(4*(L**2)), where, E = 10.5E6, I = 8.333-5, L=10 means Pcrit = 21.59 2) Nonlinear buckling with PARAM,BUCKLE,2 In f06 result file search for following message (right after eigenvalue table). *** USER INFORMATION MESSAGE 9040 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"nf_dc_page":"","_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","_lmt_disableupdate":"yes","_lmt_disable":"","jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[34,22],"tags":[],"class_list":["post-651","post","type-post","status-publish","format-standard","hentry","category-buckling","category-nastran"],"modified_by":"gantovnik","jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p8bH0k-av","jetpack_likes_enabled":true,"jetpack-related-posts":[{"id":481,"url":"https:\/\/gantovnik.com\/bio-tips\/2020\/03\/how-to-include-temperature-dependent-material-in-a-buckling-analysis\/","url_meta":{"origin":651,"position":0},"title":"#70 How to include temperature-dependent material in a buckling analysis?","author":"gantovnik","date":"2020-03-04","format":false,"excerpt":"To include temperature-dependent material in a linear buckling analysis (SOL 105), the temp(load) case control command must be placed above the first subcase; otherwise, the temperature-dependent material properties will be ignored. As in all linear solutions, the material lookup is performed only once in a run.","rel":"","context":"In "nastran"","block_context":{"text":"nastran","link":"https:\/\/gantovnik.com\/bio-tips\/category\/nastran\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":637,"url":"https:\/\/gantovnik.com\/bio-tips\/2020\/09\/100-negative-eigenvalue-in-buckling-analysis\/","url_meta":{"origin":651,"position":1},"title":"#100: Negative eigenvalue in buckling analysis","author":"gantovnik","date":"2020-09-25","format":false,"excerpt":"#100: Negative eigenvalue in buckling analysis In buckling analysis, a negative eigenvalue implies that the load that causes the structure to buckle is opposite to the direction of the applied load. A negative eigenvalue means that the critical load has a reversed direction compared to the applied load. Applying a\u2026","rel":"","context":"In "nastran"","block_context":{"text":"nastran","link":"https:\/\/gantovnik.com\/bio-tips\/category\/nastran\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":479,"url":"https:\/\/gantovnik.com\/bio-tips\/2020\/03\/what-does-a-negative-eigenvalue-in-a-buckling-analysis-mean\/","url_meta":{"origin":651,"position":2},"title":"#69 What does a negative eigenvalue in a buckling analysis mean?","author":"gantovnik","date":"2020-03-04","format":false,"excerpt":"A negative eigenvalue in a buckling analysis may imply that the buckling load is in the opposite direction of the applied load. For example, for a simply supported beam, if a tensile axial load is applied to one end, then the eigenvalue will be negative. One the other hand, for\u2026","rel":"","context":"In "nastran"","block_context":{"text":"nastran","link":"https:\/\/gantovnik.com\/bio-tips\/category\/nastran\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":666,"url":"https:\/\/gantovnik.com\/bio-tips\/2020\/09\/112-msc-nastran-sol-106\/","url_meta":{"origin":651,"position":3},"title":"#112 MSC Nastran SOL 106","author":"gantovnik","date":"2020-09-28","format":false,"excerpt":"#112 MSC Nastran SOL 106 MSC Nastran SOL 106 The nonlinear effects in structures occur due to nonlinear material behavior and large deformations. Geometric nonlinearity becomes relevant when the structure is subjected to large displacement and rotation. Geometric nonlinearity effects are prominent in two aspects: geometric stiffening due to initial\u2026","rel":"","context":"In "nastran"","block_context":{"text":"nastran","link":"https:\/\/gantovnik.com\/bio-tips\/category\/nastran\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":648,"url":"https:\/\/gantovnik.com\/bio-tips\/2020\/09\/105-how-to-remove-offsets-defined-on-shell-and-beams\/","url_meta":{"origin":651,"position":4},"title":"#105: How to remove offsets defined on shell and beams?","author":"gantovnik","date":"2020-09-25","format":false,"excerpt":"#105: How to remove offsets defined on shell and beams? Method 1: Properties > Modify > GLOBAL Method 2: Utility > Property > Property Editor Property Words to Change: 4111 Plate Offset, Only Process If Exists: Check, Action: =, Value: 0 Offsets should not be used for the beam and\u2026","rel":"","context":"In "nastran"","block_context":{"text":"nastran","link":"https:\/\/gantovnik.com\/bio-tips\/category\/nastran\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]},{"id":1558,"url":"https:\/\/gantovnik.com\/bio-tips\/2022\/09\/210-parametric-curve-in-3d-2-2-2-2-2-2-2-2-2-2-2-2-2-3-3-2-2-3-2\/","url_meta":{"origin":651,"position":5},"title":"#301 Buckling Glossary","author":"gantovnik","date":"2022-09-13","format":false,"excerpt":"Asymmetric buckling: When the structure has asymmetric geometry it may undergo stable buckling for a positive displacement parameter and unstable buckling if this displacement parameter is negative. Bifurcation point: A point along a primary equilibrium path (or curve) intersected by a secondary path. Beyond this point, the primary equilibrium path\u2026","rel":"","context":"In "python"","block_context":{"text":"python","link":"https:\/\/gantovnik.com\/bio-tips\/category\/python\/"},"img":{"alt_text":"","src":"","width":0,"height":0},"classes":[]}],"_links":{"self":[{"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/posts\/651","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/comments?post=651"}],"version-history":[{"count":0,"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/posts\/651\/revisions"}],"wp:attachment":[{"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/media?parent=651"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/categories?post=651"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gantovnik.com\/bio-tips\/wp-json\/wp\/v2\/tags?post=651"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}