Task 6 - Safety Review and Licensing On the Job Training on Stress Analysis F.M. with Finite Element analysis - Different calculation techniques + Numerical examples (ANSYS Apdl) 2/2 Davide Mazzini Ciro Santus Pisa (Italy) June 15 July 14, 2015
FM parameters with Finite Element method Content Different FE techniques ANSYS Workbench - Crack geometry element mesh preparation - K I(II,III) and J calculation - Examples ANSYS Apdl - The quarter point technique - Examples 2
CT specimen (plane) modelling P 10 kn a 30mm, W 60mm, B 12mm K P 2 (1 ) 2 3 4 I (0.886 4.64 13.32 14.72 5.64 ) 971MPa mm 3/2 B W B W a 3
CT specimen (plane) modelling 4 L3 3 L11 10 L10 L4 11 9 L2 5 L12 L13 12 Y L9 1Z X L1 L8 82 L15 14 L14 L5 15 13 L7 L16 16 L17 6 L6 7 4
CT specimen (plane) modelling Multiple and nested refinements around the crack tip Total: 304605 nodes Element size at innermost refinement: 0.012 mm 5
CT specimen (plane) modelling Planeload applied at pins: P p 833 N/mm B Load on each node on the half circle: p pi nn nn: number of nodes on the half circle 6
CT specimen (plane) modelling In-plane contraints to remove any lability No rotation No rigid motion 7
CT specimen (plane) modelling Stress singularity The maximum value just related to the innermost element size 8
CT specimen (plane) modelling Asymptotic calculation log-log scales 10 4 ANSYS Apdl y, MPa 10 3 K I = 992 MPa mm 1/2 MATLAB "polyfit": p, p p K 1 I I I 0.5(approx.) 10 % 2.2% 1 p2 log 10 (2 ) 2 1 2 K (ANSYS-MATLAB) 992 MPa mm K (ASTM) 971MPa mm 10 2 10-3 10-2 10-1 10 0 10 1 r, mm 9
Question: By switching from Pl. Strain to Pl. Stress, is the K I result expected to change? Why? 10
CT specimen (plane) modelling KCAL command 1 st step: calculate the solution as previous 2 nd step: prepare a spiderweb mesh, on a separate geometry then to run the submodeling technique 11
CT specimen (plane) modelling KCAL command ANSYS command for mid-side node at quarter point: KSCON,1,r/3,1,6 Regular mid-side elements Distinct overlying nodes (crack side) ¼ side position of mid-side nodes for crack tip elements 12
CT specimen (plane) modelling KCAL command 3 rd step: interpolate the displacements, apply to the submodel boundary, and run the solution Submodel solution Displacements derived from the full model 13
CT specimen (plane) modelling KCAL command 4 th step: define an ANSYS path and run the KCAL command Point 1: the crack tip Point 2 Point 5 Point 3 Point 4 14
CT specimen (plane) modelling CINT,SIFS command 1 st step: prepare the model as previous 15
CT specimen (plane) modelling CINT,SIFS command 2 nd step: define the crack tip for the CINT command and input other CINT options Axis extension direction Crack tip Contours Axis normal 16
CT specimen (plane) modelling CINT,SIFS command 3 rd step: just run the solution 17
CT specimen (plane) modelling CINT,SIFS command 4 th step: issue the prcint command to see the results Different results for the contours, convergence of results after the first few contours K2 values (that should be zero) give an indication of the accuracy 18
CT specimen (plane) modelling Results comparison K K K K I I I I (ASTM) 971MPa mm (ANSYS-Asympt.) 992MPa mm % 2.2% (ANSYS-KCALC) 1033MPa mm % 6.4% (ANSYS-CINT) 1034 MPa mm % 6.5% Which one could, or should, be the most accurate? Why, any possible explanation? 19
ANSYS Workbench -> ANSYS Apdl CT specimen (3D) modelling, ANSYS Wb ANSYS Apdl Save to ANSYS APDL environment 20
ANSYS Workbench -> ANSYS Apdl Exercise: Calculate K I at the crack front nodes by means of the CINT,SIFS command in the Apdl 3D environment. Same results as Workbench PreCracked-Mesh procedure? From ANSYS Workbench 21
ANSYS Workbench -> ANSYS Apdl Crack front component CM selection CINT commands 22
ANSYS Workbench -> ANSYS Apdl Mode I SIF, MPa mm 1/2 1200 1100 1000 K I (3D) 1102 MPa mm Plane model solution KI 1034 MPa mm 900 1st cont. 800 2nd cont. 3rd cont. 4th cont. 700 0 2 4 6 8 10 12 Crack front position, mm Different contours solution, coincident after the 3 rd 23
ANSYS Workbench -> ANSYS Apdl Workbench then Adpl to manually introduce the CINT command Workbench with PreMeshed-crack Mode I SIF, MPa mm 1/2 1200 1100 1000 900 1st cont. 800 2nd cont. 3rd cont. 4th cont. 700 0 2 4 6 8 10 12 Crack front position, mm Same solution, so why this difference? 24
ANSYS Workbench -> ANSYS Apdl Adpl Commands can be manually introduced in the Wb environment Results redirected to an output text file 1200 Same results as after Wb to Apdl Mode I SIF, MPa mm 1/2 1100 1000 900 1st cont. 800 2nd cont. 3rd cont. 4th cont. Pisa, June 15 July 14, 700 2015 25 0 2 4 6 8 10 12 Crack front position, mm
ANSYS Workbench -> ANSYS Apdl Workbench to Adpl, Hexahedrons mesh Workbench with PreMeshed-crack OK NO 1200 1200 Mode I SIF, MPa mm 1/2 1100 1000 900 1st cont. 800 2nd cont. 3rd cont. 4th cont. 700 0 2 4 6 8 10 12 Crack front position, mm Mode I SIF, MPa mm 1/2 1100 1000 900 1st cont. 800 2nd cont. 3rd cont. 4th cont. 700 0 2 4 6 8 10 12 Crack front position, mm 26