Fatigue Assessment with LIMIT According to Eurocode 3 Version LIMIT PDF Free Download

Documentation Fatigue Assessment with LIMIT According to Eurocode 3 Version LIMIT2014 March 2014

Overview Fatigue Assessment with LIMIT According to Eurocode 3 Motivation Stresses and section forces Stress spectra Stress-relieved welded details in compression FAT classes S-N curves Fatigue analysis Results page 2

Motivation Example: Beam under bending load Cross section: Height h = 200mm Width w = 200mm Distance between webs d = 160 mm Thickness of all sheets t = 10mm Vertical force No displacement, no rotation page 3

Motivation FEA Analysis Results: FEA code Abaqus Shell: 8 Node Direct stress in local 1 direction Von Mises equivalent stress Weld zone of interest Direct stress in local 2 direction page 4

Motivation Aim of the Investigation: Weld between flange and web is highly loaded and critical Proof of fatigue strength according to Eurocode 3 Max. vonmises stress: 25,4 MPa page 5

Motivation FAT Classes, Eurocode 3, Nominal Stress Assessment of weld section Table 8.1 to 8.10 in the Eurocode 3 Different permissible stresses depending on the loading direction: parallel to weld transverse to weld shear Toe, transverse loading, EC3 Table 8.5 Root, transvers loading, EC3 Table 8.5 Toe/root, shear, EC3 Table 8.5 We need local stresses in weld section! page 6

Motivation Local Shell Element Coordinate Systems Element coordinate systems not aligned with weld direction Abaqus default: local 1-axis in general parallel to global x-axis Ansys or Nastran default: local 1-axis depends on node numbering and interpolation functions Direct usage of stress data for assessment not possible! e.g. Abaqus Local weld direction page 7

Motivation Weld Analysis with LIMIT Basic features of LIMIT Automated detection of elements along welds based on different shell properties for flange and web Visualization of weld details relative to local weld direction page 8

Motivation Weld Analysis with LIMIT Basic features of LIMIT Checking all critical points (red circles) Base material Weld section Toes and Roots m n page 9

Stresses and Section Forces Stresses and Section Forces Basic Procedure for generating stress quantities for the fatigue analysis: Transformation of stresses to local coordinate system of weld Choosing the stress concept: Nominal stress Structural hot spot stress according to IIW Finding section forces and section moments Calculating the stresses in the weld section Setting up the stress spectra Example: web element page 11

Stresses and Section Forces Stress Data FE Stresses in integration points in local element system (Solver dependent, here ABAQUS) extrapolation of stresses to the nodes LIMIT Imported Data stresses at nodes in local element system shells (2D-tensor): Bottom side (s 11,SNEG, s 22,SNEG, s 12,SNEG ) Top side (s 11,POS, s 22,SPOS, s 12,SPOS ) weld Integration points transverse longitudinal weld directions page 12

Stresses and Section Forces Different Ways to Use Stresses Direct assessment Stresses taken at weld line Green points show stress positions Stress are equal to Finite Element shell stresses directions taken from weld orientation Similar to nominal stress approach for reduced integrated elements and coarse meshes. weld transverse longitudinal weld directions page 13

Stresses and Section Forces Different Ways to Use Stresses Offset by a certain distance (see i.e. DVS 1612) for nominal stress concepts taken at i.e. 1,5 x thickness green points are stress extraction locations (can be visualized in LIMIT Viewer). Stress interpolation within the target element using stresses at corners directions taken from weld orientation See also additional Information in document: LIMIT-Defining_Offset_Endings_Directions.pdf i.e. 1,5 x t weld transverse longitudinal weld orientation page 14

Stresses and Section Forces Different Ways to Use Stresses Stress extrapolation for IIW structural hot spot stress IIW reference points at distance of IIW, IIW_A: 0,5 x thickness and 1,5 x thickness or IIW_B: 5mm and 15mm Local stresses defined relative to extrapolation direction Extrapolation direction = transverse Longitudinal = transverse to extrapolation dir. structural hot spot stress type IIW_A 0,5 x t 1,5 x t 1,5 x t weld 0,5 x t longitudinal IIW reference points transverse weld directions page 15

Stresses and Section Forces Section Forces Transformation from s 11, s 22, t 12 to s ll, s, t li For bottom and top side of shell Integration over shell thickness Linear distribution across thickness assumed Section forces (s shear force): n II, n, s II. [N/mm] Section moments (t torsion): m ll, m, t II [N mm/mm] m II, t II n, s II 2 1 2 2 1 1 2 1 Orientations: Weld system IIW m, t II n ll, s li weld transverse li parallel page 16

Stresses and Section Forces Assessment Points within LIMIT Are used to perform the fatigue analysis for potential areas of crack initiation Are defined relative to shell normal P1 to P4. Weld Section P1, P4 toe P2, P3 root P5 and P6. Base Material Weld dimensions A-Bot welded from bottom side A-Top welded from top side Root Position (midplane to root) D-Bot welded from bottom side D-Top welded from top side page 17

Assessment Points and Stresses One sided full penetration weld Thickness = a (weld section) Stresses for base material and weld section equal! Top Shell normal Stresses and Section Forces Bottom (s ll, s, t II ) Top,P4,P5 (s ll, s, t II ) Bottom,P5 page 18

Stresses and Section Forces Assessment Points and Stresses Double sided fillet weld a = t/2 P5, P6 shell stresses transformed to local directions (II, ) P1-P4 next slide (s ll, s, t II ) Top,P6 Top Shell normal Bottom (s ll, s, t II ) Bottom,P5 (s ll, s, t II ) P4 (s ll, s, t II ) P1 (s ll, s, t II ) P3 (s ll, s, t II ) P2 page 19

Stresses and Section Forces Double sided fillet weld, Stresses in P1 to P4 a = t/2.t = sheet thickness continously welded Top t/2 t/2 Bottom Stress longitudinal to weld direction: s ll,p1,2 = s ll,bot, s ll,p3,4 = s ll,top Stress lateral to weld direction: i = 1 to 4 s,pi = n tt / (A Bot +A Top ) + m e Pi /J weld A Top A Bot P4 P3 P2 P1 Shear in weld (in plane): t II,Pi = s Il / (A Bot +A Top ) e P4 e P3 e P2 e P1 with J weld = (t+a Bot +A Top )³/12 - t³/12 m e.g.: e P1 = - t/2 - A bot n page 20

Stresses and Section Forces Assessment Points and Stresses Single sided fillet weld a = 0.7 t Shell normal Top Bottom (s ll, s, t II ) P6 (s ll, s, t II ) P4 (s ll, s, t II ) P3 page 21

Stresses and Section Forces Single sided fillet weld, Stresses in P3 and P4 t sheet thickness continously welded Top t/2 t/2 Bottom Stress longitudinal to weld direction: s ll,p3,4 = s ll,top Stress lateral to weld direction: i = 3 and 4 s,pi = n / A Top + (m - n e EXC ) e Pi /J weld e EXC = e SSW free (FKM, FAT71) e EXC = e SSW /2 constrained P4 A Top P3 e P3 = -A top /2 e P4 = A top /2 Shear in weld (in plane): t Il,Pi = s Il / A Top with J weld = A Top ³/12 e P4 e P3 e SSW m e SSW = t/2 + A top /2 (SSW Single Sided Weld) n page 22

Stresses and Section Forces Single sided fillet weld, Stresses in P6 t sheet thickness continously welded P6 stresses caculated for sheet thickness Top t/2 t/2 Bottom Stress longitudinal to weld direction: s ll, 6 = s ll,top P6 Stress lateral to weld direction: s,p6 = n / t + (m + n e EXC ) / (t²/6) e EXC = e SSW free (FKM, FAT71) e EXC = e SSW /2 constrained Shear in weld (in plane): t Il,P6 = s Il / t A Top e e e SSW m e = A top /2 e SSW = t/2 + A top /2 (SSW Single Sided Weld) n page 23

Stress Spectra Stress Spectrum Assessment Points P1 to P4: 1. Setting up the local weld coordinate system (longitudinal, transverse) 2. Scanning through all load cases of a spectrum 3. According to EC3 relevant stresses are s and t li. 4. Saving maxima and minima stresses for all components: (s, t li ) Pi 5. Calculating stress range of each component and point: Ds = s max - s min 6. Assembly of stress spectrum for transverse stress s and shear t li Ds Ds,1 Example shown with four stress blocks: Ds,2 Ds,3 Ds,4 Number of cycles page 25

Stress Spectra Stress Spectrum Assessment Points P5 and P6: 1. Cutting plane algorithm in order to find direction of maximum stress range Angle increment of 15, starting point: 0 parallel to weld Normal stress is calculated using plain stress tensor (s ll, s, t li ) Pi 2. Scanning through all load cases of a spectrum at one angle j 3. Saving maxima and minima stresses for all components: s jj 4. Calculating stress range of each component and point: Ds = s max - s min 5. Assembly of stress spectrum for s jj Ds Ds jj,1 Example shown with four stress blocks: Ds jj,2 Ds jj,3 Ds jj,4 Number of cycles page 26

Eurocode 3 Stress-relieved welded details in compression 60% of the compression is used to calculate a reduced effective stress range Can be activated in the JobManager in Keywords: *EC3_REDUCED_EFFECTIVE_STRESS_RANGE Output to the.txt-file: SPECTRUM DATA: BLOCK: 1, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, EFF.RANGE, CYCLES, RELEVANT LOAD CASES) S11 5.0888 0.0000 4.07104 0.50000E+07 1 2 page 28

Eurocode 3 Fatigue Assessment according to Eurocode 3 Basic Procedure: Selecting FAT classes for the assessment point Definition of relevant S-N curve Proof of strength Calculation of damage Calculation of degree of utilization in terms of stress range page 30

FAT Classes FAT Classes, Eurocode 3 Assessment points P1 to P4 Stresses used for weld section: s and t li Out of plane shear is not resolved over shell stresses and in general negligible Stress in longitudinal direction is not taken into account in the weld section Relevant stresses page 31

FAT Classes FAT Classes, Eurocode 3 Assessment points P1 to P4, nominal stress approach Table 8.1 to 8.10 in the Eurocode 3 Typical values for transverse loading are : Toe: FAT71 to FAT80 Root: FAT36 Typical values for shear loading are : Toe or root: FAT80 Toe, transverse loading, EC3 Table 8.5 Root, transvers loading, EC3 Table 8.5 Toe/root, shear, EC3 Table 8.5 page 32

FAT Classes Fat Classes Assessment points P5 and P6 Toe points Cutting plane algorithm in increments of 15 (0 parallel to weld, 90 is transverse to weld) Only the direct stress in the cutting plane is used. Permissible FAT values from EC3 If FAT values for loading parallel and transverse to weld orientation differ, LIMIT will interpolate for intermediate cutting angles (ellyptic interpolation) FAT parallel FAT transverse e.g. But-Weld page 33

FAT Classes FAT Classes Assessment points P5 and P6, nominal stress approach: Direct stress, FAT class interpolated depending on angle Typical values for transverse loading are: Toe: FAT71 to FAT80 Typical values for parallel loading is: Toe: FAT100 Toe, transverse loading, EC3 Table 8.5 Toe, parallel loading, EC3 Table 8.2 FAT parallel FAT transverse page 34

FAT Classes FAT Classes Assessment points P5 and P6, structural hot spot stress approach: According to IIW s found by extrapolation Direct stress and FAT class depending on angle of cutting plane Typical value for transverse loading is: Toe: FAT100 Typical values for parallel loading is: Toe: FAT100 Toe, transverse loading, EC3 Table B.1 Toe, parallel loading, EC3 Table 8.2 FAT parallel FAT transverse page 35

Overview Fatigue Assessment with LIMIT According to Eurocode 3 Motivation Stresses and section forces Stress Spectra Stress-relieved welded details in compression FAT classes S-N curves Fatigue analysis Results page 36

S-N Curves S-N-Curve Characteristic values for direct stress FAT class at 2 mio. cycles: Ds C Fatigue strenght 5 mio. cycles: Ds D = (2/5) 1/3 Ds C Cutoff at 100 mio. cycles: Ds L = (5/100) 1/5 Ds D Ds Direct Stress Ds C Ds D Ds L Number of cycles page 37

S-N Curves S-N-Curve Characteristic values for shear stress FAT class at 2 mio. cycles Dt C Cutoff at 100 mio. cycles Ds L = (2/100) 1/5 x Dt C Dt Shear Stress Dt C Dt L Number of cycles page 38

Proof of Fatigue Strength According to Eurocode 3 Two possibilities for proof 1. Damage criteria 2. Calculation of the equivalent constant amplitude stress range for 2 million cycles In case of no damage, the actual margin of safety in terms of stress is not clear! Implementation of the Eurocode 3 in LIMIT Two modes available 1. Direct damage calculation with damage as the result quantity The stress spectra are taken directly for damage calculation 2. Degree of utilization in terms of stress range A load multiplier LF is calculated Stress spectra are multiplied by LF Since the SN-curve may change its shape depending on LF, an iterative procedure is used Iteration ends when damage is equal or changes from lower 1.0 to values higher than 1.0 Fatigue Analysis page 40

Fatigue Analysis Direct Damage Procedure Scaling spectra by: g Ff Reducing SN-curve with: g Mf Ds 1 g Ff Ds Ds 1 g Ff Ds D =(2/5) 1/3 Ds C / g Mf Ds 2 g Ff Ds 3 g Ff Ds 4 g Ff Ds 2 g Ff Ds 3 g Ff Ds 4 g Ff Ds L = (5/100) 1/5 Ds D Number of cycles page 41

Fatigue Analysis Degree of utilization in terms of stress range Search for the margin of safety against a damage above 1.0 with respect to stress range! Damage calculation Scaling spectra by: g Ff and LF Reducing SN-curve with: g Mf Iterative procedure First iteration: LF = 1.0 Last iteration: LF = margin of safety 1/LF = degree of utilization Ds 1 g Ff LF Ds 2 g Ff LF Ds 3 g Ff LF Ds 4 g Ff LF Ds D =(2/5) 1/3 Ds C / g Mf Ds L = (5/100) 1/5 Ds D = 1.0 page 42

Fatigue Analysis Interaction of Direct Stress and Shear Stress Direct Damage In assessment points P1 to P4: Damage due to transverse direct stress and shear stress are added. In assessment points P5 and P6: Always in cutting plane mode only using direct stress in combination with angle dependent FAT-class. No interaction necessary. Degree of utilization in terms of stress range In assessment points P1 to P4: LF is calculated taking interaction into account according to: with: g Ff Ds E,2 = D 1/3 direct Ds C / g Mf g Ff Dt E,2 = D 1/5 shear Dt C / g Mf and D direct Damage from direct stress and D shear Damage from shear stress Interaction becomes again summation of damages for transverse stress and shear: D direct + D shear 1.0 In assessment points P5 and P6: Always in cutting plane mode only using direct stress in combination with angle dependent FAT-class. No interaction necessary. page 43

Results Results In the following section results are compared and documented for different weld types : a.) One sided full penetration weld b.) Double sided fillet weld c.) One sided fillet weld a.) b.) c.) page 45

Results Loads / Spectra Picture shows LoadManager of LIMIT FE Result Defines which time steps should be imported Names the Step 1: Load_1 Loads Is used to generate linear combinations of FE results Loads generated for fatigue analysis:» Force_1 = 4 x Load_1» Force_2 = 3 x Load_1» Force_3 = 2 x Load_1» Force_4 = 1 x Load_1» Zero = 0 x Load_1 Spectra Four blocks (BL1 to BL4) are defined each cycling between Zero and either Force_1,2,3, or 4 Each block has its own cycle number, decreasing with load height page 46

Results Comparison for different geometries Degree of Utilization (DoU) DoU = 0,85 DoU = 0,71 DoU = 4,88 page 47

Results Full penetration weld, details Critical assessment Point P3 (root) FAT classes: Parallel: FAT100 Transverse: FAT36 (conservative setting) Shear: FAT80 Root page 48

Results: Full penetration weld: LIMIT Text output EC3-VALUES: Rp02 = 235.000 gff = 1.00000 gfm = 1.00000, Yield stress, Partial safety factor fatigue load, Partial safety factor fatigue strength FATIGUE STRENGTH and CUTOFF: DSDQ = 26.5320, Fatigue strength, lateral to weld direction DSLQ = 14.5735, Cutoff lateral to weld direction DSLT = 36.5600, Cutoff shear FAT-VALUES (2 Mio Cycles): DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section FAT-VALUES (2 Mio Cycles) / gfm : DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section DIMENSIONS: t = 10.0000, Sheet thickness au = 0.00000, A-dimension, shell bottom side ao = 10.0000, A-dimension, shell top side dau = 0.00000, Root position, shell bottom side dao = -5.00000, Root position, shell top side IEX = 2, Exzentricity: constrained (50%) ALFA = 0.00000, Angle between cutting plane and weld (only valid for weld toes: Positions 5 and 6) ------------------------------------------------ RESULT OF FIRST ITERATION: LF=1 DST... SHEAR STRESS WELD SECTION STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 13.200 13.200 0.10000E+06 2 9.9000 9.9000 0.20000E+06 3 6.6000 6.6000 0.50000E+06 4 3.3000 3.3000 0.10000E+07 DSQ... STRESS IN WELD SECTION (LATERAL) DAMAGE AT DSQ*gff*LF: STEP DSQ*gff DSQ*gff*LF N_D=1 N_given DAM 1 60.824 60.824 0.41469E+06 0.10000E+06 0.24114 2 45.618 45.618 0.98297E+06 0.20000E+06 0.20347 3 30.412 30.412 0.33175E+07 0.50000E+06 0.15072 4 15.206 15.206 0.80865E+08 0.10000E+07 0.12366E-01 ------------------------------------------------ RESULT OF LAST ITERATION: LF= 1.17859 (STRESS RANGES ARE MULTIPLIED WITH MARGIN OF SAFETY) no damage for shear at LF = 1 damage for direct stress at LF = 1 SPECTRUM DATA: BLOCK: 1, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 76.284 38.142 0.10000E+06 1 2 S22 60.824 30.412 0.10000E+06 1 2 S33 0.0000 0.0000 0.0000 0 0 S12 13.200 6.6000 0.10000E+06 1 2 BLOCK: 2, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 57.213 28.607 0.20000E+06 3 2 S22 45.618 22.809 0.20000E+06 3 2 S33 0.0000 0.0000 0.0000 0 0 S12 9.9000 4.9500 0.20000E+06 3 2 BLOCK: 3, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 38.142 19.071 0.50000E+06 4 2 S22 30.412 15.206 0.50000E+06 4 2 S33 0.0000 0.0000 0.0000 0 0 S12 6.6000 3.3000 0.50000E+06 4 2 BLOCK: 4, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 19.071 9.5355 0.10000E+07 5 2 S22 15.206 7.6029 0.10000E+07 5 2 S33 0.0000 0.0000 0.0000 0 0 S12 3.3000 1.6500 0.10000E+07 5 2 DAMAGE AT DST*gff*LF: STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 13.200 15.557 0.10000E+06 2 9.9000 11.668 0.20000E+06 3 6.6000 7.7787 0.50000E+06 4 3.3000 3.8893 0.10000E+07 DAMAGE EQUIVALENT STRESS RANGE: DSTEQ2 = 0.00000, Damage equivalent range DS, 2 mio cycles DAMAGE AT DSQ*gff*LF: 1 60.824 71.686 0.25330E+06 0.10000E+06 0.39479 2 45.618 53.764 0.60041E+06 0.20000E+06 0.33310 3 30.412 35.843 0.20264E+07 0.50000E+06 0.24674 4 15.206 17.921 0.35559E+08 0.10000E+07 0.28122E-01 DAMAGE EQUIVALENT STRESS RANGE: DSQEQ2 = 36.0331, Damage equivalent range DS, 2 mio cycles ------------------------------------------------ DEGREE OF UTILIZATION (INVERSE MARGIN OF SAFETY) ALGQ = 0.848472, Degree of utilization lateral to weld ALGQPL = 0.172549, Degree of utilization lateral to weld, yielding ALGT = 0.848472E-06, Degree of utilization shear ALGTPL = 0.648595E-01, Degree of utilization shear, yielding ALGKMB = 0.848472, Combined degree of utilization no damage for shear at LF = 1.17 damage = 1 for direct stress at LF = 1.17 DoU = 0.85 page 49

Results Double sided fillet weld, details Critical assessment Point P2 (root) FAT classes: Parallel: FAT100 Transverse: FAT36 (conservative setting) Shear: FAT80 Root page 50

Results: Double sided fillet weld: LIMIT Text output EC3-VALUES: Rp02 = 235.000 gff = 1.00000 gfm = 1.00000, Yield stress, Partial safety factor fatigue load, Partial safety factor fatigue strength FATIGUE STRENGTH and CUTOFF: DSDQ = 26.5320, Fatigue strength, lateral to weld direction DSLQ = 14.5735, Cutoff lateral to weld direction DSLT = 36.5600, Cutoff shear FAT-VALUES (2 Mio Cycles): DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section FAT-VALUES (2 Mio Cycles) / gfm : DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section DIMENSIONS: t = 10.0000, Sheet thickness au = 5.00000, A-dimension, shell bottom side ao = 5.00000, A-dimension, shell top side dau = 5.00000, Root position, shell bottom side dao = 5.00000, Root position, shell top side IEX = 2, Exzentricity: constrained (50%) ALFA = 0.00000, Angle between cutting plane and weld (only valid for weld toes: Positions 5 and 6) SPECTRUM DATA: BLOCK: 1, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 76.284 38.142 0.10000E+06 1 2 S22 50.830 25.415 0.10000E+06 1 2 S33 0.0000 0.0000 0.0000 1 2 S12 13.243 6.6217 0.10000E+06 1 2 BLOCK: 2, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 57.213 28.607 0.20000E+06 3 2 S22 38.123 19.061 0.20000E+06 3 2 S33 0.0000 0.0000 0.0000 0 0 S12 9.9326 4.9663 0.20000E+06 3 2 BLOCK: 3, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 38.142 19.071 0.50000E+06 4 2 S22 25.415 12.708 0.50000E+06 4 2 S33 0.0000 0.0000 0.0000 0 0 S12 6.6217 3.3109 0.50000E+06 4 2 BLOCK: 4, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 19.071 9.5355 0.10000E+07 5 2 S22 12.708 6.3538 0.10000E+07 5 2 S33 0.0000 0.0000 0.0000 0 0 S12 3.3109 1.6554 0.10000E+07 5 2 ------------------------------------------------ RESULT OF FIRST ITERATION: LF=1 DST... SHEAR STRESS WELD SECTION STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 13.243 13.243 0.10000E+06 2 9.9326 9.9326 0.20000E+06 3 6.6217 6.6217 0.50000E+06 4 3.3109 3.3109 0.10000E+07 DSQ... STRESS IN WELD SECTION (LATERAL) DAMAGE AT DSQ*gff*LF: STEP DSQ*gff DSQ*gff*LF N_D=1 N_given DAM 1 50.830 50.830 0.71051E+06 0.10000E+06 0.14074 2 38.123 38.123 0.16842E+07 0.20000E+06 0.11875 3 25.415 25.415 0.61994E+07 0.50000E+06 0.80652E-01 4 12.708 12.708 0.10000E+07 ------------------------------------------------ RESULT OF LAST ITERATION: LF= 1.41052 (STRESS RANGES ARE MULTIPLIED WITH MARGIN OF SAFETY) DAMAGE AT DST*gff*LF: STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 13.243 18.680 0.10000E+06 2 9.9326 14.010 0.20000E+06 3 6.6217 9.3401 0.50000E+06 4 3.3109 4.6700 0.10000E+07 DAMAGE EQUIVALENT STRESS RANGE: DSTEQ2 = 0.00000, Damage equivalent range DS, 2 mio cycles DAMAGE AT DSQ*gff*LF: 1 50.830 71.697 0.25318E+06 0.10000E+06 0.39498 2 38.123 53.773 0.60013E+06 0.20000E+06 0.33326 3 25.415 35.849 0.20254E+07 0.50000E+06 0.24686 4 12.708 17.924 0.35531E+08 0.10000E+07 0.28145E-01 DAMAGE EQUIVALENT STRESS RANGE: DSQEQ2 = 36.0389, Damage equivalent range DS, 2 mio cycles ------------------------------------------------ DEGREE OF UTILIZATION (INVERSE MARGIN OF SAFETY) ALGQ = 0.708957, Degree of utilization lateral to weld ALGQPL = 0.144200, Degree of utilization lateral to weld, yielding ALGT = 0.708957E-06, Degree of utilization shear ALGTPL = 0.650730E-01, Degree of utilization shear, yielding ALGKMB = 0.708957, Combined degree of utilization no damage for shear at LF = 1 damage for direct stress at LF = 1 no damage for shear at LF = 1.41 damage = 1 for direct stress at LF = 1.41 DoU = 0.71 page 51

Results One sided fillet weld, details Critical assessment Point P3 (root) FAT classes: Parallel: FAT100 Transverse: FAT36 Shear: FAT80 Root Bending moment due to excentricity was taken into account (half excentricity with FAT36). page 52

Results: One sided fillet weld: LIMIT Text output EC3-VALUES: Rp02 = 235.000 gff = 1.00000 gfm = 1.00000, Yield stress, Partial safety factor fatigue load, Partial safety factor fatigue strength FATIGUE STRENGTH and CUTOFF: DSDQ = 26.5320, Fatigue strength, lateral to weld direction DSLQ = 14.5735, Cutoff lateral to weld direction DSLT = 36.5600, Cutoff shear FAT-VALUES (2 Mio Cycles): DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section FAT-VALUES (2 Mio Cycles) / gfm : DSCL = 100.000, 1. direction resp. parallel to weld DSCQ = 36.0000, 2. direction resp. lateral to weld DSCS = 80.0000, Shear in weld section DIMENSIONS: t = 10.0000, Sheet thickness au = 0.00000, A-dimension, shell bottom side ao = 7.00000, A-dimension, shell top side dau = 0.00000, Root position, shell bottom side dao = 5.00000, Root position, shell top side IEX = 2, Exzentricity: constrained (50%) ALFA = 0.00000, Angle between cutting plane and weld (only valid for weld toes: Positions 5 and 6) ------------------------------------------------ RESULT OF FIRST ITERATION: LF=1 DST... SHEAR STRESS WELD SECTION STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 18.826 18.826 0.10000E+06 2 14.120 14.120 0.20000E+06 3 9.4130 9.4130 0.50000E+06 4 4.7065 4.7065 0.10000E+07 DSQ... STRESS IN WELD SECTION (LATERAL) DAMAGE AT DSQ*gff*LF: STEP DSQ*gff DSQ*gff*LF N_D=1 N_given DAM 1 349.89 349.89 2178.5 0.10000E+06 45.903 2 262.42 262.42 5163.8 0.20000E+06 38.731 3 174.94 174.94 17428. 0.50000E+06 28.690 4 87.472 87.472 0.13942E+06 0.10000E+07 7.1724 ------------------------------------------------ RESULT OF LAST ITERATION: LF=0.205078 (STRESS RANGES ARE MULTIPLIED WITH MARGIN OF SAFETY) no damage for shear at LF = 1 damage for direct stress at LF = 1 SPECTRUM DATA: BLOCK: 1, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 76.284 38.142 0.10000E+06 1 2 S22 349.89 174.94 0.10000E+06 1 2 S33 0.0000 0.0000 0.0000 0 0 S12 18.826 9.4130 0.10000E+06 1 2 BLOCK: 2, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 57.213 28.607 0.20000E+06 3 2 S22 262.42 131.21 0.20000E+06 3 2 S33 0.0000 0.0000 0.0000 0 0 S12 14.120 7.0598 0.20000E+06 3 2 BLOCK: 3, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 38.142 19.071 0.50000E+06 4 2 S22 174.94 87.472 0.50000E+06 4 2 S33 0.0000 0.0000 0.0000 0 0 S12 9.4130 4.7065 0.50000E+06 4 2 BLOCK: 4, PER ROW ONE STRESS COMPONENT (11,22,33,12) COMP., STRESS RANGE, MEAN STRESS, CYCLES, RELEVANT LOAD CASES) S11 19.071 9.5355 0.10000E+07 5 2 S22 87.472 43.736 0.10000E+07 5 2 S33 0.0000 0.0000 0.0000 0 0 S12 4.7065 2.3533 0.10000E+07 5 2 DAMAGE AT DST*gff*LF: STEP DST*gff DST*gff*LF N_D=1 N_given DAM 1 18.826 3.8608 0.10000E+06 2 14.120 2.8956 0.20000E+06 3 9.4130 1.9304 0.50000E+06 4 4.7065 0.96520 0.10000E+07 DAMAGE EQUIVALENT STRESS RANGE: DSTEQ2 = 0.00000, Damage equivalent range DS, 2 mio cycles DAMAGE AT DSQ*gff*LF: 1 349.89 71.754 0.25258E+06 0.10000E+06 0.39592 2 262.42 53.816 0.59871E+06 0.20000E+06 0.33405 3 174.94 35.877 0.20206E+07 0.50000E+06 0.24745 4 87.472 17.939 0.35390E+08 0.10000E+07 0.28256E-01 DAMAGE EQUIVALENT STRESS RANGE: DSQEQ2 = 36.0680, Damage equivalent range DS, 2 mio cycles ------------------------------------------------ DEGREE OF UTILIZATION (INVERSE MARGIN OF SAFETY) ALGQ = 4.87619, Degree of utilization lateral to weld ALGQPL = 0.992586, Degree of utilization lateral to weld, yielding ALGT = 0.487619E-05, Degree of utilization shear ALGTPL = 0.925039E-01, Degree of utilization shear, yielding ALGKMB = 4.87619, Combined degree of utilization no damage for shear at LF = 0.205 damage = 1 for direct stress at LF = 0.205 DoU = 4.88 page 53

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