V Predicted Weldment Fatigue Behavior AM 11/03 1

Transcription

1 V Predicted Weldment Fatigue Behavior AM 11/03 1

2 Outline Heavy and Light Industry weldments The IP model Some predictions of the IP model AM 11/03 2

3 Heavy industry AM 11/03 3

4 Heavy industry AM 11/03 4

5 Complex weldment Web frame Side shell Panel stiffener Torsion from web frame Axial load (bending at bulkhead detail) Longitudinal Web shear Bending In-plane bending and shear fromlateral load (axial load in deck & bottom) AM 11/03 5

6 Light Industry There are about 4000 to 6000 spot welds in the average automobile. For the most part these are executed by welding robots. AM 11/03 6

7 Weld quality The civil engineering view: Everpresent small flaws virtually eliminate fatigue crack initiation life N I of weldments. AM 11/03 7

8 Weld quality and welding procedure Nominal Welding procedure A Cold laps virtually eliminate the fatigue crack initiation life (NI) AM 11/03 Ideal Welding procedure B Such weldments may have an appreciable fatigue crack initiation life (NI) 8

9 Two rather different paradigms Symbol Application Weld Quality Civil Eng. Heavy Industry Nominal One-of-a-kind Very high reliability Low: Flaws < 0.1-in Mechanical Eng. Light Industry Ideal Mass production Moderate reliability High: Flaws > 0.01-in Weldment size Weldment Complexity Big > 1-in High (redundant) Small < 1/2-in Low (simple) AM 11/03 9

10 Outline Heavy and Light Industry weldments The IP model Some predictions of the IP model AM 11/03 10

11 Fatigue of a component The fatigue life of an engineering component consists of two main life periods: Initiation or nucleation of a fatigue crack (NI) And Its growth to failure (NP) A smooth specimen AM 11/03 11

12 Initiation-Propagation (IP) Model N N a =? i N I Total Fatigue Life N P1 Nucleation and "early" growth Invisible Crack Nucleation Short Crack Growth N P2 a smooth specimen? i N P2 Growth of "long" cracks Visible Long Crack Growth N I N T = ( N N + N P1 )+ N P2 = N I + N P 2 Actual IP Model N T = N N + N P1 + N P2 The total fatigue life of a component may be considered to be the sum of three periods: crack nucleation (N N ), short crack growth (N P1 ) in which the crack closure phenomenon is a function of crack length, and long crack propagation (N P2 ) in which the crack closure phenomenon is more or less independent of crack length AM 11/03 12

13 Relative importance of N I and N P Percent Total Life (%) 100% 90% 80% 70% 60% Shear Crack Growth 50% 40% 30% 20% 10% Tensile Crack Growth Crack Nucleation The relative importance of fatigue crack initiation and propagation (SAE 1045 steel after Socie) 0% 1E+02 1E+03 1E+04 1E+05 1E+06 Total Life, Nt (cycles) AM 11/03 13

14 IP Model details Laboratory fatigue tests on smooth specimens Mechanics analysis FEA K K n n K t ε f σ f b c Set-up cycle σ ε σ m σ m K fmax Estimation of N I σ m time Estimation of N T Kfmax N T = N I + N P M k Fatigue notch size effect K t -> K fmax Laboratory tests on precracked specimens C n U a i =? Estimation of N P AM 11/03 14

15 FEM determination of notch root stresses Two weld toes, on incomplete joint penetration AM 11/03 15

16 Determining K t for use in calculating N I Highest stress region of weld toe is 20 away from the point of tangency (Can also get stresses along crack path for N P calculation from this analysis.) AM 11/03 16

17 Estimation of notch-root conditions Experimental determination of K f Analytical determination of K t ²S ²S ²S ²S unnotched component notched component 10 6 cycles notch σ,ε S,e unnotched component notched component notch root plasticity log fatigue life, N (cycles) The severity of a notch in fatigue, that is the magnitude of K f can be measured The effect of notch-size can be estimated using Peterson s equation. K f = 1 + K t 1 Alternatively, for generally elastic body, the local stress can be estimated analytically using Neuber s Rule. ( σ ε = K t S ) 2 E 1 + a r AM 11/03 17

18 Perfect weldments: concept of the worst-case notch 0.5-in. plate thickness 5083 Al weldment Looks pretty smooth with well defined radius, BUT!!!??? AM 11/03 18

19 Perfect weldments: K fmax K t = 1+ α t r Peterson s equation K f = 1 + K t a r K f r = 0, r crit = a K f max = α S u t a a( in. ) S u 2 K f max = α S u t AM 11/03 19

20 Determining K t for use in calculating N I Highest stress region of weld toe is 20 away from the point of tangency (Can also get stresses along crack path for N P calculation from this analysis.) AM 11/03 20

21 Events during the first load application Notch-root residuals after set-up cycle The difference between the monotonic (tensile)and cyclic σ ε behavior causes the notch root mean stresses to be altered during the first load application!!!! S Set-up cycle analysis AM 11/03 21 t

22 Determining?K for use in calculating N P Highest stress region of weld toe is 20 away from the point of tangency Get stresses along crack path for N P calculation from this analysis. AM 11/03 22

23 Determining P P?K(a) for the calculation of N P 2a K σ( x ) = + σ( x ) K' K K' Weight function approach x σ( x ) a σ a K( a) = Y( a) S πa Y( a) = 1 a S 1.12 σ x a f dσ a da dx x f = 1.12 a 0 2 sin 1 x π a AM 11/03 23

24 Estimating N P2 a i =????? K(a) =? U(a) =??? N P2 is estimated using crack growth data for long cracks and the Paris Power Law N P2 = a f a i da C K eff n where: a i = initial crack length, a f = crack length at failure, and C, n = material constants in the Paris Power Law. AM 11/03 24

25 IP Model example: a f =? D x a f? S You have noticed a fatigue crack of length (x) propagating from a sharp elliptical notch of depth (D) and notch root radius (r) in a semiinfinite body (component) subjected to a fluctuating remote stress which ranges from 0 to S max = 10 ksi. The current temperature is 0 F. What will the crack length be at fracture a f = (D + x f )? Applied remote stress (S max, ksi): 10 Applied remote stress range (? S, ksi) 10 Fracture 68 F (K IC, ksiv in ): 70 Fracture 0 F (K IC, ksiv in ): 45 Geometry correction factor for an edge crack (Y): 1.12 Linear-elastic fracture mechanics (LEFM): K IC = YS f (πa f ) 1/2 AM 11/03 25

26 IP Model example: N P =? D x a f? S Calculate the number of cycles to propagate the fatigue crack from an initial length (x i ) of 0.01 in to the length at which it fractures (a f ), that is, estimate the fatigue crack propagation life (N P ). Assume for the moment that the notch is really a crack of depth D, i. e.: a i = D +x i af = D + x f. Applied remote stress (S max, ksi): 10 Applied remote stress range (? S, ksi) 10 Fatigue crack propagation coefficient (C, ksiv in) 1x10-9 Fatigue crack propagation exponent (n) 3 Notch depth (D, in) 0.1 Geometry correction factor for an edge crack (Y): 1.12 Fatigue crack propagation life: N P = 1 C (1 m / 2) (Y S π) m AM 11/03 26 a f 1 m 2 ao 1 m 2

27 r IP Model example: N I =? D K t? S The fatigue crack initiation life (N I ) of the notched component can be calculated considering the elliptical notch to have a fatigue notch factor (K f ) which can be estimated using Peterson's equation which estimates size effect of the notch in fatigue from the elastic stress concentration factor (K t ). 1 / b N I = 1 ' σ f σ m 2 K f S/ 2 Notch depth (D, in.): 0.1 Notch root radius (r, in.): 0.01 Elastic stress concentration factor of an ellipse (K t ): Fatigue notch factor (K f ) - Peterson's Equation: Peterson's constant for this material (a, in.): 0.01 Applied remote stress range (? S, ksi): 10 Applied mean stress (Sm, ksi): 5 Notch root mean stress (σ m, ksi): S m K f Fatigue strength coefficient (σ f, ksi): 110 Fatigue strength exponent (b): D r 1 + K t a r AM 11/03 27

28 IP Model example: N T =? a f (in.) = 5.14 K t = 7.32 K f = 4.16 N I (cycles) = 1,051,000 62% N P (cycles) = 650,000 38% N T = N I + N P (cycles) = 1,701, % AM 11/03 28

29 Outline Heavy and Light Industry weldments The IP model Some predictions of the IP model AM 11/03 29

30 IP model results for a cruciform joint 100 S (ksi) Sr = 36 ksi MK - British Standards T = in. R = 0 10 Failures Runouts Propagation Life (ai = 0.01 in.) Total Life Fatigue Life, N (cycles) AM 11/03 30

31 IP Model predictions and experimental results 100 S (ksi.) 10 N P N T Failures R = 0.1 Runouts Fatigue Life, N (cycles) AM 11/03 31

32 IP model predictions for a cruciform weldment IP Restraint model predicts notch-root (local) stresses and strains r Sr ²Sb ²Sa Sfab ²M ²M ²S T + + ²S θ Remote Weld Done Last Mean stresses considered to be composed of applied mean (R ratio), fabrication Non Load-Carrying residual stresses Cruciform (S fab Joint )and welding residual stresses (Sr). AM 11/03 32

33 Residual stress effects - 1/2-in S Remote stress-time history Local (weld-toe) stressstrain response t Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 0 ksi RS.Fabric = 0 ksi N t = 17,170,000 N i = 17,000,000 N p = 170,000 AM 11/03 33

34 Residual stress effects - 1/2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 2,400,000 N i = 2,200,000 N p = 170,000 AM 11/03 34

35 Residual stress effects - 1/2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 36 ksi RS.Fabric = 36 ksi N t = 980,000 N i = 930,000 N p = 500,000 AM 11/03 35

36 Residual stress effects - 2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 2.0-in. K feff = 3.42 RS.Weld = 0 ksi RS.Fabric = 0 ksi N t = 360,000 N i = 260,000 N p = 100,000 AM 11/03 36

37 Residual stress effects - 2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 2.0-in. K feff = 3.4 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 340,000 N i = 240,000 N p = 100,000 AM 11/03 37

38 Residual stress effects - 2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 2.0-in. K feff = 3.4 RS.Weld = 36 ksi RS.Fabric = 36 ksi N t = 200,000 N i = 170,000 N p = 30,000 AM 11/03 38

39 Mean stress - 1/2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 2,400,000 N i = 2,200,000 N p = 170,000 AM 11/03 39

40 Mean stress - 1/2-in Mild steel Cruciform? S = 25 ksi Axial R = -1 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 7,800,000 N i = 7,700,000 N p = 170,000 AM 11/03 40

41 Mean stress - 2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 2.0-in. K feff = 3.4 RS.Weld = 36 ksi RS.Fabric = 36 ksi N t = 200,000 N i = 170,000 N p = 30,000 AM 11/03 41

42 Mean stress - 2-in Mild steel Cruciform? S = 25 ksi Axial R = -1 Flank angle = 40 Thickness = 2.0-in. K feff = 3.4 RS.Weld = 36 ksi RS.Fabric = 36 ksi N t = 210,000 N i = 180,000 N p = 32,000 AM 11/03 42

43 Size effects - 1/2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 0.5-in. K feff = 2.2 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 2,400,000 N i = 2,200,000 N p = 170,000 AM 11/03 43

44 Size effects - 2-in Mild steel Cruciform? S = 25 ksi Axial R = 0 Flank angle = 40 Thickness = 2.0-in. K feff = 3.4 RS.Weld = 36 ksi RS.Fabric = 0 ksi N t = 340,000 N i = 240,000 N p = 100,000 AM 11/03 44

45 Range of predicted behavior at 25 ksi. 100 Weld Category A B 10 C D E F Fatigue Life, N (cycles) AM 11/03 45

46 Predicted effect of weld quality Ideal: S r = 36, S fab = 36 Nominal: S fab = Fatigue Life, N (cycles) AM 11/03 46

47 Predicted effect of mean, fabrication, and residual stresses Ideal: S r = 0, S fab = 0 Ideal: S r = 36, S fab = 0 Ideal: S r = 36, S fab = 36 Nominal: S fab = 0 Nominal: S fab = Fatigue Life, N (cycles) AM 11/03 47

48 Predicted size effect /2-1/7 Ideal, Sfab=36 Ideal, Sfab=0 Nominal, Sfab=36 Nominal, Sfab=0-1/3 Ideal weldments are predicted to be more affected by the size effect than Nominal weldments Thickness, T (in.) AM 11/03 48

49 Predicted effect of S ubm Fatigue Strength at 1E+07 cycles, Sa (ksi) Hot Rolled Steel Q & T Steel Over Stressed Stress Relieved As Welded, S fab = 0 As Welded, S fab = Sy BM Trends in Ideal 1.0- in plate thickness, non-load carrying cruciform weldments fatigue strength. R = 0 Welding residual stresses = 50% of SYBM S fab SYBM Base Metal Ultimate Strength, Su BM (ksi) AM 11/03 49

50 Predicted behavior of light industry weldments. 100 Remote Stress Range, DS (ksi) 10 Ideal Nominal Toe Ground (radius = 0.1 in.) Over Stressed Stress Relieved Weld Profile (flank angle 20 o ) As Welded t = 0.5-in. (12mm); R = 0 Without Fabrication Stresses Fatigue Life, N T (cycles) AM 11/03 50

51 Predicted behavior of heavy industry weldments. 100 t = 2.0-in. (50 mm); R = 0 Remote Stress Range, DS (ksi) 10 Ideal Nominal Toe Ground (radius = 0.1 in.) Over Stressed Stress Relieved Weld Profile (flank angle 20 o ) As Welded With Fabrication Stresses AM 11/03 Fatigue Life, N T (cycles) 51

52 Modeling results Symbol Application Welding Residual Stresses Mean and Fabrication Residual Stresses Weld toe geometry Civil Eng. Heavy Industry Nominal Unimportant Important Unimportant Mechanical Eng. Light Industry Ideal Very important Very important Very important AM 11/03 52

53 Summary While not perfect, the IP model predicts the fatigue behavior of ideal weldment and thus suggests the upper bound of weldment fatigue behavior. The role of residual stresses is subtle and depends upon weldment size and weld quality. AM 11/03 53