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Ch5 epfm

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Ch5 epfm

  1. 1. Elastic-Plastic Fracture Mechanics Introduction •When does one need to use LEFM and EPFM? •What is the concept of small-scale and large-scale yielding? Contents of this Chapter •The basics of the two criteria used in EPFM: COD (CTOD), and J-Integral (with H-R-R) •Concept of K- and J-dominated regions, plastic zones •Measurement methods of COD and J-integral •Effect of Geometry Background Knowledge •Theory of Plasticity (Yield criteria, Hardening rules) •Concept of K, G and K-dominated regions •Plastic zone size due to Irwin and Dugdal
  2. 2. LEFM and EPFM LEFM •In LEFM, the crack tip stress and displacement field can be uniquely characterized by K, the stress intensity factor. It is neither the magnitude of stress or strain, but a unique parameter that describes the effect of loading at the crack tip region and the resistance of the material. K filed is valid for a small region around the crack tip. It depends on both the values of stress and crack size. We noted that when a far field stress acts on an edge crack of width “a” then for mode I, plane strain case σ σ τ π θ θ θ θ θ θ θ xx yy xy IK r R S| T| U V| W| = − + L N MMMMMMM O Q PPPPPPP 2 2 1 2 3 2 1 2 3 2 2 3 2 cos sin( ) sin( ) sin( ) sin( ) sin( ) sin( ) σ σ ν σ σzz zz xx yy= = +0 for plane stress; for plane strain( ) u u K r k k x y I RST UVW = − + + − L N MMMM O Q PPPP 2 2 1 2 2 2 1 2 2 2 2µ π θ θ θ θ2 cos ( sin ( )) sin ( cos ( ))
  3. 3. LEFM concepts are valid if the plastic zone is much smaller than the singularity zones. Irwin estimates Dugdale strip yield model: r K p I ys = 1 2 2 π σ ( ) r K p I ys = 1 8 2 ( ) σ ASTM: a,B, W-a 2.5 , i.e. of specimen dimension.≥ ( ) KI ysσ 2 rp ≤ 1 50 LEFM cont. Singularity dominated region σ σ τ π xx yy xy IK r R S| T| U V| W| = L N MMM O Q PPP2 1 1 0 For =0θ For = 2 all ijθ θ σ, =0
  4. 4. EPFM •In EPFM, the crack tip undergoes significant plasticity as seen in the following diagram. s h a r p t i p Ideal elastic brittle behavior cleavage fracture P: Applied load P : Yield loady Displacement, u Loadratio,P/Py 1.0 Fracture Blunt tip Limited plasticity at crack tip, still cleavage fracture Displacement, u Loadratio,P/Py 1.0 Fracture
  5. 5. Blunt tip Void formation & coalescence failure due to fibrous tearing Displacement, u Loadratio,P/Py 1.0 Fracture large scale blunting Large scale plasticity fibrous rapture/ductile failure Displacement, u Loadratio,P/Py 1.0 Fracture
  6. 6. EPFM cont. •EPFM applies to elastoc-rate-independent materials, generally in the large-scale plastic deformation. • Two parameters are generally used: (a)Crack opening displacement (COD) or crack tip opening displacement (CTOD). (b) J-integral. •Both these parameters give geometry independent measure of fracture toughness. δSharp crack Blunting crack y x Γ ds
  7. 7. EPFM cont. •Wells discovered that Kic measurements in structural steels required very large thicknesses for LEFM condition. --- Crack face moved away prior to fracture. --- Plastic deformation blunted the sharp crack. δ Sharp crack Blunting crack • Irwin showed that crack tip plasticity makes the crack behave as if it were longer, say from size a to a + rp -----plane stress From Table 2.2, Set , r K p I ys = 1 2 2 πσ( ) u K r ky I = + − 2 2 2 1 2 2 2 µ π θ θ sin( )[ cos ( )] θ π= u k K r y I y = +1 2 2µ π a ry+ θ π= δ πσ = =2 4 2 2 u K E y I ys Note: since k E= − + = + 3 1 2 1 ν ν µ νand ( ) δ π σ = =CTOD 4 G ys G K E I = 2
  8. 8. CTOD and strain-energy release rate • Equation relates CTOD ( ) to G for small-scale yielding. Wells proved that Can valid even for large scale yielding, and is later shown to be related to J. • can also be analyzed using Dugdales strip yield model. If “ ” is the opening at the end of the strip. δ π σ = =CTOD 4 G ys δ δ δ δ δ σ ys Consider an infinite plate with a image crack subject to a Expanding in an infinite series, σ σ∞ = δ σ π π σ σ = = ⋅2 8 u a E y ys ys lin sec( 2 ) δ σ π π σ σ π σ σ = ⋅ + ⋅ + 8 1 12 2 4ys ys ys a E [ 1 2 ( 2 ( 2 ) ) ...] If , and can be given as: In general, δ σ π σ σ = + ⋅ K E I ys ys 2 2 [1 1 6 ( 2 ) ] σ σ σ σ δ σ σys ys ysE G ≈ << =0 ( then = KI 2 ys ), δ δ σ = G m ys , m = 1.0 for plane stress; m = 2.0 for plane strain
  9. 9. Alternative definition of CTOD δSharp crack Blunting crack δBlunting crack Displacement at the original crack tip Displacement at 900 line intersection, suggested by Rice CTOD measurement using three-point bend specimen W P a rp(W-a) z Vp δ p ' ' 'δpl p p p r W a V r W a a z = − − + + ( ) ( ) displacement expandingδ
  10. 10. Elastic-plastic analysis of three-point bend specimen δ δ δ σ = + = + − − + + el pl I ys p p p K m E r W a V r W a a z 2 ( ) ( ) Where is rotational factor, which equates 0.44 for SENT specimen.δ pl • Specified by ASTM E1290-89 --- can be done by both compact tension, and SENT specimen • Cross section can be rectangular or W=2B; square W=B KI is given by δ ν σel I ys K E = −2 2 1 2 ( ) K P B W f a W I = ⋅ ( ) δpl p p p r W a V r W a a z = − − + + ( ) ( ) load Mouth opening υ p υe V,P
  11. 11. CTOD analysis using ASTM standards Figure (a). Fracture mechanism is purely cleavage, and critical CTOD <0.2mm, stable crack growth, (lower transition). Figure (b). --- CTOD corresponding to initiation of stable crack growth. --- Stable crack growth prior to fracture.(upper transition of fracture steels). Figure (c) and then ---CTOD at the maximum load plateau (case of raising R-curve). δ c δ i δ i δ m δ u load Mouth opening Pc fracture (a) (b) (c) Pi Pu Pm fracture Pi
  12. 12. More on CTOD The derivative is based on Dugdale’s strip yield model. For Strain hardening materials, based on HRR singular field. By setting =0 and n the strain hardening index based on *Definition of COD is arbitrary since A function as the tip is approached *Based on another definition, COD is the distance between upper and lower crack faces between two 45o lines from the tip. With this Definition 2 orI COD y y K J E δ σ σ = = ( ) 11 1 , n n n i y i y y n J u r u n I αε θ ασ ε + +   =  ÷   1 3 2 n y ije y y y ε σσα ε σ σ −   =  ÷   ( ) ( ),0 ,0y yu x u xδ + − = − ( ) 1 1nx +− COD n y J dδ σ =
  13. 13. Where ranging from 0.3 to 0.8 as n is varied from 3 to 13 (Shih, 1981) *Condition of quasi-static fracture can be stated as the Reaches a critical value . The major advantage is that this provides the missing length scale in relating microscopic failure processes to macroscopic fracture toughness. *In fatigue loading, continues to vary with load and is a function of: (a) Load variation (b) Roughness of fracture surface (mechanisms related) (c) Corrosion (d) Failure of nearby zones altering the local stiffness response ( ), ,n n yd d nα ε= tipδ CODδ 2 2 I y k δ σ ε ∆ ∆ =
  14. 14. 3.2 J-contour Integral • By idealizing elastic-plastic deformation as non-linear elastic, Rice proposed J-integral, for regions beyond LEFM. • In loading path elastic-plastic can be modeled as non-linear elastic but not in unloading part. • Also J-integral uses deformation plasticity. It states that the stress state can be determined knowing the initial and final configuration. The plastic strain is loading-path independent. True in proportional load, i.e. • under the above conditions, J-integral characterizes the crack tip stress and crack tip strain and energy release rate uniquely. • J-integral is numerically equivalent to G for linear elastic material. It is a path-independent integral. • When the above conditions are not satisfied, J becomes path dependent and does not relates to any physical quantities d d d d d d k σ σ σ σ σ σ σ σ σ σ σ σ 1 1 2 2 3 3 4 4 5 5 6 6 = = = = = =
  15. 15. 3.2 J-contour Integral, cont. y x Γ ds Consider an arbitrary path ( ) around the crack tip. J-integral is defined asΓ J wdy T u x ds w di i i ij ij ij = − ∂ ∂ = zz( ), σ ε ε 0 Γ It can be shown that J is path independent and represents energy release rate for a material where is a monotonically increasing with σ ij εij Proof: Consider a closed contour: Using divergence theorem: J wdy T u x dsi i i * ( ) * = − ∂ ∂zΓ J w x x u x dxdy i ij i A * ( ) * = ∂ ∂ − ∂ ∂ ∂ ∂z σ where w is strain energy density, Ti is component of traction vector normal to contour. A* Γ*
  16. 16. * * ( )i ij j A uw J dxdy x x x σ  ∂∂ ∂ = −  ∂ ∂ ∂   ∫ Evaluate ∂ ∂ = ∂ ∂ ⋅ ∂ ∂ = ∂ ∂ w x w x xij ij ij ij ε ε σ ε Note is only valid if such a potential function exists Again, σ ε ij ij w = ∂ ∂ ∂ ∂ = ∂ ∂ + ∂ ∂ = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ w x x u x u x u x x u x ij i j i j ij j i i j 1 2 1 2 σ σ [ ( ) ( )] [ ( ) ( )] , , Since σ σ σ ij ji ij j i x u x = = ∂ ∂ ∂ ∂ ( ) Recall ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ σ σ σ ij j j i j ij i x x u x x u x 0 (equilibrium) leads to ij ( ) ( ) Evaluation of J Integral ---1 w (equilibrium) leads to ,j i
  17. 17. Hence, Thus for any closed contourJ* .= 0 J* .= 0 Now consider Γ1 Γ2 Γ3 Γ4 1 2 3 4 0J J J J J= + + + = Recall J wdy w x dsi * ( )= − ∂ ∂z τ Γ On crack face, (no traction and y-displacement), thus , leaving behind Thus any counter-clockwise path around the crack tip will yield J; J is path independent. τ i dy= =0 0, J J3 4 0= = J J1 2= − Evaluation of J Integral ---2 1 2 3 4 ti ti
  18. 18. Γ' a y x 2D body bounded by Γ' In the absence of body force, potential energyΠ Π Γ = −z zwdA u ds A i i ' '' τ Suppose the crack has a vertical extension, then d da dw da dA du da ds A i iΠ Γ = −z z' ' τ (1) Note the integration is now over Γ' Evaluation of J Integral ---3 ti ti
  19. 19. Noting that d da a x a x a x x a = ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ − ∂ ∂ ∂ ∂ = −since 1 d da w a w x dA w a u x ds A i iΠ Γ = ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂z z( ) ( ) ' ' τ (2) ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ w a w a x u aij i ij j i ε ε σ ( ) Using principle of virtual work, for equilibrium, then from eq.(1), we have d da Π = 0 σ τij j A i i i x u a dA u a ds ∂ ∂ ∂ ∂ = ∂ ∂z z( ) ' ' Γ Thus, d da du dx ds dw dx dAi i A Π Γ = −z zτ ' ' Using divergence theorem and multiplying by -1 − = − = − ∂ ∂z zd da wn du dx ds wdy w x dsx i i i Π Γ Γ ( ) ' ' τ τ Evaluation of J Integral ---4 j titi ti ti ti ⋅⋅
  20. 20. Therefore, J is energy release rate , for linear or non-linear elastic material d da Π In general Potential energy; U=strain energy stored; F=work done by external force and A is the crack area. Π = − = − ∂Π ∂ U F J A and Π = a u p Evaluation of J Integral ---5 -dP *dU dU= − d∆ Displacement *U P UΠ = − ∆ = = Complementary strain energy = dP∆∫0 p Load
  21. 21. For Load Control For Displacement Control The Difference in the two cases is and hence J for both load Displacement controls are same * p dU J da = dU J da ∆ = − 0 0 0 . . . pD p p J dp dp a a or p J pd d a a ∆ ∆ ∆ ∂ ∂∆ = ∆ = ∂ ∂ ∂ ∂ = − ∆ = − ∆ ∂ ∂ ∫ ∫ ∫ ∫ J=G and is more general description of energy release rate 2 ' IK J E = Evaluation of J-Integral
  22. 22. More on J Dominance J integral provides a unique measure of the strength of the singular fields in nonlinear fracture. However there are a few important Limitations, (Hutchinson, 1993) (1) Deformation theory of plasticity should be valid with small strain behavior with monotonic loading (2) If finite strain effects dominate and microscopic failures occur, then this region should be much smaller compared to J dominated region Again based on the HRR singularity ( ) 1 1 , n I ijij y y y n J n I r σ σ σ θ ασ σ +  =  ÷   Based on the condition (2), we would like to evaluate the inner radius ro of J dominance. Let R be the radius where the J solutions are satisfied within 10% of complete solution. FEM shows that R o r 3o CODr δ;
  23. 23. •However we need ro should be greater than the forces zone (e.g. grain size in intergranular fracture, mean spacing of voids) •Numerical simulations show that HRR singular solutions hold good for about 20-25% of plastic zone in mode I under SSY • Hence we need a large crack size (a/w >0.5) . Then finite strain region is , minimum ligament size for valis JIC is • For J Controlled growth elastic unloading/non proportional loading should be well within the region of J dominance • Note that near tip strain distribution for a growing crack has a logarithmic singularity which is weaker then 1/r singularity for a 3 CODδ 25 IC y J b σ = and a R dJ J da R ≥ ∆ =
  24. 24. Williams solution to fracture problem Williams in 1957 proposed Airy’s stress function As a solution to the biharmonic equation For the crack problem the boundary conditions are Note will have singularity at the crack tip but is single valued Note that both p and q satisfy Laplace equations such that ( ) ( )R rψ θ θ= 2 2 4 2 2 2 2 1 1 0 where r r r r ψ θ ∂ ∂ ∂ ∇ = ∇ = + + ∂ ∂ ∂ 0 forrθθ θσ σ θ π= = = ± ψ ( ) ( )2 , ,r p r q rψ θ θ= + 2 2 0p q∇ = ∇ =
  25. 25. Now, for the present problem. ( ) ( ) ( )[ ] ( )[ ] ( )[ ] ( ) ( ) ( )[ ] 1 2 2 1 2 1 1 2 2 2 2 1 1 2 1 12 cos sin cos 2 sin 2 Then cos cos 2 sin sin 2 Consider only mode I solution with cos cos 2 1 2 cos cos 2 z z r p A r A r q B r A r r A B r A B r A B r A B r r λ λ λ λ λ λ λ λ θθ θ λθ λθ λ θ λ θ ψ λθ λ θ λθ λ θ ψ λθ λ θ ψ σ λ λ λθ λ θ σ + + + + + = + = + + + = + + + + + = + + ∂ = = + + + + ∂ ∂ = − ∂ ( ) ( ) ( )[ ]1 1 1 1 sin 2 sin 2 r r A Bλ ψ θ λ λ λθ λ λ θ ∂   ÷ ∂  = + + + +
  26. 26. Williams Singularity…3 Applying boundary conditions, ( ) ( ) 1 1 1 1 A +B cos 0 2 sin 0A B λπ λ λ λπ = + + =   Case (i) cos 0λπ = 2 1 , Z=0,1,2... 2 Z λ + = 1 1B 2 A λ λ = − + or, sin 0λπ = Zλ = 1 1B A= − Case (ii) Since the problem is linear, any linear combination of the above two will also be acceptable. Thus Though all values are mathematically fine, from the physics point of view, since 2 with Z= ... 3, 2, 1,0,1,2,3...Z λ = − − − − andij ijr rλ λ σ ε∝ ∝
  27. 27. Williams Singularity…4 ( ) 0 0 21 2 2 1 20 2A 2 1 0 = ij ij R ij ij r R r U r rdrd r drd λ π λ σ ε σ ε θ θ+ = ∝ ∫ ∫ ∫ ∫; Since U should be provided for any annular rising behavior and R ,0r 0 ˆas 0, 1 ( 1 makes 0)ijU r λ σ< ∞ → > − = − = 1 1 31 1 Z 2 2 2 2 1 12 3 , needs > 1. Thus =- ,0, ,1, ,2... with = Z=-1,0, positive number. The most dominant singular form =- and B r i A Also u r where λ λ λ λ + ∝ − =
  28. 28. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 5 2 1 1 2 2 31 1 2 3 2 2 0 1 Now cos cos + ... ... and where indicates the order of I ij ij ij ij r A r r A r r r θ θ σ σ θ− Φ = +   Φ + Φ = + Φ + Φ Φ % Williams Singularity…4 ( ) ( ) 0 1 Note the second term in is a non-singular and non-vanishing term. However, higher order vanish as r 0 with 2 (no sum on x) 2 ij ij I II ij ij ix jx r K A K T r σ π σ σ θ δ δ π = Φ → = = +%
  29. 29. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 5 2 1 1 2 2 31 1 2 3 2 2 0 1 Now cos cos ... ... and where indicates the order of I ij ij ij ij r A r r A r r r θ θφ φ φ σ σ θ φ φ φ − = +   + + = + +% Williams Singularity…5 ( ) ( ) 0 1 Note the second term in is a non-singular and non-vanishing term. However, higher order vanish as r 0 with 2 (no sum on x) 2 ij ij I II ij ij ix jx r K A K T r σ π σ σ θ δ δ π = Φ → = = +%
  30. 30. Williams Singularity…6 ( ) ( ) ( ) ( ) For in-plane stress components, 0 0 02 I I xx xy xx xyI I I yx yy yx yy TK r σ σ σ θ σ θ σ σ π σ θ σ θ      = +           % % % % I Second-term is generally termed as "T-stress" or "T-tensor" with For brittle crack of length 2a in x-z plane with & applied K and xx yy xx yy T a σ σ σ σ π ∞ ∞ ∞ = = T= yy xxσ σ∞ ∞ − x y 2a z
  31. 31. HRR Singularity…1 0 0 0 Hutchinson, Rice and Rosenbren have evaluated the character of crack tip in power-law hardening materials. Suppose the material is represented by Ramberg-Osgood model, ε σ σ α ε σ σ  = +  0 0 0 Reference value of stress=yield strength , strain at yield E dimensionless constant strain-hardening exponent n n σ σ ε α     − − − − 1 Note if elastic strains are negligible, then ˆ3 3 ˆ; 2 2 n y y n ij eq ij eq ij y ij y ε σ α ε σ ε σ σα σ σ ε σ σ −   =        = =    
  32. 32. HRR Singularity…2 ( ) ( ) ( ) ( ) 4 0 1 2 0 Then , , , , (similar to Williams expression)s t s f r n C r r k r φ φ σ α φ θ ρ θ φ σ φ θ ∇ + = + = ⋅ ⋅ % ( ) ( ) 1 1 0 2 0 1 0 2 0 Applying the appropriate boundary conditions , , Integration const n ij ij n n n ij ij n n EJ n I r EJ n E I r I σ σ σ θ ασ σ ε ε θ ασ + +   =   ⋅    =   ⋅  − % % ant , Dimensionless functions of n andσ ε θ−%%

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