CH-4 Plane problems in linear isotropic elasticity HUMBERT Laurent laurent.humbert@epfl.ch laurent.humbert@ecp.fr Thursday, march 18th 010 Thursday, march 5th 010 1
4.1 ntroduction Framework : linear isotropic elasticity, small strains assumptions, D problems (plane strain, plane stress) The basic equations of elasticity (appendix ) : - Equilibrium equations (3 scalar equations) : div + f = 0 ij, j + fi = 0 f : body forces (given) : Cauchy (second order) stress tensor T = ij = ji xx yx zx + + + fx = 0 x y z Explicitly, yx yy yz + + + fy = 0 3 x y z R zx zy zz + + + fz = 0 x y z symmetric y, x e z, x 3 e 3 e 1 x, x 1 Cartesian basis
- Linearized Strain-displacement relations (6 scalar equations) : 1 ε = u+ u ε ij ( T ( ) ) 1 ui = + xj u x i j Equations of compatibility: ε ε ε ε ε ε ε 11 1 3 13 1 11 + =, + + = x x1 x1 x x1 x1 x x3 x x3 ε ε ε ε ε ε ε 33 3 3 13 1 + =, + = x3 x x x3 x x1 x x3 x3 x1 ε ε ε ε 33 11 1 3 13 1 33 + =, + = x1 x3 x3 x1 x3 x1 x x3 x1 x ε ε ε because the deformations are defined as partial derivative of the displacements u 1,u,u 3
- Hooke s law (6 scalar equations) : sotropic homogeneous stress-strain relation = λ trε + μ ε = λε δ + με ij kk ij ij nversely, ν 1+ ν ε= tr + E E ν 1+ ν εij = kk δij + ij E E 15 unknowns ui, εij, ij well-posed problem find u λ, μ Lamé s constants ( 3 + ) λ ν = λ+ μ ( λ + μ ) E = μ λ μ with E > 0 1 ν 1 4
Young s modulus for various materials : 5
1D interpretation : Before elongation 1 3 traction = Eε 33 33 ε11 = ε = νε33 0 0 0 [ ] 0 0 0 but [] ε = 0 0 33 ν 33 E 0 0 = 0 ν33 E 0 0 0 33 E 6
- Navier s equations: (λ+μ)u k,ki +μu i,jj+f i =0 3 displacement components taken as unknowns u i - Stress compatibility equations of Beltrami - Michell : 1 ν ij,kk + mm,ij + fi, j + fj,i + fn,nδ ij = 0 1+ ν 1 ν ij 6 stress components considered as unknowns 7
Boundary conditions : Displacements imposed on S u u = u Surface tractions applied on S t t = n = t t f t o ω n outwards unit normal to S t u ω = S t S u S t S = u displacement and/or traction boundary conditions to solve the previous field equations 8
4. Conditions of plane strain Assume that u3 = 0 ε ε Strain components : 11 u = x u = x 1 1 1 u u ε 1 = + x x 1 1 thick plate Thus, [ ε] ε ε ε 11 1 ε 0 0 = 1 0 0 0 functions of x 1 and x only 9
Associated stress components = E 11 11 ( 1-ν) ν 1 ν 1-ν ε + ε ( + )( ) = E ( 1-ν) ν 11 1 ν 1-ν ε + ε ( + )( ) [ ] 0 11 1 = 1 0 0 0 33 E = ε 1 ν and also (!), + ( )( ) ( ) 33 = ε 11 +ε 1 1 νe 1+ ν 1-ν nverse relations, 1 + ν ε 11 = ( 1 ν) 11 ν E 1 + ν ε = ( 1 ν) ν 11 E 1 + ν ε = E 1 1 rewritten as 1 ε = E* 1 ε = ν * E* [ ν * ] 11 11 [ ] 11 E E* =, ν * = ν 1 ν 1 ν 10
- D static equilibrium equations : n t x x + + f1 = 0 x 11 1 1 + + f = 0 x 1 1 f ( x,x 1 ) : body forces Surface forces t are also functions of x 1 and x only t1 = 11n1+1n t = 1n1+n n,n 1 : components of the unit outwards vector n 11
- Non-zero equation of compatibility (under plane strain assumption) ε ε ε 11 1 + = x x1 x1 x implies for the relationship for the stresses: That reduces to by neglecting the body forces, Proof? 1
Proof: - ntroduce the previous strain expressions in the compatibility equation 1+ ν ε = ( 1 ν) ν E 11 11 1+ ν ε = ( 1 ν) ν E 11 one obtains 1 + ν ε = E 1 1 1 ( 1 ν) 11 ν + ( 1 ν) ν 11 = 1 1 x x x x (1) - Differentiate the equilibrium equation and add 11 1 + + f1 = 0 x1 x1 x 1 + + f = 0 x x1 x () - ntroduce () in (1) and simplify 13
4.3 Conditions of plane stress Condition : 33 = 13 = 3 = 0 thin plate From Hooke s law, 1 ε = E ( ν ) 11 11 1 ε = E ( ν ) 11 1+ ν ε = E 1 1 Similar equations obtained in plane strain : E* E ν* ν and also, ν ε 33 = ( 11 + ), ε 13 =ε 3 = 0 E functions of x 1 and x only 14
nverse relations, E = ε + ε 1 ν [ ν ] 11 11 E = ε + ε 1 ν [ ν ] 11 E = ε 1+ ν 1 1 [ ] 11 1 0 0 = 1 0 0 0 and the normal out of plane strain, ν ε = ε +ε 1 ν ( ) 33 11 15
ψ Airy s stress function ψ : introduce the function ψ as ψ 11 = x ψ = x 1 ψ 1 = x1 x equations of equilibrium automatically satisfied! then substitute in leads to 4 4 4 ψ ψ ψ + + = 0 x x x x 4 4 1 1 4 ψ=0 Biharmonic equation Same differential equation for plane stress and plane strain problems Find Airy s function that satisfies the boundary conditions of the elastic problem 16
4.4 Local stress field in a cracked plate : - Solution D derived by Williams (1957) - Based on the Airy s stress function Notch / crack tip r, θ : polar coordinate system Crack when θ=±π, notch otherwise 17
Local boundary conditions : θθ = rθ = 0 for θ=± α, r > 0 Remote boundary conditions Find ( r, θ ) or ( x,x ) 1 ( r, θ ) or ( x,x ) stress field u u 1 displacement field Concept of self-similarity of the stress field (appendix ) : Stress field remains similar to itself when a change in the intensity (and scale) is imposed Stress function in the form λ ( ) ( ) ψ r,θ = r φ θ, λ> 0 18
Biharmonic equation in cylindrical coordinates: 1 1 d φ λ+ ( ) + + r 0 λ φ+ = r r r r θ dθ 4 d φ d φ + 4 ( λ+ ) +λ + λ ( λ+ ) φ = 0 (1) dθ dθ Consider the form of solution ( ) a θ φ θ = e,a = cst () 1 m ( ) m λ ( ) + λ + + λ + λ+ = 0 with m = a Solutions of the quadratic equation : m = λ 1 m = ( λ+ ) a 1, =± λ =± i λ ( ) i( ) a3,4 =± λ+ =± λ+ complex conjugate roots 19
Consequently, : ( ) ( ) ( ) λθi λθi λ+ θi λ+ θi 1e e 3e 4e φθ= c + c + c + c c i (complex) constants Using Euler s formula ± i e λθ = cosλθ± isinλθ φ( θ ) = Acosλθ+ Bsinλθ+ Ccos( λ+ ) θ+ Dsin( λ+ ) θ A, B, C and D constants to be determined according to the symmetry properties of the problem! 0
Modes of fracture : A crack may be subjected to three modes More dangerous! Notch, crack Example : Compact Tension (CT) specimen : F natural crack Mode loading F F F 1
Mode loading ( ) ψ= r λ φs θ with φs ( θ ) = Acosλθ+ Ccos( λ+ ) θ symmetric part of φ( θ) Stress components in cylindrical coordinates 1 ψ 1 = + r r r ψ rr θ θθ ψ = r 1 ψ 1 ψ r θ = r θ r θ r Use of boundary conditions, ( ) r, θ = 0 θθ θ=±α φs ( α ) = 0 ( ) rθ r, θ = 0 θ=±α dφ dθ ( ) s α = 0 ( ) ( ) ( ) Acosλα + C cos λ + α = 0 A λ sin λα C λ+ sin λ+ α = 0
Non trivial solution exits for A, C if ( ) ( ) ( ) cosλα cos λ + α λsin λα λ+ sin λ+ α = 0 ( λ+ ) cosλα sin( λ+ ) α λ cos( λ+ ) α sinλα= 0 or tan α= sin λα λ+ cos λα that determines the unknowns λ For a crack, it only remains sin λ π= 0 n λ= n integer infinite number of solutions 3
Relationship between A and C For each value of n relationship between the coefficients A and C infinite number of coefficients that are written: A, C n =...,, 1, 0, 1,,... From, Aλsinλα C( λ+ ) sin( λ+ ) α = 0 n n with λ = n α =±π (crack) n π n π An sin n Cn + sin n + π = 0 π n n sin n An + Cn + = 0 0or± 1 A n = C n 4+ n n 4
Airy s function for the (mode ) problem expressed by: n n n ψ= r Ancos θ+ Cncos + θ n Reporting n C n = A n 4 + n n nθ n n ψ (r,θ) = Anr cos cos + θ n n+ 4 5
Expressions of the stress components in series form (eqs 4.3): Starting with, n nθ n n ψ (r,θ) = Anr cos cos + θ n n+ 4 and recalling that, 1 ψ 1 = + r r r ψ rr θ n ψ n nθ n n = Ar n sin + sin + θ θ n 1 ψ + n nθ n n = Ar n cos + cos + θ r θ n ( ) n ( n+ 4) 1 ψ ( + n ) n nθ n n = Ar n cos cos + θ r r n n+ 4 rr = + + + + + n n+ 4 (+ n ) n n nθ n n n n Anr 1 cos cos θ 6
From, θθ ψ = r n n (+ n ) nθ n n θθ = An + 1 r cos cos + θ n n+ 4 and using, 1 ψ 1 ψ r θ = r θ r θ r 1 ψ ( n ) n nθ n n = Ar sin + sin + θ θ + n r n ψ ( 1+ n ) n n nθ n n = Ar n sin sin θ r n + + θ 1 ψ ( + n ) n n nθ n n = Ar n sin + sin + θ r θ r n n n nθ n n n rθ = + + + + n (+ n ) Anr 1 sin 1 sin θ 7
Range of n for the physical problem? The elastic energy at the crack tip has to be bounded Ω WdΩ = ε rdr dθ Ω ij ij but, ij ( ) ( ) (+ n ) r... ε ij (+ n ) r... Ω (...) (4+ n) Wd r r dr d Ω θ Ω is integrable if 3 n 0 n 3 or 5 λ=..., 3,,, 3 8
Singular term when n = 3 or 3A θ 3θ rr = 5cos cos 4 r 3A θ 3θ θθ = 3cos cos 4 r + 3A θ 3θ rθ = sin sin 4 r + 3 λ = 3A = K π Mode - stress intensity factor (SF) : K K ( ) K 5 θ 1 3 θ f cos cos πr πr 4 4 rr = rr θ = K K 3 θ 1 3θ f ( ) cos cos πr πr 4 4 θθ = θθ θ = + K ( ) K 1 θ 1 3 θ f sin sin πr πr 4 4 rθ = rθ θ = + 9
n Cartesian components, τ xx yy xy K θ θ 3θ = cos 1 sin sin π r K θ θ 3θ = cos 1+ sin sin π r K θ θ 3θ = cos sin cos π r does not contain the elastic constants of the material applicable for both plane stress and plane strain problems : 0 plane stress zz =, τxz = τ yz = 0 ν( xx + yy ) plane strain ν is Poisson s ratio 30
Asymptotic Stress field: y θθ τ rθ rr y yy τ xy O r θ x Similarly, O r θ x xx,, τ rr θθ rθ xx, yy, τxy K = θ + θ n ij fij n ij πr n= 0 (n) ( ) A r g ( ) 1 singularity at the crack tip + higher order terms (depending on geometry) r f ij : dimensionless function of θ in the leading term A n amplitude, g ij dimensionless function of θ for the nth term 31
Evolution of the stress normal to the crack plane in mode : yy K θ θ 3θ = cos 1+ sin sin π r Stresses near the crack tip increase in proportion to K f K is known all components of stress, strain and displacement are determined as functions of r and θ (one-parameter field) 3
Singularity dominated zone : Admit the existence of a plastic zone small compared to the length of the crack 33
Expressions for the SF : Closed form solutions for the SF obtained by expressing the biharmonic function in terms of analytical functions of the complex variable z=x+iy Westergaard (1939) Muskhelishvili (1953),... Ex : Through-thickness crack in an infinite plate loaded in mode -: K = πa Units of stress length 3/ i MPa m or MN m 34
For more complex situations the SF can be estimated by experiments or numerical analysis K = Y πa Y: dimensionless function taking into account of geometry (effect of finite size), crack shape Stress intensity solutions gathered in handbooks : Tada H., Paris P.C. and rwin G.R., «The Stress Analysis of Cracks Handbook», nd Ed., Paris Productions, St. Louis, 1985 Obtained usually from finite-element analysis or other numerical methods 35
Examples for common Test Specimens P a = B W K Y W B : specimen thickness Y a = W πa tan W a πa 0. 75 +. 0 + 0. 37 1 sin πa cos W W W 3 a lim Y = 11. W aw 0 πa W K 11 πa, =. P BW Y a = W a + W a a 0886. 464. 133. 3/ + a W W 1 W 3 4 a a + 14. 7 5. 60 W W 36
Mode- SFs for elliptical / semi-elliptical cracks Solutions valid if Crack small compared to the plate dimension a c When a = c,φ = 0 Circular: K 063. πa = πa π Semi-circular: (closed-form solution) K = 07. πa 37
Associated asymptotic mode displacement field : Polar components : ( ) K 1+ ν r θ 3θ = ( ) E π ur κ 1cos cos ( ) K 1+ ν r θ 3θ = ( ) E π + + uθ κ 1sin sin Cartesian components : K r θ θ ux = cos κ 1+ sin μ π K r θ θ uy = sin κ+ 1 cos μ π with μ= ( 1+ν) E shear modulus 3 ν plane stress κ= 1+ν 3 4ν planestrain E: Young modulus ν: Poisson s ratio Displacement near the crack tip varies with r Material parameters are present in the solution 38
Ex: sovalues of the mode- asymptotic displacement: μux K 0.7 y=r sinθ 0 0.1 0.1 0.3 0. crack 0. 0.3 0.4 0.4 0.5 0.6 plane strain, ν=0.38 0.5 0.6 μuy K x= r cosθ 0.7 1 0.8 0.6 0.4 0.3 0. 0.1 y=r sinθ crack 0 0.1 0. 1 0.8 0.6 0.4 0.3 x= r cosθ 39
Mode loading Same procedure as mode with the antisymmetric part of φ( θ) ( ) ψ= r λ φa θ φa ( θ ) = Bsin λθ+ Dsin( λ+ ) θ Asymptotic stress field : K 5 θ 3 3θ sin sin πr 4 4 rr = + K 3 θ 3 3θ sin sin πr 4 4 θθ = K 1 θ 3 3θ cos πr 4 4 rθ = cos + 40
Cartesian components: τ xx yy xy K θ θ 3θ = sin + cos cos π r K θ θ 3θ = sin cos cos π r K θ θ 3θ = cos 1 sin sin π r Associated displacement field : K r θ θ ux = sin κ + 1+ cos μ π K r θ θ uy = cos κ 1 sin μ π 41
Mode loading Stress components : = rz = θz K π r K π r θ sin θ cos [ ] 0 0 rz = 0 0 θ z rz θ z 0 Displacement component : 4K r θ ( ) E π uz = 1+ ν sin 4
Closed form solutions for the SF Mode -loading : K = τ πa Mode -loading : K = τ πa 43
Principe of superposition for the SFs: With n applied loads in Mode, K total n = K i= 1 Similar relations for the other modes of fracture But SFs of different modes cannot be added! Principe of great importance in obtaining SF of complicated specimen loading configuration (i) Example: p p p p (a) (b) (c) (a) (b) πa K = K 0 =λ p f φ s Q ( ) 44
4.5 Relationship between K and G : Mode only : G = K E' E' = E Plane stress ( 1 ) E' = E / ν Plane strain When all three modes apply : K K K G = + + E' E' μ ( ) μ = E/ 1+ν Self-similar crack growth Values of G are not additive for the same mode but can be added for the different modes 45
Proof ( a) K yy ( x) = πx ΔU G = lim in load control (ch 3) Δa 0 Δ a Work done by the closing stresses : Δa ( ) Δ U = du x 0 with 1 du( x) = u y( x) yy( x) dx but, u y ( x) ( 1K ) ( a a) ( 1K ) ( a a) κ+ +Δ r κ+ +Δ Δa x μ π μ π = = slide 38, with θ = 0 and also ( ) du x = ( 1 ) K ( a ) ( ) κ+ K a+δa Δa x dx μ π x slide 3 for yy ( x) Calculating ΔU and injecting in G G = lim Δ a 0 ( 1 ) K ( a ) ( ) κ+ K a+δa πδa 4μπΔa G = ( κ+ 1K ) 8μ 46
4.6 Mixed mode fracture biaxial loading e e e 1 e 1 0 ( Q) 0 in global frame ( e 1 e ) 1 expressed in local frame ( e e ) e 1 = cosβ e1+ sinβe e = sinβ e + cosβe 1 Thus, e e e e 1 1 1 = e e 1 e e ( Q) cosβ sinβ = sinβ cosβ Q = Rotation tensor 1 R K = 1 = πa (0) 1 Stress tensor components : 11 1 0 = 0 T ( Q) ( Q) 1 1 47
11 1 = 1 1sin β 1sinβcosβ 1sinβcosβ 1cos β cos β sinβcosβ + sinβcosβ sin β Mode loading : K K (1) () = 1 cos β π a = sin β π a = K(0) cos β = RK(0) sin β e e 1 1 11 Principe of superposition : (0) ( ) K = K cos β+ Rsin β Mode loading : K (1) = cos βsinβ πa 1 = K cos β sinβ (0) K () = cosβ sinβ πa (0) (0) ( ) K = K cosβ sinβ 1 R = RK cosβ sinβ 48
Propagation criteria Mode Crack initiation when the SF equals to the fracture toughness KC K = K or G C = GC = ' E Mixed mode loading Self-similar crack growth is not followed for several material K K K K 1 ν + + = C Useful if the specimen is subjected to all three Modes, but 'dominated' by Mode General criteria: ( ) Ω K,K,K,K, β,... = 0 C C i explicit form obtained experimentally 49
Examples in Modes and m K K + = C0 1 KC KC n m, n and C 0 parameters determined experimentally Erdogan / Shih criterion (1963): Crack growth occurs on directions normal to the maximum principal stress 3θ θ θ 3θ K sin + sin + K cos + 3cos = 0 Condition to obtain the crack direction 50
4.7 Fracture toughness testing Assuming a small plastic zone compared to the specimen dimensions, a critical value of the mode- SF may be an appropriate fracture parameter : plane strain fracture toughness K C K C plane stress plane strain K C : critical SF, depends on thickness K C K > K C : crack propagation Specimen Thickness K C : Lower limiting value of fracture toughness K C Material constant for a specific temperature and loading speed G C KC = Apparent fracture surface energy kj / m E 51
How to perform K C measurements? Use of standards: - American Society of Testing and Materials (ASTM) - nternational Organization of Standardization (SO) ASTM E 399 first standardized test method for K C : CT - was originally published in 1970 - is intended for metallic materials - has undergone a number of revisions over the years - gives specimen size requirements to ensure measurements in the plateau region ASTM D 5045-99 is used for plastic materials: - Many similarities to E 399, with additional specifications important for plastics. K based test method ensures that the specimen fractures under linear elastic conditions (i.e. confined plastic zone at the crack tip) 5
Chart of fracture toughness K C and modulus E (from Ashby) Large range of K C 0.01->100 MPa.m 1/ At lower end, brittle materials that remain elastic until they fracture 53
Chart of fracture toughness K C and yield strength Y (from Ashby) Materials towards the bottom right : high strength and low toughness fracture before they yield Materials towards the top left : opposite yield before they fracture Metals are both strong and tough! 54
Typical K C values: 1MPa m = 1MN m 3/ 55
Ex Aircraft components Fuselage made of 04 alloy (Al + 4% Cu + 1% Mg) Thickness of the sheet ~ 3mm Y 350 MPa (elastic limit) K K C C 30 MPa m 100 110 MPa m ARBUS A330 Plane stress criterion with K c is typically used here in place of K C 56