Course Outline. Exam Topics

Transcription

1 Course Outline. Fundamental Postulates of Solid Mechanics. Introduction to FEA using ABAQUS 3. Math Review, introduction to tensors and index notation 4. Describing Deformations 5. Describing Forces 6. Equations of Motion Exam opics 7. Overview of material models relating stress to strain 8. Linear Elastic Stress-Strain Relations 9. Analytical Solutions for Linear Elastic Solids 0. Energy Methods for Linear Elastic Solids. Implementing the Finite Element Method for Elastic Solids. Solids with special shapes beams and plates 3. Dynamic elasticity waves and vibrations 4. Plastic stress-strain relations; 5. Solutions for elastic-plastic solids 6. Modeling failure

2 Concept Checklist. FEA analysis Be able to idealize a solid component as a 3D continuum, rod, shell or plate Understand how to choose a material model for a component or structure Be familiar with features of a finite element mesh; be able to design an suitable mesh for a component Understand the role of the FE mesh as a way to interpolate displacement fields Understand the difference between solid, shell and beam elements Understand that selecting inappropriate element types and poor mesh design may lead to inaccurate results Understand how to select boundary conditions and loading applied to a mesh Understand that for static analysis boundary conditions must prevent rigid motion to ensure that FEA will converge Understand use of tie constraints to bond meshes or to bond a rigid surface to a part Be able to analyze contact between deformable solids Be able to choose a static, explicit dynamic, or implicit dynamic analysis; Be able to calculate natural frequencies and mode shapes for elastic solids and structures Be able to interpret and draw conclusions from analysis predictions; have the physical insight to recognize incorrect predictions Be able to use dimensional analysis to simplify finite element simulations

3 FEA Analysis Features of FE Mesh Nodes: Used to track motion of points in solid Elements: Main purpose is to interpolate displacements between values at nodes. ABAQUS offers linear (nodes at corners) and quadratic (nodes at mid-sides) elements Linear Quadratic Special element types russ: Special displacement interpolation for force members Beam: Special displacement interpolation for slender member. Have rotation DOFS/moments Plate/Shell: Special displacement interpolation for thin sheets that can deform out-of-plane. Rotations/moments Materials (Some examples) Linear Elasticity: OK for most materials subjected to small loads Plasticity Metals beyond yield Hyperelasticity: Large strain reversible model used for rubbers Viscoelasticity: ime dependent material used for polymers/tissue Linear Elastic Elastic-Plastic Hyperelastic Visco-elastic

4 Boundary Conditions FEA Review Analysis Properly constrained solids We can apply. Prescribed displacements. Forces on nodes 3. Pressure on element faces 4. Body forces 5. For some elements, can apply rotations/moments For static analysis we have to make sure we stop solid from translating/rotating Contact Select. Contact algorithm (Surface/Node Based). Constitutive law for contact Soft or Hard normal contact Friction law Unconstrained horizontal motion Unconstrained rotation Unconstrained horizontal motion Incorrectly constrained solids Unconstrained vertical motion Unconstrained rotation Master/slave pairs Slave surface Master surface Nodes on slave surface are prevented from penetrating inside master surface

5 Solution Procedures FEA Review Analysis Small strain v- large strain (NLGEOM) Static analysis * * Solves Ru ( )= F u= u using Newton-Raphson iteration Explicit Dynamics: solves Implicit Dynamics: solves Mu + R( u) = F u = u * * Mu + R( u) = F u = u * * using nd order forward Euler scheme using nd order backward Euler scheme Special procedures: modal dynamics, buckling ( Linear Perturbation steps ) Using Dimensional Analysis Input data to code Dimensionless form (others are possible) δ = If we know δ = f ( P, EI, L) f( PEIL,,, ) δ P I = f (, ) 4 L EL L E,I L δ P δ hen = L PL f ( ) EI If we know problem is linear, then δ = C L PL EI for some constant C

6 Concept Checklist 3. Mathematics Understand the concepts of scalar, vector and tensor fields; understand use of Cartesian and polar basis vectors to represent vector and tensor fields Be able to compute gradient and divergence of scalar, and vector fields in Cartesian and Polar coordinates; Understand the concept of a tensor as a linear mapping of vectors; Be able to create a tensor using vector dyadic products; be able to add, subtract, multiply tensors; be able to calculate contracted products of tensors; be able to find the determinant, eigenvalues and eigenvectors of tensors; understand the spectral decomposition of a symmetric tensor; Be familiar with special tensors (identity, symmetric, skew, and orthogonal) Be able to transform tensor components from one basis to another. Be familiar with the conventions of index notation and perform simple algebra with index notation Be able to calculate the divergence of a symmetric tensor field in Cartesian or polar coordinates (eg to check the stress equilibrium equation)

7 Routine tensor operations Math Position Scalar Field r = xe = x e + x e + x e i i Vector Field v( x i ) 3 3 φ( x i ) φ gradient φ = ei gradient ensor: linear map of vectors onto vectors v= S u vi = Sijuj x i i v v= ei e x j e x 3 e 3 x x e S S S S S S S S S S= a b S u= a b u= b u a S = ab Dyadic product of vectors ( ) ( ) ( ) ij i j j General tensor as a sum of dyads Basis change formulas Vectors: ensors: ( m) () e i i vi v= v m = m ( m) () e kl = ki ij S Q S Q lj i S= S ij e i e j ( m) () e ij i j Sij i j S= S m m = e e v ( m) () e i = Qv ij j Q = m e ij i j [ Q] v v m e m e m e3 = m e m e m e m e m e m e

8 Gradients in Polar Coordinates Position r = Rsinθcosφi+ Rsinθsinφj+ Rcosθk Vector a= arer + aθeθ+ aφeφ Gradient of a scalar Gradient of a vector Divergence of a vector f f f f = er + eθ + eφ R R θ Rsinθ φ vr vr v v v θ R φ R R θ R Rsinθ φ R v v vr v v θ θ θ φ v + cotθ R R θ R Rsinθ φ R v φ vφ vφ v v + cotθ θ + R R R θ Rsinθ φ R R v ( ) R v v R v v φ v= trace v = + + θ + cotθ θ + R R R θ R Rsinθ φ Position Vector r = re + ze = rcosθi+ rsinθj+ zk r a= a e + a e + a e r r z θ θ z z Gradient of a scalar Gradient of a vector f f f f = er + eθ + ez r r θ z vr vr vθ vr r r θ r z vθ vθ vr v v + θ r r θ r z vz vz vz r r θ z v ( ) r v vr v trace θ v= v = z r r θ r z Divergence of a vector

9 Routine tensor operations ensor Operations Operations on 3x3 matrices also apply to tensors Addition U= S+ U U U3 S + S + S3 + 3 U U U = S + S + S U3 U3 U33 S3 + 3 S3 + 3 S Vector/ensor product v= Su v S S S3 u Su + Su + S3u3 v = S S S u = S u + S u + S u v3 S3 S3 S33 u3 S3u + S3u + S33u3 ensor product U= S S S S3 us + us + us 3 3 v v v3 = u u u 3 S S S 3 us us us = S S S us + u S + us v= us [ ] [ ] U U U3 3 S S S3 U U U = S S S U3 U3 U S3 S3 S S + S + S S + S + S S + S + S = S + S + 3S3 S + S + 3 S 3 S + S + 3 S 3 S + S + S S + S + S S + S + S

10 ensor Operations ranspose 3 3 S S S S S S S S S = S S S 3 3 S3 S3 S33 S3 S3 S33 Determinant ( S) ( ) us = Su ( ) AB = B A det = S S S S S S (S S S S ) + S ( S S S S ) Eigenvalues/vectors Inverse S S= I 0 0 Sm =λm Spectral decomposition for symmetric S det( S λi) = 0 3 () i () i S= λ m m i= i Identity I Symmetric tensor Skew tensors S= S S = S Proper orthogonal tensors R R = R R = I R = R det( R) = + Inner product S: S S S = S S + S S + S S +... ij ij 3 3 Outer product S S S S = SS + SS + S3S ij ji

11 Routine Index tensor Notation operations Summary Vector Index Notation ( x, x, x ) x = 3 3 x Summation convention 3 i i x i S S ensor ij S S S S S S S S S e x 3 e 3 x x λ = ab λ = ab i i λ = ab + ab+ a33 b = a b ci = Sik xk i= 3 c = Sx + Sx + S3x3 ci = Sik xk c = Sx + Sx + S3x3 = S x k= c3 = S3x + S3x + S33x3 3 3 λ = SijSij λ = SijSij λ = SS + SS + + S3S3 + S3S3 + S33S33 = S : S i= j= 3 3 Cij = Aik Bkj Cij = Aik Bkj C = AB Cij = Aki Bkj Cij = Aki Bkj C = AB k= k= Kronecker Delta 0 0 i= j δij = 0 0 x a 0 i j δ ij = a j i 0 0 x j = δij e Permutation symbol i, jk, =,,3;,3, or 3,, ijk = i, jk, = 3,,;,,3 or,3, 0 otherwise = = = = = ijk kij jki jik kji kji = 0 kki = δ δ δ δ ijk imn jm kn jn mk λ = det A λ = ijk lmn Ali Amj A nk S = S S 6 det( S) ij ipq jkl pk ql

12 Concept Checklist 4. Deformations Understand the concept and definition of a deformation gradient; be able to calculate a deformation gradient from a displacement field in Cartesian/polar coordinates; be able to calculate understand the physical significance of the Jacobian of the deformation gradient Understand Lagrange strain and its physical significance; be able to calculate Lagrange strain from deformation gradient tensor or displacement measurements. Know the definition of the infinitesimal strain tensor; understand that it is an approximate measure of deformation; be able to calculate infinitesimal strains from a displacement in Cartesian/polar coords Know and understand the significance of the compatibility equation for infinitesimal strain in D, and be able to integrate D infinitesimal strain fields to calculate a displacement field. Be able to calculate principal strains and understand their physical significance Be able to transform strain components from one basis to another Be aware of the existence of other strain measures (stretch tensors, rotation) that appear in some material models

13 Course Deformations Outline Deformation Mapping: yx (,) t Displacement Vector: ux (,) t = ux (,) t x Deformation Gradient: F= y = u+ I F ij yi u = = i + δij x x j j e e 3 e dx x Original Configuration u(x+dx) u(x) dy Deformed Configuration dy = F dx dy = F dx i ij j Jacobian: J = det( F) dv = JdV 0 Lagrange Strain: ( E= F F I ) Eij = ( Fki Fkj δij ) l l 0 l 0 = m Em = me m i ij j Can use this to find E given ll,, m 0 e l 0 e 3 e m Original Configuration for 3 (in D) directions l Deformed Configuration

14 Infinitesimal strain: ε ( u ( u) ) Course Deformations Outline u u = + ε i ij = + xj x Properties: () Approximate strain measure used only for small deformation () For small strains ε E (3) Components quantify length and angle changes of unit cube (4) m εm = miε ijmj ( l l0)/ l0 (5) trace( ε) = εkk ( dv dv0)/ dv0 D Compatibility conditions o be able to integrate strains (to find displacement) Integrating strains ε ε ε x x x x + = 0 ε ε u = u = εdx + f x x u ( ) = u = εdx + g x x Find f,g using ε ( ) u u = + x x i j e l 0 e 3 e dv 0 Compatible Incompatible m Original Configuration dv l n Deformed Configuration

15 Cauchy-Green ensors: Right: Significance: Left: C= F F B = FF Significance: Course Deformations Outline / 0 n l 0 / l mcm =l l n B = Stretch ensors: Right: U= F F Significance: m U m=l / l 0 Left: V = FF Significance: n V n= l 0 / e 3 e l e l 0 m Original Configuration l n Deformed Configuration Principal Strains and stretches () i () i Infinitesimal strain εm m Stretches Um () i () i = λ i m = e i (eigenvalues use usual method to find them) Same idea for other symmetric strain tensors Spectral decomposition In principal basis ε () () (3) { m, m, m } e 0 0 λ 0 0 ( m) ( ) ij 0 e 0 m U ij 0 λ e3 0 0 λ3 3 3 () i () i () i () i ei λi i= i= ε = m m U = m m strains are diagonal e e Undeformed m 3 m m e Deformed m e

16 Visualizing deformation and displacement gradient Polar Decomposition of Deformation Gradient: Stretch ensors: Right: U= F F Significance: m U m=l / l 0 Left: n V n= V = FF Significance: l 0 / l Define rotation tensor: R = FU = V F then: F = RU = VR Significance: all deformation gradients can be visualized as a stretch U followed by rotation R (or a rotation R followed by a stretch V) Displacement Gradient ( ( ) ) u= u+ u + u ( u) = ε+ w ( ) Infinitesimal rotation tensor Significance: (Small) displacement gradient consists of a strain plus a small rotation (in either sequence)

17 Concept Checklist 5. Forces Understand the concepts of external surface traction and internal body force; Understand the concept of internal traction inside a solid. Understand how Newton s laws imply the existence of the Cauchy stress tensor Be able to calculate tractions acting on an internal plane with given orientation from the Cauchy stress tensor Be able to integrate tractions exerted by stresses over a surface to find the resultant force Know the definition of principal stresses, be able to calculate values of principal stress and their directions, understand the physical significance of principal stresses Know the definition of Hydrostatic stress and von Mises stress Understand the use of stresses in simple failure criteria (yield and fracture) Understand the boundary conditions for stresses at an exterior surface Be aware of the existence of other generalized stress measures (nominal stress, material stress) that may appear in some material models 6. Equations of motion for solids Know the equations for linear momentum balance and angular momentum balance for a deformable solid Understand the significance of the small deformation approximation of the general equations of motion Know the equations of motion and static equilibrium for stress in Cartesian and polar coordinates Be able to check whether a stress field satisfies static equilibrium

18 Describing external and internal forces External Loading Surface raction t = t + t n= Body force (per unit mass) Internal raction Vector (n) Quantifies force per unit area at a point on internal plane raction depends on direction of normal to surface Satisfies: ( n) = n ( ) n ( ) = e ( ) n + e ( ) n + e ( ) n Cauchy ( rue ) Stress ensor t 3 3 Definition (components): = ( e ) ij j i hen: ( n) = n = n j i ij n b = lim da 0 lim dp da dv 0 dp ρdv Warning: Some texts use transpose of this definition = n Cauchy stress (force per unit deformed area) is symmetric = t da -e n R e (n) e 3 da (n) (-e ) da ij ji, so both are the same, but some other stresses eg nominal stress (force per unit undeformed area) are not, so be careful. n da dv t n tt t dp t e S n e Deformed Configuration e 3 V R b (-n) (n) n (n) t -n da (n) e

19 Stresses Principal stresses (eigenvalues of stress tensor) n () i () i () i () i = in j jk = i k > > 3 or n n (no sum on i ) If then is the largest stress acting normal to any plane Hydrostatic stress h = trace( ) / 3 / 3 kk ( ) = + + h 3 /3 Deviatoric stress ij = ij hδij Von Mises stress 3 3 e = : = ij ij {( ) ( 3) ( 3) } e = + + < Failure criterion for brittle materials (approximate) frac Yield criterion for metals (Von Mises) e < Y

20 Stresses Stresses near a boundary n = t n = t i ij j S n eg for n= e t = 0 = = 3 = 0 e t e 3 e R here are also several types of Nominal Stress (Force per unit undeformed area) Deformation gradient F I u Jacobian J = det( F) Kirchhoff stress τ= J F u = + ij = δij + x i j e x e 3 e da 0 n 0 dp (n) 0 u(x) Original Configuration t n dp (n) da Deformed Configuration Nominal (First Piola-Kirchhoff Stress) Sij JFik kj S= JF = Has property that dp= da0n0 S= dan Unsymmetric some texts use transpose Material (Second Piola-Kirchhoff Stress) Σij ik kl jl Σ= JF F = JF F Has property that dp = F dp= da n Σ (Weird but useful for material models)

21 Linear Momentum Angular Momentum ij ji Equations of motion ij + ρbi = ρai yi 3 dv ρb = ρ y y y dt 3 dv y y y dt ρb = ρ 3 dv y y y dt ρb 3 3 = ρ 3 = e S 0 x e 3 e R 0 y Original Configuration Small deformations: replace y by x (approximate, but much easier to solve) u(x) t S V R b n Deformed Configuration (n) Spherical-polar coordinates RR Rθ Rφ θ R θθ θφ φr φθ φφ θr RR Rθ e R Rφ RR RR θr cot θr φr dv θ + ( θθ + R φφ ) R R R θ R Rsinθ φ R ρbr ρ dt dv Rθ Rθ θθ φθ cot θθ θ R φφ dv θ + ρb= ρ θ + + cotθ + ρbθ = ρ dt R R R θ R Rsinθ φ R R dt R φ Rφ sinθ θφ θφ φφ dv cosθ + + ( φr + φθ ) φ ρbφ ρ R R R θ R Rsinθ φ R dt e θ θθ θφ φr φθ φφ e φ Cylindrical-polar coordinates e z rr rr dv θr zr θθ r ρbr ρ r r r z r dt θ dv θθ rθ rθ θr zθ dvθ + ρb= ρ ρbθ = ρ dt r θ r r r z dt zz rz rz θ z dv z ρbz ρ z r r r θ dt rr rθ rz θr θθ θ z zr zθ zz e r rz rz zz rθ zr θr zθ θz θθ e θ

22 Concept Checklist 7. Stress-strain relations for elastic materials subjected to small strains Understand the concept of an isotropic material Understand the assumptions associated with idealizing a material as linear elastic Know the stress-strain-temperature equations for an isotropic, linear elastic solid Understand how to simplify stress-strain temperature equations for plane stress or plane strain deformation Be familiar with definitions of elastic constants (Young s, shear, bulk and Lame moduli, Poissons ratio) Be able to calculate strain energy density of a stress or strain field in an elastic solid Be able to calculate stress/strain in an elastic solid subjected to uniform loading or temperature.

23 Stress-strain-temperature relations for elastic solids Assumptions Displacements/rotations are small we can use infinitesimal strain as our deformation measure Isotropy: Material response is independent of orientation of specimen with respect to underlying material Elasticity: Material behavior is perfectly reversible, and relation between stress, strain and temperature is linear e e e 3 E ε ε ε 33 ε ν hen + ν ν εij = ij kkδij + α δij E E E ν Eα ij = εij + εkkδij δij + ν ν ν Stress-strain response equal More generally ( ) = C ε α δ ε = S + α δ ij ijkl kl kl ij ijkl kl ij Strain Energy Density e Separate strain into elastic and thermal parts ε = ε + ε ij ij ij e + ν ν εij = ij kkδij E E εij = α δij Strain energy density e U = ijεij U U + ν ν = ijij kk jj E E E Eν = εijεij + e e e e ε jjεkk ( + ν) ( + ν)( ν)

24 Useful elasticity formulas for isotropic materials Matrix form for stress-strain law (3D) ε ν ν ε ν ν ε33 ν ν = + α ε3 E ( + ν ) ε ( + ν ) ε ( + ν ) 0 For plane strain ε ν ν 0 ( ν ) ε + ν ν 0 ( ν) α = E + + ε For plane stress ε33 = ε3 = ε3 = 0 ( ) ( )( ) 33 = 3 = 3 = 0 ε ν 0 ε ν 0 α = E + ε 0 0 ( + ν) 0 Eν ε + ε Eα 33 = +, 3 = 3 = 0 + ν ν ν ν ε ( ) α E 33 = + + ν ν ν ν ν ν ε ν ν ν ε ( ) 33 E ν ε33 Eα = 3 ( + ν)( ν) ε3 ν 0 ( ν ) ε 3 0 ε 0 ( ν ) ν ν 0 ε E Eα ν ν 0 ε = ( + ν)( ν) ν ν ε ν 0 ε E Eα ν 0 ε = ( ν ) ( ν ) 0 0 ( ν ) / ε 0 Strain Energy Density e e e U = ε + ε + 33ε33 + ε + 3ε3 + 3ε3 e e e ε = ε α ε = ε α ε = ε α

25 Relations between elastic constants

26 Concept Checklist 8. Analytical solutions to static problems for linear elastic solids Know the general equations (strain-displacement/compatibility, stress-strain relations, equilibrium) and boundary conditions that are used to calculate solutions for elastic solids Understand general features of solutions to elasticity problems: () solutions are linear; () solutions can be superposed; (3) Saint-Venants principle Know how to simplify the equations for spherically symmetric solids (using polar coords) Be able to calculate stress/strain in spherically or cylindrically symmetric solids under spherical/cylindrical symmetric loading by hand Understand how the Airy function satisfies the equations of equilibrium and compatibility for an elastic solid Be able to check whether an Airy function is valid, and be able to calculate stress/strain/displacements from an Airy function and check that the solution satisfies boundary conditions

27 Solutions for elastic solids Static boundary value problems for linear elastic solids Assumptions:. Small displacements. Isotropic, linear elastic material Given:. raction or displacement on all exterior surfaces. Body force and temperature distribution Find: [ u, ε, ] i ij ij e S 0 x e 3 e R 0 y Original Configuration S u(x) b S t V Deformed Configuration Governing Equations:. Strain-displacement relation (you can use the compatibility equation instead) ε ij = ( ui / xj + uj / xi )/ ε= u+ ( u) /. Stress-strain law E ν Eα ij = εij + εkkδij δ E ν ij () E α = ε+ trace εi I + ν ν ( ν) + ν ν ( ν) ij 3. Equilibrium + ρ0bj = 0 + ρ x 0 b= 0 i 4. Boundary conditions on external surfaces. Where displacements are prescribed. Where tractions are prescribed u n i * * = ui u= u = t n = t j ji i

28 Solutions for Review elastic solids Spherically symmetric solids Position, displacement, body force x= ReR u= ur ( ) e R b= ρ0 br ( ) er Stress/strain RR 0 0 εrr θθ 0 ε 0 εθθ φφ 0 0 ε φφ du u εrr = εφφ = εθθ = dr R Equilibrium d dr RR ( RR θθ φφ ) ρ0 + + br = 0 R E Eα RR = ν ε + νε + νε + ν ν ν ( )( ) { ( ) } RR θθ φφ E Eα θθ = φφ = εθθ + νε + ν ν ν ( Ru) ( )( ) { RR } ( ) ( ν) ( )( ) ( ν) d u du u d d + d + + = R dr dr = dr R R dr dr E α ν ν ν ρ 0bR ( ) Boundary conditions u ( a) = g u ( b) = g R a R b or ( a) = t ( b) = t RR a RR b

29 Solutions for Review elastic solids Cylindrically symmetric solids Position, displacement, body force x= re + ze u= ure + ε e r z () r zz z z b= ρ bre 0 () r Stress/strain rr 0 0 εrr θθ 0 ε 0 εθθ zz 0 0 εzz du u = = dr r εrr ε θθ rr ν ν ν εrr E Eα θθ ν ν ν ε θθ = ( + ν)( ν) ν zz ν ν ν εzz Plane strain rr E ν εrr Eα = θθ ν ν ε θθ ν Plane stress Equilibrium d dr rr + ( rr θθ ) + ρ0br = 0 r ( ) ( ν) ( )( ) u u u α + ν + ν ν + = ( ru) = ρ 0( b + ω r) r r r r r r r r E( ν) ( ν ) u u u + = ( ru) ( ) = α + ν ρ0( b + ω r) r r r r r r r r E Plane strain Plane stress Boundary conditions or u ( a) = g u ( b) = g r a r b ( a) = t ( b) = t rr a rr b

30 Airy Function solution to elasticity problems Airy Function φ φ φ φ + + = 0 x x x x 4 4 S Stress φ φ φ = = = = x x x x 33 = 0 (Plane Stress) ( ) = ν + 33 = = (Plane Strain) e t e 3 e R Airy Function Stress + + φ = 0 r r r r θ φ φ φ φ rr = + θθ = rθ = r r r θ r r r θ z e 3 θ r e θ e z er e e Strain εrr ν ν 0 rr ( ν ) ε + θθ ν ν 0 = θθ E εrθ 0 0 rθ Plane Strain εrr ν 0 rr ε θθ ν 0 = θθ E εrθ 0 0 ( + ν) rθ Plane Stress Displacement u r u r u rr θ u r r u θ u θ ε = εθθ = + ε θ = + r r r θ r θ r r

31 Proof of the Airy Representation Airy Function φ 4 φ φ φ + + = 4 x 0 x x x Stress φ φ φ = = = = x x x x 33 = 0 (Plane Stress) 33 = ( + ) (Plane Strain) 3 = 3 = 0 ν + ν ν E E + ν ν = ( + E E )( + ) + ν = β = (Plane Strain) E β = 0 (Plane Stress) ε Strain = ( + βν )( + ) ε βν ε Equilibrium x x + = 0 x + = 0 x φ φ + = 0 x x x x x φ φ + = 0 x x x x x Compatibility ε ε ε x x x x + = 0 + ν ν + ν + ( + βν ) + ( + ) = 0 E E x E x x x x x ( + βν ) 4 φ 4 φ ν φ φ 4 φ = x x + ν x x x x x x φ 4 φ φ + + = 4 x 0 x x x