Continuum Mechanics Lecture 2 Theory of elasticity and examples

Transcription

1 Continuum Mechanics Lecture 2 Theory of elasticity and examples Prof.

2 Outline - work of the stress field - principle of elasticity and elastic potential - isotropic materials and linear elasticity - stress modulus and stability of the material - the Navier equation - linear elastic boundary values problems - 3D elastic problems - 2D planar problems - elastic bars - 1D problems: strings and beams

3 The kinetic energy theorem with a stress field We consider a finite volume V with external surface S. We compute the work of external forces for an infinitesimal displacement field δu = vdt and we define the vector σ such as T i = i σ i n. δw ext = ρg δudv + T δuds V S T δuds = ( σ x n) δu x ds + ( σ y n) δu y ds + ( σ z n) δu z ds S S S S ( σ i n) δu i ds = (δui Using the divergence theorem, we have σ i )dv Using the vector identity (a b )=a( b )+ b a, we get finally: T δuds = σ δudv + Tr σδg dv δg = δ + δω, and using the fact that Tr(S A) =0, we finally obtain T δuds = σ δudv + Tr σδ dv S S V V Decomposing the gradient tensor into a symmetric and antisymmetric part, S V V V

4 The kinetic energy theorem with a stress field Rearranging the terms, we get for the work of the external forces δw ext = (ρg + σ) δudv + Tr σδ dv V V We inject the dynamical equilibrium equation: ρ g + σ = ρ dv dt Defining the total kinetic energy K = ρ 1 v vdv and using δu = vdt 2 we finally get: dk = δw ext V V V Tr σδ dv We already know from the kinetic energy theorem that dk = δw ext + δw int Identifying the term leads to the result that the work of the internal stress is given by δw int = Tr σδ dv Solid body motions have δ =0 and therefore no internal work.

5 Thermoelasticity We recall the first principle of thermodynamics, defining the total internal energy E and the internal energy density e by E = de +dk = δw ext + δq and the rate of heating/cooling received by the system by δq = V δqdv Using the kinetic energy theorem, we obtain the internal energy equation: de = δq + Tr σδ dv This equation is also valid at the continuous level, where it reads de = δq +Tr σδ V V edv Introducing the entropy S and the temperature T, the second principle of thermodynamics tells us that in general we have and that the exact equality holds true for reversible systems. ds δq T

6 Thermoelasticity For reversible systems, it is convenient to define the free energy F=E-TS. We have the simple relation df = Tr σδ dv SdT V We define the free energy density φ as F = φdv. We now define a thermoelastic material by assuming that the free energy depends only on the strain tensor and on the temperature: φ(,t) df = Tr( φ F We can now differentiate the free energy δ)dv + V T dt and identifying the terms, we get σ = φ In most applications, the displacements are slow enough so that the material remains isothermal T=constant. df = Tr(σδ)dV = δw int V and S = F T Elastic stresses are derived from a potential energy called the elastic energy. V

7 Isotropic material An isotropic material has no preferred directions for its elastic properties. In this case, the elastic energy depends only on the invariants of the strain tensor, because they define the eigenvalues in the eigenvector basis. φ() =φ(i 1,I 2,I 3 ) I 1 Recalling that, we get = 1 I I 2 i =Tr( i ) =2 Using the definition of the elastic stress tensor, we get finally φ σ = I φ I2 + 3 φ I3 2 I 3 =32 Very important consequence for the 2 tensors σ and : they have the same eigenvectors decomposition.

8 Linear elasticity and linear material From experiment, we observe that small deformations result in a linear increase of the stress intensity (and vice versa). Moreover, the transformation is reversible (the material recover its original shape). Above a given stress threshold, we enter the plastic regime: the transformation is now irreversible. The initial elongation is not recovered. Ultimately, the material breaks. The most general linear form writes σ = Γ where the new tensor is of order 4! It contains 21 independent coefficients. The elastic potential is a quadratic form φ = 1 Γ 2 Tr Γ ijkl = 2 φ ij kl

9 Linear, isotropic and elastic material The most general quadratic form depending on the strain invariants is φ() = λ 2 (I 1 ) 2 + µi 2 Differentiating with respect to the strain tensor, we obtain σ = λ Tr 1+2µ λ and µ are called the Lamé elastic coefficients. The elastic potential is given by φ() = λ 2 2 Tr + µtr( ). 2 In order for a material to be stable to small perturbations around its equilibrium state, we require the elastic energy to be a positive definite form. In the principal axis frame, we have φ( 1, 2, 3 )= λ 2 ( ) 2 + µ φ The second order derivative matrix is given by =2µδ ij + λ i j This 3x3 matrix, also noted 2µ1+λJ, should have 3 positive eigenvalues µ +3λ > 0 2µ >0 2µ >0 J =

10 Particular cases Isotropic compression For isotropic stress, the stress tensor writes as the isotropic pressure. It follows that the strain tensor is also diagonal, with From the Lamé relations, we get σ = p1 Trσ =(2µ +3λ)Tr which relates to the compressibility of the material κ>0 by 2µ +3λ dv p = 3 V = κdv V where p is Tr = u = dv V Pure shear σ = 0 σ 0 σ = σ =2µ µ is also called the «shear modulus»

11 An experimental approach: Hooke s law 2 σ 1 σ = σ = When applying an axial pull to a solid cylinder, Young has experimentally established that the resulting strain is proportional to the stress. This defines Young s modulus E = σ 1. Poisson has also observed in the same experiment that for a given axial elongation, there is a transverse contraction also proportional to the stress but smaller. This defines the Poisson coefficient ν = 2. 1 For a general stress applied in the linear regime, we can combine linearly the 3 displacements and strains, and we obtain = 1+ν E σ ν Trσ 1 E 1

12 Relations between the different elastic modulus. Traditionally, the Lamé coefficients are used to compute the stress field for a given strain, and the Young modulus and the Poisson coefficient are considered when the strain field is given as a function of the stress. In the principal axis frame, we can express the strain ( 1, 2, 3 ) as a function of the stress (σ 1, σ 2, σ 3 ) using the linear transform: 1+ν E J = 1 ν E J This is the inverse of the positive definite form that defines the elastic potential. 1 2ν E = 1 2µ +3λ > 0 1+ν E = 1 2µ > 0 From the stability condition, we get the following conditions: Experimentally, for most materials we have instead 0 < ν < 1 2 E>0 1 < ν < 1 2 The 2 sets of coefficients (λ,µ) and (E,ν) are related by the following relations: λ = νe (1 + ν)(1 2ν) µ = E 2(1 + ν) E = µ(3λ +2µ) λ + µ ν = λ 2(λ + µ)

13 It is based on the maximum shear stress The limit of elasticity Experimentally, the elastic regime breaks down when the deviatoric stress exceeds a certain threshold called the yield strength. The criterion must depends on invariants of the stress tensor. Tresca yield criterion (Henri Tresca, 1864) using the eigenvalues of the stress tensor For pure shear: σ max = σ σ min = σ σ σ y 2 It does not depend on the isotropic stress. Von Mises yield criterion (Richard von Mises, 1913) It is based on the second invariant of the deviatoric 3 stress where J2 =Tr(τ 2 ). 2 J 2 σ y Equivalently, (σ 1 σ 2 ) 2 +(σ 2 σ 3 ) 2 +(σ 1 σ 3 ) 2 2σy 2 Physical interpretation, the deviatoric elastic energy φ d () =µi2 = 1 4µ J 2 exceeds locally the threshold Max(σ i σ j ) σ y Max(σ i σ j )=σ max σ min 1 6µ σ2 y

14 Examples of stress moduli and yield strength Material Poisson Young modulus Yield strength coefficient (GPa) (MPa) Mild steel Copper Aluminum Silver Concrete Platinum Gold Glass

15 We consider a body of volume V and external surface S. We assume that a natural equilibrium state exists, for which the stress field is zero. We then apply to the body a state of external constraints (body forces, surface forces) that perturb the body from its initial state. We then reach a new equilibrium state displaced from the natural state. We have in the volume V: Elastic equilibrium problem σ + ρ g =0 ij = 1 2 ui + u j x j x i σ = λ Tr 1+2µ We define on the body surface boundary conditions, for both the stress field and the displacement field. We have on the surface S: σ ij n j = T g i j for i =1, 2, 3 u i = u g i where index g stands for «given». For a linear elastic material, a unique solution always exists. It depends only on the boundary conditions. It is called an elastic boundary value problem (BVP).

16 Boundary conditions Since the elastic properties also apply to the surface, we apply only 3 constraints. 1- Imposed displacement field on the surface u i = u g i for i =1, 2, 3 A particular case: rigid surface: u i = 0 for i =1, 2, 3 2- Imposed stress field on the surface σ ij n j = T g i for i =1, 2, 3 j Usually, one uses as natural basis (t 1,t 2, n) where (t 1,t 2 ) defines the tangential plane and n is the normal to the surface. Particular case: isotropic external pressure σ1j n j =0 σ2j n j =0 σ3j n j = p p=0 defines a free surface boundary condition. 3- Imposed tangential stress and normal displacement. σ u 3 = u g 1j n j = T g 1 σ 2j n j = T g 2 3 j j Particular case: imposed normal displacement without friction σ1j n j =0 σ2j n j =0 u 3 = u g 3

17 The superposition principle For a given set of constraints ( T g, u g, ρg), there is a unique solution (σ, ) that depends linearly on the constraints. If, from constraints 1 ( T g 1, ug 1, ρ 1g 1 ), we get solution 1 (σ 1, 1 ) and from constraints 2 ( T g 2, ug 2, ρ 2g 2 ), we get solution 2 (σ 2, 2 ), then from constraint ( T g, u g, ρg) = (α 1 T g 1 + α 2 we will get the linear combination of the 2 solution (σ, ) =(α 1 σ 1 + α 2 σ 2, α α 2 2 ) T g 2, α 1u g 1 + α 2u g 2, α 1ρ 1 g 1 + α 2 ρ 2 g 2 ) Interesting consequence: new displacement from an already displaced configuration. The displacement from natural equilibrium equilibrium 2, it the sum of the displacements from natural equilibrium equilibrium 1 and from equilibrium 1 equilibrium 2.

18 The Navier equation From the equilibrium condition, linear elasticity and strain definition, we have ij = 1 ui + u j σ + ρ g =0 σ = λ Tr 1+2µ 2 x j x i We need the following vector calculus relations (proof left to the reader). (α1) = α ( T G)= ( u) (G) = u We obtain the Navier equation, where the displacement field is the unknown. (λ + µ) ( u)+µ u + ρg =0 The final equilibrium linear problem is expressed in the rather simple form: In the volume V: (λ + µ)grad (divu)+µ u + ρg =0 On the surface S: u = u g σn = T g

19 General solutions for the Navier equation Defining the divergence of the displacement field as θ = u, we have (λ + µ) θ + µ u + ρg =0 If we take the divergence of the Navier equation, we have: (λ +2µ) θ + ρ g =0 If there is no body force, the divergence is an harmonic function and the final solution is obtained solving 3 Poisson s equations θ =0 u = λ + µ θ µ Another strategy is to look for displacement fields u = φ where φ is called the strain potential. In this case, we have θ = φ Injecting this new form into the equation for the divergence, we get: The strain potential is a biharmonic function. φ =0 In both case, the solution is found by trying different harmonic potentials and building the unique solution by constraining these potentials to fulfill the boundary conditions.

20 Summary: elastic problems in Cartesian coordinates Equilibrium equations on the stress tensor: σ xx x σ yx x σ zx x + σ xy y + σ yy y + σ zy y + σ xz z + ρg x =0 + σ yz z + ρg y =0 + σ zz z + ρg z =0 Hooke s law: = 1+ν E σ ν E Kinematic equations on the strain tensor: Trσ 1 ij = 1 2 ui + u j x j x i In order for the stress tensor (6 unknown variables) to be compatible with a displacement (3 unknown variables), it has to follow the 6 compatibility conditions: 2 xx y yy x 2 =2 2 xy x y + 2 permutations with 2 zz x y = z yz x + xz y x y and y z and z x xy z

21 Equilibrium of an elastic sphere under pressure p ext r e We assume no body forces ρg =0 p int r i Boundary conditions: - T g = p int e r for r =ri T g = p ext e r - for r =re We look for a spherically symmetric solution of the form u = u(r)e r with u(r) = φ r From the Navier equation, we have φ =0 with φ = 1 r 2 r 2 φ r r This gives the general solution: u(r) =ar + b r 2 Using the strain definition, we get φφ = θθ = a + b rr = a 2 b r 3 r 3 Using the linear elasticity law, we get with A =(3λ +2µ)a and B =2µb σ rr = A 2 B r 3 σ φφ = σ θθ = A + B r 3 Finally, using the boundary conditions, we find the complete solution A = p extr 3 e p int r 3 i r 3 i r3 e B = 1 2 (p ext p int ) r 3 er 3 i r 3 i r3 e

22 Equilibrium of an elastic sphere under pressure We look for 2 interesting limiting case: 1- the full sphere 2- the thin shell Uniform sphere We consider the case r i 0 We have A p ext and B p 2 r3 i For any radius r>0, we have for r i 0 : σ = PR 2h σ rr σ θθ = σ φφ = p ext There is a singularity at r=0, which is removed by continuity using p int = p ext The stress tensor in a uniform sphere in equilibrium is necessarily isotropic. Thin shell We consider the case r i = R and r e = R + h = R(1 + ) For simplicity, we also assume that p ext p int = P Taylor expanding, we find A P and B P 3 6 R3 To leading order, we find everywhere inside the shell σ rr 0 and σ θθ = σ φφ P 2 The shell resists to the internal pressure only by a membrane stress where

23 Equilibrium of the snow pack under its own gravity We consider an infinite inclined slab of snow of height H. The displacement field depends only on coordinate z: u =[u x (z), 0,u z (z)] z x 1 u 0 0 x 2 z H = θ g σ xx = σ yy = λ u z z σ zz =(λ +2µ) u z z σ xz 1 2 u x z 0 From the strain tensor, we deduce the stress tensor using Lamé coefficients: σ xy = σ yz =0 σ zz The equilibrium equations write: = ρg sin θ = ρg cos θ z z Boundary conditions: 1- free surface at the top 2- no displacement at the bottom u z z σ xz = µ u x z Solution with compressive stress σ zz = ρg cos θz and shear stress σ xz = ρg sin θz Maximum vertical displacement of the snow pack ρg cos θ H 2 u z = λ +2µ 2 Maximum transverse displacement (crevasse) ρg sin θ H 2 u x = µ 2 Typical value for light snow: ρ = 100 kg/m 3 E =2MPa u 1 mm

24 Spontaneous triggering of an avalanche Snow is probably one of the most brittle material (C. Sigrist PhD thesis at ETHZ). ρ σ y 80 ρ ice 2 kpa 3 ρ E 2 ρ ice GPa

25 Spontaneous triggering of an avalanche Heavy snow slab sitting on top of a weak layer. dense slab weak layer dense slab For the weak layer, we assume ρ = 100 kg/m 3 and σ y 1kPa With a slope inclination of 30 deg and a dense (wind-blown) snow pack with ρ = 500 kg/m 3, the vertical stress will exceed the yield strength if: z σ y 20 cm ρg cos θ Spontaneous avalanches are however very rare, because such unstable configurations are very difficult to form.

26 Point loading of an infinite half-space (Boussinesq) P r We apply a force P as a delta function at x=0. We want to compute the stress field and the displacement field inside the half-plane defined by z>0. Rough estimate of the solution using force balance on the z half-sphere defined by R 2 = r 2 + z 2 R σ P σ2πr 2 = 0 P δ(r = 0)2πrdr = P We now compute the exact solution due to Boussinesq. The displacement field is symmetric around the z-axis. From the Navier equation, we have θ =0 θ = u u =[u x (r, z), 0,u z (r, z)] The simplest harmonic function that satisfies the geometry θ = a z u z = λ + µ is θ From the Navier equation, we also have µ z The general solution of this inhomogeneous equation is u z = a λ + µ 2µ 1 R R 3 + b 1 R θ = 1 r r (ru r)+ u z From, we get (with the constraint that u r 0 as r 0 ) z ru r = b a 3µ + λ b a 2µ + λ z 2µ µ R aµ + λ z 3 2µ R 3 for r z 2 2πR 2

27 Point loading of an infinite half-space (Boussinesq) The constants a and b are determined using the boundary conditions. G = u r r 0 0 u z r 0 u r z u r r 0 u z z = 1 2 ur z u r 1 r u z r ur z u r r 0 0 u z z + u z r Using the Lame coefficients, we get: σ θθ = λ u +2µ u r σ rr = λ u +2µ u r σ zz = λ u +2µ u z r r z ur σ rz = µ z + u z We require that T r z = σe z =0 at z = 0 and r>0: u r σ rz (r, 0) = 0 (r, 0) = b a 2µ + λ 1 u z z µ r 2 r (r, 0) = b 1 r 2 b = a 2µ + λ 2µ We require that each plane perpendicular to z is in equilibrium with the load P. + 0 σ zz (r, z)2πrdr = P 6πa (λ + µ) + 0 with σ zz = 3a (µ + λ) z3 R 5 z 3 rdr (z 2 + r 2 ) = P 5/2 a = P We get 2π (λ + µ) σ zz = 3P 2π z 3 R 5

28 Triggering of an avalanche by a lone skier. We consider now the case of a stable weak layer embedded in a dense snow slab. We consider a skier of weight 70 kg. We get the following stress profiles. The typical failure depth will be around 30 cm. The thicker the slab, the most dangerous the avalanche will be. If the weak layer fails, one hears a typical «woof» sound. We are then suddenly exposed to a very unstable configuration: the yield strength of the compressed weak layer is now much smaller for shear stress: the dense slab will start moving downwards.

29 Two dimensional elastic problems Under certain circumstances, elastic problems can be classified as 2D. Plane stress: infinitely thin plate with free boundaries along the z-axis. Plane strain: infinitely long cylindrical body along the z-axis We first write Hooke s law in Cartesian coordinates: xx = 1 E σ xx ν E (σ yy + σ zz ) yy = 1 E σ yy ν E (σ xx + σ zz ) zz = 1 E σ zz ν E (σ xx + σ yy ) xy = 1+ν E σ xy xz = 1+ν E σ xz yz = 1+ν E σ yz Plane stress: we impose σ zz =0. We obtain a closed 2D system described by: xx = 1 E σ xx ν E σ yy yy = 1 E σ yy ν E σ xx xy = 1+ν E σ xy Plane strain: we impose zz =0. We get σ zz = ν (σ xx + σ yy ) and get also xx = 1+ν E [(1 ν)σ xx νσ yy ] yy = 1+ν E [(1 ν)σ yy νσ xx ] xy = 1+ν E They are equivalent under the transformation: ν = ν 1 ν E = E 1 ν 2 σ xy

30 Two dimensional elastic problems We now use the equilibrium equation and assume no body force. σ xx x + σ xy y =0 σ yx x + σ yy y =0 There are 2 equations with 3 unknown functions. The Ansatz σ xx = φ y σ automatically satisfies the equilibrium along x. y xy = φ y x We also have σ yy = φ x σ along y. x yx = φ x y We still need only one additional degree of freedom. We define the stress potential φ so that φ and φ y = φ x = φ. x y The stress tensor for 2-dimensional problems has the general form: σ xx = 2 φ y 2 σ yy = 2 φ x 2 σ xy = 2 φ x y

31 Two dimensional elastic problems Using Hooke s law for plane stress problems, we have: xx = 1 2 φ E y 2 ν 2 φ E x 2 yy = 1 2 φ E x 2 ν E 2 φ y 2 xy = 1+ν E 2 φ x y We have 3 independent components for the strain tensor for 2 components for the displacement field u =[u x (x, y),u y (x, y)] so we have only 1 compatibility condition: 2 xx y yy x 2 =2 2 xy x y Injecting Hooke s law into the compatibility condition, we finally get: φ = 4 φ x φ x 2 y φ y 4 =0 The stress potential is therefore a biharmonic function. It is called the function of Airy from the British astronomer Sir George B. Airy, who derived first this equation in In some limited cases, the stress potential can take the form φ = a i x p i yq i, where the polynomials have p<4 and q<4 and the coefficient are determined i using the boundary conditions (see exercises).

32 Linear loading of an infinite half-space θ r z We consider the plane strain problem of a linear loading of the half-space. We work in cylindrical coordinates. Rough estimate of the solution using force balance on the half-cylinder defined by +r σlπr = P δ(x = 0)dx = P σ P/L πr r The equilibrium equations in cylindrical coordinates are: σ rr r σ rθ r + σ rr r + σ rθ r σ θr + 1 r θ + 1 σ θθ r θ σ θθ r =0 + σ θr r =0 Trying σ θθ = φ r and σ θr = φ θ, we obtain σ rr = φ r. r r φ r = φ r + 1 φ θ r θ φ θ = 1 φ Requiring only one degree of freedom, we find and. r r θ Using Hooke s law, we again obtain φ =0(a biharmonic function). φ = 2 φ r φ r r φ where the Laplacian operator now reads r 2 θ 2

33 Linear loading of an infinite half-space We consider the following simple Airy function: The stress tensor components write σ rr =2a 1, r cos θ P L = π/2 π/2 σe r cos θrdθ = a φ = arθ sin θ π/2 π/2 σ rθ = σ θθ =0 We now work the force balance along the x axis for a cylinder of radius r. We get the final solution: σ rr = 2P/L π cos θ r 2 cos 2 θdθ The Boussinesq solution was scaling as 1/r 2. This one scales as 1/r. The line loading can be interpreted as the collective effect of many point-loading along the line. Triggering of an avalanche by a group of skiers. We consider skiers separated by 25 cm. The stress profile penetrates much deeper into the snow pack. 2 negative effects: 1- the probability to hit a weak layer is increased 2- the thickness of the released slab can be larger.

34 x 0 z Sh S0 Axial torsion on a bar with arbitrary section y The section has an arbitrary shape. No body force g =0 and free transverse surface We apply a torque +Mz on Sh and -Mz on S0. We postulate a coordinate transform based on a rotation cos ω(z) sin ω(z) 0 X = sin ω(z) cos ω(z) The displacement field u = R I x u =( αyz, αxz, 0) and the strain and stress tensors write = α 0 0 y 0 0 y 0 0 x σ = µα 0 0 x 2 y x 0 y x 0 One can check that the equilibrium equations are trivially satisfied. Boundary conditions at the top: F = σe z ds = µα ( y, x, 0) dxdy =0 S S T O = r σe z ds = M z e z = µαie z S x y z writes for T g =0 ω(z) =αz ω 1

35 Axial torsion on a bar with arbitrary section Let s compute now the transverse boundary condition: n =(n x,n y, 0) x T = σn =(0, 0,ny x n x y) n = r, y We have. This is zero only for r, 0 Only bars with circular section satisfy the transverse boundary condition. We need to postulate a more general displacement field u = α ( yz, xz, φ(x, y)) φ 0 0 x y The stress tensor writes now φ σ = µα 0 0 y + x φ x y φ y + x 0 The equilibrium equation along the z-axis writes σ zx x + σ zy y =0 which translates into a Poisson equation φ =0 φ T z = n x x + n φ The transverse boundary condition writes y y + n yx n x y =0 or in vector form φ n = nx y n y x. This is a Von Neumann boundary condition. We know in this case there exists a unique solution for this strain potential.

36 Axial torsion on a bar with arbitrary section We compute the net force on the upper surface φ We use x y = φ x x x y + x y to get, from the divergence theorem in 2D: φ F x = n 1 x + n 2 L h x The same applies to the y component. The z component is trivially zero. φ F x = µα S h x y dxdy x φ and φ =0 φ y + x φ y + n 2x n 1 y The same computation can be done for the bottom surface. dl =0

37 The torques components are and M z = µα We define J as M z = µαj. J is the torsional constant. The solution is now fully determined by the torque acting on the normal faces. It can be shown that J is always positive. Using Axial torsion on a bar with arbitrary section We compute the net torque on the surface of the bar. S h M x = M y =0 φ x y + x y ( y φ + x) 2 = x ( y φ + x)+ y φ ( y φ + x) ( x φ y) 2 = y ( x φ y)+ x φ ( x φ y) we have J = ( x φ + x) 2 +( y φ y) 2 ds>0 S h φ x y dxdy M z = µj dω z dz We used y φ ( y φ + x) = y [φ ( y φ + x)] φ xφ 2 x φ ( x φ y) = x [φ ( x φ y)] φ yφ 2 + the divergence theorem in 2D, the transverse boundary condition and φ =0

38 Axial torsion on a bar with arbitrary section Let s consider again the case of a cylindrical bar with circular section φ The Neumann boundary condition writes n 1 so that x + n φ 2 y =0 the solution of φ =0 which is φ = a ln r + b follows a=0. The displacement field is simply The corresponding stress tensor has σ θz = σ zθ = µαr The stress field on the upper face is T = µαruθ The torque on the upper face is M z = µαj with J = The stress tensor is purely deviatoric (pure shear). The eigenvalues are (µαr, 0, µαr). The Von Mises criterion writes 3µαr σy u =( αzy,αzx,0) = αzre θ The cylinder will enter the plastic regime from the outside-in, if : M z πr3 2 3 σ y S (x 2 + y 2 )dxdy = π 2 R4

39 Axial torsion on a bar with arbitrary section Let s consider a cylindrical bar with an elliptical section. The section is defined by the ellipse or a 2 + y2 =1 x =(acos θ,bsin θ) b2 The potential φ = Axy satisfies the 2D Poisson equation. φ The transverse boundary condition writes n x x + n φ y y = n xy n y x The tangent vector to the ellipse is a b t = sin θ, a2 + b2 a2 + b cos θ 2 b a The normal vector to the ellipse is n = cos θ, a2 + b2 a2 + b sin θ 2 Injecting all this into the boundary condition, we obtain A = b2 a 2 b 2 + a 2 We now need to compute J = x 2 + y 2 + x y φ y x φ ds S 2 We get J = b 2 a 2 + b 2 x 2 + a 2 y 2 ds where the moments of inertia are S I I y = x 2 ds = π x = y 2 ds = π 4 ab3 4 ba3 S x 2 J = πa3 b 3 a 2 + b 2 S

40 Axial torsion on a bar with arbitrary section We now consider an alternative strategy, based on the equilibrium equation. We postulate the following stress tensor: 0 0 σ xz σ = 0 0 σ yz σ zx σ zy 0 From the equilibrium equation, we get: σ xz z =0 σ yz z =0 σ xz x = σ yz y We define the stress potential ψ(x, y) such as σ xz = ψ and The equilibrium equations are trivially satisfied. The remaining components give u z x =2 xz u x z = 1 ψ µ y + αy u z y =2 yz u y ψ = 2αµ y z = 1 µ σ yz = ψ x The strain tensor is just given by = 1. 2µ σ Because we have xx = yy = xy =0, we deduce that the displacement in each z=constant plane is a rigid body motion, so that (u x,u y )=( αzy,αzx) We use the compatibility condition, which writes here to get the following Poisson equation ψ x αx 2 u z y x = 2 u z x y

41 Axial torsion on a bar with arbitrary section ψ We finally writes the transverse boundary conditions y n x ψ x n y =0 Recall that the unit vector tangent to the outer contour is t =( n y,n x ) so we can write on the contour that dψ = ψ ψ dx = t ds =0 The stress potential is the unique solution of the Poisson equation ψ = 2αµ with a Dirichlet boundary condition ψ = constant on the outer contour. For the ellipse, we obtain ψ = A x 2 y 2 αµ x 2 + y 2 with A = αµ a2 b a 2 + b 2 We now define ψ = αµ ψ x2 + y 2 that satisfies ψ =0 2 ψ We have σ xz = αµ y y σ yz = αµ ψ x + x Identifying the stress components as a function of the strain potential, we have: φ x = ψ φ y y = φ =0 ψ =0 ψ with x and φ n = C ψ t = C These are the Cauchy-Riemann relations for each harmonic potential. The stress and the strain potentials are harmonic conjugate function. We can define a complex potential Φ(x, y) =φ + iψ which is analytic.

42 z Sh Axial pull on a cylindrical bar The section has an circular shape. - no body force - free transverse surface - we apply a displacement u z =0 on S0 and u z = δ on Sh. We postulate the following uniform stress tensor σ = σe z e z 0 y which satisfies the equilibrium equation for constant. ν σ S0 E 0 Using Young s modulus and Poisson = 0 ν σ x E coefficient we get the following strain tensor: σ 0 0 E u = ν σ E x, ν σ E y, σ We can integrate to get E z (+ an arbitrary rigid body motion). From the boundary condition, we get the stress tensor σ = E δ. h The relative volume change is given by Tr() =(1 2ν) δ. δ h The force per unit surface on each face is T = ±E h e z The total net force is just F = ±ES δ h e z F = ES du z dz e z

43 Axial pull and torsion on a cylindrical bar We apply the principle of superposition to combine both boundary conditions and solutions for a circular section. u = αzy ν σ E x, αzx ν σ E y, σ E z In cylindrical coordinates, we have σ = σ The eigenvalues are 0 and. 2 ± 1 σ2 +4µ 2 2 α 2 r 2 The Tresca elasticity limit criterion is given by Fz 2 2 S 2 +4M z J 2 R2 σy µαr 0 µαr σ

44 z Sh Normal torque on a cylindrical bar The section has an arbitrary shape. - no body force - free transverse surface x 0 S0 We postulate y - we apply a torque +My on Sh and -My on S-h, using a force field on the surface given by T = αxez We assume that the axis of symmetry is axis 0z. We have therefore xds =0 yds =0 xyds =0 σ = αxe z e z. This stress tensor satisfies the boundary conditions, as well as the equilibrium condition inside the volume. The net force on the cylinder is zero and the torque gives α = M y. The Hooke s law gives the strain tensor = M y x [ νe x e x νe y e y + e z e z ] EI y To integrate the displacement field, we use the diagonal components of to derive u x while off diagonal terms give x = u y y = ν M y x u z EI y z = M y u i x = u j EI y x j x i S S I y S

45 Diagonal terms of the strain tensor: u x = ν M x 2 EI 2 + f x(y, z) u y = ν M EI xy + f y(x, z) u z = M EI xz + f z(x, y) Off-diagonal terms: Normal torque on a cylindrical bar u x y = u y x u x z = u z x f x y = ν M EI y f y x f x z =+M EI z f z x 2 f y x 2 =0 2 f z x 2 =0 2 f x y z = 2 f y x z 2 f x y z = 2 f z x y u y z = u z y General solutions: f y (x, z) =cx az f y z = f z y f z (x, y) =ay bx f x (y, z) = ν M y 2 EI 2 cy + M EI 2 f y z 2 =0 2 f z y 2 =0 2 f y x z = 2 f z x y 2 f y x z =0 z bz Ω =(a, b, c) Displacement field is unique up to arbitrary rigid body motions

46 We obtain the displacement field Normal torque on a cylindrical bar u x = M y EI y z 2 u y = M y EI y (νxy) u z = M y EI y ( xz) 2 + ν x2 y 2 2 Deformation of the symmetry axis: initially, the axis is defined by x = y =0 After the torque, we have u x = M y z 2. The axis becomes a parabola. EI y 2 The new axis makes an angle ω y with the horizontal axis tan ω y = dy dx = M y EI y z and the curvature is dω y ds = M y EI y Using the superposition principle, we can combine the 3 basic actions (axial torque, axial pull and normal torque) to get general solutions on cylindrical bars.

47 Curvilinear systems n b t We consider an elongated cylinder with a curvilinear coordinate system. S Integrating on the small section S, we get the force field T (s) = σtds = T t t + T n n + T b b M(s + ds) M(s)+dst T (s + ds) =0 S and the normal torque field (w.r.t point s) M(s) = r σtds = M t t + M n n + M b b S The equilibrium conditions writes (using a small line segment ds) T (s + ds) T (s)+µdsg =0 where we define the mass per unit length d We get the equilibrium equation T + µg =0 µ(s) = dm ds ds Similarly, the total torque on the line segment writes (w.r.t point s) or d M ds + t T =0

48 Curvilinear systems The displacement field is expressed as u = u t t + u n n + u b b It described the transformation of the curvilinear object from its initial state to its final state x(s)+u(s) x(s) This is the macroscopic displacement field, not to be confused with the «microscopic» or internal displacement field inside each infinitesimal bar. We define the elongation as = du ds t We define the rotation vector as ω = ω t t + ω n n + ω b b du The analog of the strain tensor is just ds = t + ω n n + ω b b Component ω t describes the internal axial torsion of the curvilinear object, it does not generate any macroscopic displacement. We use the previous solutions for elastic bars to derive curvilinear elastic laws = T t ES dω ds = M t µj t + M n EI n + M b EI b axial pull axial torque normal torques

49 Curvilinear systems Recall that the derivatives of the Frenet basis vectors are non zero. Any vector derivative is thus d T ds = + + dtt ds CT n t dtn ds + CT t T T b dtb ds + T T n b n C =1/Ris the curvature and T is the torsion of the curvilinear system. C = a a 2 + b 2 T = C =0 T =0 C =1/a T = b a 2 + b 2

50 Planar curvilinear systems We now restrict ourselves to objects and displacements in the vertical plane. This imposes T b =0but also M t = M n =0. The Frenet torsion is also zero. Using the properties of the Frenet frame, we derive the equilibrium equation in curvilinear coordinates: dt t ds T n R + µg t =0 dt n ds + T t R + µg n =0 dm ds + T n =0 The displacement field is expressed as u = u t t + u n n and we have We define The curvilinear elastic laws are just du ds =(du t ds u n R ) t +( du n ds + u t R )n = t du ds =(du t ds u n R ) ω = t du ds =(du n ds + u t R ) b = ω b = T t ES dω ds = M EI

51 These are defined by the zero torque condition M=0. We get T n =0. d T t + µg =0 ds We assume no elasticity for the beginning. We have thus From the geometry of the problem, we use: The equilibrium condition writes The solution is thus Or, more explicitly, T = T 0 cos α This is the catenary solution: Flexible ropes and strings d ds (T cos α) =0 d d ds (tan α) =µg T 0 µ = µ 0 t = (cos α, sin α) ds tan α = dy dx = y ds = dx 1+(y ) 2 y y 0 = cosh x x 0 L L (T sin α) =µg y = L = T 0 µg g =(0, g) 1+(y ) 2 L

52 Elastic ropes and strings If now we consider the elasticity, the rope gets elongated ds =(1+) ds 0 Mass conservation gives us µ = dm with. ds = µ 0 = T 1+ k = T ES 0 We find, introducing a dimensionless number ES Ly 1+(y ) = 2 1+k 1+(y ) 2 In case of small Young s modulus, the curve converges towards a parabola. Rigid ropes are more like a catenary.

53 Rigid beams: general equations Beams are defined as elastic structures with zero curvature. The rigidity allows non zero torques. dt x dx =0 (g x = 0) dt y dx + µg y =0 dm dx + T y =0 The elastic response of the beam is The displacement field is given by = T x ES du x dx = dω dx = M EI du y dx = ω The problem is fully specified by the gravity force and the boundary conditions. As for the 3D case, you can specify either the displacement field (, ω,u x,u y ) or the force field (T x,t y,m) but not both (examples follow).

54 Beams with no gravity We consider a beam, embedded in a rigid wall on the left end We apply a vertical external force on the right boundary and no normal torque. Right boundary conditions at x=l are T x =0, T y = F, M =0 F From the previous equations, we get the equilibrium equations valid inside the beam T x =0 u x = u y =0, ω =0 T y = F M = F (L x) The deformation of the beam is given by the elastic law dω dx = M =0 EI We finally obtain the displacement field from u x = u x (0) = 0 and the differential equation: d 2 u y (L x) = F dx2 EI du y (0) Using the left boundary conditions, namely u y (0) = 0, =0, dx we obtain the final solution u y (L) = F 2EI x2 (L x 3 ) The maximum amplitude is obtained at the right end with value u y (L) = FL3 3EI

55 Beams with only gravity We consider a beam, embedded in a rigid wall on the left end, with a mass per unit length µ, and a free right boundary at x=l: T x = T y = M =0 (L x)2 We get for 0<x<L: T x =0 T y = µg(x L) M = µg 2 From the elastic law, we get and dω which writes dx = M EI d 2 u y dx 2 = µg (L x) 2 EI 2 u x =u x (0) = 0 du y (0) Applying the left boundary conditions u y (0) = 0, =0, dx we obtain u y = µg 24EI x2 (x 2 4Lx +6L 2 ) The maximum displacement is obtained at x=l and is u y (L) = µg 8EI L4 If we write the total gravity force as F = µgl, we have u y (L) = FL3 8EI

56 Superposition principle on a beam We consider a beam, embedded in a rigid wall on the left end, with a mass per unit length µ, and a fixed right boundary at x=l. F The force acting on the beam is unknown. It represents the reaction from the support. We write the displacement as the sum of the 2 previous problems, which is u y = µg 24EI x2 (x 2 4Lx +6L 2 )+ F 2EI x2 (L x 3 ) We impose the new boundary condition u y (L) =0 from which we deduce the reaction from the support as well as the final solution. du Note that y (L) = 0. dx F = 3 8 µgl