CH.7. PLANE LINEAR ELASTICITY. Multimedia Course on Continuum Mechanics

CH.7. PLANE LINEAR ELASTICITY Multimedia Course on Continuum Mechanics

Overview Plane Linear Elasticit Theor Plane Stress Simplifing Hpothesis Strain Field Constitutive Equation Displacement Field The Linear Elastic Problem in Plane Stress Eamples Plane Strain Simplifing Hpothesis Strain Field Constitutive Equation Stress Field The Linear Elastic Problem in Plane Stress Eamples Lecture 1 Lecture 2 Lecture 3 Lecture 4 2

Overview (cont d) The Plane Linear Elastic Problem Governing Equations Representative Curves Isostatics or stress trajectories Isoclines Isobars Maimum shear lines Lecture 5 Lecture 6 3

7.1 Plane Linear Elasticit Theor Ch.7. Plane Linear Elasticit 4

Plane Linear Elasticit For some problems, one of the principal directions is known a priori: Due to particular geometries, loading and boundar conditions involved. The elastic problem can be solved independentl for this direction. Setting the known direction as z, the elastic problem analsis is reduced to the - plane There are two main classes of plane linear elastic problems: Plane stress Plane strain PLANE ELASTICITY REMARK The isothermal case will not be studied here for the sake of simplicit. Generalization of the results obtained to thermo-elasticit is straight-forward. 7

7.2 Plane Stress Ch.7. Plane Linear Elasticit 8

Hpothesis on the Stress Tensor Simplifing hpothesis of a plane stress linear elastic problem: 1. Onl stresses contained in the - plane are not null [ σ] z σ τ 0 τ σ 0 0 0 0 2. The stress are independent of the z direction. σ σ τ = σ = σ = τ ( t,, ) ( t,, ) ( t,, ) REMARK The name plane stress arises from the fact that all (not null) stress are contained in the - plane. 9

Geometr and Actions in Plane Stress These hpothesis are valid when: The thickness is much smaller than the tpical dimension associated to the plane of analsis: e<< L b( ) u ( ) ( ) * * The actions,t,,t and t,t are contained in the plane of analsis (in-plane actions) and independent of the third dimension, z. ( ) * t,t is onl non-zero on the contour of the bod s thickness: 10

Strain Field in Plane Stress The strain field is obtained from the inverse Hooke s Law: ν 1+ ν ε= Tr ( σ) 1 + σ E E σ τ τ As z= 0 z= 0 z= 0 = σ (,, ) ε= ε = σ (,, ) ( t,, ) σ t σ t 1 2(1 + ν ) ε = ( σ νσ ) γ = 2ε = τ E E 1 ε = ( σ νσ ) γ z = 2ε z E = 0 ν εz = ( σ + σ ) γ z = 2εz E = 0 11 And the strain tensor for plane stress is: 1 ε γ 0 2 1,, γ ε 0 2 0 0 ε z ε with εz = ( ε + ε) ( t) ν 1 ν

Constitutive equation in Plane Stress Operating on the result ields: 1 2(1 + ν ) ε = ( σ νσ ) γ = 2ε = τ E E 1 ε = ( σ νσ ) γ z = 2ε z E = 0 ν εz = ( σ + σ ) γ z = 2εz E = 0 plane stress = C E σ = ε + νε 2 ( 1 ν ) E σ = ε + νε τ = 2 ( 1 ν ) E 21 ( + ν ) γ z σ 1 ν 0 ε E σ ν 1 0 = ε 2 1 ν τ 1 ν γ 0 0 2 = ε { } = σ { } 12 Constitutive equation in plane stress (Voigt s notation) plane { σ} = C stress { ε}

Displacement Field in Plane Stress The displacement field is obtained from the geometric equations, ε (, t) = S u(, t). These are split into: Those which do not affect the displacement : u ε ( t,, ) γ = u ε ( t,, ) = u u,, = 2ε = + ( t) Integration in. u z (,, ) (,, ) u = u t u = u t Those in which u z appears: ν uz εz( t,, ) = ( ε + ε) ( t,, ) = uz( zt,,, ) 1 ν z u(, ) uz uz γ z ( t,, ) = 2εz = + = = 0 z = 0 uz(,) zt u (, ) uz uz γ z ( t,, ) = 2εz = + = = 0 z = 0 13 Contradiction!!!

The Linear Elastic Problem in Plane Stress The problem can be reduced to the two dimensions of the plane of analsis. The unknowns are: The additional unknowns (with respect to the general problem) are either null, or independentl obtained, or irrelevant: σz = τz = τ z = γ z = γ z = εz = ν ε + ε 1 ν uz u ( t,, ) { ε }( t,, ) ε { σ}( t,, ) ( zt,,, ) u u ( ) 0 does not appear in the problem ε γ σ σ τ REMARK This is an ideal elastic problem because it cannot be eactl reproduced as a particular case of the 3D elastic problem. There is no guarantee that the solution to u ( t,, ) and u ( t,, ) will allow obtaining the solution to u zt,,, for the additional geometric eqns. z ( ) 14

Eamples of Plane Stress Analsis 3D problems which are tpicall assimilated to a plane stress state are characterized b: One of the bod s dimensions is significantl smaller than the other two. The actions are contained in the plane formed b the two large dimensions. Slab loaded on the mean plane Deep beam 15

7.3 Plane Strain Ch.7. Plane Linear Elasticit 16

Hpothesis on the Displacement Field Simplifing hpothesis of a plane strain linear elastic problem: 1. The displacement field is u u = u 0 2. The displacement variables associated to the - plane are independent of the z direction. (,, ) (,, ) u = u t u = u t 17

Geometr and Actions in Plane Strain These hpothesis are valid when: The bod being studied is generated b moving the plane of analsis along a generational line. b( ) u ( ) ( ) * * The actions,t,,t and t,t are contained in the plane of analsis and independent of the third dimension, z. In the central section, considered as the analsis section the following holds (approimatel) true: u z = 0 u = 0 z u = 0 z 18

Strain Field in Plane Strain u z = 0 u = 0 z u = 0 z ε ε γ The strain field is obtained from the geometric equations: u ( t,, ) uz ( t,, ) = ε = = 0 z u ( t,, ) u ( t,, ) uz,, = γ z = + = 0 z ( t) ( t) (,, ) u ( t,, ) u ( t,, ) u t uz,, = + γ z = + = 0 z And the strain tensor for plane strain is: z ε 1 ε γ 0 2 1,, γ ε 0 2 0 0 0 ( t) REMARK The name plane strain arises from the fact that all strain is contained in the - plane. 19

Stress Field in Plane Strain Introducing the strain tensor into Hooke s Law operating on the result ields: As ( ) ( ) ( ) σ = λ ε + ε + 2Gε τ = Gγ ( σ = λtr ( ε) 1 + 2Gε) σ = λ ε + ε + 2Gε τz = Gγ z = 0 = ( λ + 2G) ε + λε σ = λ ε + ε = v( σ + σ ) τ = Gγ = 0 z z z (,, ) (,, ) (,, ) (,, ) ε = ε t ε = ε t εz = εz t γ = γ t σ = σ ( t,, ) and And the stress tensor for plane strain is: σ τ 0 σ ( t,, ) τ σ 0 with σz = ν ( σ + σ ) 0 0 σ z 20

Constitutive equation in Plane Strain Introducing the values of the strain tensor into the constitutive equation and operating on the result ields: E ( 1 ν ) ν σ = ( λ + 2G) ε + λε = ε + ε ( 1+ ν)( 1 2ν) 1 ν σ = λ Tr ( ε) 1 + 2µ ε E ( 1 ν ) ν σ = ( λ + 2G) ε + λε = ε + ε ( 1+ ν)( 1 2ν) 1 ν E τ = Gγ = γ 21 ( + ν ) = C plane strain ν 1 0 σ 1 ν ε E ( 1 ν ) ν σ = 1 0 ε ( 1 ν)( 1 2ν) 1 ν τ + γ 1 2ν 0 0 = { σ } 21 ( ν ) = { ε} plane { σ} = C strain { ε} Constitutive equation in plane strain (Voigt s notation) 21

The Lineal Elastic Problem in Plane Strain (summar) The problem can be reduced to the two dimensions of the plane of analsis. The unknowns are: u ( t,, ) { ε }( t,, ) ε { σ}( t,, ) The additional unknowns (with respect to the general problem) are either null or obtained from the unknowns of the problem: u z = 0 u u εz = γ z = γ z = τz = τ z = ( ) σ = ν σ + σ z 0 ε γ σ σ τ 22

Eamples of Plane Strain Analsis 3D problems which are tpicall assimilated to a plane strain state are characterized b: The bod is generated b translating a generational section with actions contained in its plane along a line perpendicular to this plane. The plane strain hpothesis ( εz = γ z = γ z = 0) must be justifiable. This tpicall occurs when: 1. One of the bod s dimensions is significantl larger than the other two. An section not close to the etremes can be considered a smmetr plane and satisfies: u z = 0 u = 0 z u = 0 z u u = u 0 2. The displacement in z is blocked at the etreme sections. 23

Eamples of Plane Strain Analsis 3D problems which are tpicall assimilated to a plane strain state are: Pressure pipe Continuous brake shoe Tunnel Solid with blocked z displacements at the ends 24

7.4 The Plane Linear Elastic Problem Ch.7. Plane Linear Elasticit 25

Plane problem A lineal elastic solid is subjected to bod forces and prescribed traction and displacement Actions: On : Γ σ On : Γ u On Ω : ( t,, ) ( t,, ) The Plane Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding displacements u ( t,, ), strains ε ( t,, ) and stresses σ ( t,, ). * t * u * t = * t * u ( t,, ) = * u ( t,, ) ( t,, ) ( t,, ) b b = b 26

Governing Equations The Plane Linear Elastic Problem is governed b the equations: 1. Cauch s Equation of Motion. Linear Momentum Balance Equation. (, t) ρ b(, t) σ + = ρ 0 0 2 2D 2 (, t) u t σ τ τ u z t 2 z + + + ρb = ρ 2 τ σ τ 2 z + + + ρb = ρ 2 z t τ τ σ u z t 2 z z z z + + + ρbz = ρ 2 u 27

Governing Equations The Plane Linear Elastic Problem is governed b the equations: 2. Constitutive Equation (Voigt s notation). Isotropic Linear Elastic Constitutive Equation. σ ( ), t = C : ε 2D { σ} = C { ε} σ ε With { σ} σ, { ε} = ε and τ γ PLANE STRESS E = E ν = ν PLANE STRAIN E C = 2 1 ν E E = 1 ν = 2 ν ν ( 1 ν ) 1 ν 0 ν 1 0 0 0 ( 1 ν ) 2 28

Governing Equations The Plane Linear Elastic Problem is governed b the equations: 3. Geometrical Equation. Kinematic Compatibilit. 1 ε 2 S (, t) = u(, t) = ( u + u) 2D u ε = u ε = γ u = + u This is a PDE sstem of 8 eqns -8 unknowns: u(,t) ( ) 2 unknowns ε,t 3 unknowns σ (,t) 3 unknowns Which must be solved in 2 the space. + 29

Boundar Conditions Boundar conditions in space Affect the spatial arguments of the unknowns Are applied on the contour Γ of the solid, which is divided into: Γ u Prescribed displacements on : u * (,, ) (,, ) * * u = u t = * * u = u t Γ σ Prescribed stresses on : t * (,, ) (,, ) * * t = t t = * * t = t t t * = σ n with n n = n σ σ τ τ σ 30

Boundar Conditions INTIAL CONDITIONS (boundar conditions in time) Affect the time argument of the unknowns. Generall, the are the known values at t = 0 : Initial displacements: u u(,,0) = = 0 u Initial velocit: u ( t,, ) u v u = = = u v t t= 0 (,,0) v (, ) 0 31

Unknowns The 8 unknowns to be solved in the problem are: u(, t,) u = u ε ( t,, ) ε 1 γ 2 1 γ 2 ε σ ( t,, ) σ τ τ σ Once these are obtained, the following are calculated eplicitl: PLANE STRESS PLANE STRAIN εz = ν ε + ε 1 ν ( ) ( ) σ = ν σ + σ z 32

7.5 Representative Curves Ch.7. Plane Linear Elasticit 33

Introduction Traditionall, plane stress states where graphicall represented with the aid of the following contour lines: Isostatics or stress trajectories Isoclines Isobars Maimum shear lines Others: isochromatics, isopatchs, etc. 34

Isostatics or Stress Trajectories Sstem of curves which are tangent to the principal aes of stress at each material point. The are the envelopes of the principal stress vector fields. There will eist two (orthogonal) families of curves at each point: Isostatics Isostatics σ 1 σ 2, tangents to the largest principal stress., tangents to the smallest principal stress. REMARK The principal stresses are orthogonal to each other, therefore, so will the two families of isostatics orthogonal to each other. 35

Singular and Neutral Points Singular point: characterized b the stress state σ τ = σ = 0 Neutral point: characterized b the stress state σ = σ = τ = 0 Mohr s Circle of a singular point Mohr s Circle of a neutral point REMARK In a singular point, all directions are principal directions. Thus, in singular points isostatics tend to loose their regularit and can abruptl change direction. 36

Differential Equation of the Isostatics Consider the general equation of an isostatic curve: = f ( ) 2τ 2 = σ σ 1 ( α ) = = 2 tg 2 d tgα = = d 2τ 2 tgα σ σ 1 tg α ( ) 2 Solving the 2 nd order eq.: Differential equation of the isostatics σ σ τ ( ) 2 + 1= 0 ( ) σ σ σ σ ' = ± + 1 2τ 2τ 2 (, ) φ Known this function, the eq. can be integrated to obtain a famil of curves of the tpe: = f ( ) + C 37

Isoclines Locus of the points along which the principal stresses are in the same direction. The principal stress vectors in all points of an isocline are parallel to each other, forming a constant angle θ with the -ais. These curves can be directl found using photoelasticit methods. 38

Equation of the Isoclines To obtain the general equation of an isocline with angle θ, the principal stress σ must form an angle α = θ with the -ais: Algebraic equation of the isoclines 1 ( θ ) tg 2 = σ 2τ σ ϕ (, ) θ For each value of, the equation of the famil of isoclines parameterized in function of is obtained: θ = f (, θ ) REMARK Once the famil of isoclines is known, the principal stress directions in an point of the medium can be obtained and, thus, the isostatics calculated. 39

Maimum shear lines Envelopes of the maimum shear stress (in modulus) vector fields. The are the curves on which the shear stress modulus is a maimum. Two planes of maimum shear stress correspond to each material point, and. τ ma τ min These planes are easil determined using Mohr s Circle. REMARK The two planes form a 45º angle with the principal stress directions and, thus, are orthogonal to each other. The form an angle of 45º with the isostatics. 42

Equation of the maimum shear lines Consider the general equation of a slip line 2 π tan 2 α τ = and β = α + ( β) σ σ 4 Then, 1 σ σ 2 tan β tan ( 2β ) = = = 2 tan ( 2α) 2τ 1 tan β tan ( β ) ( ), the relation = f π 1 tan 2 = tan 2α = 2 tan 2α d not 2τ = = 1 ( ) 2 d σ σ 2 = τ 1 0 σ σ = 4 ( ) 2 43

Equation of the maimum shear lines Solving the 2 nd order eq.: Differential equation of the slip lines 2τ 2τ ' = ± + 1 σ σ σ σ 2 (, ) φ Known this function, the eq. can be integrated to obtain a famil of curves of the tpe: = f ( ) + C 44

Chapter 7 Plane Linear Elasticit 7.1 Introduction As seen in Chapter 6, from a mathematical point of view, the elastic problem consists in a sstem of PDEs that must be solved in the three dimensions of space and in the dimension associated with time ( R 3 R +). However, in certain situations, the problem can be simplified so that it is reduced to two dimensions in space in addition to, obviousl, the temporal dimension ( R 2 R +). This simplification is possible because, in certain cases, the geometr and boundar conditions of the problem allow identifing an irrelevant direction (associated with a direction of the problem) such that solutions independent of this dimension can be posed a priori for this elastic problem. Consider a local coordinate sstem {,, z} in which the aforementioned irrelevant direction (assumed constant) coincides with the z-direction. Then, the analsis is reduced to the - plane and, hence, the name plane elasticit used to denote such problems. In turn, these are tpicall divided into two large groups associated with two families of simplifing hpotheses, plane stress problems and plane strain problems. For the sake of simplicit, the isothermal case will be considered here, even though there is no intrinsic limitation to generalizing the results that will be Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar obtained to the thermoelastic case. 7.2 Plane Stress State The plane stress state is characterized b the following simplifing hpotheses: 1) The stress state is of the tpe σ τ 0 not [σ] z τ σ 0. (7.1) 0 0 0 339

340 CHAPTER 7. PLANE LINEAR ELASTICITY 2) The non-zero stresses (that is, those associated with the - plane) do not depend on the z-variable, σ = σ (,,t), σ = σ (,,t) and τ = τ (,,t). (7.2) To analze under which conditions these hpotheses are reasonable, consider a plane elastic medium whose dimensions and form associated with the - plane (denoted as plane of analsis) are arbitrar and such that the third dimension (denoted as the thickness of the piece) is associated with the z-ais (see Figure 7.1). Assume the following circumstances hold for this elastic medium: a) The thickness e is much smaller than the tpical dimension associated with the plane of analsis -, e L. (7.3) b) The actions (bod forces b(,t), prescribed displacements u (,t) and traction vector t (,t) ) are contained within the plane of analsis - (its z- component is null) and, in addition, do not depend on the third dimension, b (,,t) u (,,t) b not b (,,t), Γ u : u not u (,,t), 0 Γ σ = Γ + Γ Γ e σ σ σ t (,,t) : t not t (,,t). Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar (7.4) c) The traction vector t (,t) is onl non-zero on the boundar of the piece s thickness (boundar Γσ e ), whilst on the lateral surfaces Γ σ + and Γσ it is null (see Figure 7.1). 0 Γ σ + Γ σ : t not 0. (7.5) 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Plane Stress State 341 Figure 7.1: Eample of a plane stress state. Remark 7.1. The piece with the actions defined b (7.4) and (7.5) is compatible with the plane stress state given b (7.1) and (7.2), and schematized in Figure 7.2 1. In effect, appling the boundar conditions Γ σ on the piece ields: Lateral surfaces Γ + and Γ n not n not Edge Γ e n n 0 0 0, ±1 σ, σ σ σ τ 0 0 0 σ n not τ σ 0 0 = 0, 0 0 0 ±1 0 σ τ 0 σ (,,t) n not τ σ 0 0 0 0 Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar which is compatible with the assumptions (7.4) and (7.5). n n 0 t (,,t) = t (,,t), 0 1 The fact that all the non-null stresses are contained in the - plane is what gives rise to the name plane stress. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

342 CHAPTER 7. PLANE LINEAR ELASTICITY Figure 7.2: Plane stress state. 7.2.1 Strain Field. Constitutive Equation Consider now the linear elastic constitutive equation (6.24), ε = ν 1 + ν Tr (σ)1 + E E σ = ν E Tr (σ)1 + 1 2G σ, (7.6) which, applied on the stress state in (7.1) and in engineering notation, provides the strains (6.25) 2 ε = 1 E (σ ν (σ + σ z )) = 1 E (σ νσ ) γ = 1 G τ, ε = 1 E (σ ν (σ + σ z )) = 1 E (σ νσ ) γ z = 1 G τ z = 0, Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar ε z = 1 E (σ z ν (σ + σ )) = ν E (σ + σ ) γ z = 1 G τ z = 0, (7.7) where the conditions σ z = τ z = τ z = 0 have been taken into account. From (7.2) and (7.7) it is concluded that the strains do not depend on the z-coordinate either (ε = ε (,,t)). In addition, the strain ε z in (7.7) can be solved as ε z = ν 1 ν (ε + ε ). (7.8) 2 The engineering angular strains are defined as γ = 2ε, γ z = 2ε z and γ z = 2ε z. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Plane Stress State 343 In short, the strain tensor for the plane stress case results in ε (,,t) not ε 1 2 γ 0 1 2 γ ε 0 0 0 ε z with ε z = ν 1 ν (ε + ε ) (7.9) and replacing (7.8)in(7.7) leads, after certain algebraic operations, to σ = E 1 ν 2 (ε + νε ), σ = E 1 ν 2 (ε + νε ), E and τ = 2(1 + ν) γ, which can be rewritten as σ 1 ν 0 σ = E ε 1 ν 2 ν 1 0 τ 0 0 1 ν ε γ }{{}}{{ 2 }}{{} {σ} {ε} C plane stress 7.2.2 Displacement Field (7.10) = {σ} = C plane stress {ε}. The components of the geometric equation of the problem (6.3), (7.11) Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar can be decomposed into two groups: ε (,t)= S u(,t)= 1 (u + u), (7.12) 2 1) Those that do not affect the displacement u z (and are hpotheticall integrable in R 2 for the - domain), ε (,,t)= u ε (,,t)= u γ (,,t)=2ε = u + u integration in R 2 = { u = u (,,t) u = u (,,t). (7.13) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

344 CHAPTER 7. PLANE LINEAR ELASTICITY 2) Those in which the displacement u z intervenes, ε z (,,t) = u z z = ν 1 ν (ε + ε ), γ z (,,t) = 2ε z = u z + u z = 0, γ z (,,t) = 2ε z = u z + u z = 0. (7.14) Observation of (7.1) to (7.14) suggests considering an ideal elastic plane stress problem reduced to the two dimensions of the plane of analsis and characterized b the unknowns u(,,t) not [ u u ], {ε (,,t)} not ε ε γ and {σ (,,t)} not Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar σ σ τ, (7.15) in which the additional unknowns with respect to the general problem are either null, or can be calculated in terms of those in (7.15), or do not intervene in the reduced problem, σ z = τ z = τ z = γ z = γ z = 0, ε z = ν 1 ν (ε + ε ), (7.16) and u z (,,z,t) does not intervene in the problem. Remark 7.2. The plane stress problem is an ideal elastic problem since it cannot be eactl reproduced as a particular case of a threedimensional elastic problem. In effect, there is no guarantee that the solution of the reduced plane stress u (,,t) and u (,,t) will allow obtaining a solution u z (,,z,t) for the rest of components of the geometric equation (7.14). 7.3 Plane Strain The strain state is characterized b the simplifing hpotheses u u (,,t) u not u = u (,,t). (7.17) u z 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Plane Strain 345 Again, it is illustrative to analze in which situations these hpotheses are plausible. Consider, for eample, an elastic medium whose geometr and actions can be generated from a bidimensional section (associated with the - plane and with the actions b(,t), u (,t) and t (,t) contained in this plane) that is translated along a straight generatri perpendicular to said section and, thus, associated with the z-ais (see Figure 7.3). The actions of the problem can then be characterized b b not b (,,t) u (,,t) t (,,t) b (,,t), Γ u : u not u (,,t) and Γ σ : t not t (,,t). (7.18) 0 0 0 In the central section (which is a plane of smmetr with respect to the z-ais) the conditions u u z = 0, z = 0 and u z = 0 (7.19) are satisfied and, thus, the displacement field in this central section is of the form u (,,t) u(,,t) not u (,,t). (7.20) 0 Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Figure 7.3: Eample of a plane strain state. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

346 CHAPTER 7. PLANE LINEAR ELASTICITY 7.3.1 Strain and Stress Fields The strain field corresponding with the displacement field characteristic of a plane strain state (7.20)is ε (,,t)= u, ε (,,t)= u, γ (,,t)= u + u, Therefore, the structure of the strain tensor is 3 ε z (,,t)= u z z = 0, γ z (,,t)= u z + u z = 0, γ z (,,t)= u z + u z = 0. (7.21) 1 ε 2 γ 0 ε (,,t) not 1 2 γ ε 0. (7.22) 0 0 0 Consider now the lineal elastic constitutive equation (6.20) σ = λ Tr (ε)1 + 2με = λ Tr (ε)1 + 2Gε, (7.23) which, applied to the strain field (7.21), produces the stresses σ = λ (ε + ε )+2με =(λ + 2G)ε + λε, τ = Gγ, σ = λ (ε + ε )+2με =(λ + 2G)ε + λε, τ z = Gγ z = 0, Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar σ z = λ (ε + ε ), τ z = Gγ z = 0. (7.24) Considering (7.21) and (7.24), one concludes that stresses do not depend on the z-coordinate either (σ = σ (,,t)). On the other hand, the stress σ z in (7.24) can be solved as σ z = λ 2(λ + μ) (σ + σ )=ν (σ + σ ) (7.25) 3 B analog with the plane stress case, the fact that all non-null strains are contained in the - plane gives rise to the name plane strain. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Plane Strain 347 and the stress tensor for the plane strain case results in σ (,,t) not σ τ 0 τ σ 0 with σ z = ν (σ + σ ), (7.26) 0 0 σ z where the non-null components of the stress tensor (7.26) are ( E (1 ν) σ =(λ + 2G)ε + λε = ε + ν ) (1 + ν)(1 2ν) 1 ν ε, ( E (1 ν) σ =(λ + 2G)ε + λε = ε + ν ) (1 + ν)(1 2ν) 1 ν ε, E and τ = Gγ = 2(1 + ν) γ. Equation (7.27) can be rewritten in matri form as ν 1 0 σ 1 ν ε σ E (1 ν) ν = 1 0 (1 + ν)(1 2ν) 1 ν ε 1 2ν τ 0 0 γ }{{} 2(1 ν) }{{}}{{} {σ} {ε} C plane strain Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar {σ} = C plane strain {ε}. (7.27) (7.28) Similarl to the plane stress problem, (7.20), (7.21) and (7.26) suggest considering an elastic plane strain problem reduced to the two dimensions of the plane of analsis - and characterized b the unknowns u(,,t) not [ u u ], {ε (,,t)} not ε ε γ and {σ (,,t)} not σ σ τ, (7.29) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

348 CHAPTER 7. PLANE LINEAR ELASTICITY Figure 7.4: The plane linear elastic problem. in which the additional unknowns with respect to the general problem are either null or can be calculated in terms of those in (7.29), u z = 0, ε z = γ z = γ z = τ z = τ z = 0 and σ z = ν (σ + σ ). (7.30) 7.4 The Plane Linear Elastic Problem In view of the equations in Sections 7.2 and 7.3, the linear elastic problem for the plane stress and plane strain problems is characterized as follows (see Figure 7.4). Equations 4 a) Cauch s equation Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar σ + τ + ρb = ρ 2 u t 2 τ + σ + ρb = ρ 2 u t 2 (7.31) 4 The equation corresponding to the z-component either does not intervene (plane stress), or is identicall null (plane strain). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

The Plane Linear Elastic Problem 349 b) Constitutive equation {σ} not σ σ τ, {ε} not ε ε γ ; {σ} = C {ε}, (7.32) where the constitutive matri C can be written in a general form, from (7.11) and (7.28), as { Ē = E Plane stress 1 ν 0 ν = ν C not Ē 1 ν 2 ν 1 0 0 0 1 ν Ē = E (7.33) Plane strain 1 ν 2 2 ν = ν 1 ν c) Geometric equation ε = u, d) Boundar conditions in space Γ u : u not ε = u, γ = u + u [ ] [ ] u (,,t) u, Γ σ : t not t (,,t) (,,t) t (,,t) (7.34) Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar t = σ n, σ not [ σ τ τ σ ], n not [ n n ] (7.35) e) Initial conditions u(,,t) = 0, t=0. u(,,t) = v 0 (,) (7.36) t=0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

350 CHAPTER 7. PLANE LINEAR ELASTICITY Unknowns u(,,t) not [ u u ], ε (,,t) not 1 ε 2 γ 1, σ (,,t) not 2 γ ε [ σ τ τ σ ] (7.37) Equations (7.31) to(7.37) define a sstem of 8 PDEs with 8 unknowns that must be solved in the reduced space-time domain R 2 R +. Once the problem is solved, the following can be eplicitl calculated: Plane stress ε z = ν 1 ν (ε + ε ) Plane strain σ z = ν (σ + σ ) 7.5 Problems Tpicall Assimilated to Plane Elasticit 7.5.1 Plane Stress (7.38) The stress and strain states produced in solids that have a dimension considerabl inferior to the other two (which constitute the plane of analsis -) and whose actions are contained in said plane are tpicall assimilated to a plane stress state. The slab loaded on its mean plane and the deep beam of Figure 7.5 are classic eamples of structures that can be analzed as being in a plane stress state. As a particular case, the problems of simple and comple bending in beams considered in strength of materials can also be assimilated to plane stress problems. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Figure 7.5: Slab loaded on its mean plane (left) and deep beam (right). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Problems Tpicall Assimilated to Plane Elasticit 351 7.5.2 Plane Strain The solids whose geometr can be obtained b translation of a plane section with actions contained in its plane (plane of analsis -) along a generatri line perpendicular to said section are tpicall assimilated to plane strain states. In addition, the plane strain hpothesis ε z = γ z = γ z = 0 must be justifiable. In general, this situation occurs in two circumstances: 1) The dimension of the piece in the z-direction is ver large (for the purposes of analsis, it is assumed to be infinite). In this case, an central transversal section (not close to the etremes) can be considered a smmetr plane and, thus, satisfies the conditions u u z = 0, z = 0 and u z = 0, (7.39) which result in the initial condition of the plane strain state (7.17), u u (,,t) u not = u (,,t). (7.40) u u z Eamples of this case are a pipe under internal (and/or eternal) pressure (see Figure 7.6), a tunnel (see Figure 7.7) and a strip foundation (see Figure 7.8). 0 Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Figure 7.6: Pressure tube. 2) The length of the piece in the longitudinal direction is reduced, but the displacements in the z-direction are impeded b the boundar conditions at the end sections (see Figure 7.9). In this case, the plane strain hpothesis (7.17) can be assumed for all the transversal sections of the piece. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

352 CHAPTER 7. PLANE LINEAR ELASTICITY Figure 7.7: Tunnel. Figure 7.8: Strip foundation. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Figure 7.9: Solid with impeded z-displacements. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Representative Curves of Plane Elasticit 353 7.6 Representative Curves of Plane Elasticit There is an important tradition in engineering of graphicall representing the distribution of plane elasticit. To this aim, certain families of curves are used, whose plotting on the plane of analsis provides useful information of said stress state. 7.6.1 Isostatics or stress trajectories Definition 7.1. The isostatics or stress trajectories are the envelopes of the vector field determined b the principal stresses. Considering the definition of the envelope of a vector field, isostatics are, at each point, tangent to the two principal directions and, thus, there eist two families of isostatics: Isostatics σ 1, tangent to the direction of the largest principal stress. Isostatics σ 2, tangent to the direction of the smallest principal stress. In addition, since the principal stress directions are orthogonal to each other, both families of curves are also be orthogonal. The isostatic lines provide information on the mode in which the flu of principal stresses occurs on the plane of analsis. As an eample, Figure 7.10 shows the distribution of isostatics on a supported beam with uniforml distributed loading. Definition 7.2. A singular point is a point characterized b the stress state σ = σ and τ = 0 and its Mohr s circle is a point on the ais σ (see Figure 7.11). A neutral point is a singular point characterized b the stress state Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar σ = σ = τ = 0 and its Mohr s circle is the origin of the σ τ space (see Figure 7.11). X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

354 CHAPTER 7. PLANE LINEAR ELASTICITY Figure 7.10: Isostatics or stress trajectories on a beam. Figure 7.11: Singular and neutral points. Remark 7.3. All directions in a singular point are principal stress directions (the pole is the Mohr s circle itself, see Figure 7.11). Consequentl, the isostatics tend to loose their regularit in singular points and can brusquel change their direction. 7.6.1.1 Differential Equation of the Isostatics Consider the general equation of an isostatic line = f () and the value of the angle formed b the principal stress direction σ 1 with respect to the horizontal direction (see Figure 7.12), tan(2α)= 2τ = 2tanα σ σ 1 tan 2 α tanα = d not = d Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar 2τ σ σ = 2 1 ( ) 2 (7.41) ( ) 2 + σ σ τ 1 = 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Representative Curves of Plane Elasticit 355 Figure 7.12: Determination of the differential equation of the isostatics. and solving the second-order equation (7.41) for, the differential equation of the isostatics is obtained. Differential equation of the isostatics = σ (σ ) σ σ 2 ± + 1 2τ 2τ }{{} ϕ (,) (7.42) If the function ϕ (,) in (7.42) is known, this equation can be integrated to obtain the algebraic equation of the famil of isostatics, = f ()+C. (7.43) The double sign in (7.42) leads to two differential equations corresponding to the two families of isostatics. Eample 7.1 A rectangular plate is subjected to the following stress states. σ = 3 ; σ = 2 3 3 2 ; τ = 3 2 ; τ z = τ z = σ z = 0 Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Obtain and plot the singular points and distribution of isostatics. Solution The singular points are defined b σ = σ and τ = 0. Then, { σ = 3 = 0 τ = 3 2 = 0 = = 0 = = 0 = σ = 2 3 3 2 = 0 { σ = 3 σ = 2 3 3 2 = 2 3 = = 0 X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

356 CHAPTER 7. PLANE LINEAR ELASTICITY Therefore, the locus of singular points is the straight line = 0. These singular points are, in addition, neutral points (σ = σ = 0). The isostatics are obtained from (7.42), = d d = σ σ 2τ ± (σ ) σ 2 + 1, 2τ which, for the given data of this problem, results in d d = d d = = integrating = 2 2 = C 1. = C 2 Therefore, the isostatics are two families of equilateral hperboles orthogonal to each other. On the line of singular points = 0 (which divides the plate in two regions) the isostatics will brusquel change their slope. To identif the famil of isostatics σ 1, consider a point in each region: Point (1,0): σ = σ 2 = 1; σ = σ 1 =+2; τ = 0 (isostatic σ 1 in the -direction) Point ( 1,0): σ = σ 1 =+1; σ = σ 2 = 2; τ = 0 (isostatic σ 1 in the -direction) Finall, the distribution of isostatics is as follows: Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Representative Curves of Plane Elasticit 357 7.6.2 Isoclines Definition 7.3. Isoclines are the locus of the points in the plane of analsis along which the principal stress directions form a certain angle with the -ais. It follows from its definition that in all the points of a same isocline the principal stress directions are parallel to each other, forming a constant angle θ (which characterizes the isocline) with the -ais (see Figure 7.13). 7.6.2.1 Equation of the Isoclines The equation = f () of the isocline with an angle θ is obtained b establishing that the principal stress direction σ 1 forms an angle α = θ with the horizontal direction, that is, Algebraic equation of the isoclines 2τ tan(2θ)= σ σ }{{} ϕ (,) This algebraic equation allows isolating, for each value of θ, (7.44) = f (,θ), (7.45) which constitutes the equation of the famil of isoclines parametrized in terms of the angle θ. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar Figure 7.13: Isocline. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

358 CHAPTER 7. PLANE LINEAR ELASTICITY Remark 7.4. Determining the famil of isoclines allows knowing, at each point in the medium, the direction of the principal stresses and, thus, the obtainment of the isostatics can be sought. Given that isoclines can be determined b means of eperimental methods (methods based on photoelasticit) the provide, indirectl, a method for the eperimental determination of the isostatics. 7.6.3 Isobars Definition 7.4. Isobars are the locus of points in the plane of analsis with the same value of principal stress σ 1 (or σ 2 ). Two families of isobars will cross at each point of the plane of analsis: one corresponding to σ 1 and another to σ 2. Note that the isobars depend on the value of σ 1, but not on its direction (see Figure 7.14). 7.6.3.1 Equation of the Isobars The equation that provides the value of the principal stresses (see Chapter 4) implicitl defines the algebraic equation of the two families of isobars = f 1 (,c 1 ) and = f 2 (,c 2 ), Algebraic equation of the isobars σ 1 = σ (σ ) + σ σ 2 + + τ 2 2 2 = const. = c 1 }{{} ϕ 1 (,) σ 2 = σ (σ ) + σ σ 2 + τ 2 2 2 = const. = c 2 }{{} ϕ 2 (,) Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar (7.46) which leads to { = f 1 (,c 1 ) = f 2 (,c 2 ) (7.47) X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Representative Curves of Plane Elasticit 359 Figure 7.14: Isobars. 7.6.4 Maimum Shear Stress or Slip Lines Definition 7.5. Maimum shear stress lines or slip lines are the envelopes of the directions that, at each point, correspond with the maimum value (in modulus) of the shear (or tangent) stress. Remark 7.5. At each point of the plane of analsis there are two planes on which the shear stresses reach the same maimum value (in module) but that have opposite directions, τ ma and τ min. These planes can be determined b means of the Mohr s circle and form a 45 angle with the principal stress directions (see Figure 7.15). Therefore, their envelopes (maimum shear stress lines) are two families of curves orthogonal to each other that form a 45 angle with the isostatics. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar 7.6.4.1 Differential Equation of the Maimum Shear Lines Consider β is the angle formed b the direction of τ ma with the horizontal direction (see Figure 7.16). In accordance with Remark 7.5 5, β = α π ( = tan(2β)=tan 2α π ) = 1 4 2 tan(2α), (7.48) 5 Here, the trigonometric epression tan(θ π/2)= cotθ = 1/tanθ is used. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

360 CHAPTER 7. PLANE LINEAR ELASTICITY Figure 7.15: Maimum shear stress planes. where α is the angle formed b the principal stress direction σ 1 with the horizontal direction. Consequentl, considering the general equation of a slip line, = f (), the epression (7.48) and the relation tan(2α)=2τ /(σ σ ) ields tan(2β)= 1 tan(2α) = σ σ = 2tanβ 2τ 1 tan 2 β tanβ = d not = = d (7.49) σ σ = 2 2τ 1 ( ) 2 = ( ) 2 4τ 1 = 0. σ σ Solving the second-order equation in (7.49) for provides the differential equation of the maimum shear stress lines. Differential equation of the ma. shear stress or slip lines Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar ( = 2τ ) 2 2τ ± + 1 σ σ σ σ }{{} ϕ (,) (7.50) If the function ϕ (,) in (7.50) is known, this differential equation can be integrated and the algebraic equation of the two families of orthogonal curves (corresponding to the double sign in (7.50)) is obtained. X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Representative Curves of Plane Elasticit 361 Figure 7.16: Maimum shear stress or slip lines. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

362 CHAPTER 7. PLANE LINEAR ELASTICITY Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Problems and Eercises 363 PROBLEMS Problem 7.1 Justif whether the following statements are true or false. a) If a plane stress state has a singular point, all the isoclines cross this point. b) If a plane stress state is uniform, all the slip lines are parallel to each other. Solution a) A singular point is defined as: { σ1 = σ 2 τ = 0 The stress state is represented b a point. Therefore, all directions are principal stress directions and, given an angle θ which can take an value, the principal stress direction will form an angle θ with the -ais. Then, an isocline of angle θ will cross said point and, since this holds true for an value of θ, all the isoclines will cross this point. Therefore, the statement is true. b) A uniform stress state implies that the Mohr s circle is equal in all points of the medium, therefore, the planes of maimum shear stress will be the same in all points. Then, the maimum shear stress lines (or slip lines) will be parallel to each other. In conclusion, the statement is true. Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

364 CHAPTER 7. PLANE LINEAR ELASTICITY Problem 7.2 A rectangular plate is subjected to the following plane stress states. 1) σ = 0 ; σ = b > 0 ; τ = 0 2) σ = 0 ; σ = 0 ; τ = m, m > 0 Plot for each state the isostatics and the slip lines, and indicate the singular points. Solution 1) The Mohr s circle for the stress state σ = 0;σ = b > 0;τ = 0 is: Then, the isostatics are: Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Problems and Eercises 365 And the slip lines are: There do not eist singular points for this stress state. 2) The Mohr s circle for the stress state σ = 0;σ = 0;τ = m, m > 0 is: Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

366 CHAPTER 7. PLANE LINEAR ELASTICITY Then, the isostatics and singular points are: And the slip lines are: Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961

Problems and Eercises 367 EXERCISES 7.1 A rectangular plate is subjected to the following plane strain state: σ = σ τ = a σ = b (a > 0, b > 0) Plot the isostatics and the slip lines, and indicate the singular points. 7.2 Plot the isostatics in the transversal section of the clindrical shell shown below. Assume a field of the form: u r = Ar + B r ; A > 0, B > 0 u θ = 0 u z = 0 Continuum Mechanics for Engineers Theor and Problems X. Oliver and C. Agelet de Saracibar X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theor and Problems doi:10.13140/rg.2.2.25821.20961