Ch.7. Plane Linear Elasticity

CH.7. PLANE LINEAR ELASTICITY

Multimedia Course on Continuum Mechanics Overview

 Plane Linear Elasticity Theory Lecture 1  Plane Stress  Simplifying Hypothesis  Strain Field Lecture 2  Constitutive Equation  Displacement Field  The Linear Elastic Problem in Plane Stress Lecture 3  Examples  Plane Strain  Simplifying Hypothesis  Strain Field  Constitutive Equation Lecture 4  Stress Field  The Linear Elastic Problem in Plane Stress  Examples

2 Overview (cont’d)

 The Plane Linear Elastic Problem Lecture 5  Governing Equations  Representative Curves  Isostatics or stress trajectories  Isoclines  Isobars Lecture 6  Maximum shear lines

3 7.1 Plane Linear Elasticity Theory

Ch.7. Plane Linear Elasticity

4 Plane Linear Elasticity

 For some problems, one of the principal directions is known a priori:  Due to particular geometries, loading and boundary conditions involved.  The elastic problem can be solved independently for this direction.  Setting the known direction as z, the elastic problem analysis is reduced to the x-y plane PLANE ELASTICITY

 There are two main classes of plane linear elastic problems:  Plane stress REMARK  Plane strain The isothermal case will not be studied here for the sake of simplicity. Generalization of the results obtained to thermo-elasticity is straight-forward.

7 7.2 Plane Stress

Ch.7. Plane Linear Elasticity

8 Hypothesis on the Stress Tensor

 Simplifying hypothesis of a plane stress linear elastic problem: 1. Only stresses “contained in the x-y plane” are not null

στ 0 x xy σ ≡ τσ []xyz xy y 0  0 00

2. The stress are independent of the z direction.

σσxx= ()xyt,, σσ= yy()xyt,, REMARK The name “plane stress” arises ττxy= xy ()xyt,, from the fact that all (not null) stress are contained in the x-y plane.

9 Geometry and Actions in Plane Stress

 These hypothesis are valid when:

 The thickness is much smaller than the typical dimension associated to the plane of analysis: eL<<

 The actions bx () , t , ux * () , t and tx * () , t are contained in the plane of analysis (in-plane actions) and independent of the third dimension, z.

 tx * () , t is only non-zero on the contour of the body’s thickness:

10 Strain Field in Plane Stress

 The strain field is obtained from the inverse Hooke’s Law:

1 2(1+ν ) εx =−==() σx νσ y γxy2 ε xy τ xy νν 1+ EE εσσ=−+Tr () 1 EE 1 σ εy =−==() σy νσ x γxz20 ε xz z=0 E τ xz=0 ν τ =0 εz =−+() σσx y γyz ==20 ε yz yz E σσ=  As xx()xyt,, εε= ()xyt,, σσyy= ()xyt,,

 And the strain tensor for plane stress is: 1 εγ x xy 0 2 ν 1 ε=−+ εε ε()xyt,, ≡ γε0 with z ()xy 2 xy y 1−ν  00ε z 11 Constitutive equation in Plane Stress

 Operating on the result yields: E +ν σ = ε+ νε 1 2(1 ) x 2 xy εx =−==() σx νσ y γxy2 ε xy τ xy 1 −ν EE() E 1 σ = ε+ νε εy =−==() σy νσ x γxz20 ε xz y 2 yx E ()1 −ν ν ε=−+ σσ γ == ε E z ()x y yz20 yz τγ= E plane xy xz stress 21()+ν = C

σxx10 νε  E  σν= 10  ε yy1−ν 2  τ 1−νγ xy 00 xy  = {}σ 2 = {}ε

Constitutive equation plane stress in plane stress {}{}σε=C ⋅ (Voigt’s notation) 12 Displacement Field in Plane Stress

 The displacement field is obtained from the geometric equations, ε∇ ()() x ,, tt = S ux . These are split into:

 Those which do not affect the displacement u z : ∂u ε ()xyt,, = x x ∂x ∂u u= u() xyt,, ε ()xyt,, = y Integration xx y ∂y in . uyy= u() xyt,, ∂u ∂u γε ()xyt,,= 2 =x + y xy xy ∂∂yx

 Those in which u z appears:

ν ∂uz εz()xyt,, =−+=()εεxy(,,)xyt ⇒uz(,,,) xyzt 1−∂ν z

∂ux(, xy ) ∂∂ u zz u  γεxz ()xyt,,= 2xz = +==0 Contradiction !!! ∂z ∂∂ xx      =0   ⇒ u(,) zt ∂u(, xy ) ∂∂uu z γε()xyt,,= 2 =y +==zz0 yz yz    ∂z  ∂∂ yy =0  13 The Linear Elastic Problem in Plane Stress

 The problem can be reduced to the two dimensions of the plane of analysis.  The unknowns are: ε σ u x x ≡ x   u()xyt,,  {}()ε xyt,, ≡ ε y {}()σ xyt,, ≡ σ y uy γ τ xy xy  The additional unknowns (with respect to the general problem) are either null, or independently obtained, or irrelevant: REMARK στz= xz = τ yz = γ xz = γ yz = 0 This is an ideal elastic problem because it ν εz=−+() εεxy cannot be exactly reproduced as a particular 1−ν case of the 3D elastic problem. There is no does not appear u xyzt,,, guarantee that the solution to u x () xyt ,, and z () in the problem u y () xyt ,, will allow obtaining the solution to

u z () xyzt ,,, for the additional geometric eqns.

14 Examples of Plane Stress Analysis

 3D problems which are typically assimilated to a plane stress state are characterized by:  One of the body’s dimensions is significantly smaller than the other two.  The actions are contained in the plane formed by the two “large” dimensions.

Slab loaded on Deep beam the mean plane

15 7.3 Plane Strain

Ch.7. Plane Linear Elasticity

16 Hypothesis on the Displacement Field

 Simplifying hypothesis of a plane strain linear elastic problem: 1. The displacement field is

u x u = uy  0

2. The displacement variables associated to the x-y plane are independent of the z direction.

uxx= u() xyt,,

uyy= u() xyt,,

17 Geometry and Actions in Plane Strain

 These hypothesis are valid when:

 The body being studied is generated by moving the plane of analysis along a generational line.

 The actions bx () , t , ux * () , t and tx * () , t are contained in the plane of analysis and independent of the third dimension, z.

 In the central section, considered as the “analysis section” the following holds (approximately) true:

uz = 0 ∂u x = 0 ∂z ∂u y = 0 ∂z

18 uz = 0 ∂u x = 0 ∂z

Strain Field in Plane Strain ∂uy = 0 ∂z

 The strain field is obtained from the geometric equations: ∂u() xyt,, ∂u εε()xyt,, = x =z = 0 xz∂∂xz ∂u() xyt,, ∂u() xyt,, ∂u εγ()xyt,, =y =x +=z 0 y ∂y xz ∂∂zx ∂u() xyt,, ∂∂u() xyt,, u() xyt,, ∂u γγ()xyt,, =x +yy= +=z 0 xy ∂y ∂ x yz ∂∂zy

 And the strain tensor for plane strain is:

1 εγ x xy 0 2 1 REMARK ε()xyt,, ≡ γε0 2 xy y The name “plane strain” arises  0 00 from the fact that all strain is  contained in the x-y plane.

19 Stress Field in Plane Strain

 Introducing the strain tensor into Hooke’s Law ( σ = λ Tr () ε 1 + 2ε G ) and operating on the result yields:

σx= λε() x ++ ε y 2GG εx τxy= γ xy

σy= λε() x ++ ε y 20GG εy τxz= γ xz = =++()λ2G εyx λε σλεεz=() x += y vG() σσx + y τyz= γ yz =0

εεxx= ()xyt,,  As εε= ()xyt,, yy σσ= ()xyt,, εεzz= ()xyt,,

γγxy= xy ()xyt,,

στx xy 0  And the stress tensor  σ ()xyt,, ≡ τσxy y 0 with σ= νσ + σ for plane strain is:  z() xy 00σ z

20 Constitutive equation in Plane Strain

 Introducing the values of the strain tensor into the constitutive equation and operating on the result yields: E ()1−ν ν σ=+() λ2G ε += λε ε + ε x xy()()1+−νν 12x 1 −ν y σεε=λµTr ()1 + 2 E ()1−ν ν σ=+() λ2G ε += λε ε+ ε y yx()()1+−νν 12y 1 − ν x E τγ=G = γ xy xy 21()+ν xy

plane = Cstrain ν 10 plane  1−ν  strain σεxx{}{}σε=C ⋅  E ()1−ν ν  σε yy= 10 ()()1+−ν 12 νν 1 − Constitutive equation τγ xy 12− ν xy in plane strain 00 = {}ε = {}σ 21()−ν (Voigt’s notation)

21 The Lineal Elastic Problem in Plane Strain (summary)

 The problem can be reduced to the two dimensions of the plane of analysis.  The unknowns are:

ε σ u x x ≡ x   u()xyt,,  {}()ε xyt,, ≡ ε y {}()σ xyt,, ≡ σ y uy γ τ xy xy

 The additional unknowns (with respect to the general problem) are either null or obtained from the unknowns of the problem:

uz = 0

εγz= xz = γ yz = τ xz = τ yz = 0

σz= νσ() xy + σ

22 Examples of Plane Strain Analysis

 3D problems which are typically assimilated to a plane strain state are characterized by:  The body is generated by translating a generational section with actions contained in its plane along a line perpendicular to this plane.

 The plane strain hypothesis ( εγ z = xz = γ yz = 0 ) must be justifiable. This typically occurs when: 1. One of the body’s dimensions is significantly larger than the other two. Any section not close to the extremes can be considered a symmetry plane and satisfies:

uz = 0 ∂u ux x = 0  ∂z u = uy  ∂uy 0 = 0  ∂z 2. The displacement in z is blocked at the extreme sections.

23 Examples of Plane Strain Analysis

 3D problems which are typically assimilated to a plane strain state are:

Pressure pipe Continuous brake shoe

Solid with blocked z Tunnel displacements at the ends

24 7.4 The Plane Linear Elastic Problem

Ch.7. Plane Linear Elasticity

25 Plane problem

 A lineal elastic solid is subjected to body forces and prescribed traction and displacement Actions: * tx () xyt,, On Γ : t* = σ * ty () xyt,, * ux () xyt,, On Γ : u* = u * uy () xyt,,

bx () xyt,, On Ω : b =  by () xyt,,

 The Plane Linear Elastic problem is the set of equations that allow obtaining the evolution through time of the corresponding displacements u () xyt ,, , strains ε () xyt ,, and stresses σ () xyt ,, .

26 Governing Equations

 The Plane Linear Elastic Problem is governed by the equations: 1. Cauchy’s Equation of Motion. Linear Momentum Balance Equation. ∂2ux(),t ∇σ ⋅+()()x,,ttρρ bx = 00∂ t 2 2D

∂∂στ∂τ ∂2u x +xy +xz +=ρρb x ∂∂∂xyzx ∂ t2 ∂∂∂τ στ ∂2u xy+ y + yz +=ρρb y ∂∂∂xyzy ∂ t2 ∂τ ∂τ ∂∂σ 2u xz +yz +zz +=ρρb ∂∂∂xyzz ∂ t2

27 Governing Equations

 The Plane Linear Elastic Problem is governed by the equations: 2. Constitutive Equation (Voigt’s notation). Isotropic Linear Elastic Constitutive Equation. σε()x,:t = C 2D

{}{}σε=C ⋅

σ x ε x 10ν   E  With {} σ ≡  σ y , {} ε =  ε y and C = ν 10 1−ν 2  τ γ  xy xy 00() 1−ν 2 E = E 2 PLANE EE= PLANE 1−ν ν STRESS νν= STRAIN ν = ()1−ν

28 Governing Equations

 The Plane Linear Elastic Problem is governed by the equations: 3. Geometrical Equation. Kinematic Compatibility. 1 ε()()x,,tt= ∇S ux =( u ⊗+⊗∇∇ u) 2

This is a PDE system of ∂u ε = x 8 eqns -8 unknowns: x ∂x ∂u ux(),t 2 unknowns ε = y y ∂y ε()x,t 3 unknowns ∂u ∂ux y σ ()x,t 3 unknowns γ xy = + ∂∂yx Which must be solved in 2 the  × + space.

29 Boundary Conditions

 Boundary conditions in space  Affect the spatial arguments of the unknowns  Are applied on the contour Γ of the solid, which is divided into:  Prescribed displacements on Γ u : ** uxx= u() xyt,, u* =  **= uyy u() xyt,,

 Prescribed stresses on Γ σ : nx n =  n ** y txx= t() xyt,, * t* = tn=σ ⋅ with ** t= t xyt,, στx xy yy() σ ≡ τσ xy y

30 Boundary Conditions

 INTIAL CONDITIONS (boundary conditions in time)  Affect the time argument of the unknowns.  Generally, they are the known values at t = 0 :  Initial displacements: ux u0()xy, ,0 = = uy  Initial velocity:

∂u()xyt,, uxxv ≡===uv ()xy, ,0  ()xy, ∂ u v 0 t t=0 yy

31 Unknowns

 The 8 unknowns to be solved in the problem are: 1 εγ u x xy x 2 στx xy u(,xyt ,)=  ε( xyt,,) ≡ σ ( xyt,,) ≡ u τσ y 1 xy y γεxy y 2

 Once these are obtained, the following are calculated explicitly: ν PLANE STRESS εz=−+( εεxy) 1 −ν

PLANE STRAIN σz= νσ( xy + σ)

32 7.5 Representative Curves

Ch.7. Plane Linear Elasticity

33 Introduction

 Traditionally, plane stress states where graphically represented with the aid of the following contour lines:  Isostatics or stress trajectories  Isoclines  Isobars  Maximum shear lines  Others: isochromatics, isopatchs, etc.

34 Isostatics or Stress Trajectories

 System of curves which are tangent to the principal axes of stress at each material point .  They are the envelopes of the principal stress vector fields.  There will exist two (orthogonal) families of curves at each point:  Isostatics σ 1 , tangents to the largest principal stress.  Isostatics σ 2 , 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 by the stress state

σσxy=

τ xy = 0  Neutral point: characterized by the stress state

σστx= y = xy = 0

Mohr’s Circle of REMARK a singular point In a singular point, all directions are principal directions. Thus, in singular points isostatics tend to Mohr’s Circle of loose their regularity and can a neutral point abruptly change direction.

36 Differential Equation of the Isostatics

 Consider the general equation of an isostatic curve: y= fx()

2τ 2 tgα α =xy = tg() 2 2 σσxy−−1 tg α dy tgα = = y′ dx

2τ ′ σσ− xy 2 y ′′2 xy = 2 ()yy+ −=10 σσxy− 1− ()y′ τ xy

φ ()xy, Known this function, nd  Solving the 2 order eq.: the eq. can be integrated

2 to obtain a family of σσ−  ()xy σσxy− curves of the type: Differential equation y '1=−±+ of the isostatics 22ττ xy xy y= fx() + 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 x-axis.

 These curves can be directly found using photoelasticity methods.

38 Equation of the Isoclines

 To obtain the general equation of an isocline with angle θ , the

principal stress σ 1 must form an angle αθ = with the x-axis: ϕ ()xy, 2τ Algebraic equation tg() 2θ = xy For each value of θ , the equation of of the isoclines the family of isoclines parameterized σσxy− in function of θ is obtained: y= fx(),θ

REMARK Once the family of isoclines is known, the principal stress directions in any point of the medium can be obtained and, thus, the isostatics calculated.

39 Maximum shear lines

 Envelopes of the maximum shear stress (in modulus) vector fields.  They are the curves on which the shear stress modulus is a maximum.  Two planes of maximum shear stress correspond to each material

point, τ max and τ min .  These planes are easily 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. They form an angle of 45º with the isostatics.

42 Equation of the maximum shear lines

 Consider the general equation of a slip line y = fx ( ), the relation 2τ xy π π 1 tan 2 α = and βα= + tan( 2βα) = tan 2 −=− α σσxy− 4 2 tan 2  Then, 1 σσ− 2 tan β tan( 2β ) =−=−=xy tan( 2ατ) 2 1− tan 2 β σσ− 2 y′ xy −=xy not 2 dy 2τ xy 1− y′ tan (β ) = = y′ ( ) dx

2 4τ ( yy′′) −xy −=10 σσxy−

43 Equation of the maximum shear lines

nd  Solving the 2 order eq.: φ ( xy, ) Known this function, the eq. can be integrated Differential 2 to obtain a family of 22ττ equation of the y '1=±+xy xy curves of the type: σσ−− σσ slip lines xy xy y= fx( ) + C

44 Chapter 7 Plane Linear Elasticity

7.1 Introduction As seen in Chapter 6, from a mathematical point of view, the elastic problem consists in a system of PDEs that must be solved in the three dimensions of space and in the dimension associated with time R3 × R+ . However, in certain situations, the problem can be simpliﬁed so that it is reduced to two dimensions in space in addition to, obviously, the temporal dimension R2 × R+ . This sim- pliﬁcation is possible because, in certain cases, the geometry and boundary conditions of the problem allow identifying 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 system {x,y,z} in which the aforementioned irrelevant direction (assumed constant) coincides with the z-direction. Then, the analysis is reduced to the x-y plane and, hence, the name plane elasticity used to denote such problems. In turn, these are typically divided into two large groups associated with two familiesTheory of simplifying and hypotheses, Problemsplane stress problems and plane strain problems. For the sake of simplicity, the isothermal case will be considered here, even thoughContinuum there is no intrinsic Mechanics limitation to generalizing for Engineers the results that will be obtained to the thermoelastic© X. case. Oliver and C. Agelet de Saracibar

7.2 Plane Stress State The plane stress state is characterized by the following simplifying hypotheses: 1) The stress state is of the type ⎡ ⎤ σ τ ⎢ x xy 0 ⎥ [σ ] not≡ ⎢ τ σ ⎥ . xyz ⎣ xy y 0 ⎦ (7.1) 00 0

339 340 CHAPTER 7. PLANE LINEAR ELASTICITY

2) The non-zero stresses (that is, those associated with the x-y plane) do not depend on the z-variable,

σx = σx (x,y,t) , σy = σy (x,y,t) and τxy = τxy (x,y,t) . (7.2)

To analyze under which conditions these hypotheses are reasonable, consider a plane elastic medium whose dimensions and form associated with the x-y plane (denoted as plane of analysis) are arbitrary and such that the third dimension (denoted as the thickness of the piece) is associated with the z-axis (see Figure 7.1). Assume the following circumstances hold for this elastic medium: a) The thickness e is much smaller than the typical dimension associated with the plane of analysis x-y, e L . (7.3) b) The actions (body forces b(x,t), prescribed displacements u∗ (x,t) and traction vector t∗ (x,t) ) are contained within the plane of analysis x-y (its z- component is null) and, in addition, do not depend on the third dimension, ⎡ ⎤ ⎡ ⎤ b (x,y,t) u∗ (x,y,t) ⎢ x ⎥ ⎢ x ⎥ not≡ ⎢ ( , , ) ⎥ , Γ ∗ not≡ ⎢ ∗ ( , , ) ⎥ , b ⎣ by x y t ⎦ u : u ⎣ uy x y t ⎦ 0 − ⎡ ⎤ (7.4) t∗ (x,y,t) ⎢ x ⎥ + − ∗ not ⎢ ∗ ⎥ Γσ = Γ Γ Γ e ≡ ( , , ) . σ σ σ : t ⎣ty x y t ⎦ Theory and Problems− c) The traction vector t∗ (x,t) is only non-zero on the boundary of the piece’s Γ e Γ + Γ − thicknessContinuum(boundary σ ), Mechanics whilst on the lateral for surfaces Engineersσ and σ it is null (see Figure 7.1). © X. Oliver and C. Agelet de Saracibar ⎡ ⎤ 0 ⎢ ⎥ + − ∗ not ⎢ ⎥ Γσ Γσ : t ≡ ⎣ 0 ⎦ . (7.5) 0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Plane Stress State 341

Figure 7.1: Example of a plane stress state.

Remark 7.1. The piece with the actions deﬁned by (7.4) and (7.5)is compatible with the plane stress state given by (7.1) and (7.2), and schematized in Figure 7.21. In effect, applying the boundary conditions Γσ on the piece yields: + − • Lateral surfaces Γσ and Γσ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 σ τ 0 0 0 ⎢ ⎥ ⎢ x xy ⎥⎢ ⎥ ⎢ ⎥ not≡ , σ · not≡ = , n ⎣ 0 ⎦ n ⎣ τxy σy 0 ⎦⎣ 0 ⎦ ⎣ 0 ⎦ ±1 000 ±1 0 Theory and Problems e • Edge Γσ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ Continuumn Mechanicsσx τxy for0 Engineersn t (x,y,t) ⎢ x ⎥ © X. Oliver⎢ and C. Agelet⎥⎢ x de⎥ Saracibar⎢ x ⎥ not≡ , σ ( , , )· not≡ ⎢ ⎥ = , n ⎣ ny ⎦ x y t n ⎣ τxy σy 0 ⎦⎣ ny ⎦ ⎣ty (x,y,t) ⎦ 0 000 0 0 which is compatible with the assumptions (7.4) and (7.5).

1 The fact that all the non-null stresses are contained in the x-y plane is what gives rise to the name plane stress.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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), ν 1 + ν ν 1 ε = − Tr (σ )1 + σ = − Tr (σ )1 + σ , (7.6) E E E 2G which, applied on the stress state in (7.1) and in engineering notation, provides the strains (6.25)2

ε = 1 (σ − ν (σ + σ )) = 1 (σ − νσ ) γ = 1 τ , x E x y z E x y xy G xy 1 1 1 ε = (σ − ν (σ Theory+ σ )) = ( andσ − νσ Problems) γ = τ = 0 , (7.7) y E y x z E y x xz G xz ν ε = 1 (σ − ν (σ + σ )) = − (σ + σ ) γ = 1 τ = , Continuumz z x Mechanicsy x fory Engineersyz yz 0 E © X. OliverE and C. Agelet deG Saracibar where the conditions σz = τxz = τyz = 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 (x,y,t)). In addition, the strain εz in (7.7) can be solved as ν ε = − (ε + ε ) . (7.8) z 1 − ν x y

2 The engineering angular strains are deﬁned as γxy = 2εxy, γxz = 2εxz and γyz = 2εyz.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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 1γ ⎢ x xy 0 ⎥ ⎢ 2 ⎥ ν ε ( , , ) not≡ 1 ε = − (ε + ε ) x y t ⎢ γ ε 0 ⎥ with z x y (7.9) ⎣ 2 xy y ⎦ 1 − ν 00εz and replacing (7.8)in(7.7) leads, after certain algebraic operations, to

σ = E (ε + νε ) , σ = E (ε + νε ) , x − ν2 x y y − ν2 y x 1 1 (7.10) E and τ = γ , xy 2(1 + ν) xy which can be rewritten as ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ σ ν ε x ⎢ 1 0 ⎥ x ⎢ ⎥ E ⎢ ⎥ ⎢ ⎥ ⎢ σ ⎥ = ⎢ ν 10⎥ ⎢ ε ⎥ =⇒ {σ } = C plane · {ε} . ⎣ y ⎦ − ν2 ⎣ ⎦ ⎣ y ⎦ stress 1 1 − ν τxy 00 γxy 2 {σ } plane {ε} C stress (7.11)

7.2.2 Displacement Field Theory and Problems The components of the geometric equation of the problem (6.3), 1 Continuumε (x,t)= Mechanics∇Su(x,t)= (u ⊗ for∇ + ∇ Engineers⊗ u) , (7.12) © X. Oliver and2 C. Agelet de Saracibar can be decomposed into two groups:

1) Those that do not affect the displacement uz (and are hypothetically inte- grable in R2 for the x-y domain), ⎫ ∂ ⎪ ε ( , , )= ux ⎪ x x y t ⎪ integration ∂x ⎪ 2 ⎬ in R = ( , , ) ∂ ux ux x y t ε ( , , )= uy =⇒ . (7.13) y x y t ⎪ = ( , , ) ∂y ⎪ uy uy x y t ∂u ∂u ⎪ γ (x,y,t)=2ε = x + y ⎭ xy xy ∂y ∂x

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 344 CHAPTER 7. PLANE LINEAR ELASTICITY

2) Those in which the displacement uz intervenes,

∂uz ν εz (x,y,t) = = − (εx + εy) , ∂z 1 − ν ∂ux ∂uz γxz (x,y,t) = 2εxz = + = 0 , (7.14) ∂z ∂x ∂uy ∂uz γyz (x,y,t) = 2εyz = + = 0 . ∂z ∂y Observation of (7.1) to (7.14) suggests considering an ideal elastic plane stress problem reduced to the two dimensions of the plane of analysis and characterized by the unknowns     εx σx not ux not not u(x,y,t) ≡ , {ε (x,y,t)} ≡  εy  and {σ (x,y,t)} ≡  σy , (7.15) uy γxy or Engineersτxy 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, ν σz = τxz = τyz = γxz = γyz = 0 , εz = − (εx + εy) , 1 − ν (7.16) and uz (x,y,z,t)Mechanicsdoes not intervene f in the problem.

Remark 7.2. The plane stress problem is an ideal elastic problem since it cannot be exactlyTheory reproduced and as Problems a particular case of a three- dimensional elastic problem. In effect, there is no guarantee that the solution of the reduced plane stress ux (x,y,t) and uy (x,y,t) will al- lowContinuum obtaining a solution uz (x,y,z,t) for the rest of components of the geometric equation© X. (7.14 Oliver). and C. Agelet de Saracibar

7.3 Plane Strain The strain state is characterized by the simplifying hypotheses     ux ux (x,y,t) not u ≡  uy  =  uy (x,y,t)  . (7.17) uz 0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Plane Strain 345

Again, it is illustrative to analyze in which situations these hypotheses are plau- sible. Consider, for example, an elastic medium whose geometry and actions can be generated from a bidimensional section (associated with the x-y plane and with the actions b(x,t), u∗ (x,t) and t∗ (x,t) contained in this plane) that is translated along a straight generatrix perpendicular to said section and, thus, associated with the z-axis (see Figure 7.3). The actions of the problem can then be characterized by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ b (x,y,t) u∗ (x,y,t) t∗ (x,y,t) ⎢ x ⎥ ⎢ x ⎥ ⎢ x ⎥ not≡ ⎣ ( , , ) ⎦, Γ ∗ not≡ ⎣ ∗ ( , , ) ⎦ Γ ∗ not≡ ⎣ ∗ ( , , ) ⎦. b by x y t u : u uy x y t and σ : t ty x y t (7.18) 0 0 0

In the central section (which is a plane of symmetry with respect to the z-axis) the conditions ∂u ∂u u = 0 , x = 0 and y = 0 (7.19) z ∂z ∂z are satisﬁed and, thus, the displacement ﬁeld in this central section is of the form ⎡ ⎤ u (x,y,t) ⎢ x ⎥ ( , , ) not≡ . u x y t ⎣ uy (x,y,t) ⎦ (7.20) 0

Theory and Problems

Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 7.3: Example of a plane strain state.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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

∂u ∂u ε (x,y,t)= x , ε (x,y,t)= z = 0 , x ∂x z ∂z ∂u ∂u ∂u ε (x,y,t)= y , γ (x,y,t)= x + z = 0 , (7.21) y ∂y xz ∂z ∂x ∂u ∂u ∂u ∂u γ (x,y,t)= x + y , γ (x,y,t)= y + z = 0 . xy ∂y ∂x yz ∂z ∂y

Therefore, the structure of the strain tensor is3 ⎡ ⎤ 1 ε γ 0 ⎢ x 2 xy ⎥ not ⎢ 1 ⎥ ε (x,y,t) ≡ ⎢ γ ε 0 ⎥ . (7.22) ⎣ 2 xy y ⎦ 000

Consider now the lineal elastic constitutive equation (6.20) σ = λ Tr (ε)1 + 2μεε = λ Tr (ε)1 + 2Gε , (7.23) which, applied to the strain ﬁeld (7.21), produces the stresses

σ = λ (ε + ε )+ με =(λ + )ε + λε , τ = γ , x x y 2 Theoryx 2G andx Problemsy xy G xy σy = λ (εx + εy)+2μεy =(λ + 2G)εy + λεx , τxz = Gγxz = 0 , (7.24) σ =Continuumλ (ε + ε ) , Mechanics for Engineersτ = Gγ = 0 . z x y © X. Oliver and C. Ageletyz de Saracibaryz Considering (7.21) and (7.24), one concludes that stresses do not depend on the z-coordinate either (σ = σ (x,y,t)). On the other hand, the stress σz in (7.24) can be solved as λ σ = (σ + σ )=ν (σ + σ ) (7.25) z 2(λ + μ) x y x y

3 By analogy with the plane stress case, the fact that all non-null strains are contained in the x-y plane gives rise to the name plane strain.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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 τ 0 ⎢ x xy ⎥ σ ( , , ) not≡ ⎢ ⎥ σ = −ν (σ + σ ) , x y t ⎣ τxy σy 0 ⎦ with z x y (7.26)

00σz where the non-null components of the stress tensor (7.26) are E (1 − ν) ν σ =(λ + 2G)ε + λε = ε + ε , x x y (1 + ν)(1 − 2ν) x 1 − ν y E (1 − ν) ν σ =(λ + 2G)ε + λε = ε + ε , (7.27) y y x (1 + ν)(1 − 2ν) y 1 − ν x E and τ = Gγ = γ . xy xy 2(1 + ν) xy Equation (7.27) can be rewritten in matrix form as ⎡ ⎤ ⎡ ⎤ ν ⎡ ⎤ σ ⎢ 1 − ν 0 ⎥ ε ⎢ x ⎥ ⎢ ν 1 ⎥ ⎢ x ⎥ ⎢ ⎥ E (1 − ν) ⎢ ⎥ ⎢ ⎥ ⎣ σ ⎦ = ⎢ 10⎥ ⎣ ε ⎦ ⇒ y (1 + ν)(1 − 2ν) ⎣ 1 − ν ⎦ y 1 − 2ν τxy 00 γxy ( − ν) 2 1 {σ } {ε} (7.28) plane TheoryC strain and Problems

Continuum Mechanicsplane for. Engineers ©{ X.σ } Oliver= C strain and· {ε C.} Agelet de Saracibar

Similarly 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 analysis x-y and characterized by the unknowns ⎡ ⎤ ⎡ ⎤ ε σ ⎢ x ⎥ ⎢ x ⎥ ( , , ) not≡ ux , {ε ( , , )} not≡ {σ ( , , )} not≡ , u x y t x y t ⎣ εy ⎦ and x y t ⎣ σy ⎦ (7.29) uy γxy τxy

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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),

uz = 0 , εz = γxz = γyz = τxz = τyz = 0 and σz = ν (σx + σy). (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 Fig- ure 7.4).

Equations 4 Theory and Problems a) Cauchy’s equation Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar ∂σ ∂τ ∂ 2u x + xy + ρb = ρ x ∂ ∂ x ∂ 2 x y t (7.31) ∂τ ∂σ ∂ 2u xy + y + ρb = ρ y ∂x ∂y y ∂t2

4 The equation corresponding to the z-component either does not intervene (plane stress), or is identically null (plane strain).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 The Plane Linear Elastic Problem 349

b) Constitutive equation ⎡ ⎤ ⎡ ⎤ σ ε ⎢ x ⎥ ⎢ x ⎥ {σ } not≡ , {ε} not≡ {σ } = C · {ε} , ⎣ σy ⎦ ⎣ εy ⎦ ; (7.32)

τxy γxy where the constitutive matrix C can be written in a general form, from (7.11) and (7.28), as E¯ = E ⎡ ⎤ Plane stress ν ν¯ = ν ⎢ 1 ¯ 0 ⎥ E¯ ⎢ ⎥ ⎧ C not≡ ⎢ ν¯ 10⎥ ⎪ (7.33) 2 ⎨⎪ E 1 − ν¯ ⎣ − ν ⎦ E¯ = 1 ¯ 1 − ν2 00 Plane strain ⎪ 2 ⎩⎪ ν ν¯ = 1 − ν c) Geometric equation

∂u ∂u ∂u ∂u ε = x , ε = y , γ = x + y (7.34) x ∂x y ∂y xy ∂y ∂x

d) Boundary conditions in space ∗ ( , , ) ∗ ( , , ) ∗ not ux x y t ∗ not tx x y t Γ : u ≡ , Γσ : t ≡ u ∗Theory( , , ) and Problems∗ ( , , ) uy x y t ty x y t (7.35) Continuum Mechanicsσ τ for Engineersn t∗ = σ · n ,© X.σ Olivernot≡ x andxy C., Ageletn not≡ dex Saracibar τxy σy ny e) Initial conditions . u(x,y,t) = 0 , u(x,y,t) = v0 (x,y) (7.36) t=0 t=0

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 350 CHAPTER 7. PLANE LINEAR ELASTICITY

Unknowns ⎡ ⎤ ε 1γ u ⎢ x xy ⎥ σ τ ( , , ) not≡ x , ε ( , , ) not≡ ⎣ 2 ⎦, σ ( , , ) not≡ x xy (7.37) u x y t x y t 1 x y t uy γ ε τxy σy 2 xy y

Equations (7.31)to(7.37) deﬁne a system of 8 PDEs with 8 unknowns that must be solved in the reduced space-time domain R2 × R+. Once the problem is solved, the following can be explicitly calculated: ν Plane stress → εz = (εx + εy) 1 − ν (7.38) Plane strain → σz = ν (σx + σy)

7.5 Problems Typically Assimilated to Plane Elasticity 7.5.1 Plane Stress The stress and strain states produced in solids that have a dimension consider- ably inferior to the other two (which constitute the plane of analysis x-y) and whose actions are contained in said plane are typically assimilated to a plane stress state. The slab loaded on its mean plane and the deep beam of Figure 7.5 are classic examples of structures that can be analyzed as being in a plane stress state. As a particular case, the problems of simple and complex bending in beams considered in strength of materials can also be assimilated to plane stress problems. Theory and Problems

Continuum Mechanics for Engineers © 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.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems Typically Assimilated to Plane Elasticity 351

7.5.2 Plane Strain The solids whose geometry can be obtained by translation of a plane section with actions contained in its plane (plane of analysis x-y) along a generatrix line perpendicular to said section are typically assimilated to plane strain states. In addition, the plane strain hypothesis εz = γxz = γyz = 0 must be justifiable. In general, this situation occurs in two circumstances: 1) The dimension of the piece in the z-direction is very large (for the purposes of analysis, it is assumed to be infinite). In this case, any central transversal section (not close to the extremes) can be considered a symmetry plane and, thus, satisfies the conditions ∂u ∂u u = 0 , x = 0 and y = 0 , (7.39) z ∂z ∂z which result in the initial condition of the plane strain state (7.17), ⎡ ⎤ ⎡ ⎤ u u (x,y,t) ⎢ x ⎥ ⎢ x ⎥ not≡ = . u ⎣ uy ⎦ ⎣ uy (x,y,t) ⎦ (7.40)

uz 0 Examples of this case are a pipe under internal (and/or external) pressure (see Figure 7.6), a tunnel (see Figure 7.7) and a strip foundation (see Fig- ure 7.8).

Theory and Problems

Continuum Mechanics for Engineers © 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 by the boundary conditions at the end sections (see Figure 7.9). In this case, the plane strain hypothesis (7.17) can be assumed for all the transversal sections of the piece.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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. Theory and Problems

Continuum Mechanics for Engineers © 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.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Representative Curves of Plane Elasticity 353

7.6 Representative Curves of Plane Elasticity There is an important tradition in engineering of graphically representing the distribution of plane elasticity. To this aim, certain families of curves are used, whose plotting on the plane of analysis 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 by 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 exist two families of isostatics: − Isostatics σ1, tangent to the direction of the largest principal stress. − Isostatics σ2, 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 flux of principal stresses occurs on the plane of analysis. As an example, Figure 7.10 shows the distribution of isostatics on a supported beam with uniformly distributed loading.

Deﬁnition 7.2. A singular point is a point characterized by the stress state Theory and Problems σx = σy and τxy = 0 andContinuum its Mohr’s circle is Mechanicsa point on the axis forσ (see Engineers Figure 7.11). © X. Oliver and C. Agelet de Saracibar A neutral point is a singular point characterized by the stress state

σx = σy = τxy = 0 and its Mohr’s circle is the origin of the σ − τ space (see Fig- ure 7.11).

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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). Conse- quently, the isostatics tend to loose their regularity in singular points and can brusquely change their direction.

7.6.1.1 Differential EquationTheory of the Isostatics and Problems Consider the general equation of an isostatic line y = f (x) and the value of the σ angle formedContinuum by the principal Mechanics stress direction for1 with Engineers respect to the horizontal direction (see Figure 7.12©), X. Oliver and C. Agelet de Saracibar ⎫ ⎪ 2τxy 2tanα ⎪ tan(2α)= = ⎬ σ − σ − 2α 2τ 2y x y 1 tan ⇒ xy = ⇒ ⎪ σ − σ − ( )2 dy not ⎭⎪ x y 1 y tanα = = y (7.41) dx σ − σ (y)2 + x y y − 1 = 0 τxy

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Representative Curves of Plane Elasticity 355

Figure 7.12: Determination of the differential equation of the isostatics. and solving the second-order equation (7.41) for y, the differential equation of the isostatics is obtained.

2 Differential σx − σy σx − σy equation → y = − ± + 1 2τ 2τ (7.42) of the isostatics xy xy ϕ (x,y)

If the function ϕ (x,y) in (7.42) is known, this equation can be integrated to obtain the algebraic equation of the family of isostatics, y = f (x)+C . (7.43)

The double sign in (7.42) leads to two differential equations corresponding to the two families of isostatics.

Example 7.1 – A rectangularTheory plate and is subjected Problems to the following stress states. σ = − 3 σ = 3 − 2 τ = 2 τ = τ = σ = Continuumx x ; y 2 Mechanicsx 3xy ; xy for3x y ; Engineersxz yz z 0 Obtain and plot the singular© X. Oliver points and and distribution C. Agelet of de isostatics. Saracibar

Solution

The singular points are deﬁned by σx = σy and τxy = 0 . Then, ⎧ ⎪ ⎪ σ = −x3 = 0 ⎪ x = 0 =⇒ x ∀y ⎨ σ = 3 − 2 = 2 y 2x 3xy 0 τxy = 3x y = 0 =⇒ ⎪ 3 ⎪ σx = −x ⎪ y = 0 =⇒ =⇒ x = 0 ⎩ 3 2 3 σy = 2x − 3xy = 2x

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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 x = 0. These singular points are, in addition, neutral points (σx = σy = 0). The isostatics are obtained from (7.42), 2 dy σ − σ σ − σ y = = − x y ± x y + 1 , dx 2τxy 2τxy which, for the given data of this problem, results in ⎧ ⎪ ⎨⎪ dy = x x2 − y2 = C dx y =⇒ integrating =⇒ 1 . ⎪ − xy = C ⎩⎪ dy = y 2 dx x Therefore, the isostatics are two families of equilateral hyperboles orthogonal to each other. On the line of singular points x = 0 (which divides the plate in two regions) the isostatics will brusquely change their slope. To identify the family of isostatics σ1, consider a point in each region:

− Point (1,0): σx = σ2 = −1; σy = σ1 =+2; τxy = 0 (isostatic σ1 in the y-direction)

− Point (−1,0): σx = σ1 =+1; σy = σ2 = −2; τxy = 0 (isostatic σ1 in the x-direction) Finally, the distribution of isostatics is as follows: Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Representative Curves of Plane Elasticity 357

7.6.2 Isoclines

Deﬁnition 7.3. Isoclines are the locus of the points in the plane of analysis along which the principal stress directions form a certain angle with the x-axis.

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 x-axis (see Figure 7.13).

7.6.2.1 Equation of the Isoclines The equation y = f (x) of the isocline with an angle θ is obtained by establishing that the principal stress direction σ1 forms an angle α = θ with the horizontal direction, that is,

2τ Algebraic equation tan(2θ)= xy of the isoclines σ − σ (7.44) x y ϕ (x,y)

This algebraic equation allows isolating, for each value of θ,

y = f (x,θ) , (7.45) which constitutes the equation of the family of isoclines parametrized in terms of the angle θ. Theory and Problems Continuum Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

Figure 7.13: Isocline.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 358 CHAPTER 7. PLANE LINEAR ELASTICITY

Remark 7.4. Determining the family 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 by means of experimental methods (methods based on photoelasticity) they provide, indirectly, a method for the experimental determination of the isostatics.

7.6.3 Isobars

Deﬁnition 7.4. Isobars are the locus of points in the plane of analysis with the same value of principal stress σ1 (or σ2).

Two families of isobars will cross at each point of the plane of analysis: one corresponding to σ1 and another to σ2. Note that the isobars depend on the value of σ1, 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) im- plicitly deﬁnes the algebraic equation of the two families of isobars y = f1 (x,c1) and y = f2 (x,c2), ⎧ Theory and Problems ⎪ ⎪ σ + σ σ − σ 2 ⎪ x y x y 2 ⎪ σ1 = + + τ = const. = c1 ⎪ 2 2 xy Continuum⎪ Mechanics for Engineers Algebraic ⎪ © X. Oliver and C. Agelet de Saracibar ⎨ ϕ ( , ) equation 1 x y ⎪ (7.46) of the ⎪ isobars ⎪ σ + σ σ − σ 2 ⎪ σ = x y − x y + τ2 = const. = c ⎪ 2 2 2 xy 2 ⎪ ⎩⎪ ϕ2 (x,y) which leads to y = f (x,c ) 1 1 (7.47) y = f2 (x,c2)

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Representative Curves of Plane Elasticity 359

Figure 7.14: Isobars.

7.6.4 Maximum Shear Stress or Slip Lines

Deﬁnition 7.5. Maximum shear stress lines or slip lines are the envelopes of the directions that, at each point, correspond with the maximum value (in modulus) of the shear (or tangent) stress.

Remark 7.5. At each point of the plane of analysis there are two planes on which the shear stresses reach the same maximum value (in module) but that have opposite directions, τmax and τmin. These planes can be determined by means of the Mohr’s circle and form a 45◦ angle with the principal stress directions (see Figure 7.15). Therefore, their envelopesTheory (maximum and Problemsshear stress lines) are two families of curves orthogonal to each other that form a 45◦ angle withContinuum the isostatics. Mechanics for Engineers © X. Oliver and C. Agelet de Saracibar

7.6.4.1 Differential Equation of the Maximum Shear Lines

Consider β is the angle formed by the direction of τmax with the horizontal direction (see Figure 7.16). In accordance with Remark 7.55, π π 1 β = α − =⇒ tan(2β)=tan 2α − = − , (7.48) 4 2 tan(2α)

5 Here, the trigonometric expression tan(θ − π/2)=−cotθ = −1/tanθ is used.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 360 CHAPTER 7. PLANE LINEAR ELASTICITY

Figure 7.15: Maximum shear stress planes.

where α is the angle formed by the principal stress direction σ1 with the horizontal direction. Consequently, considering the general equation of a slip line, y = f (x), the expression (7.48) and the relation tan(2α)=2τxy/(σx − σy) yields ⎫ σ − σ β ( β)=− 1 = x y = 2tan ⎬⎪ tan 2 2 tan(2α) 2τxy 1 − tan β =⇒ dy not ⎭⎪ tanβ = = y dx (7.49)

σ − σ τ − x y = 2y =⇒ ( )2 − 4 xy − = . 2 y y 1 0 2τxy 1 − (y ) σx − σy

Solving the second-order equation in (7.49) for y provides the differential equation of the maximum shearTheory stress lines. and Problems Differential Continuum Mechanicsτ for Engineersτ 2 equation of the© X. Oliver= − 2 andxy C.± Agelet2 xy de Saracibar+ y 1 (7.50) max. shear stress σx − σy σx − σy or slip lines ϕ (x,y)

If the function ϕ (x,y) 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.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Representative Curves of Plane Elasticity 361

Figure 7.16: Maximum shear stress or slip lines.

Theory and Problems

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

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 363

PROBLEMS

Problem 7.1 – Justify 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:

The stress state is represented by a point. σ1 = σ2 τ = 0

Therefore, all directions are principal stress directions and, given an angle θ which can take any value, the principal stress direction will form an angle θ with the x-axis. Then, an isocline of angle θ will cross said point and, since this holds true for any value of θ, all the isoclines will cross this point. Therefore, the statement is true. Theory and Problems b) A uniform stress state implies that the Mohr’s circle is equal in all points of the medium,Continuum therefore, the planes Mechanics of maximum for shear Engineers stress will be the same in all points. Then, the maximum© X. Oliver shear stress and lines C. Agelet (or slip lines)de Saracibar will be parallel to each other. In conclusion, the statement is true.

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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) σx = 0 ; σy = b > 0 ; τxy = 0

2) σx = 0 ; σy = 0 ; τxy = my, 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 σx = 0;σy = b > 0;τxy = 0 is:

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 365

And the slip lines are:

There do not exist singular points for this stress state.

2) The Mohr’s circle for the stress state σx = 0;σy = 0;τxy = my, m > 0 is:

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory 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:

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961 Problems and Exercises 367

EXERCISES

7.1 – A rectangular plate is subjected to the following plane strain state:

σx = σy

τxy = ax

σy = 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 cylindrical shell shown below. Assume a ﬁeld of the form:

⎧ ⎪ B ⎨⎪ u = Ar + ; A > 0, B > 0 r r ⎪ uθ = 0 ⎩⎪ uz = 0

Theory and Problems

X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems doi:10.13140/RG.2.2.25821.20961