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Chapter 7: Constitutive Equations Definition:

In the previous chapters we’ve learned about the definition and meaning of the concepts of and strain. One is an objective measure of load and the other is an objective measure of . In , one talks about the rate-of-deformation as opposed to simply strain (i.e. deformation alone or by itself). We all know though that deformation is caused by loads (i.e. there must be a relationship between stress and strain). A relationship between stress and strain (or rate-of-deformation ) is simply called a “”. Below we will describe how such equations are formulated.

Constitutive equations between stress and strain are normally written based on phenomenological (i.e. experimental) observations and some assumption(s) on the physical behavior or response of a material to loading. Such equations can and should always be tested against experimental observations.

Although there is almost an infinite amount of different materials, leading one to conclude that there is an equivalently infinite amount of constitutive equations or relations that describe such materials behavior, it turns out that there are really three major equations that cover the behavior of a wide range of materials of applied interest. One equation describes stress and small strain in and called “Hooke’s law”. The other two equations describe the behavior of fluidic materials.

Hookean Elastic : We will start explaining these equations by considering Hooke’s law first. Hooke’s law simply states that the stress tensor is assumed to be linearly related to the strain tensor. More specifically that every strain component depends on all independent stress components or vice versa (i.e. every stress component depends on all strain components). Such a relationship can be written concisely as: σ = ij Cijkl ekl σ where ij is the stress tensor, ekl is the strain tensor, and Cijkl is a fourth-order tensor whose components are essentially proportionality coefficients (which is clearer upon expansion of coefficients). Such constants are also called the “elastic constants” or “moduli” of the solid material and are independent of the magnitude of stress or strain at a material point.

The question might arise: How does one arrive at such a relationship? The answer is by looking at some known phenomena and extrapolating on it to cover more general behavior of material deformation. Specifically and more simply, consider a rod/bar of material cut out from a board of the same material along one direction (call it the x- direction). Alternatively, one may have cut out this rod along the y-direction and so forth. When one applies a tensile or extension , say at the ends of this rod, we all know from elementary experience and scientific knowledge that the stress σ (force/end area) is

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proportional to the strain e (change in length/length) via a proportionality constant, call it E, such that we can write σ = Ee Indeed, a lot of us know this equation as Hooke’s law and call the constant E either Young’s modulus, modulus of , or the stiffness of the bar. The truth is that this last equation is a special simple 1-D form of the true Hooke’s law given by the tensorial and more general relationship above (which is inherently 3-D). At this juncture, it is worth noting that Hooke’s law in this form is the result of experimental observation especially valid for small strains in the material.

To be more precise, and utilizing knowledge from earlier chapters, σ here is really σxx and e is really exx. Because of the specific directionality of the problem, and hence the x subscripts, E can also be written more appropriately as Ex (i.e. we should have written σ = xx Exexx ) The reason for Ex having a special value when performing this tension test along the x-axis, is that it can be envisioned in the most general case that if we pull on the rod in any other directions, say along y-direction, then we must have a different proportionality constant, i.e. E ≠ Ex. Indeed, thinking generally we can write in this σ = particular case yy E y eyy .

The same argument can be made for the z-direction, for example. Ex, Ey and Ez are elastic constants in the sense above, i.e. they relate stresses to strain in the material. These three

elastic constants are indeed related to the constants of the fourth-order tensor Cijkl as well

be illustrated later (by the way Cijkl is also called the “stiffness tensor”).

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F x Figure: a bar in tension

In the above light, one can see that the general and true form of Hooke’s law is simply an envisioned generalization and extension of the 1-D form where it is assumed that stresses (whether normal or shear) are related (or coupled) not just to one component of strain (normal or shear) but to all of them in general (and visa versa, every strain component is related to all six independent stress components in general). The exact form of the constitutive equation (Hooke’s law for example with linear dependence on strain and not nonlinear quadratic or other dependence) depends on the material and the assumption used in constructing the equation.

Upon expansion of indices of the tensorial equation above, one gets σ = + + + + + + + + ij Cij11e11 Cij22e22 Cij33e33 Cij12e12 Cij13e13 Cij23e23 Cij21e21 Cij31e31 Cij32e32 For example,

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σ = + + + + + + + + 11 C1111e11 C1122e22 C1133e33 C1112e12 C1113e13 C1123e23 C1121e21 C1131e31 C1132e32

Notice that barring any special conditions or restriction, Cijkl has 3×3×3×3 = 81 components or elastic constants to be determined for a general material. This is a rather large number of constants that no real material has and can be reduced, based on physical arguments, as follows. First, eij is a , i.e. eij = eji. Hence, we can write

σ = = = ij Cijklekl Cijlk elk Cijlk ekl − = ⇒ ekl (Cijkl Cijlk ) 0 = ⇒ Cijkl Cijlk

Based on the last relationship, Cijkl now has 3×3×6 = 54 independent constants.

σ σ = σ Similarly, ij is symmetric, i.e. ij ji , hence σ = = σ = ij Cijklekl ji C jiklekl − = ⇒ ekl (Cijkl C jikl ) 0 = ⇒ Cijkl C jikl This reduces the number of independent constants to 6×6 = 36.

Note that since Cijkl is a fourth-order tensor, it transforms accordingly: ′ = β β β β Cijkl im jn ks lt Cmnst

Since Cijkl has only 36 independent constants, we can conveniently then write Hooke’s law in notation as: σ  C C C C C C  e  11  1111 1122 1133 1123 1113 1112  11 σ  C C C C C C  e   22   2211 2222 2233 2223 2213 2212  22  σ    33  C3311 C3322 C3333 C3323 C3313 C3312  e33  =   σ     23  C2311 C2322 C2333 C2323 C2313 C2312 2e23  σ  C C C C C C 2e  13  1311 1322 1333 1323 1313 1312  13 σ      12  C1211 C1222 C1233 C1223 C1213 C1212 2e12  where

 e11  e11   e  e   22   22   e33  e33  =   γ  2e23   23    γ  2e13 13   γ  2e12   12  , and where the 6×6 matrix of elastic constants is also called the “stiffness matrix” C.

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Further reductions in the number of independent elastic constants from 36 can be achieved as follows: The work dU done by the stress components σij acting on a unit cube of elastic material when the deformation is increased so that the strain tensor components increase by deij, this work is given by: = σ dU ij deij (which has units of work per unit volume) The total work in the process would be given by U:

eij U = σ de ∫ ij ij = eij 0 In 1-D, this is the area under the stress-strain line or curve:

1-D stress-strain line

σ 11 Area = U

e11 de11

The quantity U is also called the “strain energy density” since its units are energy units per unit volume. From the last equation, one can see that ∂U = σ ∂ ij eij = σ σ = Now since dU ij deij and ij Cijkl ekl ∂U ⇒ dU = C e de = de ijkl kl ij ∂ ij eij

∂U ⇒ = C e ∂ ijkl kl eij Differentiating the last equation gives: ∂ ∂U = C ∂ ∂ ijkl ekl eij Interchanging indices order in the last equation gives: ∂ ∂U = C ∂ ∂ klij eij ekl From the last two equations, one can see that: = Cijkl Cklij

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The last equation means that the stiffness matrix C is symmetric. This further reduces the number of independent elastic constants to 21 from 36.

A material with 21 independent constants in Cijkl is called an “anisotropic material” or a “generally-anisotropic material”.

A “Monoclinic” material is a material that exhibits symmetry with respect to one plane. We take this plane to be the x1x2 plane without loss of generality. The elastic constants of the material do not change under the coordinate change: ′ = ′ = ′ = − x1 x1 , x2 x2 , x3 x3 x 3 ′ x2 , x2

′ x1 , x1

′ x3

The transformation matrix in this case is: 1 0 0  β = (β ) = 0 1 0  ij   0 0 −1 Under this transformation, we must have: ′ = Cijkl Cijkl

Check: C′ = β β β β C 1111 1i 1 j 1k 1l ijkl ′ = β β β β = ⇒ C1111 11 11 11 11C1111 C1111

The above relationship is satisfied for any arbitrary C1111 . ′ Consider now C1123 . It is supposed to be equal to C1123 under the transformation above, ′ = i.e. C1123 C1123

Check: C′ = β β β β C 1123 1i 1 j 2k 3l ijkl ′ = β β β β = − ⇒ C1123 11 11 22 33C1123 C1123 ′ Since C1123 can NOT be BOTH equal to and the negative of C1123 UNLESS = = ′ = C1123 0 ⇒ C1123 C1123 0

By similar arguments, we can show that 7 additional constants are equal to zero so that the number of elastic constants drops to 13 from the 21 describing a generally-anisotropic material:

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C C C 0 0 C   1111 1122 1133 1112  C2211 C2222 C2233 0 0 C2212  C C C 0 0 C  C =  3311 3322 3333 3312   0 0 0 C2323 C2313 0   0 0 0 C C 0   1323 1313    C1211 C1222 C1233 0 0 C1212 

An “orthotropic” material has symmetry of its elastic properties with respect to two orthogonal planes. Suppose that these planes are the x1x2 and x2x3 planes. The is then one that shows symmetry with respect to the transformation: 1 0 0  β = (β ) = 0 1 0  ij   0 0 −1 And the transformation: −1 0 0 β = (β ) =  0 1 0 ij    0 0 1 ′ = − ′ = ′ = Which corresponds to x1 x1 , x2 x2 , x3 x3 . The first of these transformations above has been taken care of already and has reduced the number of independent elastic constants to 13. Starting with this number of constants in C, we apply the transformation equations to the second of the transformations above, to get: C′ = β β β β C 1112 1i 1 j 1k 2l ijkl ′ = β β β β = − ⇒ C1112 11 11 11 22C1112 C1112 But because of symmetry, we are supposed to have ′ = C1112 C1112 = From the last two equations, we conclude that C1112 0. = = = Similarly, C2212 C3312 C1323 0.

Hence, for an orthotropic material, C has only 9 independent elastic constants: C C C 0 0 0   1111 1122 1133  C2211 C2222 C2233 0 0 0  C C C 0 0 0  C =  3311 3322 3333   0 0 0 C2323 0 0   0 0 0 0 C 0   1313     0 0 0 0 0 C1212  Note that, solely by virtue of geometry, a material that has two planes of symmetry automatically has three planes of symmetry as well. In fact, applying a third plane of

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symmetry to the last reduced C matrix above, will NOT reduce the number of independent elastic constants any further because of that fact.

A “Tetragonal” material is an orthotropic material that has the exact same properties along two axes and different properties along the third axis. Let the third axis be the x1 axis. Hence a 90o rotation for this orthotropic material about this axis should not alter the elastic constants, i.e. the elastic constants should remain unchanged under the ′ = ′ = ′ = − transformation x1 x1 , x2 x3 , x3 x2 described by 1 0 0 = β =   β ( ij ) 0 0 1 0 −1 0 Applying the transformation equations, one gets ′ = β β β β C2222 2i 2 j 2k 2lCijkl ′ = β β β β ⇒ C2222 23 23 23 23C3333 = C3333 = Which should equal C2222 . Hence, C2222 C3333 = = Similarly, C1122 C1133 ,C1212 C1313

Hence, a tetragonal material has six independent constants: C C C 0 0 0   1111 1122 1122  C2211 C2222 C2233 0 0 0  C C C 0 0 0  C =  2211 3322 3333   0 0 0 C2323 0 0   0 0 0 0 C 0   1212     0 0 0 0 0 C1212 

A “cubic” material is a tetragonal material that is invariant to an additional change of ′ = ′ = − ′ = coordinates described by x1 x2 , x2 x1 , x3 x3 In matrix form, this transformation corresponds to  0 1 0 β = (β ) = −1 0 0 ij    0 0 1 Again, one can show that ′ = β β β β C1111 1i 1 j 1k 1l Cijkl ′ = β β β β ⇒ C1111 12 12 12 12C2222 = C2222 = Which should equal C1111 . Hence, C2222 C1111 = = Similarly, C2233 C1122 ,C2323 C1212

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C C C 0 0 0   1111 1122 1122  C1122 C1111 C1122 0 0 0  C C C 0 0 0  ⇒ C =  1122 1122 1111   0 0 0 C1212 0 0   0 0 0 0 C 0   1212     0 0 0 0 0 C1212  which has three independent constants.

Finally, a material is called an “isotropic” material if it has the same elastic properties in all directions, i.e. an infinite number of them. In other words, the elastic constants Cijkl should be invariant under any coordinate transformation. For example, and without loss ′ = 1 + 1 ′ = − 1 + 1 ′ = of generality, under the transformation x1 x1 x2 , x2 x1 x2 , x3 x3 2 2 2 2  1 1   0  2 2  = β = − 1 1  Or β ( ij ) 0  2 2   0 0 1     Again, using the transformation equations, one gets: ′ = β β β β C1212 1i 2 j 1k 2lCijkl = β β β β + β 2 j 1k 2l ( 11C1 jkl 12C2 jkl ) = β β β β + β + β β β β + β 11 1k 2l ( 21C11kl 22C12kl ) 12 1k 2l ( 21C21kl 22C22kl ) = β β β β + β + β β β β + β 11 21 2l ( 11C111l 12C112l ) 11 22 2l ( 11C121l 12C122l ) + β β β β + β + β β β β + β 12 21 2l ( 11C211l 12C212l ) 12 22 2l ( 11C221l 12C222l ) = β 2 β β + β + β β β β + β 11 21( 21C1111 22C1112 ) 11 21 12 ( 21C1121 22C1122 ) + β 2 β β + β + β β β β + β 11 22 ( 21C1211 22C1212 ) 11 22 12 ( 21C1221 22C1222 ) + β β β β + β + β 2 β β + β 12 21 11( 21C2111 22C2112 ) 12 21( 21C2121 22C2122 ) + β β β β + β + β 2 β β + β 12 22 11( 21C2211 22C2212 ) 12 22 ( 21C2221 22C2222 ) − = C1111 C1122 2 C − C Which should equal C . Hence, C = 1111 1122 . 1212 1212 2

An isotropic material has thus two independent elastic constants. These constants are usually denoted as = λ = µ = µ + λ C1122 , C1212 , C1111 2 The constants λ and µ are called Lamé constants.

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Therefore, we can write C as: λ + 2µ λ λ 0 0 0     λ λ + 2µ λ 0 0 0   λ λ λ + 2µ 0 0 0  = = C (Cijkl )    0 0 0 µ 0 0   µ   0 0 0 0 0     0 0 0 0 0 µ 

Notice that in index notation, we can write the isotropic tensor Cijkl as = λδ δ + µ δ δ + δ δ Cijkl ij kl ( ik jl il jk ) Indeed, any isotropic fourth-order tensor can be represented using the last equation, which involves only two independent constants λ and µ.

With C from above, one can write the generalized Hooke’s Law for an isotropic material as follows: σ = λ + µ + λ + λ = λ + + + µ = λ + µ 11 ( 2 )e11 e22 e33 (e11 e22 e33 ) 2 e11 I 1 2 e11 σ = λ + µ 22 I 1 2 e22

σ = λI 1 + 2µ e 33 33 σ = µ 23 2 e23 σ = µ 13 2 e13 σ = µ 12 2 e12 where I 1 is the first invariant of the strain tensor, the volumetric strain or the dilatation

( I 1 = ekk).

The last equations (i.e. Hooke’s Law for an elastic isotropic material) can be written in shorthand indicial notation as follows: σ = λ δ + µ ij ekk ij 2 eij Where the symbol µ was replaced by G, a practice common in the theory of elasticity.

One can invert the last equation and write the strains in terms of the stresses as follows: 1 λδ e = (σ − ij σ ) ij 2G ij 2G + 3λ kk σ = = , where kk I1 the first invariant of the stress tensor.

Relationship Between Lamé Constants and E (Young’s modulus) and ν (Poisson’s ratio) Consider a state of pure tension in a material. Such a state of stress can be described by σ  11 0 0 σ = (σ ) =  0 0 0 ij    0 0 0

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This state can be reached, for example, in a rod with a uniform tension force acting over its end cross-sectional areas, where the rod axis is aligned with the x or x1 direction.

To find the strains corresponding to this state of stress in an isotropic material, we use the last equations λ σ + λ = 1 σ − (1) σ = 11  G  e11 ( 11 11 )   2G 2G + 3λ G  2G + 3λ  Also, λ −σ λ = 1 − (1) σ = 11   = e22 (0 11 )   e33 2G 2G + 3λ 2G  2G + 3λ  By definition, Young’s modulus E (also known as “the modulus of elasticity in tension”) is defined as: σ E = for one-dimensional stress states like the one discussed here e σ ⇒ E = 11 e11 σ G(2G + 3λ) ⇒ E = 11 = σ  G + λ  G + λ 11   G  2G + 3λ  Also, Poisson’s ratio v is, by definition, the negative of the ratio between the lateral strain e22 = e33 and the axial strain e11: σ  λ  11   e 2G 2G + 3λ λ E = − 22 =   = e σ  G + λ  2(G + λ) 11 11   G  2G + 3λ  The last two equations can be solved for λ and G in terms of E and ν as follows: Eν E λ = , G = µ = (1+ν )(1− 2ν ) 2(1+ν )

The case of Pure Shear In this case, σ τ  0 12 0 0 0 σ = (σ ) = σ 0 0 = τ 0 0 ij  12     0 0 0 0 0 0 From Hooke’s Law for isotropic materials, we get σ = = γ 12 2Ge12 G 12 σ ⇒ G = 12 = γ 12 angle change Thus G is the ratio of the shear stress to the change in an originally right angle. For this reason, G is known as the “modulus of rigidity” or “”.

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Hydrostatic Tension or Compression In this case, the stress tensor is given by σ 0 0 σ = (σ ) = 0 σ 0, σ = 3σ ij   kk 0 0 σ 

If σ > 0 then we have pure hydrostatic tension If σ < 0 then we have pure hydrostatic compression

Applying Hooke’s Law, we get: λ λ λ = σ − (1) σ = σ − σ = σ  1 −  = σ  2G  2Ge11 11 kk (3 ) 3   3   2G + 3λ 2G + 3λ  3 2G + 3λ   3(2G + 3λ)  σ = + λ (2G 3 )e11 Similarly, we get for e22 and e33: σ = (2G + 3λ)e 22 σ = + λ (2G 3 )e33 Adding the last three equations: σ = + λ + + = + λ 3 (2G 3 )(e11 e22 e33 ) (2G 3 )I 1 Define p as the average or mean of the normal stresses at a point (similar to hydrostatic in fluids) σ + σ + σ I p = 11 22 33 = 1 3 3 σ + σ + σ = = σ 3 (the second equation is valid in the case of pure hydrostatic tension or compression)

We can therefore write: 2G + 3λ σ = e = (λ + 2G / 3)e 3 kk kk σ ⇒ = λ + 2G / 3 = constant = K ekk The constant K is known as the “ of Elasticity”. The inverse of K (or K-1) is a measure of material incompressibility.

It can be shown that K = E/3(1-2ν), therefore if ν = 1/2 then K = ∞ or K-1 = 0 and this material is called an incompressible material since basically ekk = 0.

We also notice that from K = E/3(1-2ν), we must have ν < 1/2 , otherwise this will produce a physically unintuitive negative K. Also, from G = E/2(1+ν), we must have ν > -1, otherwise we will also have a physically unintuitive negative G.

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Finally, we write Hooke’s Law for an isotropic material in terms of E and ν instead of λ and µ as follows: E σ = ((1− 2ν )e +ν e δ ) ij (1+ν )(1− 2ν ) ij kk ij Also, 1 e = ((1+ν )σ −νσ δ ) ij E ij kk ij

Before ending this Chapter, we say that Hooke’s Law in the above forms is valid for any orthogonal curvilinear in addition to Cartesian coordinates. The indices i and j stand for x, y or z in Cartesian coordinates, r, θ or z in cylindrical coordinates, and r, φ, and θ in spherical coordinates.

Example: Wood is generally considered to be an orthotropic material. For example, a Douglas fir strain-stress relationship is, relative to material axes (x,y,z) 6 = σ − σ − σ 10 exx 87 xx 34.8 yy 43.5 zz 6 = − σ + σ − σ 10 eyy 34.8 xx 1305 yy 609 zz 6 = − σ − σ + σ 10 ezz 43.5 xx 609 yy 1740 zz

6 = σ 10 exy 696 xy 6 = σ 10 exz 290 xz 6 = σ 10 eyz 3045 yz where the x-axis is longitudinal, the y-axis is radial in the tree, and the z-axis is tangent to the growth rings of the tree. The unit of stress is MPa.

At a point in a Douglas fir log, the nonzero components of stress are: σ = σ = σ = − σ = xx 7MPa, yy 2.1MPa, zz 2.8MPa, xy 1.4MPa (a) Determine the principal stresses and orientation of the principal axes of stress. (b) Determine the strain components. (c) Determine the principal strains and the orientation of the principal axes of strain.

Solution: σ = σ = (a) First, we note that yz xz 0 To find the principal stresses, we either use the characteristic equation σ 3 − σ 2 −σ − = I1 I2 I3 0

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, where = σ + σ + σ = I1 xx yy zz 6.3 σ σ σ σ xx xy σ σ yy yz I = − − xx xz − = 12.74 2 σ σ σ σ σ σ xy yy xz zz yz zz σ σ σ xx xy xz = σ σ σ = = − I3 xy yy yz σ 35.672 σ σ σ xz yz zz we can then solve the cubic equation for the principal stresses. σ = − However, a more clever approach is to notice that zz 2.8 MPa is already a principal stress since no shear stresses act in the z-direction. Thus also, n=(0,0,1) is a principal stress direction.

To find the other two principal stresses, we use long division

σ 2 − 9.1σ +12.74 σ + 2.8 σ 3 − 6.3σ 2 −12.74σ + 35.672 Hence, we can write (σ + 2.8)(σ 2 − 9.1σ +12.74) = 0 The roots of the quadratic term in the last equation are: 9.1± (9.1)2 − 4(1)(12.74) σ = = 7.37MPa or 1.73MPa (2)(1) Hence, σ = σ = σ = − 1 7.37MPa, 2 1.73MPa, 3 2.8MPa

To find the principal stress directions, we already know that = N3 (0,0,1) is one.

The other two directions must lie in the xy-plane normal to N3 . To find these two directions, it is simpler to use Mohr’s circle for this purpose (where θ is positive for counter-clockwise directions):

σ 2 xy 2.8 tan 2θ = = = 0.5714 σ −σ xx yy 4.9 ⇒ θ = 14.9o or 14.9o + 90o = 104.9o

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(b) To find the strain components, we simply substitute the stress values in the first equations above to get: 6 = + − + − − 10 exx (87)(7) ( 34.8)(2.1) ( 43.5)( 2.8) = × −6 ⇒ exx 657.72 10 = − + + − − × −6 = × −6 eyy [( 34.8)(7) (1305)(2.1) ( 609)( 2.8)] 10 4202.1 10

= − + − + − × −6 = − × −6 ezz [( 43.5)(7) ( 609)(2.1) (1740)( 2.8)] 10 6455.4 10 = × −6 = × −6 exy (696)(1.4) 10 974.4 10 = = exz eyz 0

= = (c) Since exz eyz 0 ⇒ the z-axis is a principal axis of strain and ezz is a principal strain value. To find the other two principal strains, we use long division just as before = = = = − and arrive at e1 0.00445, e2 0. 00041 with e3 ezz 0. 006455

Again, similar to part (a), the principal strain directions in the xy-plane can be found via: γ 2exy xy 1948.8 tan 2θ = = = = −0.5498 − − − exx eyy exx eyy 3544.38 ⇒ θ = −14.4o or −14.4o + 90o = 75.6o

The main thing to notice here is that the principal strain directions are not coincident with the principal stress directions. They are only coincident for the case of an isotropic material.

Example For an isotropic material, prove that σ ′ = ′ ij 2Geij σ ′ ′ Where ij and eij are the stress and strain deviator , respectively. Solution: For an isotropic material, we know that Hooke’s law takes on the following form σ = λ δ + ij eαα ij 2G eij And the stress and strain deviator tensors are defined as σ e σ ′ = σ − αα δ , e′ = e − αα δ ij ij 3 ij ij ij 3 ij Now, we would like to show that σ e σ − αα δ = 2G(e − αα δ ) ij 3 ij ij 3 ij Lets check: From Hooke’s law above for isotropic materials σαα = 3λ eαα + 2G eαα Hence,

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(3λ eαα + 2G eαα ) eαα σ − δ = 2G(e − δ ) ij 3 ij ij 3 ij 2 eαα ⇒ σ − λ eαα δ − G eαα δ = 2Ge − 2G δ ij ij 3 ij ij 3 ij σ = λ δ + ⇒ ij eαα ij 2Geij Q

Example If for an isotropic material, = λδ δ + µ δ δ + δ δ Cijkl ij kl ( ik jl il jk ) Prove that C is isotropic (isotropic means that it is invariant under orthogonal coordinate transformation), knowing that β β = δ ik jk ij

Solution: For C to be isotropic, we must have ′ = Cijkl Cijkl where ′ = β β β β Cijkl im jn ks lt Cmnst Check: ′ = β β β β λδ δ + µ δ δ + δ δ Cijkl im jn ks lt ( mn st ( ms nt mt ns )) ′ = λβ β β β + µ β β β β + β β β β Cijkl in jn kt lt ( is ks jn ln it lt jn kn ) ′ = λδ δ + µ δ δ + δ δ Cijkl ij kl ( ik jl il jk )

Note that the (or lack of it, also called “”) of a material is due to or stems from its microstructure. If the microstructure, i.e. the building units of the material, is randomly oriented in space then the material will behave as an isotropic material (approximately speaking since no material is totally or perfectly random). If the microstructure is totally random then the material has infinitely many symmetry planes, which is another definition of isotropy. However, if the material has some definite structure (i.e. specific and finite number of symmetry planes) then the material will be anisotropic, i.e. it will behave differently when tested using a different (e.g. tested along different tension or compression axes). Depending on the number of symmetry planes, the material can be cubic, tetragonal, orthotropic, etc. A good example of cubic materials is single crystal iron, aluminum or copper. Hence a single crystal from these materials is an anisotropic material (i.e. gives a different stress-strain response when tested along different axes) and is specifically, for these cases, of cubic symmetry. However, these materials also (and most typically) exist in polycrystalline form where many crystals (also called “grains”) are joined together with boundaries that are called “grain boundaries”. Such grains are typically few hundred nanometers to few hundred micrometers in size and are randomly oriented with respect to one another as a result of typical manufacturing or growth processes. Hence although each single grain, within its dimensions or length scale, is anisotropic in its behavior, the collection of randomly

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oriented anisotropic crystals produce a polycrystalline material that is essentially isotropic (to a more or less extent depending on the degree of randomness of the grain orientation within the polycrystal). Note that this resulting isotropic behavior is only obtained at a larger length scale than one or very few single crystals, i.e. at a scale of millimeters or larger as this scale is important to house/include/incorporate an enough number of single grains, from a statistical point of view, to render the collection random as a whole. Hence, when discussing isotropy or anisotropy of materials, the length scale under consideration is a very important factor in the discussion.

Newtonian : In a , to arrive at a constitutive relationship connecting stress at a fluid particle and deformation (or rate-of-deformation as alluded to earlier since this deformation is happening so rapidly), it is assumed that a fluid particle undergoing motion in this type of fluids feels a hydrostatic pressure p (which is basically the weight of the column on top of it or as determined from boundary conditions) and frictional that dissipates its energy and tries to slow it down. In a Newtonian fluid, it is assumed that the frictional stresses are proportional to the rate-of-deformation tensor. Based on all of the above, the stress felt a fluid particle would be given by: σ = − δ + ij p ij DijklVkl

, where the fourth-order tensor Dijkl is termed the “tensor of coefficients” of the fluid, where for a Newtonian fluid Dijkl is assumed to possibly depend on temperature but not on stress or the rate of deformation. Recall that the rat-of-deformation tensor was ∂ 1  ∂v v j  defined previously as V =  i +  . The term − pδ represents the state of stress ij  ∂ ∂  ij 2  x j xi  = possible in a fluid at rest (since in this case Vkl 0 in this case) thus p should be a positive scalar since the minus sign in front of it produces a state of hydrostatic compression which physically meaningful for a fluid.

The static pressure p is assumed to depend on the density and temperature of the fluid according to an . For example, the equation of state in the case of an ideal (one that does not have internal dissipation or in it thus producing perfectly elastic collisions between the gas molecules, and one in which no attractive forces exist between molecules and whose average kinetic energy per molecule is proportional to the temperature in Kelvin) is given by: p = RT ρ , where R is the universal gas constant, and ρ is the gas mass density. For an ideal gas or fluid, an equation of state in the form of f ( p, ρ,T ) = 0 is possible. Such equations of state are typically found by experimentation. Such pressure determination through an equation of state as above is classic in the theory of thermodynamics.

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In the case of an “incompressible fluid”, the density of the fluid is constant, i.e. ρ = constant which is the equation of state in this case. In this case, the pressure is not determined thermodynamically but rather from mechanical boundary conditions. An example of this is a hydraulic press; here the incompressible fluid assumes any pressure emanating from the force applied to the piston.

σ As it turns out most known fluids can be assumed to be isotropic, because ij and Vij are symmetric and the fluid microstructure is made up of randomly distributed particles, which would imply that Dijkl is a symmetric tensor. Since this fourth-order tensor is a symmetric tensor, it can be represented tensorially in terms of two material constants λ and µ as = λδ δ + µ δ δ + δ δ Dijkl ij kl ( ik jl il jk )

This is similar to the elastic constants tensor Cijkl for an isotropic solid discussed before, although the meaning of the two constants here differ from before.

Using the last equation, we can thus write σ = − δ + λ δ + µ ij p ij Vkk ij 2 Vij Contracting the last equation, we get σ (3λ + 2µ) kk = − p + V 3 3 kk It is assumed for a fluid that the mean stress depends only on the pressure. This means that 3λ + 2µ = 0 (this is known as “Stokes’ condition”)

Thus the constitutive equation becomes 2 σ = − pδ + 2µV − µV δ ij ij ij 3 kk ij A fluid that obeys the last constitutive equation is termed a “Stokes fluid”, for which only one material constant µ, called “the coefficient of viscosity”, is needed in the equation.

For an incompressible viscous fluid, Vkk = 0 and hence we have σ = − δ + µ ij p ij 2 Vij

For a “nonviscous” incompressible fluid, the viscosity µ is set to zero, and we have σ = − δ ij p ij

The constant µ, which is called “the coefficient of viscosity”, was obtained by Newton as he observed that in fluid flows with relatively low (i.e. smooth flowing), an applied shear stress τ to a water slap of some thickness would produce a gradient in the direction normal to the shearing plane, i.e. du/dy, such that these two are proportional (see figure). The proportionality constant in between the two was termed

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“µ”, i.e. du τ = µ dy The units of µ” are N.sec/m2 or sometimes it is expressed in units of poise (after the scientist Poiseuille) which is equal to 0.1 N.sec/m2.

y τ

x

τ

Notice that the fluid velocity, also called the “systemic velocity”, is really the local average of the velocities of the molecules.

Before going any forward, note that du τ = µ dy can be deduced from 2 σ = − pδ + 2µV − µV δ ij ij ij 3 kk ij

This is because for the sketch above, the only non-zero stress component is: σxy which is set equal to τ in the figure. Substituting i = 1 and j = 2 in the last constitutive equation above, we get: 2  1  ∂v ∂v/  dv du σ = − pδ/ + 2µV − µV δ/ = 2µ  1 + 2  = µ 1 = µ 12 12 12 kk 12   ∂ ∂  3  2  x2 x1  dx2 dy Which is the same as what Newton has put forth. Note that we usually denote v1 as u, v2 as v, etc. Also, the partial derivative on the right-hand side of the last equation was converted to total derivative since u depended on y only according to the sketch above.

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