Overview of Elasticity of Materials/Anisotropic Response

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This section covers linear anisotropic response.

At this point, we have developed our notation for describing stress and strain, and have written expressions relating the two in terms of material parameters such as E, G, and ν. This is all within the assumption of a homogeneous, isotropic solid. This is often times a fairly good approximation, and you are likely to recognize the expressions derived here. Hopefully, this section will provide an insight to mechanics of solids as well as other materials-focused topics.

We will begin by looking at an example. Consider a brittle material that is hot at first and then suddenly quenched to low temperature. This can cause thermal shock and breakage. So the part is quenched from Tmax to Tmin, where ΔT=TmaxTmin.

It is known that, without proof, the thermal stress is

σTH=αΔTE(12ν) Template:Spaces[1]

Where α is the coefficient of thermal expansion.

Where is this from? We know that this is the hydrostatic stress:

σii=E12νεkk

Which can be rewritten as

σ11+σ22+σ33=E12ν(ε11+ε22+ε33)

Where

σm=13σii

Then it must be approximated that

13Δ=αΔT

For spherical, circular flaws, or cracks with radius r, the energy is

UTOT=UoUSTRAIN+USURF Template:Spaces[2]

Where UTOT is the total energy of the system and Uo is the energy of the stress-free and crack-free system with volume Vo.

The strain energy is

USTRAIN=VoσTH22ENσTH22E43πr3 Template:Spaces[3]

Where N is the number of cracks.

Where is this from? Our expression for strain energy when normal stress is applied is

UoNORMAL=12σ112E

Where UoNORMAL is the energy per unit volume.

Therefore, the first term in Equation 3 is simply the energy due to the thermal strain. The second term is the strain relieved when N number of cracks of volume 43πr3 open (we are subtracting this term because the strain is being relieved):

N(43πr3)σTH22E

Is this reasonable? This is probably slightly underestimated. Finally, the last term is

USURF=NGcπr2 Template:Spaces[4]

Where Gc is the toughness energy required to create new surface (units of J/m2). Note for ideal brittle materials, Gc=2γ.

In this example, isotropic elasticity is used, and in all likelihood is a reasonable approximation for bulk polycrystalline or amorphous solids. Sometimes we cannot assume an isotropic solid. This is most common when dealing with systems where single crystals are studied such as the one in Figure 1. Looking at Figure 1, even if |P1|=|P3|, pulling in direction |P1| will encounter different resistance (constant of proportionality in Hooke's Law) than pulling in direction |P2|.

Figure 1: Example of a single crystal system where bonds behave in spring-like fashion.

The anisotropic expression of Hooke's Law is a tensor relation:

σij=Cijklεkl Template:Spaces[5]
εij=Sijklσkl Template:Spaces[6]

Where Cijkl is the stiffness or elastic constant and Sijkl is the elastic compliance.

Here, the primary focus will be on Cijkl, but the two behave similarly. The elastic constant tensor is a fourth rank tensor that connects two second rank tensors. In Equation 5, the right hand side is a double sum over k and l, resulting in 9 terms in the sum. Since there are 9 expressions for σij, there is a total of 81 elements in 𝑪. This seems like a large number, but are they unique? We know σ and ε are symmetric, therefore

σij=σjiCijkl=Cjikl

And

εkl=εlkCijkl=Cijlk

This reduces the number of elastic constants from 81 to 36 unique values. To further simplify, we must consider the elastic energy. We know that the energy of the system is

Uo=12σijεij Template:Spaces[7]
Figure 2: Example of a homogeneous elastic stress-strain curve. The area underneath the curve is highlighted to show the total elastic energy.

As depicted in Figure 2.

For homogeneous elastic loading, we can write

dUo=σijdεij Template:Spaces[8]

Which is the superposition of

dUo=σ11dε11+σ12dε12+σ13dε13+σ21dε21+σ22dε22+σ23dε23+σ31dε31+σ32dε32+σ33dε33 Template:Spaces[9]

Now we will consider starting in an initial state σ=ε=0 and straining ε11. This increases the internal energy, the energy stored in bonds, by

dw1=σ11dε11=C1111ε11dε11 Template:Spaces[10]

Where σ11=C1111ε11 from Equation 5 (all strain except ε11 are zero).

Integrating this

0w1dw1=0ε11C1111ε11dε11

Becomes

w1=12C1111ε112 Template:Spaces[11]

Now apply a second strain deformation ε22 and integrate:

0w2dw2=0ε22σ22dε22=0ε22(C2222ε22+C2211ε11)dε22

Using Equation 5 and the fact that all ε=0 except ε11 and ε22. This becomes

w2=12C2222ε222+C2211ε11ε22 Template:Spaces[12]

Then the total work is

wTOT=w1+w2=12C1111ε112+12C2222ε222+C2211ε11ε22 Template:Spaces[13]

Imagine now we reverse the order; first applying ε22 then ε11.

w1=12C2222ε22
w2=12C1111ε112+C1122ε22ε11
wTOT=12C1111ε112+12C2222ε222+C1122ε11ε22 Template:Spaces[14]

In the linear regime superposition holds, therefore the order is not important and

wTOT=wTOTC2211=C1122

This is generalized to

Cijkl=Cklij

Which reduces the number of unique values to 21. This is the fewest number of elastic constants that must be specified for an arbitrary crystal.

Crystal Symmetry Simplifications

Fortunately, tensor relations are tied to the symmetry of a crystal. For high symmetry crystals, such as cubic, only 3 unique values are needed and further many zeroes exist. The following rules can be utilized to simplify the notation:

  1. All Ciiii=C1111
  2. All Ciijj=C1122
  3. All Cijij=C2323
  4. All other Cijkl=0 (not in Equations 1-3).

This makes the situation much simpler. Next, we will simplify the notation using Voigt notation as described in the table below.

𝒊𝒋 pair 11 22 33 23 31 12
m 1 2 3 4 5 6

This is often (not always) found in textbooks and manuscripts. In this notation, work is

dw=σ1dε1+σ6dε6+σ5dε5+σ6dε6+σ2dε2+σ4dε4+σ5dε5+σ4dε4+σ3dε3dw=σ1dε1+σ2dε2+σ3dε3+2σ4dε4+2σ5dε5+2σ6dε6

Using Voigt notation, we can now write the stress-strain relation

σ=𝑪ε

In 2D instead of the original 4D format.

[σ1σ2σ3σ4σ5σ6σ4σ5σ6]=[C11C12C13C14C15C16C14C15C16C12C22C23C24C25C26C24C25C26C13C23C33C34C35C36C34C35C36C14C24C34C44C45C46C44C45C46C15C25C35C45C55C56C45C55C56C16C26C36C46C56C66C46C56C66C14C24C34C44C45C46C44C45C46C15C25C35C45C55C56C45C55C56C16C26C36C46C56C66C46C56C66][ε1ε2ε3ε4ε5ε6ε4ε5ε6] Template:Spaces[15]

In this format, it is still possible to transform the vectors and tensor 𝑪 together. This can further be shortened to

[σ1σ2σ3σ4σ5σ6]=[C11C12C13C14C15C16C12C22C23C24C25C26C13C23C33C34C35C36C14C24C34C44C45C46C15C25C35C45C55C56C16C26C36C46C56C66][ε1ε2ε3γ4γ5γ6] Template:Spaces[16]

However, this has shear strain components (γ), which means that in this representation, C is not a tensor but a matrix. This is because the rotational properties of tensors are not satisfied. We can see how these two are related by writing out one of the stresses from Equation 15:

σ1=C11ε1+C12ε2+C13ε3+C14ε4+C15ε5+C16ε6+C14ε4+C15ε5+C16ε6
σ1=C11ε1+C12ε2+C13ε3+C142ε4+C152ε5+C162ε6

Where γ4=2ε4, γ5=2ε5, and γ6=2ε6.

Applying our simplifications due to crystalline symmetry we can write for a cubic crystal:

CijCUBIC=[C11C12C12000C12C11C12000C12C12C11000000C44000000C44000000C44] Template:Spaces[17]

Returning to our definition of strain energy (Equation 7) we can write this in Voigt notation:

Uo=12σiεi Template:Spaces[18]

And we can write the stress-strain relation (Equation 5) as

σj=Cjkεk Template:Spaces[19]

Which allows us to write

Uo=12Cijεiεj Template:Spaces[20]

Which is a double sum over i and j. Putting CijCUBIC from Equation 17 into Equation 20 yields

UoCUBIC=12C11(ε12+ε22+ε32)+2C44(ε42+ε52+ε62)+C12(ε1ε2+ε2ε3+ε3ε1) Template:Spaces[21]

Here are a few useful relationships for cubic crystals which are given without proof:

C11=S11+S12(S11S12)(S11+2S12) Template:Spaces[22]
C12=S12(S11S12)(S11+2S12) Template:Spaces[23]
C44=1S44 Template:Spaces[24]
1E=S112[(S11S12)12S44](l2m2+m2n2+l2n2) Template:Spaces[25]

Here l, m, and n are direction cosine from x1, x2, and x3 and are the direction of applied uniaxial loading. Mapping the anisotropic elastic properties to the isotropic limit, we have

S11=1E Template:Spaces[26]
S12=νE Template:Spaces[27]
S44=1G Template:Spaces[28]

Yet for a truly isotropic medium, only 2 material parameters are needed. We find

S44=2(S11S12) Template:Spaces[29]

And

C12=λ Template:Spaces[30]
C11=2G+λ Template:Spaces[31]
C44=12(C11C12) Template:Spaces[32]
A=2C44C11C12 Template:Spaces[33]

Where A is the Zener anisotropy ratio (A=1 for isotropic material).Template:BookCat