Elasticity/Constitutive example 4

From testwiki
Jump to navigation Jump to search

Example 4

Given:

A monoclinic crystal has inversion symmetric about the 𝐞^1-𝐞^2 plane. Therefore, the material properties do not change for a mirror-reflection through this plane. The stress-strain relations must therefore remain unchanged under this transformation. The transformation matrix [L] for this for the mirror inversion is given by

[L]=[100010001]

Show:

If we apply this transformation to the stress and strain tensors, then the stiffness matrix of the material (in Voigt notation) is

[C]=[C11C12C1300C16C21C22C2300C26C31C32C3300C36000C44C450000C54C550C61C62C6300C66]

Solution

In 3×3 matrix form, the strain tensor is given by

ε=[ε11ε12ε13ε21ε22ε23ε31ε32ε33]

The transformation rule for a second order tensor 𝑨 is

[A]'=[L][A][L]T

Applying this transformation to the strain tensor, we have

[ε11'ε12'ε13'ε21'ε22'ε23'ε31'ε32'ε33']=[100010001][ε11ε12ε13ε21ε22ε23ε31ε32ε33][100010001]=[100010001][ε11ε12ε13ε21ε22ε23ε31ε32ε33]=[ε11ε12ε13ε21ε22ε23ε31ε32ε33]

In engineering notation (Voigt notation),

[ε]=[ε11ε22ε332ε232ε312ε12]T=[ε1ε2ε3ε4ε5ε6]T

Therefore, the transformed strain tensor can be written as

[ε]'=[ε1'ε2'ε3'ε4'ε5'ε6']T=[ε1ε2ε3ε4ε5ε6]T

The expression for the strain energy density of a linear elastic material imposes a constraint on the components of the stiffness tensor in the presence of planes of material symmetry. This constraint is

Cij(εiεjεi'εj')=0(i,j=16)

where Cij are the components of the 6×6 matrix that contains the independent components of the stiffness tensor.

Therefore,

C11(ε1ε1ε1'ε1')+C12(ε1ε2ε1'ε2')+C13(ε1ε3ε1'ε3')+C14(ε1ε4ε1'ε4')+C15(ε1ε5ε1'ε5')+C16(ε1ε6ε1'ε6')+C21(ε2ε1ε2'ε1')+C22(ε2ε2ε2'ε2')+C23(ε2ε3ε2'ε3')+C24(ε2ε4ε2'ε4')+C25(ε2ε5ε2'ε5')+C26(ε2ε6ε2'ε6')+C31(ε3ε1ε3'ε1')+C32(ε3ε2ε3'ε2')+C33(ε3ε3ε3'ε3')+C34(ε3ε4ε3'ε4')+C35(ε3ε5ε3'ε5')+C36(ε3ε6ε3'ε6')+C41(ε4ε1ε4'ε1')+C42(ε4ε2ε4'ε2')+C43(ε4ε3ε4'ε3')+C44(ε4ε4ε4'ε4')+C45(ε4ε5ε4'ε5')+C46(ε4ε6ε4'ε6')+C51(ε5ε1ε5'ε1')+C52(ε5ε2ε5'ε2')+C53(ε5ε3ε5'ε3')+C54(ε5ε4ε5'ε4')+C55(ε5ε5ε5'ε5')+C56(ε5ε6ε5'ε6')+C61(ε6ε1ε6'ε1')+C62(ε6ε2ε6'ε2')+C63(ε6ε3ε6'ε3')+C64(ε6ε4ε6'ε4')+C65(ε6ε5ε6'ε5')+C66(ε6ε6ε6'ε6')=0

For a monoclinic material, replacing the transformed strain components by the equivalent original strain components, we get

C14(ε1ε4+ε1ε4)+C15(ε1ε5+ε1ε5)+C24(ε2ε4+ε2ε4)+C25(ε2ε5+ε2ε5)+C34(ε3ε4+ε3ε4)+C35(ε3ε5+ε3ε5)+C41(ε4ε1+ε4ε1)+C42(ε4ε2+ε4ε2)+C43(ε4ε3+ε4ε3)+C46(ε4ε6+ε4ε6)+C51(ε5ε1+ε5ε1)+C52(ε5ε2+ε5ε2)+C53(ε5ε3+ε5ε3)+C56(ε5ε6+ε5ε6)+C64(ε6ε4+ε6ε4)+C65(ε6ε5+ε6ε5)=0

or,

2C14ε1ε4+2C15ε1ε5+2C24ε2ε4+2C25ε2ε5+2C34ε3ε4+2C35ε3ε5+2C41ε4ε1+2C42ε4ε2+2C43ε4ε3+2C46ε4ε6+2C51ε5ε1+2C52ε5ε2+2C53ε5ε3+2C56ε5ε6+2C64ε6ε4+2C65ε6ε5=0

Using the symmetry of the stiffness matrix, we have

4C14ε1ε4+4C15ε1ε5+4C24ε2ε4+4C25ε2ε5+4C34ε3ε4+4C35ε3ε5+4C46ε4ε6+4C56ε5ε6=0

Since the strains can be arbitrary, the above condition is satisfied only if

C14=C15=C24=C25=C34=C35=C46=C56=0

Therefore, the stiffness matrix is given by

[C]=[C11C12C1300C16C21C22C2300C26C31C32C3300C36000C44C450000C54C550C61C62C6300C66]

Hence shown.

Template:Subpage navbar