140
3 Continuum Mechanics and Nonlinear Elasticity
Are these equations consistent with Hooke’s law? To answer this question, it
is easiest to check against Hooke’s law expressed in the form (Timoshenko and
Goodier 1969)
σ = 2μE + λ(tr E)I,
where the shear modulus μ and
λ =
2μν
1 − 2ν
are Lamé parameters, with ν being Poisson’s ratio. Inverting this equation to obtain
the strains as functions of stress gives equations that may be more familiar to some
readers. Since tr E = E 11 + E 22 , the above stress-strain relations derived from the
given strain-energy density function are indeed Hooke’s law.
In the above example, symmetry of the stress tensor follows from symmetry of
the strain tensor. It is important to note that we did not set E 12 = E 21 prior to taking
derivatives of W . If we had, then the shear term in Eq. (3.226) would become W s =
2μE 2
12 , giving σ 12 = ∂W s /∂E 12 = 4μE 12 , which is twice the correct result given
above. Thus, symmetry of the strain tensor should be enforced after differentiating
W .
Transverse Isotropy
A transversely isotropic material contains a single axis of symmetry at each point.
Skeletal muscle often is treated as transversely isotropic because the muscle fibers
are aligned along its length, with transverse cross sections appearing isotropic
(Fig. 3.25b). In general, the orientation of the symmetry axis can change from point
to point. For example, a bar with parallel but wavy fibers is also locally transversely
isotropic (Fig. 3.25c). Materials of this type have isotropic properties orthogonal to
the fibers, but different properties in the fiber direction.
Transversely isotropic tissues are often modeled as composites consisting of
aligned fibers embedded in isotropic matrix. In this case, the strain-energy density
function can be taken in the form
W = W m (I 1 , I 2 , I 3 ) + W f (I 4 , I 5 ),
(3.227)
where W m and W f are contributions of the matrix and fibers, respectively. One of the
isotropic forms given above is often used for W m , while W f is assumed to depend
on the two additional strain invariants (Spencer 1984)
I 4 = N f · C · N f = N f · (I + 2E) · N f = λ
2
f
I 5 = N f · C
2
· N f = N f · (I + 2E)
2
· N f ,
(3.228)
3 Continuum Mechanics and Nonlinear Elasticity
Are these equations consistent with Hooke’s law? To answer this question, it
is easiest to check against Hooke’s law expressed in the form (Timoshenko and
Goodier 1969)
σ = 2μE + λ(tr E)I,
where the shear modulus μ and
λ =
2μν
1 − 2ν
are Lamé parameters, with ν being Poisson’s ratio. Inverting this equation to obtain
the strains as functions of stress gives equations that may be more familiar to some
readers. Since tr E = E 11 + E 22 , the above stress-strain relations derived from the
given strain-energy density function are indeed Hooke’s law.
In the above example, symmetry of the stress tensor follows from symmetry of
the strain tensor. It is important to note that we did not set E 12 = E 21 prior to taking
derivatives of W . If we had, then the shear term in Eq. (3.226) would become W s =
2μE 2
12 , giving σ 12 = ∂W s /∂E 12 = 4μE 12 , which is twice the correct result given
above. Thus, symmetry of the strain tensor should be enforced after differentiating
W .
Transverse Isotropy
A transversely isotropic material contains a single axis of symmetry at each point.
Skeletal muscle often is treated as transversely isotropic because the muscle fibers
are aligned along its length, with transverse cross sections appearing isotropic
(Fig. 3.25b). In general, the orientation of the symmetry axis can change from point
to point. For example, a bar with parallel but wavy fibers is also locally transversely
isotropic (Fig. 3.25c). Materials of this type have isotropic properties orthogonal to
the fibers, but different properties in the fiber direction.
Transversely isotropic tissues are often modeled as composites consisting of
aligned fibers embedded in isotropic matrix. In this case, the strain-energy density
function can be taken in the form
W = W m (I 1 , I 2 , I 3 ) + W f (I 4 , I 5 ),
(3.227)
where W m and W f are contributions of the matrix and fibers, respectively. One of the
isotropic forms given above is often used for W m , while W f is assumed to depend
on the two additional strain invariants (Spencer 1984)
I 4 = N f · C · N f = N f · (I + 2E) · N f = λ
2
f
I 5 = N f · C
2
· N f = N f · (I + 2E)
2
· N f ,
(3.228)