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In this chapter a number of concepts are introduced to describe a linear elastic body. In particular, the displacement vector, strain tensor, and stress tensor fields are introduced to define a linear elastic body which satisfies the strain-displacement relations, the equations of motion, and the constitutive relations. Also, the compatibility relations, the general solutions of elastostatics, and an alternative definition of the displacement field of elastodynamics are discussed. The stored energy of an elastic body, the positive definiteness and strong ellipticity of the elasticity fourth-order tensor, and the stress-strain-temperature relations for a thermoelastic body are also discussed.

1 Deformation of an Elastic Body

A material bodyB is defined as a set of elements x, called particles, for which there is a one-to-one correspondence with the points of a region \(\kappa (\mathrm{{B}})\) of a physical space; while a deformation of B is a map \(\kappa \) of B onto a region \(\kappa (\mathrm{{B}})\) in \({E}^3\) with \(\mathrm{{det}}\;(\nabla \kappa )\,{>}\,0\). The point \(\kappa (\mathbf{x })\) is the place occupied by the particle x in the deformation \(\kappa \), and

$$\begin{aligned} \mathbf{u }(\mathbf{x })=\kappa (\mathbf{x })-\mathbf{x } \end{aligned}$$
(2.1)

is the displacement of x.

If the mapping \(\kappa \) depends also on time \({t}\in [0,\infty )\), such a mapping defines a motion of B, and the displacement of x at time \(t\) is

$$\begin{aligned} \mathbf{u }(\mathbf{x },\mathrm{{t}})=\kappa (\mathbf{x },{t})-\mathbf{x } \end{aligned}$$
(2.2)

By the deformation gradientand the displacement gradientwe mean the tensor fields \(\mathbf F =\nabla \kappa \) and \(\nabla \mathbf u \), respectively. A finite strain tensor D is defined by

$$\begin{aligned} \mathbf{D }=\frac{1}{2}(\mathbf{F }^{\mathrm{{T}}} \mathbf{F }-\mathbf{1 }) \end{aligned}$$
(2.3)

or, equivalently, by

$$\begin{aligned} \mathbf{D }=\mathbf{E }+\frac{1}{2}(\nabla \mathbf{u })(\nabla \mathbf{u }^\mathrm{{T}}) \end{aligned}$$
(2.4)

where

$$\begin{aligned} \mathbf{E }=\frac{1}{2}(\nabla \mathbf{u }+\nabla \mathbf{u }^\mathrm{{T}})=\widehat{\nabla }\mathbf{u } \end{aligned}$$
(2.5)

The tensor field E is called an infinitesimal strain tensor.

An infinitesimal rigid displacement of B is defined by

$$\begin{aligned} \mathbf{u }(\mathbf{x })=\mathbf{u }_0 +\mathbf{W }(\mathbf{x }-\mathbf{x }_0) \end{aligned}$$
(2.6)

where \(\mathbf{u }_0,\mathbf{x }_0 \) are constant vectors and W is a skew constant tensor.

An infinitesimal volume change of B is defined by

$$\begin{aligned} \delta \mathrm{{v}}(\mathrm{{B}})=\int \limits _\mathrm{{B}} {\mathrm{{div}}\;\mathbf{u }\;\mathrm{{dv}}} \end{aligned}$$
(2.7)

while

$$\begin{aligned} \mathrm{{div}}\;\mathbf{u }=\mathrm{{tr}}\,\mathbf{E } \end{aligned}$$
(2.8)

represents a dilatation field.

If a deformation is not accompanied by a change of volume, that is, if \(\delta \mathrm{{v}}(\mathrm{{P}})=0\) for every \(\mathrm{{P}}\subset \mathrm{{B}}\), the displacement u is called isochoric.

Kirchhoff Theorem. If two displacement fields \(\mathbf{u }_1\) and \(\mathbf{u }_2\) correspond to the same strain field E then

$$\begin{aligned} \mathbf{u }_1 -\mathbf{u }_2 =\mathbf{w } \end{aligned}$$
(2.9)

where w is a rigid displacement field.

A homogeneous displacement field is defined by

$$\begin{aligned} \mathbf{u }(\mathbf{x })=\mathbf{u }_0 +\mathbf{A }(\mathbf{x }-\mathbf{x }_0 ) \end{aligned}$$
(2.10)

where A is an arbitrary constant tensor and \(\mathbf{u }_0,\;\mathbf{x }_0\) are constant vectors. Clearly, if A is skew, (2.10) represents a rigid displacement, while for an arbitrary A

$$\begin{aligned} \mathbf{u }(\mathbf{x })=\mathbf{u }_1 (\mathbf{x })+\mathbf{u }_2 (\mathbf{x }) \end{aligned}$$
(2.11)

where \(\mathbf{u }_1 (\mathbf{x })\) is a rigid displacement field and \(\mathbf{u }_2 (\mathbf{x })\) is a displacement field corresponding to the strain \(\mathbf{E }=\mathrm{{sym}}\,\mathbf{A }\). The displacement \(\mathbf{u }_2 (\mathbf{x })\) of the form

$$\begin{aligned} \mathbf{u }_2 (\mathbf{x })=\mathbf{E }\,(\mathbf{x }-\mathbf{x }_0) \end{aligned}$$
(2.12)

corresponds to a pure strain from\(\mathbf{x }_0\).

Let \(\mathrm{{e}}>0\) and let n be a unit vector. Then by substituting \(\mathbf{E }=\mathrm{{e}}\,\mathbf{n }\otimes \mathbf{n }\) into (2.12) we obtain a simple extension of amount e in the direction n; and by substituting \(\mathbf{E }=\mathrm{{e}}\,\mathbf{1 }\) into (2.12) we obtain uniform dilatation of amount e. Finally, let \(g>0\) and let m be a unit vector perpendicular to n. Then substituting \(\mathbf{E }=\mathrm{{g}}\,[\mathbf{m }\otimes \mathbf{n }+\mathbf{n }\otimes \mathbf{m }\,]\) into (2.12) we obtain a simple shear of amount g with respect to the pair (m,n).

Decomposition of a strain tensor E into spherical and deviatoric tensors

$$\begin{aligned} \mathbf{E }=\mathbf{E }^{(\mathrm{{s}})}+\mathbf{E }^{(\mathrm{{d}})} \end{aligned}$$
(2.13)

where

$$\begin{aligned} \mathbf{E }^{(\mathrm{{s}})}=\frac{1}{3}(\mathrm{{tr}}\,\mathbf{E })\,\mathbf{1 } \end{aligned}$$
(2.14)

is called a spherical part of E, and \(\mathbf{E }^{(\mathrm{{d}})}=\mathbf{E }-\mathbf{E }^\mathrm{{(s)}}\) is called a deviatoric part of E. Clearly,

$$\begin{aligned} \mathrm{{tr}}\,(\mathbf{E }^{(\mathrm{{d}})})=0 \end{aligned}$$
(2.15)

2 Compatibility

Theorem

Let \(\mathrm{{B}}\subset {E}^3\) be simply connected. If u is a displacement field corresponding to a strain field E on B, that is, if

$$\begin{aligned} \mathbf{E }=\frac{1}{2}(\nabla \mathbf{u }+\nabla \mathbf{u }^\mathrm{{T}})\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(2.16)

then E satisfies the equations of compatibility

$$\begin{aligned} \mathrm{{curl}}\;\mathrm{{curl}}\;\mathbf{E }=\mathbf{0 }\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(2.17)

Conversely, let E be a symmetric tensor field that satisfies the equations of compatibility (2.17), then there exists a displacement field u on B such that u and E satisfy (2.16).

In components the equations of compatibility (2.17) take the form

$$\begin{aligned} {E}_{{ij},{kl}} +{E}_{{kl},{ij}} -{E}_{{ik},{jl}} -{E}_{{jl},{ik}} =0 \end{aligned}$$
(2.18)

An alternative form of (2.17) reads

$$\begin{aligned} \nabla ^2\mathbf{E }+\nabla \nabla (\mathrm{{tr}}\;\mathbf{E })-2\widehat{\nabla }(\mathrm{{div}}\;\mathbf{E })=\mathbf{0 } \end{aligned}$$
(2.19)

3 Motion and Equilibrium

Let \(S\) be a surface in B with unit normal n. Let B be subject to a deformation, and let \(\mathbf{s }_\mathrm{{n}} =\mathbf{s }_\mathrm{{n}} (\mathbf{x },t)\) denote a force per unit area at x and for \({t}\ge 0\) exerted by a portion of B on the side \(S\) toward which n points on a portion of B on the other side of \(S\). The force \(\mathbf{s }_\mathrm{{n}} \) is called the stress vector at (x, \(t\)), while a second-order tensor field \(\mathbf{S }=\mathbf{S }(\mathbf{x },\mathrm{{t}})\) such that

$$\begin{aligned} \mathbf{S }\,\mathbf{n }=\mathbf{s }_\mathrm{{n}} \quad \mathrm{{on}}\quad {S}\times [0,\infty ) \end{aligned}$$
(2.20)

is called a time-dependent stress tensor fieldon \({S}\times [0,\infty ).\)

The equilibrium equations of elastostatics

$$\begin{aligned} \mathrm{{div}}\;\mathbf{S }+\mathbf{b }=\mathbf{0 } \end{aligned}$$
(2.21)
$$\begin{aligned} \mathbf{S }=\mathbf{S }^\mathrm{{T}} \end{aligned}$$
(2.22)

Equation (2.21) expresses the balance of forces, and Eq. (2.22) expresses the balance of moments; and b in (2.21) is the body force vector.

The Beltrami representation of S

$$\begin{aligned} \mathbf{S }=\mathrm{{curl}}\;\mathrm{{curl}}\;\mathbf{A } \end{aligned}$$
(2.23)

where A is a symmetric tensor field, or

$$\begin{aligned} \mathbf{S }=-\nabla ^2\mathbf{G }+2\widehat{\nabla }(\mathrm{{div}}\;\mathbf{G })-(\mathrm{{div}}\;\mathrm{{div}}\;\mathbf{G })\,\mathbf{1 } \end{aligned}$$
(2.24)

where G is a symmetric tensor field.

Self-equilibrated stress field

If \(\mathbf{S }=\mathbf{S }^\mathrm{{T}}\) on B, and

$$\begin{aligned} \int \limits _\mathrm{{S}} \mathbf{Sn }\;\mathrm{{da}} =\mathbf{0 } \end{aligned}$$
(2.25)
$$\begin{aligned} \int \limits _\mathrm{{S}} \mathbf{x \times (\mathbf{Sn })\;{\mathrm{{da}}}} =\mathbf 0 \end{aligned}$$
(2.26)

for every closed surface S in B, then S is called a self-equilibrated stress field.

One can show that S given by (2.23) is a self-equilibrated stress field, and S given by (2.23) is complete in the sense that for any self-equilibrated S there is a symmetric tensor A such that (2.23) is satisfied.

The Beltrami-Schaefer representation of S

$$\begin{aligned} \mathbf{S }=\mathrm{{curl}}\;\mathrm{{curl}}\;\mathbf{A }+2\widehat{\nabla } \mathbf{h }-(\mathrm{{div}}\;\mathbf{h })\,\mathbf{1 } \end{aligned}$$
(2.27)

where A is a symmetric tensor field and h is a harmonic vector field on B.

4 Equations of Motion

 

$$\begin{aligned} \mathrm{{div}}\;\mathbf{S }+\mathbf{b }=\rho \,{\ddot{\mathbf{u }}}\quad \mathrm{{on}}\quad \mathrm{{B}}\times [0,\infty ) \end{aligned}$$
(2.28)

where \(\rho \) is density and b is the body force vector field.

Kinetic energy of B for \(t \ge 0\)

$$\begin{aligned} \mathrm{{K}}(\mathrm{{t}})=\frac{1}{2}\,\int \limits _\mathrm{{B}} {\rho \,{\dot{\mathbf{u }}}^2} \mathrm{{dv}} \end{aligned}$$
(2.29)

Stress power of B for \(t \ge 0\)

$$\begin{aligned} \mathrm{{P}}(\mathrm{{t}})=\,\int \limits _\mathrm{{B}} {\mathbf{S }\cdot \dot{\mathbf{E }}} \mathrm{{dv}} \end{aligned}$$
(2.30)

A dynamic processis identified with a triplet \([\mathbf{u },\mathbf{S },\mathbf{b }]\) that satisfies the equations of motion (2.28).

Theorem

An array of functions \([\mathbf{u },\mathbf{S },\mathbf{b }]\) is a dynamic process consistent with the initial conditions

$$\begin{aligned} {\mathbf{u }}({\mathbf{x }},0)={\mathbf{u }}_0 ({\mathbf{x }}),\quad {\dot{\mathbf{u }}}({\mathbf{x }},0)={\dot{\mathbf{u }}}_0 ({\mathbf{x }})\quad \mathrm{{for}}\quad {\mathbf{x }}\in {\overline{\mathrm{{B}}}} \end{aligned}$$
(2.31)

if and only if

$$\begin{aligned} \mathrm{{i}}*\mathrm{{div}}\;\mathbf{S }+\mathbf{f }=\rho \,\mathbf{u }\quad \mathrm{{on}}\quad {\overline{\mathrm{{B}}}}\times [0,\infty ) \end{aligned}$$
(2.32)

where

$$\begin{aligned} \mathbf{f }(\mathbf{x },{ t})=\mathrm{{i}}*\mathbf{b }(\mathbf{x },{t})+\rho (\mathbf x )\,[\mathbf u _0 (\mathbf{x })+{t}\,{\dot{\mathbf{u }}}_0 (\mathbf{x })] \end{aligned}$$
(2.33)

and

$$\begin{aligned} \mathrm{{i}}=\mathrm{{i}}({t})={t} \end{aligned}$$
(2.34)

The function \(\mathbf{f }=\mathbf{f }(\mathbf{x },{t})\) given by (2.33) is called pseudo-body force field.

Clearly, since \(\rho >0\), Eq. (2.32) provides an alternative definition of the displacement vector \(\mathbf{u }=\mathbf{u }\)(x,t) related to the stress tensor \(\mathbf{S }=\mathbf{S }\)(x,t).

5 Constitutive Relations

A body B is said to be linearly elastic if for every point \(\mathbf{x }\in \mathrm{{B}}\) there is a linear transformation \({\mathbf{\mathsf{{C}} }}\) from the space of all symmetric tensors E into the space of all symmetric tensors S, or

$$\begin{aligned} \mathbf{S }={\mathbf{\mathsf{{C}} }}\,[\mathbf E ] \end{aligned}$$
(2.35)

In components

$$\begin{aligned} {S}_{ij} ={C}_{ijkl} {E}_{kl} \end{aligned}$$
(2.36)

The tensor \({\mathbf{\mathsf{{C}} }}={{\mathbf{\mathsf{{C}} }}}(\mathbf{x })\) is called the elasticity tensor field on B. It follows from Eq. (1.54) that

$$\begin{aligned} {C}_{ijkl} =(\mathbf{e }_{i} \otimes \mathbf{e }_{j})\cdot {{\mathbf{\mathsf{{C}} }}}\,[(\mathbf{e }_{k} \otimes \mathbf{e }_{l} )] \end{aligned}$$
(2.37)

and, since S and E are symmetric, we postulate that

$$\begin{aligned} {C}_{ijkl} ={C}_{jikl} ={C}_{ijlk} \end{aligned}$$
(2.38)

The elasticity tensor \({{\mathbf{\mathsf{{C}} }}}\) is also assumed to be invertible, that means that a restriction of \({{\mathbf{\mathsf{{C}} }}}\) to the space of all symmetric tensors is invertible. The elasticity tensor on the space of all tensors cannot be invertible since its value on every skew tensor is zero.

The invertibility of \({{\mathbf{\mathsf{{C}} }}}\) means that there is a fourth-order tensor \({{\mathbf{\mathsf{{K}} }}}={{\mathbf{\mathsf{{K}} }}}(\mathbf{x })\) such that

$$\begin{aligned} {{\mathbf{\mathsf{{K}} }}}={{\mathbf{\mathsf{{C}} }}}^{-1} \end{aligned}$$
(2.39)

Then equivalent form of (2.35) is

$$\begin{aligned} \mathbf{E }={{\mathbf{\mathsf{{K}} }}}\,[\mathbf{S }] \end{aligned}$$
(2.40)

The tensor \({{\mathbf{\mathsf{{K}} }}}={{\mathbf{\mathsf{{K}} }}}(\mathbf{x })\) is called the compliance tensor.

The fourth-order tensor \({{\mathbf{\mathsf{{C}} }}}\) is symmetric if and only if

$$\begin{aligned} \mathbf{A }\cdot {{\mathbf{\mathsf{{C}} }}}[\mathbf B ]=\mathbf{B }\cdot {\mathbf{\mathsf{{C}} }}[\mathbf{A }] \end{aligned}$$
(2.41)

for any symmetric tensors A and B.

In components the symmetry of \({{\mathbf{\mathsf{{C}} }}}\) means that

$$\begin{aligned} {C}_{ijkl} ={C}_{klij} \end{aligned}$$
(2.42)

The tensor \({{\mathbf{\mathsf{{C}} }}}\) is positive semi-definiteif

$$\begin{aligned} \mathbf{A }\cdot {{\mathbf{\mathsf{{C}} }}}[\mathbf{A }]\ge 0 \end{aligned}$$
(2.43)

for every symmetric tensor A.

The tensor \({{\mathbf{\mathsf{{C}} }}}\) is positive definite if

$$\begin{aligned} \mathbf{A }\cdot {{\mathbf{\mathsf{{C}} }}}[\mathbf{A }]>0 \end{aligned}$$
(2.44)

for every symmetric nonzero tensor A.

The compliance tensor \({{\mathbf{\mathsf{{K}} }}}\) enjoys the properties similar to those of the elasticity tensor \({{\mathbf{\mathsf{{C}} }}}\) [see, Eqs. (2.38) and (2.42)–(2.44)].

By an anisotropic elastic body we mean the body for which the tensor \({{\mathbf{\mathsf{{C}} }}}\) possesses in general 21 different components.

6 Isotropic Elastic Body

For an isotropic elastic body the Eqs. (2.35) and (2.40), respectively, take the form

$$\begin{aligned} \mathbf{S }=2\mu \,\mathbf{E }+\lambda \,(\mathrm{{tr}}\,\mathbf{E }) \end{aligned}$$
(2.45)

and

$$\begin{aligned} \mathbf{E }=\frac{1}{2\mu } \left[ {\mathbf{S }-\frac{\lambda }{3\lambda +2\mu } (\mathrm{{tr}}\,\mathbf{S })\;\mathbf{1 }} \right] \end{aligned}$$
(2.46)

where \(\lambda \ \mathrm{{and}}\ \mu \) are Lamé moduli subject to the constitutive restrictions

$$\begin{aligned} \mu >0,\quad 3\lambda +2\mu >0 \end{aligned}$$
(2.47)

An alternative form of Eqs. (2.45) and (2.46), written in terms of Young’s modulus \(E\) and Poisson’s ratio \(\nu \), reads

$$\begin{aligned} \mathbf{S }&=\frac{{E}}{1+\nu } \left[ \mathbf{E }+\frac{\nu }{1-2\nu } (\mathrm{{tr}}\,\mathbf{E })\;\mathbf{1 } \right] \end{aligned}$$
(2.48)
$$\begin{aligned} \mathbf{E }&=\frac{1}{{E}}\left[ {(1+\nu )\,\mathbf{S }-\nu \,(\mathrm{{tr}}\,\mathbf{S })\;\mathbf{1 }} \right] \end{aligned}$$
(2.49)

where

$$\begin{aligned} {E}>0\quad \mathrm{{and}}\quad -1<\nu <1/2 \end{aligned}$$
(2.50)

Strain energy density of B

$$\begin{aligned} \mathrm{{W}}(\mathbf{E })=\frac{1}{2}\mathbf{E }\cdot {{\mathbf{\mathsf{{C}} }}}\,[\mathbf{E }] \end{aligned}$$
(2.51)

Stress energy density of B

$$\begin{aligned} {\widehat{\mathrm{{W}}}}({\mathbf{S }})=\frac{1}{2}{\mathbf{S }}\cdot {{\mathbf{\mathsf{{K}} }}}\,[{\mathbf{S }}] \end{aligned}$$
(2.52)

The tensor \({{\mathbf{\mathsf{{C}} }}}\) is said to be strongly elliptic if

$$\begin{aligned} \mathbf{A }\cdot {{\mathbf{\mathsf{{C}} }}}[\mathbf{A }]>0 \end{aligned}$$
(2.53)

for every A of the form

$$\begin{aligned} \mathbf{A }=\mathbf{a }\otimes \mathbf{b } \end{aligned}$$
(2.54)

where a and b are arbitrary nonzero vectors.

7 The Cauchy Relations

An anisotropic elastic body obeying, in addition to the symmetry relations (2.38) and (2.42), the restrictions

$$\begin{aligned} {C}_{ijkl} ={C}_{ikjl} \end{aligned}$$
(2.55)

is said to be of the Cauchy type.

8 Constitutive Relations for a Thermoelastic Body

For an anisotropic body subject to an uneven heating the constitutive relations take the form

$$\begin{aligned} \mathbf{S }={{\mathbf{\mathsf{{C}} }}}\,[\mathbf{E }]+{T}\mathbf{M } \end{aligned}$$
(2.56)

and

$$\begin{aligned} \mathbf{E }={{\mathbf{\mathsf{{K}} }}}\,[\mathbf{S }]+{T}\mathbf{A } \end{aligned}$$
(2.57)

where

$$\begin{aligned} {T}=\theta -\theta _0, \quad \theta _0 >0 \end{aligned}$$
(2.58)

is a temperature change, \(\mathbf{M }=\mathbf{M }^{T}\) is called the stress-temperature tensor,\(\mathbf{A }=\mathbf{A }^{T}\) is called the thermal expansion tensor, \(\theta \) is the absolute temperature,and \(\theta _0\) is a reference temperature.

Since relations (2.56) and (2.57) are equivalent

$$\begin{aligned} {{\mathbf{\mathsf{{K}} }}}={{\mathbf{\mathsf{{C}} }}}^{-1}\quad \mathrm{{and}} \quad \mathbf{A }=-\,{{\mathbf{\mathsf{{K}} }}}[\mathbf{M }] \end{aligned}$$
(2.59)

for an isotropic body

$$\begin{aligned} \mathbf{S }=2\mu \,\mathbf{E }+\lambda \,(\mathrm{{tr}}\,\mathbf{E })-(3\lambda +2\mu )\alpha \,{T}\,\mathbf{1 } \end{aligned}$$
(2.60)

and

$$\begin{aligned} \mathbf{E }=\frac{1}{2\mu } \left[ {\mathbf{S }-\frac{\lambda }{3\lambda +2\mu }(\mathrm{{tr}}\,\mathbf{S })\;\mathbf{1 }} \right] +\alpha \,{T}\,\mathbf{1 } \end{aligned}$$
(2.61)

where \(\alpha \) is the coefficient of thermal expansion,

or

$$\begin{aligned} \mathbf{S }=\frac{{E}}{1+\nu }\left[ {\mathbf{E }+\frac{\nu }{1-2\nu }(\mathrm{{tr}}\,\mathbf{E })\;\mathbf{1 }} \right] -\frac{{E}}{1-2\nu }\alpha \,{T}\,\mathbf{1 } \end{aligned}$$
(2.62)

and

$$\begin{aligned} \mathbf{E }=\frac{1}{{E}}\left[ {(1+\nu )\,\mathbf{S }-\nu \,(\mathrm{{tr}} \ \mathbf{S })\;\mathbf{1 }} \right] +\alpha \,{T}\,\mathbf{1 } \end{aligned}$$
(2.63)

9 Problems and Solutions Related to the Fundamentals of Linear Elasticity

Problem 2.1.

Show that if u is a pure strain from \(\mathbf{x }_0 \), then u admits the decomposition

$$\begin{aligned} \mathbf{u }=\mathbf{u }_1 +\mathbf{u }_2 +\mathbf{u }_3 \end{aligned}$$
(2.64)

where \(\mathbf{u }_1,\ \mathbf{u }_2,\ \mathrm{{and}}\ \mathbf{u }_3\) are simple extensions in mutually perpendicular directions from \(\mathbf{x }_0\).

Solution.

Since u represents a pure strain from \(\mathbf{x }_{0},\ \mathbf{u }\) takes the form [see definition of \(\mathbf{u }_2\) in (2.12)]

$$\begin{aligned} \mathbf{u }=\mathbf{E }(\mathbf{x }-\mathbf{x }_{0}) \end{aligned}$$
(2.65)

where E is the strain tensor corresponding to u. Now, by the decomposition spectral theorem [see Eq. (1.45) in which \(\mathbf{T }=\mathbf{E }\) and \(\lambda _{ i}={ e_i}\)]

$$\begin{aligned} \mathbf{E }=\sum \limits _{{ i}=1}^{3}{ e_{i}}\ \mathbf{n }_{ i}\otimes \mathbf n _{ i} \end{aligned}$$
(2.66)

where \(\mathbf{n }_{ i}\) is a principal direction corresponding to a principal value \(e_{i}\) of E.

Substituting (2.66) into (2.65) we obtain

$$\begin{aligned} \mathbf{u }=\sum \limits _{{ i}=1}^{3}{ e_{i}} (\mathbf{n }_{ i}\otimes \mathbf{n }_{ i})(\mathbf{x }-\mathbf{x }_{0}) \end{aligned}$$
(2.67)

Since for two arbitrary vectors a and b

$$\begin{aligned} (\mathbf{a }\otimes \mathbf{a })\mathbf{b }=(\mathbf{a }\cdot \mathbf{b })\mathbf{a } \end{aligned}$$
(2.68)

therefore, Eq. (2.67) is equivalent to

$$\begin{aligned} \mathbf{u }=\mathbf{u }_{1}+\mathbf{u }_{2}+\mathbf{u }_{3} \end{aligned}$$
(2.69)

where

$$\begin{aligned} \mathbf{u }_{ i}={ e_{i}}[\mathbf{n }_{ i}\cdot (\mathbf{x }-\mathbf{x }_{0})]\mathbf{n }_{ i}\ (\mathrm{{no}\ sum}) \end{aligned}$$
(2.70)

Since \(\mathbf{u }_{ i}\) represents a simple extension of magnitude \({ e_{i}}\) in the direction of \(\mathbf{n }_{ i}\) [see Eq. (2.12)], and \(\mathbf{n }_{1},\ \mathbf{n }_{2}\), and \(\mathbf{n }_{3}\) are orthogonal, Eq. (2.69) is equivalent to (2.64). This completes proof of (2.64).

Problem 2.2.

Show that u in Problem 2.1 admits an alternative representation

$$\begin{aligned} {\mathbf{u }}={\mathbf{u }}_{\mathrm{{d}}} +{\mathbf{u }}_{\mathrm{{c}}} \end{aligned}$$
(2.71)

where \(\mathbf{u }_\mathrm{{d}}\) is a uniform dilatation from \(\mathbf{x }_0\), while \(\mathbf{u }_\mathrm{{c}}\) is an isochoric pure strain from \(\mathbf{x }_0\).

Solution.

We rewrite E of Problem 2.1 as

$$\begin{aligned} \mathbf{E }=\mathbf{E }^{({ s})}+\mathbf{E }^{({ d})} \end{aligned}$$
(2.72)

where

$$\begin{aligned} \mathbf{E }^{({ s})}=\frac{1}{3}\mathbf{1 }({\mathrm{tr}}\ \mathbf{E }),\quad \mathbf{E }^{({ d})}=\mathbf{E }-\frac{1}{3}\mathbf{1 }({ \mathrm{tr}}\ \mathbf{E }) \end{aligned}$$
(2.73)

Then Eq. (2.65) of Problem 2.1 takes the form

$$\begin{aligned} \mathbf{u }=\mathbf{u }_{ d}+\mathbf{u }_{ c} \end{aligned}$$
(2.74)

where

$$\begin{aligned}&\mathbf{u }_{d}=\mathbf{E }^{({ s})}(\mathbf{x }-\mathbf{x }_{0}) \end{aligned}$$
(2.75)
$$\begin{aligned}&\mathbf{u }_{ c}=\mathbf{E }^{({d})}(\mathbf{x }-\mathbf{x }_{0}) \end{aligned}$$
(2.76)

It follows from Eqs. (2.73) and (2.75) that \(\mathbf{u }_{ d}\) represents a uniform dilatation of magnitude \({ e}=\frac{1}{3}(\mathrm{{tr}}\ \mathbf{E })\), while the condition \(\mathrm{{tr}}\ \mathbf{E }^{({ d})}=0\) implies that \(\mathbf{u }_{ c}\) represents an isochoric pure strain. This completes solution to Problem 2.2.

Problem 2.3.

Show that if u is a simple shear of amount \(\gamma \) with respect to the pair (m, n), where m and n are perpendicular unit vectors, then u admits the decomposition

$$\begin{aligned} \mathbf{u }=\mathbf{u }_1 +\mathbf{u }_2 \end{aligned}$$
(2.77)

where \(\mathbf{u }_1\) is a simple extension of amount \(\gamma \) in the direction \(\frac{1}{\sqrt{2} }(\mathbf{m }+\mathbf{n })\), and \(\mathbf{u }_2\) is a simple extension of amount \(-\gamma \) in the direction \(\frac{1}{\sqrt{2} }(\mathbf{m }-\mathbf{n })\).

Solution.

Since u represents a simple shear of amount \(\gamma \) with respect to \((\mathbf{m }\), \(\mathbf{n })\), then the strain tensor corresponding to u takes the form [see the definition of a simple shear below Eq. (2.12)]

$$\begin{aligned} \mathbf{E }=\gamma \ (\mathbf{m }\otimes \mathbf{n }+\mathbf{n }\otimes \mathbf{m }) \end{aligned}$$
(2.78)

Let \(\lambda \) and a denote a principal value and a principal vector of E, respectively. Then

$$\begin{aligned} \gamma \ (\mathbf{n }\cdot \mathbf{a }) \mathbf{m }+\gamma (\mathbf m \cdot \mathbf{a }) \mathbf{n }-\lambda {\mathbf{a }}=\mathbf 0 \end{aligned}$$
(2.79)

It is easy to check that Eq. (2.79) has the three eigensolutions

$$\begin{aligned} \mathbf{a }_{1}&=\mathbf{m }\times \mathbf{n },\qquad \qquad \lambda _{1}=0 \end{aligned}$$
(2.80)
$$\begin{aligned} \mathbf{a }_{2}&=\frac{1}{\sqrt{2}}(\mathbf{m }+\mathbf{n }),\quad \lambda _{2}=\gamma \end{aligned}$$
(2.81)
$$\begin{aligned} \mathbf{a }_{3}&=\frac{1}{\sqrt{2}}(\mathbf{m }-\mathbf{n }),\quad \lambda _{3}=-\gamma \end{aligned}$$
(2.82)

Therefore, using the solution (2.67) of Problem 2.1 we find that Eq. (2.77) holds true. This completes solution of Problem 2.3.

Problem 2.4.

Let u and E denote a displacement vector field and the corresponding strain tensor field defined on \({\overline{\mathrm{{B}}}}\). Show that the mean strain \({\widehat{\mathbf{E }}}(\mathrm{{B}})\) is represented by the surface integral

$$\begin{aligned} {{\widehat{\mathbf{E }}}}(\mathrm{{B}})=\frac{1}{{{v}}(\mathrm{{B}})}\int \limits _{\partial \mathrm{{B}}} {\mathrm{{sym}}\;(\mathbf{u }\otimes \mathbf{n })\,{da}} \end{aligned}$$
(2.83)

where \(v\)(B) is the volume of B.

Solution.

The mean strain \(\widehat{\mathbf{E }}({B})\) is defined by

$$\begin{aligned} \widehat{\mathbf{E }}({B})=\frac{1}{ {{v}}(B)}\ \int \limits _{ B} \mathbf{E }\ {d{v}} \end{aligned}$$
(2.84)

In components we obtain

$$\begin{aligned} \widehat{E}_{ ij}({B})=\frac{1}{ {{v}}(B)}\ \int \limits _{ B} { E_{ij}\ d{{v}}} \end{aligned}$$
(2.85)

Since

$$\begin{aligned} {E_{ij}}=\frac{1}{2}({ u_{i,j}}+{ u_{j,i}}) \end{aligned}$$
(2.86)

therefore, by the divergence theorem,

$$\begin{aligned} \int \limits _{ B}{ E_{ij}}\ { d{ {v}}}=\frac{1}{2}\int \limits _{ B}({ u_{i,j}}+{ u_{j,i}}){ d{ {v}}}=\frac{1}{2}\int \limits _{\partial { B}}({ u_{i}\ n_{j}}+{ u_{j}\ n_{i}}){ da} \end{aligned}$$
(2.87)

Equations (2.87) and (2.85) imply that Eq. (2.83) holds true, and this completes solution to Problem 2.4.

Problem 2.5.

Show that if \(\mathbf{u }=\mathbf 0 \) on \(\partial \mathrm{{B}}\) then

$$\begin{aligned} \int \limits _\mathrm{{B}} (\nabla \mathbf{u })^2{d{{v}}}\le 2\int \limits _\mathrm{{B}} {\left| \mathbf{E } \right| ^2{d {{v}}}} \end{aligned}$$
(2.88)

where E is the strain tensor field corresponding to a displacement field u on B.

Solution.

We recall the relation

$$\begin{aligned} \varvec{\nabla } \mathbf{u }=\mathbf{E }+\mathbf{W } \end{aligned}$$
(2.89)

where

$$\begin{aligned} \mathbf{E }=\frac{1}{2}(\varvec{\nabla } \mathbf{u }+\varvec{\nabla } \mathbf{u }^\mathrm{{T}}) \end{aligned}$$
(2.90)

and

$$\begin{aligned} \mathbf{W }=\frac{1}{2}(\varvec{\nabla } \mathbf{u }-\varvec{\nabla } \mathbf{u }^\mathrm{{T}}) \end{aligned}$$
(2.91)

Since \(\mathbf{E }\cdot \mathbf W =0\), Eq. (2.89) implies that

$$\begin{aligned} |\varvec{\nabla } \mathbf{u }|^{2}=|\mathbf{E }|^{2}+|\mathbf W |^{2} \end{aligned}$$
(2.92)

and it follows from Eqs. (2.90) and (2.91), respectively, that

$$\begin{aligned} |\mathbf{E }|^{2}=\frac{1}{2}[(\varvec{\nabla } \mathbf{u })^{2}+(\varvec{\nabla } \mathbf{u })\cdot (\varvec{\nabla } \mathbf{u }^\mathrm{{T}})] \end{aligned}$$
(2.93)

and

$$\begin{aligned} |\mathbf W |^{2}=\frac{1}{2}[(\varvec{\nabla } \mathbf{u })^{2}-(\varvec{\nabla } \mathbf{u })\cdot (\varvec{\nabla } \mathbf{u }^\mathrm{{T}})] \end{aligned}$$
(2.94)

Hence,

$$\begin{aligned} |\mathbf{E }|^{2}-|\mathbf W |^{2}=(\varvec{\nabla } \mathbf{u })\cdot (\varvec{\nabla } \mathbf{u }^\mathrm{{T}}) \end{aligned}$$
(2.95)

Now

$$\begin{aligned} (\varvec{\nabla } \mathbf{u })\cdot (\varvec{\nabla } \mathbf{u }^\mathrm{{T}})&={ u_{i,j}\ u_{i,j}^{T}}={ u_{i,j}\ u_{j,i}}\nonumber \\&={ (u_{i,j}\ u_{j}),_{i}}-{ u_{i,ji}\ u_{j}}\nonumber \\&={ (u_{i,j}\ u_{j}),_{i}}-{ (u_{i,i}\ u_{j}),_{j}}+({ u_{i,i}})^{2}\nonumber \\&={ (u_{j,i}\ u_{i}}-{ u_{i,i}\ u_{j}),_{j}}+({ u_{i,i}})^{2}\nonumber \\&=\mathrm{{div}}[(\varvec{\nabla } \mathbf{u })\mathbf{u }-(\mathrm{{div}}\ \mathbf{u })\mathbf{u }]+(\mathrm{{div}}\ \mathbf{u })^{2} \end{aligned}$$
(2.96)

Therefore, integrating Eq. (2.96) over \(B\), using the divergence theorem, and the homogeneous boundary condition: \(\mathbf{u }=\mathbf{0 }\) on \(\partial { B}\), we obtain

$$\begin{aligned} \int \limits _{ B}(\varvec{\nabla } \mathbf{u })\cdot (\varvec{\nabla } \mathbf{u }^\mathrm{{T}}){d{ {v}}}=\int \limits _{ B}(\mathrm{{div}}\ \mathbf{u })^{2}{ d{{v}}} \end{aligned}$$
(2.97)

Equations (2.95) and (2.97) imply that

$$\begin{aligned} \int \limits _{B}(|\mathbf{E }|^{2}-|\mathbf W |^{2}){ d{ {v}}}=\int \limits _{ B}(\mathrm{{div}}\ \mathbf{u })^{2}{ d{{v}}} \end{aligned}$$
(2.98)

and it follows from Eq. (2.92) that

$$\begin{aligned} \int \limits _{B}(|\mathbf{E }|^{2}+|\mathbf{W }|^{2}){d{{v}}}=\int \limits _{ B}|\varvec{\nabla } \mathbf{u }|^{2}{ d{{v}}} \end{aligned}$$
(2.99)

Therefore, by adding Eqs. (2.98) and (2.99), we obtain

$$\begin{aligned} 2\int \limits _{B}|\mathbf{E }|^{2}{ d{{v}}}=\int \limits _{ B}|\varvec{\nabla } \mathbf{u }|^{2}{ d{{v}}}+\int \limits _{ B}(\mathrm{{div}}\ \mathbf{u })^{2}{ d{{v}}} \end{aligned}$$
(2.100)

and Eq. (2.100) leads to the inequality

$$\begin{aligned} 2\int \limits _{ B}|\mathbf{E }|^{2}{ d{{v}}}\ge \int \limits _{ B}|\varvec{\nabla } \mathbf{u }|^{2}{ d{{v}}} \end{aligned}$$
(2.101)

This completes solution of Problem 2.5.

Problem 2.6.

 

  1. (i)

    Let E be a strain tensor field on \({E}^3\) defined by the matrix

    $$\begin{aligned} \mathbf{E }=\frac{{N}}{{E}}\;\left[ \begin{array}{ccc} 1&{}0&{}0 \\ 0&{}-\nu &{}0 \\ 0&{}0&{}-\nu \\ \end{array} \right] \end{aligned}$$
    (2.102)

    where \(E\), \(N\), and \(\nu \) are positive constants. Show that a solution u to the equation \(\mathbf{E }={\widehat{{\nabla }}}\mathbf{u }\) on \({E}^3\) subject to the condition \(\mathbf{u }(\mathbf 0 )=\mathbf 0 \) takes the form

    $$\begin{aligned} \mathbf{u }=\left[ {\frac{{ N}}{{ E}}{ x}_1, \;-\nu \frac{{N}}{{ E}}{ x}_2, \;-\nu \frac{{N}}{{E}}{ x}_3} \right] ^\mathrm{{T}} \end{aligned}$$
    (2.103)
  2. (ii)

    Let E be a strain tensor field on \({E}^3\) defined by the matrix

    $$\begin{aligned} \mathbf{E }=\frac{{M}}{{E}{I}}{x}_1 \; \left[ \begin{array}{ccc} \nu &{} 0 &{} 0 \\ 0 &{} \nu &{} 0 \\ 0 &{} 0 &{} -1 \\ \end{array} \right] \end{aligned}$$
    (2.104)

    where \(M, E, I\), and \(\nu \) are positive constants. Show that a solution u to the equation \(\mathbf{E }=\widehat{\nabla } \mathbf{u }\) on \({E}^3\) subject to the condition \(\mathbf{u }(\mathbf 0 )=\mathbf 0 \) takes the form

    $$\begin{aligned} \mathbf{u }=\frac{{M}}{{E}{I}}\left[ {\frac{1}{2}({x}_3^2 +\nu \,{x}_1^2 -\nu \,{x}_2^2 ),\;\;\nu \,{x}_1 x_2, \;\;-{x}_1 {x}_3 } \right] ^\mathrm{{T}} \end{aligned}$$
    (2.105)

Solution.

 

  1. (i)

    Using (2.103) we find that \(\mathbf{u }(\mathbf{0 })=\mathbf{0 }\) and

    $$\begin{aligned} \varvec{\nabla } \mathbf{u }=\frac{ N}{ E}\left[ \begin{array}{ccc} 1 &{} 0 &{} 0\\ 0 &{}-{ \nu } &{} 0\\ 0 &{} 0 &{}-{ \nu }\end{array}\right] \end{aligned}$$
    (2.106)

    Since \(\varvec{\nabla } \mathbf{u }=\varvec{\nabla } \mathbf{u }^\mathrm{{T}}\), the equation

    $$\begin{aligned} \widehat{\varvec{\nabla }}\mathbf{u }=\mathbf{E } \end{aligned}$$
    (2.107)

    in which E is given by (2.102) is identically satisfied. This completes a proof of (i).

  2. (ii)

    Using (2.105) we obtain \(\mathbf{u }(\mathbf{0 })=\mathbf 0 \) and

    $$\begin{aligned} \varvec{\nabla } \mathbf{u }=\frac{ M}{ EI}\left[ \begin{array}{ccc} { \nu x}_{1} &{}-{ \nu x}_{2}&{}{x}_{3}\\ { \nu x}_{2} &{}{ \nu x}_{1} &{} 0\\ -{x}_{3} &{} 0&{}-{x}_{1}\end{array}\right] \end{aligned}$$
    (2.108)

    Hence

    $$\begin{aligned} \varvec{\nabla } \mathbf{u }^\mathrm{{T}}=\frac{M}{ EI}\left[ \begin{array}{ccc} { \nu x}_{1} &{}{ \nu x}_{2}&{}-{x}_{3}\\ -{ \nu x}_{2} &{}{ \nu x}_{1} &{} 0\\ {x}_{3} &{} 0&{}-{x}_{1}\end{array}\right] \end{aligned}$$
    (2.109)

    and

    $$\begin{aligned} \widehat{\varvec{\nabla }} \mathbf{u }=\frac{ M}{ EI}\left[ \begin{array}{ccc} { \nu x}_{1} &{} 0 &{} 0\\ 0&{}{ \nu x}_{1} &{} 0\\ 0 &{} 0 &{}-{x}_{1}\end{array}\right] \end{aligned}$$
    (2.110)

    Equation (2.110) implies that u given by (2.105) satisfies the equation

    $$\begin{aligned} \widehat{\varvec{\nabla }} \mathbf{u }=\mathbf{E } \end{aligned}$$
    (2.111)

    where E is given by (2.104). This completes proof of (ii).

Problem 2.7.

Given a stress tensor S at a point A, find: (i) the stress vector s on a plane through A parallel to the plane \(\mathbf{n }\cdot \mathbf{x }-\mathrm{{vt}}=0\quad (\left| \mathbf{n } \right| =1,\;\mathrm{{v}}>0,\;\mathrm{{t}}\ge 0)\), (ii) the magnitude of s, (iii) the angle between s and the normal to the plane, and (iv) the normal and tangential components of the stress vector s.

Answers. (i) \(\mathbf{s }=\mathbf{Sn }\); (ii) \(\left| \mathbf{s } \right| =\left| \mathbf{Sn } \right| \); (iii) \(\cos \;\theta =\mathbf{s }\cdot \mathbf{n }/\left| \mathbf{s } \right| \); (iv) \(\mathbf{s }=\mathbf{s }_\mathrm{{n}} +\mathbf{s }_\tau \), where \(\mathbf{s }_\mathrm{{n}} =(\mathbf{n }\cdot \mathbf{s })\,\mathbf{n }\) and \(\mathbf{s }_\tau =\mathbf{n }\times (\mathbf{s }\times \mathbf{n })\).

Solution.

Solution to Problem 2.7 is presented by the answers (i)–(iv).

Problem 2.8.

Let \(\{\mathbf{e }_{i}\}\) be an orthonormal basis for a stress tensor S, and let \(\{\mathbf{e }_{i}^*\}\) be an orthonormal basis formed by the eigenvectors of S. Then a tensor \(\mathbf{S }^{*}\) obtained from S by the transformation formula from \(\{\mathbf{e }_{i}\}\) to \(\{\mathbf{e }_{i}^{*} \}\) takes the form

$$\begin{aligned} \mathbf{S }^{*} =\lambda _1 \,\mathbf{e }_1^{*} \otimes \mathbf{e }_1^{*} +\lambda _2 \,\mathbf{e }_2^{*} \otimes \mathbf{e }_2^{*} +\lambda _3 \,\mathbf{e }_3^{*} \otimes \mathbf{e }_3^{*} \end{aligned}$$
(2.112)

where \(\lambda _i \) is an eigenvalue of S corresponding to the eigenvector \(\mathbf{e}_{i}^{*}\). Show that the function

$$\begin{aligned} \mathrm{{g}}(\mathbf{n }^{*})=\left| \mathbf{s _\tau ^{*}} \right| =\left| {\mathbf{n }^{*} \times (\mathbf{S }^{*} \mathbf{n }^{*} \times \mathbf{n }^{*} )} \right| \end{aligned}$$
(2.113)

representing the tangent stress vector magnitude with regard to a plane with a normal \(\mathbf{n }^{*}\) in the \(\{\mathbf{e }_{i}^{*}\}\) basis, assumes the extreme values

$$\begin{aligned} \left| {\mathbf{s }_\tau ^*} \right| _1 =\frac{1}{2}\left| {\lambda _2 -\lambda _3 } \right| \end{aligned}$$
(2.114)
$$\begin{aligned} \left| {\mathbf{s }_\tau ^*} \right| _2 =\frac{1}{2}\left| {\lambda _3 -\lambda _1 } \right| \end{aligned}$$
(2.115)

and

$$\begin{aligned} \left| {\mathbf{s }_\tau ^*} \right| _3 =\frac{1}{2}\left| {\lambda _1 -\lambda _2 } \right| \end{aligned}$$
(2.116)

at

$$\begin{aligned} \mathbf{n }_1^{*}&=[0,\;\;\pm 1/\sqrt{2}, \;\;\pm 1/\sqrt{2} ]^\mathrm{{T}}\end{aligned}$$
(2.117)
$$\begin{aligned} \mathbf{n }_2^*&=[\;\pm 1/\sqrt{2}, \;\;0,\;\;\pm 1/\sqrt{2} ]^\mathrm{{T}} \end{aligned}$$
(2.118)

and

$$\begin{aligned} \mathbf{n }_3^{*} =[\;\pm 1/\sqrt{2}, \;\;\pm 1/\sqrt{2},\; 0]^\mathrm{{T}} \end{aligned}$$
(2.119)

respectively. Hence, if \(\lambda _1 >\lambda _2 >\lambda _3 \) then the largest tangential stress vector magnitude is

$$\begin{aligned} \left| {\mathbf{s }_\tau ^*} \right| _2 =\frac{1}{2}\left| {\lambda _3 -\lambda _1 } \right| \end{aligned}$$
(2.120)

and this extreme vector acts on the plane that bisects the angle between \(\mathbf{e }_1^{*}\) and \(\mathbf{e }_3^{*}\).

Solution.

It follows from (iv) of Problem 2.7 that

$$\begin{aligned} \mathbf{s }^{*}=\mathbf{s }_{ n}^{*}+{\mathbf{s }_{\tau }^{*}} \end{aligned}$$
(2.121)

where

$$\begin{aligned} \mathbf s ^{*}=\mathbf{S }^{*}\mathbf{n }^{*} \end{aligned}$$
(2.122)

and

$$\begin{aligned} \mathbf{s }_{ n}^{*}=(\mathbf{s }^{*}\cdot \mathbf{n }^{*})\mathbf{n },\quad \mathbf{s }_{\tau }^{*}=\mathbf{n }^{*}\times (\mathbf{s }^{*}\times \mathbf{n }) \end{aligned}$$
(2.123)

Using (2.112), (2.122) and (2.123), we obtain

$$\begin{aligned} \mathbf{s }^{*}=\lambda _{1}{ n}_{1}^{*} \mathbf{e }_{1}^{*}+\lambda _{2}{ n}_{2}^{*}\mathbf{e }_{2}^{*}+\lambda _{3}{ n}_{3}^{*}\mathbf{e }_{3}^{*} \end{aligned}$$
(2.124)

and

$$\begin{aligned} \mathbf{s }^{*}\cdot \mathbf{n }^{*}=\lambda _{1}({ n}_{1}^{*})^{2}+\lambda _{2}({ n}_{2}^{*})^{2}+\lambda _{3}({ n}_{3}^{*})^{2} \end{aligned}$$
(2.125)

Since \(\mathbf{s }_{ n}^{*}\cdot \mathbf{s }_{\tau }^{*}=0\), by squaring (2.121), we get

$$\begin{aligned} |\mathbf{s }^{*}|^{2}=\left| \mathbf{s }_{ n}^{*}\right| ^{2}+\left| \mathbf{s }_{\tau }^{*}\right| ^{2} \end{aligned}$$
(2.126)

Now, introduce the function

$$\begin{aligned} f(\mathbf{n }^{*})&=\left| \mathbf{s }_{\tau }^{*}\right| ^{2}=|\mathbf{s }^{*}|^{2}-\left| \mathbf{s }_{ n}^{*}\right| ^{2} =\lambda _{1}^{2}({ n}_{1}^{*})^{2}+\lambda _{2}^{2}({ n}_{2}^{*})^{2}+\lambda _{3}^{2}({ n}_{3}^{*})^{2}\nonumber \\&\qquad -\left[ \lambda _{1}({ n}_{1}^{*})^{2}+\lambda _{2}({ n}_{2}^{*})^{2}+\lambda _{3}^{2}({ n}_{3}^{*})^{2}\right] ^2 \end{aligned}$$
(2.127)

If there is an extremum of \({ f}={ f}(\mathbf{n }^{*})\), treated as a function of \({ n}_{1}^{*},\ { n}_{2}^{*}\), and \({n}_{3}^{*}\), it is also an extremum of \({ g}={g}(\mathbf{n }^{*})=\sqrt{{ f}(\mathbf{n }^{*})}\).

To find the extreme values of \({ f}={ f}{(\mathbf{n }^{*})}\) subject to the condition \(|\mathbf{n }^{*}|=1\) we solve the algebraic equation for \(\mathbf{n }^{*}\)

$$\begin{aligned} \frac{\partial }{\partial n_{ i}^{*}}\ [{ f}(\mathbf{n }^{*})-{ t}(|\mathbf{n }^{*}|^{2}-1)]=0 \end{aligned}$$
(2.128)

where \(t\) is a Lagrangian multiplier. In expanded form Eq. (2.128) takes the form

$$\begin{aligned} \left[ \lambda _{1}^{2}-2\lambda _{1} (\mathbf{s }^{*}\cdot \mathbf{n }^{*})-{ t}\right] { n}_{1}^{*}=0 \end{aligned}$$
(2.129)
$$\begin{aligned} \left[ \lambda _{2}^{2}-2\lambda _{2} (\mathbf{s }^{*}\cdot \mathbf{n }^{*})-{ t}\right] { n}_{2}^{*}=0 \end{aligned}$$
(2.130)
$$\begin{aligned} \left[ \lambda _{3}^{2}-2\lambda _{3} (\mathbf{s }^{*}\cdot \mathbf{n }^{*})-{ t}\right] { n}_{3}^{*}=0 \end{aligned}$$
(2.131)

where \((\mathbf{s }^{*}\cdot \mathbf{n }^{*})\) is given by (2.125). It can be verified that the unit vectors \(\mathbf{n }_{1}^{*}, \mathbf{n }_{2}^{*}\), and \(\mathbf{n }_{3}^{*}\), given by Eqs. (2.117), (2.118), and (2.119), respectively, satisfy Eqs. (2.129)–(2.131) with \({ t}=\lambda _{2}\lambda _{3}\). In addition, by substituting \(\mathbf{n }_{1}^{*}, \mathbf{n }_{2}^{*}\), and \(\mathbf{n }_{3}^{*}\) into (2.127), we obtain Eqs. (2.114), (2.115), and (2.116), respectively. Also, the vector \(\mathbf{n }_{2}^{*}\) that is normal to the surface element on which the largest tangential stress vector \(\left( \mathbf{s }_{\tau }^{*}\right) _{2}\) acts bisects the angle between \(\mathbf{e }_{1}^{*}\) and \(\mathbf{e }_{3}^{*}\). This completes solution of Problem 2.8.

Fig. 2.1
figure 1figure 1

The wedge region

Full size image

Problem 2.9.

Let \({D}=\{\mathbf{x }:\;{x}_1 \ge 0,\;\;{x}_1 \tan \,\theta \ge {x}_2 \ge 0\}\) be a two-dimensional wedge region shown in the Fig. 2.1, and let \({S}_{\alpha \beta } ={S}_{\alpha \beta } (\mathbf{x }),\ [\mathbf{x }=({x}_1, {x}_2 );\ \alpha , \beta =1,2]\) be a symmetric tensor field on \(D\) defined by

$$\begin{aligned} {S}_{11} ={d\;x}_2 +{e\,x}_1 -\rho \,g {x}_1, \quad {S}_{22} =-\gamma {x}_1, \quad {S}_{12} ={S}_{21} =-{e\,x}_2 \end{aligned}$$
(2.132)

where \(d, e, g, \rho \), and \(\gamma \) are constants [\(\mathrm{{g}}>0,\;\rho >0,\;\gamma >0]\). (i) Show that

$$\begin{aligned} \mathrm{{div}}\;\mathbf{S }+\mathbf{b }=\mathbf{0 }\quad \mathrm{{on}}\quad {D} \end{aligned}$$
(2.133)

where

$$\begin{aligned} \mathbf{b }=[\rho \,{g},\;0]^\mathrm{{T}}\quad \mathrm{{on}}\quad {D} \end{aligned}$$
(2.134)
  1. (ii)

    Using the transformation formula from the system \({x}_\alpha \) to the system \({x}^{\prime }_\alpha \)[see Eq. (1.157) in Problem 1.8] find the components \({S}^{\prime }_{\alpha \beta } \) in terms of \({S}_{\alpha \beta }\), and show that

    $$\begin{aligned} {S}^{\prime }_{12} =0\quad \mathrm{{and}}\quad {S}^{\prime }_{22} =0\quad \mathrm{{for}}\;\;{x}_2 ={x}_1 \tan \,\theta \end{aligned}$$
    (2.135)

    provided

    $$\begin{aligned} {e}=\frac{\gamma }{\tan ^2\theta },\quad \mathrm{{and}}\quad {d}=\frac{\rho \,{g}}{\tan \,\theta }-\frac{2\gamma }{\tan ^3\theta } \end{aligned}$$
    (2.136)
  2. (iii)

    Give diagrams of \({S}_{11}\;\mathrm{{and}}\;\;{S}_{12} \) over a horizontal section \({x}_1 ={x}_1^0 =\mathrm{{constant}}\).

  3. (iv)

    Give a diagram of \({S}_{22} \) over the vertical section \({x}_2 =0\).

Solution.

To show (i) we note that \({ S}_{\alpha \beta }={ S}_{\alpha \beta }({x}_{1}, { x}_{2})\) given by Eq. (2.132) satisfies the equilibrium equation

$$\begin{aligned} { S}_{\alpha \beta , \beta }+{ b}_{\alpha }=0\quad { \mathrm{{on}}\ D} \end{aligned}$$
(2.137)

since

$$\begin{aligned} { S}_{1\beta , \beta }=-\rho \,{ g},\quad { S}_{2\beta , \beta }=0\quad { \mathrm{{on}}\ D} \end{aligned}$$
(2.138)

for arbitrary constants \(d\), \(e\), \(g\), \(\rho \), and \(\gamma \). To show (ii) we use the transformation formulas [see Eq. (1.157) in Problem 1.8]

$$\begin{aligned} {S}_{11}^{\prime }&={ S}_{11} \cos ^{2}\theta +{S}_{12} \sin 2\theta +{ S}_{22}\sin ^{2}\theta \end{aligned}$$
(2.139)
$$\begin{aligned} { S}_{12}^{\prime }&=\frac{1}{2} ({ S}_{22}-{ S}_{11}) \sin 2\theta +{ S}_{12}\ \cos 2\theta \end{aligned}$$
(2.140)
$$\begin{aligned} { S}_{22}^{\prime }&={ S}_{11}\ \sin ^{2}\theta -{ S}_{12}\ \sin 2\theta +{ S}_{22}\cos ^{2}\theta \end{aligned}$$
(2.141)

[see Fig. 2.1].

The components \({ S}_{11},\ { S}_{12}\), and \({ S}_{22}\) taken on the line \({ x}_{2}={ x_{1}}\tan \theta \) assume the forms

$$\begin{aligned} { S}_{11} ({ x_{1}, x_{1}}\tan \theta )&=({ d}\tan \theta +{ e}-\rho { g}){ x}_{1} \end{aligned}$$
(2.142)
$$\begin{aligned} { S}_{12} ({ x_{1}, x_{1}}\tan \theta )&=-({ e}\tan \theta ){ x}_{1} \end{aligned}$$
(2.143)
$$\begin{aligned} { S}_{22} ({ x_{1}, x_{1}}\tan \theta )&=-\gamma \,{ x}_{1} \end{aligned}$$
(2.144)

Therefore, substituting (2.142)–(2.144) into the RHS\(^{\prime }\) of (2.140) and (2.141), and equating the results to zero, we obtain the algebraic equations for the unknown constants \(e\) and \(d\), provided \(\gamma \) and \(\rho { g}\) are prescribed

$$\begin{aligned}&{ e}(\sin ^{2}\theta +\tan \theta \ \sin 2\theta )+{ d}\ \tan \theta \ \sin ^{2}\theta \nonumber \\&\quad =\gamma \ \cos ^{2}\theta +\rho { g}\ \sin ^{2}\theta \end{aligned}$$
(2.145)
$$\begin{aligned}&{e}(\sin 2\theta +2\tan \theta \ \cos 2\theta )+{ d}\ \tan \theta \ \sin 2\theta \nonumber \\&\quad =-\gamma \ \sin 2\theta +\rho { g}\ \sin 2\theta \end{aligned}$$
(2.146)

Dividing Eq. (2.145) by \(\sin ^{2}\theta \) and Eq. (2.146) by \(\sin 2\theta \) and introducing the notation

$$\begin{aligned} \tan \theta =u \end{aligned}$$
(2.147)

we obtain

$$\begin{aligned} 3{ e}+{ du}&=\frac{\gamma }{{ u}^{2}}+\rho { g}\nonumber \\ (2-{ u}^{2}){ e}+{ du}&=-\gamma +\rho { g} \end{aligned}$$
(2.148)

It is easy to check that a unique solution (\(e\), \(d\)) of Eqs. (2.148) takes the form (2.136) that is

$$\begin{aligned} { e}=\gamma /{ u}^{2},\quad { d}=\rho { g/u}-2\gamma /{ u}^{3} \end{aligned}$$
(2.149)

This completes proof of (ii).

Finally, when \({ x}_{1}={ x}_{1}^{0}={ \mathrm{{const}},\ S}_{11}\) and \({S}_{12}\) are represented by straight lines on the planes \(({ x_{2}, S_{11}})\) and \(({ x_{2}, S_{12}})\), respectively, and \({ S}_{22}\) at \({ x}_{2}=0\) is represented by a straight line passing through the origin 0 as shown in Fig. 2.1. This completes solution to Problem 2.9.

Problem 2.10.

Let B denote a cylinder of length \(l\) and of arbitrary cross section, suspended from the upper end and subject to its own weight \(\rho \mathrm{{g}}\). Then the stress tensor \(\mathbf{S }=\mathbf{S }(\mathbf{x })\) on B takes the form

$$\begin{aligned} \mathbf{S }=\;\left[ {\begin{array}{c@{\quad }c@{\quad }c} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \rho g {x}_3 \\ \end{array}} \right] \end{aligned}$$
(2.150)

since, in this case, the body force vector field is given by \(\mathbf{b }=[0,\ 0,\ -\rho \,{g}]^\mathrm{{T}}\), and \(\mathrm{{div}}\;\mathbf{S }+\mathbf{b }=\mathbf{0 }\ \mathrm{{on}}\ \mathrm{{B}}\). The stress vector s associated with S on \(\partial \mathrm{{B}}\) has the following properties: \(\mathbf{s }=[0,0,\rho \,{g}\,l\,]^\mathrm{{T}}\) on the end plane \({x}_3 =l\); and \(\mathbf{s }=\mathbf{0 }\) on the plane \({x}_3 =0\) and on the lateral surface of the cylinder since \(\mathbf{n }=[{n}_1, {n}_2, 0]^\mathrm{{T}}\) on the surface. Assuming that the cylinder is made of a homogeneous isotropic elastic material, the associated strain tensor field E takes the form [see Eqs. (2.49)]

$$\begin{aligned} \mathbf{E }=\frac{\rho \,{g}\;{x}_3 }{{E}}\;\left[ {\begin{array}{cccl} -\nu &{} 0 &{} 0 \\ 0 &{} -\nu &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array}} \right] \end{aligned}$$
(2.151)

where \(E\) and \(\nu \) are Young’s modulus and Poisson’s ratio, respectively.

  1. (i)

    Show that a solution u of the equation

    $$\begin{aligned} \mathbf{E }={\widehat{\nabla }}\mathbf{u } \quad \mathrm{{on}} \quad \mathrm{{B}} \end{aligned}$$
    (2.152)

    subject to the condition

    $$\begin{aligned} \mathbf{u }(0,0,l)=\mathbf{0 } \end{aligned}$$
    (2.153)

    takes the form

    $$\begin{aligned} \mathbf{u }=\frac{\rho \,g}{E} \left[ {-\nu {x}_1 {x}_3, \;\;-\nu {x}_2 {x}_3 ,\;\;\frac{\nu }{2}({x}_1^2 +\;{x}_2^2 )+\frac{1}{2}({x}_3^2 -l^2)} \right] ^\mathrm{{T}} \end{aligned}$$
    (2.154)
  2. (ii)

    Plot \({u}_3 ={u}_3 (0,0,{x}_3 )\) over the range \(0\le {x}_3 \le l\).

Fig. 2.2
figure 2figure 2

The cylinder of arbitrary cross section

Full size image

Solution.

To solve the problem we use (2.154) and obtain

$$\begin{aligned} \varvec{\nabla } \mathbf{u }=\frac{\rho { g}}{{ E}}\left[ \begin{matrix} -{ \nu x}_{3} &{} 0 &{}-{ \nu x}_{1}\\ 0 &{}-{ \nu x}_{3} &{}-{ \nu x}_{2}\\ { \nu x}_{1} &{}{ \nu x}_{2} &{}{x}_{3}\\ \end{matrix}\right] \end{aligned}$$
(2.155)

and

$$\begin{aligned} \varvec{\nabla } \mathbf{u }^\mathrm{{T}}=\frac{\rho { g}}{{ E}}\left[ \begin{array}{ccc} -{ \nu x}_{3} &{} 0 &{}{ \nu x}_{1}\\ 0 &{}-{ \nu x}_{3} &{}{ \nu x}_{2}\\ -{ \nu x}_{1} &{}-{ \nu x}_{2} &{}{x}_{3}\\ \end{array}\right] \end{aligned}$$
(2.156)

Hence

$$\begin{aligned} \widehat{\varvec{\nabla }}\mathbf{u }=\frac{\rho { g}}{{ E}}\left[ \begin{array}{ccc} -{ \nu x}_{3} &{} 0 &{} 0\\ 0 &{}-{ \nu x}_{3} &{} 0\\ 0 &{} 0 &{}{x}_{3}\\ \end{array}\right] \end{aligned}$$
(2.157)

Therefore, u given by (2.154) satisfies (2.152). Also, it is easy to prove that u satisfies (2.153). Finally, \({u}_{3}={ u}_{3}(0, 0, {x}_{3})\) is represented by a parabolic curve restricted to the interval \(0\le { x}_{3}\le \ell \). This completes solution to Problem 2.10.

Problem 2.11.

For a transversely isotropic elastic body each material point posses an axis of rotational symmetry, which means that the elastic properties are the same in any direction on any plane perpendicular to the axis, but they are different than those in the direction of the axis. If the \(x_3\) axis coincides with the axis of symmetry, then the stress-strain relation for such a body takes the form

$$\begin{aligned} \left[ {\begin{array}{l} {S}_{11} \\ {S}_{22} \\ {S}_{33} \\ {S}_{32} \\ {S}_{31} \\ {S}_{12} \\ \end{array}} \right] =\left[ {\begin{array}{cccccc} {c}_{11}&{} {c}_{12} &{} {c}_{13} &{} 0 &{} 0 &{}0 \\ {c}_{12}&{} {c}_{11} &{} {c}_{13} &{} 0 &{} 0 &{} 0 \\ {c}_{13} &{} {c}_{13} &{} {c}_{33} &{} 0 &{} 0 &{}0\\ 0&{}0&{}0&{}{c}_{44}&{}0&{}0 \\ 0 &{} 0 &{} 0 &{} 0 &{} {c}_{44} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} ({c}_{11} -{c}_{12} )/2 \\ \end{array}} \right] \;\left[ {\begin{array}{l} {E}_{11} \\ {E}_{22} \\ {E}_{33} \\ 2{E}_{32} \\ 2{E}_{31} \\ 2{E}_{12} \\ \end{array}} \right] \end{aligned}$$
(2.158)

where S and E are the stress and strain tensors, respectively, and five numerically independent moduli \({c}_{11}, {c}_{33}, {c}_{12}, {c}_{13},\ \mathrm{{and}}\ {c}_{44}\) are related to the components \({C}_{ijkl}\) of the fourth-order elasticity tensor \({\mathbf{\mathsf{{C}} }}\) by [see Eq. (2.35)]

$$\begin{aligned} {c}_{11} ={C}_{1111},\quad {c}_{12} ={C}_{1122},\quad {c}_{13} ={C}_{1133},\quad {c}_{33} ={C}_{3333},\quad {c}_{44} ={C}_{1313} \end{aligned}$$
(2.159)

Show that if the axis of symmetry of a transversely isotropic body coincides with the direction of an arbitrary unit vector e, then the stress-strain relation takes the form

$$\begin{aligned} \mathbf{S } ={{\mathbf{\mathsf{{C}} }}}\,[\mathbf{E }]&=({c}_{11} -{c}_{12} )\mathbf{E }+\{{c}_{12} (\mathrm{{tr}}\,\mathbf{E })-({c}_{12} -{c}_{13} )\mathbf{e }\cdot (\mathbf{Ee })\}\,\mathbf{1 } \nonumber \\&\quad \; -({c}_{11} -{c}_{12} -2c_{44} )\{\mathbf{e }\otimes (\mathbf{Ee })+(\mathbf{Ee })\otimes \mathbf{e }\}\\&\quad \; -\{({c}_{12} -{c}_{13} )(\mathrm{{tr}}\,\mathbf{E })-({c}_{11} +{c}_{33} -2{c}_{13} -4c_{44} )\mathbf{e }\cdot (\mathbf{Ee })\} \mathbf{e }\otimes \mathbf{e }\nonumber \end{aligned}$$
(2.160)

Solution.

For a transversely isotropic body in which the axis of symmetry coincides with an arbitrary unit vector e, the stress–strain relation takes the formFootnote 1

$$\begin{aligned} \mathbf{S }={{\mathbf{\mathsf{{C}} }}} [\mathbf{E }] \end{aligned}$$
(2.161)

where S and E are the stress and strain tensors, respectively, and the elasticity tensor \({\mathbf{\mathsf{{C}} }}\) is given by

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}&={ c}_{33} {\mathbf{\mathsf{{C}} }}^{(1)}+({ c}_{11}+{ c}_{12}) {\mathbf{\mathsf{{C}} }}^{(2)} +\sqrt{2}\ {c}_{13} ({\mathbf{\mathsf{{C}} }}^{(3)}+{\mathbf{\mathsf{{C}} }}^{(4)})\nonumber \\&\quad +({c}_{11}-{ c}_{12}) {\mathbf{\mathsf{{C}} }}^{(5)}+2\ { c}_{44} {\mathbf{\mathsf{{C}} }}^{(6)} \end{aligned}$$
(2.162)

In Eq. (2.162) the tensors \({\mathbf{\mathsf{{C}} }}^{({ a})},\ { a}=1, 2, 3, 4, 5, 6\), are defined by

$$\begin{aligned} \begin{array}{l} {C}_{ ijkl}^{(1)}={ A_{ij}\ A_{kl}},\quad { C}_{ ijkl}^{(2)}=\frac{1}{2}\ { B_{ij}\ B_{kl}}\\ [7pt] {C}_{ ijkl}^{(3)}=\frac{1}{\sqrt{2}}\ { A_{ij}\ B_{kl}},\quad {C}_{ ijkl}^{(4)}=\frac{1}{\sqrt{2}}\ { B_{ij}\ A_{kl}}\\ [7pt] {C}_{ ijkl}^{(5)}=\frac{1}{2} ({ B_{ik}\ B_{jl}}+{ B_{il}\ B_{jk}}-{ B_{ij}\ B_{kl}})\\ [7pt] {C}_{ ijkl}^{(6)}=\frac{1}{2} ({ A_{ik}\ B_{jl}}+{ A_{il}\ B_{jk}}+{ A_{jk}\ B_{il}}+{ A_{jl}\ B_{ik}}) \end{array} \end{aligned}$$
(2.163)

where

$$\begin{aligned} { A_{ij}}={ e_i\ e_j},\quad { B_{ij}}=\delta _{ij}-{ e_i\ e_j} \end{aligned}$$
(2.164)

and \({ e_{i}}\) are the components of e in the coordinates \(\{{ x_{i}}\}\).

Using (2.163) and (2.164), we obtain

$$\begin{aligned} { C_{ijkl}^{(1)}\ E_{kl}}={ e_i\ e_j\ e_k\ e_l\ E_{kl}} \end{aligned}$$
(2.165)

or in direct notation

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{(1)}[\mathbf{E }]=[\mathbf{e }\cdot (\mathbf{Ee })] \mathbf{e }\otimes \mathbf{e } \end{aligned}$$
(2.166)

Similarly, by (2.163) and (2.164), we get

$$\begin{aligned} { C_{ijkl}^{(2)}\ E_{kl}}=\frac{1}{2}(\delta _{ ij}-{ e_i\ e_j})(\delta _{ kl}-{ e_k\ e_l}){ E_{kl}} \end{aligned}$$
(2.167)

or

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{(2)}[\mathbf{E }]=\frac{1}{2}(\mathbf{1 }-\mathbf{e }\otimes \mathbf{e })[\mathrm{{tr}}\ \mathbf{E }-\mathbf{e }\cdot (\mathbf{Ee })] \end{aligned}$$
(2.168)

Also, using (2.163) and (2.164), we obtain

$$\begin{aligned} \sqrt{2} \bigg ({ C_{ijkl}^{(3)}}+{ C_{ijkl}^{(4)}}\bigg ){E_{kl}} ={ e_i\ e_j}(\delta _{ kl}-{ e_k\ e_l}){ E_{kl}}+(\delta _{ ij}-{ e_i\ e_j}){ e_k\ e_l\ E_{kl}} \end{aligned}$$
(2.169)

or

$$\begin{aligned} \sqrt{2} ({\mathbf{\mathsf{{C}} }}^{(3)}[\mathbf{E }]+{\mathbf{\mathsf{{C}} }}^{(4)}[\mathbf{E }]) =[\mathbf{e }\cdot (\mathbf{Ee })]\mathbf{1 }+[{\mathrm{{tr}}}\ \mathbf{E }-2\mathbf{e }\cdot (\mathbf{Ee })]\mathbf{e }\otimes \mathbf{e } \end{aligned}$$
(2.170)

and

$$\begin{aligned} { C_{ijkl}^{(5)}\ E_{kl}}&={ E_{ij}}-{ e_i\ e_k\ E_{jk}}-{ e_j\ e_k\ E_{ ik}}+{ e_i\ e_j\ e_k\ e_l\ E_{kl}} \nonumber \\&\quad -\frac{1}{2}(\delta _{ ij}-{ e_i\ e_j})(E_{ kk}-{ e_k\ e_l\ E_{kl}}) \end{aligned}$$
(2.171)

or

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{(5)}[\mathbf{E }]&=\mathbf{E }-\mathbf{e }\otimes (\mathbf{Ee })-(\mathbf{Ee })\otimes \mathbf{e } +\mathbf{e }\otimes \mathbf{e }(\mathbf{e }\cdot \mathbf{Ee })\nonumber \\&\quad -\frac{1}{2}(\mathbf{1 }-\mathbf{e }\otimes \mathbf{e })({ \mathrm{{tr}}}\ \mathbf{E }-\mathbf{e }\cdot \mathbf{Ee }) \end{aligned}$$
(2.172)

and

$$\begin{aligned} { C_{ijkl}^{(6)}}\ { E_{kl}}={ e_i\ E_{jk}\,e_k}+{ E_{ik}\ e_k\ e_j}-2{ e_i\ e_j\ e_k\ e_l\ E_{kl}} \end{aligned}$$
(2.173)

or

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{(6)}[\mathbf{E }]=\mathbf{e }\otimes \mathbf{Ee }+(\mathbf{Ee })\otimes \mathbf{e } -2\mathbf{e }\otimes \mathbf{e }[\mathbf{e }\cdot (\mathbf{Ee })] \end{aligned}$$
(2.174)

Therefore, substituting \({\mathbf{\mathsf{{C}} }}\) from (2.162) into (2.161) and using (2.166), (2.168), (2.170), (2.172), and (2.174), we obtain (2.160). Note that the representation (2.160) coincides with (2.158) if \(\mathbf{e }=(0, 0, 1)\). This can be proved by substituting \(\mathbf{e }=(0, 0, 1)\) into (2.160).

Problem 2.12.

Show that the stress-strain relation (2.160) in Problem 2.11 is invertible provided

$$\begin{aligned} {c}\equiv ({c}_{11} +{c}_{12} ){c}_{33} -2{c}_{13}^2 >0,\quad {c}_{11} >\left| {{c}_{12} } \right| ,\quad {c}_{44} >0 \end{aligned}$$
(2.175)

and that the strain-stress relation reads

$$\begin{aligned} \mathbf{E }&={{\mathbf{\mathsf{{K}} }}}[\mathbf{S }]=({c}_{11} -{c}_{12})^{-1}\mathbf{S }+\frac{1}{2}\left[ \left\{ {c}^{-1}{ c}_{33} -({c}_{11} -{c}_{12})^{-1}\right\} (\mathrm{{tr}}\,\mathbf{S })\right. \nonumber \\&\left. \qquad \qquad \quad \;\; -\left\{ {c}^{-1}({c}_{33} +2{c}_{13} )-({c}_{11}-{c}_{12})^{-1}\right\} \mathbf{e }\cdot (\mathbf{Se })\right] \,\mathbf{1 }\nonumber \\&\qquad \qquad \quad \;\; -\left\{ ({c}_{11} -{c}_{12} )^{-1}-\frac{1}{2}{c}_{44}^{-1} \right\} \{\mathbf{e }\otimes (\mathbf{Se })+(\mathbf{Se })\otimes \mathbf{e }\}\\&\qquad \qquad \quad \;\; -\frac{1}{2}\left[ \{{c}^{-1}({c}_{33} +2{c}_{13} )-({c}_{11} -{c}_{12})^{-1}\}\,(\mathrm{{tr}}\,\mathbf{S }) \right. \nonumber \\&\qquad \qquad \quad \;\; \left. -\{{c}^{-1}(2\,{c}_{11} +{c}_{33} +2{c}_{12} +4c_{13})+({c}_{11} -{c}_{12} )^{-1}-2{c}_{44}^{-1} \}\;\mathbf{e }\cdot (\mathbf{Se })\right] \;\mathbf{e }\otimes \mathbf{e }\nonumber \end{aligned}$$
(2.176)

Solution.

To show that (2.160) in Problem 2.11 is invertible if the inequalities (2.175) are satisfied, and the inverted formula takes the form (2.176), consider the fourth-order tensor

$$\begin{aligned} {\mathbf{\mathsf{{A}} }}={ a}_1{\mathbf{\mathsf{{C}} }}^{(1)}+{ a}_{2}{\mathbf{\mathsf{{C}} }}^{(2)}+{ a}_{3}({\mathbf{\mathsf{{C}} }}^{(3)}+{\mathbf{\mathsf{{C}} }}^{(4)}) +{ a}_{5}{\mathbf{\mathsf{{C}} }}^{(5)}+{ a}_{6}{\mathbf{\mathsf{{C}} }}^{(6)} \end{aligned}$$
(2.177)

where \({ a_{1},\ a_{2},\ a_{3},\ a_{5}}\), and \({ a_{6}}\) are scalars that satisfy the inequalities

$$\begin{aligned} { a}_1>0,\quad { a}_2>0,\quad { a_1a_2}-{ a}_3^2>0,\quad {a}_5>0,\quad { a}_6>0 \end{aligned}$$
(2.178)

and \({\mathbf{\mathsf{{C}} }}^{({ a})}\ ({ a}=1, 2, 3, 4, 5, 6)\) are the fourth-order tensors defined by (2.163) and (2.164) in Problem 2.11.

Then, there is \({\mathbf{\mathsf{{A}} }}^{-1}\) in the form

$$\begin{aligned} {\mathbf{\mathsf{{A}} }}^{-1}=\left( { a_1a_2}-{ a}_3^2\right) ^{-1}\{{{ a}_{2}{\mathbf{\mathsf{{C}} }}^{(1)}+{ a}_{1}{\mathbf{\mathsf{{C}} }}^{(2)}-{ a}_{3}({\mathbf{\mathsf{{C}} }}^{(3)}+{\mathbf{\mathsf{{C}} }}^{(4)})}\} +{ a}_{5}^{-1}{\mathbf{\mathsf{{C}} }}^{(5)}+{ a}_{6}^{-1}{\mathbf{\mathsf{{C}} }}^{(6)} \end{aligned}$$
(2.179)

such that

$$\begin{aligned} {\mathbf{\mathsf{{A}} }} {\mathbf{\mathsf{{A}} }}^{-1}={\mathbf{\mathsf{{A}} }}^{-1} {\mathbf{\mathsf{{A}} }}={\mathbf{\mathsf{{1}} }} \end{aligned}$$
(2.180)

where \({\mathbf{\mathsf{{1}} }}\) is the fourth-order identity tensor with components

$$\begin{aligned} { I_{ijkl}}=\frac{1}{2}(\delta _{ ik}\, \delta _{ jl}+\delta _{ jk}\, \delta _{ il}) \end{aligned}$$
(2.181)

To prove (2.180) we use (2.163) and (2.164) of Problem 2.11 to obtain the \(6\times 6\) tensor matrix

$$\begin{aligned}{}[{\mathbf{\mathsf{{C}} }}^{({ a})}\ {\mathbf{\mathsf{{C}} }}^{({ b})}]=\left[ \begin{array}{c@{}c@{}c@{}c@{}c@{}c}{\mathbf{\mathsf{{C}} }}^{(1)} &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(3)} &{} \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 }\\ \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(2)} &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(4)} &{} \mathbf{0 } &{} \mathbf{0 }\\ \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(3)} &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(1)} &{} \mathbf{0 } &{} \mathbf{0 }\\ {\mathbf{\mathsf{{C}} }}^{(4)} &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(2)} &{} \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 }\\ \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(5)} &{} \mathbf{0 }\\ \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 } &{} \mathbf{0 } &{} {\mathbf{\mathsf{{C}} }}^{(6)}\end{array}\right] \end{aligned}$$
(2.182)

as well as the identity

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{(1)}+{\mathbf{\mathsf{{C}} }}^{(2)}+{\mathbf{\mathsf{{C}} }}^{(5)}+{\mathbf{\mathsf{{C}} }}^{(6)}={\mathbf{\mathsf{{1}} }} \end{aligned}$$
(2.183)

Calculating the tensor \({\mathbf{\mathsf{{A}} }}\ {\mathbf{\mathsf{{A}} }}^{-1}\), by using Eqs. (2.177), (2.179), (2.182), and (2.183) we obtain

$$\begin{aligned} {\mathbf{\mathsf{{A}} }}\ {\mathbf{\mathsf{{A}} }}^{-1}={\mathbf{\mathsf{{1}} }} \end{aligned}$$
(2.184)

Similarly, using Eqs. (2.177), (2.179), (2.182), and (2.183) we get

$$\begin{aligned} {\mathbf{\mathsf{{A}} }}^{-1}\ {\mathbf{\mathsf{{A}} }}={\mathbf{\mathsf{{1}} }} \end{aligned}$$
(2.185)

Now, by letting

$$\begin{aligned} \begin{array}{l} { a}_{1}={ c}_{33},\quad { a}_{2}={ c}_{11}+{ c}_{12},\quad { a}_{3}=\sqrt{2}\ { c}_{13}\\ [5pt] { a}_5={ c}_{11}-{ c}_{12},\quad { a}_{6}=2{ c}_{44} \end{array} \end{aligned}$$
(2.186)

in Eq. (2.177) we obtain \({\mathbf{\mathsf{{A}} }}={\mathbf{\mathsf{{C}} }}\), and the inequalities (2.178) reduce to those of (2.175). Also, Eq. (2.179) reduces to

$$\begin{aligned} {\mathbf{\mathsf{{C}} }}^{-1}={\mathbf{\mathsf{{K}} }}&={ c}^{-1}\{({ c}_{11}+{ c}_{12}) {\mathbf{\mathsf{{C}} }}^{(1)}+{ c}_{33}{\mathbf{\mathsf{{C}} }}^{(2)} -\sqrt{2}\ { c}_{13} ({\mathbf{\mathsf{{C}} }}^{(3)}+{\mathbf{\mathsf{{C}} }}^{(4)})\}\nonumber \\&\quad \; +({ c}_{11}-{ c}_{12})^{-1}{\mathbf{\mathsf{{C}} }}^{(5)}+2^{-1}{c}_{44}^{-1}{\mathbf{\mathsf{{C}} }}^{(6)} \end{aligned}$$
(2.187)

where

$$\begin{aligned} { c}=({ c}_{11}+{ c}_{12}){ c}_{33}-2{ c}_{13}^{2}>0 \end{aligned}$$
(2.188)

Therefore, the strain–stress relation reads

$$\begin{aligned} \mathbf{E }={\mathbf{\mathsf{{K}} }}[\mathbf{S }]&={ c}^{-1}\{({ c}_{11}+{ c}_{12}){\mathbf{\mathsf{{C}} }}^{(1)}[\mathbf{S }]+{ c}_{33}{\mathbf{\mathsf{{C}} }}^{(2)}[\mathbf{S }] -\sqrt{2}\ { c}_{13}({\mathbf{\mathsf{{C}} }}^{(3)}[\mathbf{S }]+{\mathbf{\mathsf{{C}} }}^{(4)}[\mathbf{S }])\}\nonumber \\&\quad \;+({ c}_{11}-{ c}_{12})^{-1}{\mathbf{\mathsf{{C}} }}^{(5)}[\mathbf{S }] + 2^{-1}{ c}_{44}^{-1}{\mathbf{\mathsf{{C}} }}^{(6)}[\mathbf{S }] \end{aligned}$$
(2.189)

Finally, if Eqs. (2.166), (2.168), (2.170), (2.172), and (2.174) of Problem 2.11 in which E is replaced by S are taken into account, Eq. (2.189) reduces to (2.176).

Problem 2.13.

Prove that the inequalities (2.175) in Problem 2.12 are necessary and sufficient conditions for the elasticity tensor \({\mathbf{\mathsf{{C}} }}\) (compliance tensor \({\mathbf{\mathsf{{K}} }})\) to be positive definite. This means that the strain energy density (stress energy density) of a transversely isotropic body is positive definite if and only if the inequalities (2.175) in Problem 2.12 hold true.

Solution.

Define the fourth-order tensor \({\mathbf{\mathsf{{H}} }}\) by

$$\begin{aligned} {\mathbf{\mathsf{{H}} }}&=\sqrt{{ a}_{1}} { a}_1^{\prime }\left( { a}_{1}^{\prime \ 2}+{ a}_{3}^{2}\right) ^{-1/2}{\mathbf{\mathsf{{C}} }}^{(1)} +\sqrt{{ a}_{2}} { a}_2^{\prime }\left( { a}_{2}^{\prime \ 2}+{ a}_{3}^{2}\right) ^{-1/2}{\mathbf{\mathsf{{C}} }}^{(2)} \nonumber \\&\quad \; +{ a}_{3}({ a}_1^{\prime }+{ a}_{2}^{\prime })^{-1/2} ({\mathbf{\mathsf{{C}} }}^{(3)}+{\mathbf{\mathsf{{C}} }}^{(4)}) +\sqrt{{ a}_5}\ {\mathbf{\mathsf{{C}} }}^{(5)}+\sqrt{{ a}_6}\ {\mathbf{\mathsf{{C}} }}^{(6)} \end{aligned}$$
(2.190)

where

$$\begin{aligned} { a}_{1}^{\prime }={ a}_{1}+({ a_1\ a_2}-{ a}_3)^{1/2},\quad { a}_2^\prime ={ a}_{2}+({ a_1\ a_2}-{ a_3})^{1/2} \end{aligned}$$
(2.191)

and \({ a_{1},\ a_{2},\ a_{3},\ a_{5}}\), and \({ a_{6}}\) satisfy the inequalities (2.178) of Problem 2.12.

Then using the matrix equation (2.182) of Problem 2.12 we obtain

$$\begin{aligned} {\mathbf{\mathsf{{H\ H}} }}={\mathbf{\mathsf{{A}} }} \end{aligned}$$
(2.192)

where \({\mathbf{\mathsf{{A}} }}\) is the fourth-order tensor given by (2.177) of Problem 2.12. Hence, we get

$$\begin{aligned} {\mathbf{\mathsf{{A}} }}[\mathbf{E }]={\mathbf{\mathsf{{H}} }}({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }])\quad \forall \ \mathbf{E }=\mathbf{E }^\mathrm{{T}}\ne \mathbf{0 } \end{aligned}$$
(2.193)

and

$$\begin{aligned} \mathbf{E }\cdot {\mathbf{\mathsf{{A}} }}[\mathbf{E }]=\mathbf{E }\cdot {\mathbf{\mathsf{{H}} }}({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }]) \end{aligned}$$
(2.194)

or

$$\begin{aligned} \mathbf{E }\cdot {\mathbf{\mathsf{{A}} }}[\mathbf{E }]=({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }])\cdot ({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }]) \end{aligned}$$
(2.195)

Since

$$\begin{aligned} ({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }])\cdot ({\mathbf{\mathsf{{H}} }}\,[\mathbf{E }])>0\quad \forall \ \mathbf{E }=\mathbf{E }^\mathrm{{T}}\ne \mathbf{0 } \end{aligned}$$
(2.196)

therefore, expressing \({ a_{1},a_{2},a_{3},a_{5}}\), and \({ a}_{6}\) in terms of \({ c_{11}, c_{12}, c_{13}, c_{33}}\), and \({ c}_{44}\) [see Eqs. (2.186) of Problem 2.12], we conclude that if (2.175) of Problem 2.12 is satisfied then

$$\begin{aligned} \mathbf{E }\cdot {\mathbf{\mathsf{{C}} }}\,[\mathbf{E }]>0\quad \forall \ \mathbf{E }=\mathbf{E }^\mathrm{{T}}\ne \mathbf{0 } \end{aligned}$$
(2.197)

that is, the elasticity tensor \({\mathbf{\mathsf{{C}} }}\) is positive definite.

To prove that (2.197) implies (2.175) of Problem 2.12, that is, that (2.175) of Problem 2.12 are also necessary conditions, we take advantage of the fact that E in (2.197) is an arbitrary second-order symmetric tensor, and select the following choices

$$\begin{aligned} \mathbf{E }_{1}&=\alpha \mathbf{e }\otimes \mathbf{e }+\frac{\beta }{\sqrt{2}}(\mathbf{1 }-\mathbf{e }\otimes \mathbf{e }) \end{aligned}$$
(2.198)
$$\begin{aligned} \mathbf{E }_{2}&=\mathbf{e }\otimes \mathbf{c }+\mathbf{c }\otimes \mathbf{e } \end{aligned}$$
(2.199)
$$\begin{aligned} \mathbf{E }_{3}&=\mathbf{c }\otimes (\mathbf{e }\times \mathbf{c })+(\mathbf{e }\times \mathbf{c })\otimes \mathbf{c } \end{aligned}$$
(2.200)

where \(\alpha \ \mathrm{{and}}\ \beta \) are the real numbers, and c is an arbitrary unit vector orthogonal to e.

Then, substituting (2.198), (2.199), and (2.200) into (2.197), respectively, we obtain

$$\begin{aligned} \mathbf{E }_{1}\cdot {\mathbf{\mathsf{{C}} }}\,[\mathbf{E }_{1}]&={ c}_{33}\alpha ^{2}+2\sqrt{2}\ { c}_{13}\,\alpha \beta + ({ c}_{11}+{ c}_{12})\beta ^{2} \end{aligned}$$
(2.201)
$$\begin{aligned} \mathbf{E }_{2}\cdot {\mathbf{\mathsf{{C}} }}\,[\mathbf{E }_{2}]&=4 { c}_{44} \end{aligned}$$
(2.202)
$$\begin{aligned} \mathbf{E }_{3}\cdot {\mathbf{\mathsf{{C}} }}\,[\mathbf{E }_{3}]&=2\ ({ c}_{11}-{ c}_{12}) \end{aligned}$$
(2.203)

Since the RHS of (2.201) is positive for non-vanishing numbers \(\alpha \) and \(\beta \), we obtain

$$\begin{aligned} \Delta =8 { c}_{13}^{2}\,\alpha ^{2}-4({ c}_{11}+{ c}_{12}){ c}_{33}\alpha ^{2}<0 \end{aligned}$$
(2.204)

or

$$\begin{aligned} { c}=({ c}_{11}+{ c}_{12}){ c}_{33}-2 { c}_{13}^{2}>0 \end{aligned}$$
(2.205)

and

$$\begin{aligned} { c}_{11}+{ c}_{12}>0,\ \quad { c}_{33}>0 \end{aligned}$$
(2.206)

Also, Eqs. (2.202) and (2.203) together with the positiveness of \({\mathbf{\mathsf{{C}} }}\) imply that

$$\begin{aligned} { c}_{44}>0,\quad { c}_{11}-{ c}_{12}>0 \end{aligned}$$
(2.207)

Since the inequalities (2.205)–(2.207) are equivalent to the inequalities (2.175) of Problem 2.12, the solution to Problem 2.13 is complete.

Problem 2.14.

Consider a plane \(\mathbf{n }\cdot \mathbf{x }-\mathrm{{vt}}=0\quad (\left| \mathbf{n } \right| =1,\;\mathrm{{v}}>0,\;\mathrm{{t}}\ge 0)\). Let S be the stress tensor obtained form Eq. (2.160) of Problem 2.11 in which \(0<\mathbf{e }\cdot \mathbf{n }<1\), and the strain tensor E is defined by

$$\begin{aligned} \mathbf{E }=\mathrm{{sym}}\;(\mathbf{n }\otimes \mathbf{a }) \end{aligned}$$
(2.208)

where a is an arbitrary vector orthogonal to n. Let \(\mathbf{S }^\bot \;\mathrm{{and}}\;\;\mathbf{S }^{\vert \vert }\) represent the normal and tangential parts of S with respect to the plane [see Problem 1.4 in which T is replaced by S]. Show that

$$\begin{aligned} \mathbf{S }&=({c}_{11} -{c}_{12} )\,\mathrm{{sym}}\,(\mathbf{n }\otimes \mathbf{a })-({c}_{11} -{c}_{12} -2{c}_{44} )\,\cos \theta \;\mathrm{{sym}}\,(\mathbf{e }\otimes \mathbf{a })\end{aligned}$$
(2.209)
$$\begin{aligned} \mathbf{S }^\bot&=[({c}_{11} -{c}_{12} )\,\sin ^2\theta +2{c}_{44} \cos ^2\theta \,]\,\mathrm{{sym}}\;(\mathbf{n }\otimes \mathbf{a })\end{aligned}$$
(2.210)
$$\begin{aligned} \mathbf{S }^{\vert \vert }&=({c}_{11} -{c}_{12} -2{c}_{44} )\,\cos \theta \;\mathrm{{sym}}\,[(\mathbf{n }\,\cos \theta -\mathbf{e })\otimes \mathbf{a }] \end{aligned}$$
(2.211)

where

$$\begin{aligned} \cos \theta =\mathbf{e }\cdot \mathbf{n },\quad \quad 0<\theta <\pi \,/2 \end{aligned}$$
(2.212)

Solution.

If we let \(\mathbf{E }=\mathrm{{sym}}(\mathbf{n }\otimes \mathbf{a })\) into Eq. (2.160) of Problem 2.11, we obtain (2.209). To obtain (2.210) and (2.211) we use the formulas:

$$\begin{aligned} \mathbf{S }^{\bot }=2\,\mathrm{{sym}}(\mathbf{n }\otimes \mathbf{Sn })-(\mathbf{n }\cdot \mathbf{Sn })\mathbf{n }\otimes \mathbf{n } \end{aligned}$$
(2.213)

and

$$\begin{aligned} \mathbf{S }^{\Vert }=(\mathbf{1 }-\mathbf{n }\otimes \mathbf{n })[\mathbf{S }(\mathbf{1 }-\mathbf{n }\otimes \mathbf{n })] \end{aligned}$$
(2.214)

By substituting \(\mathbf{S }\) from Eq. (2.209) into Eqs. (2.213) and (2.214), we obtain (2.210) and (2.211), respectively. This completes solution to Problem 2.14.

Problem 2.15.

Show that for a transversely isotropic elastic body the stress energy density

$$\begin{aligned} {\widehat{\mathrm{{W}}}}(\mathbf{S })=\frac{1}{2}\mathbf{S }\cdot {\mathbf{\mathsf{{K}} }} [\mathbf{S }] \end{aligned}$$
(2.215)

corresponding to the stress tensor given by Eq. (2.209) of Problem 2.14 takes the form

$$\begin{aligned} {\widehat{\mathrm{{W}}}}(\mathbf{S })=\frac{1}{2}\mathbf{a }^2\,\left[ {\frac{1}{2}({c}_{11} -{c}_{12} )\,\sin ^2\theta +{c}_{44} \cos ^2\theta } \right] \end{aligned}$$
(2.216)

Also, show that

$$\begin{aligned} {\widehat{\mathrm{{W}}}} (\mathbf{S })={\widehat{\mathrm{{W}}}}(\mathbf{S }^\bot )-{\widehat{\mathrm{{W}}}}(\mathbf{S }^{\vert \vert }) \end{aligned}$$
(2.217)

where \({\widehat{\mathrm{{W}}}}(\mathbf{S }^\bot )\) and \({\widehat{\mathrm{{W}}}}(\mathbf{S }^{\vert \vert })\) represent the “normal” and “tangential” stress energies, respectively, given by

$$\begin{aligned} {\widehat{\mathrm{{W}}}}(\mathbf{S }^\bot )=\frac{1}{2}\mathbf{S }^\bot \cdot {{\mathbf{\mathsf{{K}} }}}[\mathbf{S }^\bot ]={\widehat{\mathrm{{W}}}}(\mathbf{S })\left[ {1+\frac{1}{8}{c}_{44}^{-1} ({c}_{11} -{c}_{12} )^{-1}({c}_{11} -{c}_{12} -2{c}_{44} )^2\sin ^22\theta } \right] \end{aligned}$$
(2.218)

and

$$\begin{aligned} {\widehat{\mathrm{{W}}}}(\mathbf{S }^{\vert \vert })=\frac{1}{2}\mathbf{S }^{\vert \vert }\cdot {{\mathbf{\mathsf{{K}} }}}[\mathbf{S }^{\vert \vert }]={\widehat{\mathrm{{W}}}}(\mathbf{S })\left[ {\frac{1}{8}{c}_{44}^{-1} ({c}_{11} -{c}_{12} )^{-1}({c}_{11} -{c}_{12} -2{c}_{44} )^2\sin ^22\theta } \right] \end{aligned}$$
(2.219)

Here, \(\mathbf{S }^\bot \;\mathrm{{and}}\;\;\mathbf{S }^{\vert \vert }\) are given by Eqs. (2.210) and (2.211), respectively, of Problem 2.14.

Solution.

If we substitute S from Eq. (2.209) of Problem 2.14 into Eq. (2.215) we arrive at (2.216). Next, using Eqs. (2.208) and (2.210) of Problem 2.14 we obtain (2.218); and using Eqs. (2.208) and (2.211) of Problem 2.14, we get Eq. (2.219). Finally, subtracting (2.219) from (2.218) we obtain Eq. (2.217). This completes solution to Problem 2.15.

Problem 2.16.

Let \(\varphi (\theta )={\widehat{\mathrm{{W}}}}(\mathbf{S }^{\vert \vert })/{\widehat{\mathrm{{W}}}}(\mathbf{S }^\bot )\), where \({\widehat{\mathrm{{W}}}}(\mathbf{S }^\bot )\) and \({\widehat{\mathrm{{W}}}}(\mathbf{S }^{\vert \vert })\) denote the normal and tangential stress energy densities, respectively, of Problem 2.15. Show that

$$\begin{aligned} {\mathop {\max }\limits _{\theta \,\in \,[\,0,\,\pi \,/2\,]}} \;[\varphi (\theta )]=\varphi (\pi / 4)=\frac{\mathrm{{A}}^2}{1+\mathrm{{A}}^2} \end{aligned}$$
(2.220)

where

$$\begin{aligned} \mathrm{{A}}=\frac{1}{2\sqrt{2}}\frac{\left| {{c}_{11} -{c}_{12} -2{c}_{44} } \right| }{({c}_{11} -{c}_{12} )^{1/2}{c}_{44}^{1/2} } \end{aligned}$$
(2.221)

Note. When the body is isotropic we have

$$\begin{aligned} {c}_{11} ={c}_{33} =\lambda +2\mu , \quad {c}_{12} ={c}_{13} =\lambda , \quad {c}_{44} =\mu \end{aligned}$$
(2.222)

where \(\lambda \) and \(\mu \) are the Lamé material constants. In this case Eq. (2.221) reduces to \(\mathrm{{A}}=0\), which means that for an isotropic body the tangential stress energy corresponding to the stress (2.211) of Problem 2.14 vanishes.

Solution.

Note that, by using Eqs. (2.218) and (2.219) of Problem 2.15, we obtain

$$\begin{aligned} \varphi (\theta )=\widehat{ W}(\mathbf{S }^{\Vert })/\widehat{ W}(\mathbf{S }^{\bot }) =\frac{{ A}^{2}\sin ^{2}2\theta }{(1+{ A}^{2}\sin ^{2}2\theta )} \end{aligned}$$
(2.223)

where \(A\) is given by Eq. (2.221). Equation (2.223) implies (2.220), and this completes solution of Problem 2.16.

Problem 2.17.

Let u, E, and S denote the displacement vector, strain tensor, and stress tensor fields, respectively, corresponding to a body force b and a temperature change \(T\). Suppose that the fields u, E, and S satisfy the equations

$$\begin{aligned} \mathbf{E }&=\widehat{\nabla }\mathbf{u }\quad \mathrm{{on}}\quad \mathrm{{B}}\end{aligned}$$
(2.224)
$$\begin{aligned} \mathrm{{div}}\;\mathbf{S }+\mathbf{b }&=\mathbf{0 }\quad \mathrm{{on}}\quad \mathrm{{B}}\end{aligned}$$
(2.225)
$$\begin{aligned} \mathbf{S }&={\mathbf{\mathsf{{C}} }}[\mathbf{E }]+{T} \mathbf M \quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(2.226)

where B is a bounded domain in \({E}^3\); while \({\mathbf{\mathsf{{C}} }}\) and M denote the elasticity and stress-temperature tensors, respectively, independent of \({\mathbf{x }}\in {{\overline{\mathrm{{B}}}}}\). Also, suppose that an alternative equation to Eq. (2.226) reads

$$\begin{aligned} \mathbf{E }={\mathbf{\mathsf{{K}} }}[\mathbf{S }]+{T}\mathbf{A }\quad \mathrm{{on}}\quad \mathrm{{B}} \end{aligned}$$
(2.227)

where \({\mathbf{\mathsf{{K}} }}\) and A represent the compliance and thermal expansion tensors, respectively. Let \(\widehat{\mathrm{{f}}}=\widehat{\mathrm{{f}}}\,(\mathrm{{B}})\) denote the mean value of a function \(\mathrm{{f}}=\mathrm{{f}}(\mathbf{x })\) on B

$$\begin{aligned} \widehat{\mathrm{{f}}}\,(\mathrm{{B}})=\frac{1}{{{v}}(\mathrm{{B}})}\;\int \limits _\mathrm{{B}} {\mathrm{{f}}(\mathbf{x })\,{ d{{v}}}} \end{aligned}$$
(2.228)

where \({v}\)(B) stands for the volume of B. Show that

$$\begin{aligned} {\widehat{\mathbf{E }}}\,(\mathrm{{B}})=\frac{1}{{{v}}(\mathrm{{B}})}\;\int \limits _{\partial \mathrm{{B}}} {\mathrm{{sym}}\,(\mathbf{u }\otimes \mathbf{n }){da}} \end{aligned}$$
(2.229)

and

$$\begin{aligned} {\widehat{\mathbf{S }}}\,(\mathrm{{B}})=\frac{1}{{{v}}(\mathrm{{B}})}\left[ \; {\int \limits _{\partial \mathrm{{B}}} {\mathrm{{sym}}\;(\mathbf{x }\otimes \mathbf{Sn })\;{ da}} +\int \limits _\mathrm{{B}} \mathbf{sym \,(\mathbf{u }\otimes \mathbf{b })\;\,d{{v}}}} \right] \end{aligned}$$
(2.230)

where n is the outward unit normal on \(\partial \mathrm{{B}}\). Also, show that

$$\begin{aligned} {\widehat{\mathbf{E }}}(\mathrm{{B}})={{\mathbf{\mathsf{{K}} }}}[{\widehat{\mathbf{S }}} (\mathrm{{B}})]+{\widehat{{T}}}(\mathrm{{B}})\mathbf{A } \end{aligned}$$
(2.231)

and

$$\begin{aligned} {\widehat{\mathbf{S }}}(\mathrm{{B}})={{\mathbf{\mathsf{{C}} }}}[{\widehat{\mathbf{E }}} (\mathrm{{B}})]+{\widehat{{T}}}(\mathrm{{B}})\mathbf{M } \end{aligned}$$
(2.232)

Solution.

Equation (2.229) is identical with Eq. (2.83) of Problem 2.4. Therefore, a proof of (2.229) is the same as that of (2.83) of Problem 2.4. To show (2.230), we note that Eq. (2.225) implies the tensorial equation

$$\begin{aligned} \mathbf{x }\otimes \mathrm{{div}}\ \mathbf{S }+\mathbf{x }\otimes \mathbf{b }=\mathbf{0 } \end{aligned}$$
(2.233)

or in components

$$\begin{aligned} { x_{i}\ S_{jk,k}}+{ x_{i}\,b_{j}}=0 \end{aligned}$$
(2.234)

An equivalent form of Eq. (2.234) reads

$$\begin{aligned} ({ x_{i}\ S_{jk}}),_{k}-{ S_{ji}}+{ x_{i}\,b_{j}}=0 \end{aligned}$$
(2.235)

Integrating Eq. (2.235) over \(B\) and using the divergence theorem we obtain

$$\begin{aligned} \int \limits _{\partial { B}}{ x_{i}\ S_{jk}\,n_{k}\ da}-\int \limits _{ B}\ { S_{ji}\,d{{v}}}+\int \limits _{ B}{ x_{i}\ b_{j}\,d{{v}}}=0 \end{aligned}$$
(2.236)

Finally, taking into account the symmetry of \({ S_{ij}}\), and applying the operator sym to Eq. (2.236); and dividing (2.236) by \({v}\)(B), we obtain (2.230). Also applying the mean value operator to Eqs. (2.227) and (2.226), we obtain (2.231) and (2.232), respectively; since the fourth-order tensors \({\mathbf{\mathsf{{C}} }}\) and \({\mathbf{\mathsf{{K}} }}\), and the second-order tensors M and A are independent of x. This completes solution of Problem 2.17.

Problem 2.18.

The volume change \(\delta {{v}}(\mathrm{{B}})\) associated with the fields u, E, and S in Problem 2.17 is defined by [see Eqs. (2.7)–(2.8)]

$$\begin{aligned} \delta {{v}}(\mathrm{{B}})={{v}}(\mathrm{{B}})\,\mathrm{{tr}}\,{{\widehat{\mathbf{E }}}}(\mathrm{{B}}) \end{aligned}$$
(2.237)

Show that

$$\begin{aligned} (\mathrm{i})\delta {{v}}(\mathrm{{B}})=0,\quad {\widehat{\mathbf{S }}}(\mathrm{{B}})={\widehat{{T}}}(\mathrm{{B}})\,\mathbf{M }\quad \mathrm{{if}}\quad \mathbf{u }=\mathbf{0 }\quad \mathrm{{on}}\;\;\partial \mathrm{{B}} \end{aligned}$$
(2.238)

and

$$\begin{aligned} (\mathrm{ii})\widehat{\mathbf{S}}(\mathrm{{B}})&=\mathbf{0 },\quad \widehat{\mathbf{E}}(\mathrm{{B}})=\widehat{T}(\mathrm{{B}})\mathbf{A},\quad \delta {{v}}(\mathrm{{B}})={{v}}(\mathrm{{B}})\widehat{T}(\mathrm{{B}})\,\mathrm{{tr}}\,\mathbf{A} \nonumber \\ \mathrm{{if}}\;\;\mathbf{S }\mathbf{n}&=\mathbf{0 }\;\;\mathrm{{on}}\;\;\partial \mathrm{{B}}\;\;\mathrm{{and}}\;\;\mathbf{{b}}=\mathbf{0 }\;\;\mathrm{{on}}\;\;\mathrm{{B}} \end{aligned}$$
(2.239)

Note. Equations (2.239) imply that the volume change \(\delta {{v}}(\mathrm{{B}})\) of a homogeneous isotropic thermoelastic body with zero stress vector on \(\partial \mathrm{{B}}\) and zero body force vector on B subject to a temperature change \(T\) on B is given by

$$\begin{aligned} \delta {{v}}(\mathrm{{B}})=3\,\alpha \,{\widehat{{T}}}(\mathrm{{B}})\,{{v}}(\mathrm{{B}}) \end{aligned}$$
(2.240)

where \(\alpha \) is the coefficient of linear thermal expansion of the body.

Solution.

If \(\mathbf{u }=\mathbf{0 }\) on \(\partial { B}\) then it follows from Eq. (2.229) of Problem 2.17 that \({\widehat{\mathbf{E }}}({ B})=\mathbf{0 }\). This together with Eq. (2.232) of Problem 2.17 and Eq. (2.237) implies (i).

To show (ii) we note that if \(\mathbf{S }\mathbf{n}=\mathbf{0 }\) on \(\partial { B}\) and \(\mathbf{b }=\mathbf{0 }\) on \(\overline{B}\) then, by virtue of (2.230) of Problem 2.17 we obtain

$$\begin{aligned} \widehat{\mathbf{S }}({ B})=\mathbf{0 } \end{aligned}$$
(2.241)

Hence, using (2.231) of Problem 2.17 we get

$$\begin{aligned} \widehat{\mathbf{E }}({ B})=\widehat{ T}({ B})\mathbf{A } \end{aligned}$$
(2.242)

Finally, taking the trace of (2.234) and using (2.237) we obtain

$$\begin{aligned} \delta { {{v}}(B)}={ {{v}}(B)}\ \widehat{ T}({ B}){ \mathrm{{tr}}}\ \mathbf{A } \end{aligned}$$
(2.243)

This completes proof of (ii). The result (2.240) follows from the fact that in a homogeneous isotropic body

$$\begin{aligned} {\mathrm{{tr}}}\ \mathbf{A }=3\alpha \end{aligned}$$
(2.244)

This completes solution of Problem 2.18.