 2.3 Equations of motion and equilibrium for deformable solids

In this section, we generalize Newton’s laws of motion (conservation of linear and angular momentum) to a deformable solid.

2.3.1 Linear momentum balance in terms of Cauchy stress Let ${\sigma }_{ij}$ denote the Cauchy stress distribution within a deformed solid.  Assume that the solid is subjected to a body force ${b}_{i}$, and let ${u}_{i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{v}_{i}$ and ${a}_{i}$ denote the displacement, velocity and acceleration of a material particle at position ${y}_{i}$  in the deformed solid.

Newton’s third law of motion (F=ma) can be expressed as

Written out in full

$\begin{array}{l}\frac{\partial {\sigma }_{11}}{\partial {y}_{1}}+\frac{\partial {\sigma }_{21}}{\partial {y}_{2}}+\frac{\partial {\sigma }_{31}}{\partial {y}_{3}}+\rho {b}_{1}=\rho {a}_{1}\\ \frac{\partial {\sigma }_{12}}{\partial {y}_{1}}+\frac{\partial {\sigma }_{22}}{\partial {y}_{2}}+\frac{\partial {\sigma }_{32}}{\partial {y}_{3}}+\rho {b}_{2}=\rho {a}_{2}\\ \frac{\partial {\sigma }_{13}}{\partial {y}_{1}}+\frac{\partial {\sigma }_{23}}{\partial {y}_{2}}+\frac{\partial {\sigma }_{33}}{\partial {y}_{3}}+\rho {b}_{3}=\rho {a}_{3}\end{array}$

Note that the derivative is taken with respect to position in the actual, deformed solid. For the special (but rather common) case of a solid in static equilibrium in the absence of body forces

$\frac{\partial {\sigma }_{ij}}{\partial {y}_{i}}=0$

Derivation: Recall that the resultant force acting on an arbitrary volume of material V within a solid is

${P}_{i}=\underset{A}{\int }{T}_{i}\left(n\right)dA+\underset{V}{\int }\rho \text{\hspace{0.17em}}{b}_{i}\text{\hspace{0.17em}}dV$

where T(n) is the internal traction acting on the surface A with normal n that bounds V.

The linear momentum of the volume V is

${\Lambda }_{i}=\underset{V}{\int }\rho \text{\hspace{0.17em}}{v}_{i}dV$

where v is the velocity vector of a material particle in the deformed solid. Express T in terms of ${\sigma }_{ij}$ and set ${P}_{i}=d{\Lambda }_{i}/dt$

$\underset{A}{\int }{\sigma }_{ji}{n}_{j}dA+\underset{V}{\int }\rho \text{\hspace{0.17em}}{b}_{i}\text{\hspace{0.17em}}dV=\frac{d}{dt}\left\{\underset{V}{\int }\rho \text{\hspace{0.17em}}{v}_{i}dV\right\}$

Apply the divergence theorem to convert the first integral into a volume integral, and note that one can show (see Appendix D) that

$\frac{d}{dt}\left\{\underset{V}{\int }\rho \text{\hspace{0.17em}}{v}_{i}dV\right\}=\underset{V}{\int }\rho {a}_{i}dV$

so

$\underset{V}{\int }\frac{\partial {\sigma }_{ji}}{\partial {y}_{j}}dV+\underset{V}{\int }\rho \text{\hspace{0.17em}}{b}_{i}\text{\hspace{0.17em}}dV=\underset{V}{\int }\rho {a}_{i}dV⇒\underset{V}{\int }\left(\frac{\partial {\sigma }_{ji}}{\partial {y}_{j}}+\rho \text{\hspace{0.17em}}{b}_{i}-\rho \text{\hspace{0.17em}}{a}_{i}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}dV=0$

Since this must hold for every volume of material within a solid, it follows that

$\frac{\partial {\sigma }_{ji}}{\partial {y}_{j}}+\rho {b}_{i}=\rho {a}_{i}$

as stated.

2.3.2 Angular momentum balance in terms of Cauchy stress

Conservation of angular momentum for a continuum requires that the Cauchy stress satisfy

${\sigma }_{ji}={\sigma }_{ij}$

i.e. the stress tensor must be symmetric.

Derivation: write down the equation for balance of angular momentum for the region V within the  deformed solid

$\underset{A}{\int }y×T\text{\hspace{0.17em}}dA+\underset{V}{\int }y×\rho \text{\hspace{0.17em}}bdV=\frac{d}{dt}\left\{\underset{V}{\int }y×\rho \text{\hspace{0.17em}}vdV\right\}$

Here, the left hand side is the resultant moment (about the origin) exerted by tractions and body forces acting on a general region within a solid.  The right hand side is the total angular momentum of the solid about the origin.

We can write the same expression using index notation

$\underset{A}{\int }{\in }_{ijk}{y}_{j}{T}_{k}dA+\underset{V}{\int }{\in }_{ijk}{y}_{j}{b}_{k}\rho \text{\hspace{0.17em}}dV=\frac{d}{dt}\left\{\underset{V}{\int }{\in }_{ijk}{y}_{j}{v}_{k}\rho \text{\hspace{0.17em}}dV\right\}$

Express T in terms of ${\sigma }_{ij}$ and re-write the first integral as a volume integral using the divergence theorem

$\begin{array}{l}\underset{A}{\int }{\in }_{ijk}{y}_{j}{T}_{k}dA=\underset{A}{\int }{\in }_{ijk}{y}_{j}{\sigma }_{mk}{n}_{m}dA=\underset{V}{\int }\frac{\partial }{\partial {x}_{m}}\left({\in }_{ijk}{y}_{j}{\sigma }_{mk}\right)dV\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{V}{\int }{\in }_{ijk}\left({\delta }_{jm}{\sigma }_{mk}+{y}_{j}\frac{\partial {\sigma }_{mk}}{\partial {x}_{m}}\right)dV\end{array}$

We may also show (see Appendix D) that

$\frac{d}{dt}\left\{\underset{V}{\int }{\in }_{ijk}{y}_{j}{v}_{k}\rho \text{\hspace{0.17em}}dV\right\}=\underset{V}{\int }{\in }_{ijk}{y}_{j}{a}_{k}\rho \text{\hspace{0.17em}}dV$

Substitute the last two results into the angular momentum balance equation to see that

$\begin{array}{l}\underset{V}{\int }{\in }_{ijk}\left({\delta }_{jm}{\sigma }_{mk}+{y}_{j}\frac{\partial {\sigma }_{mk}}{\partial {x}_{m}}\right)dV+\underset{V}{\int }{\in }_{ijk}{y}_{j}{b}_{k}\rho \text{\hspace{0.17em}}dV=\underset{V}{\int }{\in }_{ijk}{y}_{j}{a}_{k}\rho \text{\hspace{0.17em}}dV\\ ⇒\underset{V}{\int }{\in }_{ijk}{\delta }_{jm}{\sigma }_{mk}=-\underset{V}{\int }{\in }_{ijk}{y}_{j}\left(\frac{\partial {\sigma }_{mk}}{\partial {y}_{m}}+\rho {b}_{k}-\rho {a}_{k}\right)dV\end{array}$

The integral on the right hand side of this expression is zero, because the stresses must satisfy the linear momentum balance equation.  Since this holds for any volume V, we conclude that

$\begin{array}{l}{\in }_{ijk}{\delta }_{jm}{\sigma }_{mk}={\in }_{ijk}{\sigma }_{jk}=0\\ ⇒{\in }_{imn}{\in }_{ijk}{\sigma }_{jk}=0\\ ⇒\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{mk}{\delta }_{nj}\right){\sigma }_{jk}=0\\ ⇒{\sigma }_{mn}-{\sigma }_{nm}=0\end{array}$

which is the result we wanted.

2.3.3 Equations of motion in terms of other stress measures

In terms of nominal and material stress the balance of linear momentum is

$\nabla \cdot S+{\rho }_{0}b={\rho }_{0}a\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{S}_{ij}}{\partial {x}_{i}}+{\rho }_{0}{b}_{j}={\rho }_{0}{a}_{j}$

$\nabla \cdot \left[\Sigma \cdot {F}^{T}\right]+{\rho }_{0}b={\rho }_{0}a\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial \left({\Sigma }_{ik}{F}_{jk}\right)}{\partial {x}_{i}}+{\rho }_{0}{b}_{j}={\rho }_{0}{a}_{j}$

Note that the derivatives are taken with respect to position in the undeformed solid.

The angular momentum balance equation is

$⥄⥄⥄⥄⥄⥄⥄⥄F\cdot S={\left[F\cdot S\right]}^{T}⥄⥄⥄⥄⥄⥄⥄\Sigma ={\Sigma }^{Τ}$

To derive these results, you can start with the integral form of the linear momentum balance in terms of Cauchy stress

$\underset{A}{\int }{\sigma }_{ji}{n}_{j}dA+\underset{V}{\int }\rho \text{\hspace{0.17em}}{b}_{i}\text{\hspace{0.17em}}dV=\frac{d}{dt}\left\{\underset{V}{\int }\rho \text{\hspace{0.17em}}{v}_{i}dV\right\}$

Recall (or see Appendix D for a reminder) that area elements in the deformed and undeformed solids are related through

$dA{n}_{i}^{}=J{F}_{ki}^{-1}{n}_{k}^{0}d{A}_{0}$

In addition, volume elements are related by $dV=Jd{V}_{0}$.  We can use these results to re-write the integrals as integrals over a volume in the undeformed solid as

$\underset{A0}{\int }{\sigma }_{ji}J{F}_{kj}^{-1}{n}_{k}^{0}d{A}_{0}+\underset{V0}{\int }\rho \text{\hspace{0.17em}}{b}_{i}\text{\hspace{0.17em}}Jd{V}_{0}=\frac{d}{dt}\left\{\underset{V0}{\int }\rho \text{\hspace{0.17em}}{v}_{i}Jd{V}_{0}\right\}$

Finally, recall that ${S}_{ij}=J{F}_{ik}^{-1}{\sigma }_{kj}$ and that $J\rho ={\rho }_{0}$ to see that

$\underset{A0}{\int }S{}_{ki}{n}_{k}^{0}d{A}_{0}+\underset{V0}{\int }\rho {\text{\hspace{0.17em}}}_{0}{b}_{i}\text{\hspace{0.17em}}d{V}_{0}=\frac{d}{dt}\left\{\underset{V0}{\int }\rho {\text{\hspace{0.17em}}}_{0}{v}_{i}d{V}_{0}\right\}$

Apply the divergence theorem to the first term and rearrange

$\underset{V}{\int }\left(\frac{\partial {S}_{ji}}{\partial {x}_{j}}+{\rho }_{0}\text{\hspace{0.17em}}{b}_{i}-{\rho }_{0}\text{\hspace{0.17em}}\frac{d{v}_{i}}{dt}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}d{V}_{0}=0$

Once again, since this must hold for any material volume, we conclude that

$\frac{{S}_{ij}}{\partial {x}_{i}}+{\rho }_{0}{b}_{j}={\rho }_{0}{a}_{j}$

The linear momentum balance equation in terms of material stress follows directly, by substituting into this equation for ${S}_{ij}$ in terms of ${\Sigma }_{ij}$

The angular momentum balance equation can be derived simply by substituting into the momentum balance equation in terms of Cauchy stress ${\sigma }_{ij}={\sigma }_{ji}$