Chapter 2

 

Governing Equations

 

 

 

2.4 Work done by stresses; Principle of Virtual Work

 

In this section, we derive formulas that enable you to calculate the work done by stresses acting on a solid.  In addition, we prove the principle of virtual work $–$ which is an alternative way of expressing the equations of motion and equilibrium derived in Section 2.3.  The principle of virtual work is the starting point for finite element analysis, and so is a particularly important result.

 

 

2.4.1 Work done by Cauchy stresses

 

Consider a solid with mass density ${\rho }_0$ in its initial configuration, and density $\rho$ in the deformed solid. Let ${\sigma }_{ij}$ denote the Cauchy stress distribution within the solid.  Assume that the solid is subjected to a body force $b_i$ (per unit mass), and let $u_i,v_i$ and $a_i$ denote the displacement, velocity and acceleration of a material particle at position $y_i$  in the deformed solid. In addition, let

$D_{ij}=\frac{1}{2}\left(\frac{\partial v_i}{\partial y_j}+\frac{\partial v_j}{\partial y_i}\right)$

denote the stretch rate in the solid.

 

The rate of work done by Cauchy stresses per unit deformed volume is then ${\sigma }_{ij}{D}_{ij}$.  This energy is either dissipated as heat or stored as internal energy in the solid, depending on the material behavior.

 

We shall show that the rate of work done by internal forces acting on any sub-volume V bounded by a surface A in the deformed solid can be calculated from

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}+\frac{d}{dt}\left\{{\displaystyle \underset{V}{\int }\frac{1}{2}\rho v_i{v}_{i}dV}\right\}$

Here, the two terms on the left hand side represent the rate of work done by tractions and body forces acting on the solid (work done = force x velocity).  The first term on the right-hand side can be interpreted as the work done by Cauchy stresses; the second term is the rate of change of kinetic energy. 

 

Derivation: Substitute for $T_i^{(n)}$ in terms of Cauchy stress to see that

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{A}{\int }{n}_j{\sigma }_{ji}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}$

Now, apply the divergence theorem to the first term on the right hand side

$\dot{r}={\displaystyle \underset{V}{\int }\frac{\partial }{\partial y_j}\left({\sigma }_{ji}{v}_i\right)dV}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}$

Evaluate the derivative and collect together the terms involving body force and stress divergence

$\dot{r}={\displaystyle \underset{V}{\int }\left\{{\sigma }_{ji}\frac{\partial v_i}{\partial y_j}+\left(\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i\right)v_i\right\}dV}$

Recall the equation of motion

$\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i=\rho a_i$

and note that since the stress is symmetric ${\sigma }_{ij}={\sigma }_{ji}$

${\sigma }_{ji}\frac{\partial v_i}{\partial y_j}=\frac{1}{2}\left({\sigma }_{ij}+{\sigma }_{ji}\right)\frac{\partial v_i}{\partial y_j}=\frac{1}{2}{\sigma }_{ij}\left(\frac{\partial v_i}{\partial y_j}+\frac{\partial v_j}{\partial y_i}\right)={\sigma }_{ij}{D}_{ij}$

to see that

$\dot{r}={\displaystyle \underset{V}{\int }\left\{{\sigma }_{ij}{D}_{ij}+\rho a_i{v}_i\right\}dV}$

Finally, note that

$\begin{array}{l}{\displaystyle \underset{V}{\int }\rho a_i{v}_{i}dV}={\displaystyle \underset{V_0}{\int }{\rho }_o\frac{d{v}_i}{dt}{v}_{i}d{V}_0=}{\displaystyle \underset{V_0}{\int }{\rho }_o\frac{1}{2}\frac{d}{dt}\left(v_i{v}_i\right)d{V}_0}\\ =\frac{d}{dt}\left({\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0\left(v_i{v}_i\right)d{V}_0}\right)=\frac{d}{dt}\left({\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0\left(v_i{v}_i\right)d{V}_0}\right)=\frac{d}{dt}{\displaystyle \underset{V_{}}{\int }\frac{1}{2}\rho \left(v_i{v}_i\right)dV}\end{array}$

Finally, substitution leads to

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}+\frac{d}{dt}\left\{{\displaystyle \underset{V}{\int }\frac{1}{2}\rho v_i{v}_{i}dV}\right\}$

as required.

 

 

2.4.2 Rate of mechanical work in terms of other stress measures

 

 The rate of work done per unit undeformed volume by Kirchhoff stress is ${\tau }_{ij}{D}_{ij}$

 The rate of work done per unit undeformed volume by Nominal stress is $S_{ij}{\dot{F}}_{ji}$

 The rate of work done per unit undeformed volume by Material stress is ${\Sigma }_{ij}{\dot{E}}_{ij}$

 

This shows that nominal stress and deformation gradient are work conjugate, as are material stress and Lagrange strain.

 

In addition, the rate of work done on a volume $V_0$ of the undeformed solid can be expressed as

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V_0}{\int }{\tau }_{ij}{D}_{ij}d{V}_0}+\frac{d}{dt}\left\{{\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0{v}_i{v}_{i}d{V}_0}\right\}$

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V_0}{\int }{S}_{ij}{\dot{F}}_{ji}d{V}_0}+\frac{d}{dt}\left\{{\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0{v}_i{v}_{i}d{V}_0}\right\}$

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V_0}{\int }{\Sigma }_{ij}{\dot{E}}_{ij}d{V}_0}+\frac{d}{dt}\left\{{\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0{v}_i{v}_{i}d{V}_0}\right\}$

 

Derivations: The proof of the first result (and the stress power of Kirchhoff stress) is straightforward and is left as an exercise.  To show the second result, note that $T_i^{(n)}dA=n_j^0{S}_{ji}d{A}_0$ and $dV=Jd{V}_0$ to re-write the integrals over the undeformed solid; then and apply the divergence theorem to see that

$\dot{r}={\displaystyle \underset{V_0}{\int }\frac{\partial }{\partial x_j}\left(S_{ji}{v}_i\right)d{V}_0}+{\displaystyle \underset{V0}{\int }\rho b_i{v}_{i}Jd{V}_0}$

Evaluate the derivative, recall that $J\rho ={\rho }_0$ and use the equation of motion

$\frac{S_{ij}}{\partial x_i}+{\rho }_0{b}_j={\rho }_0\frac{d{v}_j}{dt}$

to see that

$\dot{r}={\displaystyle \underset{V_0}{\int }{S}_{ji}\frac{\partial v_i}{\partial x_j}d{V}_0}+{\displaystyle \underset{V0}{\int }{\rho }_0\frac{d{v}_i}{dt}{v}_{i}d{V}_0}$

Finally, note that $\partial v_i/\partial x_j=\left(\partial {\dot{u}}_i/\partial x_j\right)={\dot{F}}_{ij}$ and re-write the second integral as a kinetic energy term as before to obtain the required result.

 

The third result follows by straightforward algebraic manipulations $–$ note that by definition

$S_{ij}{\dot{F}}_{ji}={\Sigma }_{ik}{F}_{jk}^{}{\dot{F}}_{ji}$

Since ${\Sigma }_{ij}$ is symmetric it follows that

${\Sigma }_{ik}{F}_{jk}^{}{\dot{F}}_{ji}=\frac{1}{2}\left({\Sigma }_{ik}+{\Sigma }_{ki}\right)F_{jk}^{}{\dot{F}}_{ji}={\Sigma }_{ik}\frac{1}{2}\left(F_{jk}^{}{\dot{F}}_{ji}+F_{ji}^{}{\dot{F}}_{jk}\right)={\Sigma }_{ik}{\dot{E}}_{ik}$

 

 

2.4.3 Rate of mechanical work for infinitesimal deformations

 

For infintesimal motions all stress measures are equal; and all strain rate measures can be approximated by the infinitesimal strain tensor $\epsilon$.  The rate of work done by stresses per unit volume of either deformed or undeformed solid (the difference is neglected) can be expressed as ${\sigma }_{ij}{\dot{\epsilon }}_{ij}$, and the work done on a volume $V_0$ of the solid is

$\dot{r}={\displaystyle \underset{A}{\int }{T}_i^{(n)}{v}_{i}dA}+{\displaystyle \underset{V}{\int }\rho b_i{v}_{i}dV}={\displaystyle \underset{V_0}{\int }{\sigma }_{ij}{\dot{\epsilon }}_{ij}d{V}_0}+\frac{d}{dt}\left\{{\displaystyle \underset{V_0}{\int }\frac{1}{2}{\rho }_0{v}_i{v}_{i}d{V}_0}\right\}$

 

 

 

2.4.4 The principle of Virtual Work

 

The principle of virtual work forms the basis for the finite element method in the mechanics of solids and so will be discussed in detail in this section.

 

Suppose that a deformable solid is subjected to loading that induces a displacement field $u(x)$, and a velocity field $v(x)$.  The loading consists of a prescribed displacement on part of the boundary (denoted by $S_1$ ), together with a traction t (which may be zero in places) applied to the rest of the boundary (denoted by $S_2$ ).  The loading induces a Cauchy stress ${\sigma }_{ij}$.  The stress field satisfies the angular momentum balance equation ${\sigma }_{ij}={\sigma }_{ji}$.

 

The principle of virtual work is a different way of re-writing partial differential equation for linear moment balance

$\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i=\rho \frac{d{v}_i}{dt}$

in an equivalent integral form, which is much better suited for computer solution.

 

To express the principle, we define a kinematically admissible virtual velocity field $\delta v(y)$, satisfying  $\delta v=0$ on $S_1$.  You can visualize this field as a small change in the velocity of the solid, if you like, but it is really just an arbitrary differentiable vector field.  The term `kinematically admissible’ is just a complicated way of saying that the field is continuous, differentiable, and satisfies $\delta v=0$ on $S_1$ - that is to say, if you perturb the velocity by $\delta v(y)$, the boundary conditions on displacement are still satisfied.

 

In addition, we define an associated virtual velocity gradient, and virtual stretch rate as

$\delta L_{ij}=\frac{\partial \delta v_i}{\partial y_j}\delta D_{ij}=\frac{1}{2}\left(\frac{\partial \delta v_i}{\partial y_j}+\frac{\partial \delta v_j}{\partial y_i}\right)$

 

The principal of virtual work may be stated in two ways.

 

First version of the principle of virtual work

 

The first is not very interesting, but we will state it anyway.  Suppose that the Cauchy stress satisfies:

1.      The boundary condition $n_i{\sigma }_{ij}=t_j$ on $S_2$

2.      The linear momentum balance equation

$\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i=\rho \frac{d{v}_i}{dt}$

Then the virtual work equation

${\displaystyle \underset{V}{\int }{\sigma }_{ij}\delta D_{ij}dV+{\displaystyle \underset{V}{\int }\rho \frac{d{v}_i}{dt}\delta v_i}dV-{\displaystyle \underset{V}{\int }\rho b_i\delta v_{i}dV-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

is satisfied for all virtual velocity fields.

 

Proof:  Observe that since the Cauchy stress is symmetric

${\sigma }_{ij}\delta D_{ij}=\frac{1}{2}{\sigma }_{ij}\left(\frac{\partial \delta v_i}{\partial y_j}+\frac{\partial \delta v_j}{\partial y_i}\right)=\frac{1}{2}\left({\sigma }_{ji}\frac{\partial \delta v_i}{\partial y_j}+{\sigma }_{ij}\frac{\partial \delta v_j}{\partial y_i}\right)={\sigma }_{ji}\frac{\partial \delta v_i}{\partial y_j}$

Next, note that

${\sigma }_{ji}\frac{\partial v_i}{\partial y_j}=\frac{\partial }{\partial y_j}\left({\sigma }_{ji}\delta v_i\right)-\frac{\partial {\sigma }_{ji}}{\partial y_j}\delta v_i$

Finally, substituting the latter identity into the virtual work equation, applying the divergence theorem, using the linear momentum balance equation and boundary conditions on $\sigma$ and $\delta v(y)$ we obtain the required result.

 

Second version of the principle of virtual work

 

The converse of this statement is much more interesting and useful.  Suppose that ${\sigma }_{ij}$ satisfies the virtual work equation

${\displaystyle \underset{V}{\int }{\sigma }_{ij}\delta D_{ij}dV+{\displaystyle \underset{V}{\int }\rho \frac{d{v}_i}{dt}\delta v_i}dV-{\displaystyle \underset{V}{\int }\rho b_i\delta v_{i}dV-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

for all virtual velocity fields $\delta v(y)$.  Then the stress field must satisfy

3.      The boundary condition $n_i{\sigma }_{ij}=t_j$ on $S_2$

4.      The linear momentum balance equation

$\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i=\rho \frac{d{v}_i}{dt}$

The significance of this result is that it gives us an alternative way to solve for a stress field that satisfies the linear momentum balance equation, which avoids having to differentiate the stress.  It is not easy to differentiate functions accurately in the computer, but it is easy to integrate them.  The virtual work statement is the starting point for any finite element solution involving deformable solids.

 

Proof: Follow the same preliminary steps as before, i..e.

${\sigma }_{ij}\delta D_{ij}=\frac{1}{2}{\sigma }_{ij}\left(\frac{\partial \delta v_i}{\partial y_j}+\frac{\partial \delta v_j}{\partial y_i}\right)=\frac{1}{2}\left({\sigma }_{ji}\frac{\partial \delta v_i}{\partial y_j}+{\sigma }_{ij}\frac{\partial \delta v_j}{\partial y_i}\right)={\sigma }_{ji}\frac{\partial \delta v_i}{\partial y_j}$

${\sigma }_{ji}\frac{\partial v_i}{\partial y_j}=\frac{\partial }{\partial y_j}\left({\sigma }_{ji}\delta v_i\right)-\frac{\partial {\sigma }_{ji}}{\partial y_j}\delta v_i$

and substitute into the virtual work equation

${\displaystyle \underset{V}{\int }\left\{\frac{\partial }{\partial y_j}\left({\sigma }_{ji}\delta v_i\right)-\frac{\partial {\sigma }_{ji}}{\partial y_j}\delta v_i\right\}dV+{\displaystyle \underset{V}{\int }\rho \frac{d{v}_i}{dt}\delta v_i}dV-{\displaystyle \underset{V}{\int }\rho b_i\delta v_{i}dV-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

Apply the divergence theorem to the first term in the first integral, and recall that $\delta v=0$ on $S_1$, we see that

$-{\displaystyle \underset{V}{\int }\left\{\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i-\rho \frac{d{v}_i}{dt}\right\}\delta v_{i}dV}+{\displaystyle \underset{S_2}{\int }\left({\sigma }_{ji}{n}_j-t_i\right)\delta v_i}dA=0$

Since this must hold for all virtual velocity fields we could choose

$\delta v_i=f(y)\left\{\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i-\rho \frac{d{v}_i}{dt}\right\}$

where $f(y)=0$ is an arbitrary function that is positive everywhere inside the solid, but is equal to zero on $S$.  For this choice, the virtual work equation reduces to

$-{\displaystyle \underset{V}{\int }f\left(y\right)\left\{\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i-\rho \frac{d{v}_i}{dt}\right\}\left\{\frac{\partial {\sigma }_{ki}}{\partial y_k}+\rho b_i-\rho \frac{d{v}_i}{dt}\right\}dV}=0$

and since the integrand is positive everywhere the only way the equation can be satisfied is if

$\frac{\partial {\sigma }_{ji}}{\partial y_j}+\rho b_i=\rho \frac{d{v}_i}{dt}$

Given this, we can next choose a virtual velocity field that satisfies

$\delta v_i=\left({\sigma }_{ji}{n}_j-t_i\right)$

on $S_2$.  For this choice (and noting that the volume integral is zero) the virtual work equation reduces to

$+{\displaystyle \underset{S_2}{\int }\left({\sigma }_{ji}{n}_j-t_i\right)\left({\sigma }_{ki}{n}_k-t_i\right)}dA=0$

Again, the integrand is positive everywhere (it is a perfect square) and so can vanish only if

${\sigma }_{ji}{n}_j=t_i$

as stated.

 

 

 

2.4.5 The Virtual Work equation in terms of other stress measures.

 

It is often convenient to implement the virtual work equation in a finite element code using different stress measures. 

 

To do so, we define

1.      The actual deformation gradient in the solid $F_{ij}={\delta }_{ij}+\frac{\partial u_i}{\partial x_j}$

2.      The virtual rate of change of deformation gradient  $\delta {\dot{F}}_{ij}=\frac{\partial \delta v_i}{\partial y_k}{F}_{kj}=\frac{\partial \delta v_i}{\partial x_j}$

3.      The virtual rate of change of Lagrange strain $\delta {\dot{E}}_{ij}=\frac{1}{2}\left(F_{ki}\delta {\dot{F}}_{kj}+\delta {\dot{F}}_{ki}{F}_{kj}\right)$

In addition, we define (in the usual way)

1.      Kirchhoff stress  ${\tau }_{ij}=J{\sigma }_{ij}$

2.      Nominal (First Piola-Kirchhoff) stress  $S_{ij}=J{F}_{ik}^{-1}{\sigma }_{kj}$

3.      Material (Second Piola-Kirchhoff) stress  ${\Sigma }_{ij}=J{F}_{ik}^{-1}{\sigma }_{kl}{F}_{jl}^{-1}$

 

In terms of these quantities, the virtual work equation may be expressed as

${\displaystyle \underset{V0}{\int }{\tau }_{ij}\delta D_{ij}d{V}_0+{\displaystyle \underset{V_0}{\int }{\rho }_0\frac{d{v}_i}{dt}\delta v_i}d{V}_0-{\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta v_{i}d{V}_0-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

${\displaystyle \underset{V0}{\int }{S}_{ij}\delta {\dot{F}}_{ji}d{V}_0+{\displaystyle \underset{V_0}{\int }{\rho }_0\frac{d{v}_i}{dt}\delta v_i}d{V}_0-{\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta v_{i}d{V}_0-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

${\displaystyle \underset{V0}{\int }{\Sigma }_{ij}\delta {\dot{E}}_{ij}d{V}_0+{\displaystyle \underset{V_0}{\int }{\rho }_0\frac{d{v}_i}{dt}\delta v_i}d{V}_0-{\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta v_{i}d{V}_0-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}dA=0$

Note that all the volume integrals are now taken over the undeformed solid $–$ this is convenient for computer applications, because the shape of the undeformed solid is known.  The area integral is evaluated over the deformed solid, unfortunately.  It can be expressed as an equivalent integral over the undeformed solid, but the result is messy and will be deferred until we actually need to do it.

 

 

 

2.4.6 The Virtual Work equation for infinitesimal deformations.

 

For infintesimal motions, the Cauchy, Nominal, and Material stress tensors are equal; and the virtual stretch rate can be replaced by the virtual infinitesimal strain rate

$\delta {\dot{\epsilon }}_{ij}=\frac{1}{2}\left(\frac{\partial \delta v_i}{\partial x_j}+\frac{\partial \delta v_j}{\partial x_i}\right)$

There is no need to distinguish between the volume or surface area of the deformed and undeformed solid.  The virtual work equation can thus be expressed as

${\displaystyle \underset{V0}{\int }{\sigma }_{ij}\delta {\dot{\epsilon }}_{ij}d{V}_0+{\displaystyle \underset{V_0}{\int }{\rho }_0\frac{d{v}_i}{dt}\delta v_i}d{V}_0-{\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta v_{i}d{V}_0-{\displaystyle \underset{S_2}{\int }{t}_i\delta v_i}}}d{A}_0=0$

for all kinematically admissible velocity fields.

 

 

As a special case, this expression can be applied to a quasi-static state with $v_i=0$. Then, for a stress state ${\sigma }_{ij}$ satisfying the static equilibrium equation ${\sigma }_{ij}/d{x}_i+{\rho }_0{b}_j=0$ and boundary conditions ${\sigma }_{ij}{n}_j=t_i$ on $S_2$, the virtual work equation reduces to

${\displaystyle \underset{V0}{\int }{\sigma }_{ij}\delta {\epsilon }_{ij}d{V}_0}={\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta u_{i}d{V}_0+{\displaystyle \underset{S_2}{\int }{t}_i\delta u_{i}dA}}$

In which $\delta u_i$ are kinematically admissible displacements components $(\delta u_i=0$ on S₂) and $\delta {\epsilon }_{ij}=\left(\partial \delta u_i/x_j+\partial \delta u_j/x_i\right)/2$.

 

Conversely, if  the stress state ${\sigma }_{ij}$ satisfies ${\displaystyle \underset{V0}{\int }{\sigma }_{ij}\delta {\epsilon }_{ij}d{V}_0}={\displaystyle \underset{V0}{\int }{\rho }_0{b}_i\delta u_{i}d{V}_0+{\displaystyle \underset{S_2}{\int }{t}_i\delta u_{i}dA}}$ for every set of kinematically admissible virtual displacements, then the stress state ${\sigma }_{ij}$ satisfies the static equilibrium equation ${\sigma }_{ij}/d{x}_i+{\rho }_0{b}_j=0$ and boundary conditions ${\sigma }_{ij}{n}_j=t_i$ on $S_2$.