principle of virtual work

Section 2.5 principle of virtual work

Starting from equilibrium equations (2.3.10) it is possible to reach the definition of the work deriving from the surrounding environment, external work, and the work carried out by the forces inside the body, internal work. We obtain a principle stating the equality of these two works known as the Principle of virtual work. This result is obtained by reformulating the differential equations (2.3.10) in an integral format. An operation that can be simply interpreted as an alternative reformulation of the starting differential equations but which also leads to a result that has the important mechanical interpretation mentioned above.

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Subsection 2.5.1 from the differential formulation to the integral formulation

Let us evaluate the scalar product between Eq. (2.3.10) and a generic vector field \(\vec{v}\) and let us integrate the result over the assigned body \(\body\) (for continuity, this operation could be carried out on any subset of \(\body\)). Because Eq. (2.3.10) must to be satisfied in each \(\vec{x}\) of \(\body\text{,}\) then

\begin{equation} \int_{\body} \left( \text{div}\tens{\sigma} + \vec{b} \right) \cdot \vec{v} \,dv = 0\,,\quad \forall\vec{v}\,,\tag{2.5.1} \end{equation}

where \(\vec{v} \) is a continuous and differentiable vector field defined on the body \(\body\text{.}\) As already shown in the previous video about divergence it is possible to write

\begin{equation*} \text{div}\left(\tens{\sigma}\vec{v}\right) = \text{div}\tens{\sigma} \cdot \vec{v} + \tens{\sigma} : \tens{\nabla v}\,, \end{equation*}

from which

\begin{equation*} \text{div}\tens{\sigma} \cdot \vec{v} = \text{div}\left(\tens{\sigma}\vec{v}\right) - \tens{\sigma} : \tens{\nabla v} \,. \end{equation*}

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Remark 2.5.1. important remark.

In the previous steps the derivatives are transferred from tensor \(\tens{\sigma} \) to vector \(\vec{v} \text{.}\) The derivatives on \(\tens{\sigma} \) are made with respect to coordinates \(x_1 \text{,}\) \(x_2 \) and \(x_3 \) related to the current configuration, therefore in this context the gradient of the displacement is expressed by

\begin{equation} \mat{\nabla v} = \left[\begin{array}{ccc} \regulardiff{u_1}{x_1} \amp \regulardiff{u_1}{x_2} \amp \regulardiff{u_1}{x_3}\\ \regulardiff{u_2}{x_1} \amp \regulardiff{u_2}{x_2} \amp \regulardiff{u_2}{x_3}\\ \regulardiff{u_3}{x_1} \amp \regulardiff{u_3}{x_2} \amp \regulardiff{u_3}{x_3} \end{array}\right]\,,\tag{2.5.2} \end{equation}

expression to be not confused with similar Eq. (1.4.5) where the derivatives are performed with respect to the position coordinates, \(X_1\text{,}\) \(X_2\text{,}\) \(X_3\text{,}\) of the reference configuration.

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Now the starting integral equation can be rewritten as follows

\begin{equation} \int_{\body} \left(\text{div}\left(\tens{\sigma}\vec{v}\right) - \tens{\sigma} : \tens{\nabla v} + \vec{b} \cdot \vec{v} \right) \,dv= 0\,,\quad \forall\vec{v}\,.\tag{2.5.3} \end{equation}

The first integral, by applying the divergence theorem and the Cauchy stress tensor theorem, can be written also as

\begin{equation*} \int_{\body} \text{div}\left(\tens{\sigma}\vec{v}\right)\, dv = \int_{\partial\body} \tens{\sigma}\vec{v} \cdot \vec{n}\,ds = \int_{\partial\body} \vec{v} \cdot \tens{\sigma}\vec{n}\,ds = \int_{\partial\body} \vec{t} \cdot \vec{v}\,ds \,. \end{equation*}

Then Eq. (2.5.3) becomes

\begin{equation} \int_{\partial\body} \vec{t} \cdot \vec{v}\,ds - \int_{\body} \tens{\sigma} : \tens{\nabla v}\,dv + \int_{\body} \vec{b} \cdot \vec{v} \,dv = 0\,,\quad \forall\vec{v}\,.\tag{2.5.4} \end{equation}

Using decompostion of \(\tens{\nabla v}\)

\begin{equation*} \tens{\nabla v} = \text{sym}\tens{\nabla v} + \text{skew}\tens{\nabla v} \end{equation*}

and symmetry condition on \(\tens{\sigma}\text{,}\) the following result is obtained

\begin{equation*} \tens{\sigma} : \tens{\nabla v} = \tens{\sigma} : \text{sym}\tens{\nabla v}\,. \end{equation*}

Finally the following expression of the integral form of the indefinite equations of equilibrium (2.3.10) can be formulated:

\begin{equation} \int_{\partial\body} \vec{t} \cdot \vec{v}\,ds + \int_{\body} \vec{b} \cdot \vec{v} \,dv = \int_{\body} \tens{\sigma} : \text{sym}\tens{\nabla v}\,dv \,,\quad \forall\vec{v}\,.\tag{2.5.5} \end{equation}

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Subsection 2.5.2 mechanical interpretation

The vector field \(\vec{v} \) can be interpreted both as a displacement field that the body actually undergoes under the action of the assigned loads and as a generic variation of the actual displacement field or as a virtual displacement field.

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The first member of the equality (2.5.5) represents the work done by the surrounding environment on the body \(\body \text{,}\) that is the external work.

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The second member of the equality (2.5.5) contains the internal work, that is the work carried out by the stress tensor \(\tens{\sigma} \text{.}\)

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Without the need to make any preliminary assumptions, we arrive at the definition of the kinematic counterpart \(\text{sym}\tens{\nabla v} \) on which the stress tensor \(\tens{\sigma} \) does work. Thus the meaning of the symmetric part of the gradient of the displacement is widened because it comprises, as already shown in the chapter about kinematics, also the definition of the infinitesimal strain tensor.

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A formal statement of the result obtained is as follows: "Given a body, or any subset of a body, satisfying the equilibrium condition with respect to the applied loads, i.e Eq. (2.3.10) is satisfied, then also Eq. (2.5.5) is valid. This is the Principle of virtual work". Moreover Eq. (2.3.10) is not only a necessary condition but also is sufficient for satisfaction of Eq. (2.5.5), i.e. the statement can be reversed as follows: "Given a body, or any subset of a body, satisfying the equivalence of the internal and external works, Eq. (2.5.5), then Eq. (2.3.10) is satisfied".

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In the case of displacement field \(\vec{v} \) coincident with a rigid displacement field \(\vec{r} \text{,}\) then the condition \(\vec{v} = \vec{r} \) provides
\begin{equation*} \text{sym}\tens{\nabla v} = \text{sym}\tens{\nabla r} = \tens{0} \end{equation*}
and then the Principle of virtual work redices to
\begin{equation} \int_{\partial\body} \vec{t} \cdot \vec{r}\,ds + \int_{\body} \vec{b} \cdot \vec{r} \,dv = 0 \,,\quad \forall\vec{r}\,.\tag{2.5.6} \end{equation}
This result establishes that the applied loads, if in equilibrium, perform zero work on any rigid motion.

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In the dynamic field the vector field \(\vec{v} \) is a velocity field. Therefore we will not speak of works but of external power and internal power. Furthermore, the external power will equal the sum of internal power and variation of kinetic energy.

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