Virtual Work

Whilst the calculation by geometry of the displacement for the very simplest truss of Figure (2.2) is simple enough, it can get quite complex for real life trusses. A much more systematic approach is by Virtual Force. Some further definitions are required before we can state the principle of Virtual Work and then Virtual Forces as well as Virtual Displacements. Fortunately these definitions are really simple to state and simple to apply once the initial shock is overcome. It is necessary to refer collectively to all forces acting on a given truss: the applied loads, the reactions and the internal forces that the pins transmit onto the members. We call this collection of forces a Force Field.

When all forces are in equilibrium with reactions and the forces by the pins onto the bars are in equilibrium with the forces that the bars exert onto the members, this collection is an Equilibrium Force Field.

The displacements of a truss and all bar extensions or contractions are defined by the displacement of the pins. We call this collection a Displacement Field. When the displacements satisfy all conditions of geometry, they constitute a Geometrically Compatible Displacement Field.

Unless noted otherwise, we will refer to Equilibrium Force Field simply as the Force Field. Similarly, we shall refer to Geometrically Compatible Displacement Field simply as Displacement Field.

We have defined the Work Product as the sum of forces multiplied by their displacements. We shall separate the Work Product into two parts, The Internal Work Product, denoted by $U$ and the External Work Product, denoted by $W$.

The forces exerted onto the bars by the pins cause the internal work product. Since each bar has an axial force N, which acts at both ends of the bar, but in opposite directions, and the extension of the bar is given by the difference of the displacements measured in the direction of the bar at its two ends, each bar contributes $N \cdot e$ to the internal work product. This is expressed by the following:

$$U = \Sigma N \cdot e$$ (2.4)

Similarly, if all the forces, including the reactions, are denoted by P, and their corresponding displacements as u, the equation for the external work product can be written as follows:

$$W = \Sigma P \cdot u$$ (2.5)

It is fairly obvious that the work product of an equilibrium force field and a compatible displacement field must be zero. Some clarification of the internal force field, however, is necessary.

Let us denote the internal force field that the bars exert onto the pins by $\bar U'$. Then the above paragraph of describing equilibrium in terms of work product can be written as

$$W + U' = 0 \textrm{ or } W = - U'$$ (2.6)

Since the forces exerted by the pins onto the bars are in opposite direction, but of the same magnitude as the forces of the bars onto the pins, $U' = -U$ and, therefore,

$$W = U$$ (2.7)

This is the Equation of Work Product employed throughout this text.

Basically, the rather long winded talk about U' and U is an attempt to underline the different forms of possible representation of internal work product, each the same, but with different sign. Which one is used is largely a matter of taste - in this text the Equation (2.7), which equates the internal work product with the external work product, is used.