Now that we are able to work out the support forces for the beam itself, let us consider the effect of an additional load upon the beam. We may consider our mathematical load positioned at various points upon the beam, and to have various magnitudes.
The diagram shows a 'point' load of magnitude V positioned at a distance d from the left end of the beam. The beam is taken as uniform for simplicity, so its weight acts at its mid-point, M.
This will form the mathematical model for a load such as a gymnast on a beam or train passing over a bridge.
We can (intuitively) assume the following:
- The larger the combined weight of the load and the beam, the larger the support forces at the bank will need to be.
- The sum of the support forces at the bank will equal the combined weight of the load and the beam.
- If the combined load is nearer one bank, that bank will bear a larger fraction of the load.
Doing the calculations
In order for us to perform actual calculations, we must assign values to some of the variables.
We therefore take the beam to have a uniform weight distribution of 1 per unit length and a total length (and hence also weight) of 10 (arbitrary units). This simplifies the formulas, but can easily be changed to any value we want later on, when we have actual data to deal with.
The only things that you can change in the activity are the values of the load (V) and its position (d).
In the next section, we will actually carry out the calculations to derive formulas for the sizes of Fa and Fb, as they were done on the previous section. The formulas for Fa and Fb will contain d and V, as these are the ‘input’ parameters for the model. We will therefore be in a position to analyse the effects of changes in d and V with complete precision.