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module:Beam calculations

page:The effect of a load on the beam

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 effect of a load on a beam

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.

Initial thoughts

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).



Text only version

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    If a load of magnitude 10 is positioned at one end of the beam (d = 0, say), in what proportion do the two banks bear its weight? (Remember that the weight of the beam is still shared equally between the two banks.)
    If the load moves to the centre of the bridge, how is the burden of its weight split between the two banks?
    If d = 3, how much of the 10 units of the load is carried by the bank nearest it, and how much is carried by the more distant bank?
    In general, if V = 10 and the load is positioned a distance d from the left-hand end, write down a formula for Fa and Fb.
    See if your formula gives the correct values for Fa and Fb when V = 10 and d = 6.4.
    Two challenges 
    1. Make Fa = 17.16 and Fb = 8.84, by varying the values of V and d. A little thought should give you the value of V first.
    2. Now take V = 20.
    Find where the load should be placed on the beam in order that the ratio of the forces at the banks is 1: 2 with Fa being the smaller.
    Try to explain your answer in terms of the shares of the beam's weight and the load that each bank carries.


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    When d = 0, the total support forces at the banks are in the ratio of 15 : 5 or 3 : 1, but when we take out the component of 5 at each end that is carrying the weight of the beam itself, we see that the entire weight of the additional load is being carried by the bank on which the load is resting. No great surprise there.
    When d = 5, the total support forces are equal, and the additional load is obviously being split equally between the two banks.
    When d = 3, the total support forces are 12 : 8, and the load is shared in the ratio of 7 : 3
    In general, Fa takes half the weight of the beam plus a share (10 – d) of the additional load.
    i.e. Fa = 5 + (10 – d) = 15 - d
    Then Fb = 20 – Fa = 5 + d
    When d = 6.4, Fa = 8.6 and Fb = 11.4
    Two challenges:
    Adding Fa and Fb gives V + W = 22.4, so V = 12.4
    Then d = 2.4
    Load should be placed so that d = 7.5
    This can be explained by the fact that both banks always carry half the weight of the beam itself, and then carry a fraction of any additional load, in inverse proportion to the distances of this load from that bank.
    When d = 7.5, the load is three quarters of the way across the beam, so its weight will be carried by the banks in the ratio of 1 : 3 – i.e. 5 : 15 for a load of size 20.
    Adding these components to the 5 for the weight of the beam gives 10 : 20 overall… the required ratio we were asked for.


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.

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