Michael Wood footbridge over motorway M5

MichaelWood35HC.jpg (130950 bytes)This photograph shows the footbridge over the M5 motorway at Michael Wood services. It is based on Warren trusses with inclined supports. The approach ramps are constructed in the same way. The members are all of square section steel, making for a clean and neat appearance.

This page contains calculations based on a simplified model of the bridge. The members are all assumed to be pin jointed, unlike those in the real bridge, which are welded together. Firstly we consider a simple truss with vertical supports.

The next picture shows the forces in the members graphically. The vertical axis represents force. The white line represents the zero of force. The red lines represent the top chord, which is in compression, while the blue ones represent the bottom chord in tension. The purple lines represent the sloping members, which are alternately in tension and compression.

MWTrussB.gif (2805 bytes)

The next diagram shows the same data, but with compression as positive and tension as negative.

MWTrussBB.gif (2937 bytes)

Next we incline the supports from the vertical.

MWTrussC.gif (2855 bytes)

The effect is to reduce the tension in the lower chord. In the next picture we align the supports with the sloping struts in the truss. The forces in the outer parts of the lower chord are now (of course) zero (blue lines).

MWTrussD.gif (2970 bytes)

Finally, we incline the supports even more, and the outer parts of the lower chord now experience negative tension, that is, they are now in compression. Were this a concrete bridge, we could regard this as a kind of prestressing for those parts.

MWTrussE.gif (3092 bytes)

Next we add the two-panel extensions at each end of the truss, but we ignore any forces from the ramps. In this case the extensions act as cantilevers.

Again we start with vertical supports and then try sloping ones. By comparing the results below with the first diagram, repeated below, we can see the effects of the extensions.

Simple beam

Beam with extensions

MWTrussF.gif (3477 bytes)

Beam with extensions and sloping supports

MWTrussG.gif (3506 bytes)

We could go even further with these extensions, by anchoring them in some way at their extremities, and we could apply upward or downward forces at these anchorages to modify the forces in the main beam. We could even break the original truss in two places, creating a cantilever bridge with a suspended beam at the centre.

These diagrams hint at the versatility of multiple beams, where one can employ separate beams or continue them across the piers. For a long bridge with many piers, it is possible to pull the top chords together at each pier, and push the bottom chords apart. This case differs from the case treated above, because the required forces do not occur automatically. In the bridge we just looked at, the forces were generated by the overhanging weights, but in joining two beams, the generated forces depend very critically on the lengths of the members used to make the join. This occurs because of the great stiffness of the metal.


MWTrussThreeBeamsA.gif (5901 bytes)


Treatment of multi-span beam bridges

Consider a bridge with numerous beams, such as a truss bridge, of which three spans are shown above. The red and blue lines represent the forces in the lower chords and upper chords respectively, the white line representing the zero of force. In such a bridge, no use is made of the proximity of the trusses at the piers. We will next consider a similar bridge in which the top chords are joined, so that there is continuity along the top. We will also assume that the beams touch at the bottom. One span is shown below. The compression and tension forces are now plotted with opposite sign for clarity. The resemblance of the force diagram to the shape of a lenticular truss or of Brunel's Royal Albert bridge at Saltash is uncanny. Is this merely coincidental?

MWTrussContBeam0.gif (3469 bytes)

Next we set up the joining members at the top to be in tension, with a force F, and we set up a corresponding compressive force F between the bottom chords of the trusses.  If the weight of one truss panel is W, the applied force F in the diagram below is given by F = 8W/3.  Note the reduction in all the forces in the chords.

MWTrussContBeam2667.gif (3416 bytes)

Next we consider the case where F = 16W/3. The reduction is even greater, and we see that in the members at the ends, tension and compression have been interchanged.

MWTrussContBeam5333.gif (3352 bytes)

Finally, we set F = 8W.

MWTrussContBeam8000.gif (2801 bytes)

Thus by making use of the proximity of the beams, we have been able to reduce the forces in the chords while adding only a very small amount of extra weight. Easy, isn't it?

No, it isn't. It's not at all easy. There is a problem. When a Warren truss is lifted on to the piers, while the weight is slowly released by the crane, all the members deflect in exactly the right way to transfer the weight of all the parts to the piers, leaving the truss in equilibrium at all points. But when we try to add the extra member with the required force, we realise that the force in this member is a very strong function of its length. This is expressed by the very high value of Young's modulus.

In order to create the required force with sufficient accuracy, we would have to build the new member with unattainable accuracy. For example, suppose that the extra tie is 5 metres long, with a square cross section of 3 cm x 3 cm, and that we need to create a force of 10 tonnes weight with an inaccuracy of less than 0.1 tonne weight, ie 1% of the total. Let us also assume a Young's modulus of 200 GN per square metre.

10 tonnes = 10000 kg, and 10 tonnes weight = 10000 x 9.8 Newton = 98000 N,

and 0.1 tonne weight = 980 N.

Let the required maximum inaccuracy in length be m metres, caused by the 980 N change in force.

Young's modulus YM = stress / strain.

so YM = (Force/area) / (m/length)

m = force x length / (YM x area)

= 980 x 5 / (200,000,000,000 x 0.03 x 0.03)

= 4900 / 180,000,000 = 49 / 1,800,000

= 0.000027 m = 0.027 mm on a 5 metre length = 0.00054 %.

This clearly impractical, and would probably be invalidated by changes in temperature.

This situation is called static indeterminacy, meaning that if a structure contains more members than a certain value, we cannot simply calculate all the stresses. What is happening is that the distance between the ends of the top chords is being defined by two different parts of the structure, and something has to give. What gives is our design for the forces: in order to get the thing to fit, all the forces in the structure have to differ from those we wanted. This can be dangerous, especially if buckling is possible.

In order to control the stresses in indeterminate structures, we have to include measures to control the stresses during construction. During the building of a cable-stayed bridge, for example, the attachments would include a means of applying and measuring tension.

Another detail about continuous beams - if we make the beams deeper near the supports we can reduce the forces that connect the beams. Such beams are called haunched beams.

See also static indeterminacy. See also beam to arch.