217. The frame may be further simplified by the following method of taking away double joints.

If to the frame A, B, C, ... the point D be connected by three bars DA, DB, DC, then the addition is over-rigid and either the bar AB, or BC may be taken away. Hence if we add a point and three bars we can take away a bar joining two of the points to which the new bars go*. Conversely we may take away a double joint if we add a bar joining two of the points to which the old bars went, the rigidity of the frame

Fig. 137.

being unaffected by the operation.

Any frame may thus be reduced step by step by the removal of single and double joints until it becomes sufficiently simple to see by inspection whether it be just rigid, over-rigid, or non-rigid.

218. If we take the frame, Fig. 138, and reduce it in this manner step by step, it will be found to be over-rigid. The separate stages are shewn by the successive figures, the dotted lines indicating the bars added.

Fig. 138.

The last figure is evidently over-rigid.

*The frame will not be perfectly rigid if the point D lies on a certain conic through ABC. If it does the point D will allow of infinitely small displacements, just as it would if joined by two bars to two fixed points A and B so that it lies in a straight line with them. In either case the stresses are not determinate.

In the frame, Fig. 139, the cutting away is done in the order indicated, the dotted lines again shewing the new bars added; the second figure gives the result of the process; it is just rigid.

3

Fig. 139.

219. In many frames actually in use it will be found, on numbering the spaces in the usual way, that several of the bars are denoted by more than one pair of numbers and consequently the stress diagram will contain more lines than are really necessary.

To shew how the unnecessary work that this entails may to some extent be avoided, we will find the stresses in the various bars of the Bollman Bridge Truss.

220. The Bollman Truss. The spaces being numbered as shewn, it is seen that AC has the numbers 7, 10 and 9, 8.

ED has 10, 9; 87 and 06. Consequently the stress diagram will have more lines than are really necessary. We will first shew how the stresses may be found and then how the frame may be supposed modified in order to reduce the number of lines drawn to the smallest number possible.

The spaces being numbered as in fig. (1), the loads 12, 23 are drawn to scale and the line 13 bisected at O to give the reactions at the supports. Since the bars 1, 10 and 27 are horizontal, the bar AC must transmit the whole

load 12 vertically downwards to the point C. We have then at C to decompose the load 12 in known directions CE and CF, hence through 0 in stress diagram draw 99' and 88' parallel to 09 and 08 in frame; if now we draw a

(3) Modified frame with only one crossing of bars

(4) Stress Diagram

Fig. 140.

line parallel to 12 cutting these lines such that the intercept is equal in length to the load 12, we shall obtain the vector-polygon for the point C. This can be done in two ways as seen from fig. (2). That the one on the left (098) is correct and the one on the right (08'9') incorrect is seen when we take the next step. 1, 10 is horizontal, so is 27; 10, 9 and 87 must give the same stress, and this can only be the case when 8 is below 9 in the stress diagram. The rest of the diagram offers no new difficulty.

In order to have less lines in the stress diagram we may modify the framework as in fig. (3) for purposes of numbering. It will be noticed that by withdrawing the

point D within the triangle ECF, the number of unpinned crossings is reduced to one and the number of spaces to eight; this reduces the number of stresses given by the vector-polygon in duplicate to two.

221. Cross Girders. In actual bridge girders we very often find a double set of sloping bars and no vertical ones (cp. the Warren and N girders). The figure will give an idea of the kind of girder alluded to; they are generally known as Cross-girders.

First of all it should be noticed that such frames are

over-rigid. This may be seen by applying the method of $217. Removing the three bars which meet at A and inserting a new bar between B and C, we see that B and C are fixed relatively to E and F without the additional bar, hence the frame is over-rigid. This being the case the stresses in the bars are indeterminate unless the stress in one bar is known. Some assumption is therefore necessary in order to solve the problem, the one usually made is that the stresses in the end bars AD and A'D' are due to the loads at C, G', C', and not to those at F and F". This is equivalent to saying that we may suppose the girder divided up into two elementary girders as shewn in sketch, when it is evident that the loads at FF" cannot affect the stress in AD or A'D'.

With this assumption, the stresses may be found by drawing the stress diagram for the original frame in the

usual manner. We may also find the stresses from the elementary girders, and where one bar forms part of both frames, the stresses determined must be added together.

222. General Method of Trial for Complicated Frames. In complicated frames it is sometimes impossible to find a starting point for our stress diagram, owing to there never being less than three bars meeting at a point.

In such cases we fall back on the method of trial; it will be seen that after two trials at the most we can draw our stress diagram correctly.