We have to first determine the position of the pole 0 of the vector-polygon. This may be done by aid of the theory of moments explained in §§ 170, 171, thus—

If we imagine 04 cut through, then we must apply a certain force along the bar to maintain equilibrium with

the forces in 45, 67, 08, 78. We know that for equilibrium the sum of the moments must be zero for all poles. Taking moments about B, the unknown moments of 78 and 08 vanish.

All the forces except 04 and all the distances are known, hence 04 can be determined.

For instance, suppose the dip to be 20 ft., the number of bars to be as in figure and 8 ft. apart and that each bar bears a load of 5 tons, then if x is the unknown force in 04,

x x 20 5 (24+ 16 +8)

or x = 12.

Hence the tension in 04 is 12 tons.

We now know 04 in tons, this must be set off along

the horizontal through 4 to the same scale as the loads. The lines 05, 06... of the vector-polygon can now be drawn, and the corresponding links drawn for the chain.

(2) If the number of tie-rods be odd there will be no middle link to the chain, but the stresses in the two lowest links can be determined in a manner similar to the first case. Suppose P the lowest point and that the chain is cut through just to the right of P. The force that would have to be applied along the cut link to maintain equilibrium can be supposed resolved into two components, vertical and horizontal. The magnitude of the first is half the load on the middle tie-rod, and by taking moments about B we can, as in the first case, find the horizontal component. The stress in one link of the chain being thus found we can find the position of the pole of the vectorpolygon and thence the positions of the various links.

202. The suspension bridge problem is a particular case of the string or link polygon (funicular polygon).

...

Suppose a string is fastened to two fixed points A and B and that forces act at the points C, D, E, and F, then the string will take up some position of equilibrium and the forces (see figure) 12, 23, will be in equilibrium with the reactions at A and B. The stress diagram for this figure offers no difficulty. It should however be remarked that the funicular polygon cannot be drawn at random, for if a given set of forces be applied at definite points along the string, the string will take up a perfectly definite shape. This is brought out well when we come to draw the stress diagram, for the point O can be determined by considering the equilibrium of either the point C, D, E or F. Hence given the form of the string A, C, D, E, F, B, the force 12 acting at C and the directions of the forces at D, E, F, the magnitude of these forces could have been predicted.

The following is one of the many problems that the string polygon gives rise to.

Given two fixed points A and B from which the string is to be suspended and the lines of action and the magnitudes of a number of weights to be suspended from it, then if in addition the direction of the string at A or B be given, the string polygon can be constructed.

The process would be as follows-let 12, 23... be the forces, find by moments or the link-polygon the position

of the resultant R, then since the direction of one of the components at A and B is supposed given, the direction of the other can be found and therefore the position of the pole 0. Suppose direction of string at A is given by AC, then join the intersection C of this with R to the point B and through the points 1 and 5 of the stress diagram draw 10 and 50 parallel to AC and CB to determine 0, then by joining 0 to 2, 3, 4, 5 the directions of the string in the spaces 2, 3 and 4 are found.

The reactions at A and B are given by the lengths of the lines 01 and 50, while 02, 03, ... give the tensions in the corresponding parts of the funicular polygon.

The apparatus necessary for the practical demonstration of these and similar problems being of the simplest character, the student should not rest satisfied until he has verified these and similar constructions.

203. For a Roof-girder as shown in Fig. 131, we find the stresses due to the three loads W as follows.

(It should be remarked that although such bars as 18 and 27 are in a straight line, yet we suppose them jointed at the point where 78 meets them.)

Since the loads are vertical and symmetrical, we know that the supporting forces at A and B must be each 3 W equal to and upwards.

2

Start then with the force 01 upwards and consider the stresses in the bars 18 and 80 round A. Through 1 of the vector-polygon draw 18 parallel to the bar 18 and through 0 draw 08 parallel to the bar 08. Hence since 018 is a closed vector-polygon, the bar 18 pushes at A and is therefore in compression, 80 pulls at A and is therefore in tension.

It is no use at the present stage trying to find the forces at C, for we only know the force in 08 and by symmetry that in 05, and we cannot find the components of these in three collinear directions, viz. 78, 67, 56. A

similar argument applies to the forces at D. At E however we know the force in 18 and also W, and we can find the components in two given directions.

In the vector-polygon then set off 12 downwards to represent W, then 27 || to bar 27 and 87 || to bar 87, this determines the point 7 and the sense of the polygon for E is 12781.

Hence 27 pushes at E and is therefore in compression, 78 pushes and is in compression, 81 pushes and is in compression as found before.

Now proceed to the point D, set off 23 vertically downwards to represent W, 76 || to bar 76 and 36 || to bar 36, these determine the point 6, and the sense of the vector-polygon for D is 23672. Hence 67 pulls at D and that bar is in tension.