order with respect to the joint on which the bar acts as we have taken the letters on each side of the external forces acting on the frame. Then along the line in the stress diagram which is named after the bar, from the first letters towards the second gives the way of the stress's action with respect to the joint under consideration. If the way is towards the joint, the stress in the bar is compression or push, and the bar is called a strut. If the way is away from the joint, the stress is a tension or pull, and the bar is called a tie. Figs. 184 and 185 show a somewhat complicated case. The loads are

not now vertical. Loads which are not vertical are usually the result of wind pressure on roofs, etc. In this case the load line is no longer vertical, but broken into units of various lengths and directions, according to the value and direction of the loads to which they are respectively parallel. ABCDE is the load line. The two lines BH and CH fix the point H, while GH and GD fix the point G similarly. We are also assuming, for demonstration, that one end of the frame rests upon rollers, like the large bridge trusses. In this case we obviously have a vertical reaction and want to find its value, but at A we have a reaction of which we know neither value nor direction. Then by

ED, having settled the point E, draw EF vertical, because we know it as a vertical reaction. Then the line FA gives both the direction and magnitude of the reaction at A.

In big roofs we also very often employ a roller bearing at one end. In all cases the free rafter should be turned towards the prevailing wind (usually west in Britain), because by neglecting to do so unnecessary stresses are set up in the truss.

In figs. 186 and 187 we show the diagrams for a roof in which the wind pressure is taken into account. Such a form of roof is not suited for practical construction, but the form of the frame does not materially alter the stresses therein, as they depend merely on the load. It is assumed to be securely

anchored at both ends. We do not know the direction of the reactions, and have to find them. The reader should now be able to understand the stress diagram without further explanation.

The best way to treat hanging loads which we referred to above is to produce them vertically upwards in the same direction as that of their action, and from the points at which they act until the lines cut the rafters. Then simply treat the frame as having extra loads applied at these points. See fig. 188.

Take a roof truss having a load line, say, similar to fig. 189.

We want to

find the reaction on the supports. If the loading is symmetrical the middistance X between CD will give this.

Suppose, now, we have a roof truss of the bowstring type, we want to draw a stress diagram. In this case we always assume the wind pressure acting tangential to the curved surface. At the top we shall probably have a horizontal tie, as in fig. 190. It is wise to find the line which gives the stress on the rod early in the diagram. It will make the rest of the work come

easier.

We will now describe what is called the funicular polygon, as applied to a loaded beam, and which, in such cases, is often called the bendingmoment diagram. Take the beam loaded as in fig. 191. Letter it the same way as we have described. Draw the load line ae, and take any point O at a convenient distance from ae. This point is called a pole. Join o with

ao

abcde. Take any point K directly below R1, draw KL parallel to cutting the load line AB produced. In the same way draw LM, MN, etc., and complete the diagram. This diagram is then called the funicular polygon, while the diagram in fig. 192 is called the polar diagram. It has already

been explained that the BM at any section of a beam is the moment of the resultant of all the forces which act on one side or other of the section, then by means of the diagrams we can ascertain the bending moment at any point in a frame.

To do this we multiply the ordinate of the point by the polar distance.

2,Tons

8

9

6.8

3

5-7

3

In fig. 191, at the point y, xy is the ordinate and Oh drawn horizontal is the polar distance... the bending moment at the point x = xy × Oh. There is another way of determining the stresses in the members of a structure by what are known as Ritter's methods of sections. Its fundamental principle is that the equality of moments may be extended to any number of forces, but the sum of the moments of the forces tending to cause motion in one direction must always be equal to the sum of the moments tending to cause motion in the opposite direction. Their treatment is almost outside the scope of this work, but interested readers would do well by referring to Professor Jamieson's Applied Mechanics, vol. ii. p. 623. (They are much used by designers and draughtsmen of structural steelwork.)

FIGS. 200 and 201.

We cannot very well say anything more in general reference to graphic statics as applied to structures. The principle is the same in all of them; they want a little care and forethought, and, above all, proper notation. We shall now give a few more typical examples to illustrate our previous discussion.

Case I. A Latticed Girder with a uniform Load on the top Boom (see figs. 193, 194, 195, 196).-On the vertical line XX set off to scale the units of load at each apex. As the girder has 8 divisions, these equal total load. Draw from each point so set off lines parallel to the top and bottom booms, and diagonal lines parallel to the bracing. Then the stresses can be measured as usual by the scale of tons. In fig. 196, draw BC equal and parallel to XY, and draw BA at any convenient angle. Make B1 = load on 1, 1-2 loads on 2, and so on up to point 7. Join points 7 and C, and through the points 1, 2, 3, 4, 5, 6 draw lines parallel to it to cut the line BC as shown. Then draw the parallel lines and start at C, drawing the diagonals to each of them in turn till we finish up. These lines then will give us the greatest stresses due to the maximum travelling load.

10 Tons

=

10

2

13

fig 204

18

12

10

FIG. 202.

FIG. 203.

Case II. A Lattice Girder with a Load unequally distributed (see figs. 197, 198, 199). Find the reactions on the supports in the usual manner, and lay off these loads on the vertical lines. Draw the stress lines parallel to each member as shown. Owing to the heavy loads on the right-hand side of the girder, the stress on MN becomes compressive. We should then, in all probability, introduce an extra member, shown by the dotted line.

Case III. A Braced Beam (as shown in figs. 200 and 201).—The diagrams are quite self-explanatory.

Case IV. Stress Diagram for a Lattice Stanchion. In order to work out the stress diagram, the central load of 20 tons must be divided into two loads of 10 tons each, acting at the outer angles of the stanchion. The frame diagram will then be as shown in fig. 202, from which the stress diagram, fig. 203, may be easily worked out, while fig. 204 shows part of the stress

diagram enlarged to four times its original size. In calculating the members it should be noted that, although the member 2-5 has no stress from the WL 20 × 2 × 12 reciprocal diagram, it will have to resist a BM of 4

4

=

120 ton-ins. If the stanchion consists of uprights and bracing all on one

plane at right angies to the bracing, and if the stanchion is a 4-sided one and each side braced as shown, the corner members must be calculated as columns between the points of connection with the braces.

As a good example of what might be an anyday occurrence in an engineer's office, we give the following worked out very fully.

Required the scantlings of a timber bridge of the form in fig. 205. The distances between girders 7 ft., the load 6 small waggons 6 ft. long, weighing 2 tons each, such as might be required on a large cleansing depôt of a big city. Figs. 206 and 207 show the diagrams.