The designs at Figs. 59 to 63 are distinguished from those at Figs. 64 to 67 by having a horizontal tie-beam. In the latter group, and in all designs similarly destitute of the horizontal tie at the foot of the rafters, the strains are much greater than in those having the tie, unless the truss be protected by exterior resistance, such as may be afforded by competent buttresses.
To the uninitiated it may appear preferable, in Fig. 64, to extend the inclined ties to the rafters, as shown by the dotted lines. But this would not be beneficial; on the contrary, it would be injurious. The point of the rafter where the tie would be attached is near the middle of its length, and consequently is a point the least capable of resisting transverse strains. The weight of the roofing itself tends to bend the rafter; and the inclined tie, were it attached to the rafter, would, by its tension, have a tendency to increase this bending. As a necessary consequence, the feet of the rafters would separate, and the ridge descend.
In Fig. 65 the inclined ties are extended to the rafters; but here the horizontal strut or straining beam, located at the points of contact between the ties and rafters, counteracts the bending tendency of the rafters and renders these points stable. In this design, therefore, and only in such designs, it is permissible to extend the ties through to the rafters. Even here it is not advisable to do so, because of the increased strain produced. (See Figs, 77 and 79.) The design in Fig. 64, 66, or 67 is to be preferred to that in Fig. 65.
204. - Force Diagram: Load upon Each Support. - By a comparison of the force diagrams hereinafter given, of each of the foregoing designs, we may see that the strains in the trusses without horizontal tie-beams at the feet of the rafters are greatly in excess of those having the tie. In constructing these diagrams, the first step is to ascertain the reaction of, or load carried by, each of the supports at the ends of the truss. In symmetrically loaded trusses, the weight upon each support is always just one half of the whole load.
205. - Force Diagram for Truss in Fig. 59. - To obtain the force diagram appropriate to the design in Fig. 59, first letter the figure as directed in Art. 195, and as in Fig. 68. Then draw a vertical line, E F (Fig. 69), equal to the weight W at the apex of roof; or (which is the same thing in effect) equal to the sum of the two loads of the roof, one extending on each side of W half-way to the foot of the rafter. Divide E F into two equal parts at G. Make G C and G D each equal to one half of the weight N. Now, since E G is equal to one half of the upper load, and G D to one half of the lower load, therefore their sum, E G+ G D = E D, is equal to one half of the total load, or to the reaction of each support, E or F. From D draw DA parallel with D A of Fig. 68, and from E draw E A parallel with E A of Fig. 68. The three lines of the triangle A E D represent the strains, respectively, in the three lines converging at the point A D E of Fig.68. Draw the other lines of the diagram parallel with the lines of
Fig. 68, and as directed in Arts. 195 and 197. The various lines of Fig. 69 will represent the forces in the corresponding lines of Fig. 68; bearing in mind (Art. 195.) that while a line in the force diagram is designated in the usual manner by the letters at the two ends of it, a line of the frame diagram is designated by the two letters between which it passes. Thus, the horizontal lines A D, the vertical lines A B, and the inclined lines A E have these letters at their ends in Fig. 69, while they pass between these letters in Fig. 68.
206. - Force Diagram for Truss in Fig. 60. - For this truss we have, in Fig. 70, a like design, repeated and lettered as required. We here have one load on the tie-beam, and three loads above the truss: one on each rafter and one at the ridge. In the force diagram, Fig. 71, make G H, H J and J K, by any convenient scale, equal respectively to the weights G H, H J, and J K of Fig. 70. Divide G K into two equal parts at L. Make L E and L F each equal to one half the weight E F (Fig. 70). Then G F is equal to one half the total load, or to the load upon the support G (Art. 205). Complete the diagram by drawing its several lines parallel with the lines of Fig. 70, as indicated by the letters (see Art. 205), commencing with G F, the load on the support G (Fig. 70). Draw from F and G the two lines FA and G A parallel with these lines in Fig. 70. Their point of intersection defines the point A. From this the several points B, C, and D are developed, and the figure completed. Then the lines in Fig. 71 will represent the forces in the corresponding lines of Fig. 70, as indicated by the lettering. (See Art. 195.)