Fig. 985 represents the diagram for a "queen post" truss ( w/12 being equivalent to 10 cwt. and w/6 20 cwt), which should explain itself; abed a being the reciprocal of A, bcefb of B, dcegd of E, f h j g e f of C, hkjh of D; the others being similar, as the truss is equally loaded.

A larger truss, with more framing, is depicted in Fig. 986, with diagram, from which it will be gathered that a bed a is the reciprocal of A, b c e f b that of B, d c e g d of C, f e g h j f of D; while jhklj and d g k h g d represent E and F respectively. The other part has been left out intentionally, to save confusion; it is practically a repetition of the other half.

Queen Post Truss PracticalBuildingConstruction01 788

Fig. 985.

Fig. 987 represents a similar truss, with a cambered tie rod, the diagram for which will illustrate what a difference it makes compared with the straight rod The student must work out, or rather "identify the reciprocals" for himself. It is precisely similar to Fig. 986.

Queen Post Truss PracticalBuildingConstruction01 789

Fig. 986.

Queen Post Truss PracticalBuildingConstruction01 790

Fig. 987.

The next form of roof truss treated will be as Fig. 988; and as this is a truss in common use, the writer will take this opportunity of dealing with unequal loading, hereinafter using this truss.

Having located our weights, and ascertained the reactions in the usual way, as hereinbefore explained, we draw the diagram as shown in Fig. 988, from the vertical load line, and parallel to the members of the truss; the reciprocals formed thereby representing the joints as follows: - acgoa = A; cghdc=B; dhtled=C;othgo=F;otlko = G;elkft = D; and bfkob = E; all of them closing together as required. The strains can be found by scaling off the corresponding members in the diagram in the ordinary way.

Hitherto no distinction has been made between lines in compression and those in tension, because the distinction was too apparent; but they are shown in Figs. 988 and 989 by thick and thin lines respectively.

Special Loading

It will be seen that in Fig. 989 two loads of 20 cwt. are hung from points F and G of the truss in addition to the previous loading, which, of course, alters the diagram considerably as shown, and increases the reactions which remain equal - i.e., 60 cwt. each, as compared with 40 cwt. on Fig. 988.

On the usual line of loads mark off the loads on the truss, neglecting the suspended loads, and from each end of the line set up and down the reactions to scale, making a v and by, which cross each other, as will be noticed. This done, we proceed in the usual manner, drawing eg from C parallel to A B, and from V (the reaction point) drawing v g parallel to A F, making aegva the reciprocal of A; while further parallels will give us cghdc for B, dhtled for C, and oygh to for F,as compared with othgo on Fig. 988, the latter containing four forces, while this contains five. The other joints at D E and G are similar to A B and F, while the line to is common to both halves of the truss and the joints F and G.

Unsymmetrical Loading

When we come to unequal and unsymme-ttrical loading we meet with a difficulty at the very beginning; inasmuch as we have hitherto used no means, except calculations, to ascertain the reactions; and this is done by means of a funicular and polygon of forces.

A line is set up, just as a "load" line - viz., as Z Y (see Fig. 990) - and on it are marked off the loads, as so, op, and py, equivalent to W1, W2, and W3, on the truss; after which " any" point X is taken, somewhere level with the centre of the load line ZY, but no less than its length away, and we join X 0, Xp, Xy, which are called vectors; then drop perpendiculars from the points of loading and reaction on the truss, after which we draw 1.2, 2.3, and 3.4, parallel to Xp, Xo,Xs, on the funicular to connect these perpendiculars; and by joining 1.4 we complete the polygon of forces; and parallel to the latter we draw X R on the funicular from X to the load line, the intersection determining the magnitude of the reactions R1 and R2.

The reactions are then transferred to the ordinary load line, from which the diagram is drawn in the usual way; the resulting reciprocals abeoa, befcb, cftkdc, oefto, otko, dkod, representing the forces at A, B, C, D, E, and F. The diagram shows that there is no strain on œ G, and that the strains on C G and G F are alike.

Having mastered these few examples, the student ought to be able to deal with all ordinary strains on building structures.

Unsymmetrical Loading PracticalBuildingConstruction01 791

Fig. 988.

Unsymmetrical Loading PracticalBuildingConstruction01 792

Fig. 989.

Unsymmetrical Loading PracticalBuildingConstruction01 793

Fig. 990.