Coming to framed structures, the student must thoroughly understand, and never lose sight of the fact, that " the forces acting at any point, to be in equilibrium (which is the foundation of the structure), must complete and form a triangle or polygon whose sides are parallel to the directions of the forces, and therefore are equivalent to their magnitudes."
This theorem is the foundation of all calculations and diagrams, as will hereinafter be illustrated (vide Figs. 979 to 984), though these reciprocals (as they are sometimes called) will, of course, be pointed out, as the diagrams are constructed, in the usual way.
The first procedure is to locate the weight at various points, and then ascertain the reactions (as should always be the first thought in whatever class of work), the polygon of forces (as Fig.- 990) being used to determine the necessary reactions. But when the loading is vertical and equal on both slopes of the roof, each reaction will be equal to one half the total load, so that the polygon of forces becomes a straight line, obviating further trouble, which, it will hereinafter be noticed, is present in unequal loading.
Taking a "king post" truss as our first form, we proceed as follows: -
Set up any line vertically in any position as a line of loads, and assuming that the truss carries 4 tons weight, it will be divided as - w/4 w/4 w/8 w/8 and w/8, on diagram 978; so that we shall have to support 1 ton of it at each purlin and the ridge, with 1/2 ton at each of the eaves to equally divide the 8 tons. Therefore the reactions will be:-
Starting from A on the line of loads, we set off to scale A B = w1/8; B C = w2/4; C D = w3/4; D E = w4/4; EF = w5/8 and, bisecting A F in G, we have A G and G F equivalent to R1 and R2 - i.e, 40 cwt. each, measured from each end.
From B we then draw B H, and from G G H, parallel to Z Y and Y R, respectively; giving us A B H G A as the reciprocal of the forces at the joint Y, while B H and H G give us (according to scale) the strains on Y V and Y R respectively.
Next, taking the joint V, we draw H J and J C parallel to V R and V Z, giving us B C J H B as the reciprocal of V, with H J and J C equivalent to the strains on V R and V Z.
Following with joint R, we draw J K parallel to R Z, making G H K J H G reciprocal to R, with J K equivalent to R Z, which gives us the strains on half the members; and since the loading is equal, the strains are equal on the other half, which is completed on the diagram; and all that now remains to be done is to scale off the strains and calculate the material required to resist them in the usual way.
Figs. 979 to 984 give the polygons of forces at the various joints in equilibrium; the arrows indicating the directions, for all forces must go one way round the polygon, beginning at any one point; though they must continue until they close to be in equilibrium.
Each Fig.ure is a "reciprocal" of the joint of which it carries the corresponding letters, and each line forming the Fig.ure represents a force acting at the joint as A B (Fig. 979) represents that at the eaves, B H that on the lower part of the rafter, H G that on part of the tie-rod, and A B the upward reaction of the support.
This subject will require careful study and reasoning, for it must be clearly understood and mastered, as this principle is, as it were, the very foundation of our work.
The amounts of the strains are Fig.ured on the polygons, which are all drawn to scale similar to the diagram 978. This must always be done most carefully and accurately, for slight inaccuracies often end in large errors, on account of the dependence of one line on another.
It will be noticed that the weights and reactions play their part and belong to these polygons, for they are known forces, which close the sides, and where they occur they have been marked accordingly.