Fig. 251a.

If, however, one foot only were bolted down and the other foot were placed on rollers, to allow lor expansion and contraetion, we should have to make, two diagrams, with the wind from left and right respectively, as it makes a decided difference to the strains as shown in Figures 250 and 251. In Figure 250 the wind is from the right with rollers under the left or opposite foot of truss; in Figure 25] the wind is from the left and the rollers are under the left (or same) foot of. truss. It will be readily seen that in both cases the (left) roller foot will not oppose any tendency to slide due to the wind, the right foot must therefore be bolted down to resist this tendency in both cases. We can readily see therefore that in Figure 251, the member HF or 0 E must be a brace to keep the truss from sliding towards the right foot, whereas in Figure 250, it must be a tie to hold back the truss from sliding away from the right foot. The strain diagrams will show this to be the case. To obtain the reactions we proceed as before, so far as the central wind pressure (1/2) and lines P 2 and 2 Q are concerned, but it is evident that the reaction P will be vertical, while the reaction Q will no longer be in direct opposition to the wind as shown by dotted line, as it is deflected from this line, owing to the horizontal (sliding) strain to be taken up from the other foot. We therefore draw in Figure 250a the load line ab c d as before, also the reactions d ox := a and o, a =p. Through a draw the vertical projection a o of a ol - (that is draw o, 0 horizontally and a o vertically) -then will o a be the amount of vertical reaction p and d o the direction and amount of vertical reaction q. The rest of the strain diagram is easily drawn, remembering to take the foot joint first and then to skip to the apex joint.

Truss with rollers.

To analyze Figure 251, it must be treated in exactly the same way and it offers no difficulty. We notice that Figure 250 has strains similar to those in Figure 240 excepting the upper part of rafter on the wind side, which is now in tension. In Figure 251, however, things are greatly changed. The upper end of rafter becomes a tie, both ties become struts and the short strut G F (or JH in Figure 240) now becomes a tie.

Wind strains and steady loads can be calculated from one strain in diagram, as shown in Figure 252 and Figure 252a. This is a combination of the conditions obtaining in Figures 240 and 249 and the strain diagram of their respective strain diagrams, Figures 243 and 219a. By simply following around the arrows, as there shown (first obtaining the amounts of A B and K O) the diagram offers no difficulty whatever. If an actual example were Figured out, first separately, as shown in Figures 240 and 249, and then in combination, as shown in Figure 252, the result would be the same, remembering to make all compressive strains positive and all tension strains negative and to obtain the arithmetical result of the consequent additions or subtractions. A practical example will be given in the next chapter.

Wind opposite rollers.

Wind on roller side.

Wind and load in one diagram.

Fig. 252 a.

Where the roof rafter or top chord of a truss is not in one straight line, but is made-up of a series of differently inclined lines, as in a

FIC.253a broken roof, or in a circular or arched truss, the wind pressure will, of course, be at different angles too, each at right angles to its respective surface. We can in such a case, work out separately the reaction due to the wind pressure on each surface, and then obtain the resultant

Wind on broken roof.

(reaction) of all these lesser ones, as explained on p. 71, Vol. I, or we can combine them all into one diagram. In Figure 253, we have a mansard roof with wind pressure from the right. We draw the neutral axis (or central lines) of wind pressure on each rafter prolong them to their intersections with the horizontal and can thus Figure out the respective reactions due to each. In the strain diagram Figure 253a we now draw c e parallel and equal to total wind pressure (3) on rafter FD; and a c parallel and equal to total wind pressure (1) on rafter B G. Draw e a and it will be the direction and amount (sum of) the actual reactions. Now to get each reaction separately make eo11= to reaction q of wind pressure (3) on F D; and c o1 = to reaction q of wind pressure on G D; of course o11 c and o1a will equal the reactions p due to the respective wind forces. Or we might divide e c and c a so that e o11 : o11 c = P4 : 4 Q and co1: o1 a = P 2 : 2 Q We now draw o11 o parallel c a and if we have drawn correctly o o, must be parallel e c. Having found o the rest of the diagram presents no difficulty. We make ab = AB or pressure at apex; bc = BC pressure at joint C due to wind on GB; cd=CD pressure on same joint C due to wind on FD and finally d e = D E pressure at foot and proceed with the other lines.

Were we to make a strain diagram for a steady load on all joints we should find similar strains on all members except I H. This is a compression member of the truss, but becomes subjected to tension when the wind blows from the right.