The tops of all floor-beams are at the same height, and the bottoms of the intermediate floor-beams must be on a level with the bottom of the first section of the lower chord (see Fig. 17.4). This requires that the bottom of the intermediate floor-beams shall be 91 inches below the center line of pins, and this brings the top of the floor-beams (441 — 91) = 39 inches above the center line of the pins. Since the end floor-beam is 524 inches deep, back to back of angles, the lower flange will be (52+ 39) = 131 inches below the center line of pins. In case the end floor-beam does not rest directly upon the bearing plate or the masonry plate, the intervening space is filled out with a grillage of iron bars or a cast-steel pedestal, as shown in Figs. 197 and 19S.
The small plates upon the side of the shoe, going entirely around the pin, are called the shoe hinge-plates. These do not take any stress, and require only sufficient rivets to hold them in position. They are used during erection to keep the end-post in line; and after erection their function is to keep the end-post on the shoe, and to prevent it from having any upward motion due to the vibration of the structure.
The rivets through the vertical legs of the shoe angles are in double bearing in the ˘-inch angles, in single bearing in the vertical plate, and in double shear. A rivet in double shear has a less value than in bearing in the plates. This value is 14 440 pounds, and therefore the number of shop rivets required through the vertical legs of the angles is: The rivets which go through the horizontal leg of the angle and through the cap plate and cap angles, do not take stress. The num ber of rivets put in is that demanded by the detailing, the rivets in the horizontal legs of the angles usually staggering with those in the verti cal legs. The cap plate tends to keep the vertical plates in line, and to keep out the dust and dirt and other deteriorating influences of the elements.
Wherever the rivet-heads tend to interfere with other members or project beyond surfaces which are required to be flat—as, for example, the bottom of the masonry or bearing plates—they must be countersunk (see Carnegie Handbook, p. 191, and "Steel Construc tion," Part III, p. 192).
The space for the anchor bolts, that for the connection angles, and that for the hearing of the end floor-beam, require that the total width of the masonry plate for the fixed end shall be 2 X + 141 +2 X 6 + + 3 + 12 = 3 feet 71 inches.
The design of the roller end requires that the length of the masonry bearing, the size of the vertical plates and angles, and also the number of rivets shall be the same as that for the fixed end. The width of the masonry plate is determined by the length of the rollers and their connections at the end.
The rollers (60) are required to be 6 inches in diameter, and the unit-stress (19) per linear inch is 6 X 600 = 3 600 pounds, which requires: This is for the reaction of the bridge alone; and in addition to this, there are required for the floor-beam reaction: The total number of linear inches is 93.5 + 29.0 = 122.5; and if 5 rollers are used, they must be at least 122.5 = 24.5 inches long. The masonry plate is only 28 inches long, and therefore cylindrical rollers cannot be used, since they would occupy a space 30 inches or over. Segmental rollers (see Fig. 199) must be used.
The determination of the sizes of the angles which go at the end of the rollers, and also of the guide-plates, is a matter of judg ment and experience. Those sizes indicated in Fig. 198, repre sent good engineering practice, and will be used.
The distance from the center line of pins to the top of the ma sonry can now be determined, and is 1G;++6+;=231 inches.
On account of putting in sufficient connections and angles as shown in Fig. 198, the masonry plate must be considerably wider than that theoretically determined. According to Fig. 198, the total width must be as follows, and the width should be computed in two parts, as the plate is not symmetrical about the center line of the truss: From center line to outer edge: =3})+3 = 1 ft. 11} in., (say, 1 ft. 11 in.).
From center line to inner edge: 12}+ 1 +6+ 1 + 12 +I+21+3} = 3 ft. Of in., say,3ft. 0in.
Total width 4 ft. 11 in.
Allowing guide-plates and guide-bars of dimensions as shown in Fig. 19S, and assuming a inch as clearance at the ends, the total length of the rollers is: (4 ft. 11 in. ) — 2 (3 + 31 + I + } + 2) = 44.5 inches. This shows them to be amply long enough, as only 22 inches is theoretically required. Here, as in most cases for single-track spans up to 200 feet in length, the width of the masonry plate is determined by the detail and not by the unit hearing stress.