Fig. 179 shows the cross-section of this post. The radius of gyration which was used above was the radius of gyration of the chan nels about an axis perpendicular to their web. The radius of gyra tion of the entire section about an axis perpendicular to the web will be the same as that of one channel. In order to have the sections safe, the radius of gyration about the axis B—B must be equal to or greater than the other. The radius of gyration about the axis B—B can be in creased or decreased by spacing the channels. The exact distance which will make the two rectangular radii of gyration equal may be determined by the methods of "Strength of Materials," or it may be found in columns 14 and 15 of the Carnegie Handbook, p. 102. For any particular case it is equal to the value given in column 14, plus four times that given in column 15. For the channels under consideration, it is equal to 7.07 + 4 X 0.704 = 9.89 inches. Any increase in this distance will only tend to increase the radius of gyration about the axis B—B, and will make the post safer about that axis.
Fig. 179 shows a diaphragm. The web of this diaphragm cannot be less than inch, and the size of the angles cannot be less than 31 by 3 by $-inch, as this is the least allowed by the Specifications. The function of this diaphragm is to transfer one-half of the floor-beam reaction to the outer side of the post. The rivets which connect the angles to the diaphragm web are shop rivets, and (see design of floor beam) must be = 9 in number. The rivets which connect the diaphragm angle with the outer channel of the post are also shop rivets, and are 137 10 in number, 5 on each side. The 2 X 7 220 same rivets which connect the floor-beam to the post go through the diaphragm angle on that side of the diaphragm next to the cen ter of the bridge, and must therefore be field rivets and take the entire floor-beam reaction. These must be 137 23 in number, 6 013 12 on each side. The exact distance, back to back of the channels of the post, cannot be determined until after the top chord has been de signed, since the post must slide up in the top chord and also leave room on each side for the diagonal members of the truss. The width is determined by the packing of the members at joint (see Fig.174), and is found to be 94 inches. Since this is less than that required above, the post must be examined for bending about an axis parallel to the web of the channels.
According to the methods of "Mechanics" and "Strength of Materials," with the help of the Carnegie Handbook, p. 102, the moment of inertia about this axis is found to be 286.42, and the ra dius of gyration 3.96. The unit allowable compressive stress is then computed to be 9 580 pounds per square inch, and the required area 163 600 = 17.10 square inches, which, being less than 17.64, shows
9 580 the section to be safe.
This member is connected to the top chord at its upper end by a 5-inch pin. The total stress is 163 600 pounds, and the total bear 63 600 ing area required is 4 - = 6.8 square inches, or 3.4 square inches for each side (19). The total thickness of the bearing area for each side is 3c4 = 0.68 inch. The thickness of the web of a 12-inch 5 30-pound channel is 0.513 inch, which leaves 0.68 — 0.513 = 0.167 inch as the thickness of the pin-plate, but it must be made inch according to the Specifications. Fig. 180 shows the arrangement of the plates and the rivets.
The sum total of the pin-plates and the channel web is 0.888 inch, and therefore on one side the stress transferred to the pin by means of the pin-plate, which is 0.375 inch, is 1 X 0.375X 163 600 2 0.888 = 34 600 pounds. This plate will tend to shear off the rivets between it and the channel web, and therefore 7 = 5 shop rivets are required. 7 The stress that is shown on the stress sheet is the stress in the post above the floor-beam. The stress in that part below the floor beam is equal to the vertical component of the diagonal in the panel ahead of the post in question. In this case it is the vertical com ponent of the stress in and is equal to 242 000 pounds, and this requires a total bearing area of 242 000 = 10.1 square inches, and 10.1 24 000 a total thickness of 2 X 5 = 1.01 inches on each side, the pin being 5 inches in diameter. From this total thickness must he subtracted the thickness of the web of the channel, and this leaves 1.01 — 0.513 = 0.497 inch as the total thickness of the pin-plates required. This shows that we must use one 1-inch plate. The total thickness of the bearing area is now 0.513 + 0.50 = 1.013 inches.
Each plate takes a total stress of 0.50 X 242 00059 700 1.013 2 pounds; and the joint being weak in shear, the number of rivets required will be 59 700 = 9 rivets in single shear. The detail will be 7 220 similar to that in Fig. 180.
The distance, back to back of the channels in this post, will probably not be greater than 12 inches, and this will make the dis tance between rivet lines about 9 inches. According to (44), the end tie-plates must be at least 9 inches long and of course 12 inches wide. The thickness cannot be less than = 0.18 inch, but they will be made inch (36). Between the tie-plates the channels will be con nected by means of lattices. The Specifications (45) require that they should not be less than 21 inches in width and (1.414 X 9) = 0.318 (say ) inch in thickness. Table XXV gives the thickness of lacing bars for any distance between rivets.