Design of a Through Pratt Railway-Span 70

inches, square, inch, total, net, thickness and section

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The total net section through the pin-hole at (26) must be 11 X 18.17 = 22.7 square inches, or 11.35 square inches for one side. The plate which is to increase the section must be on the out side, since the intermediate post and the two I-bars of member must go inside. The gross width of this plate is 111 inches (see Fig. 174), and the net width is 2w = 11f — 5 = 61 inches. The net area through the pin is: Two 31 by 31 by 1-in. angles = 4.96 square inches.

One 18 by 1-in. plate = 9 — 5 X I = 6.50 " Total = 11.46 square inches Since this is greater than the 11.35 required, no plate will be necessary to fulfil (26) in this respect.

Sufficient bearing area must be provided at this point. The total stress is 234 200 pounds, the total bearing area required is 234 200 000= 9. 76 square inches, and the total thickness for one side is 24 9.76 = 976 inches. Since the thickness of the web is 4 inch, the 2X5 pin-plates must be 0.976 — 0.50 = 0.476inch (say 1 inch) thick. A -i-inch pin-plate must be used, and as the total thickness of the bearing area is now 1.00 inch, this pin-plate will take X = 58 550 pounds. The joint is weakest in shear, and will therefore require 7 22 2 0 0 — 8 -r (say 9) shop rivets. 7 In case it is necessary to put the member on the outer side of then the outer legs of the upper angles must be cut off to allow to pass. This will decrease the section by an amount (31 — X a = 1.20 square inches. Considering the pin-plate, which is (l81 — 2 X 31) — 4 = 11; inches, the f inch being allowed for clearance between the edges of its flange angles, the total net section through the pin-hole on one side will be: This is greater than 11.35 as required, and is therefore safe.

The distance from the center of the pin to the end must now be determined (26). The total net section of the body of the member is 18.17 square inches, or 9.09 square inches for one side, and the thickness of the web and the pin-plate is 1 inch. The distance from the pin to the end of the member is then 9 = 9f inches, and distance to the center of the pin is 91 + 2 = 11s, say 12 inches (see Fig. 174). Rivets should be countersunk where necessary to prevent interference with I-bars. For signs, see "Steel Construction," Part III, p. 192, and Carnegie Handbook, p. 191.

At point of this member, the pin is 64 inches in diameter, and, as previously mentioned, the legs of the angles are cut (see Fig. 175).

The total bearing area required for one side is 2.9.76 = 4.88, and the required thickness is 46.2.88 r = 0.781 inch. Subtracting the thickness of the +-inch web from this gives 0.281 inch. A pin-plate inch thick must be used.

The net area through the pin (26) must be 11.35 square inches.

This net area, remembering that the angle legs are cut and therefore their area is that of a bar 3i by *-inch, computed for one side, is as follows: Two Angles, legs cut, 31 by 31 by f-in. = 2.62 square inches One Web 18 X i — 6; X 1 sq. in. = 5.88 " " One Pin-Plate 111 X f — 61 X *sq. in. = 1.87 ` ` " Total 10.37 square inches.

This shows the section to be deficient, and the thickness of the pin plate must be made i inch. This gives a net area through the pin of 11.62 square inches.

The distance between rivet lines (see Fig. 173) is 174 inches, and (44) the tie-plates must be 174 (say 18) inches long, and their thickness 17. 25— 0.346 inch (say inch).

The lattice bars (45) must be 21 inches wide, and (47) must be double. From (45) and Table XXV, page 219, the thickness must be inch, the distance c being 17.25 X secant 45° = 2 ft. in.

The design of the hip vertical IJ,L, is also made in accordance with (80) of the Specifications. It will be assumed that the section consists of one 8 by --inch plate, and four 31 by 33 by *-inch angles, since this is the lightest section that may be used according to the Specifications, the 8-inch plate being chosen as it gives some clearance between the inner edges of the legs of the angles.

The total stress in the member is 141 600 pounds, and the unit stress is 16 000 pounds per square inch, thus requiring a net area of 8.85 square inches. The plate gives a net area of 2.20 square inches, and the four angles give a net area of 6.88 square inches, making a total of 9:08 square inches, one rivet-hole being taken out of each angle, and two out of the web, at any particular section. The net area is somewhat greater than that required; but must be used, as this is the minimum sec tion allowed by the Specifications. Fig.

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