# Design of a Through Pratt Railway-Span 70

## inches, cover-plate, section, center, plates, angles, required and inch

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The design of the first section of the top chord will now be made. Here, as in the case of the first sections of the lower chord, the diameter of the head of the greatest I-bar determines the width of the plates in the section The head of the 7-inch I-bar which constitutes the member is 171 inches, and, allowing a clearance of 1 inch on either side of the head, the total depth inside the chord should be 18/ inches. As in the case of the lower chord, plates 18 inches wide may be used.

The size of the angles to be chosen is a matter of judgment. Usually any size should be chosen at first, and the preliminary design will indicate at once what size should have been taken. For this case, 31 by 3' by i-inch will be assumed at first.

For sections of this character, the radius of gyration is approxi mately equal to 0.4h, in which h is the height, or rather the width, of the side plate. The approximate radius of gyration is r = 0.4 X 18 = 7.2 inches, and the length is equal to one panel length, or 21 feet. The allowable unit of stress (16) is: The required area is 41439_5,20° = 33.2 square inches. The correct proportion for sections of this character is that 0.4 of the total area should be taken up by the web. The area of the web would then be 0.4 X 33.2 = 13.28 square inches, and the thickness would be 13.28 = 0.37 inch. According to this, a i-inch plate should be used, 2 X 18 but (42) requires that it shall be 130 0.483 inch or thicker.

Therefore an 18 by ij-inch plate must be used for the web.

The correct proportion for sections of this character is that the width between plates should be about \$ the width of the side plates.

This will give the required width between plates equal to X 18 = 15.75 inches. The cover-plate (42) must not be thinner than 40 the distance between the connecting rivet lines. The rivet lines are, in this case, 15.75 + 2 X 2 = 19.75 inches apart, and therefore the thickness of the cover-plate cannot be less than 19 = 0.494 inch.

The cover-plate will therefore be taken as > inch thick. The width of the cover-plate (see Fig. 184) mast he about 15.75 + 2 X 31 + ' = 231 inches (say 23 inches). The cover-plate will be taken 23 by The center line of pins will be taken at the center line of the web, and the center of gravity of the section will be assumed as inch above this. In order that the center of gravity may be near that assumed, the moment of the cover-plate about the assumed center of gravity axis should be about equal to the moment of the flats about the same axis. The

the assumed axis is: 23 X}(9.0-0.5+0.25+0.25)= 23X 9 2 and the moment of the flats about the • same axis is: A (9.0 + 0.5 + 0.25 + 0.5) = 10.25 A, in which A is the area in square inches of both of the fiats. Equating these two expressions, and solving for A, there results: Assuming the flats to be 4 inches wide, the thickness on each side will be 1.25 inches. As this is too thick to punch, the flats on each side will be composed of two 4 by k-inch plates.

The total area is: But the required area is 32.2 square inches, which is considerably less than the area above given, and which does not include the angles and hence we can use the smallest size angles, which are those pre viously assumed. The area of each of these angles is 2.48 square inches, thus making the total area of the section 39.5 + 4 X 2.48 = 49.42 square inches. This is considerably in excess of the area as required according to the formula for compression; but it is the least allowed by the Specifications. Note that this is the case where (42), instead of the formula for compressive stress, is the ruling factor in the determination of the section.

The center of gravity of the approximate section must now be determined, the moment of inertia and the radius of gyration about the neutral axis must be computed, and the required area must be determined by using this radius of gyration as computed. If the required area as determined with the actual radius of gyration is Tess than the approximate area, then the thickness of the angles or the plates must be increased and the section then examined for its radius of gyration and required area. If the area is sufficient, the section is used; if not, another recomputation is in order.

In the determination of the center of gravity of the section, the moment is taken about the top of fhe cover-plate. The moments are computed as follows: Cover-plate (23 X §) X 1 2.88 Webs 2 (18 X I) X (9 + :) 175.60 Top angles 2 (2.48) X (1.01 + 1) 7.50 Lower angles 2 (2.48) X (} + 1 + 18 + f — 1.01) 89.30 Flats 2 (4 X 11) X 19n 196.25 471.53 The center of gravity is now found to be 39.50l 4 X 2.48 = 9.55 inches from the top of the cover-plate. The distance from the top of thy cover-plate to the middle line of the web is 9 + h + z = 9.75 inches, and this leaves a distance of 9.75 — 9.55 = 0.2 inch from the center line of the web to the neutral axis. This distance is gen erally represented by the letter e, and it is known"as the eccentricity of the section.

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