000 for Railway Bridges 10

rivets, method, girder, inches, approximate, vertical and plate

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As a comparison with the foregoing results, it will be well to note the pitch as determined by the approximate method, using the distance between centers of gravity of flanges. At the ends, we have 60,000 = 1,820 33 Pitch = 1 820 = 2.78 inches.

It will be seen that this approximate method gives some closer pitch than the more exact formula.

2. Given a girder 30 inches by inch, 301 inches back to back of flange angles. The flange section is made up of two angles 4 x 4 x inch. The end shear is 42,000 pounds. Determine the pitch of rivets by the approximate method.

3. Given a girder 42 feet long, loaded with a uniformly distrib uted load of 7,000 pounds per linear foot. If the web is 42 inches by inch, and the flange section at the end is made up of two angles 6 x 6 x inch, and 1 plate 14 x inch, and distance back to back of angles is 421 inches, (a) determine the pitch of horizontal rivets through web; (b) determine the pitch of vertical rivets through flange plates. Give two solutions of (a) and (b), using the exact formula and the approximate method based on distance between centers of gravity of flanges.

Answers—(a) 1 inches by the approximate method. inches by the exact method.

(b) 31 inches by the approximate method.

32: inches by the exact method.

Note that where pitch of vertical rivets through cover-plates is determined by the approximate method, they are simply ascrmed as alternating with the horizontal rivets. If there is only one line of horizontal rivets through flange angle and web, and one line of vertical rivets, then, by the approximate method, the vertical rivets through cover-plates would come centrally in the space between the horizontal rivets. If there are two lines of horizontal rivets, and one line of vertical, the vertical rivets would still alternate with the inner line of horizontal rivets, or center over the outer line of horizontal rivets. This would hold good so long as the spacing in this way did not exceed 6 inches, or sixteen times the thickness of plate. If this were the case, then the vertical rivets would be made to center over each line of hori zontal rivets. The same practice as regards vertical rivets would be followed in case both horizontal and vertical legs had two lines of rivets. The formula for exact determination of rivet pitch shows that

the above approximate methods are within the limits which would be determined if the exact method was usel.

Shop Details of Girders. Fig. 256 is a shop detail of a simple plate girder of one web. It will be noted that the detail covers only one-half the girder. Where the girder is exactly symmetrical about the center line, it would be a waste of time to draw up both halves. In such cases it is sufficient to mark the center line and mark the draw ing so that it will be clear that the other half is the same. In some cases where there is only a slight difference, as at the ends between the two halves, it is still unnecessary to detail more than half the girder; in such cases a special detail of the end which is different should be added.

This girder rests on a brick wall at each end; and therefore the end stiffeners are placed over the outer edge of bearing plate, as shown. A wall rests on top of the girder, and the intermediate stiffeners are to support the flange when the main pier lines come down, and to stiffen the web for the concentrated beam loads.

A girder such as this would probably come into the drafting room for details with only such information as is given in Fig. 257.

In many cases, even the loading on the girder might not be given. In such case, it would have to be calculated from the general plans showing amount and distribution of floor and wall loads. If the loads had been uniformly distributed, details might have been made by determining the capacity of the girder, as noted below.

The first point to be determined is the size of the bearing plate. The reaction is 65,000 pounds; and, allowing a safe bearing on the stone template of 25 tons per square foot, this requires about 1.30 square feet. A plate 12 by 16 inches, therefore, will be sufficient. Applying the formula given on page 97 of Part II, the required thick ness is found to be .26 inch; a steel plate inch thick is used here, although 1.-inch plate might have been used.

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