1. As an illustration of the use of the exact formula for pitch of rivets, the following problem will be worked out:

Take the girder given in the problem illustrating the cutting-off of flange plate. This girder 40 feet long has a web-plate 36 inches by 3/8 inch; and section of flange at end consists of two angles 6 x 6 x § inch. At point 10 feet from end section, are two angles 6 x 6 x 3/8 inch, and 2 plates 14 x 3/8 inch.

Determine first the pitch of rivets at end where the shear is 60,000 pounds. The formula is: s = SQ/I

The first step is to determine position of center of gravity of flange. As there are no cover-plates, this is taken directly from "Cambria" and is 1.64 inches.

The web is 36 inches; but in all girders where flange plates are used, the depth back to back of angles is 1/4 or 1/2 inch more than the depth of web, in order to allow for any variation in the depth of plate. In this case it will be taken as 36 1/4 inches back to back of angle.

Q = 2 X 4.36 X 16.49 = 143.8 I=4Y 15.39 = 61.6

Steel Construction Problems 4 0500278

s = 60,000x143.8/6,259.6 = 1,375

Pitch =5,060/1,375 = 3.69 inches.

Something less than this would be actually taken - probably 2 3/4 or 3 inches.

To determine the pitch at point 10 feet from end, we have to calculate the neutral axis of the flange as follows: Angles 2 X 4.36 X 2.39 = 20.9 10.5 X .38 = 4.0 24.9 24.9 -4- 19.22 = 1.3 inches from back of cover-plate to neutral axis.

Steel Construction Problems 4 0500279

S = 30,000x338/13,390 = 757

Pitch of rivets = 5,060/757 = 6.68 inches.

The maximum pitch is as stated 6 inches. At this point the actual pitch would be made somewhat less - say, 5 1/2 inches.

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 flange angles. The flange section is made up of two angles 4 x 4 x 3/8 inch. The end shear is 42,000 pounds. Determine the pitch of rivets by the approximate method.

Steel Construction Problems 4 0500280

Fig. 255.

60,000/33 = 1,820

Pitch = 5,060/1,820 = 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 5/16 inch, 30 1/4 inches back to back of

OFFICE BUILDING FOR CHICAGO & NORTHWESTERN RAILWAY COMPANY, CHICAGO, ILL.

OFFICE BUILDING FOR CHICAGO & NORTHWESTERN RAILWAY COMPANY, CHICAGO, ILL.

Frost & Granger, Architects; E. C. & R. M. Shankland, Engineers. The 52,500-lb. Girder Shown on Page 262, being Grouted with Concrete. This is Done on Girders Put in the Ground to Protect Same fromCorrision. The outsides of the Girders are also Encased in Concrete for the Same Purpose.

3. Given a girder 42 feet long, loaded with a uniformly distributed load of 7,000 pounds per linear foot. If the web is 42 inches by 7/16 inch, and the flange section at the end is made up of two angles 6x6x1/2 inch, and 1 plate 14 x 1/2 inch, and distance back to back of angles is 42 1/4 inches, (a) determine the pitch of horizontal rivets through web; (6) determine the pitch of vertical rivets through flange plates. Give two solutions of (a) and (6), using the exact formula and the approximate method based on distance between centers of gravity of flanges.

Answers - (a) 1 5/8 inches by the approximate method. 1 15/16 inches by the exact method. (6) 3 1/4 inches by the approximate method. 3 3/8 inches by the exact method.

Note that where pitch of vertical rivets through cover-plates is determined by the approximate method, they are simply assumed 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 horizontal 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 used.

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 drawing 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.