# 000 for Railway Bridges 10

## rivets, inches, horizontal, shear, rivet, distance and pitch

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Having determined the horizontal shear per linear inch, the spacing becomes the value of one rivet divided by this horizontal unit shear, or V s • For the vertical rivets through flange angles and cover-plates, the same formula applies, except that Q is taken as the statical moment of the cover-plates only about the neutral axis.

The above exact method is not the one generally followed in spacing rivets, because it is not generally necessary to space the rivets so nearly to the exact theoretical distance. It is quite a common custom to space these horizontal flange rivets by assuming that the horizontal shear is equal to the vertical shear at the section divided by the distance between the centers of gravity of the flanges. This gives spaces somewhat less than would be required by the formula above.

The vertical rivets through cover-plates are made to alternate with the horizontal rivets; and in general, if there are sufficient hori zontal rivets, this method will give sufficient vertical rivets. In doubt ful cases, the exact method should be used.

It is customary to vary the spacing of the rivets about every two or three feet, or, in long girders, at intervals somewhat greater. This involves, of course, the determination of the shear at each point where a change in pitch is made.

The minimum distance in a straight line between rivets, is three times the diameter of the rivet; if i-inch rivets are used, the minimum distance, therefore, is 21 inches. This is shown by Fig. 250. This fact many times determines the size of flange angles to be used. In some cases the horizontal shear determining the pitch of rivets is so great that the distance between rivets becomes less than three times the diameter of the rivet. The flange stress might make it possible to use perhaps an angle with a 4-inch leg; in order to get in rivets enough to take the shear, however, it becomes necessary to use an angle having a 6-inch leg so as to use two lines of rivets. In such a case the horizontal distance between center lines of rivets may be 11 inches, and still the direct distance between the rivets will not be under 21 inches. Fig. 251 illustrates this.

1. Determine the pitch at end of girder having a reaction of 60,000 pounds, with web-plate 30 inches deep and LI. inch thick.

Assume effective depth between center of gravity of flanges, 28 60,000 inches; then approximate horizontal shear per linear inch = 28 2,142.

The bearing value of a Finch rivet on *-inch plate is 5,060; there 5,060 fore pitch = 142 = 2.35 or inches.

2. Given the same web as above, with an end reaction of 95,000 pounds, determine pitch at end.

95,000 Here 28 = 3,400 = Horizontal shear per linear inch; and 5,060 = 3,400 1.49 or 14 inches.

This makes it necessary to use an angle deep enough to give two lines of rivets either a 5-inch or a 6-inch leg. If the pitch between rivet lines is 21 inches, and horizontally between rivets inches, then the actual distance between rivets is about 211 inches, which is more than three times the diameter of the rivet. Where the top flange of a girder is loaded directly, as by a heavy wall, it becomes necessary to calculate the rivets for direct shear as well as horizontal shear. The combined stress on the rivet must not exceed its value, and therefore a spacing somewhat less than that determined for horizontal shear above must be used. This can best be illustrated by a problem.

3. Given a girder having a web-plate 36 inches by a inch, with an end reaction of 75,000 pounds, and loaded directly on top flange 75,000 girder, with 3,000 pounds per foot of gir = 2,206 = horizontal 34 shear per inch. Assume a pitch of 21 inches; then 2,206 X 2.25 = 4,963 = Horizontal stress on rivet; 3,000 12= 250 = Direct vertical shearing force per inch, and 250 X 2.25 = 560 = Direct vertical load on rivet.

These forces act on the rivet as indicated by Fig. 252. The resultant, therefore, is the square root of the sum of the squares of these two forces, and equals 4,994. As the value of the rivet is 5,060, this is about the nearest even pitch which could be used for these combined stresses.

The maximum straight distance between rivets which can be used is 6 inches, or sixteen times the thickness of the thinnest metal riveted. For a flange having angles, therefore, 5 inches would be the maximum pitch; or,.if a finch cover-plate were used, 4 inches would be the maximum in rivets through these cover-plates.

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