1. Given a girder 50 feet long, having a flange section of two angles 6 x 4 x 1/2, and 2 cover-plates 10 x § inch. Construct a parabola on this length as an axis, and determine the distances between the points where from diagram each cover-plate could be left off.

2. In above girder, determine actual length of cover-plates required by using the formula for cutting off plates.

3. Given a girder 40 feet long between centers of bearing, loaded with 120,000 pounds concentrated at four points equally distant. Determine the bottom flange section, and length of cover-plates.

Solution. Max M. = 30,000 X 8 X 3 X 12 = 8,640,000 inch-pounds. Assume web 36 inches deep, and effective depth as 34 inches; then flange stress = 254,000 pounds. This, at 15,000 pounds' fiber

. 254,000 stress, requires = 16.95 square inches.

In this, as in all calculations of girders, a great many sections could be chosen. In all problems the student must use his own judgment as to just what shapes to use in order to make up the section. Take

 2 angles 6 x 6 x 3/8 = 7.22 (two holes out) 2 plates 14 x 3/8 = 9 (two holes out) 16.22

Note that in deducting area of rivet-holes from bottom flange, the hole is considered 1 inch in diameter, even though 3/4-inch rivets are used. If smaller rivets were used, this might reduce the assumed diameter of hole to 3/4 inch.

From the manner in which this girder is loaded, it will be seen that the points at which the plate can be left off will be near the concentrated loads. Omitting both plates will leave a net area of 7.28 square inches; this corresponds to a flange stress of 7.2S X 15,000 = 109,200 pounds; and to a bending moment, assuming the same effective depth as at the center, of 109,200 X 34 = 3,712,800 inch-pounds. The reaction is 60,000 pounds; and it is therefore seen that the point corresponding to this moment is between the reaction and the first load. Its position is found as - 3,712,800/60,000= 61.88 inches = 5 feet 1 7/8 inches.

If this first plate is carried 1 foot 6 inches beyond this point, then its total length becomes 32 feet 7 1/2 inches.

At the point where the second plate is dropped, the net area is 12.10 square inches. This corresponds to a flange stress of 12.10 X 15,000 = 181,500 pounds; and to a bending moment of 181,500 X 34 = 6,160,000 inch-pounds.

The bending moment at the load nearest the reaction is 60,000 X 8 X 12 = 5,760,000 inch-pounds.

The moment between this load and the next load increases by an amount equal to 60,000 - 30,000, multiplied by the distance from the load. That is, at a point x distance from the last load, the moment will have increased (60,000-30,000) X x X 12 inch-pounds.

The bending moment which the angles and one cover-plate can carry was found to be 6,160,000 inch-pounds. The moment at first load is 5,76,000/400,000 - = allowable increase to point where second cover is required.

The distance from this first load to the point where it will be necessary to add the second cover-plate, is found, therefore, to be

400,000/30,000x12 = 1.12 feet.

As this is so near the point at which the load is applied, it would be better to add a little more than 1 foot 6 inches to this distance, in order to carry the plate a little beyond where the concentrated load occurs. This would make it necessary to increase slightly the length of the first cover from what was previously determined. These plates might be fixed, therefore, as 26 feet long and 34 feet long, respectively.

Spacing of Flange Rivets. The purpose of the rivets through the flange is to provide for the horizontal shear. There is a definite relation between the horizontal shear and the vertical shear at a given point, which is expressed by the formula s = SQ/I in which s = Horizontal shear per linear inch;

S = Total vertical shear at section;

Q = Statical moment of the flange about the neutral axis of the girder; and I = the moment of inertia of the whole section of the girder about its neutral axis. Having determined the horizontal shear per linear inch, the spacing becomes the value of one rivet divided by this horizontal unit shear, or d = 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 horizontal rivets, this method will give sufficient vertical rivets. In doubtful 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 3/4-inch rivets are used, the minimum distance, therefore, is 2\ 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

Fig. 250.

Fig. 251 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 1 1/4 inches, and still the direct distance between the rivets will not be under 2 1/4 inches. Fig. 251 illustrates this.