Cutting Off Flange Plate. In heavy girders the flanges are made of angles with cover-plate. Sometimes only one plate is required; at other times four or more will be needed. As the maximum moment is the moment determining the flange section, and this usually varies from point to point, it will be seen that for economy the number of plates should be proportioned to the varying moment. Where the girder is loaded uniformly, the bending moment is a maximum at the center of the span, and varies toward the ends as the offsets to a par abola. A convenient way, therefore, to determine for such a case where to stop the different plates, is to lay off to scale the span, and on this axis construct a. parabola, making the ordinate at the center represent the required area, from the = A. A convenient method of constructing the parabola will be to lay off the offsets, which are determined at different points by the formula From this diagram, the point at which an area equal to one of the plates can be dropped off, will be found by drawing a horizontal line at a distance down equal to the area of the plate at the same scale as the center ordinate. Where this line cuts the line of the parabola, will be the exact length of plate re quired. Sufficient length should be added at each end to enable rivets enough to be used to develop in single shear the stress in the plate. Usually this will be about 1 foot 6 inches at each end.
Another method of determining where to drop off plates when the load is uniformly distributed, is to use the formula in which x = Distance from center to point where area of plate is not required; A, = Area of plate to be cut off; A = Total required flange area at center, and f la L = Span.
When the loads are concentrated, and the moment does not vary uniformly from point to point, the only way is to calculate the moment at different points, and proportion the flange and at these points in the same manner as at the center.
1. Given a girder 50 feet long, having a flange section of two angles 6 x 4 and 2 cover-plates 10 x b 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 re quired 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 stress, requires 254000 — 15,000 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 judg ment as to just what shapes to use in order to make up the section.
Take Note that in deducting area of rivet-holes from bottom flange, the hole is considered 1 inch in diameter, even though -inch rivets are used. If smaller rivets were used, this might reduce the assumed diameter of hole to inch.
From the mariner 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 con centrated loads. Omitting both plates will leave a net area of 7.28 square inches; this corresponds to a flange stress of 7.28 X 15,000 = 109,200 pounds; and to a bending moment, assuming the same effect ive 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 3,712,800 load. Its position is found as 60,000 — 61.SS inches = 5 feet 1iinches.
If this first plate is carried 1 foot 6 inches beyond this point, then its total length becomes 32 feet 71 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 = (1,160,000 inch-pounds.
The bending moment at the load nearest the reaction is 60,000 X S 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 5,760,000 load is — allowable increase to point where 400,000 second cover is required.
The distance from this first load to the point where it will be necessary to acid the second cover-plate, is found, therefore, to be 400,000 30,000X 12 = 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 rela tion between the horizontal shear and the vertical shear at a given point, which is expressed by the formula s = 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.