1. Determine by the method previously described the bottom flange section of a girder 28 inches deep between centers of gravity of flanges, and having a bending moment of 3,500,000 inch-pounds. The flange is to be proportioned to carry the whole bending moment. Use fiber stresses given for building.

2. In the above problem, if the top flange is unsupported laterally for 20 feet, determine the section of top flange required, using the formula given for reducing allowable compression stress.

3. Given a girder 35 feet long between centers of bearings, and carrying a uniformly distributed load of 2,000 pounds per linear foot. Assume a web 36 inches deep and 34 inches between centers of gravity of flanges. Determine bottom flange section without making any allowance for the portion of bending moment carried by the web.

4. In the above girder, redesign bottom section on the basis that the web is not spliced and that it bears a portion of the bending moment.

5. Determine the thickness of web required in above girder.

6. If the girder was 40 feet long, 42 inches deep, and loaded with 4,000 pounds per linear foot, determine the thickness of web if no stiffeners are to be used. Assume flange angles are 6 inches by 6 inches by 1/2 inch.

7. Determine thickness of web in above girder which could be used with stiffeners, and determine spacing of stiffeners required.

Solution. In this case the shear at end is S0,000 pounds. From the diagram for spacing of stiffeners, it will be seen that any thickness of web from 5/16 inch up could be used. Where stiffeners are used to prevent buckling of web, it is more economical to use a 5/16-inch web than a 3/8-inch. If the girder was 60 inches deep, probably it would not be well to use less than 3/8-inch web, even with stiffeners. In this case assume a 5/16 by 42-inch web. Area is therefore 13.12 square inches, and fiber stress is 6,150 pounds.

From the diagram it is seen that a 5/16-inch web with this stress per square inch requires stiffeners about 16 1/2 inches back to back. This then determines the space of first stiffener from those over the bearing plate. Assume two spaces the same as this, and then determine shear at point say 3 feet 6 inches from the end bearing. This is found to be 80,000 - (4,000 X 3.5) = 66,000 pounds. The stress here is about 5,075 pounds per square inch of web. From the diagram, this is seen to require stiffeners 20 inches apart. Assume two more spaces at 20 inches, and calculate shear, which is found to be 52,600 pounds. This gives a fiber stress of 4,050 pounds per square inch of web, and requires stiffeners 24 inches apart. Take three spaces at this distance, and calculate shear, which is found at this point to be 28,600. This gives a stress of 2,200 pounds per square inch of web. From the diagram the spacing of stiffeners for this fiber stress, in a 5/16-inch web, is found to be 36 inches. This distance, however, is greater than the clear distance between flange angles, which is 30 inches, and indicates, therefore, that at this point the web is strong enough without being stiffened by angles.

If it is desired to see whether or not two spaces at 24 inches, instead of three as above taken, would have been sufficient, the shear at this point can be calculated. This is found to be 36,600 pounds, or 2,800 pounds per square inch of web. This is seen to require stiffeners 31 inches apart. This is greater than the distance between flange angles, and indicates that the last stiffener could be omitted. However, it is better to carry the stiffeners a little beyond the actual point where the diagram would indicate that they could be dropped; so that it would be better to use the last stiffener, as originally determined. The spacing of stiffeners at each end of girder is of course made the same where the load is uniformly distributed.

Size of Stiffeners,, Stiffeners for concentrated loads and for reactions should have sufficient area to take the whole load or reaction at this point. Stiffeners used to prevent buckling are not generally calculated, but are made either 3 x 3 x 3/8 inch or 4 x 3 x § inch. When stiffeners are fitted to the flanges, the outstanding leg should be made large enough to come nearly out to the edge of the flange angle. If the flange angle is 6 by 6, the stiffener would be perhaps 5 by 3 1/2.

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 parabola. 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 formula M/fh = A. A convenient method of constructing the parabola will be to lay off the offsets, which are determined at different points by the formula y = h[1-(x2/1/2)] as illustrated by Fig. 249.

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 required. 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; A1 = Area of plate to be cut off; A = Total required flange area at center,

Steel Construction Problems 1 0500271Steel Construction Problems 1 0500272

Fig. 249.

= M/fh; and

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.