This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.
Web-plates, besides resisting direct vertical shear, must resist buckling. To avoid failure by buckling before the full shearing strength of the plate is realized, stiffeners, composed of angles, as shown in Fig. 20, must be used. Some engineers require the use of stiffeners when the unit shear exceeds the unit stress allowed by the following formula: s= in which s = allowable unit stress; d = depth of girder, between flanges, in inches; t = thickness of web-plate in inches. A more convenient rule provides that stiffeners be used when the thickness is less than 1/50 the clear distance between vertical legs of flange angles. For example, in a girder with a 3/8" X 36" web and 6" X 6" flange angles, the distance between vertical legs of angles is 36 -(2X6) =24 in.; 1/50of 24 in. = .48, say 1/2 in.; the web-plates being but 3/8 in. thick, stiffeners must be used.
Stiffeners should never be omitted over the end supports, and should be provided under all concentrated loads. The distance between centers of intermediate stiffeners is usually made equal to the depth of girder. Under no condition, however, should stiffeners be placed more than 5 ft. apart.
End stiffeners should be considered as columns transmitting the entire load upon the web to the supports. The size of intermediate stiffeners is not usually calculated; they are usually made the same as the end ones, or lighter. Angles used for stiffeners should not be less than 3" X 3" X 5/16"; on shallow girders, however, with light loads, it might be economical to use 2 1/2" X 2 1/2" X 5/16" angles; but smaller ones should never be used. Stiffeners should extend over the vertical legs of the flange angles, being either swaged out to fit, as shown at (a) and (6), Fig. 23, or, preferably, provided with filling pieces, as shown in Fig. 20.
Using 4 end stiffeners (2 each side), and a safe unit compressive stress of 13,000 lb., what size angles should be used for stiffeners of the girder shown in Fig. 22.
The required sectional area is 120,000 / 13,000 .= 9.23 sq. in.; 9.23 sq. in. / 4 = 2.31 sq. in., the area required for each angle. From Table XIV, a 4" X 4" X 5/16" angle has a sectional area of 2.40 sq. in., which is ample.
The flanges of a girder - called top and bottom chords in open or lattice girders - include the flange plates and angles, as shown in Fig. 20. Frequently, one-sixth the sectional area of the web plate is included as part of each flange, although, in many cities, the building ordinance will not allow this to be done. If the sixth is so included, the web should never be spliced at the point of greatest bending, and all splices should be so designed that, by proper placing of the rivets, the strength of the web included in the one-sixth be reduced as little as possible.
In a simple girder, the top flange is subjected to compression, and the bottom one to tension. It is the practice, for economy in construction, to make the flanges alike, only the lower flange being calculated. The bottom flange being in tension, the area of the rivet holes are deducted, so that the net area at the point of least strength may be obtained.
The horizontal stresses on the flanges are resisted by the fibers in one flange acting with a moment about the center of gravity of the other. This moment is equal to their total strength multiplied by the distance between centers of gravity of the flanges. Thus, Fig. 24 shows a plate girder pin connected at a, and a chain in tension, representing the lower flange, at b. It is evident that the reactions tend to turn the girder about a, and rupture the chain, which resists this tendency with a lever arm d equal to the distance bet ween the centers of gravity of the flanges. The depth is, in practice, however, usually taken to be the depth of the web. To find the net area of the lower flange, let M1 be the resisting moment in foot-pounds; M, the bending moment in foot-pounds; s, the allowable unit tensile stress; a, the net area of flange in square inches; and d, the depth of girder in feet; then s X a X d = M1 which must be equal to M; from winch a = M /(sxd).
The net sectional area of the lower flange of a plate girder is equal to the greatest bending moment in foot-pounds divided by the product of the allowable unit tensile stress in pounds and the depth of the girder in feet.
In proportioning the flanges, the sectional area of the flange plates should equal, approximately, that of the flange angles. This is not possible in heavy work, and the best that can be done is to use the heaviest sections obtainable for the flange angles.
A steel girder is 6 ft. deep, and 80 ft. span, and the load 3,000 lb. per lineal foot, (a) What net flange area is required, using a safe unit tensile stress of 15,000 lb. per sq. in.? (6) Of what size sections should the flange be made?
(a) The entire load is 80 X 3,000 = 240,000 lb. From Table XXVI, page 113, the bending moment is
M=WL /8=(240,000x80) / 8=2,400,000 ft lb.
The net flange area a=M/(sxd)=2,400,000/(15,000x6)=26.6 sq. in
(b) Assume the section shown in Fig. 26. The area of a 6" X 6" X 5/8" angle being, approximately, 7 sq. in., the sectional area of the flange is, Two 6" X 6" X 5/8" angles = 7X2 = 14.00 sq. in. Three 14" X 7/16" plates = 14 X 7/16 X 3 = 18.37 sq. in.
Total = 32.37 sq. in. From the total area deduct the metal cut out for rivet holes, which, taken as 1/8 in. larger than the rivet, are 1 in. in diameter. The web is not considered in the flange section. The areas deducted are, therefore,
Four 1" holes through 5/8" metal = 2 1/2 sq. in. Two 1" holes through 1 15/16" metal = 3 7/8 sq. in.
Total = 6 3/8 sq. in.
The net area of the flange is 32.37 - 6.37 = 26 sq. in., which, although a little less than the net area required, 26.6 sq. in., could be used.