## Length Of Flange Plates

The bending moment on a simple beam varies throughout the length; and, to design girders economically, the net flange area should vary with the bending moment. This condition is fulfilled by using flange plates of different lengths, each extending only as far as needed to provide the net section required. It is good practice to continue the inner plate the entire length of the girder, in order to stiffen it laterally. The lengths of flange plates in girders uniformly loaded may be obtained approximately by the formula:

L1 =

L1 being the required length of the plate in feet; L, the length of girder in feet; a, the net sectional area of all plates to and including the plate in question, beginning with outside one; and A, the total flange area. The 2 ft. is added to allow for riveting.

### Example

Fig. 26 shows a flange section of a girder. The span being 60 ft., what should be the length of each flange plate?

### Solution

Area of a 5"X 5"X 3/4"angle (see Table XIV, page 85) is 7.11 sq. in. Area of each plate is 3/8 in. X 12 in. = 4.5 sq. in.; the rivet holes are 3/4 + 1/8 = 7/8 in., and the area cut out by a 7/8" hole through a \" plate is .328 sq. in.; hence, the net area of each plate is 4.5 sq. in. - (.328 sq. in. X 2) = 3.844 sq. in.; or, for 3 plates, 3.844 X 3 = 11.532 sq. in. The net area of the two angles is (7.11 sq. in. X 2) - (.656 X 4) = 11.596sq. in. Therefore, the net area of flange will be 11.532+11.596=23.128 sq.in.

Fig. 26.

By the formula, the length of outside plate

L1 = = 26.42, say 28 ft. 6 in.

Similarly, the length of the intermediate plate

L1 = = 36.56 ft., say 36 ft. 8 in.

The inner plate is continued the entire length of the girder, so as to stiffen it laterally.

Fig. 27 shows a graphical method of finding the lengths of flange plates for a girder carrying any number of concentrated loads. Lay off a b to any scale, equal to the span, and locate on it the point of application of each load, as c, d, and e. Having calculated the bending moment under each load, lay it off, to a convenient scale, on the corresponding perpendicular to a b, through c, d, and e, as cf, d g, and e h. Draw the lines affg. gh, and hb. Divide eh, the line representing the greatest bending moment, into as many equal parts as there are square Inches in the net section required. (For a method of dividing a space into equal parts, see Geometrical Drawing, page 55.) Take er equal to as many parts as there are square inch the net sectional area of flange angles; also, r s equal to net area of the first plate, etc. Draw parallels to a b through r,s,t, etc. Make i k equal to a b; where i k, etc. intersects af and bh. erect perpendiculars p l and qm, etc. Then l m, measured to same scale as a b, is the length of the first flange plate; n o, that of the second, etc. The lengths thus found should be in-creased 1 ft. at each end, to allow for riveting. The lengths of plates for uniformly loaded girders may be similarly found. Having calculated the maximum bending moment, lay it off to scale, perpendicular to the middle of the span. The curve representing the bending moment is in this case a parabola, and may be drawn through the point just named and the ends of the span, by the method given in Geometrical Drawing on page 58. The remainder of the work is like that shown in Fig. 27.

Fig. 27.