139/19 1/4 = 7,22 tons.

We now take Table XLIII, pass down the 60' 0" vertical line till we strike the curve 36" and then pass to the left to find the above value 7,22 tons. We strike the curve on the fifth horizontal line from the top and passing to the left find that we cannot find any such value as 7,22 tons, in other words the flange will have to be thicker than two inches. The value of 2" we find is 4,8 tons, leaving us 7,22 - 4,8 = 2,42 tons to care for in addition to the 2" thickness; this, we find (still on the fifth horizontal line) is under the thickness marked 1 or our flanges will have to be exactly 3" thick at the centre of girder.

In regard to the web, should we decide to make a box girder, each web should be at least one-half the calculated thickness or 5/16 thick.

In practice it would be better to make them a little heavier, for such heavy girders, say about 3/8 thick each.

Value of four angles.

Value of flange.

Example II. A single web riveted plate girder is of 59',0" span. At four feet from left support, and thence every five feet to five feet from right support it carries a concentrated load of 19500 pounds, the third loads from each end, being increased (by columns} to 91000 pounds. These loads include the allowance for weight of girder. The web must not be more than 36" deep. Detail the girder.

This girder is one of some twenty-five used by the writer in a large public building in New York City, hence the limitation as to depth of girder.

Detailing a riveted girder.

1Many engineers deduct in addition to size of rivet, 1-16" for punching and 1-16" for reaming, which in our case would make the rivet holes 1" instead of 7/8".

Plate And Box Girders 20046Plate And Box Girders 20047

Fig.204.

Fig. 204 REAR HALF

Fig. 204 REAR HALF

FG206.ELEVATION OF FRONT HALF

FG206.ELEVATION OF FRONT HALF

Plate And Box Girders 20050Plate And Box Girders 20051

The working-drawings, as they were furnished to the contractors by the writer, are given in Figures 204 to 210 inclusive.1 The only objection (on the score of cost) was to the length of some of the flange plates, but this could not be avoided, as the level of beams resting on the girders could not be disturbed, and there was not room enough between beams to get in the necessary length of cover-plates for splices.

The reader will readily see that this is practically the same example as the former one {Example I), so that we need not refer to the Tables for preliminary designing. We know then from the Tables that the girder, at the centre, will need to have a 36" x5/8" web, a3"x21" flange, four 6" x 6" x 7/8" angles, and that we will use 7/8" rivets.

We will now see whether this is confirmed by calculation. We will first use the graphical method (see Chapter VII (Graphical Analysis Of Transverse Strains)) on account of the large number of loads.

In Figure 202 (p. 121) we lay off our vertical load line m a, where m l-=.l k=jh = h g, etc., = 19500 pounds, and kj = d c = 91000 pounds at pounds-scale. We would select the pole distance xy - (k/f) = 12000 pounds, but that it will make the moment-of-resistanee curve too deep for convenience. We will, therefore, decide to make the distance x y =

10. (k/f) = 120000 pounds at pounds-scale. We shall, therefore, have to multiply all the verticals through the moment-of-resistanee curve in Figure 199 by ten to get their actual values. We draw the moment-of-resistance curve Figure 199 (see Chapter VII (Graphical Analysis Of Transverse Strains)) and find its base line A M. As our loads are not symmetrical on the beam, being four feet distant at one end and five feet at the other, we draw in Figure 202 x yl parallel A M of Figure 199, and find our reactions

Size by Tables.

Curve of moment of resistance.

y1,m

=

q

175000

pounds and

ay1

-

p

182500

pounds.

The greatest bending-moment will be at load wVI where the greatest vertical v through the moment-of-resistance curve (Figure 199 on p. 121)measures 276 inches (by inch-scale). Nothaving used the pole distance x y= (k/f) in Figure 202 we must multiply this by ten to get the actual required moment of resistance which would be = 2760. For the same reason we cannot use Formula (98) to calculate the flange thickness and therefore refer to Formula (36) and have for thickness of flange at centre

Createst bending-moment.

1 It is to be regretted that some of the illustrations are necessarily very small; the reader is advised to use a magnifying glass.

= 2760/37 -16,4/19,25 = 3,02

Or we need, as found by Tables, a flange thickness of three inches at the centre of girder. In the above formula, it should be explained, 2760 was the required moment of resistance at the point (that is at load wv1,) for which we were calculating flange thickness; 37 represented the approximate total depth of girder, allowing say for one half-inch plate to each end of girder; 19,25 was the net width of flange, after deducting two rivet holes; and 16,4 was the net area of cross-section (after deducting four rivet holes) of two 6" x 6" x 7/8" angles.

We will decide to use six half-inch thick plates in each flange and must next decide where to break them off. Accordingly we use Formula (99), and have value of two angles = 16,4 .36 = 590,4 or of the whole required moment of resistance (2760) the angles furnish an amount =590; now, as our moment of resistance curve is only of one-tenth the required depth, we divide this by 10 and make O T1= 59 at inch-scale. We now divide T1 Tinto six equal parts and draw the parallel lines to base A M, their intersections