N, 0,P, Q, R, and S with the curve are the points at which the respective plates can be broken off.

We shall, however, carry the first plate NO the entire length, and as this plate is spliced inside of the curve, we shall have to carry plate No. 5 over the joint to make up for the lost section, and we shall have to carry both plates Nos. 4 and 5 sufficiently far to the left of the splice to get in the necessary number of rivets to equal the value of cut plate.

It will also be necessary to prolong some of the plates to get in the necessary rivets beyond their points of contact with curve. Thus between O and N we must get enough rivets to equal in value plate No. l. or else prolong plate beyond N; between P and 0 enough rivets to equal plate No. 2, or else prolong plate beyond 0: and so on till in the last plate No. 6 we must get enough rivets between T and S to equal plate No. 6. Now these plates are all of equal value = 1/2" x 19 1/4" = 9 5/8 square inches of cross-section, which multiplied by (k/f) = 12000 pounds gives the real strain s on the rivets, or s = 9 5/8.12000 = 115500 pounds. We will, therefore, lay out rivets enough, in each flange, between S and the end A to take this strain five times, gradually decreasing the pitch towards the end of girder. We will then carry each plate sufficiently far beyond the length required by curve, to get its respective number of rivets.

Required flange thickness.

Whereto diminish flange thickness.

Number of flange rivets required.

Now the value of rivets (7/8") in flange will be for shearing (single area) =4800 pounds each. For bearing and bending the rivets will evidently get their value from the 1/2" plate, this being thinner than the angles, and we have bearing value = 5250 pounds per rivet. Either of the above can be found by calculation, or from Tables XXXV and XXXVIII. For bending we have from Table XXXVIII for a 7/8 rivet, the safe bending-moment = 990 or say 1000 pounds-inch.

The actual greatest bending-moment will be, Formula (25)

Spacing flange rivets.

Value of each flange rivet.

m = u.(1/2)/2 = u/4 and as the safe m = 1000 pounds-inch, we have

1000 = u/4 or u =4000 pounds.

The value of the rivets against bending-(4000 pounds each; - being their least value, will control the design. Each cover-plate therefore requires 115500/4000 = 29 rivets, and from S to end we shall require 145 rivets in each flange. From S to T we require only 29 rivets, but they will have to be spaced more frequently to comply with the rule for greatest pitch, or Formula (107), accordingly the pitch of the latter should not exceed = 16. 1/2= 8 inches.

Figure 207 shows a plan of the top flange of left half of girder.

Plate No. 6 might have stopped at S, but is carried two rivets further to avoid breaking under a beam, which rested on the girder at this point. The splice of plate No. 1 has been made just to the left of R, where plate No. 5 might have stopped ; we must, therefore, carry plates Nos. 5 and 4 at least 29 rivets beyond the splice, which has been done. Plate No. 4 might have stopped two spaces nearer R, but for the splice; Plate No. 3 we stop at the right number of rivets to the left of Q, and plate No. 2 to the left of P. Plate No. 1, which might stop 29 rivets to the left of O, we decide to carry to the end.

The countersunk rivets shown come under beam or column ends. The blank spaces were to bolt column plates to. The blank spaces for beams were marked on later from memoranda in the contractor's shop.

Having detailed our flange, which will be the same both for top and bottom flange, we will now consider the web.

The size of web we settle from Formula (124) and have for thickness b, assuming that there will be no more than six rivet-holes in any vertical section, or d = 36 - 6.7/8 = 30 3/4"; b =182500/30 3/4.8000 = 0,74 or nearly 3/4". The web, however, was made 5/8" as the above was required only at the one extreme end and through its rivet-holes. The effect of decreasing the breadth of web being, of course, to raise the actual shearing per square inch at this point to a little over 9000 pounds per square inch.

We next decide where stiffeners are required; we use Formula (127) and have y = 12000. 5/8. 36/1+0,0003.362/(5/8)2

= 135000 pounds,

Or we require stiffeners from the end to the point where the vertical shearing is less than 135000 pounds.

By referring to Figure 200, we find this would be about ten feet from the end.

The stiffeners, however, were placed more frequently, as shown in Figure 206, both for looks, and as there was some danger of heavier loads being placed on the centre of girders, which would, of course, increase the vertical shearing near centre. These stiffeners were made of 6" x 6" x 1/2" angle irons, with 6" x 7/8" x 24" filler plates behind them, so as not to bend the stiffeners. The filler plates being cheaper than would be the cost of blacksmith work involved in bending these angles around the vertical legs of flange angles. Their upper and lower ends were "milled" off and made to bear firmly. Now as to value of rivets through web, we should have for bearing 5/8". 7/8" 12000 = 6562 pounds; for shearing, being in double shear, twice the value previously found for single shear or, 2.4800 = 9600 pounds; and for bending we have a 7/8" circular beam of 5/8" span. The safe bending-moment we previously found to be 1000 pounds-inch, the actual bending-moment is u.l/8, therefore u.5/8/8 = 1000 and u=12800 pounds, Or the value against bending would be 12800 pounds. As the bearing value (6562 pounds) is the smallest we will use that in determining the number of rivets in web.

Splicing the flange-

Thickness of web.

Where stiffeners are required.

Value of web rivets.

Fig. 208.