The total end shear is 117 800 pounds, and this is assumed to be carried by the two pairs of end-stiffener angles, each carrying one half. This amount would require lighter angles than the angles used for intermediate stiffeners. It is the customary practice to make them the same size and thickness as the intermediate stiffeners, additional strength being allowed in order to withstand the effects of the end cross-frame when in action.
The bearing determines the number of rivets required in each pair of stiffeners. The number required is e0p = 12 shop X 4 Q20 rivets.
Some engineers arbitrarily choose the stiffeners regardless of the shear, enough rivets, however, being put in the end stiffeners to take up all the shear; and the spacing in the intermediate stiffeners is made the same. One noted engineering firm determines its stiffeners according to the following: No rational method has as yet been determined for ascertaining the stresses in the stiffeners of plate-girders. Results obtained by placing extensometers on the stiffeners of actual plate-girders appear to indicate that the stresses are very small, in fact in most cases not being greater than 1 500 or 2 000 pounds per square inch.
I. Design, according to Cooper's Specifications, the end stiffeners _f the shear is 150 000 pounds, the distance back to back of angles is 6 feet 6!, inches, the web_ 1 inch thick, and the flange angle 6 by 6 by i-inch. Use fillers.
2. Design the intermediate stiffeners for the girder of Problem 1, above, where the shear is equal to 75 000 pounds. Use crimped stiffener angles. Note that in this case the angles lie close against the web, no filler bars used in between.
76. The Web Splice. Web splices are required because of the fact that wide plates cannot be rolled sufficiently long. Web splices should be as few as possible, and good practice demands that they be • placed at the same points as the stiffener angle.
The tables on page 30 of the Carnegie Handbook give the extreme length of plates which can be procured for any given width. The length of plates for widths which are not given in these tablcs,should be taken equal to the length of the next plate given whose width is less than that of the desired plate. From the first table it is seen that a 74 by -mch plate can be rolled up to 400 inches, or, 33 feet 4 inches, in length. Therefore, if the girder under consideration is spliced
at the center, the web plates will be required to be 61 ft in. feet 10.1 inches, which value does not exceed the 33 feet 4 inches as given above.
According to Articles 46 and 71 of the Specifications, a plate must be placed on each side of the web as shown in Fig. 159, and enough rivets placed in each side to take the total shear. The total thickness of both plates, and also their length, must be sufficient to stand the total shear, but must not be less than inch.
The total shear at the center of the girder under consideration (see Fig. 134, p. 150) is 28 600 pounds.
The area required in each of the two splice plates is 28 600 2 X 9000 = 1.59 square inches; and as their length is 62.25 inches, the thickness must be 1.59 _ 62.25 = 0.0255 inch, but they must be made s inch thick according to the Specifications. The width should be somewhat greater than twice the width of the stiffener angle leg. This would make the width in this case about 10 inches.
The bearing governs the number of rivets required in this case, and they are 28 600 _ 4 920 = 5.S1, say 6, shop rivets. More rivets than this will be required by practical considerations, as indicated by Article 54 of the Specifications or in order to make the spacing in the stiffener angle the same as that in the other stiffeners. This detail is to be left to the draftsman, the required number only being put on the stress sheet In cases of the gross area of the web is considered as efficient flange area, then provision must be made in the splice for the bending moment which the web takes. A very economical and efficient splice is shown in Fig. 160. The horizontal plates take the stress due to the moment, and the vertical plates take the stress due to the shear.
The web equivalent is 3.47 square inches and the total moment is 1 615 000 pound-feet, which is composed of 275 000 pound feet due to dead load and 1 340 000 pound-feet due to live load. Therefore that proportion of the 3.47 which is taken up by the dead load is: 275 000 X 3.47 = 0.59 square inch; 1 615 000 and that pmportion taken up by the live load is: 1 340 000 1 615000 X 3.4i = 2.88 square inches.