If the panel points are to be located at the stiffeners, the number of panels is a function of the depth of the girder (see Specifications, Articles 47 and 48). In this case the number of panels is given by: Span in inches.
_ N Depth of girder in inches' no fraction being considered. As an example, let it be required to determine the number of panels in a girder 85 feet center to center of bearings, the depth being 904 inches back to back of angles. Then, 85_X 12 N = - = 11.3, or, panels.
90.25 Each panel will then be 92.8 inches long. This, according to Article 47 of the Specifications, being greater than 5 feet, would not be allowed as a space between two stiffeners; but one stiffener can be placed in between, and then the panel points will come at every other stiffener. The cross-frames should be five in number.
The arrangement of panels and cross-frames is shown in Fig. 149. Here the cross-frames are marked C. F., and the broken lines represent the lower lateral system.
In case an even number of panels were desired, then ten would be the number chosen and the general arrangement would be as shown in Fig. 150. The length of a panel would be 85 X 12 10 = 102 inches, or 8 feet 6 inches, which would allow of one stiffener in between and still keep the stiffener spacing within the limit of 5 feet.
The cross-frames at the ends of the span are designated as end cross-frames, and those in between are designated as intermediate cross-frames.
In case the spacing of the stiffeners is not required to be such as to coincide with the panel points of the lateral bracing, the panel length will depend upon the spacing of the girders, being equal to or greater than the spacing in order to keep the angle which the diagonals make with the girder Less than 45 degrees. In this case, Span in feet 1C Distance center to center of girders in feet For the girder considered on page 174, the number of panels would be 85 11.3—or, say, 11 panels—if odd numbers were to be used, 75 and 12 if even numbers were to be desired.
For the case in hand, the panel points of the bracing will be taken at the stiffeners, and an even number of panels will be used. Then, The arrangement of the panels and cross-frames, and also the maxi mum stresses in the diagonals, are shown in Plate II, the stresses being determined according to Article 50, Part I, `Bridge Analysis," and Article 24 of the Specifications. All the wind is taken as acting on
one side of the bridge; and no overturning effect, either on the girder or on the train, is considered. Also, note that the wind stresses in the flanges are not considered. Should the student determine these, he will find them too small to be considered according to Specifications, Article 39.
Before designing the lateral diagonals which consist of one or two angles, Articles 31, 33, 34, 35 (last portion), 38, 40, 63, and 83 of the Specifications should be care fully studied. The upper lateral bracing is to be designed first. Carnegie Handbook, pp. 109 to 119, is to be used.
The member must be designed for a compressive stress equal to 23.20 + 0.8 X 20.6 = —39.68, and a tensile stress of 20.6-I- 0.8X20.6 = +37.08. The length of the diagonal measured from center to center of girders is v'6.5' +6.2- = 9 feet, or 108 inches. In reality the length is not 108 inches, as the cover-plate takes off a certain amount, as shown in Fig. 151. The true length, which is to be taken as a column length in designing, is 108 — 2y, and y is readily computed to be 9.70 inches, thus making the true length 88.6 inches. The least allowable rectan gular radius of gyration is obtained from the relation that the greatest value of r = 120, and therefore the least value of r = = 0.74.
It will be assumed that a 6 by 4 by angle with an area of. 5.31 square inches will be sufficient. Here the length equals 88.60 inches, and the least rectangular radius of gyration is 1.14; hence, The required area is 38 3608 330 = 4.73 square inches. As the angle _assumed has an area of 5.31 square inches, which is considerably greater than the 4.73 square inches required, this angle cannot be used, and other assumptions must be made until the area of the angle assumed and the required area as computed are equal or very nearly so. A 6 by 4 by 1-inch angle with an area of 4.75 square inches will now be assumed. The length is 88.60 inches as before, and the least rectangular radius of gyration is 1.15. The unit-toad P = 8 340 pounds per square inch, and the required area is38 340 360 _ 4.72 square • 8 inches. As the area of the angle assumed' and the required area as computed are very close, this sized angle will be used.