The top part of the portal bracing will consist of two. angles. Two angles 31 by 3 by *-inch will be assumed and examined to determine if the area is sufficient. The length of this strut is the distance center to center of trusses, and is equal to 17 X 12 = 204 inches. The least radius of gyration is therefore = 1.70. The radius of gyration of the two angles assumed is 1.72 when referred to an axis parallel to the shorter leg when the two angles are placed back to back and one-half inch apart. The unit-stress is now computed: P = (to 000 70 X 1; 9 625 pounds per square inch.
The required area is 27 200 5 9625 = 2.825 square inches. This is con siderably less than the area given by the two angles; but as these are the minimum angles allowable, they must be used. Since the stress in this case is less than in the previous -case, and since. the angles used are the same, it is evident that these angles are safe in tension. The number of rivets is determined by the bearing in the i-inch connection plates, and is: As in the case of the lateral strut, this member should be connected by both legs of the angle in order to reduce the size of the connection plates. Fig. 190 gives the details of the portal cing and its method of connection to the post. The full circles represent shop rivets, and the blackened circles resent field rivets. Some engineers connect the portal bracing to the top cover-plate of the post. This produces an excessive eccentricity in the end-post and is bad practice.
Those members of the portal bracing which do not take any stress will be made of single angles, and the size of these angles will be taken 3 by 3 by -inch.
90. The Transverse Bracing. This bracing will be the same general style as the portal bracing, except that the top member will consist of two angles placed at a distance apart equal to the depth of the top chord, and these angles will be joined together by lacing. As in the case of portal bracing, those members which do not take stress will be made of one angle 31 by 3 by 1--inch.
The general outline is shown in Fig. 191, and the stresses are com puted from (10) and by the methods of Article 54, Part I. In design ing this top member, the top angle only is supposed to take the stress. The length in this case is 204 inches. Two 31 by 3 by i-inch angles will be assumed as sufficient and will be examined. These angles
give a total area of 4.60 square inches. In examining these it will be found that they are amply sufficient, in fact so much so that it will be better to see if one single angle at the top will not be better. According to the length, the smallest radius of gyration which can be used is 1.7. In looking over the tables of angles, it is seen that the first angle to fulfil this condition is a 6 by 31 by i-inch, and it has a radius of gyration of 1.94. The allowable unit-stress is computed as follows: P = (16 000 70 X i 10 780 pounds per square inch; and the required area is 780 = 0.85 square inch. This is con siderably smaller than the area of the angle, which is 3.97 square inches; but since this is the smallest possible angle which will fulfil the conditions of the Specifications, and since it is much smaller than the two angles as first assumed, it will be used: Fig. 192 gives a cross section of this member. Since this angle is joined to the cover-plate by one leg, the joints will be weak in single shear, and the number of rivets required will be: 2 X 9 100 =2 shop rivets, or 7220X 1} 2 X 0 100 3 field rivets. 6013X 1} According to (45), the width of the latticing must be 2i inches; and according to Table XXV, the thickness must be 776 inch, the distance c be ing 1 foot 11 inches.
The length of the knee-bracing is 144 inches; but on account of the small stress, one angle will be used. One 4 by 3 by i-inch angle, with an area of 2.4S square inches and a radius of gyra tion 1.26, will be assumed as sufficient. The radius of gyration is greater than the minimum allowable, which is 1.2. The allowable unit-stress is: P = ( 16 000 70 X 1 26) 13 = 10 000 pounds per square inch: The required area is 3000 = 1.23 square inches. The required area is much less than the given area; but this angle must be used, since it is the only one allowed on account of its radius of gyration. Two of the minimum sized angles might have been used; but their total area, 4.60 square inches, is much in excess of that of the angle used.
This angle must be examined for tension. The net area is 2.4S (j I-) X = 2.1 square inches. The required net area in tension is 12 300= 0.615 square inch, which shows this angle 16 000X 14 to be amply sufficient.