For Exteriors, See Page 250.

## Table XI. Radii Of Gyration Of Angles Placed Back To Back Equal Legs Size (Inches) T1 r2 2 X 2 X 3/16 0.93 0.98 2 X 2 X 7/16 0.98 1.03 2½ X 2½ X 1 1.14 1.19 2½ X 2½ X ½ 1.19 1.24 3 X 3 X 1 1.34 1.39 3 X 3 X 5/8 1.41 1.46 3½ X 3½ X 3/8 1.56 1.61 3½ X 3½ X 13/16 1.65 1.70 4 X 4 X 5/16 1.76 1.80 4 X 4 X 1.3/16 1.85 1.89 6 X 6 X 3/16 2.58 2.63 6 X 6 X 7/8 2.66 2.70

Unequal Legs Size (Inches) r1 r2 r2 r2 2½ X 2 X 3/16 0.88 0.92 1.19 1.24 2½ X 2 X ½ 0.94 0.99 1.25 1.30 3 X 2½ X ¼ 1.09 1.13 1.40 1.45 3 X 2½ X 9/16 1.15 1.20 1.46 1.51 3½ X 2½ X ¼ 1.04 1.09 1.67 1.72 3½ X 2½ X 11/16 1.13 1.18 1.75 1.80 3½ X 3 X 5/16 1.30 1.35 1.61 1.66 3½ X 3 X 13/16 1.40 1.45 1.71 1.76 4 X 3 X 5/16 1.25 1.30 1.88 1.93 4 X 3 X 1 3 1.35 1.40 1.97 2.02 5 X 3 X 5/16 1.17 1.22 2.42 2.47 5 X 3 X 1 3 1.27 1.32 2.52 2.57 5 X 3½ X 3/8 1.42 1.46 2.36 2.41 5 X 3½ X 7 1.51 1.56 2.45 2.50 6 X 3½ X 3/8 1.34 1.39 2.90 2.95 6 X 3½ X 1 1.44 1.49 3.00 3.05 6 X 4 X 3/8 1.58 1.62 2.83 2.87 6 X 4 X 7 1.67 1.71 2.92 2.97

r0 = in all cases, the radius of gyration of one angle referred to neutral axis parallel to the horizontal leg as shown above.

For Member L0 U1: Two angles 3½ by 3 by 5/16-inch, long legs spaced back to back, and ¼ inch apart, will be assumed. The least radius of gyration is 1.10, and the length is 8.9 feet. The area of this section is 2 X 1.93 = 3.86 square inches. The unit allowable compressive stress is:

P = 24 000 - [110 X 12 X 8.9] / 1.10 = 13 400 Pounds.

The required area is 48 800 ÷ 13 400 = 3.65 square inches. Since the angles given are of somewhat larger area than that required, it might be well to examine the next smallest angle.

Two angles 3½ by 2½ by 5/16-inch, with a radius of gyration

1.11 and a total area of 3.56 square inches, will be assumed. The unit allowable compressive stress is:

P = 24 000 - 112 X 12 X 8.9 / 1.11 = 13 510 pounds.

The required area is 48 800 ÷ 13 510 = 3.61 square inches. Since the required area is greater than the given area, it shows that these angles are too small. Two angles 3½ by 3 by 5/16-inch will therefore be used for this member, and also for all the members of the top chord, since it is more economical to run the same size throughout than to change the section and make splices at all the upper chord panel points.

For Member U2 L2: The length of this member is easily computed from similar triangles, and is found to be 8.9 feet. Two angles 2½ by 2 by 5/16-inch, with the long legs back to back, give a total area of 1.62 square inches and a radius of gyration of 0.78. The unit-stress is computed and found to be 8 950 pounds. The required area is 11 150 ÷ 8 950 = 1.25 square inches. These two angles would be used, but the least allowable radius of gyration is 8.9 X 12 ÷ 120 = 0.89. This is seen to be considerably greater than the radius of gyration given above, and therefore these angles cannot be used, according to Specifications. By consulting the tables, it is seen that two angles 3 by 2½ by ¼-inch are the smallest angles that will give a radius of gyration nearest to the required amount (0.89) and still be standard size angles. Angles marked with a star in the tables are special angles, and can be procured only at a cost greatly in excess of the others, and then only with great delay in delivering except when large quantities are ordered. It may be said that standard angles should never be used.

For Members U1 Ll and U3 L3: The length of these members is 4.45 feet. The radius of gyration must therefore not be less than 4.45 X 12 ÷ 120 = 0.45. One angle 2\ by 2 by ¼-inch, with an area of 1.06 square inches and a least rectangular radius of gyration of 0.59, will be assumed. The.allowable unit compressive stress is:

P = 24 000 -[ 110 X 12 X 8.9/ 2 ] / 0.59 = 14 050 pounds.

The required area is 5 580 ÷ 14 050 = 0.40 square inch. The angle chosen gives a much larger area than that required; but since it is the smallest one allowed by the Specifications, it must be used.

Many designers do not place a limit on the value of the radius of gyration, but simply use the compressive formula, and any section whose radius of gyration will bring the required area near to its own area. This should not be the case, since the formula here given is not applicable when the value of l ÷ r is greater than 120.

Top and Bottom Lateral Bracing. Since the stresses in the lateral bracing are not susceptible of a well-defined mathematical analysis, it cannot be rationally designed. Experience indicates that it should be as in Article 7. The lower chord bracing will therefore consist of single angles 3 by 2½ by 5/16 inch; and the upper chord bracing, of 3 by 3 by 5/16 inch angles. This bracing should not be placed between every truss, but should be placed as indicated on the stress sheet, Plate I. If one ¾-inch rivet is taken out of the section of the bottom lateral bracing, it will give a net area of 1.62 - 0.27 = 1.35 square inches; this could withstand a stress of 1.35 X 15 000 X 1.25 = 27 000 pounds, which is the stress the bracing is assumed to carry, and which is to be used in determining the number of rivets for the connection. The stress in the top lateral bracing may be assumed to be the same.

Determination of Number of Rivets Required. It is to be remembered that ⅝-inch rivets are to be used in the 2½ and 2-inch legs of the angles, and ⅜-inch rivets in all larger legs. Field rivets are to have a value equal to f of a shop rivet. Connection plates ¼ inch thick are to be used in all cases, except where the number of rivets required will be greater than 10. In such cases, use a ⅜-inch connection plate. The correct number of field rivets may be determined by multiplying the required number of shop rivets by 4/3.

Whenever two angles back to back join on a plate, the number of rivets is governed by the bearing on the connection plate; and when one angle is joined to a plate, the number of rivets is governed by single shear if the rivet is ⅝ inch in diameter, and by single shear if the rivet is ¾ inch in diameter and the plate is over ¼ inch thick. The bearing and shearing value of the rivets are taken from Table X, p.47.

Lower End of L0 U1: Rivets ¾-inch. Plate ⅜-inch.

48 800 ÷ 5 630 = 9 shop rivets required. Upper End of U3 U4 : Rivets ¾-inch. Plate ⅜-inch.

40 500 ÷ 5 630 = 8 shop or 10 field rivets Upper End of U4 L3: Rivets |-inch. Plate ⅜-inch.

18 725 ÷ 4 690 = 4 shop or 6 field rivets.

Lower End of L2 L3: Rivets ⅝-inch. Plate ¼-inch.

12 174 ÷ 3 130 1 shop rivets. Each End of U2 L2: Rivets ⅝-inch. Plate ¼-inch.