For Member L2, L5: The required net area is 24 950 ÷ 15 000 = 1.67 square inches. From Carnegie Handbook, p. 115, two angles 2½ by 2 by ¼-inch give a gross area of 2 X 1.06 = 2.12 square inches; and taking out two 5/8-inch rivets, the net area is 2.12 - ¼ (⅝ + ⅛) X 2 = 1.74 square inches. This coincides very closely with the required area, and this angle will be used. Even if this angle should have been in excess of the required area, it would still be necessary to use it, since it is the smallest angle and of the least thickness allowed.

For Member L3 U4: The required net area is 18 725 ÷ 15 000 = 1.25 square inches. Two angles 2\ by 2 by ¼-inch give a gross area of 2.12 square inches, and a net area of 1.74 square inches, as above computed. Although they give an area considerably larger than that required, nevertheless they must be used, since they are the smallest allowed.

For Members L1 U2 and U2 L3: The required net area is 6 240 ÷ 15 000 = 0.42 square inch. One angle 2½ by 2 by ¼-inch gives a gross area of 1.06 square inches. The amount to deduct from this is ¼ X (⅝ + ⅛) = 0.19 square inch, one ⅝-inch rivet-hole being taken from the section. This gives a net area of 1.06 - 0.19 = 0.87 square inch, which shows this angle to be sufficient.

Since the member U4M has no other use than to prevent the bottom chord from sagging, it will be made of the lightest angle allowed. It will therefore be made of one angle 2½ by 2 by ¼-inch.

The member L2 L3 is made of the same section as the member L3 U4 since this is more economical than to change the section and to make a splice at L3. It will be made of two angles 2½ by 2 by ¼-inch.

Design of the Compression Members. The general method of procedure in the design of compression members is, first, to assume a cross-section, and then to determine the unit compressive stress allowable by inserting the length of the member and the radius of gyration of the assumed section in the formula given for the unit allowable compressive stress; then divide the stress in the member by the unit allowable compressive stress determined as above. This will give the required area. If this required area is equal to, or slightly less than, the area of the cross-section assumed, the section assumed will be the correct one. If the required area as computed above is greater than the area of the section, then a larger section must be assumed and the operation repeated. Usually only two operations are required in order to obtain a section whose area is correct. It should be noted that the area of the rivet-holes is not deducted from the section in compression members, since the rivet fills up the rivet-hole and makes a section as strong in compression as it was in the first place. Care should be taken to assume a section whose radius of gyration is equal to or greater than the length of the member divided by 120. This is due to the fact that l ÷ r should not be greater than 120. Compression members of roof trusses for the usual spans are made of two angles placed back to back. The radius of gyration of such a section is equal to the radius of gyration of one angle, if it is referred to an axis perpendicular to the legs which are placed together. If it is referred to an axis through the center of the section and parallel to the legs which are placed together, it is equal to some value other than the radius of gyration of one angle. The radii of gyration for pairs of angles placed either directly back to back or a small distance apart, are given on pages 144 to 146 of the Carnegie Handbook and in Table XI, and should be used in the design. The value of the radius of gyration for sizes of angles other than those given, may be obtained by interpolation.

For example, let it be required to determine the radius of gyration of two 5 by 3½ by ½-inch angles placed ½ inch apart and back to back, the 5-inch legs being horizontal (see p. 146, Carnegie Handbook). Since this value is not given in the tables, it must be interpolated from the values given for r2 for the above sized angle, which are 5/16 inch and ⅞ inch thick. The difference between the two thicknesses given is ⅞ - 5/16 = 9/16 inch. The difference between the two values given for the radius of gyration is 2.55 - 2.44 = 0.11. This gives a difference of .11 ÷ 9 = 0.0122 for each 1/16 inch difference in thickness in the angle. The difference between the thickest angle and the angle under consideration is ⅞ - ½ = ⅜, or 6/16. Therefore the amount to be subtracted from the radius of gyration of the thickest angle is 6 X 0.0122 = 0.0732; and the radius of gyration for two angles placed back to back as above stated is 2.55 - 0.07 = 2.48. In case one angle is used for a member in compression, the least rectangular radius of gyration must be used; and if two angles are employed, placed back to back, care should be exercised to use the least radius of gyration; and if the angles have unequal legs, those legs should be placed back to back, which will make the rectangular radii of gyration as nearly equal as possible. The values of the radii of gyration will indicate whether the short legs or the long legs should be placed together. The tables given in the Carnegie Handbook give the radii of gyration for angles spaced at distances ½ inch and ¾ inch apart; but since the connection plates of roof trusses are usually ¼ inch or ⅜ inch thick, the values of the radii of gyration should be given for angles spaced \ inch and f inch apart. Such values are given in Table XL

HOUSE AT ST. PAUL, MINN.

John T. Comes, Architect, Pittsburg, Pa.

Cost of House, \$7,500. For Exteriors, See Page 250.

HOUSE AT ST. PAUL, MINN.

John T. Comes, Architect, Pittsburg, Pa.