Columns. Only two columns will be designed: one having only a part of the roof to support, and one that supports this column and the load on the upper floor. The first is designated A on Plate III, page [23], and the second B.
Column A. The load due to the weight of truss, wind and snow is 24000 lbs. A column composed of 4 angles 3½ × 2½ × 3
16–in. and 2 × ½–in. lacing will be used. The cross-section is shown in the figure
. The moment of inertia about the axis AB = 20.4 × 4 = 81.6 inches, and the distance C, from the center of gravity of the cross section to the most extreme fibre, = 4. The bending moment, M, caused by the wind on the roof = 1,442,000 inch pounds. Substituting these values in the formula M = SI ÷ C and solving for S, we obtain a value of 7100 lbs. per. sq. in. The stress per. sq. in. due to the weight of the trusses = 3300 lbs. Therefore the total stress = 7100 + 3300 = 10400 lbs. per. sq. in. The allowable stress = 16000 − (45 l
v). l
v = 13.7. Therefore the allowable stress = 11400 lbs. per. sq. in.
Column B. The dead load caused by column A is 12 tons, and the load on the column due to the second floor is 41.5 tons, making a total of 53.5 tons. The length of column is 12 ft. Try a column composed of four 3 × ½ in. Z bars laced. Half of the wind pressure on the windward side above the floor is transmitted by the roof and the lateral bracing to the columns on the leeward side of the building, and half is carried directly by the columns on the windward side. The wind pressure to be resisted by 10 columns = 46 × 30 × 20 = 27,600 lbs. The columns being fixed at the base, the total moment of the wind = 27,600 × 12 × 6 = 1,987,200 in. lbs. and the moment resisted by one column = 198,720 in. lbs. I ÷ C for this column = 35.1. By substitution in the equation M = SI ÷ C, S = 2300 lbs. per sq. in (approx.). The area of the column = 9.31 sq. in.; and the stress in the column due to the dead load = 83000 ÷ 9.31 = 8900 lbs. per. sq. in.
The other columns will be stressed less than this one; but this section will be used throughout for the columns on the lower floor.
Wind Bracing. Only the method of designing member AD (see Plate III, page [23]) will be explained. The wind pressure to be transmitted = 4800 lbs. The secant of the angle of inclination = 1.06. Therefore the stress in AD = 4800 × 1.06 = 5100 lbs. A ¾–in. round rod will be used.
In the same way the sizes of the members CE, EF, and FG are determined.
Foundation. The maximum load for the column is about 55 tons. The foundation will be built on piling, as experiments made by F. J. Llewellyn—Engineering News, May 11 1899—show that the safe load for the soil at New Orleans is only about 700 lbs. per. sq. ft. Nine piles will be used in supporting each column. The depth of pile necessary to safely support the load will be found by driving a few trial piles and using what is known as the Engineering News formula (Baker’s Masonry Construction page 245) P′ = 2Wh ÷ d + 1, in which P′ = the safe load in tons; and d is the penetration in inches under the last blow. W is the weight of the hammer in tons; and h h is the fall in feet. The piles will be of good quality straight-grained white oak, and before being driven, the entire bark will be stripped off. No pile less than 15 in. at the top will be used. The piles will be spaced 3 ft. center to center. A detail drawing of the foundation is shown in Plate III, page [23]. The concrete used in the foundation will be composed of 1 part Louisville Natural cement and 4 parts of sandstone broken to pass a 2½–in. ring, the part passing a ½–in. ring being screened out. The concrete is assumed to have a compressive strength of 10 tons per sq. ft., and then the area of the cast iron base to support the column will be 55 ÷ 10 = 5.5 sq. ft. A base 30 inches square will be used.
The first floor will be composed of 6–inches of concrete made as that for the foundation resting directly on the ground. Over this will be spread ½–in. of neat Portland cement to give the floor an even surface.