89. Design of Columns. All the preceding examples relat ing to columns were on either (1) computing the factor of safety of a given loaded column, or (2) computing the safe load for a given column. A more important problem is to design a column to sustain a given load under given conditions. A complete dis cussion of this problem is given in a later paper on design. We show here merely how to compute the dimensions of the cross section of the column after the form of the cross-section has been decided upon.
In only a few cases can the dimensions be computed directly (see example 1 following), but usually, when a column formula is applied to a certain case, there will be two unknown quantities in it, A and r or d. Such cases can best be solved by trial (see examples 2 and 3 below).
Eaample. 1. What is the proper size of white pine column to sustain a load of 80,000 pounds with a factor of safety of 5, when the length of the column is 22 feet ? We use the parabola formula (equation 13). Since the safe load is 80,000 pounds and the factor of safety is 5, the breaking load P is 80,000 X 5 = 400,000 pounds.
The unknown side of the (square) cross-section being denoted by d, the area A is Hence, substituting in the formula, since / 22 feet = 264 inches, we have Hence = 176.73, or d = 13.3 inches.
2. size of cast-iron column is needed to sustain a load of 100,000 pounds with a factor of safety of 10, the length of the column being 14 feet ? We shall suppose that it has been decided to make the cross section circular, and shall compute by Rankine's formula modified for cast-iron columns (equation 10'). The breaking load for the
column would be or, reducing by dividing both sides of the equation by 10,000, and then clearing of fractions, we have 100 i1 + 1 = 8A.
L There are two unknown quantities in this equation, d and A, and we cannot solve directly for them. Probably the best way to pro ceed is to assume or guess at a practical value of d, then solve for A, and finally compute the thickness or inner diameter. Thus, let us try d equal to 7 inches, first solving the equation for A as far as possible. Dividing both sides by 8 we have The area of a hollow circle whose outer and inner diameters are d and respectively, is 0.7854 — Ilence, to find the inner diameter of the column, we substitute 7 for d in the last expres sion, equate it to the value of A just found, and solve for d,. Thus, 0.7854 (49 — = 21.5: hence 21.5 49 (112 = 0.7854 = and =-- 49 — 27.37 = 21.63 or 4.65.
This value of d makes the thickness equal to (7 — 4.65) = 1.175 inches, which is safe. It might be advisable in an actual case to try d equal to 8 repeating the computation.* 1. What size of white oak column is needed to sustain a load of 45,000 pounds with a factor of safety of 0, the length of the column being 12 feet.
Ans. d = 81, practically a 10 X 10-inch section