Examples. 1. A 40-pound 10-inch steel I-beam column

8 feet long sustains a load of 100,000 pounds, and its ends are flat. Compute its factor of safety according to the methods of this article.

The first thing to do is to compute the ratio l ÷ r for the column, to ascertain whether the parabola formula or Euler's formula should be used. As shown in example 1 of the preceding article, l ÷ r = 106⅔. This ratio being less than the limiting value, 190, of the table, we should use the parabola formula. Hence, since the area of the cross-section is 11.76 square inches (see Table C, page 72),

P/11.76=42,000 - 0.62 (106 ⅔) 2; or, P = 11.76 [42,000 - 0.62 (106⅔)2] = 410,970 pounds.

This is the breaking load according to the parabola formula; hence the factor of safety is

410,970/100,000=4.1

2. A white pine column 10 X 10 inches in cross-section and 18 feet long sustains a load of 40,000 pounds. What is its factor of safety?

The length is 18 feet or 216 inches; hence the ratio l ÷ d = 21.6, and the parabola formula is to be applied. Now, since A = 10 X 10 = 100 square inches,

P/100=2,500 - 0.6 X 21.62; or, P = 100 (2,500 - 0.6 X 21.62) = 222,000 pounds.

This being the breaking load according to the parabola formula, the factor of safety is

222,000/40,000=5.5

3. What is the safe load for a long-leaf yellow pine column 12 X 12 inches square and 30 feet long, the factor of safety being 5?

The length being 30 feet or 360 inches, l/d=360/12=30; hence the parabola formula should be used. Since A = 12 X 12 = 144 square inches,

P/144=4,000 - 0.8 X 302; or, P = 144 (4,000 - 0.8 X 302) = 472,320 pounds.

This being the breaking load according to the parabola .Formula, the safe load is

472,320/5= 94,465 pounds.

## Examples For Practice

1. A 40-pound 12-inch steel I-beam 10 feet long is used as a flat-ended column. Its load being 100,000 pounds, compute its factor of safety by the formulas of this article.

Ans. 3.8

2. A white oak column 15 feet long sustains a load of

30,000 pounds. Its section being 8x8 inches, compute the factor of safety by the parabola formula.

Ans. 6.6

3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and has square ends; the least radius of gyration of its cross-section is 3.1 inches; and the area of its cross-section is 24.5 square inches. Compute the safe load for the column by the formulas of this article, using a factor of safety of 4.

Ans. 224,500 pounds.

4. A short-leaf yellow pine column 14 X 14 inches in section is 20 feet lono-. "What load can it sustain, with a factor of safety of 6 ?

Ans. 101,100 pounds.

88. "Broken Straight-Line" Formula. A large steel company computes the strength of its flat-ended steel columns by two formulas represented by two straight lines AB and BC, Fig. 52. The formulas are

P/A= 48,000,

P/A=68,400-228 l/r, and

P, A, l, and r having the same meanings as in Art. 83.

The point B corresponds very nearly to the ratio l ÷ r = 90. Hence, for columns for which the ratio l÷ r is less than 90, the first formula applies; and for columns for which the ratio is greater than 90, the second one applies. The point C corresponds to the ratio b ÷ r = 200, and the second formula does not apply to a column for which b ÷ r is greater than that limit. Fig. 52.

The ratio I ÷ r for steel columns of practice rarely exceeds 150, and is usually less than 100.

Fig. 53 is a combination of Figs. 49, 50, 51 and 52, and represents graphically a comparison of the Rankine, straight-line, Euler, parabola-Euler, and broken straight-line formulas for flat-ended mild-steel columns, It well illustrates the fact that our knowledge of the strength of columns is not so exact as that, for example, of the strength of beams. Fig. 53.

89. Design of Columns. All the preceding examples relating 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 discussion 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).

Example. 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 d2. Hence, substituting in the formula, since l = 22 feet = 264 inches, we have

400,000/d2=2,500-0.6 2642/d2.

Multiplying both sides by d2 gives

400,000 = 2,500 d2 - 0.6 X 2642, or 2,500 (d 2 = 400,000 + 0.6 x 2642 = 441,817.6.

Hence d 2 = 176.73, or d = 13.3 inches.

2. What 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

100,000 X 10 = 1,000,000 pounds. The length is 14 feet or 168 inches; hence the formula becomes

1,000,000/A=80,000/1682

1+1682/800d2 or, reducing by dividing both sides of the equation by 10,000, and then clearing of fractions, we have

100 [1 +1682/800d2] =8A.

There are two unknown quantities in this equation, d and A, and we cannot solve directly for them. Probably the best way to proceed 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

A=100/8[1+1682/800d2], and, combining,

A = 12.5 + 441/d2.

Now, substituting 7 for d, we have

A = 12.5 + 441/49= 21.5 square inches.

The area of a hollow circle whose outer and inner diameters are d and d respectively, is 0.7854 (d2 - d2). Hence, to find the inner diameter of the column, we substitute 7 for d in the last expression, equate it to the value of A just found, and solve for d . Thus,

0.7854 (49 - d12) = 21.5. hence

49-d12 =21.5/0.7854=27. 37; and d12 = 49 - 27.37 = 21.63 or d1 = 4.65.

This value of d makes the thickness equal to

½ (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.*

## Example For Practice

1. "What size of white oak column is needed to sustain a load of 45,000 pounds with a factor of safety of 6, the length of the column being 12 feet.

Ans. d = 8½, practically a 10 X 10-inch section