83. Rankine's Column Formula. When a perfectly straight column is centrally loaded, then, if the column does not bend and if it is homogeneous, the stress on every cross-section is a uniform compression. If P denotes the load and A the area of the cross-section, the value of the unit-compression is P ÷ A.

On account of lack of straightness or lack of uniformity in material, or failure to secure exact central application of the load, the load P has what is known as an "arm" or "leverage " and bends the column more or less. There is therefore in such a column a bending or flexural stress in addition to the direct compressive stress above mentioned; this bending stress is compressive on the concave side and tensile on the convex. The value of the stress per unit-area (unit-stress) on the fibre at the concave side, according to equation 6, is Mc ÷ I, where M denotes the bending moment at the section (due to the load on the column), c the distance from the neutral axis to the concave side, and I the moment of inertia of the cross-section with respect to the neutral axis. (Notice that this axis is perpendicular to the plane in which the column bends.)

The upper set of arrows (Fig. 47) represents the direct compressive stress; and the second set the bending stress if the load is not excessive, so that the stresses are within the elastic limit of the material. The third set represents the combined stress that actually exists on the cross-section. The greatest combined unit-stress evidently occurs on the fibre at the concave side and where the deflection of the column is greatest. The stress is compressive, and its value S per unit-area is given by the formula,

S = P/A + M.c /I

Now, the bending moment at the place of greatest deflection equals the product of the load P and its arm .(that is, the deflection). Calling the deflection d, we have M = Pd', and this value of M, substituted in the last equa tion, gives

S=P /A + pdc / I.

Let r denote the radius of gyration of the cross-section with respec to the neutral axis. Then I = Ar2 (see equation 9); and this value, substituted in the last equation, gives

S = P /A + Pdc /Ar2 = p/A (1 + dc/r2

According to the theory of the stiffness of beams on end supports, deflections vary directly as the square of the length l, and inversely as the distance c from the neutral axis to the remotest fibre of the cross-section. Assuming that the deflections of columns follow the same laws, we may write d = k (b2 ÷ c), where k is some constant depending on the material of the column and on the end conditions. Substituting this value for d in the last equation, we find that

Fig. 47.

S= P/A (1+k l2/r2)

P/A = S / 1+k l2/r2 and P = SA / 1+K l2/r2

(10)

Each of these (usually the last) is known as "Rankme's formula."

For mild-steel columns a certain large steel company uses S = 50,000 pounds per square inch, and the following values of k:

1. Columns with two pin ends, k = 1 ÷ 18,000.

With these values of S and k, P of the formula means the ultimate load, that is, the load causing failure. The safe load equals P divided by the selected factor of safety - a factor of 4 for steady loads, and 5 for moving loads, being recommended by the company referred to. The same unit is to be used for I and r.

Cast-iron columns are practically always made hollow with comparatively thin walls, and are usually circular or rectangular in cross-section. The following modifications of Rankine's formula are sometimes used:

For circular sections, P/A = 80,000 / 1 + l2/800d2

, . For rectangular sections, P/A =80.000 / 1+l2/1,000 d2

(10')

In these formulas d denotes the outside diameter of the circular sections or the length of the lesser side of the rectangular sections. The same unit is to be used for l and d.

Examples. 1. A 40-pound 10-inch steel I-beam 8 feet long is used as a flat-ended column. Its load being 100,000 pounds, what is its factor of safety ?

Obviously the column tends to bend in a plane perpendicular to its web. Hence the radius of gyration to be used is the one with respect to that central axis of the cross-section which is in the web, that is, axis 2 - 2 (see figure accompanying table, page 72). The moment of inertia of the section with respect to that axis, according to the table, is 9.50 inches4; and since the area of the section is 11.76 square inches, r2 = 9.50/11.76= 0.808.

Now, l = 8 feet = 96 inches; and since k = 1 ÷ 36,000, and S = 50,000, the breaking load for this column, according to Rankine's formula, is

P =(50,000X11.76)/1+962=446,790 pounds

1+ 962/(36,000 x 0.808)

Since the factor of safety equals the ratio of the breaking load to the actual load on the column, the factor of safety in this case is

446,790/100,000 =4.5(approx.).

2. What is the safe load for a cast-iron column 10 feet long with square ends and a hollow rectangular section, the outside dimensions being 5x8 inches; the inner, 4x7 inches; and the factor of safety, 6 ?

In this case l = 10 feet = 120 inches; A = 5x8-4X7 = 12 square inches; and d = 5 inches. Hence, according to formula 10', for rectangular sections, the breaking load is

P=(80,000 X 12)1202 = 610,000 pounds.

1+ 1202/1,000 X52

Since the safe load equals the breaking load divided by the factor of safety, in this case the safe load equals

610,000/6 = 101,700 pounds.

3. A channel column (see Fig. 46, b) is pin-ended, the pins being perpendicular to the webs of the channel (represented by A A in the figure), and its length is 16 feet (distance between axes of the pins). If the sectional area is 23.5 square inches, and the moment of inertia with respect to AA is 386 inches4 and with respect to BB 214 inches4, what is the safe load with a factor of safety of 4 ?

The column is liable to bend in one of two ways, namely, in the plane perpendicular to the axes of the two pins, or in the plane containing those axes.