COLUMNS 309. Methods of Reinforcement. The laws of mechanics, as well as experimental testing on full-sized columns of various structural materials, show that very short columns, or even those whose length is ten times their smallest diameter, will fail by crushing or shearing of the material. If the columns are very long, say twenty or more times their smallest diameter, they will probably fail by bending, which will produce an actual tension on the convex side of the column. The line of division between long and short columns is practically very uncertain, owing to the fact that the center line of pressure of a column is frequently more or less eccentric because of irregularity of the bearing surface at top or bottom. Such an eccentric action will cause buckling of the column, even when its length is not very great. On this account, it is always wise (especiall• for long columns) to place reinforcing bars within the column. The reinforcing bars consist of longitudinal bars (usually four, and sometimes more with the larger columns), and bands of small bars spaced from 6 to 18 inches apart vertically, bind together the longitudinal bars. The longitudinal bars are used for the purpose of providing the necessary transverse strength to prevent buckling of the column. As it is practically impossible to develop a satisfactory theory on which to compute the required tensional strength in the convex side of a column of given length, without making assumptions which are themselves of doubtful accuracy, no exact rules for the sizes of the longitudinal bars in a column will be given. The bars ordinarily used vary from inch square to 1 inch square; and the number is usually four, unless the column is very large (400 square inches or larger) or is rectangular rather than square. It has been claimed by many, that longitudinal bars in a column may actually be a source of danger; since the buckling of the bars outward may tend to disinte grate the column. This buckling can be avoided, and the bars made mutually self-supporting, by means of the bands which are placed around the column. These bands are usually 1-inch or round or square bars. The specifications of the Prussian Public Works for 1004 require that these horizontal bars shall be spaced a distance not more than 30 times their diameter, which would be V, inches for 1-inch bars, and 11 inches for bars. The bands in the column are likewise useful to resist the bursting tendency of the column, especially when it is short. They will also reinforce the column against the tendency to shear, which is the method by which failure usually takes place. The angle between this plane of rupture and a
plane perpendicular to the line of stress, is stated to be 60°. If, therefore, the bands are placed at a distance apart equal to the smallest diameter of the column, any probable plane of rupture will intersect one of the bands, even if the angle of rupture is somewhat smaller than 60°.
The unit working pressure permissible in concrete columns is usually computed at from 350 to 500 pounds per square inch. The ultimate compression for transverse stresses for 1:3:5 concrete has been taken at 2,000 pounds per square inch. With a factor of 4, this gives a working pressure of 500 pounds per square inch; but the ultimate stress in a column of plain concrete is generally less than 2,000 pounds per square inch. Tests of a large number of 12 by 12 inch plain concrete columns showed an ultimate compressive strength of approximately 1,000 pounds per square inch; but such columns generally begin to fail by the development of longitudinal cracks. These would be largely prevented by the use of lateral reinforcement or bands. Therefore the use of 500 pounds per square inch as a working stress for columns which are properly reinforced, may be considered justifiable although not conservative.
310. Design of Columns. It may be demonstrated by theoreti cal mechanics, that if a load is jointly supported by two kinds of material with dissimilar elasticities, the proportion of the loading borne by each will be in a ratio depending on their relative areas and moduli of elasticity. The formula for this may be developed as follows: The actual linear compression of the concrete equals that of the steel; therefore, C s= E, From this equation, since r = , we may write the equation re=s Solving the above equation for C, we obtain: Example 1. A column is designed to carry a load of 160,000 pounds. If the column is made 18 inches square, and the load per square inch to be carried by the concrete is limited to 400 pounds, what must be the ratio of the steel, and how much steel would be required? Answer. A column 18 inches square has an area of 324 square inches. Dividing 160,000 by 324, we have 494 pounds per square inch as the total unit-compression upon the concrete and the steel, which is C in the above formula. Assume that the concrete is 1 : 3:5 concrete, and that the ratio of the moduli of elasticity (r) is therefore 12. Substituting these values in Equation 41, we have: 494 — 400 400 (12 — .0214.