tricity is greater than that given by Formula (14), the procedure for reinforcedconcrete members subject to tension on a part of the section should be followed. Example 5. A column with rectangular cross-section is 15 ft in height. It has section-dimensions and reinforcements as shown in Fig. 8, and sustains an N = 150000 lb. Fig. 8. Reinforced-Concrete Member. Rectangular Cross-Section. Symmetrical Reinforcement eccentric load of 150 000 lb, acting 3 in from the median axis. It is required to determine the MAXIMUM AND MINIMUM FIBER-STRESSES IN THE CONCRETE, when n = 15. In this Example the GREATEST ECCENTRICITY which will cause no tension in the section is, by Formula (14), t2 + 12(n 1)pa2 (24)2 + 12 × 14 × 0.0178 × (9)2 = e = Using the diagram in Fig. 12 for the solution of this example 4.54 in e Beginning at the bottom of the diagram with the value of, tracing vertically 1.8% and then to the left, N Median Axis = 580 lb per sq in As the diagram of Fig. 12 is plotted for a value of a = 0.4t, that is, an insulation measured from the center of the reinforcement equal to one tenth the thickness of the section, which is somewhat less than that of this particular example, the result obtained is only approximate. For CIRCULAR SECTIONS with SYMMETRICAL REINFORCEMENT, Fig. 9, if r = radius of section to the center of the steel, and if the concrete outside of the core is neglected Ic+ (n-1)I, Example 6. A column with circular cross-section is 15 ft in height, has a core 24 in in diameter and is reinforced with twelve 1-in round rods. It has an eccentric load of 225 000 lb acting 3 in from the center. It is required to determine the MAXIMUM AND MINIMUM FIBER-STRESSES IN THE CONCRETE, when n = 15. 5. Reinforced-Concrete Members. Tension over Part of Section. When the minimum stress computed by Formula (15) is a tensile stress and has a numerical value greater than the ALLOWABLE WORKING UNIT STRESS for plain concrete, it is often possible to increase either the sectional area or the percentage of reinforcement until the unit stress falls within the allowable limit. It is usually more practicable, however, to disregard the tensile resistance of the concrete and to apply sufficient steel to take care of the entire tensile stress. Median Axis Tension f" and Section - Area of Steel-Ag da B N Compression fc and f StressDiagram Section - Area of Steel-A's Reinforced-Concrete Member. Rect- (16) (17) Since N is the resultant of the forces acting normal to the section, for a condition of equilibrium it must equal the sum of the stresses in the two materials, or Again, since the bending moment upon the section must equal the sum of the moments of the areas of the concrete and steel about the point O where the median axis intersects the line BC representing, in projection, the surface of the section, From these four equations of Formulas (16) to (19), are determined the values of k, fe, fs, and f', in any given problem when the section-area, reinforce Median Axis N Neutral Axis ment, M, and N are known. The work, however, is somewhat tedious as it is. necessary to solve a cubic equation in order to obtain the value of k, and in most cases the use of a SYMMETRICAL REINFORCEMENT is advisable under the conditions ordinarily encountered in building design. Tension fe and fs Compression fe and f's StressDiagram Section Fig. 11. Reinforced-Concrete Member. Rectangular Cross-Section. Symmetrical Rein forcement t the equation for the moment is 2' Again, since the moment, M, is equal to the force, N, multiplied by the eccentricity, e, the second member of Formula (20) multiplied by e is equal to the second member of Formula (21). From this equality the following formula containing k is derived: The diagrams of Fig. (13) are drawn to give the values of k for any known PERCENTAGE OF REINFORCEMENT, e and RATIO Having determined k, the cor t responding value of the term F, used for convenience to designate the expression npa2 k k2 is taken from the diagrams of Fig. 14 or Fig. 15. The STRESS IN THE CONCRETE is then determined by the formula, The STEEL-STRESSES, f', and f, are given by Formulas (16) and (17). Example 7. A column with rectangular cross-section is 15 ft in height, and has a thickness t, of 16 in and the breadth b, of 36 in. The value of n is 15, the distance of the reinforcement from the surface is 11⁄2 in, and a is 6 in. It is required to determine the MAXIMUM AND MINIMUM FIBER-STRESSES in the concrete and steel, when there is 0.5% of vertical reinforcement, and when the column sustains an eccentric load of 115 000 lb, acting 4 in from the median axis e = Beginning at the top of the diagram in Fig. 13, with the value of tracing ť vertically downward to the intersection of the curve for p 0.5 %, and then to the left, it is found that k = 0.81. With this value of k, beginning at the bottom of the diagram in Fig. 14, tracing vertically upward to the curve for Р 0.5 o/c, and then to the left, it is found that F = = 0.109 458 lb per sq in = [15 X 458] X 1.5 = 0.81 X 16 6080 lb per sq in (compression) |