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Since the concrete of the original beam carries no tension its outline is shown by dotted lines below the neutral axis, where it serves simply to transmit shearing stresses. Knowing the value of x the rest of the problem follows directly. Although no formulas have been written, all the means are now at hand for the design and investigation of reinforced concrete beams of any shape provided the section is symmetrical about the plane of bending. It is much easier to solve problems by the simple methods that follow logically and simply from the outline just given than it is to use formulas, unless plots and tables are at hand to use with the formulas.

Rectangular Concrete Beams with Tension Reinforcement

Three problems arise regarding rectangular concrete beams reinforced in tension only: (1) the design of a beam to carry a stated bending moment at given stresses; (2) investigation of the maximum stresses at a given section subjected to a known bending moment; (3) investigation of the maximum permissible bending moment for a given beam with certain limiting stresses given. It is better to consider the cases of investigation first. The problem of design is by far the most common in practice. The beam here shown (Fig. 12a) carries a bending moment of 40,000 ft.-lbs.; E =30,000,000 lbs./sq. in.; E = 2,000,000 lbs./sq. in. What are the maximum fiber stresses? Following the argument of the last article the transformed section is that here given (Fig. 12b) The neutral axis is located by noting that the statically moment of the compression area about that axis (which passes through the center of gravity) equals the statically moment of the tension concrete area about the same axis, giving the expression

(10x) = 30(20 �x) 2+6x= 120 Example 6. What is the maximum moment that can be carried by the beam of Example 5, the limiting fiber stresses being = 16,000 lbs./sq. in. and f, = 650 Ibs./sq. in.? n = 15.2 As before, transform the section (Fig. 12b), locate the neutral axis and determine the arm of the resisting couple, a = 17.2 in. If f, has its maximum value of 650 Ibs./sq. in., C equals X 650 X 10 X 8.4 = 27,300 lbs.; if f5 has its maximum value T equals 16,000 X 2 = 32,000 lbs. If this last value is attained C also equals 32,000 lbs. and f exceeds the limit of 650 lbs./sq. in. It is necessary to limit C and T to 27,300 lbs. each, thus using the steel at a lower stress than the permissible, and the limiting moment accordingly is 27,300 X 17.2 x= 39,100 ft.-lbs. Another, but slightly longer, method of carrying through this problem is to complete the stress diagram by assuming either unit stress as realized. This method of solving a quadratic equation is logical and easily remembered, while solving by use of a memorized formula is a dangerous and foolish strain on that useful mental function, the memory. The standard notation uses the letter n to designate the ratio of the module and it will be employed for that purpose henceforth tripling the actual steel area by a larger value of the ratio of module. Tests show that the use of this higher value is proper and also that the common theory of design gives conservative results. It should be kept in mind that, since the modulus of elasticity for the concrete has been assumed constant, it is not possible to compute the breaking strength or the effect of loads that cause high concrete stresses by the methods outlined here. For such problems it becomes necessary to consider the actual shape of the stress-strain curve for concrete.