RECTANGULAR BEAMS WITH TENSION REINFORCEMENT 105. Flexure Formulas.—The common or straight-line formulas for reinforced concrete beams are based upon the ordinary theory of flexure, and involve the following assumptions: (1) A section of the beam that is plane before bending remains plane when bent.
(2) The modulus of elasticity .of concrete is constant within the limits of safe unit stresses.
(3) The concrete resists compression only, all tensions being carried by the steel.
(4) Initial stresses due to expansion or contraction of the con crete are negligible.
These assumptions greatly simplify the computations and are found experimentally to be sufficiently accurate within the limits of stresses used in ordinary beam design; they are not applicable to ultimate loads and can be used only for working loads and work ing stresses.
The following notation will be used: h= total depth of beam; b= breadth of beam; d= depth of center of gravity of steel below compression face of beam, or effective depth of beam; of neutral axis below compression face of beam; jd = distance between centers of tension and compression; A = area of cross-section of steel; p = rat io of steel area to effective area of beam (p = A/bd); fa = unit tensile stress on steel; unit compressive stress on concrete at compression face; Es= modulus of el st icity of steel; Es=modulus of elasticity of concrete; n= ratio of moduli, Es; C = total compression in concrete; T=totn1 tension in steel; 31= resisting moment or bending moment.
Assuming that a plane section before bending remains plane after bending, we have (see Fig. 45), Combining (1) and (2) we have and solving for k, The centroid of compressive stresses is at a distance kd/3 from the compressive face of the beam, and From the foregoing it is readily seen that the ratio of the unit stresses on the steel and concrete, and the values of k, j and p are interdependent. If the unit stresses and value of n be assumed, k and the required percentage of steel may be found from Formulas (1) and (2). If the percentage of steel be known and the ratio n assumed, the values of k and the ratio fs/ff, may be found from (4) and (2).
The resisting moment of the beam is due to the couple formed by the tensions and compressions and is equal to either of them into the arm of the couple: Formulas (6) and (7) give a means of determining the moment of resistance of a beam of known dimensions and safe unit stresses, while from (8) the necessary dimensions may be found to resist any given bending moment with assumed unit stresses in steel and con crete.
Examples.—The problems arising in the use of these formulas are of two kinds—the design of beams to carry certain loads; the investigation of existing beaus to determine the loads they may safely carry, or the unit stresses resulting from given loads. The following examples illustrate the use of the formulas for these pur poses: 1. A reinforced concrete beam is to carry a bending moment of 152,000 in.-lb. The safe unit stresses upon concrete and steel are 600 and 14,000 respectively. n=15. Find dimensions for the beam and area of steel required.
(1) gives We may now assume a value for either b or d, or fix a relation be tween them. Assuming inches, we have and d= 13.7 inches. A = pbd = .00S4 X S X13.7= .92 Taking d as 131 inches, if the concrete extends 11 inches below the steel, the total depth, inches.
2. A concrete beam is 9 inches wide and 16 inches sleep, and is reinforced with four 1-inch round steel bars, with centers 2 inches above the lower surface of the beam. The safe unit stresses for the concrete and steel are 700 and 14,000 respectively. n=15. What is the safe bending moment for the beam? area of steel is and This shows that if a stress of 14,000 lb/in.2 be brought upon the steel, the stress upon the concrete will be greater than 700 Hence the safe moment is that which causes a stress of 700 in the concrete, or applying (7) 3. If the beam in the preceding example is subjected to a bend ing moment of 225,000 in.-lb., what are the maximum stresses upon the steel and concrete? in the preceding case, we find A =1.767 and 106. Tables.—The labor of computation in the use of the above formulas may be considerably lessened by tabulation of values of some of the terms involved.