Example1. Assume flint a 5-inch slab is supporting a load on beams spaced 8 feet apart, the beams having a span of 20 feet. Assume that the moment of the beam has been computed as 900,000 inch-pounds. What will be the dimensions of the beam if the concrete is not to have a compression greater than 500 pounds per square inch and the tension of the steel is not to be greater than 16,000 pounds per square inch?

Answer. There are an indefinite number of solutions to this problem. There are several terms in Equation 35 which are mutually dependent; it is therefore impracticable to obtain directly the depth of the beam on the basis of assuming the other quantities; therefore it is only possible to assume figures which experience shows will give approximately accurate results, and then test these figures to see whether all the conditions are satisfied. Within limitations, we may assume the amount of steel to be used, and determine the depth of beam which will satisfy the other conditions together with that of the assumed area of steel. For example, we shall assume that six 7/8-inch square bars having an area of 4.59 square inches will be a suitable reinforcement for this beam. We shall also assume as a trial figure that x = 1.5. Substituting these values in the second formula of Equation 35, we may write the second formula:

900,000 = 4.59 X 16,000 (d-1.5).

Solving for d, we find that d = 13.75. If we test this value by means of Equation 36, we shall find that, substituting the values of t, c, r, and s in Equation 36, the resulting value of d equals 18.33. This shows that if we make the depth of the beam only 13.75, the neutral axis will probably be within the slab, and the problem comes under Case 3, to which we must apply Equation 29. Dividing the area of the steel, 4.50, by (b' X d), we have the value of p equals .00348. Interpolating with this value of p in Table XV, we find that when r = 12, k = .250; kd = 3.44; x = 1.15; and d - x = 12.0. Substituting these values in Equation 29, we find that the moment 900,000 = 2,082c, or that c = 432 pounds per square inch. This shows that the unit-compression of the concrete is safely within the required figure. Substituting the known values in the second part of Equation 29, we find that the stress in the steel s equals about 15,500 pounds per square inch.

Example 2. Assume that a floor is loaded so that the total weight of live and dead load is 200 pounds per square foot; assume that the T-beams are to be 5 feet apart, and that the slab is to be 4 inches thick; assume that the span of the T-beams is 30 feet. We therefore have an area of 150 square feet to be supported by each beam, which will give a total load of 30,000 pounds on each beam., The moment at the center of such a beam will therefore be equal to the total load, multiplied by one-eighth of the span (expressed in inches), and the moment is therefore 1,350,000 inch-pounds. As a trial value, we shall assume that the beam is to be reinforced with six 3/4-inch bars, which have an area of 3.37 square inches. Substituting this value of the area in the second part of Equation 35, and assuming that s = 16,000 pounds per square inch, we find as an approximate value for d - x, that it will equal 25 inches. This is very much greater than the value of d that would be found from substituting the proper values in Equation 36, so that we know at once that the problem must be solved by the methods of Case 1. For a 4-inch slab, the value of x must be somewhere between 1.33 and 2.0. As a trial value, we may call it 1.5, and this means that d will equal 26.5. Assuming that this slab is to be made of concrete using a value for r = 12, we know all the values in Equation 34, and may solve for kd, which we find to equal 5.54 inches. As a check on the approximations made above, we may substitute this value of kd, and also the value of t in Equation 33, and obtain a more precise value of x, which we find to equal 1.62. Substituting the value of the moment and the other known quantities in the upper formula of Equation 35, we may solve for the value of c, and obtain the value that c = 352 pounds per square inch. This value for c is so very moderate that it would probably be economy to assume a lower value for the area of the steel, and increase the unit-compression in the concrete; but this solution will not be here worked out.