In Formula (8), the denominators fspj and Zfckj are constant for any particular values of f, and fc, and may be represented by a single term, In Table VII values of j, k, p and R are given for several values of f, and fc when n=15.
Table VIII gives values of the same quantities when n=12.
In Table IX values of A' fc, k, j and IA are given for various values of p, with n=12 and n=15.
Table X gives the areas and weights of square and round steel bars of the sizes commonly used in reinforced concrete work.
Examples.—The use of the tables will be illustrated by solving a few probleins.
(4) A reinforced concrete beam is to resist a bending moment of 315,000 in.-lb. The safe unit stresses upon the concrete and steel are 650 and 16,000 respectively. n=15. Find dimensions for the beam and area of steel required.
Solution.—From Table VII, for and ff =650, we find R=10S and p=.007S. Substituting this value of R in (9), ?l7/R= 315000 10S = 2917. If we assume b inches, we have and d= 17 inches.
If the center of the steel be 2 inches from the surface of the con crete, the total depth of the beam, h= 17+2=19 inches.
The area of steel required, .1 =pbd = .0078X 10X 17 = 1.33 in.' Example 5.—A concrete beam is 10 inches wide and 18 inches deep and is reinforced with four :;-inch round steel bars, with centers 2 inches above the lower surface of beam. The safe unit stresses for concrete and steel are 700 and 16,000 respectively. n=15. What is the safe bending moment for the beam? Table Xi, four --inch round bars have area of 1.77 and p= 1.77/160= .0116. From Table 11i, for p=.0116 and n=15, we find and jk/2=1SS.
When fs=16,000, This is greater than the allowable stress on concrete, and the safe bending moment is that which causes a stress of 700 on the concrete, or by (7) 700X.1SSX1OX1GX16=336,900 in.-lb.
Example 6.—A concrete beam 9 inches wide and 15 inches deep is reinforced with five --inch round bars of steel, with centers 2 inches above lower face of beam. The beam carries a bending moment of 155,000 in.-lb. If n=12, find the unit stresses on the steel and con crete.
the concrete on the two sides, and and T2 the corresponding ten sions in the steel. The difference of tensions T must he commu nicated to the concrete and carried as horizontal shear to the com pression side of the beam.
The intensity of the horizontal shear at any point is equal to the intensity of the vertical shear at the same point, as in any beam. If r is the shearing stress upon unit area and b the width of the beam, the total shear upon any horizontal section below the neutral axis is For equilibrium of the forces acting upon the portion of the beam of length S, as shown in Fig. 4, and the two couples are also equal, or Vs= jd. Equating these values of and reducing, we find.
In designing reinforced concrete beams, it is usual to adopt a limiting value for v and make the section of the beam large enough so that the safe value of v as given by the above formula shall not he exceeded. The Joint Committee on Concrete recommends that the maximum value of r shall not exceed 6 per cent of the ultimate corn pressive strength. This gives v=120 as a safe value for ordinary concrete as commonly used in structural work (with crush ing strength of about 2000 at thirty clays).
Rectangular beams, reinforced for tension only, are usually suffi ciently strong to resist direct shearing stresses when properly designed for flexural stresses. The areas of such beams are not affected by providing for shear, although they may sometimes need reinforcement against diagonal tension.
108. Diagonal Tension.—The intensity of the horizontal shear at any point in a beam is equal to the intensity of the vertical shear at the same point. Li Fig. 47 let ABCD be an extremely small prism in a reinforced beam, the vetical and horizontal dimensions of which, parallel to the side of the beam, are represented by s, and the thick ness normal to the side by b. If v represent the unit shearing stress, the shear acting upon each of the four sides of the prism is rsb. If the two forces meeting at B and the two meeting at D be combined into resultants, Td, there will result two equal and opposite forces producing tension in a diagonal direction upon the prism. The value of this tension is , and it is distributed over an area, bs/cos 45°. The unit tension due to shear is The unit diagonal tension due to shear is therefore equal to the unit shear and acts at an angle of with the axis of the beam.