This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

These will be computed in accordance with straight-line formulae. There are three possible cases, according as the neutral axis is: (1) below the bottom of the slab (which is the most common case, and which is illustrated in Fig. 105); (2) at the bottom of the slab; or (3) above it. All possible effect of tension in the concrete is ignored. For Case 1, even the compression furnished by the concrete between the neutral axis and. the under side of the slab is ignored. Such compression is of course zero at the neutral axis; its maximum value at the bottom of the slab is small; the summation of the compression is evidently small; the lever arm is certainly not more than 2/3y; therefore the moment clue to such compression is insignificant compared with the resisting moment due to the slab. The computations are much more complicated; the resulting error is a very small percentage of the true figure, and the error is on the side of safety.

Fig. 105. Compression Stress Diagram for T-Beam.

286 Case 1. If c is the maximum compression at the top of the slab, and the stress-strain diagram is rectilinear, as in Fig. 105, then the compression at the bottom of the slab is kd - t . The average kd

compression =1/2 (c + c kd - t) = c (kd - 1/2 t) The total comkd kd

pression equals the average compression multiplied by the area b't; or,

C = As = b't c/kd (kd - 1/2t) .(32)

The center of gravity of the compressive stresses is evidently at the center of gravity of the trapezoid of pressures. The distance x of this center of gravity from the top of the beam is given by the formula: x = t/3 3kd - 2t / 2kd-1 ... (33)

It has already been shown in Article 204, that:

εc /εs = cr / s = kd /d-kd

Combining this equation with Equation 32, we may eliminate c /s, and obtain a value for kd: kd = Ard + 1/2 b't2 /Ar+b't ..(34)

If the percentage of steel is chosen at random, the beam will probably be over-reinforced or under-reinforced. In general it will therefore be necessary to compute the moment with reference to the steel and also with reference to the concrete, and, as before with plain beams (Equation 29), we shall have a pair of equations:

Mc = C (d-x) | = b't (c / kd - 1/2 t) (d - x) | . .(35) |

Ms = As (d - x) | = pb'ds (d-x) |

If we place kd - t in the equation above Equation 34, and solve for d, we have a relation between d, c, s, r, and t, which holds when the neutral axis is just at the bottom of the slab. The equation becomes: d = t (cr + s) ..(36) cr

A combination of dimensions and stresses which would place the neutral axis exactly in this position, is improbable, although readily possible; but Equation 36 is very useful to determine whether a given numerical problem belongs to Case 1 or Case 3. When the stresses s and c in the steel and concrete, the ratio r of the elasticities, and the thickness t of the slab are all determined, then the solution of Equation 36 will give a value of d which would bring the neutral axis at the bottom of the slab. But it should not be forgotten that the com-pression in the concrete (c) and the tension in the steel will not simultaneously have certain definite values (say c = 500, and s = 16,000) unless the percentage of steel has been so chosen as to give those simultaneous values. When, as is usual, some other percentage of steel is used, the equation is not strictly applicable, and it therefore should not be used to determine a value of d which will place the neutral axis at the bottom of the slab and thus simplify somewhat the numerical calculations. For example, for c = 500, s = 16,000, r = 12, and t = 4 inches, d will equal 14.67 inches. Of course this particular depth may not satisfy the requirements of the problem. If the proper value for d is less than that indicated by Equation 36, the problem belongs to Case 3; if it is more, the problem belongs to Case 1.

The diagram of pressure is very similar to that in Fig. 105, except that it is a triangle instead of a trapezoid, the triangle having a base c and a height kd which is less than t. The center of compression is at 1/3 the height from the base, or x = 1/3 kd. Equations 25 to 29 are applicable to this case as well as to Case 2, which may be considered merely as the limiting case to Case 3. But it should be remembered that b' refers to the width of the flange or slab, and not to the width of the stem or rib.

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