Assume that Fig. 229 represents a portion of the cross-section of an arch at any point, the particular portion having a total depth h equal to the thickness of the arch at that point, and a unit-width b, which is presumably less than the total length of the arch parallel to its axis. Assume that there is reinforcement in both the top and bottom of this section, and that the reinforcement is placed at a distance of 1/10 of the thickness of the arch from the top and the bottom, or from the extrados and the in-trados. The moment of inertia of the plain concrete without any reinforcement would evidently equal 1/12 bh3. A transverse stress on such a section will cause the bars on one side of the section (say the bottom) to be in tension, while those in the top will be in compression. As already developed in the treatment of columns (Part III, Article 310), the steel which is in compression will develop a compressive stress which is in proportion to the ratio of the moduli of elasticity of the steel and the concrete; and we may therefore consider that the area of steel in compression at the top (calling its area A) is the equivalent of an area of concrete equal to Ar, in which r is as usual Es ÷ Ec. The exact position of the neutral axis in a section which is reinforced both in compression and in tension, depends upon the percentage of steel which is used; but when the percentage is as large as it usually is, the neutral axis is not far from the center of the section; and since it very much simplifies the computations to consider it at the center of the section, it will be so considered, and the moment of inertia of the steel and concrete combined may be expressed by the equation:

I = 1/12 bh3 + 2Ar(.4h)2..(55)

If, in any numerical problem, it is considered preferable to place the steel so that the distance of its center of gravity from the surface of the concrete is greater or less than 0.1 h, a corresponding change must be made in the second term of the right-hand side of Equation 55.

Example. Assume that p = .015, and that the thickness of the arch h equals 15 inches. For a unit-section 12 inches wide, the area of the concrete would be bh, which equals 180 square inches. Then 180 X.015= 2.70 = 2A, since A is the area at either top or bottom. Therefore A= 1.35.

Assume that r = 12; .4h = .4 X 15 = 6; then,

I = 1/12 X 12 X 153 + 2 X 1.35 X 12 X 62 = 3,375 + 1,166 = 4,541.

It should be noted from Equation 55, that when, as is usual, the area of the steel in the extrados and intrados remains constant, while the thickness of the arch varies, the increase in the moment of inertia is not strictly according to the cube of the depth, but increases in accordance with two terms, one of which varies as the cube of the depth, and the other as the square of the depth. To illustrate the discrepancy, let us assume that the depth of the arch at the abutment is 10 per cent greater than the depth at the crown; or that, applying it to the above numerical case, the depth at the crown is 15 inches, and at the abutment the depth is 16.5 inches. Then, since b, A, and r remain the same in Equation 55, the value of the moment of inertia for the abutment would be:

I = 1/12 X 12 X 16.53 + 2 X 1.35 X 12 X 6.62 = 5,903.

Using the approximate rule that I varies as the cube of h, we find that:

I = 4,541 x (1.10)3 = 6,044, which is about two per cent in excess of the value found from Equation 55. Computing the moment of inertia similarly on the assumption that the depth at the abutment is increased 50 per cent, so that it equals 22.5 inches, we find that the approximate rule will give a moment of inertia which is nearly 8 per cent in excess of the actual. Therefore, when the increase in the depth of the rib from the crown to the abutment is comparatively small, we may adopt the approximation that the moment of inertia increases as the cube of the depth. When the variation is greater, the inaccuracy will not permit the utilization of the simplified forms which this approximation allows.

## 427. Value Of N

Still another simplification may be made, on the assumption that the moment of inertia varies as the cube of the depth, and also that we may increase the depth of the rib as desired. Assume that the depth of the rib is increased so that at any point n = ds ÷ dx (see Fig. 230); ds is always greater than dx, and n is a ratio varying from one upward. Then, on the assumption that: n = I/Ic = h3/hc3' we may compute a series of values for h in terms of the height at the center hc which will correspond to various angles a. For each angle, we find the ratio between h and hc that will correspond to the value which n has for that particular angle on the basis that n=ds/dx. If we substitute this value of ds/dx. in Equations 52, 53, and 54, we shall have the following Equations:

Fig. 229. Arch Cross-Section.

Fig. 230. Determination of Value of n.

The values of h which cause n to vary in this way, are as given in the tabular form:

 a n=ds/dx 0° 1.000 1.00hc 15° 1.035 1.01 30° 1.155 1.05 40° 1.305 1.09 45° 1.414 1.12 50° 1.550 1.16 55° 1.743 1.20
 a n=ds/dx 60° 2.000 1.26hc 65° 2.366 1.33 70° 2.924 1.43 75° 3.864 1.57 80° 5.759 1.79 85° 11.474 2.25 90° Infinity Infinity

Therefore, when all sections have the same moment of inertia, and n is uniformly 1, use Equations 52, 53, and 54, ignoring the n. When an increase in depth of section, as indicated above, will fulfil the ultimate requirements, there is an advantage of simplicity in making the sections accordingly, and using the Equations 56, 57, and 58. When it proves necessary to vary the sections according to some different law, n must be determined at frequent intervals, spaced by a uniform ds, and the summations of Equations 52, 53, and 54 determined. The remainder of this method follows out the assumption that n varies as ds/dx, or that dx = ds/n