It is another principle of the science of summations, that if we have a series of terms whose summation equals zero, and also have another series of terms whose summation equals zero, but whose terms are made up of the difference of two terms, one of which corresponds in each case to the terms of the first summation, then we may say that the summation of the other corresponding terms is likewise zero. For example, the first one of Equations 50 consists of a series of terms which may be rewritten: The first one of Equations 51 is the summation of a series of terms, each with the form In each of these summations the different terms corresponding to the variable values of z' exactly correspond. We may therefore say that the summation of a series z" of corresponding terms, each one of the form —ds, will exactly equal zero; and we may therefore write Equation 52 as given below. We may also combine the second part of Equation 50 with the second part of Equation 51 in a similar manner, and obtain Equation 53 as given below It will be found more convenient to separate the third part of Equation 50 into two summations, one of z" which consists of a series of terms —yds, and the other of a series of terms ds; and since the difference of these ' n mations equals zero, then the summations must equal each other, and we may therefore write Equation 54: An infinite number of equilibrium polygons may he drawn which will satisfy Equation 52 and 53. Au equilibrium polygon may be drawn by trial, and the values of the summations for arch side of Equation 54 may be determined. But since the position of the line via is definitely determined by Equation 51, then the value of the right side of Equation 54 is fixed, and we only need to alter the pole dis tance of the trial equilibrium polygon in the inverse ratio of the re quired change in z", and then Equation 54 will be satisfied. Since changing all the values of z" in the same ratio does not alter the satis faction of Equations 52 and 53, the changing of the pole distance does not vitiate the previous work. The value for the true pole distance is thus obtained, by which the true curve PQR may be graphically drawn out. We may then determine the moment at any point, which is the product of Hz for any point of the curve, in which z is the vertical distance between the center line of the arch rib and the finally determined equilibrium polygon, and H is the pole distance corre sponding to that polygon.
It will be shown later that the thrust and the shear for any point of the curve equal the projection onto the tangent and normal respec tively of the proper ray of the force diagram. It should be noted that the above equations apply only to symmetrical arch ribs which have their abutments at the same level. Under such conditions, Equation 53 is always satisfied when Equation 52 is satisfied. In the very rare eases where arches have to be designed on a different basis, some of the simplifications given above cannot be utilized, and the work becomes far more complicated. The solution of these very rare cases will not be given here.
426. Moment of Inertia of any Section. Assume that Fig. 229 represents a portion of the cross-section of an arch at any point, the particular portion having a total depth it 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 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 IA'. 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 alreadydeveloped 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 The exact position of the neutral axis in a section which is i;inforced 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: / = + (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 .015, and that the thickness of the arch h equals 15 inches. For a unit-section 12 inches wide, the area of the con crete would be bh, which equals 1S0 square inches. Then 1S0 X .015= 2.70=2A, since A is the area at either top or bottom. Therefore A= 1.35.
It should be noted from Equa don 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: / = x 12 X 16.53 + 2 X 1.35 X 12 x = 5,903.
Using the approximate rule that I varies as the cube of h, we find that: = 4,541 X = 6,044, which is about two per cent in excess of the value found from Equation 55. Computing the mo ment 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 rude 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 approxi mation that the moment of inertia increases as the cube of the depth. When the variation is greater, the inaccuracy will not permit the util ization of the simplified forms which this approximation allows.