An inspection of the diagram shows what might be expected, that the maximum moment occurs on the right-hand side of the arch, nearly under the center of the live load and very near to load 16. At this point the vertical distance z, between the equilibrium polygon and the center of the arch rib, is 0.55 foot, or 6.6 inches. The pole distance O3n, scaled at 5,000 pounds per inch, indicates 22,650 pounds; therefore the moment at that point equals 22,650 X 6.6 = 149,490 inch-pounds. It may be observed, also, that the moment at the abutment scales exactly the same quantity, as nearly as it can be measured, but the moment is of contrary sign. In other words, the intrados will be in tension under load 16, and in compression at the two abutments. If this arch were reinforced with 3/4-inch bars spaced 12 inches apart, in both the top and the bottom, there would then be two such bars in each section of the arch one foot wide; and the area A (see Equation 55) would be the area of one such bar, or 0.56 square inch. Since the depth of the arch at the abutment (h) is 21.25 inches, then Ah equals 8.5; and, by substituting these quantities in Equation 55, we find that the moment of inertia is:

I = 1/12 X 12 X 21.253 + 2 X 0.56 X 12 X 8.52 = 9,593+ 971 = 10,564 bi-quadratic inches.

Then, transposing the equation M = pI/e into the equation p = Me/I, we may substitute for M the value 149,490; for e the half-depth of the beam, 10.625; and for I the value found above, 10,564; and find that the unit-stress in the concrete at the abutment equals about 151 pounds per square inch.

For this particular case of loading, the moment at the crown is almost zero, since it may be observed from the drawing, that the special equilibrium polygon crosses the center line of the rib about 18 inches at the right of the center. This crossing indicates a point of contraflexure where there is no moment. Also, since the equilibrium polygon is below the center line of the rib at the crown, it indicates that such moment as there is for this loading is negative; or, in other words, the tension is on the upper side of the rib. This same kind of moment exists on the entire left-hand side of the arch for this loading. It should also be observed that there is another point of contraflexure a few feet from the right-hand abutment. It will be shown later that the thrust at the abutment has a greater intensity per square inch than the maximum compression or tension due to moment. This practically means that the compression side of the arch is subjected to the combined compressions due to thrust and moment; while, on the other side, the thrust more than neutralizes the tension, and actually relieves it altogether.

Near the crown of the arch, the thrust is not so great, and will not wholly neutralize the tension due to moment. In order to withstand the various stresses, the rib must have a larger cross-section than would be required for moment alone. This means that the equation developed in Article 271, Part III, M = 62 bd2, can be utilized only in a roundabout way. For example, the moment at the abutment was computed above as 149,490 inch-pounds. But a section 12 inches wide and 19.125 inches deep to the reinforcement, should withstand a moment of 272,300 inch-pounds with 0.43 per cent of steel. The above moment is only 55 per cent of this. Therefore 55 per cent of the steel, or .55 X .00436bd = .002366cd of steel, could safely carry the tension. But the actual ratio of steel adopted was .00245. As shown above, the steel at the abutment is unnecessary for transverse moment and for this condition of loading. Nearer the crown, the moment is less; but the relief to tension afforded by thrust is very much less, and the steel has much more to do. With other conditions of loading, it will also be different; and therefore the same amount of steel 3/4-inch bars spaced 12 inches, in both top and bottom, is used throughout.