The ordinates z" are measured from um to the trial equilibrium polygon when it has been shifted not only so that its closing line is horizontal, but also shifted vertically (up or down) so that its line v'"m'" corresponds with vm; but it is unnecessary to draw it in that way, since we may measure the ordinates from the transposed line v rn', because we know that they are in each case the equal of the ordinates as they would be if the transposition had been actually made. But the lengths of these ordinates below vm' have already been determined; and since we know that v'" nil" is 3.44 feet vertically below vm', we need only change the ordinates z'" by 3.44 (taking care of the algebraic sign), and we then obtain the values of z" as given in the tabular form. Multiplying each value of z" by the corresponding value of y, we obtain the various values (plus and minus) for yz" as given in the last column of the tabular form. The algebraic sum of these quanti ties is 4• 232.88.
Since this value is less than the value of the summation of yz', we must select a smaller value of H, so that the values of z" will be proportionately larger. We must therefore use a pole distance which shall be smaller than 0 21 in the ratio of 252.SS to 392.12. To solve this graphically, we must draw an indefinite line Its, and lay off the distance ns equal to 392.12, at any convenient scale. Laying off a distance nt at the same scale so that it equals 232.SS; we may then join s and and draw a line from I parallel to sO„ obtaining the point which is the required pole of the special equilibrium polygon.
436. Locating the True Equilibrium Polygon. We know that the segments of the true equilibrium polygon must be parallel with the rays of the force diagram which has its pole at 0„ and also that it must pass through the points h and k on the line vm. The point k lies between loads 4 and 5; therefore we draw through the point h a line parallel to the ray of the force diagram from to the point on the load line between load 4 and load 5. Similarly, since the point k lies between loads 16 and 17, we may draw through k a segment of the equilibrium polygon which is parallel to the ray from to the point on the load line which is between load 16 and load 17. In order to avoid inaccuracy, the segments of the equilibrium polygon should be drawn from these two segments each way toward the crown of the arch and each way toward the abutments. As a check on the work, these separate sections of the equilibr:um polygons should accurately meet at the top of the arch, and they should also reach the last verticals through 0 and B at points which are on the same hori zontal line. It is merely a coincidence that these points 0" and B"
are almost exactly at the same level as the lowest point of the skew backs.
437. Maximum Moment under this Loading. 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 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 w:11 be in tension under load 16, and in compression at the two abutments. If this arch were reinforced with 1-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 .4k equals 8.5; and, by substituting these quantities in Equation 55, we find that the moment of inertia is: 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 sub jected 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.