COMPLETE SOLUTION OF NUMERICAL PROBLEM ELASTIC ARCH 431. Dimensions of Arch. We shall apply the above prin ciples to the design of a segmental arch having a span of 60 feet and a rise of 15 feet. To find the radius for the intrados which will fulfil these conditions, we may note from Fig. 232 that the angle O'B'C' is measured by one-half of the arc O'C', and therefore O'B'C' is one-half of the angle a; but the angle O'B'C' is an angle whose natural tangent equals 15 30, or precisely 0.5. The angle whose tangent has this value is 26° 34', and therefore a equals 53° 5'. To find the radius, we must divide the half-span (30) by the sine of 53° S', and we find that the radius equals 37.50 feet.
For the depth of the keystone we can employ only empirical rules. The depth as computed from Equation 47 would call for a keystone depth of about 27 inches, which would be proper for an ordinary masonry arch; but considering recent successful practice in rein forced-concrete arches, and the far greater reliability and higher permissible unit-stresses which may be adopted, we may select about of this —or, say, 15 inches—as the depth of the arch ring at the crown.
We shall compute t_he depth the arch ring should have at various points, according to the tabular form in Article 427, so that the mo ment of inertia will vary in the ratio d ds , which will make Equations x 56 to 58 applicable. An arc of 1 degree equals .0175 of the radius, and therefore an arc of 1 degree on a circle with a radius of 37.5 feet will have a length of .6545 foot. At a distance of 15 degrees from the center, or at a distance of 9.82 feet, the depth is one per cent greater than the depth at the center, or it is 18.18 inches deep.
At 30 degrees from the center, or at a distance of 19.63 feet measured on the 'arc, the depth is 5 per cent greater, or it is 18.90 inches deep. At a distance of 40 degrees from the center, or 26.18 feet measured on the arc, the value of it is 0 per cent greater, or it is 19.62 inches. At 45 degrees from the center, or 29.45 feet measured on the arc, h is 12 per cent greater than the center depth, or the depth is 20.16 inches. At 50 degrees from the center, or 32.72 feet measured on the arc, it is 16 per cent greater, or its thickness is 20.88 inches. At the abutment, which is 53° 8' from the center, the thickness should be (by interpolation) about 18 per cent greater than at the center, or it should be 21.24 inches, or say 214 inches.
Laying off these various distances from the center on the in trados, and measuring radial distances at each point to represent the proper thickness at the several points, we may join the various points and obtain the curve of the extrados. Bisecting each one of these
several arch thicknesses will give us a series of points which are points on the center line of the arch rib. We thus find that the actual span may be considered as 61.40 feet, and that the rise is scaled at 15.25 feet.
432. Position of vat. When the center line of the rib is a parab ola, we may 1* off yin by drawing it at a height of the rise above the line OB. Even when it is a circle, it is comparatively easy to com pute with mathematical accuracy the height of OK, by computing the area of the segment OCR, and dividing it by the length of the line OB. As previously explained in Article 428, the two-thirds rule may be used even for circular arcs, when the arch is very flat. In this partic ular case, the two-thirds rule is far from applicable; and since for multi-centered curves no such rule is applicable, the general method will be here given. We divide the span (61.40) into 20 equal parts.
Practically this is most easily done by setting a pair of dividers by trial so that 10 equal spaces may be stepped off in the length of the half-span. From these division points on the line OB, erect perpendic ulars to the center line of the arch rib OCB. The area of a curve hounded by a straight line at the bottom, and which has vertical and equally spaced ordinates, may be computed with very close accuracy by the adoption of Simpson's -rule. If represents the ordinate at the beginning of the curve (and in this case = 0), while y, up to yr, represent the lengths of the several ordinates (y” being the last ordi nate, and, in this case, also equal to 0; and 1i being equal to the even number of divisions, in this case 20), then the area may be expressed by the formula : Applying this rule, we find that the area will be 640.50 square feet. Dividing this by the span, 61.40, we find that the height of vm above the line OB will be 10.59 feet. The approximate two-thirds rule would give us 10.17. Making a rough interpolation in the tabular form of Article 428, we could say that for an angle 2 a equal to 106° 16', the quantity to be added to the result by the two-thirds rule would be approximately 5 per cent. Adding 5 per cent to 10.17, we would have 10.68, which gives a rough check with the far more accurate value just found.