408. Numerical Illustration. The above principles will be applied to the case of an arch having a span of 20 feet and a rise of 3 feet (see Fig. 223). If this arch is to be a circular or segmental arch, the radius which will fulfil these conditions may be computed as illustrated in Fig. 222. We may draw a horizontal line, at some scale, which will represent the span of 20 feet. At the center of this line we may erect a perpendicular which shall be 3 feet long (at the same scale). Joining the points a and c, and bisecting ae at d, we may draw a line from the bisecting point, which is dicular to ae, and this must pass through the center of the required arc. A vertical line through c will also pass through the center of the required arc, and their tersection will give the point o. As a graphical check on the work, a circle drawn about a as a center, and with oc as a radius, should also pass through the points a and b. Since some prefer a numerical solution to determine the radius for a given span and rise, the ius for this case may be computed as follows: The line (le equals the square root of the sum of the squares of the half-span and the / rise, which equals 1 , but the angle cae = angle aod, and, from similar triangles, we may write the proportion: This equals numerically in the above case, 109 6 = 1S.17.
Applying the above rule for the depth of the keystone, we would find for this case that the depth should be: Since the total pressure on the voussoirs is always greater at the ment than at the crown, the depth of the stones near the end of the arch should be somewhat greater than the depth of the keystone. We shall therefore adopt, in this case, the dimensions of IS inches for the depth of the keystone, and 2 feet for the depth at the skewback.
409. Plotting the Reduced Load Line. We shall assume that the earth or cinder fill on top of the arch has a thickness of one foot at the crown, and that it is level on top. We shall also assume that the arch ring is composed of stones which weigh 160 pounds per cubic foot, and we shall therefore consider 160 pounds per cubic foot as the unit-weight in determining the reduced load line. From the ex tremities of the extrados, draw verticals until they intersect the upper line of the earth fill. For convenience we shall divide the horizontal distance between these verticals into 11 equal parts, each to he about 2 feet wide. Draw verticals through these points of division down to the extrados; then draw radial lines from the extrados to the intrados. These lines are drawn radially from a point approximately half way between the center of the extrados and the center of the intrados. This means that the joints, instead of being exactly perpendicular to either the extrados or intrados, have a direction which is a coin promise between the two. The discrepancy is greatest at the abut ments, and approaches zero at the crown. This will divide the arch ring into 11 voussoirs, together with a keystone at the center or crown. Assuming that the earth fill weighs 100 pounds per cubic foot, the lines of division between the 11 sections of the earth fill should each be reduced to va or 5 of its actual depth. If we further assume that the pavement is a little over six inches thick, and that its weight is equivalent to six inches of solid stone, we may add a uniform ordinate equal to six inches in thickness (according to the scale adopted), and this gives the total dead load on the arch. We shall assume further a
live load amounting to 200 pounds per square foot over the whole bridge. This is equivalent to g o of a foot, or 1 foot 3 inches, of solid masonry over the whole arch. This gives the reduced load line for the condition of loading that the entire arch is loaded with its maximum load.
As another condition of loading, we shall assume that the above load extends only across one-half of the arch. We shall probably find that, owing to the eccentricity of this form of loading, the stability of the arch is in much greater danger than when the entire arch is loaded with a maximum load.
We shall also consider the condition which would be found by running a twenty-ton road roller over the arch. A complete test of all the possible stresses which might be produced under this condition would be long and tedious; but we may make a first trial of it by finding the stresses which would be produced by placing the road roller at one of the quarter-points of the arch—.-a position which would test the arch almost, if not quite, as severely as any other possible position. Owing to the very considerable thickness of earth fill, as well as the effect of the pavement, the load of the roller is distributed in a very much unknown and very uncertain fashion over a con siderable area of the haunch of the arch. The extreme width of such a roller is eight feet; the weight on each of the rear wheels is approxi mately 12,000 pounds. We shall assume that the weight of each rear wheel is distributed over a width of three feet and a length of four feet, so that the load on the top of the arch under one of the wheels may be considered at the rate of 1,000 pounds per square foot over an area of 12 square feet. For the unit-section of the arch one foot wide, this means a load of 4,000 pounds loaded on two voussoirs which are four feet in total length. The front roller of the road roller comes between the two rear rollers, and therefore would affect but little, if any, the particular arch ring which we are testing. Not only is it improbable that there would be a full loading of the arch simul taneously with that of a road roller, but it is also true that a full loading would add to the stability of the arch. Yet, in order to make the worst possible condition, we shall assume that the part of the arch which has the road roller is also loaded for the remainder of its length with a maximum load of 200 pounds per square foot; this item alone will take care of the effect of the front roller. A load of 1,000 pounds per square foot is the equivalent of a loading of 6 feet 3 inches of stone; and therefore, if we draw over voussoirs Nos. 3 and 4 a parallelo gram having a vertical height above the line equal to 6 feet 3 inches of stone, and consider a reduced live-load line 15 inches deep = 1.25 = 1 foot 3 inches) over the remainder of that half-span, we have the reduced load line for the third condition of loading.