It is also required that the thickness at bottom of abutment gh, shall not be less than two-thirds of the height ag.
if the abutment is of rough rubble, 6 inches is added to the thickness to in sure full thickness in every part.
These rules usually give arches which are amply strong for heavy railway service and heavier than necessary for highway bridges. For structures of small span, however, when voussoir or plain concrete arches are used, the saving effected by paring them down is small, and rather heavy work is common practice.
162. Investigation of Stability.—After assuming dimensions for the arch ring and abutments, the stability of the arch is investigated by the methods outlined in Section 156. The stability of the abut ment is tested by continuing the line of thrust and determining whether it cuts the base of the abutment within the middle third. The sufficiency of the foundation for the abutment must also be examined and footings provided which will properly distribute the pressure over the soil upon which it is to be placed. The following example will outline the method of procedure.
Example.—A highway arch is to have a span of 40 feet and a rise of 10 feet. It is to carry a moving load of 200 pounds per square foot.
The depth of fill at crown is 2 feet. The weight of earth fill is 100 and of masonry 150 pounds per cubic foot.
We will try a segmental arch. By the Douglas rule, the thickness at crown would be 1.4 feet. By Trautwine's formula, it would be 1.95 feet. Make the crown thickness 18 inches. By the Douglas rule the thickness at springing would be between 1.5 and 2 times the crown thickness. We will try 30 inches. Draw the arch ring as shown in Fig. 87, and divide it into equal parts by radial lines. The line z-t represents the roadway and verticals from the points where the radial divisions cut the extrados divide the earth fill into parts resting upon the sections of the arch ring. These loads, including the weights of the sections of arch ring, are now computed, and their vertical lines of action determined.
In finding the loads, it is often convenient to draw the reduced load contour, which is obtained by reducing the height of the sections so that the volume contained by them may be considered to weigh the same per unit as the arch ring. Thus if the earth fill weighs 100 pounds and masonry 150 pounds per cubic foot, the height ax is made two-thirds of az, and the other verticals are reduced propor tionately, giving the volume a-x-u-g, which has the same weight at 150 pounds as the earth fill at 100 pounds. In the same way x-y-v-u represents the live load which would come upon half the arch ring reduced to 150 pounds per cubic foot. In the example, the loadings given represent live load extending over the left half of the arch, dead load only upon the right half.
The horizontal thrusts against the arch ring are sometimes com puted by assuming that the unit horizontal thrust bears a definite proportion (usually about one-quarter) to the unit vertical thrust. Thus in Fig. 87, if the vertical load upon the section a-b is 5085 pounds the horizontal component of the load on the section is In the example, the horizontal components upon the two lower divi sions on each side arc used, those upon the upper divisions being too small to affect the results appreciably. The horizontal components
of the loads are not usually considered in a problem of this kind unless the rise of the arch is large as compared with the span.
Having computed the loads, a line of pressure may now be drawn through any three points in the arch ring. Assume that it is to pass through the lower third point of the joint a on the loaded side, the middle point at the crown, and the upper third point at the joint n on the unloaded side.
The load line is first plotted on a convenient scale by laying off the loads which come upon the various sections in succession, n-m, m-1, etc.; n-a is now the resultant of all the loads upon the arch ring. A pole 0' is assumed and the strings O'a, O'b, etc., drawn.
The equilibrium polygon, shown in broken lines, may now be drawn. Starting from A, the lower third point on joint a, with a line parallel to the string O'a to an intersection with the line of action of the load upon the section a-b. From this intersection, draw a line parallel to O'b to intersection with the line of action of the load on b-c, and continue it until a parallel to O'n is intersected in N' upon a line through N parallel to the resultant n-a.
Connect N' with A, and from 0' draw a line parallel to N'-A to intersection J with the resultant n-a of the loads, thus dividing the resultant into two reactions, n-J and J-a, which would exist at the ends of the span if the horizontal thrust of the arch be neglected. Join the points A and N and from J draw a line parallel to A-N. A pole lying upon this line will give an equilibrium polygon passing through A and N.
The distance of the pole from J must now be determined to cause the equilibrium polygon to pass through the middle of the crown joint. The line g-a in the force polygon, is the resultant of the loads upon the left half of the arch. From the middle of the crown section, draw G-G', parallel to g-a, to intersection with the trial equilibrium polygon. Connect A-G' and A-G. From 0' draw O'k parallel to G'A to intersection with g-a in k, and from k draw k-O parallel to AG. The point 0 where KO intersects JO is the new pole.
From A, the new line of thrust may now be drawn with sides par allel to the strings, Oa, Ob, etc. This passes through the points G and N.
By inspection we see that the line of thrust, as thus drawn, is everywhere within the middle of the arch ring. The thrust upon the joint at a is represented by the length of the line O-a=27000 pounds, and the maximum unit compression is The unit compression upon any other joint may be found in the sane manner.
The resultant pressure h' upon the base of the abutment is found by combining the weight IV of the abutment with the thrust O-a of the arch against the abutment. The footing under t he base of the abutment should be so designed as properly to distribute the load over the foundation soil.