415. Design of Abutments. The force diagram of Fig. 223, which shows the pressures between the voussoirs of the arch, also gives, for any condition of loading, the pressure of the last voussoir against the abutment. A glance at the diagram shows that the maxi mum pressure against the abutment comes against the left-hand abutment under the third condition of loading, when the concen trated load is on the left-hand side of the arch. Although the first condition of loading does not create so great a pressure against the left-hand abutment, yet the angle of the line of pressure is somewhat flatter, and this causes the resultant pressure on the base of the abut ment to be slightly nearer the rear toe of the abutment. It is therefore necessary to consider this case, as well as that of the third condition of loading.
An abutment may fail in three ways: (1) by sliding on its founda tions; (2) by tipping over; and (3) by crushing the masonry. The possibility of failure by crushing the masonry at the skewback may be promptly dismissed, provided the quality of the masonry is reason ably good, since the abutment is always made somewhat larger than the arch ring, and the unit-pressure is therefore less. The possi bility of failure by the crushing of the masonry at the base, owing to an intensity of pressure near the rear toe of the abutment, will be dis cussed below. The possibility that the abutment may slide on its foundations is usually so remote that it hardly need be considered. The resultant pressure of the abutment on its subsoil is usually nearer to the perpendicular than the angle of friction; and in such a case, there will be no danger of sliding, even if there is no backing of earth behind the abutment, such as is almost invariably found.
The test for possible tipping over or crushing of the masonry due to an intensity of pressure near the rear toe, must be investigated by determining the resultant pressure on the subsoil of the abutment. This is done graphically by the method illustrated in Fig. 221. This is an extension of the arch problem already considered. The line be gives the angle of the skewback at the abutment, while the lines of force for the pressures induced by the first and third conditions of loading have been drawn at their proper angle. In common with the general method used in designing an arch, it is necessary to design first an abutment which is assumed to fulfil the conditions, and then to test the design to see whether it is actually suitable. The cross section abcde has been assumed as the cross-section of solid masonry for the abutment. The problem therefore consists in finding the amount and line of action of the force representing the weight of the abutment. It will be proved that this force passes through the point and it therefore intersects the pressure on the abutment for the first condition of loading, at the point k. The weight of a section of the abutment one foot thick (parallel with the axis of the arch), is computed (as detailed below) to weigh 19,500 pounds, while the pressure of the arch is scaled from Fig. 223 as 16,350 pounds. Laying of these forces on these two lines at the scale of 5,000 pounds per inch, we have the resultant, which- intersects the base at the point vi, and which scales 31,350 pounds. Similarly, the resultant of the weight of the abutment and the line of pressure for the third condition of loading intersects the base at the point n, and scales 33,600 pounds. These pressures on the base will be dis cussed later.
The line of action and the amount of the weight of a section of the ment, are mined as follows: The center of ity of the pentagon abcdc is determined by dividing the tagon into three mentary triangles, abe,bcc, and cde. We may consider be as a base which is mon to the triangles abe and bee. By secting the base be and drawing lines to the vertices a and c, and secting these lines to the vertices, we determine the points o, and which arc the centers of gravity, respectively, of the two triangles. The center of gravity of the combination of the two triangles must lie on the line joining o, and o„ and must be located on the line at distances from each end which arc inversely proportional to the areas of the angles. Since the triangles have a common base be, their areas are portional to their altitudes of and ye. In the diagram at the side, we may lay off in succession, on the horizontal line, the distances ye and at On the vertical line, we lay off a distance equal to By joining the lower end of this line with the right-hand end of the line of, and then drawing a parallel line from the point between gc and af, we have divided the distance o,o, into two parts which are proportional to the two altitudes af and ye. Laying off the shorter of these distances toward the triangle abe (since its greater altitude shows that it has the greater area), we have the position of o„ which is the center of gravity of the two triangles combined. The area abce is measured by one-half the product of eb and the sum of of and ye. The triangle cde is measured by one-half the product of the base ed by the altitude ch. If we lay off be as a vertical line in the side diagram, and alsothe line ed as a vertical line, and join the lower end of ed with the line which represents the sum of ye and of, and then draw a line from the lower end of be, parallel with this other line, we have two similar triangles from which we may write the proportion: ed : (ge + af) :: be : f'g'c' Since the product of the means equals the product of the extremes, we find that (ye af) X be = cd X a'f'y'c'; but (ye + of) X he .= the combined area of the two triangles, and therefore the line dr y'c' is the height of an equivalent triangle whose base equals ed; therefore the area of these two combined triangles is to the area of the triangle cdc as the equivalent altitude drg'c' is to the altitude ch of the triangle cde. By bisecting the base cd, and drawing a line from the bisecting point to the point c, and trisecting this line in the point o„ we have the center of gravity of the triangle ede. The center of gravity of the entire area, therefore, lies on the line and at a distance from which is inversely proportional to the areas of the two combined triangles and the triangle cdc. These areas are proportional to the altitudes as determined above; therefore, by laying off in the side diagram the line and drawing a line from its lower extremity to the right-hand extremity of the line eh, and then drawing a parallel line from the point between a'f'g'c' and eh, we divide the line op, into two parts which are proportional to these altitudes. The line ch. is the greater altitude, and the triangle cdc has the greater area; therefore the point is nearer to the point than it is to the point o„ and the shorter of these two sections is laid off from the point This gives the point o„ which is the center of gravity of the entire area of the abutment.