This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
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 maximum pressure against the abutment comes against the left-hand abutment under the third condition of loading, when the concentrated 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 abutment to be slightly nearer the rear toe of the abutment. It is therefore necessary to consider this well as that of the third condition of loading.
An abutment may fail in three ways: (1) by sliding on its foundations; (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 reasonably good, since the abutment is always made somewhat larger than the arch ring, and the unit-pressure is therefore less. The possibility 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 discussed 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. 224. 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 o5, 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 off 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 to, 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 discussed later.
The line of action and the amount of the weight of a unit-section of the abutment, are determined as follows: The center of gravity of the pentagon abcde is determined by dividing the pentagon into three ele-mentary triangles, abe, bce, and cde. We may consider be as a base which is common to the triangles abe and bee. By bisecting the base be and drawing lines to the vertices a and c, and trisecting these lines to the vertices, we determine the points o1 and o2, which are 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 o1 and o2, and must be located on the line at distances from each end which are inversely proportional to the areas of the triangles. Since the triangles have a common base be, their areas are proportional to their altitudes af and gc. In the diagram at the side, we may lay off in succession, on the horizontal line, the distances gc and af.
Fig. 224. Forces Acting on Abutments.
On the vertical line, we lay off a distance equal to o1o2. By joining the lower end of this line with the right-hand end of the line af, and then drawing a parallel line from the point between gc and af, we have divided the distance o1o2 into two parts which are proportional to the two altitudes af and gc. 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 o3, 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 af and gc. The triangle cde is measured by one-half the product of the base ed by the altitude ch. If we lay off bc as a vertical line in the side diagram, and also the line ed as a vertical line, and join the lower end of ed with the line which represents the sum of gc and af, 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: (gc + af):: be: a'j'g'c'.
Since the product of the means equals the product of the extremes, we find that (gc + af) x be = ed x a'f'g'c'; but 1/2 (gc + af) x be = the combined area of the two triangles, and therefore the line a'f'g'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 cde as the equivalent altitude a'f'g'c' is to the altitude ch of the triangle cde. By bisecting the base ed, and drawing a line from the bisecting point to the point c, and trisecting this line in the point o4 we have the center of gravity of the triangle cde. The center of gravity of the entire area, therefore, lies on the line o3o4 and at a distance from o4 which is inversely proportional to the areas of the two combined triangles and the triangle cde. These areas are proportional to the altitudes as determined above; therefore, by laying off in the side diagram the line o3o4 and drawing a line from its lower extremity to the right-hand extremity of the line ch, and then drawing a parallel line from the point between a'f'g'c' and ch, we divide the line o3o4 into two parts which are proportional to these altitudes. The line ch is the greater altitude, and the triangle cde has the greater area; therefore the point o5 is nearer to the point o4 than it is to the point o3, and the shorter of these two sections is laid off from the point o4. This gives the point o5, which is the center of gravity of the entire area of the abutment.
The actually computed weight of a unit-section of the abutment is determined by multiplying the sum of a'fg'c' and ch by the base ed. Since this masonry is assumed to weigh 160 pounds per cubic foot, the product of these scaled distances, measured at the scale of \ inch equals one foot (which was the scale adopted for the original drawing), shows that the section one foot thick has a weight of 19,500 pounds. Laying off this weight from the point k, and laying off the pressure for the first condition of loading, 16,350 pounds, at the scale of 5,000 pounds per inch, and forming a parallelogram on these two lines, we have the resultant of 31,350 pounds as the pressure on the base of the abutment, that pressure passing through the point m.
The intersection of the weight of the abutment with the line of pressure for the third condition of loading, is a little below the point k; and we similarly form a parallelogram which shows a resulting pressure of 33,600 pounds, passing through the base at the point n. It is usually required that such a line of pressure shall pass through the middle third of the abutment; but there are other conditions which may justify the design, even when the line of pressure passes a little outside of the middle third.
The point n is 2.85 feet from the point e. According to the theory of pressures enunciated in Article 405, it may be considered that the pressure is maximum at the point e, and that it extends backward toward the point d for a distance of three times en, or a distance of 8.55 feet. This would give an average pressure of 3,930 pounds per square foot, or, since the pressure at the toe is twice the average pressure, 7,860 pounds per square foot on the toe. Such a pressure might or might not be greater than the subsoil could endure without yielding. Since this pressure is equivalent to about 55 pounds per square inch, there should be no danger that the masonry itself would fail; and, if the subsoil is rock or even a hard, firm clay, there will be no danger in trusting such a pressure on it.
Another very large item of safety which has been utterly ignored, but which would unquestionably be present, is the pressure of the earth back of the abutment. The effect of the back-pressure of the earth would be to make the line which represents the resultant pressure on the subsoil more nearly vertical, and to make it pass much more nearly through the center of the base ed. This would very much reduce the intensity of pressure near the point e, and would reduce very materially the unit-pressure on the subsoil. Cases, of course, are conceivable, in which there might be no back-pressure of earth against the rear of the abutment. In such cases, the ability of the subsoil to withstand the unit-pressure at the rear toe of the abutment (near the point c) must be more carefully considered. In order that the investigation shall be complete, it should be numerically determined whether the lower pressure, 31,350 pounds, passing through the point m, might produce a greater intensity of pressure at the point e than the larger pressure passing through the point n.
The abutment described above is the general form which is adopted very frequently. The front face cd is made with a batter of one in twelve. The line ba slopes backward from the arch on an angle which is practically the continuation of the extrados of the arch. The total thickness of the abutment de must be such that the line of pressure will come nearly, if not quite, within the middle third. The line ea generally has a considerable slope, as is illustrated. When the subsoil is very soft, so that the area of the base is necessarily very great, the abutment is sometimes made hollow, with the idea of having an abutment with a very large area of base, but which does not require the full weight of so much masonry to hold it down; and therefore economy is sought in the reduction of the amount of masonry. Since such a hollow abutment would require a better class of masonry than could be used for a solid block of masonry, it is seldom that there is any economy in such methods. 'Since the abutment of an arch invariably must withstand a very great lateral thrust from the arch, there is never any danger that the resultant pressure of the abutment on the subsoil will approach the front toe of the arch, as is the case in the abutment of a steel bridge, which has little or no lateral pressure from the bridge, but which is usually subjected to the pressure of the earth behind it. These questions have already been taken up under the subject of abutments for truss bridges, in Part II.