When the load is uniformly distributed over the entire arch, the load is symmetrical, and we need to consider only one-half of the arch. The sections of the load line for the force diagram corresponding to this condition of loading, must be drawn as explained in detail in Article 397. Since the arch is quite flat, the loading is considered to be entirely vertical. Since the load is symmetrical and the abutments are at the same elevation, we need only draw a horizontal line from the lower end of the half- load line, and select on it a trial position (o1 for the pole, drawing the rays as previously explained; the trial equilibrium polygon passes through the center vertical at the point a'. Drawing a horizontal line from a' until it intersects the first line (produced; of the trial equilibrium polygon, and drawing through it a vertical line, we have the line of action of the resultant (R1 of all the forces on that half of the arch. If we draw through a, the center of the keystone, a horizontal line, its intersection with R1 gives a point in the first line (produced) of the true equilibrium polygon. A line from the upper end of the load line parallel to this first section of the true equilibrium polygon, intersects the horizontal line through the middle of the load line at o1', which is the position of the true pole. Drawing the rays from the true pole to the load line, and drawing the segments of the true equilibrium polygon parallel to these rays, we may at once test whether the true equilibrium polygon always passes through the middle third of each joint. As is almost invariably the case, it is found that for full loading, the true equilibrium polygon passes within the middle third at every joint.

Fig. 223. Stresses in a 20 Foot Arch. Reproduced from an original drawn at scale of 1/2 inch = 1 foot.

Fig. 223. Stresses in a 20-Foot Arch. Reproduced from an original drawn at scale of 1/2 inch = 1 foot.

The student should carefully check over all these calculations, drawing the arch at the scale of one-half inch to the foot, and the load line of the force diagram at the scale of 1,500 pounds per inch; then the rays of the true equilibrium polygon will represent at that scale the pressure at the joints. Dividing the total depth of any joint by the pressure found at that joint, gives the average pressure. In the case of the joint at the crown, the total pressure at the joint is 13,900 pounds. The depth of the joint is 1.5 feet, and the area of the joint is 216 square inches; therefore the average unit-pressure is 64 pounds per square inch; if it is assumed that the line of pressure passes through either edge of the middle third, then the pressure at the edge of the joint is twice the average, or is 128 pounds per square inch. This is a very low pressure for any good quality of building stone.

Similarly, the maximum pressure at the skewback is scaled from the force diagram as 16,350 pounds; but since the arch is here two feet thick, and the area is 288 square inches, it gives an average pressure of 57 pounds per square inch. Since this equilibrium polygon is supposed to start from the center of this joint, this represents the actual pressure.

Usually it is only a matter of form to make the test for uniform full loading. Eccentric loading nearly always tests an arch more severely than uniform loading. The ability to carry a full uniform load is no indication of ability to carry a partial eccentric loading, except that if the arch appeared to be only just able to carry the uniform load, it might be predicted that it would probably fail under the eccentric load. On the other hand, if an arch will safely carry a heavy eccentric load, it will certainly carry a load of the same intensity uniformly distributed over it.

411. Test for the Second Condition, or Loading of Maximum Load over One=Half of the Arch. Since the arch has a dead load over the entire arch, and a live load over only one-half of the arch, the load line for the entire arch must be drawn. The load line for the loaded half of the arch will be identical with that already drawn for the previous case. The load line for the remainder of the arch may be similarly drawn. This case is worked out by precisely the same general method as that already employed in the similar case given in detail in Article 410. As in that article, we select a trial pole which in general will give an oblique closing line for the equilibrium polygon. This closing line must be brought down to the horizontal by the method already explained in Article 400; then a second trial must be made in order to shift the polygon so that it shall pass through the middle third at the crown joint. This line should pass through the middle of the crown joint; then the real test is to determine how it passes through the haunches of the arch. As in the previous case, the total pressure at any joint will be determined by the corresponding lines in the force diagram, and the unit-pressure at the joint may be determined from the area of the joint and the position of the line of force with respect to the center of the joint. Even though a line of force passed slightly outside of the middle third, it would not necessarily mean that the arch will fail, provided that the maximum intensity of pressure, determined according to the principles enunciated in Article 405, does not exceed the safe unit-pressure for the kind of stone used.

An inspection of the force diagram with the pole at o2', shows that the rays are all shorter than those of the force diagram for the first condition of loading - with pole at o1'. This means that the actual pressure at any joint is less than for the first case; but since the true equilibrium polygon for- this case does not pass so near the center of the joints as it does for the first condition of loading, the intensity of pressure at the edges of the joints may be higher than in the first case. However, since the equilibrium polygon for this second case is always well within the middle third at every joint, and since even twice the average joint pressure for the first case is well within the safe allowable pressure on any good building stone, we may know that the second condition of loading will be safe, even without exactly measuring and computing the maximum intensity of pressure produced by this loading.