The loads on each voussoir are scaled from the reduced load line according to the various conditions of loading. The area between the two verticals over each voussoir is measured with all necessary ac curacy by multiplying the horizontal width between the verticals by the scaled length of the perpendicular which is midway between the verticals. The weight of the voussoir itself inav be computed as accurately as necessary, by multiplying the radial thickness by the length between the joints as measured on the curve lying half-way between the intrados and the extrados.
For example, the load for full loading of the arch which is over voussoir No. 1, is measured as follows: The width between the per pendiculars is 2.0 feet; the height measured on the middle vertical is 4.05 feet; the area is therefore 8.10 feet, which, multiplied by 160, equals 1,296 pounds, which is the load on this voussoir for every foot of width of the arch parallel with the axis. The radial thickness of voussoir No. 1 is 1.90 feet, and the length is 2.15 feet; this gives an area of 4.0S5 feet, which, multiplied by 160, equals 653.6 pounds. The weight of the voussoir is therefore almost exactly one-half that of the live and dead loads above it; therefore the resultant of these two weights will be almost precisely one-third of the distance between the center of this stone and the vertical through the center of the load ing. By drawing this line, we have the line of action of the resultant of these two forces, and this value is the sum of 1,296 and 654, or 1,950 pounds.
In order to simplify the figure, the arrows representing the lines of force of the loading on the voussoir and the weight of the voussoir have been omitted from the figure, and only their resultant is drawn in. It was of course necessary to draw in these forces in pencil and obtain the position of the resultant, as explained in Fig. 221; and then, for simplicity, only the resultant was inked in.
The loads on the other voussoirs are computed similarly. The numerical values for the loads on the various voussoirs (including the weights of the voussoirs), are tabulated as follows: For this first condition of loading, the total loads for voussoirs Nos. 7, S, 9, 10, and 11 will be the same as those for voussoirs 5, 1, 3, 2, and 1 respectively.
The loads for the second condition of loading are found by using the same load on the first five voussoirs, but with only half of the live load on voussoir No. 6, which means that the load for the first con dition of loading (1,322 pounds) is reduced by 200 pounds, making it 1,122 pounds. Voussoirs Nos. 7 to 11 are each reduced by 400 pounds.
The total load for each voussoir is as tabulated below.
The loads for the third condition of loading are found by using the same loads as were employed for the second condition, except that for voussoirs Nos. 3 and 4, 1,600 pounds should be added to each load. These loads are also tabulated below: Fig. 223 was originally drawn at the scale of )i inch = 1 foot, and with the force diagram at the scale of 1,500 pounds per inch. The photographic reproduction has of course changed these scales some what. The student should redraw the figure at these scales, and should obtain substantially the same final results.
410. Drawing the Load Line for the First Condition of Loading. 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-kmd line, and select On it a trial position 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 (R,) 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 R, 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 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.