We shall assume that the earth or cinder fill on top of the arch has a thickness of one foot at the crown, and that it is level on top. We shall also assume that the arch ring is composed of stones which weigh 160 pounds per cubic foot, and we shall therefore consider 160 pounds per cubic foot as the unit-weight in determining the reduced load line. From the extremities of the extrados, draw verticals until they intersect the upper line of the earth fill. For convenience we shall divide the horizontal distance between these verticals into 11 equal parts, each to be about 2 feet wide. Draw verticals through these points of division down to the extrados; then draw radial lines from the extrados to the intrados. These lines are drawn radially from a point approximately halfway between the center of the extrados and the center of the intrados. This means that the joints, instead of being exactly perpendicular to either the extrados or intrados, have a direction which is a compromise between the two. The discrepancy is greatest at the abutments, and approaches zero at the crown. This will divide the arch ring into 11 voussoirs, together with a keystone at the center or crown. Assuming that the earth fill weighs 100 pounds per cubic foot, the lines of division between the 11 sections of the earth fill should each be reduced to 100/160 or 5/8 of its actual depth. If we further assume that the pavement is a little over six inches thick, and that its weight is equivalent to six inches of solid stone, we may add a uniform ordinate equal to six inches in thickness (according to the scale adopted), and this gives the total dead load on the arch. We shall assume further a live load amounting to 200 pounds per square foot over the whole bridge. This is equivalent to 200/160 of a foot, or 1 foot 3 inches, of solid masonry over the whole arch. This gives the reduced load line for the condition of loading that the entire arch is loaded with its maximum load.

As another condition of loading, we shall assume that the above load extends only across one-half of the arch. We shall probably find that, owing to the eccentricity of this form of loading, the stability of the arch is in much greater danger than when the entire arch is loaded with a maximum load.

We shall also consider the condition which would be found by running a twenty-ton road roller over the arch. A complete test of all the possible stresses which might be produced under this condition would be long and tedious; but we may make a first trial of it by finding the stresses which would be produced by placing the road roller at one of the quarter-points of the arch - a position which would test the arch almost, if not quite, as severely as any other possible position. Owing to the very considerable thickness of earth fill, as well as the effect of the pavement, the load of the roller is distributed in a very much unknown and very uncertain fashion over a considerable area of the haunch of the arch. The extreme width of such a roller is eight feet; the weight on each of the rear wheels is approximately 12,000 pounds. We shall assume that the weight of each rear wheel is distributed over a width of three feet and a length of four feet, so that the load on the top of the arch under one of the wheels may be considered at the rate of 1,000 pounds per square foot over an area of 12 square feet. For the unit-section of the arch one foot wide, this means a load of 4,000 pounds loaded on two voussoirs which are four feet in total length. The front roller of the road roller comes between the two rear rollers, and therefore would affect but little, if any, the particular arch ring which we are testing. Not only is it improbable that there would be a full loading of the arch simul-taneouslv with that of a road roller, but it is also true that a full loading would add to the stability of the arch. Yet, in order to make the worst possible condition, we shall assume that the part of the arch which has the road roller is also loaded for the remainder of its length with a maximum load of 200 pounds per square foot; this item alone will take care of the effect of the front roller. A load of 1,000 pounds per square foot is the equivalent of a loading of 6 feet 3 inches of stone; and therefore, if we draw over voussoirs Nos. 3 and 4 a parallelogram having a vertical height above the dead-load line equal to 6 feet 3 inches of stone, and consider a reduced live-load line 15 inches deep (200/160 = 125 = 1 foot 3 inches) over the remainder of that half-span, we have the reduced load line for the third condition of loading.

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 accuracy 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 may 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 perpendiculars 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 4.085 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 loading. 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:

First Condition Of Loading

Voussoir No.

Load

Weight of Voussoir

Total

1

and

11

1,296

654

1,950

2

,,

10

1,135

592

1,727

3

,,

9

1,010

528

1,538

4

,,

8

927

483

1,410

5

,,

7

880

456

1,336

6

867

455

1,322

For this first condition of loading, the total loads for voussoirs Nos. 7, 8, 9, 10, and 11 will be the same as those for voussoirs 5, 4, 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 condition 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:

Second Condition Of Loading

Voussoir No.

Total Load

1

1,950

2

1,727

3

1,538

4

1,410

5

1,336

6

1,122

7

936

8

1,010

9

1,138

10

1,327

11

1,550

Third Condition Of Loading

Voussoir No.

Total Load

1

1,950

2

1,727

3

3,138

4

3,010

5

1,336

6

1,122

7

936

8

1,010

9

1,138

10

1,327

11

1,550

Fig. 223 was originally drawn at the scale of \ 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 somewhat. The student should redraw the figure at these scales, and should obtain substantially the same final results.