73. Inverted Arches

Inverted arches are sometimes built under and between the bases of piers, as shown in Fig. 36, with the idea of distributing the weight of the piers over the whole length of the footings. This method is objectionable - first, because it is nearly impossible to prevent the end piers of a series from being pushed outward by the thrust of the arch, as shown by the dotted line, and second, it is generally impossible with inverted arches to make the areas of the different parts of the foundation proportional to the load to be supported. It is much better to build the piers with separate footings, projecting equally on all four sides of the pier and proportioned to the load supported by the piers. The intermediate wall may either be supported by steel beams or arches, as preferred.

In some instances, however, when building on comparatively soft soils, and where it is impracticable to use spread footings, inverted arches may be advantageously used, especially when it is necessary to reduce the height of the footing to a minimum.

73 Inverted Arches 10037

Fig. 36.

If it is decided to use inverted arches, the foundation bed should be leveled and a footing built over the whole bed to a depth of at least 12 to 18 inches below the bottom of the arch. Concrete is much the best material for this footing, although brick or stone may be used if found more economical. The upper surface of the footing should be accurately formed to receive the arch. The arch should be built of hard brick, laid in cement mortar, and generally in separate rings or rowlocks, and should abut against stone or concrete skewbacks, as shown in Fig. 37.

It is better to build the arches before putting in the skewbacks, and for the latter 1 to 6 Portland cement concrete possesses special advantages, as the concrete can be deposited between the ends of the arches and rammed evenly and simultaneously, thus giving a solid and uniform bearing against the ends of the arches, tending to prevent unequal settlement and cracking.

74. Above the concrete skewback a solid block of stone should be placed if it can be readily obtained. The thickness of the arch ring should be at least 12 inches, and heavy iron plates or washers should be set in the middle of the concrete skewbacks and connected with iron or steel rods, to take up the thrust of the end arches. The "rise" of the arch, or distance R, Fig. 37, should be equal to from one-fourth to one-sixth of the span. The sectional area of the arch should equal the result obtained by the following formula:

73 Inverted Arches 10038

Fig. 37.

Section of arch in sq. inches = Total load on arch (in lbs.) X span / 8 X R X 10 and the area of the tie rods should equal

Total load on arch (in lbs ) X span / 8XRX850 for wrought iron and

Total load on arch X span / 8 X R X 1050 for steel, the span being measured in feet, and the distance R in inches.

The load on the arch will be equal to the span multiplied by the pressure per lineal foot imposed on the soil. The latter will be obtained by dividing the load on the piers by the distance between centres of piers.

75. Example

It is desired to use inverted arches between the piers of a three-story building, resting on a soil whose bearing power cannot be safely estimated at over 3,000 pounds per square foot The piers are of stone, 4 feet long, 22 inches thick, and 14 feet apart from centres. Each pier supports a total load of 98,000 pounds. What should be the sectional area of the arch, and of the rods in end spans ?

Answer. - The span of the arch will be 10 feet, and the distance R about one-fifth of 10 feet, or 24 inches. The load per lineal foot on the soil will equal 98,00014, or 7,000 pounds. The footing under the arch must therefore be 2 feet 4 inches wide to reduce the pressure to 3,000 pounds per square foot. The width of the arch itself we will make 22 inches, or two and one-half bricks. The total load on the arch will equal 10X7,000, or 70,000 pounds

The sectional area of the arch must therefore equal 70000X10 / 8 X 24 X 10 or 354 square inches.

As the width is 22 inches, the depth must equal 35422, or 16 inches, which will require four rowlocks or rings.

The sectional area of the ties must equal, for wrought iron, 70000 X 10 / 8X24X850 or 4.3 square inches.

In this case it will be better to use two rods of 2.15 square inches in area, or say two 1-inch rods.

All cast iron work in the foundation should be coated with hot asphalt, and the rods should be dipped in linseed oil while new and hot and afterward painted one heavy coat of oxide of iron or red lead paint.