a = 160000/2.12000 = 6 2/3

Or we would use a band, say, 5" x 1 1/3".

If dovetailed dowels of stone are used, as shown in Fig. 115, there should be one, of course, in every vertical joint. The dowels should be large enough not to tear apart at A D, nor to shear off at A E, nor to crush A 13. Similarly, care should be taken that the area of C G + B F is sufficient to resist the tension, and of H C to resist the shearing.

The strain on A D will, of course, be tension and equal to one-half the horizontal thrust; the same formula holds good, therefore, as for

Thrust of dome.

Metal bandsStrength of dowels.

metal bands, except, of course, that we use for (t/f) its value for whatever stone we select. Supposing the dome to be built of average marble, we should have from Table V, (t/f) = 70, therefore, a = 160000/2.70 = 1143

Fig. II5.

Now if the dome is built in courses 2' 6" or 30" high, the width of dovetail at its neck A D would need to be:

AD = 1143/30 = 38 inches.

As this would evidently not have sufficient area at G C and B F we must make; A D smaller, and shall be compelled to use either a metal dowel, or some stronger stone. By reference to Table V we find that for bluestone,

(t/c) = 140, therefore: a = 160000/2.140 = 571

This would almost do for the lower joint, which is 60" wide, for if we made: G C + B F = 38" and

A D = 19" it would leave 60 - 38 + 19 = 3" or, say, 1 1/2" splay; that is, D H = 1 1/2". Had we used iron, or even slate, however, there would be no trouble.

The shearing strain on cither A E or H C will, of course, be equal to one-quarter of the horizontal thrust or b = h/4.d. (g/f) (66)

Where b = the width of one-half the dowel, in inches (A E) Where h = the horizontal thrust at the joint, in pounds Where d = the height of the course, in inches.

Where (g/f) = the safe resistance to shearing, per square inch, of either dowel or stone voussoir (whichever is weaker).

When the shearing stress of a stone is not known, we can take the tension instead. Thus in our case as the marble is the weaker, we should use (g/f) = 70, therefore

AE = HC= 160000/4.30.70 = 19 inches.

The compression on A B need not be figured, of course, for while the strain is the same as on A E, the area of A B is somewhat larger, and all stones resist compression better than shearing.

When a plank is " cambered," sprung up, and the ends securely confined, it becomes much stronger transversely than when lying flat. The reason is very simple, as it now acts as an arch, and forces the abutments to do part of its work. Such a plank can be calculated the same as any other arch.

Quite a curiosity in construction, somewhat in the above line, has been recently introduced in New York by a Spanish architect. He builds floor-arches but 3" thick, of 3 successive layers of 1 "-thick tiles, up to 20' span, and more. His arches have withstood safely test-loads of 700 pounds a square foot. The secret of the strength of his arches consists in their following closely the curve of pressure, thus avoiding tension in the voussoirs, as far as possible. But even were this to exist, it could not open a joint without bodily tearing off several tiles and opening many joints, as shown in Fig. 116, owing to the fact that each course is thoroughly bonded and breaks joints with the course below; besides this, each upper layer is attached to its lower layer by Portland-cement mortar. Specimens of these tiles have been tested for the writer and were found to be as follows: Compression (12 days old) c = 2911 pounds or say ( c/f ) - 300 pounds.

Tension (10 days old) t = 287 pounds or say (t/f) = 30 pounds.

The shearing stress of the mortar-joints is evidently greater than the tension, as samples tested tore across the tile and could not be sheared off.

The modulus of rupture (about 10 days old) was

Cambered plank arch.

Spanish tile arches.

Fig. 116

Tile Floor Arch, 177 k= 91 pounds or, say, (k/f) = 10 pounds.

Of course older specimens would prove much stronger-There is only one valid objection that the writer has heard so far against these arches, and this is, that in case of any uneven settlement they might prove dangerous; as, of course, the margin in which the curve of pressure can safely shift in case of changed surroundings is very small. The writer does not think the objection very great, however, as settlements are apt to ruin any arch and should be carefully provided against in every case. The arches have some very great advantages. The principal one, of course, is their lightness of construction and saving of weight on the floors, walls and foundations. Then, too, in most cases iron beams can be entirely dispensed with, the arches resting directly on the brick walls; of course, there must be weight enough on the wall to resist the horizontal thrust, or else iron tie-rods must be resorted to. The former is calculated as already explained for retaining walls. If tie-rods are used, they are calculated as explained above for the example of the 7" flat floor arch. An example of these 3"-tile arches may be of interest.

Example IX. A segmental 3" tile arch, built as explained above, has 20' clear span with a rise at the centre of 20". The floor is loaded at the rate of 150 pounds per square foot. Is the arch safe?

We divide the arch into five parts or voussoirs, each voussoir carrying 2' of floor = 300 pounds; we make (Fig. 118) ab = 300 pounds, be = 300 pounds, etc., and find

Example of tile arch.

Fig 117.

the horizontal pressures (Fig. 117) g1 h1,g2 h1, etc. The last one g5 h5 is the largest and scales 4500 units or pounds. We now make (Fig.