292. Before proceeding to investigate the mode of framing and stiffness of centres, the point must be determined at which the arch-stones first begin to press upon them, and also the pressure at different periods of the formation of the arch.

It has been found by experiment that a stone placed upon an inclined plane does not begin to slide until that plane has an inclination of about 30 degrees from the horizontal plane,* and until a stone would slide upon its joint or bed, it is obvious that it would not press upon the centre. Also, when a hard stone is laid with a bed of mortar it will not slide until the angle becomes from 34 to 36 degrees. A soft stone bedded in mortar will stand when the angle which the joint makes with the horizon is 45 degrees, if it absorbs water quickly, because in that case the mortar becomes partially set.† Similar results have been obtained by other experimentalists; therefore we may consider the pressure in general to commence at the joint which makes an angle of about 32 degrees with the horizon.

This angle is called the angle of repose, and if we suppose the pressure to be represented by the radius, the tangent of this angle will represent the friction; hence, considering the pressure as unify, the friction will be 0.625. Perronet estimates the friction at 0.8; ‡ but it is erring on the safe side to take the lower result.

293. The next course above the angle of repose will press upon the centre, but only in a small degree, and the pressure will increase with each succeeding course. The relation between the weight of an arch-stone and its pressure upon the centre, in a direction perpendicular to the curve of the centre, may be determined from the following equation: W (sin. a - f cos. a) = P.

* Rondelet, 'L'Art de Batir.' † Idem.

‡ " Memoire sur le Cintrement et le Decintrement des Ponts," in the 'Memoirs of the Academy of Sciences' at Paris for 1773.

Where W is the weight of the arch-stone, P = the pressure upon the centre, f = the friction, and a = the angle which the plane of the lower joint of the arch-stone makes with the horizon.

294. When the angle which the joint makes with the horizon is..............................

34 degrees

P

=

.04 W.

"

"

"

36

"

P

=

.08 W.

"

"

"

38

"

P

=

.12 W.

"

"

"

40

"

P

=

.17 W.

"

"

"

42

"

P

=

.21 W.

"

"

"

44

"

P

=

.25 W.

"

"

"

46

"

P

=

.29 W.

"

"

"

48

"

P

=

.33 W.

"

"

"

50

"

P

=

.37 W.

"

"

"

52

"

P

=

.40 W.

"

"

"

54

"

P

=

.44 W.

"

"

"

56

"

P

=

.48 W.

"

"

"

58

"

P

=

.52 W.

"

"

"

60

"

P

=

.54 W.

But when the plane of the joint becomes so much inclined hat a vertical line passing through the centre of gravity of he arch-stone does not fall within the lower bed of the stone, he whole weight of the arch-stone may be considered, with-out material error, as resting upon the centre. We have thus an easy method of estimating the weight upon a centre, at any period of the construction, or when any portion of the arch-stones is laid, as well as when the whole weight it has to sustain is upon it. 295. As an example, let it be required to determine the pressure of the arch-stones upon 20 degrees of the centre, counting from the joint which makes an angle of 32 degrees with the horizon.

Rule. - Take out of the Table in the last Art. the decimals opposite every second degree for the first 20 degrees, that is, from 32 to 52 degrees, and add them together. Multiply the sum thus found by the weight of a portion of the arch-stones comprehended between 2 degrees; the product will be equal to the pressure of 20 degrees of the arch upon the centre.

Suppose the frames of the centre to be 5 feet from middle to middle, and the depth of the arch-stones to be 4 feet; also, that the space comprehended between 2 degrees of the arch measured at the middle of the depth of the stone is 1.5 foot. The solid content will be found to be 30 cubic feet; and if the weight of a cubic foot of the stone be 150 lbs., the weight of 2 degrees will be 30 x 150 = 4500 lbs.

Then adding together the decimals for 20 degrees, that is, from 32 degrees to 52, the sum is 2.26. This sum multiplied by the weight of 2 degrees, or 4500 lbs., gives 10,170 lbs. for the pressure of 20 degrees upon one rib of the centre.

296. It will be seen from the Table that the pressure increases very slowly until the joint begins to make a considerable angle with the horizon; and it is of importance to bear this in mind in designing centres, because the strength should be directed to the parts where the strain is greatest. For instance, at the point where the joint makes an angle of 44 degrees with the horizon, the arch-stone only exerts a pressure of one-fourth of its weight upon the centre; where the angle of the joint is 58 degrees, the pressure exceeds half the weight; but near to the crown the stones rest wholly upon the centre. Now it would be absurd to make the centre equally strong at each of these points; besides, by such a method there would not be the means of applying the strength where it is really required, owing to the interference of ties and braces, that are only an encumbrance to the framing.

When the depth of the arch-stone is about double its thick ness, the whole of its weight may he considered to rest upon the centre when the joint makes an angle of about 60 degrees with the horizon. If the length be less than twice the thickness, it may be considered to rest wholly upon the centre when the angle is below 60 degrees; and if the length exceed twice the thickness, the angle will be considerably above 60 degrees before the whole weight will press upon the centre.