Page images
PDF
EPUB

that necessary to make it slide, and, reciprocally, it will be overturned when less force necessary to produce that effect than to make it slide.

1366. II. When the parallelopiped is placed on an inclined plane, it will slide so lon as the vertical QS drawn from its centre of gravity does not fall without the base Al Hence, to ascertain whether a parallelopiped ABCD with a rectangular base (fig. 564.) will slide down or overturn; from the point B we must raise the perpendicular BE: if it pass out of the centre of gravity, it will slide; if, on the contrary, the line BE passes within, it will overturn.

Fig. 564.

1367. If the surfaces of stones were infinitely smooth, as they are supposed to be in the application of the principles of mechanics, they would begin to slide the moment the plane upon which they are placed ceases to be perfectly horizontal; but as their surfaces are fu of little inequalities which catch one another in their positions, Rondelet found, by r peated experiments, that even those whose surfaces are wrought in the best manner do no begin to slide upon the best worked planes of similar stone to the solids until such plane are inclined at angles varying from 28 to 36 degrees. This difficulty of moving one ston upon another increases as the roughness of their surfaces, and, till a certain point, as thei weight for it is manifest, 1st, That the rougher their surfaces, the greater are the ir equalities which catch one another. 2d. That the greater their weight, the greater is th effort necessary to disengage them; but as these inequalities are susceptible of bein broken up or bruised, the maximum of force wanting to overcome the friction must b equal to that which produces this effect, whatever the weight of the stone. sd. That thi proportion is rather as the hardness than the weight of the stone.

:

1368. In experiments on the sliding of hard stones of different sizes which weighed from 2 to 60 lbs., our author found that the friction which was more than half the weigh for the smaller was reduced to a third for the larger. He remarked that after each experi ment made with the larger stones a sort of dust was disengaged by the friction. In sof stones this dust facilitated the sliding.

1369. These circumstances, which would have considerable influence on stones of a grea weight, were of little importance in the experiments which will be cited, the object bein to verify upon hard stones, whose mass was small, the result of operations which the theor was expected to confirm. By many experiments very carefully made upon hard freeston well wrought and squared, it was found, 1st, That they did not begin to slide upon a plan of the same material equally well wrought until it was inclined a little more than 30 degrees 2d. That to drag upon such stone a parallelopiped of the same material, a little more tha half its weight was required. Thus, to drag upon a level plane a parallelopiped 6 in. long 4 in. wide, and 2 in. thick, weighing 4 lbs. 11 oz., (the measures and weights are French as throughout *), it was necessary to employ a weight equal to 2 lbs. 7 oz. and 4 drs 3d. That the size of the rubbing surface is of no consequence, since exactly the same fore is necessary to move this parallelopiped upon a face of two in. wide as upon one of 4.

1370. Taking then into consideration that by the principles of mechanics it is proved that to raise a perfectly smooth body, or one which is round upon an homogeneous plan inclined at an angle of 30 degrees, a power must be employed parallel to the plane whic acts with a force rather greater than half its weight, we may conclude that it requires a much force to drag a parallelopiped of freestone upon an horizontal plane of the sam material as to cause the motion up an inclined plane of 30 degrees of a round or infinitel polished body.

1371. From these considerations in applying the principles of mechanics to arches compose of freestone well wrought, a plane inclined at 30 degrees might be considered as one upo which the voussoirs would be sustained, or, in other words, equivalent to an horizontal plane 1972. We shall here submit another experiment, which tends to establish such an hype thesis. If a parallelopiped C ( fig. 565.) of this stone be placed between two others, BD, RS, whose masses are each double, upon a plane of the same stone, the parallelopiped C is sustained by the friction alone of the vertical surfaces that touch it. This effect is a consequence of our hypothesis; for, the inequalities of the surfaces of bodies being stopped by one another, the parallelopiped C, before it can fall, must push aside the two others, BD, RS, by making them slide along the horizontal plane of the same material, and for that purpose a force must be employed equal to doub the weight sustained.

The Paris pound = 7561 Troy grains.

Ounce 472.5625.

Dram or gros = 59:0703.

Grain = 0:8204.

Fig. 565.

And as the English avoirdupols pound = 7000 Troy grains, it contains 8538 Paris grains
The Parts foot of 12 inches = 12-7977 English inches.

The Paris line one-twelfth of the foot.

1973. If to this experiment the principles of mechanics be applied, considering the plane of 30 degrees inclination as a horizontal plane, the vertical faces ED FR may be considered as inclined planes of 60 degrees. On this hypothesis it may be demonstrated by mechanics, that to sustain a body between two planes forming an angle of 60 degrees (fig. 566.), the resistance of each of these planes must be to half the weight sustained as HD is to DG, as the radius is to the sine of 30 degrees, or as 1 is to 2.

H

D

Fig. 566.

EQUILIBRIUM OF ARCHES.

1374. The resistance of each parallelopiped represented by the prism ABDE (fig. 565.) being equal to half their weight, it follows that the weight to be sustained by the two prisms should equal one quarter of the two parallelopipeds taken together, or the half of one, which is confirmed by the experiment. This agreement between theory and practice determined Rondelet to apply the hypothesis to models of vaults composed of voussoirs and wedges disunited, made of freestone, with the utmost exactness, the joints and surfaces nicely wrought, as the parallelopipeds in the preceding example. 1375. The first model was of a semicircular arch 9 inches diameter, comprised between two concentric semi-circumferences of circles 21 lines apart. It was divided into 9 equal voussoirs. This arch was 17 lines deep, and was carried on piers 2 inches and 7 lines thick. It was found, by gradually diminishing the piers, which were at first 2 inches and 10 Ap lines thick, that the thickness first named was the least which could be assigned to resist the thrust of the voussoirs.

1376. The model in question is represented in fig. 567., whereon we have to observe, - 1st. That the first voussoir, I, being placed on a level joint, not only sustains itself, but is able to resist by friction an effort equal to one half of its weight. 2d. That the second voussoir, M, being upon a joint inclined 20 degrees, will also, through friction, sustain itself; and that, moreover, these two voussoirs would resist, previous to giving way on the joint AB, an horizontal effort equal to one half of their weight. 3d. That the third voussoir, N, standing on a joint inclined at 40 degrees, would slide if it were not retained by a power PN acting in an opposite direction. 4th. That taking, according to our hypothesis, an inclined plane of 30 degrees, whereon the stones would remain in equilibrium as an horizontal one, the inelined point of 40 degrees may be considered as an inclined plane of 10 degrees, supposing the surfaces infinitely smooth. 5th. That the effort of the horizontal power which holds this voussoir in equilibrium upon its joints will be to its weight as the sine of 10 degrees is to its cosine, as we have, in the section on Mechanics, previously shown. (1255 et seq.)

[graphic]

Fig. 567.

1377. The model of the vault whereon we are speaking being but 9 inches, or 108 lines in diameter, by 21 lines for the depth of the voussoirs, that is, the width between the two concentric circumferences, its entire superficies will be 4257 square lines, which, divided by 9, gives for each voussoir 473 square lines. Then, letting the weight of each voussoir be expressed by its superficies, and calling P the horizontal power, we have

P: 473:: sin. 10° : cosin. 10°;

Or, P: 473::17365: 98481; which gives P-83

The fourth voussoir, being placed upon a bed inclined at 60 degrees, will be considered as standing on a plane inclined only at 30 degrees, which gives, calling Q the horizontal power which keeps it on its joint,

Q: 473:: sin. 30°: cosin. 30°.

Or, Q: 473::50000 : 86603=273.

1378. The half-keystones, being placed on a joint inclined 80 degrees, are to be considered as standing on an inclined plane of 50, the area of the half key which represents its weight being 236. If we call R the horizontal power which sustains it on its joint, we shall have the proportion

R: 236: sin. 50: cosin. 50;

or, R: 236:76604 64279; which gives R-281.

1379. Wishing to ascertain if the sum of these horizontal efforts, which were necessary to keep on their joints the two voussoirs N, O, and the half-keystone, was capable of thrusting away the first voussoir upon its horizontal joint AB, the half arch was laid down upon a level plane of the same stone without piers, and it was proved that to make it give way an horizontal effort of more than 16 ounces was required, whilst only 10 were neces

sary to sustain the half-keystone and the two voussoirs N, O. The two halves of the arch united bore a weight of 5 lbs. 2 oz. before the first voussoirs gave way.

1380. To find the effect of each of these voussoirs when the arch is raised upon its pier let fall from the centres of gravity N, O, S of these voussoirs the perpendiculars Në, Oo, S in order to obtain the arms of the levers of the powers P, Q, R, which keep them in thei places, tending at the same time to overturn upon the fulcrum T the pier which carrie the half arch, and we have their effort

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

Total effort in respect of the fulcrum, 163898-804

1381. The pier resists this effort, 1st, by its weight or area multiplied by the arm of the lever determined by the distance Tu from the fulcrum T to the perpendicular let fall from the centre of gravity G upon the base of the pier. 2d. By the weight of the half arcl multiplied by the arm of its lever VY determined by the vertical LY let fall from the centre of gravity L, and which becomes in respect of the common fulcrum T=Tt o VB-BY, in order to distinguish BY, which indicates the distance of the centre of gravity of the half arch (and which is supposed known because it may be found by the rules giver in 1275. et seq.) from the width VB that the pier ought to have to resist the effort o the half arch sought. In order to find it, let P, the effort of the arch above found, b 163898 804.

I
2

=3

ar
2

[blocks in formation]

1382. The area of the pier which represents its weight multiplied by the arm of the lever will be ax x That of the half arch multiplied by its arm of lever will be shown by VB+ BY, where x + c will be br+ bc, whence the equation P= + bx + bc,

ar2

2

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

26

a

to eliminate a2, we have

the second power; but as x2+

square of half the known quantity which multiplies the second term; by adding this

b2 a29

2br b2
+ =
a a2

b2 2p-2bc

+

The

a

[ocr errors]
[blocks in formation]

which is to each side of the equation, we have x2+ square, first member by this means having become a perfect square whose root is a + a

+ which becomes, by transferring to the other side of the

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

in which being only in the first member of the equation, its value is determined from the known quantities on the other side. Substituting, then, the values of the known quantities, we have

[blocks in formation]

which gives r=28] lines instead of 2 inches and 5 lines, which was assigned to the piers that they might a little exceed equilibrium in their stability.

Proof of the above Method by another Method of
estimating Friction.

1383. A proof of the truth of the hypothesis in the preceding section is to be found in the method proposed by Bossut in his Treatise on Mechanics.

Let the voussoir N (fig. 568.) standing on an inclined plane be sustained by a power Q acting horizontally. From the

Fig. 568.

centre of gravity let fall the vertical Nn, which may be taken to express the weight of the voussoir. This weight may be resolved into two forces, whereof one, Ne, is parallel to the joint, and the other Na is perpendicular to it. In the same manner the power Q expressed by QN in its direction may be resolved into two forces, whereof Nf will be parallel to the joint and the other Nd perpendicular to it. Producing the line from the joint HG, drawing the horizontal line GI and letting fall the vertical HI, consider the line HG as an inclined plane whose height is HI and base IG. Then the force Nc with which the voussoir will descend will be to the weight as the height HI of the inclined plane is to its length HG. Calling p the weight of the voussoir, we then have Nc=px and the force Na which presses against the plane as the base of the

HG
HI'

IH

IG

IG
GH'

plane IG is to its length, which gives the force Nu=p × TIG 1384. Considering, in the same way, the two forces of the power Q which retain the voussoir on the inclined plane, we shall find the parallel force Nf=Q × and the perpendicular force Nd = Qx HG The force resulting from the two forces Na, Nd, which press against the joint, will be expressed by p× HG + Q× GH; and as the voussoir only begins to slide upon a plane whose inclination is greater than 30 degrees, the friction will be to the pressure as the sine of 30 degrees is to its cosine, or nearly as 500 is to 866, or of its expression. Calling this ratio n, we shall, to express the friction, have

(px

IG

× GH+ Q ×

IG

IG

x n.

H

As the friction prevents the voussoir sliding on its joint, in a state of equilibrium, we shall have the force Nf equal to the force Nc, less the friction; from which results the equation—

[blocks in formation]

All the terms of which equation having the common divisor HG, it becomes

QxIG=px HI−(p × IG - Q × III) × n ;

and, bringing the quantities multiplied by Q to the same side of the equation, we have

Q× IG + (Q× IH) x n =px HI-(px IG) × n; which becomes
2x (IG+ n x IH=px (HI−n × IG); whence results

Qx(

Q=px

[ocr errors]

x

which is the formula for each voussoir, substituting the

values for the expression.

1385. Thus for the third voussoir N (fig. 567.) placed on an inclined plane of 40 degrees, HI which represents the sine of the inclination will be 643, and its cosine represented by IG, 766, the expression of the friction n will be go, or nearly. The weight of the voussoir expressed by its area will be 473, which several values being substituted in the formula, we have

[blocks in formation]

which gives Q=836, the expression of the horizontal force P, which will keep the voussoir N in equilibrium on its joint instead of 834, which was the result of the operation in the preceding subsection.

[ocr errors][merged small]

1386. The same formula Q=P × IG+XIH gives for the voussoir M on an inclined joint

of 60 degrees, whose sine HI is 866 and cosine IG 500, Q=473 ×

[blocks in formation]

instead of 273.3, which was the result of the operation in the preceding section. 1387. For the half-keystone, the sine HI, being of 80 degrees, will be expressed by 985, and its cosine IG by 174; the half-keystone by 236, and the friction by

985-1 × 174

The formula now will be Q=236 × 174+18×985' which gives Q=282·2, instead of 231 found by the other method. These slight differences may arise from suppressing the two last figures of the sines, and some remainders of fractions which have been neglected. Multiplying these values of the powers which keep the voussoirs in equilibrium upon their beds by the several arms of the levers, as in the preceding calculations, their energy will be as follows:

[blocks in formation]

For the total force in respect of the fulcrum T=16385156.

Which is the value of p, and being substituted for it in the formula x=

[ocr errors]

as well as the values of the other letters, which are the same as in the preceding example, we have

[blocks in formation]

for the thickness of the piers, instead of 28ines found by the preceding operation.

Application of the Principles in the Model of a straight Arch.

1 2 3

G

1388. The second model to which the application of the preceding methods was made was a straight arch of the same sort (fig. 569. ), whose opening between the piers was 9 inches. The arch was 21 lines high and 18 lines thick. It was divided into 9 wedges, whose joints were concentric. To determine the section of the joints, the diagonal FG was drawn on the face of the half arch, and from its extremity F touching the pier, the perpendicular FO meeting O in the vertical, passing through the middle of the opening of the piers, all the sections meeting in this point O. Each of the sections of the piers which support the arch forms an angle of 21° 15' with the vertical, and of 68° 45' with the horizon.

1389. In considering each of the wedges of the half arch as in the preceding method, it will be found that in order to retain the voussoir A on the joint IF (of the pier) which forms with the horizontal line NF an angle of 68° 45', we have

For the horizontal force

second B

third C

fourth D

half-keystone

Fig 569.

Fig. 570.

[blocks in formation]

The height of the piers being 195 lines to the underside of the arch, and 216 to the top of the extrados, it follows that the arm of the lever, which is the same for all the wedges, is 206, from which we derive for the thrust P of the formula,

[blocks in formation]

b which expresses the area of the half arch=1219; c which expresses the distance of its centre of gravity from the vertical Fn=24, and the height of the pier a=216. stituting these values in the formula, we shall have

[blocks in formation]

Now, sub

Experiment gives 44 lines for the least width of the piers upon which the model will stand. But it is right to observe that from the impossibility of the joints being perpendicular to the intrados, the forces of the wedges press in a false direction on each other, as will be seen by the lines Fa, 1c, 2e, 3g, perpendicular to the joints against which the forces are directed, so that such an arch will only stand when the perpendicular FG does not fall within the thickness of the arch; and, indeed, this sort of arch is only secure when it comprises an arc whose thickness is equal to the section upon the piers IF, as shown in fig. 570.

Observations on the Way in which Stones forming an Arch act to
support one another.

1390. Let the semicircular arch AHCDNB (fig. 571.) consist of an infinite number of voussoirs acting without friction, and only kept in their places by their mutual forces acting on each other. It will follow

1. That the first voussoir, represented by the line AB, having its joints sensibly parallel and horizontal, will act with its whole weight in the vertical direction IE to strengthen the pier.

F

B

Fig. 571.

N

G

D

L

« PreviousContinue »