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2. That the vertical voussoir CD, which represents the keystone, having also its joints knsibly parallel, will act with its whole weight horizontally to overturn the semi-arches and piers which carry them.

3. That all the other voussoirs between these two extremes will act with the compound forces Gn, nm, ml, Kl, Kh, hg, gf, § T, which may each be resolved into two others, whereof one is vertical and the other horizontal : thus the compound force Kh is but the result of the vertical force 4h, and the horizontal force 4K.

4. That the vertical force of each voussoir diminishes from T to G, where, for the keystone CD, it becomes nothing, whilst the horizontal forces continually increase in an inrerse ratio ; so that the voussoir HN, which is in the middle, has its vertical and horizontal furces equal.

5. That in semi-circular arches whose extradosses are of equal height from their intradosses, the circumference passing through the centre of gravity of the voussoirs may represent the sum of all the compound forces with which the voussoirs act upon one another in sustaining themselves, acting only by their gravity.

6. That if from the points T and Go the vertical TF and horizontal GF be drawn meeting in the point F, the line TF will represent the sum of the vertical forces which assist the stability of the pier, and FG the sum of the horizontal forces which tend to overthrow it.

7. That if through the point K the horizontal line IKL be drawn between the parallels FT and CO, the part IK will represent the sum of the horizontal forces of the lower part AHNB of the vault, and KL those of the upper part HCDN.

8. The lower voussoirs between T and K being counterpoised by their vertical forces, the part of the arch AHNB will have a tendency to fall inwards, turning on the point B, whilst the voussoirs between K and G being counterpoised by their horizontal forces, the part HCDN of the arch will re-act upon the lower part by its tendency to turn upon the point A.

9. The horizontal forces of the upper part of the arch shown by KL acting from L towards K, and those of the lower part shown by IK opposite in direction to the former, that is, from I to K, being directly opposed, would counterpoise each other if they were equal, and the arch would have no thrust; but as they are always unequal, it is the difference of the forces which occasions the thrust, and which acts in the direction of the strongest power.

10. If we imagine the width BO of a semi-arch constantly to diminish, its height remaining the same, the sum of the horizontal forces will diminish in the same ratio, so that when the points B and O are common, the horizontal force being annihilated, nothing remains but the vertical force, which would act only on the pier, and tend to its stability, thrust vanishing, because, instead of an arch, it would, in fact, be nothing more than a continued pier.

1. İf, on the contrary, the height OD diminishes, the width BO remaining the same, the curve B and D would, at last, vanish into the right line BO, and the arch would become a straight one. In this case, the vertical forces which give stability to the pier being destroyed, all that remains for sustaining the arch are the horizontal forces which will act with the whole weight of the arch ; whence this species of arches must be such as exert most thrust, and circular arches hold a middle place between those which have no thrust, and flat arches, whose thrust is infinite, if the stones whereof they are formed could slide freely on one another, and their joints were perpendicular to their lower surfaces, as in other arches.

12. The inconveniences which result from making the joints of fat arches concentric have been before noticed. If the stones could slide freely on one another, as they only act in a false direction, their forces could never either balance or destroy one another.

13. A vast number of experiments made by Rondelet, upon fifty-four models of arches of different forms and extradosses, divided into an equal and unequal number of voussoirs, showed that the voussoirs acted rather as levers than as wedges, or as bodies tending to slide upon one another.

14. As long as the piers are too weak to resist the thrust of the voussoirs, many of them unite as one mass, tending to overturn them on a point opposite to the parts where the joints open.

15. Arches whose voussoirs are of even number exert more thrust than those which are of unequal number, that is, which have a keystone.

16. In those divided into uneven numbers and of unequal size, the larger the keystone the less is their thrust, so that the case of the greatest thrust is when a joint is made at the serter, as in the case of arches whose voussoirs are divided into equal numbers.

17. A semicircular arch divided into four equal parts has more thrust than one divided into nine equal voussoirs.

18. Arches including more than a semicircle have less thrust than those of a similar span, the intradosses and extradosses being of similar forms.

19. Thrust does not increase as the thickness of an arch increases ; so that, cateris paribus, an arch of double the thickness has not double the thrust.

Fig. 572

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other; thus transferring IK from K to m, the difference mL multiplied by the thickness horizontal direction KH, the arm of the lever is determined by the perpendicular PH raised from the fulcrum P of the lever to the direction of the thrust, so that its effort will be er. PS, determined by a vertical let fall from the centre of gravity Q, which gives for the

1. By its weight represented by the surface EPR PR multiplied by the arm of the lever

2. By the sum of the vertical efforts of the upper part of each arch, represented by MK x AB acting at the point K, the arm of their lever in respect of the fulcrum P of the pier being KH.

3. By the sum of the vertical efforts of the lower part represented by IT multiplied by But as in this equation neither PR (= BE) nor PS nor KH noi TE is known, we must

20. A semicircular arch whose extrados is equally distant throughout from, or, in other words, concentric with, the intrados, when divided into four equal parts, will only stand when its depth is less than the eighteenth part of its diameter, even supposing the abutments immoveable.

21. Whenever, in an arch of voussoirs of equal depth, a right line can be drawn from its outer fulcrum to the centre of the extrados of the keystone (fig. 572.), fracture does not occur in the middle of the haunches if the piers are of the same thickness as the lower part of the arch.

22. Arches whose thickness or depth diminishes as they rise to the vertex have less thrust than those whose thickness is equal throughout.

23. Semicircular and segmental arches whose extrados is an horizontal line have less thrust than others.

24. As long as the piers in the models were too weak to resist the thrust, it was possible to keep them in their places by a weight equal to double the difference between the thrust and resistance of one pier, acting by a string suspended passing through the joints in the middle of the haunches, or by a weight equal to that difference placed above each middle joint of the arches, as in fig. 572.

From these experiments and many others, a formula has been deduced to determine the thickness of piers of cylindrical arches of all species whose voussoirs are of equal depth, whatever their forms ; and to this we shall now introduce the reader,

Method. 1391. Having described the mean circumference GST (figs. 573, 574.), from the points G and T draw the tangents to the curve meeting in the point F.

From this point draw the secan, FO cutting it in the point K. This point is the place of the greatest effort, and of the consequent failure, if the thickness of the piers is too weak to resist the thrust.

1392. Through the point K, between the parallels TF and GO, draw the horizontal line IKL, which will represent the sum of the horizontal forces as will the vertical TF express the vertical forces ; the mean circumference GKT will

express the compound forces.

1393. The arches having an equal thickness throughout the part IK of the horizontal line multiplied by the thickness of the arch will express the horizontal effect of the lower part of either arch, and KL multiplied by the same thickness will express that of the upper part.

These two forces

(See 1398, et seq.) acting in opposite directions will partly destroy each of the vault will be the expression of the thrust. pressed by mL * AB ~ PH.

This will be resisted resistance of the pier the expression EP - PR x PS. AB acting on the point T has for the arm of its lever TE.

mL x AB ~ PH = PEx PR PS + MK x AB x KH+ IT x AB TE.

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F M

IIr.

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S

P SR

Fig. 373.

Fig. 374

This force acting at the point K in the

Hence, if equilibrium exist,

resort to an algebraic equation for greater convenience, in which

The effect of the thrust in the expression mL * AB
The height of the pier PE
EH=TI=KL=KV
PH
EB - PR

Tx AB

2

2

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2x

PS
The sum of the vertical forces of the upper part or

MK * AB
The sum of the forces of the lower part ITX
The part iK of the horizontal IKL
TB equal to half the thickness of the arch
The arm of the lever KH

That of TE
Chus the first equation becomes pa + pd = + m (e xx)+ n (1-6,
Or pa x pd = 4:2

+ mr + mc+n: - ne.
Transferring the unknown quantities to the second side of the equation, we shall

have + mx + nx + = pa + pd + ne – me. Multiply all the terms by 2, and divide by a, in order to get rid of r’, and we have 2 +

2(m+n). Making m+n=b, and adding to each member for the purpose of extracting

the root of the first member, We have to +

= 2p+

2pd + 2ne - 2mc Extracting the root, 3+

2pd + Pne -2mc And lastly, I=V2p+ 2pd +2ne – 2mc 4 52 1394. This last equation is a formula for finding the thickness of all sorts of arches whose voussoirs are of equal depth, which we will now apply to fig. 573. · The model was 36 inches and 3 lines in span.

The arch consisted of two concentric circles, and it was divided into four equal parts, a vertical joint being in the middle, the two others being inclined at angles of 45 degrees. The piers whereon it was placed were 40 inches and 4 lines high, and on a very exact measurement the values were as follow: PE (a in the formula) was

40:333 EH - TI=KL= KV (d in the formula)

13.876 ML AB (p in the formula) representing the thrust or 8.127 * 3 24.381 2p

48.762
2pd=48.762 x 13.876

676.621
2MK® AB , KH represented by 2mc (=5749 x 3 x 4.249)
2ne, which is IT x AB « AB (= 13.876 x 3 x 3)
b=m+n=(MK + IT) AB (=19.625 x 3)
a=EP, the height of the pier being 40.333, will be

1.459

+

82 +

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2p +

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;

b

73.282 124.824 58.875

b

58.875

or 40.333

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b

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One-2mc

2pd+? Substituting these values in the formula 1=V 2p+

h2

tä? we have r=V 48.762 + 676621+124-894–73.282

+2:128-1.459; which gives r= 5•8, or 5 inches 94 lines for the thickness of the piers to resist the thrust ot the archi, supposing it to be perfectly executed. But, from the imperfection of the execution of the model, it was found that the piers required for resisting the thrust a thickness of 6 inches and s lines.

1395. When the piers of the model were made 7 inches thick the arch on its central joint was found capable of supporting a weight of three pounds, being equal to an addition of 8 superficial inches beyond that of the upper parts of the arch which are the cause of the thrust, and this makes the value of 2p in the formula 56-762 instead of 48.762,

787 629+ 194.828-86-458 and changes the equation to I-V 56-762 +

2.430-1.55; from which we should obtain z = 7.366 inches, or 7 inches 3} lines, exhibiting a singular agree. ment between theory and practice. Rondelet gives another method of investigating the preceding problem, of which we do not think it necessary to say more than that it agrees with that just exhibited so singularly that the result is the same. It is dependent on the places of the centres of gravity, and therefore not so readily applicable in practice as that which has been just given.

Second Erperiment. 1396. Fig. 567., in a preceding page, is the model of an arch in freestone, which has been before considered. It is divided into nine equal voussoirs, whose depth to the extrados is 91 lines, and whose interior diameter is 9 inches.

40.333

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lines, or a little more than 164 lines. The model of this arch would not however stand on piers less than 17 lines thick.

In taking the root of double the thrust the result is 18lines, as it is also by the geo. netrical method.

Application to the Pointed Arch. 1401. The model which fig. 575. represents was of the same height and width as the last, and the voussoirs were all of equal thickness. Having laid down all the lines on the figure as before, we shall find iK of the formula to be

2pd-2mc. 62

wherein

F

M

Hr

I=V2p+

b

+

n

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20
94
14
63
23
9

and 2p

KL
mL
IT, represented by d,
MK
AB
mL * AB, represented by p in the formula, will
be 14 x 9

1 26

252 2pd will be 252 x 63, which gives

=15876 m, which is KM * AB or 23 x 9,

207 2m= 414, 2me=414 x 20

8280 The height of the pier, represented by a, being 120, we have 2pd2mc b, or FT x AB, will be 86 x 9=774; whence == 120=6:45, and a = 41-60. Substituting these values in the formula

I= V252 +63-8 +41 -6 -6.45= 12:46 lines for the thickness of the pier. In taking the square root of double the thrust the thickness comes out 15.88 lines, as it does by the geometrical method. Experiments showed that the least thickness of piers upon which the model would stand was 14 lines.

Fix. 575.

15876-8220

= 6 8; 120

b

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thus having found Bm=22}, we have the value of p=224 * 9=201 ; and 2p=402.

140:3. This inodel was of the same dimensions as the preceding : b, which represents Tf * AB, will be 769.5; a will be 6:41, and :41:11. These values substituted in the formula give

X= 1402 +41•11-6.41 =1464 lines. | 404. Experiment determined that the pier ought not to be less than 16 lines, and the geometrical method made it 20.05.

The foilowing table shows the experiments on six different models.

769:5 120

PSR

Fig. 576

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