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2. That the vertical voussoir CD, which represents the keystone, having also its joints sensibly 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, fT, 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 4 K.

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 inverse ratio; so that the voussoir HN, which is in the middle, has its vertical and horizontal forces 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 G 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 dif ference 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.

11. If, 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 flat 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 vertex, 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.

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.

Fig. 572.

Method.

F M

M

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1391. Having described the mean circumference GXT (figs. 573, 574.), from the point. 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.

R

m

E

Fig. 573.

P SR
Fig. 574

(See 1598, et seq.)

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 acting in opposite directions will partly destroy each other; thus transferring IK from K to m, the difference mL multiplied by the thickness of the vault will be the expression of the thrust. This force acting at the point K in the 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 expressed by mL x AB × PH.

This will be resisted —

1. By its weight represented by the surface EP × PR multiplied by the arm of the lever PS, determined by a vertical let fall from the centre of gravity Q, which gives for the resistance of the pier the expression EP × PRx PS.

2. By the sum of the vertical efforts of the upper part of each arch, represented by MK × 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 AB acting on the point T has for the arm of its lever TE. Hence, if equilibrium exist,

x

mL x AB × PH PEx PRx PS+ MK × AB × KH + IT × AB × TE. But as in this equation neither PR (= BE) nor PS nor KH not TE is known, we must resort to an algebraic equation for greater convenience, in which

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Thus the first equation becomes pu + pd= + m (c × x) + n (x − e),

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Transferring the unknown quantities to the second side of the equation, we shall have + mx + nx + = pa + pd + ne — mc.

ar2
2

Multiply all the terms by 2, and divide by a, in order to get rid of r2, and we
have x2 +
2(m+n)<
pd+2nc-2mc
=2p+

a

a

;

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a2 a

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

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

ML x AB (p in the formula) representing the thrust or 8·127 x 3
2p

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2pd=48.762 x 13.876

40.333

13.876

24.381

48.762

676-621

73.282

124.824

58.875

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2 MKAB KH represented by 2mc (=5749 x 3 x 4.249)
2ne, which is ITx AB AB ( = 13·876 × 8 × 3)
b=m+n=(MK + IT) × AB (=19·625 × 3)

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Substituting these values in the formula x =√2p+

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676 621+124-824-73-282 we have r=48-762 + which gives x=5·8, or 5 inches 9 lines for the thickness of the piers to resist the thrust of the arch, 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 3 lines.

40.333

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+124 828-86-458 and changes the equation to x=√56·762+ +2.430-1.55; from which we should obtain x = 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 Experiment.

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.

1997. Having drawn the lines heretofore described, we shall find mL× AB expressed in the formula by

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Hence 2pd

2ne, which is twice the vertical effort of the lower part of
the arch, multiplied by AB, will be 456 x 21 x 21,
which gives

2me, which indicated twice the vertical effort of the upper
part, multiplied by iK, will be 18.9 x 21 x 2 x 8.4, which
gives

a, which represents the height of the piers, being 195, and
b=m+n=64·5 × 21 =1354 5,

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And all these values being substituted in the formula, will give

560-70 1121-40

45.60

5113.584

20109.60

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6667.92

6.94

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1398. Let the mean curve TKG of the arch (whatever its form) be traced as in figs. 573, 574., the secant FO perpendicularly to the curve of the arch, and through the point K, where the secant cuts the mean curve, having drawn the horizontal line IKL, and raised from the point B a vertical line meeting the horizontal IKL in the point i, make Km equal to iK, and set the part mL from B to h, and then the double thickness of the arch from B to n. Let hn be divided into two equal parts at the point d, from which as a centre with a radius equal to half hn, describe the semi-circumference of a circle which will cut in E the horizontal line BA prolonged. The part BE will indicate the thickness to be given to the piers of the arches to enable them to resist the thrust.

1399. The truth of the method above given depends upon the graphic solution of the following problem: To find the side BE of a square which shall be equal to a given surface mL x 2e; an expression which is equivalent to 2p, and we have already seen that r = √2p was a limit near enough; hence we may conclude that the thickness BE obtained by the geometrical method will be sufficiently near in all cases.

Experiments on surmounted Arches.

1400. The interior curve of fig. 574. is that of a semi-ellipsis 81 lines high; it is divided into four parts by an upright joint in the crown and two others towards the middle of the haunches determined by the secant FO, perpendicular to the interior part of the curve. Having traced the mean circumference GKT, the horizontal IKL, and the vertical Bi, we shall find

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The effect of the thrust indicated by KL-iK=mL will be
199, which gives for the expression p of the formula

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c, that is, i K, being 174 lines, we have 2me=171 x 174 × 2, which
gives

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175.5

351 0 23341.5

171 0

5899.50

The height of the piers a

120.00

b, which expresses the sum of the vertical efforts m+n, will be
equal to MK + IT × AB or 19 +66 × 9, which gives

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fines, or a little more than 16 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 18 lines, as it is also by the geonetrical 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 Hr

of the formula to be

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The height of the pier, represented by a, being 120, we have

= 8280
2pd-2mc

a

=

252

15876

=

207

Fig. 575.

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= al 41 60. Substituting

x= √252 +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.

Application to a surmounted Catenarean Arch.

1402. The lines are all as in the preceding examples (fig. 576.). The whole arch acts on the pier in the direction FT, which is resolved into the two forces Tf and Tm, and the formula, as before, is

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thus having found Bm=22, we have the value of p=22} x9=201; and 2p=402.

1403. This model was of the same dimensions as the preceding: b, which represents Tfx AB, will be 769-5; will be 6-41, and 769-5 4111. These values substituted in the formula give

b

1:20

x= √402+41·11-6-41-1464 lines.

1404. Experiment determined that the pier ought not to be less than 16 lines, and the geometrical method made it 20.05. The following table shows the experiments on six different models.

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