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the forces N, P, and R, the force R disappears from the equation, and we have

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b and h being the lever arms of the forces P and N in reference to the centre of pressure O.

We see therefore that when N and its point of application O are known, we can find the centre of pressure O' of the section KK'; or if 0 and O' are known, we can find N. But for a given position of O and O' or a given value of N, all the other parts of the curve of pressure will be determined. For in respect to a joint KK, to which corresponds a weight P' of arch and surcharge, we know N and the position of P', and therefore we can find the magnitude and position of R', the resultant of N and P', by the equations

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Consequently we can find the centre of pressure O" of the joint K1K1', just as we found Oʻ the centre of pressure of the joint KK'.

By taking different positions of O and O', we shall be able to trace a series of curves of pressure of the arch. If the centre of pressure of such joint for any one of these curves of pressure be at a distance from the edge of the joint which is consistent with the stability of the joint according to Art. 75, then the conditions of stability of the arch will be satisfied.

If several of the curves of pressure satisfy the conditions of stability of the arch, then the assumed dimensions of the arch are more than sufficient to resist the pressure to which it will be subjected. It will be necessary in such case to recommence the tracing of other curves of pressure, from a new hypothesis with respect to the positions of O and O. The instability of the structure would be indicated by some of the centres of pressure falling near the edges of the joints as in Art. 75.

117. On this subject the following extracts from Captain Woodbury's able Treatise on the Arch, are interesting.

Let us suppose that the pier or abutment of an arch is abso

lutely immovable, and that the material of the arch is susceptible of but very little compression. Then as the arch below the weakest joint at the reins is supposed to be immovable, being firmly attached to the pier or abutment, and the masonry above this joint is by hypothesis nearly incompressible, it is hence clear that the pressure at the key and at the joint of fracture will be distributed along these joints when the arch is in equilibrium. In the most perfect condition of stability, the resultant of the horizontal forces along the vertical joint at the key will pass through the middle of that joint. In like manner the resultant of all the normal forces acting along the weakest joint will pass through the middle of that joint.

Let us now suppose the thickness of the pier to be gradually diminished until its top begins to move away from the arch. The crown of the arch will now begin to settle, the reins will spread out, the curve of pressure, moreover, will approach the extrados at the key, and the intrados at the reins. Finally, when the pier has been sufficiently reduced, the curve of pressure will pass through the extremities of the specified joints at the key and at the reins.

Thus we see that, by mere external changes in the pier, the curve of pressure passes by degrees from the position of most perfect stability to that of instability, and consequent fracture of the arch.

In this final condition, the thrust of the arch is the horizontal force which, applied at the extrados of the key, is just sufficient to prevent the rotation of the segment above the joint of fracture, round the intrados of that joint, this force acting with a lever arm equal to the vertical distance between the extrados at the key, and the intrados at the joint of fracture. In the condition, however, of a perfect stability, the acting thrust is the horizontal force which, applied at the middle of the key, is just sufficient to prevent the rotation of the upper segment, round the middle of the joint at the reins, this force acting with a lever arm equal to the vertical distance between the middle points of these joints.

118. PROP. To find the condition of the stability of any arch with horizontal surcharges not symmetrical.

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In this case the reaction N (Art. 118), being no longer perpendicular to the vertical line HH', drawn through the crown of the arch, is resolved into two components, the one N" (Fig. 57) parallel, and the other N' perpendicular, to the vertical line HH'. Then taking three points O, O', O", on three joints of the arch, it is required to determine the curve of pressure of which O, O', O', form a part, O being the point of application of the oblique reaction N between the parts of the arch and surcharge on each side of the crown.

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In order to simplify the investigation, we will suppose O' and O' to be symmetrical, each at a distance d from the vertical line passing through O, and at a distance h from the horizontal line through O.

Then if P and P' be the weights of the parts of the arch and surcharge to the left and right of the crown, and b, b′ the distances of O' and O" from the directions of P and P', we have first, by taking moments about O', and then about O" the following equations:

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We may now construct the resultant N of these forces, and

the method is the same as in Art. 118. If the curve which is traced does not fulfil the necessary conditions of resistance, we make other hypotheses with the points O, O', O', just as in Art. 118.

119. The calculation of the weight P, both in magnitude and position, may be effected very readily by means of Thomas Simpson's Formula.*

* Simpson's Rule is proved in the following manner :— Let Aabc C be a portion of a curvilinear area bounded by the curve abc and the straight lines Aa, AC, and Cc. Then, as a parabola whose equation is of the form

b

D

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can be made to pass through any number of points, by giving to A, B, C, &c., suitable values, it is assumed that the points a, b, c, lie in a parabola, of which bB is a diameter. Consequently

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parallelogram a E.

Add

AacC to the second member of the right-hand side of this equation, and deduct it from the first member of the same; then we have

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Let Aa, Bb, Cc, be equidistant ordinates of the curve abc. Put a = a, Bb b, Cc = c, and AB = BC 8. Then remembering that the sum of the ordinates Aa and Cc is double of Bb, since AC is bisected in B, we have by (1),

Area of curvilinear figure Aabc C = 38 (a + c) + 38 × 26

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If we take c with two other equidistant ordinates d and e, we get, in like manner,

Curvilinear area = (c + 4 d + e) d.

Wherefore, by addition, the area of the curvilinear figure of which a and e are the extreme ordinates, and b, c, d, intermediate ordinates, is

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For this purpose the line IC =

= 1 (fig. 56) is divided into

a number of equal parts n, and the lengths of ordinates parallel to CHO, and intercepted between C'T' and H'K'K, are measured.

Let yo, y, be the extreme ordinates;

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= Sy dx = { yo+ 3 + 4 (Y1+ya+.. Yms)+2(Ya+Ya+.... Yms)} . . (1).

= Sydx=3n

If w be the weight of a unit of the stone, then

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The weight P, moreover, acts at a distance X from the line CHO, given by the equation

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yo being situated on the line CHO and y„ = KI' = y.

Now the part within the brackets is equal to the sum of the extreme ordinates a and e, twice the odd ordinate c, and four times the even ordinates b and d. The same result is obtained with any number of equidistant ordinates. Hence Simpson's Rule, which is enunciated thus :—

Having measured a number of equidistant ordinates of a curvilinear figure, to the sum of the extreme ordinates, add four times the sum of the even ordinates, and twice the sum of the odd ones; multiply the sum of all these by onethird of the distance between two ordinates, and the product will be the area of the curvilinear figure.

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