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At the point of rupture the line of resistance meets the intrados: therefore at this point

Also, generally, sufficient dimensions of the arch being supposed, the line of resistance touches the intrados at this point;

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Assuming Y to be the value of 9 at the point of rupture, and substituting it for 6 in the five preceding equations, we may eliminate between them the four quantities p, R, r, p, or the four p, R, r, Y. There will result an equation involving the quantities P and Y in the one case, and P and p in the other.

Now if the force P be supplied by the pressure of another opposite

and equal semi-arch, it has been shewn, (see Memoir on the Theory of the Arch, Vol. V. Part III.) that if the masonry be supposed perfect, P is a minimum in respect to the variable p ; moreover if the masonry

be supposed to possess those yielding properties which obtain in practice,

and which shew themselves in the settlement of the arch, then p = R:

according to either of these conditions, P and Y may therefore be de


The values of P and p becoming thus known, they may be substituted in equation (42), and the equation to the line of resistance will thus be completely determined.

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Let now the force P be supplied by the opposite pressure of an equal semi-arch, then on the hypothesis made, P is a minimum function of Y.

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From which equation Y may be determined. Also by equation (50),

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Y, P. p are thus completely determined, and all the circumstances of the equilibrium of the circular arch, thus loaded, are known.

If there be no loading, and the two semi-arches be parts of the same continuous cylindrical mass, X = 0, and 9 = 0.

Therefore, by equation (52), Y = 0. In this case, therefore, the point of rupture is in the crown of the arch (Fig. 18), at the intrados, also by equation (53), P = r"; a –3 a”}; therefore, by equation (54), p = r. Substituting these values of P and p in equation (46), we obtain for the equation to the line of resistance in the unloaded circular arch,

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Let }, be the angular distance from the crown, at which the line of resistance meets the ea trados, (Fig. 17), as Y is that at which it meets the intrados. Therefore, by the preceding equation,

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\! determined from this equation will measure the greatest semi-arch, which being unloaded, can be made to stand.

To determine the inclination (b of the resultant P, to the vertical, corresponding to the angle 6, we have

P. sin q = horizontal force on segment = P = r^{a-4a'},
P. cos b = vertical ........................ = mass of segment = }r”;a" +2a39;

- _2}1 – #a"; ... tan q = {a + 2}6 .........(57). Suppose the arch to be supported upon upright piers of a given

breadth a, (Fig. 18), and let it be required to determine what is the greatest height (z) to which they can be carried.

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It is evident, from equation (50), that as X is increased Y increases; that is, the points of rupture descend continually upon the arch as it is more loaded. The experiments of Professor Robison on chalk models are explained by this fact”,

* Having constructed chalk models of the voussoirs of a circular arch, and put them together, he loaded the arch upon its crown, increasing the load until it fell. The first tendency in the chalk to crush was observed at points of the intrados, equidistant from the crown on either side, but near it; these points were manifestly those where the line of resistance first touched the intrados. As the load was increased, the tendency to crush exhibited itself continually at points more distant from the crown—that is, the points of rupture descended,

Vol. VI. PART III. 3 Q

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