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When x=0, or the load is placed on the crown of the arch, Y_(ta’+a) vers. Y—2(Ja’ + $a") – (ta'+a)Y..... (162).

tan. -1 cot. When - ( tan. -^ cot. Y ) = 0, becomes infinite; an infinite load is therefore required to give that value to the angle of rupture which is determined by this equation. Solved in respect to ta

., it gives,

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y (7) + V ()*+(2+)

Co... (463).

2+1 No loading placed upon the arch can cause the angle of rupture to exceed that determined by this equation.

THE LINE OF RESISTANCE IN A CIRCULAR ARCH WHOSE

VOUSSOIRS ARE EQUAL, AND WHOSE LOAD IS DISTRIBUTED OVER DIFFERENT POINTS OF ITS EXTRADOS.

............

338. Let it be supposed that the pressure of the load is

wholly vertical, and such that any portion FT of the extrados sustains the weight of a mass GFTV imme. diately superincumbent to it, and bounded by the straight line GV inclined to the horizon at the an. gler; let, moreover, the weight of each cubical unit of the load be equal to that of the same unit of the material of the arch, multiplied by the constant factor u; then, representing AD by R3, ACF bro, ACT by 0, and DZ by 2, we have,

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A SEGMENTAL ARCH WHOSE EXTRADOS IS HORIZONTAL.

339. As the simplest case, let us first I suppose DV horizontal, the material of

A the loading similar to that of the arch, La and the crown of the arch at A, so

that i=0, u =1, and O=0. Substituting the values of Y and Yx (equations 464, 465) which result from these suppositions, in equation (455), solving that equation in respect to -, and re

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ducing, we have,

=

* See Note 2, at end of Part IV.-ED.

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In the case in which the line of resistance passes through the bottom of the key-stone, so that a=0, equation (466) becomes

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A GOTHIC ARCH, THE EXTRADOS OF EACH SEMI-ARCH BEING

A STRAIGHT LINE INCLINED AT ANY GIVEN ANGLE TO THE
HORIZON, AND THE MATERIAL OF THE LOADING DIFFERENT
FROM THAT OF THE ARCH.

310. Proceeding in respect to this general case of the stability of the circular arch, by precisely the same steps as in the preceding simpler case, we obtain from equation (455),

Y Y
P (fa'ta’ta) (cos.0-cos.*)-(ha+a)(4-0)sin. Yt - sin. Y

{cos. ¥-(1+2) cos. 0}

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... (470)

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the preceding equations by assuming a series of values of ¥, and calculating the corresponding values of 3 for each giren value of a, l, ll, o. The tabulated results of such a series of calculations would show the values of y corresponding to given values of a, b, c, M, . These values of y being substituted in equation (470), the corresponding values of the horizontal thrust would be determined, and thence the polar equation to the line of resistance (equation 454).

A CIRCULAR ARCH HAVING EQUAL VOUSSOIRS AND SUSTAINING

THE PRESSURE OF WATER.

341. Let us next take a case of oblique pressure on the

extrados, and let us suppose it to be

the pressure of water, whose surface B stands at a height BR above the sum

mit of the key-stone. The pressure of this water being perpendicular to the extrados will everywhere have its direction through the centre C, so that its

motion about that point will vanish, in and Y-Xy=0; moreover, by the

principles of hydrostatics,* the vertical component Y of the pressure of the water, superincumbent to the portion AT of the extrados, will equal the weight of that mass of water, and will be represented by the formula (464), if we assume i=0. The horizontal component X+ of the pressure of this mass of water is represented by the formula

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* See Hydrostatics and Hydrodynamics, p. 30, 31. + See Note 3, end of Part IV.-ED.

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