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When x=0, or the load is placed on the crown of the

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)=

Y

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tan. -λ cot. Y=0, becomes infinite;

2

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 tan. it gives,

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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

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but TV=MZ-(MT+VZ), and MZ= CD=R+R6, MT= R cos. 0, VZ=DZ tan. = R sin. 0 tan. . Therefore MT+ VZ= R cos. + R sin. 0 tan. = R {cos. O cos. +sin. O sin. } sec. = R cos. (0-1) sec. 1;

0

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.. Y=weight of mass GFTV=R {1+ß-sec. cos.(0—1)}

cos. 0d0 =μR3 { (1+6) (sin. 0—sin. ☺)—† sec. {sin. (20—6) — sin. (2 0 −1)} — † (0 — ~) } . . . (464).*

Y = moment of GFTV=R {(1+6)-sec. cos. (0— 1}

sin. 0 cos. 0d0=R' {(1+B) (cos. '-cos. '0)-(cos. ' cos. '0)-tan. (sin. '0-sin. 'O)} . . . 465).*

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(1-a) (1+a)(1+3) sin.+(1+a) (1-2a) cos.'+(ta2+}a3-1)cos.sin.++

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0 (see note, page 438.), and λ = a, and

1

(1+ α)2

(1-2a) cos. { (1 − a) (1 + 6) + (1+a) (1—2a)} cos. * +

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

Р

:= {(1 + a)3(1 + B)(1 − a) (1+cos. ¥)—!(1+a)' (1—2a)

(1+cos. Y) cos. —cot.

༡༠

+}=0.... (465);

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(1+a)'(1-2a) cos. '+(1+a)2 {(1—a)+(4—5a)} cos. +

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− {(1 + a)3 (1—a)(1+ß)+z+a2(1+ja)} = 0 . . . . (469.)

...

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.

340. 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),

P

Yx Y

(ta+a+a) (cos.e-cos.)-(a+a) (-)sin. + sin.

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in which equation the values of Y and Y are those determined by substituting for in equations (464) and (465).

dᏢ Differentiating it in respect to Y, assuming0 (note, p. 438.), and λ=a, we obtain

(a+‡ƒɑ2 —ƒƒœ3—fa') cos. ☺ sin. ¥—(a2+a) sin. ¥ cos. Y—

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termined by equations (464) and (465) the following equa tion will be obtained after a laborious reduction: it deter mines the value of Y:

A+B cos. Y-C cos.' Y+D cos.' Y+E sin. Y

F sin. Y cos. Y-G sin.' Y-H cot. Y+

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A=4(1+a)3 { ŝ(1+a) tan ɩ sin2 © —(1 +§) }2−(1+a) cos* @}

—¡(1+a) cos.' © } +(2a+a’—a’—ĝa′) cos. ©

B=(1+a)3 {2μ(1—a2) (1+B) cos. ℗—(1—μ)} +1.
C=μ(1+a)2 {(1—a) (1+ß)+(1+a) (1—2a) cos. ☺}.
D=(1+a)(1-2a).

E=(1+a)(1-2a) tan. =D tan. .

F=μ(1+a) (1—2a) tan. cos. =E(1+a) cos. ☺.
G=(1+a)(1-2a) tan. =D tan. .

H=u(1+a)3 {2(1+B)—sec. ɩ cos. (~—¿)} sin. 2☺.

I=1—(1—μ) (1+a)2.

K=(1+a) cos. .

L=μ(1+a)3 {2(1+6)—sec. cos. (-)}sin. .

Tables might readily be constructed from this or any of

the preceding equations by assuming a series of values of Y, and calculating the corresponding values of B for each given value of a,,, . The tabulated results of such a series of calculations would show the values of corresponding to given values of a, B, 1,, . These values of 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

T

THE PRESSURE OF WATER.

V

D

A

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 stands at a height SR above the summit 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, and Yx-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 t=0. The horizontal component X+ of the pressure of this mass of water is represented by the formula

X=PR' {1+6—cos. } sin. êdê=μ(1+a)",3 {(1+6) (cos. © —

....

(473).

cos. ) — (cos. '—cos. ')} Assuming then =0, we have (equation 464), in respect to that portion of the extrados which lies between the crown and the points of rupture,

Y

=μ(1+a)2 {(1+6) sin. — sin. 2 -JY},

X

and (equation 473) =" (1+a)" {(1+3) vers. ¥—} sin. *},

*See Hydrostatics and Hydrodynamics, p. 30, 31. t See Note 3, end of PART IV.—ED.

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