Y When -(tan.xcot. )=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 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. E D A B 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 immediately superincumbent to it, and bounded by the straight line GV inclined to the horizon at the angle; 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 μ; then, representing AD by Rß, ACF by ✪, ACT by 0, and DZ by z, we have, but TV=MZ-(MT+VZ), and MZ=CD=R+Rß, MT= Rcos. 4, VZ-DZ tan..=R sin. tan... Therefore MT+ VZ= R cos. + R sin. @ tan. =R {cos. cos. + sin. ◊ sin. } sec. = R cos. (0-1)sec.; 0 ... TV=R{1+ß—cos. (9—1) sec. }; also, z=DZ-R sin. ; dz ... area GFTV=STV &de=R3s {1+ß—cos. (8—1) sec. 1} cos. {d}; Rf{ . Y=weight of mass GFTV=μ R (1+3-sec. cos. (0-1)} cos. Od0= Ꮎ μR2 { (1+3) (sin. 0—sin. 0) — ‡ sec. ¿{ sin. (2 0—1)—sin.(20—1)} −1(0—0)} · · (464). 0 Yx=momentum of GFTV=μR3s{(1+3) -sec. cos. (0-1)} sin. 0 cos. fd= μR3(1+3) (cos. 20-cos. 20)-(cos. 30-cos. 30)-tan. (sin. 30-sin.30)}.. (465). P _ {( 1 − a ) ( 1 + œ )2 (1+6) sin. ?¥+} (1+œ)2 (1−2œ) cos. 3¥+(}æ2+}æ3 −}) cos. ¥—] Y sin. Y+} . (466). 0 (see note, page 471.), and λ = a, and 3 (1−2×) cos. 3¥— {(1 − a) (1+3)+(1+a)(1-2)} cos.+{+2 (1-a2) (1+)} cos. ¥ 1 + (1+a)3 {1-(1+a) cos. Y n sin. -(1-a) (1+3)-3(1+a)? ..(467). In the case in which the line of resistance passes through Р the bottom of the key-stone, so that λ=0, equation (466) becomes ́2 = } ( 1 +∞)2 (1 +8)(1 −∞) (1+cos. ¥)−}(1+œ)2 (1−2œ)(1+cos. ¥)cos. YIY cot. JY+}=0...(468); (1+a) (1-2a) cos. 'v+(1+a)2 {(1−a)ẞ+}(4—5a)}cos. + -{(1+a)2(1—a)(1+ß)+z+a2(1+a)}=0.... (469). Р A GOTHIC ARCH, THE EXTRADOS OF EACH SEMI-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), Yx ̧(4a3+a2+a) (cos. O—cos. ¥)—(§a2+a) (¥—0) sin. ¥+ {cos.―(1+\) cos. O} in which equation the values of Y and Ya are those determined by substituting for in equations (464) and (465). Differentiating it in respect to Y, assuming dP =0 (note, p. 471.), and λ=α, we obtain 'a+fx2-fæ3— fœ1) cos. ℗ sin. ¥−(fœ2+a) sin. ¥ cos. Y − (fa2+a) {1−(1+a) cos. O cos. Y}(Y-O) Y Yr ra (-{1-(1+α) cos. Y cos. ☺}+- sin. ¥+ {cos. ¥−(1+x)cos. } { d¥ 1d(Yr) sin. Y dY termined by equations (464) and (465), the following equation will be obtained after a laborious reduction: it determines the value of ¥: A+B cos. -C cos. 2y+D cos. 3y+ Esin.v-Fsin. Y cos. Y-G sin. -H cot. Y where A=μ(1+a)2{}(1+a) tan.. sin.3©−(1 +ß) {2—(1+a)cos.2O} −}(1+a) cos.3 ©} +(2a + a2— a3 — §a1) cos. ©. B=(1+a)2 {2μ(1—a2) (1+ß) cos. − − (1 −μ)} +1. E=μ(1+a)2(1-2a) tan. =3D tan... F=μ(1+a)3(1-2a) tan. cos. =E(1+a) cos. . G=}μ(1+a)2(1—2a) tan. =D tan.. H= ¦ μ(1+a)3 {2(1+ß)—sec. ¡ cos. (~—¡)} sin. 20. L=μ(1+a)2 {2(1+ẞ)— sec. cos. (-)} sin. . Tables might readily be constructed from this or any of the preceding equations by assuming a series of values of ¥, and calculating the corresponding values of ẞ for each given value of a,1, μ, O. The tabulated results of such a series of μ, calculations would show the values of corresponding to given values of a, ß, 1, μ, ☺. 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 C 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 ẞR above the summit of the key-stone. The pressure of this water being perpendicular to the extrados will every where have its direc tion through the centre C, so that its momert about that point will vanish, and Yx-Xy=0; more over, 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 = =0. The horizontal component X of the pressure of this mass of water is represented by the formula X=R2 Y T2 cƒ{1+3=cos. 0} sin. &d=μu (1+m)?,? {(1+8) (cos. ©-cos. 6) - † (cos, 20-cos. 20)} . . (473). 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 „2=μ(1+a)2 {(1 + ß) sin V—‡sin.2V —↓V}, X and (equation 473) =μ(1+a)2 ((1+ẞ) vers. Y— sin. 2Y}, 2 sin.Y cos. Y=μ(1+a)2 {(1+6)vers.Y-Ysin...... (474) Substituting this value in equation (455), making Yx-Xy=0, solving that equation in respect to and making=1+^, we have r Pa-(1+a) sin. — {a+a2+}a3—μ(1+a) (1+3)} vers. A+vers. Y (475). If, instead of supposing the pressure of the water to be borne by the extrados, we suppose it to take effect upon the intrados, tending to blow up the arch, and if ß represent the height of the water above the crown of the intrados, we shall obtain precisely the same expressions for X and Y as before, except that must be substituted for (1+a)r, and X and Y Y must be taken negatively; in this case, therefore, sin.T 2.2 * See Hydrostatics and Hydrodynamics, p. 30, 31. |