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86 THE CIVIL ENGINEER AND ARCHITECT’S JOURNAL. [March 1, 1368. M H=M P-J-a-iq- The origin in each curve is a point of “ con trary flexure.” Proceeding as in the elliptic dome, I suppose the joint E F of the masonry to be horizontal, and in this case I take N as acting at D, and take moments of N and P about F, on the outer con tour. In order to find the centre of gravity of the wedge cut out of the dome by two planes which make the angle </> at the central axis O C, and lying between E F and C D, I use, as before, the formula 7 P-rh p—p p Z - Pj-Po ’ 1 2 Z being the distance from O C of the centre of gravity of the wedge. The domical shell being the difference of the two solids of revolution made by the two curves revolving about O C, K F is the radius described by any point F ; KF=KL+LF=r 1 +y, Then P 1 =|d . <// lrf-y) 2 dx = 8 -^ nV { 1+2 sin (J) + sin 2 (f ) J d 0) Tnlririrt Plizx li-rvai4-« -fnrvm 4-zi — ■ —1_ WB 11HVP. Pi^^/CA+yi) 3 = S _L^ ri V j lj-3 sin (*->)+3 ein 2 Q) + siu 3 (^) } =’—+• 1 le+IH “(?.) ( 8+3 “ C:>) 4 ! So also We can now find z for any given values of r t and r 2 . In order to find the point on the dome where N is greatest, I take the example in which r,=ll ft., r 2 =10 ft., the span of the dome being 4r 2 or 40 ft., and the thickness AB 2 ft. Calling x and y the lever arms of P and N about E, we have x=KE—z=r.,(l + sin(® 2 ))—a 2/=DK=I?rr 2 +a: 2 , N =P^. Calculating N in this example for various values of x, and x 2 , I find that N is greatest when E F is 2j ft. above A B, or when “'=1-34355, -2=1-3208, so that the above expressions become ’•i n P x —P 2 =S . </> (1-9057 rf—l-8614r 2 3 ) P^—P^ 2 =8 . </> (2-0198r x 4 —l‘96497r 2 4 ) „_2-0198r 1 4 —l’96497r 2 4 ~ l-9057r 1 3 —D8614r 2 3 a:=l-9689r 2 —.z, y=2-8916r 3 !’=?!—P 2 =S (-06651r x 3 —-06536r 2 3 ) N=P?=8 (•06651r 1 3 --06536r$ 1 To compare the thrust of a dome of this contour with that of the domes previously considered, let the span be 20 ft., A B = 1 ft. ; then r,=5-5 ft., r 2 =5 ft., and the greatest value of N is found from the foregoing formulae to be N=62.5. The Equation of Stability, which in this dome is 2N&=P . a _|_Q , 2+F . c, can now be easily formed, and the thickness of the pier calculated. The value of the moment about S of the lower part from E F to A B, which we call F . c, is found as before from F=F 1 —F 2 =8 (-015726T 3 —-017273r 2 3 ) c=2r 2 +t— •598105rJ—-6564r., 4 Z F x —F 2 = -45054r 1 3 —-49485r 2 3 a=tf- -03109r 2 , &=H+ - 25r 2 t 3 4-30( 2 +9-75t—146-23=0 or, t—2 very nearly. The following Table exhibits the principal results obtained from the investigations of the foregoing papers. All the domes, except the last, are supposed to be of the uniform thickness throughout of 1 ft. In the last, the thickness is 1 ft. at the springing, but gets rather less towards the top. The domes are supposed to be built of material weighing 125 lbs. to the cubic foot. The thrust is calculated for the 180th part of the whole dome, being the portion cut out by two planes which make an angle of 2° at the axis. Form of Dome. Span. Position of weakest Joint. Horizontal Thrust at weakest Joint. 1 Hemisphere 20 ft. Makes angle with spring ing line of 20° 92-01 lbs. 2 Gothic, value of angle being 10° Ditto Ditto ditto Ditto ditto 17° 88'7 „ 3 Ditto ditto Ditto ditto 22 V Ditto Ditto ditto Ditto ditto 13£° 80-418 „ 4 Ditto ditto Ditto ditto 30° Ditto Ditto ditto Ditto ditto 11° 77'27 „ 5 Parabolic,hsight from springing equals the half-span Ditto At springing line 79'6 „ 6 Elliptical, axis major vertical, ratio of major axis to minor axis as 6 : 5 Ditto One-third of length of semi-major axis above springing line 90-867 „ 7 Ogival, contour the “curve of sines”... Ditto One sixteenth of span above springing line 62'5 „ Hitherto I have assumed that the domes are of uniform thickness throughout; I will now consider the spherical dome in which the thickness at the crown is one half that at the springing. Fig. 8 is a section of a dome of this form, AFC being a quadrant of the circle whose radius, Z A or Z C, is R ; I ED being the quadrant of the circle whose radius, XI or X D, is r. Then A B=R—r, very nearly, C D=| (R—r). DIB will be rather more than a quadrant, its centre being at X. I shall here take the joint EF horizontal, as in discussing the elliptical dome, and suppose N to act at C, taking moments of N and P about E. Putting, as in the elliptical dome, P=P X —P 2 , and using the formula 7_Pr s 'l— Pi-V' P being the weight of the wedge EF C D, K G _r, the distance from the axis of its centre of gravity. If we call K F=y u ZK=a: 1 , KE=t/,, X K=x 2 , then we have x 2 ^x l —\ (R—r). P 1= &^ 2 . dx =®^/ E (R 2 -x2) dx =^ljR 2 (R-x 1 )-l(R 3 -V) | so also P 2 =L^ > | r 2 (r—as,)—1 (r 3 —x.p) Ppy^Lj?/ K (R2—x 2 y dx 3 x, =V 1 M- “- 1 MW®? I
