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THE CIVIL ENGINEER AND ARCHITECT’S JOURNAL. 84 [March 1, 186S. or, 3- Thence we find N = P which is the horizontal thrust at E. H’ I now proceed as before to calculate N for various values of 0 from the above formula), in order to find the point at which N is greatest, and which will therefore be the weakest part of the dome. And in order to compare the thrust in the parabolic dome with that obtained for the spherical, I take the portion of the parabola in which the height, D Z, equals the half span B Z. Let the span be 20 ft, then BZ = 10ft.=DZ, and m=2-5 ft., 4m=10 ft.; and let X=1 ft. Substituting these values in the foregoing expressions, and calculating N for various values of 0, •43302 R-H24222r 5=H+‘1908r, a=t+’01837r F . c=-0011038 (BA—r 3 ) (t+s)—’000407 (R 4 —r 4 ). Combining these expressions with the value of Q . 2 previously found, we can form the Equation of Stability for any given values of R and r, and thence obtain the value of t. To compare the thrust of this form of dome with that of the spherical of equal span, let the span (2s) be 20 ft.; then R=21 - 5, r=20-5 ; which being substituted in the formulae, we obtain N = 77-27 lbs. I now apply the system of investigation to the dome whose inner surface is a paraboloid of revolution, the thickness of the shell being uniform throughout. Let Fig. 5 represent the para bolic section of one half of the dome ; E F any joint in the masonry which is in the normal or radius of curvature, and let R be the centre of curvature belonging to point E. If we call From the formula (C) we have the value of the greatest thrust N=80-418 lbs., which I found in the spherical dome to be 92 lbs. To obtain the requisite thickness («) of the pier, we have 2N3 = 8661’32 P . a=415-99<+189-65 F . c=180-73t—61-42 Q. 2=43-63275 (< 3 +30J 2 ) Hence the Equation of Stability is 8661’32=43-63275 (P+30t 2 )+596-72<+128-23 4 vi 3—- < 3 +30< 2 f-13'7«—195-56=0, from which we obtain <=2-259 ft. In the spherical dome the requisite thickness of the pier was sound to be <=2-45 ft. As we have fonnd that the weakest part of the dome is at a point 13|° above the springing line, it follows that this is the position in which an iron belt must be placed to produce the greatest effect in counteracting the thrust of the dome. If this is done, the thickness of the pier may be considerably diminished, and need not greatly exceed the strength necessary for support ing the superincumbent weight acting vertically downwards. It is evident, from the nature of the investigation, that the position of the point E, at which N is greatest, will vary according to the “ pitch ” of the section, that is, according to the value given to the angle a; for the smaller we make a, the more nearly does the dome approach the spherical form. I will now apply the foregoing formula) to the case of a dome in which the angle a is 10°. Substituting this value in the expressions for P . a: and y, and calculating N = P- for several values as 0, I find that the greatest value of N is obtained when 0 is 63°, or a-j-0 is 73°; so that the weakest point of this dome is at the joint which makes an angle of 17° with the springing line, E0B = 17°. If we substitute this value for 0 in the various formula;, we find P = -0058358 (R 3 —r 3 ), a=<+ - O437r -1)04567 (RS—r 3 ))-—-002514 (R 4 —r 4 ) JS_6 -4924 R-p-2r 6 = H+-29237r F . c = -0028033 (R 3 —r 3 ) (<+s)—-0008868 (R 4 —r‘) from which the Equation of Stability can be formed when R and r are given. To compare the greatest thrust in this dome with that of the former domes of equal span, I have calculated N when the span (2s) is 20 ft., thickness of dome 1 ft. Here we have R = 13-2 ft., r=12-2 ft., then N = 88‘7 lbs. If the angle a is greater than that assumed in the first example, the position of the weakest joint will be lower down in the arch, and nearer the springing. Let a be 30°, then by calculating the thrust N=P- for several values of 0, when R = ll and r=10, y as before, I find that the maximum value of N is obtained when 0 is 49° ; and «+0 being 79° in this case, the weakest joint is situated 11° above the springing. Hence the formula) for this dome become =— p m 2 X- 4" + 2 ™ 3 X A —™ X f™ 3 — 4- 4 5 c 5 c 4 V 3 / c 2 \ 4 f X 3 \ 9 39 -A sc—X 2 f 2m 2 — A. ) d+ | m 3 x-2 tan’ 0 + A X 3 tan s y -|- 4 m 3 X 3 tan 0—m X (m 3 —X 3 ) log . cos (~ — 0. The denominator of z is 0 f d0 o A 07? X X 3 =| m 3 X (sec’ 0—1)+ -A-t a n 3 ^+“3 ( 1— cos #)_log. sec 0 We have now to find the value of the moments of N and P about E, and we will suppose N to act at n half way between C and D ; E G=x=2m . tan 0—z, NE= y =^=g + ^=m.tan 3 0 + ^, n 9 X ® S 4m 3 X-s +2mX 2 +- 4- — s—4m 3 X— B —m X 2 tan 3 0 ) c 6 c 3 3 c“ t P =-002888 (R 3 —r 3 ) N==g -001387 (R 3 —r 3 )r—-000777 (R 4 —;- 4 ) KE=y', KD=+, C D=X, and m the distance of the focus from D : then by the property of the parabola, we have y' 3 =4m. P. Let E R=p (the radius of curvature) making the angle 0 with the vertical, then Since the subnormal O K=2m, therefore E0=2m . sec 0. Also y=2m . tan 0. Therefore we have p‘= 4(m+m tan 2 0) 3 , or p=2m . sec 3 0. Let P be the weight of the solid wedge cut out of the dome from E F to CD by two planes making at the axis the angle </> with each other; and let ef be an elementary solid portion, whose weight is dP ; then, calling Re=r, we have dP = 6 . Ro . d0 . dr . Or sin 0 . df OR=p—OE=p—2 m . sec 0 = 2m (sec 3 0—sec 0) = 2m ---4w- ' ‘ cos 3 0 Oc = Re—0R = r--2m cos 3 0 Oe . sin 0 = r . sin 0—2m . tan 3 0 . •. P=8 Jffr (r . sin 0—2m . tan 3 0) dr . d0 . d</>. Let a be the centre of gravity of the wedge whose section is EF CD ; G its projection on the line K E ; then, if we call KG we have fffr ( r • 8444 —~ m - tan’0) 2 cos f . dr . d0 . df> fffr (r sin 0—2m . tan’ 0) dr . d0 . df> sin -!</) ffr (r . sin 0—2m . tan 3 0)‘ dr . d0 — 2</> A 4 ' ( r • 8444 — 2m ’ tan - 3 c ^ r ■ And since </> is a small angle, we can put ^^^ = 1. The limits of integration are r=2m . sec 3 0, and r=2m sec 3 +X, so that the numerator of z becomes (putting s for sin 0 and c for cos 0) 0 f d0 ( . 3 . s 2 . . . „ s 2 . „ s 2 X 4 „ „ s 4 -o j 8)+ X —a +6m 3 X- +2m X 3 -r +~r s 2 —16m° X -a- ( c 9 c b c 3 1 4 c 9 — —8m 2 x 3 _ 3 m \s A. +8m 3 x A g + 2m 2 X 2 tan® 0 j