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38 THE CIVIL ENGINEER AND ARCHITECT’S JOURNAL. [February 1,1865— And if we substitute for 0 its value, R‘— r 3 s ,= 7351 b 3 ^- 2 (r+<) 2 —r 3 2 (3n+3r<+< 3 ) . “3 (r+<) 2 —r 2- 3 (2r+<) ’ i q = Sff=OS—Off=r-f-t—Off _3 (r+Q (2r+<)—2 (3r 3 +3rt+Q 3 (2r+<) _ 3rt+f "3 (2r+t)’ We therefore obtain for the value of Qg, 3r_|_£ Qg =-017453 8, (2r+<) t. H = •0058177 3 t . i‘ (3r+<) • H. And. c = OR—c 1 = <4-r—z 1 . Hence we find F. c=-00398 3 (R 3 —r 3 ) <+'00398 3 { r (R’-r 3 )—7351 (R*-r*} (6) We have now to find the moment about S of the weight of the pier. Let Q be the weight of the pier, 8 1 the weight of a cubic foot of its material; then Q=-M>. ( (r+«)’-n) • H = ^3 1 (2r+<) t. H = •017453 3, (2r+<) < • H. Let figure 2 represent the plan of the pier, g its centre of gravity, ST=< its thickness. Then by the integral calculus we find 0y=4(l±^x^; or, putting ^=1, 3 ( r 4“£)— r 2 i- ’ 2 weight of the portion of the “drum” on which the rib stands, and 7 = the distance from S of a vertical from its centre of gravity; we have then (in equilibrium) the equation N . 3=P . a+Q . g+F . c , (3) which I call the Equation for Equilibrium. From this equation we can find the thickness (<) of the pier necessary to produce equilibrium. In order, however, that we may have stability in the structure, we must multiply Nt by the coefficient of stability, which we may take as 2. The equation from which to find the value of t then becomes 2N . 3=P . a+Q . g+F . c, .... (4) which I will call the Equation of Stability. I now proceed to find the values of P . a, Q . q, and F . c. The value of P was previously found to be ■R3 J P=3 . </> (1—cos <9) =+=±; o and if we substitute for 0 and 6 their values, P =-007656 3 (R 3 —r 3 ). Also u=Se=<+r (1—sin #) = <+’O6O31 r; so that we find for the value of P . a, Pu=-007656 3 (R 3 —r 3 ) <+’0004618 3 . (R 3 —ri) r. . . (5) The value of F is found by means of the integral (A), taking the limits from 9=9 to 9= - ; which gives us F=3 . d>. cos 5-’—’ * 3 = •00398 3 (R 3 —r"). Let be the distance from O on OA of the perpendicular dropped from the centre of gravity of the portion of the rib whose weight is F. Then the integral (B) gives us, by taking the same limits of 9 as i'or.F, 3R< _4~ g +l shi 29 g i~ 8 R 3 —r 3 cos 9 We can now, by adding together the several quantities (5),. (6), (7), and equating to the value of NS as found from (1) and (2), form the equation for equilibrium (3), which will be a cubic equation in respect of t. We shall thence find the value of < necessary to produce equilibrium in the structure, and, by equating the same quantities to 2N6, form the equation for stability (4), from which we get the value of < necessary to pro duce stability in the structure. The following examples will serve to show how readily these formulae can be applied, and the use which the practical archi tect may make of them. Example 1.:— Let r=10 ft., R=ll ft., 3=125 lbs., 3, = 150 lbs., H=50 ft. From(l) . . . N = 92-0092; „ (2) . . . b = 53-4202; therefore NS =4915 and 2N3 =9830. From (5) . . . . P . u= 316-767 t +191-06975 ; „ (6) . . . . F. c= 164-6725 <— 50-54 ; „ (7) . . . . Q. g = 1308-98 < 2 + 43'63275 <’. Hence the equation for equilibrium (3) is 4915=43-63275 f+lSOS-OS <’+481-4395 <+140-52975, which reduces to < 3 +30< 2 +ll-03<—109-42=0. We can solve this by means of Homer’s process, and find <=1 -7 ft- for equilibrium. The equation for stability (4) becomes 9830=43-63275< 3 +1308-98< 3 +481-4395<+140-52975 which reduces to < 3 +30<’+l 1 -03<—222-069=0, from which we obtain, by Horner’s process <=2-45 ft., for stability. Example 2 :— Let r=30 ft., R=33 ft., H=80 ft., 3 and 8, as before. From (1) . . . N = 2483-84; „ . . . b = 90-2606; therefore NS =224192-8, and 2N3 =448385-6. From (5) . . . . P . a= 8552’71 < +15476’65 : J5 (6) • ■ . . F . c= 4446-16 < + 4094-17 ; (7)- . . Q.g= 69.8124 < 3 + 6283'116 < 2 Adding and reducing, the equation for equilibrium becomes < 3 +90< 3 +186-2<—-3048.31 =0, whence <=4-83 ft. for equilibrium. And the equation for stability is < 3 +90<’+186 -2<—6159 -7=0, whence <=7’07 ft. for stability. Example 3 :— I will apply the formulas to the case of the dome of Sultan Mohammed’s tomb at Beejapore, described in Mr. Fergusson’s “ Handbook of Architecture.” One peculiarity of this structure is that the inner face of the dome is set 5 ft. 6 in. within the inner face of the wall on which it rests, as shown in fig. 3 ; so that in calculating the values of a, &c., we must add 5-5 ft. to the thickness <; hence we shall get a proportionately less value for < from the resulting equations. In this dome, r=62 ft., R=72 ft., H=116 ft.; and we will suppose, as before, that 3=125 lbs., 3 1 = 150 lbs. From (1) N = 31132; „ (2) b = 137-20524: therefore N6 =4271473-522. and 2N3 =8542947-044.