186 



Mr. Tarn on the Btahility of Domes, [May 31, 



And if we substitute for its value, 



^^=•7351 ^ 



And c=OR— ^1 = ^+?*— ^j. 

 Hence we find 



F. c= -00398 a (R3-r3)^ + -00398 a {r(R3-r3) -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, the weight of a cubic foot of its ma- 

 terial ; then 



Q=|0 . a, ((r + ty-r') . H=i^ a, (2r + t . H 

 = -017453 a, {2r^t)t.B.. 

 Let figure 2 represent the plan of the pier, g its centre of gravity, 

 ST=^ its thickness. Then by the integral calculus we find 

 ^ 4 (r + #)^ — sin-i-0 ,,. sin-l-(i)_, 



^^^sM^^^'TT ^ or,puttmg-^-l, 

 *2 



=? (.r+tf-r^ _ 2{^r' + 3rt + t'') ^ 

 ^Ir+ty-r^ 3(2r+0 ' 

 q=^g=0^-0g^r+t-0g 

 _ q (r + (2r + - 2 (3r^ + ^rt + f) 

 3 (2r+0 



__3rH-^ 

 3(2r+0* 



We therefore obtain for the value of Qg', 



Q2=-017453a,(2r + O^-Il " 



3(2r+0 



= -0058177 a,. ^2(3r + 0.H (7) 



We can now, by adding together the several quantities (5), (6), (7), 

 and equating to the value of 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 t 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 t necessary to 

 produce stability in the structure. 



The following examples will serve to show how readily these formulae can 

 be applied, and the use which the practical architect may make of them. 



Example 1 : — 



Let r=10 ft., R=ll ft., a=1251bs., a,= 1501bs., H=50ft, 

 From (1) . . . N = 92 0092; 

 „ (2) . . . 6 = 53-4202; 



therefore 



N6 =4915, 



and 



=9830. 



