OP PRESSURE AND EQUILIBRIUM. 



^S 



ft given isosceles triangle, and the other two 

 meeting in its vertex, a circle being circum- 

 scribed round the triangle, and perpendicu- 

 lars erected from the quadrisections of the 

 base, the lower t)eam on each side must be 

 directed to the nearest intersection of the 

 perpendiculars with the circle, and the upper 

 one must be in a chord of equal length. 



C 



The two upper 



beams act against 

 each other in a ho- 

 rizontal direction 

 B K C; H only, consequently 



the horizontal thrust of the lower beams must be equal to 

 that of the upper beams, produced by their own weight 

 only, while the thrust of the lower beams is derived not 

 only from their own weight, but also from that -of the 

 upper beams, acting at their extremities ; but the horizontal 

 effect of the weight of the upper beam is to its weight as 

 balf AB to BC, since the centre of gravity may be supposed 

 to act on a lever of half the length of AB, and the hori- 

 Eontal force on a lever of the length BC (290) ; but the ho- 

 riiontal thrust of AD is equal to that of AC, and is to the 

 force acting vertically at A as DE to AE ; and the force 

 acting vertically at A is the whole weight of AC and half 

 the weight of AD, which is three times as much as the 



DE 

 weight acting vertically at C ; consequently — ; must be 



Aci 



equal to — — ; now the triangle ADE is similar to FDG, 



and ABC to HFG, since the angle DFH=DCH, and GFH 

 =DFH— DFGzzDFH— (DIG-1DF)=:DCH— (BCH— 



ACD)=DCB4.ACD=ACB, therefore £f=^,and-^ 



ACj rG . 3BC 



_GH 3DG DG 



— -T7;— — -— -iz:-—^, as IS required for the equilibrium. 



307. Theorem. When an arch is com- 

 posed of blocks acting on each other without 

 friction, the weight of the arch must increase 

 at each step as the portion of the vertical 

 tangent cut off by lines drawn from a given 

 point in a direction parallel to that of the 

 joints. 



The thrust in the direction AB, by which the block A 

 is supported, must be to its weight as AB to BC, or as 

 DE to EF, and to the horizontal thrust, as AB to AC, 



or DE to DF : and for 

 the same reason the 

 weight of any other 

 part FG must be to the 

 horizontal thrust as HI 

 to IG, or as FK to 

 FD : but the horizontal 

 thrust is equal throughout the arch, being propagated froir* 

 the abutments, since the weight of the blocks, acting in a 

 vertical direction, can neither increase nor diminish it ; 

 and it may therefore always be represented by the line DF, 

 while FE, EK, represent the weight of the arch and of its 

 parts ; and it will be equal to the weight of a portion of the 

 length of the radius DF and of the depth of the block AC, 

 as is obvious from considering the cfTect of the upper block 

 acting as a wedge. 



308. Theorem. A spherical dome of 

 equable thickness, having its joints in the di- 

 rection of theradii, may remain in equilibrium 

 if its height do not exceed 392 thousandths 

 of the radius. 



The action of the weight of the dome resembles that of 

 a wedge, pressing on each horizontal course with a force 

 which is to its weight as the radius to the sine of the 

 angular distance from the vertex x, and its pressure is sup- 

 ported by the weight of the course, acting also as a wedge ; 

 this weight is first reduced by the inclination of the joint in 

 the ratio of the cosine of the angular distance from the 

 vertex y, to the radius, and its effect is increased in the 

 ratio of the lengthof the wedge to its base, or of the radius 

 to the breadth of the course : the effect will therefore be 

 equal to the weight of a portion equal in breadth to the 

 radius, reduced by the obliquity of the joint in the ratio of 

 1 to the cosine j. While therefore the weight of a circum- 

 ference of the breadth y is greater than that of the dome 

 increased in the ratio of 1 to x, the course will retain the 

 incumbent dome in equilibrium ; but when it it in a smaller 

 proportion, the course will be forced outwards, unless it be 

 restrained by external pressure ; and the limit will be when 

 the weight of the dome is equal to that of a cylindrical sur^ 

 face of the breadth xy, and of the radius x. Now the 

 spherical surface is equal to a cylindrical surface of the 

 breadth 1— j and of the radius 1, therefore xxy~l — y, 



(1— i/y) j=i— !/. (i+y)y=Uy+yy+i=hy=->/i—i 



=.61803. 



— 2 



SO9. Theorem. In order that a spherical 

 dome of the span 2x, may stand without ex- 

 ternal pressure, the thickness must Ikj in- 



