136 



he anffle Tip I. and the angles Amp,Blp, are right angles ; there- 

 fore, tlie length B /, is equal to the length A jh, and therefore, the 

 pressure at A, is equal to tlie pressure at 15, and similarly the 

 pressure ou the wliole edge fi A, is equal to the pressure on the 

 whole edge /"v. Therefore, in hoth cases, diagrams lu and 11, the 

 the pressure will he sustained in a similar nianuer, hy the base 

 cfgli. So that if the resultant jjressure at p, diagram 10, is equal 

 to one ton per scpiare inch of the surface efyli, and does not 

 crush the particles in that surface, tlien, if tlie resviltant at p, 

 diagram 11, is equal to one ton pei square inch of the surface 

 efg A, the particles in that surface will not be crushed. 



if in either of the cases. Diagrams 10 and 11, the portion /j/cjr, 

 be added, it is evident that the pressure on the base cfgli, will not 

 be increased. And therefore, if a stone, as in diagram 12, lie 

 acted on by a pressure, the resultant of which, and the weight 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



LMay, 



D'agram 12. 



of the stone, passes through a point p, in the base e ikh : Draw a 

 line AC, through the point p, to the edges eh, andik; then 

 measure off, on the line p C, a portion p B, equal to the length p A, 

 and draw the line/B if, perpendicular to the line A C. Then, if 

 the resultant pressure at the point ;), divided hy the number of 

 square inches in the surface efgh, is not greater than the pressure 

 per square inch, that (by the experiment in diagram 10) the mate- 

 rial was found capable of bearing, then the stone wiU not fail 

 when acted upon by the given pressure. 



It is of course implied, that no natural fault, or laminated 

 structure, of the stone, sliould cause it to yield, it being evident 

 that the judgment of the engineer must be called into requsition, 

 to guard against such a catastrophe. 



Art. H.— The method here proposed, for the determination of 

 the proper section, &c., of arches, or for discussing the stability of 

 arches already designed, the limited strength of the material being 

 taken into consideration, is founded on the above-mentioned prin- 

 ciple. 



Section IV. 



On the conditions of stability of an arch, acted upon by forces 

 of any amount, applied in any position and direction in the plane 

 of the section ; or of an arch, whose form is not similar on both 

 sides of the crown. 



Art. 15. — In an arch under the conditions stated at the head of 

 this section, the first point of rupture is not necessarily at the 

 crown, and it is this which constitutes the difficulty of the ques- 

 tion. It may here he remarked, that when the terms first, second, 

 and third points of rupture are used, it is not meant that the 

 failure of an arch commences at the first point, and then spreads 

 to the second point of rupture, and so on ; for theoretically speak- 

 ing, the structure will fail at all those points at the same time ; 

 but by the first point of rupture is merely meant the point of rup- 

 ture first determined, and by the second point of rupture, the 

 point of rupture secondly determined, as by the process detailed 

 in the preceding sections, one point of rupture being already 

 known. 



Art. Ifi. Problem 3. — To find the first point of rupture in an 

 arch acted upon by any given pressures, in any given position, 

 and the arch itself being of any gi^•en shape. The method pro- 

 posed to solve this prolilem will, it is thouglit, be more easily 

 shown, liy reference to the example of the arclied vault in the 

 previous sections, than by a general diagram and demonstration. 



Example. — Let the arched vault shown in diagram 13, of 80 feet 

 span, and whose depth of voussoir at the crown is equal to one- 

 ninth of tlie radius, be acted on by a pressure, equal to the weight 

 of a portion of the arch of 20° length of intrados, and one foot in 

 length of transverse section, and applied vertically to the extrados, 

 at a distance of 30° from the crown. Required the first point of 

 rupture, under these conditions ; the materials being incompressi- 

 ble. First, suppose for trial, some point R, in the extrados, to be 

 the first point of rupture, say 28° from the crown, as in diagram 13. 



Draw a tangent T„ R, T,, to the extrados at R,. Then if R„ 

 be the first point of rupture, R, T,, and R, T.^, represent the di- 

 rection of the pressure of one portion of the arch on the other 



Diugratn 13. 



80.0 ■span,. 



loiosowso 



=1 



Scale ofweights of Degrees. 



10 



Scale of Feet. 



when the arch is about to fail at that point ; for then the line of 

 resistance touches the extrados at R,. Also R, being the first 

 point of rupture, and the arch being about to fail, the pressure of 

 the lower portion of the arch, in the direction To R', must be 

 equal to the pressure of the upper portion, in the direction T, R, ; 

 for the pressures must be in equilibrium about the point R„ and if 

 one is preponderating, then the arch has already failed somewhere 

 else, and the voussoirs, if about to turn on their edges at the point 

 R„ are moving in the direction of the preponderating pressure. 

 On the supposition that R, is the first point of rupture, find the 

 second points of rupture, Rj, on the right and left hand sides of 

 the arch, in the same manner as described in Section II. Problem 2. 



In this case, the second point of rupture on the right hand side, 

 is at the intrados, at 51° degrees from the crown ; for the vertical 

 line, drawn through the centre of gravity of the mass R, N, R, Nj, 

 intersects the tangent to the extrados, at the crown R, T,, at the 

 point O, and the tangent to the intrados at R„ also passes through 

 the same point. Also the second point of rupture, on the left 

 hand side, is at the intrados, 6-t° distant from the crown ; for the 

 direction of the resultant of the weight of the mass R, N, RjNj, 

 and the force impressed on the arch, intersects the tangent to the 

 extrados at Ri, and the tangent to the intrados at Rj, at the same 

 point O. 



Next, construct the parallelograms of pressure, as shown in 

 diagram 13 ; in that for the left side, the vertical line marked 

 R W I, represents the resultant of the weiglit of the mass, 30'°, 

 and the impressed forces equal to the weight of 20° of the arch ; 

 and these being pressures in the same direction, this resultant 

 equals the weight of 56° of the arch, and the side of the parallelo- 

 gram R W I, is equal to 56 parts of the scale. The directions of 

 the other two pressures are sufficient to determine the parallelo- 

 gram, by which it appears, that the pressure of the lower portion 

 of the arch on the point R„ in the direction Tj R,, is equal to the 

 weight of 40° of the arch. 



In the same manner it will be found that the pressure of the 

 upper portion of the arch on the point R,, in the direction T, R,, 

 is ecpial to the weight of 50° of the arch. But the pressure in the 

 direction Tj R^ is equal to th' weight of 40° of the arch, therefore 

 the pressure at the point R,, of the upper portion of the arch, 

 u])on the lower, is greater than that of the lower portion and its 

 impressed force, on the upper. Therefore R, is not the first point 

 of rupture. 



Take therefore, for trial, some other point nearer the crown of 

 the arch, for the point of rupture ; for it is evident, that as the 

 point of rupture approaches towards the crown, the pressure on it 

 from the left side, willbegreater; and that fromtherightsidebe less. 

 Let this second trial point be 15° from the crown ; repeat with re- 

 gard to this point, a process similar to that above described, and as 

 shown in the diagram ; and it will be found, that the pressure from 



