1849.] 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



135 



sue a precisely similar method to that described for R„ as shown in 

 the figure. 



Then, since the tangent Ro T„, produced, does not pass through 

 the point of intersection I=, but is less inclined to the vertical 

 than the line R„ L,, the point of rupture is above R^- Also since 

 the line R^, T^, more nearly coincides with the line R„ I„, than 

 the line R, T,, with the line Ri I,, the point of rupture is nearer 

 to R,., than to R,. 



One more subsequent trial generally suffices to determine the 

 correct point, which, in this example, is the point R,. For the 

 tangent R., T,, produced, passes through the point I,, which is 

 the point of intersection of the direction of the weight of the 

 mass A D R, N, B,,, and the pressure of the opposite of the arch. 

 Therefore, if the arch fails by the sinking of the crown, the second 

 point of rupture is R,. 



The second case is now to he considered : AVhere will be the 

 second point of the rupture, if the arch fails by the rising of the 

 crown ? Draw the horizontal line D L, which w ill, in this case, 

 represent the position and direction of the pressure of the oppo- 

 site side of the arch. Let the point N, be tried ; then if N, be 

 the point of rupture, the tangent to the extrados at N, N, P,, 

 will, if produced, pass through the point Q,, which is the point of 

 intersection of tlie directions of the pressures of the opposite side 

 of the arch, represented by the line D L, and of the w eight of 

 the mass A D R, N, B,, represented by the vertical lineW, I,. 

 But N, P, intersects the line D L, far from the point Q,. Also 

 if the point N, be tried, it will be found that the tangent N^ P3 

 is far distant from the point of intersection Q, ; and in like man- 

 ner it will be found, that at no other point above N,, will these 

 conditions be fulfilled, except at the point C. Therefore the arch 

 will not fail by the rising of the crown. Therefore the arch will, 

 if it fails, fail by the sinking of the crown and the sjireading of 

 the haunches ; and the point R, is the second point of rupture. 



Art. 10. The second part of the Problem. — It is re<juired to 

 determine the amount of pressure at the crown, and at the second 

 point of rupture. 



Construct a scale of equal parts, as in Problem I, each division 

 representing some unit of pressure, as pounds, hundred-weights, 

 or tons. Through the point of their intersection 1 3, produce the 

 lines Rj I,,, and W, I3 ; then on the line W,, I,, produced, mea- 

 sure off the distance, I, F, containing as many equal parts of the 

 scale as these units of weight, in the mass A D R„ N;, B,, and 

 from the point F, draw a line parallel to C E, intersecting the line 

 R3 I3, produced at the point H. Then, by the well-known prin- 

 ciple of the parallelogram of pressures, the line F H contains as 

 many equal parts of the scale as there are units in the pressure 

 of the opposite side of the arch on the crown at C, and the diago- 

 nal of the parallelogram H I3, contains as many equal parts as 

 there are units in the pressure on the point R3. Thus, in Ex- 

 ampe 1, if the weight of the mass A D R,, N3 B3, is 3 tons 3 cwt., 

 then the pressure at the crown will be 1 ton 2 cwt., and the pressure 

 at the point R„, 3 tons 5 cwt. 



Art. 11. — Thus the resultant pressure on one of the blocks of 

 the structure is determined in direction, position, and amount, 

 which is the datum required in Problem 1 ; and therefore, that 

 problem may be applied and the line of resistance be traced, as in 

 the example in Art. 3, through the whole structure, commencing 

 either from the crown, or from tlie second point of rupture ; and 

 this line will represent the resultant pressures at every part of the 

 structure, Vhen it is on the balance between standing and falling, 

 that is, when it is in the condition of unstable equilibrium. 



If the line of resistance, at any point, passes without the boun- 

 dary of the voussoirs, the structure will unquestionably fail. If it 

 touches the extrados, or intrados, at other points, and at the base, 

 then the structure is in the condition of inistable equilibrium. If 

 the line of resistance passes through the base of the structure, 

 some distance within the mass, then the arch has a certain degree 

 of stability, which may be tested, as described in Art. 6, by the 

 methods given in the following Sections. 



The stability of the structure, with regard to the first condition 

 of failure (Art. 1, diagram 2), has to be considered, and is at 

 once determined by inspecting the line of resistance, drawn as 

 described in the foregoing examples. If at any part of the struc- 

 ture a joint is made, in such a direction, that a perpendicular 

 drawn from it shall be inclined from the tangent to the line of re- 

 sistance, at that point, at an angle greater than the limiting angle 

 of resistance of the surfaces of contact, the structure will fail at 

 that place ; if, however, this is not the case at any position in 

 the arch or pier, then the structure will not fail by the slipping of 

 the blocks one upon the surface of the other, and the first condi- 

 tion of failure will not be fulfilled. 



Section III. 



On the conditions of stability of an arch, the form of which, 

 and the pressui-es sustained liy it, as regards position, direction, 

 and amount, are similar, on either side of the crown of the arch ; 

 the limited strength of tlie materials being taken into considera- 

 tion. 



Art. 12. — By reference to Art. 6 it will be seen, that it is there 

 proposed, that the conditions of stability in an arch should first 

 be discussed on the supposition that the materials were incom- 

 pressible, and that then it should be examined in what respect 

 those conditions were modified by the limited strength of the 

 materials used in building. The first part of this proposition has 

 been considered in Section II. It is the purpose of this section to 

 consider the second part. 



The arches in the examples in the last section could not stand 

 if they were built of any material at present known, because at 

 the points of rupture, the resultant pressures act at the extreme 

 edge of the voussoirs, and therefore all the pressure has to be re- 

 sisted by these extreme edges, or by a single line, which cannot be 

 the case, unless the material is incompressible. So that in all 

 practical cases of arches, even the condition of unstable equili- 

 brium cannot be attained, unless the position of the line of resist- 

 ance is some distance within the section of the arch. The ques- 

 tion which then arises is, how near to the intrados or extrados can 

 the line of resistance pass, without causing the failure of the ma- 

 terials .'' 



Art. 13. — Experiments to determine the strength of stones to 

 resist compression, have for the most part been made by the ap- 

 plication of pressures on cubes of the stone, in a direction perpen- 

 dicular to the face of the cube, as in diagram 10. The resultant 

 of this pressure, and the weight of the stone, acts in the direction 



Diagram 10. 



Diagram 11. 



of the axis of the cube. Its point of application being in the centre 

 of the base at p ; so that if any line be drawn through this point p, to 

 the edges of the block, as the line A B, the portion p A\% equal 

 to the portion p B : and as, by the principle of the equality of 

 moments, the pressure on the point A, multiplied by the length 

 Ap, is equal to the pressure on the point B, multijdied by the 

 length Bp; since the length Ap, is equal to the length B p, the 

 pressure on the point A, is equal the pressure on the point B ; and 

 similarly the pressure on the w hole edge of the stone e h, is equal 

 to the pressure on the opposite edge/g. 



Now let the block of stone, as shown in diagram 11, be acted 

 upon by a pressure whose direction is inclined to the axis of the 

 block, but which is applied in such a position, that the resultant 

 of it, and the weight of the block, acts through the point p, in 

 the centre of the base. Draw any line A B, through the point ;>, 

 to the edges eh, and fg, and draw another line through p, in the 

 direction of the resultant, and from the points A and B, draw 

 lines Am, B /, perpendicular to this line. Then, by the principle 

 of the equality of moments, the pressure on A, multiplied by the 

 length Am, is equal to the pressure on B, multiplied by the length 

 B /. But since p B, is equal top A, the angle Apm, is equal to 



