It is evident that the liability of the arch to failure by the slipping of its voussoirs, is less as its depth is less as compared to its length. In order the more effectually to protect the arch against it, the voussoirs are sometimes cut of the forms shown by the dotted lines in the preceding figure, their joints converging to a point. The pressures upon the points A and B are dependent upon the form of that portion of the arch which lies between those points, and independent of the forms of the voussoirs which compose it; these pressures, and the conditions of the equilibrium of the piers which support the arch, remain therefore unchanged by this change in the forms of the voussoirs. 310. To determine the conditions of the equilibrium of the upright piers or columns of masonry which form the abutments of a straight arch, supposing them to be terminated, as shown in the figure, on a different level from the extrados CD of the arch, let b be taken to represent the elevation of the top of the pier above the point A; then will Lb b tan. a, or (equation 407), represent the distance AG H (p. 383), or the value of k-a). Substituting for k in equation (377) and also the values of P sin. P Cos. a, from equations (409) and (405), we have (412); which is the equation to the line of resistance of the pier, a representing its thickness, b the height of its summit above the springing A of the arch, L the length of the arch, the weight of a cubic foot of the material of the arch or abutment (supposed the same). The conditions of the stability may be determined from this equation as in the preceding articles. If the arch be uniformly loaded, the value of , given by equation (410) must be substituted for . 311. THE CENTRE OF GRAVITY OF A BUTTRESS WHOSE FACES ARE INCLINED AT ANY ANGLE TO THE VERTICAL. Let the width AB of the buttress at its summit be repre sented by a, its width CD at the base by b its vertical height AF by c, the inclination of its outer face or extrados BC to the vertical by a,, that of its intrados AD by Let H represent the centre of gravity of the parallelogram ADEB, and K that of the triangle BCE, and G that of the but tress; draw HM, GL, KN, perpendiculars upon AF. Then representing GL by and observing that the area ADEB is represented by de, the area EBC by (b−a)c, and the area ADCB by (a+b)e, ac.HM+4(b—a)cKN_2a. HM+(b—a)KN (a+b) c a+b Now HM-Hh+hM=4a+je tan. a,=(a+c tan. a,), KN KI+lk+kN=31(b-a)+a+c tan. a,= 46+2a+2e tan. a); Substituting these values and reducing, λ_(a2+ab+b2)+(a+2b) e tan. a, 3(a+b) (413). b=CD=CF-DF=c tan. a,+a-c 'tan. a, ; also (a2+ab+b2) =(b−a)2+3ab=c(tan.a,—tan.a,)+3ac(tan.a,—tan.«,)+3a”, (a+2b) e tan. a,= {2c (tan. a,-tan. a,)+3a} e tan. a, =2c (tan. a-tan. ɑ,) tan. ɑ, +3ɑc tan. a,; :: (a2 + ab + b2) + ( a +26) e tan. a,=c2 (tan.*,-tan. 'a,) 312. THE LINE OF RESISTANCE IN A BUTTRESS. Let LM represent any horizontal section of the buttress, K/M BA P TK a vertical line through the centre of gravity of that portion AMLB of the but tress which rests upon this section. Produce LM to meet the vertical AE in V, and let KVλ and AV; then is the value of determined by substituting for e in equation (414). Let PO be the direction in which a single pressure Pis applied to overturn the buttress, Take This equation is, of course, to be adapted to the case in which the inclination of AD is on the other side of the vertical, as shown by the dotted line Ad by making a, and therefere tan. a negative. OS to represent P in magnitude and direction, and ON tc represent the weight of the portion AMLB of the buttress; complete the parallelogram SN, and produce its diagonal OR to Q; then will OR evidently be the direction of the resultant pressure upon AMLB, and Q a point in the line of resistance. Let VQ=y, AG=k, /GOT=1, weight of each cubic foot of material; and let the same notation be adopted in other respects as in the last article. By similar triangles, QK RI OK OI QK=QV-KV=y—^, OK=TK-TO-TK-TG cot. GOT=x-(+) cot., OI=ON+NI=¿μAV(AB+LM)+RN cos. RNI= = P sin. 'x—(λ+k) cot. ¿ ̄ ̄x{2a+x(tan. a,—tan. «,)}+P cos. ¿ i Transposing and reducing, y= {2a+x (tan. a,-tan. a,)} +P (x sin. -k cos. ¿ ; but substituting x for c in equation (414), and multiplying both sides of that equation by the denominator of the fraction in the second member, and by the factor x, we have {2a+x (tan. a, -tan. a,)} = u'a tan. a,+±μxa2; 1 (tan.,-tan. 'ɑ,)+ y= ux (tan.a,-tan.3a2)+ux1a tan. a1 +μxa2+2P(x sin. —k cos. ¿) μx {2a+x(tan.atan.a2)}+2P cos. .(415); which is the equation to the line of resistance in a buttress. If the intrados AD be vertical, tan. a, is to be assumed =0. If AD be inclined on the opposite side of the vertical to that shown in the figure, tan. a, is to be taken negatively. The line of resistance being of three dimensions in æ, it follows that, for certain values of y, there are three possible values of the curve has therefore a point of contrary flexure. The conditions of the equilibrium of the buttress are deter mined from its line of resistance precisely as those of the wall. Thus the thickness a of the buttress at its summit being given, and its height c, and it being observed that the distance CE is represented by a+c tan. a,, the inclination a, of its extrados to the vertical may be determined, so that its line of resistance may intersect its foundation at a given distance m from its extrados, by solving equation (415) in respect to tan. a,, having first substituted c for a and a+c tan. a-m for y; and any other of the elements determining the conditions of the stability of the buttress may in like manner be determined by solving the equation (the same substitutions being made in it) in respect to that element. 313. A WALL OF UNIFORM THICKNESS B A SURE OF A FLUID. * SUSTAINING THE PRES If E be taken to represent the surface of the fluid, IK any section of the wall, and EP two thirds the depth EK; then will P be the cen tre of pressure of EK, the tendency of the fluid to overturn the portion AKIB of the wall being the same as would be produced by a single pressure applied perpendicular to its surface at P, and being equal in amount to the weight of a mass of water whose base is equal to EK, and its height to the depth of the centre of gravity of EK, or to EK. Let AK=x, AE=e, weight of each cubic foot of the fluid=,; ..P=(x—e). 1(x− e) μ‚= 1⁄2) (x—e)3 Let the direction of P intersect the axis of the wall in O; let it be represented in magnitude by OS; take ON to represent the weight of the portion AKIB of the wall; complete the parallelogram SN, and produce its diagonal to meet IK in Q; then will Q be a point in the line of resistance. Let QM=y, AB-a, weight of each cubic foot of QM RN = material of wall. By similar triangles, MO NO Now * Treatise on "Hydrostaties and Hydrodynamics," by the author of this work, Art. 38. p. 26. QM=y, MO PK=}EK=}(x−e), RN=OS=P= , and observing that the fraction represents the ratio o of the specific gravities of the material of the wall and the fluid, we have which is the equation to the line of resistance in a wall of uniform thickness, sustaining the pressure of a fluid. 314. To determine the thickness, a, of the wall, so that its height, h, being given, the line of resistance may intersect its foundation at a given distance, m, within the extrados. Substituting, in equation (416), h for x, and a-m for y, and solving the resulting equation in respect to a, we obtain 1 Equation (416) may be put under the form y= Όσα x2 (1–3); whence it is apparent that y increases continually with ; so that the nearest approach is made by the line of resistance, to the extrados of the pier, at its lowest section. m therefore represents, in the above expression, the modulus of stability (Art. 286). 315. The conditions necessary that the wall should not be overthrown by the slipping of the courses of stones on one another. The angle SRO represents the inclination of the resultant pressure upon the section IK to the perpendicular; the proposed condition is therefore satisfied, so long as SRO is less than the limiting angle of resistance 9. |