Page:Encyclopædia Britannica, Ninth Edition, v. 2.djvu/348

Rh 328 ARCH supporting points P and P. Now, as this tendency is horizontal its intensity cannot be changed by the load acting only downwards, and must remain the same throughout the structure, wherefore the actual pressure at P must be found by combining this fixed horizontal thrust with the downward pressure equal to the weight of the bridge from A to P. If, then, we draw ah horizontally to represent this constant thrust, and ap upwards to represent the weight of this portion of the arch, the line ph must, according to the law of the composition of pressures, indicate both in direction and in intensity the actual strain at the point P. This pressure must be perpendicular to the joint of the stones, and must therefore be parallel to the straight line drawn to touch the curve at P. Hence, if the form of the inside of the arch, or the intrados as it is called, be prescribed, we can easily discover the law of the pressures at its various parts ; thus, to find the strain at the point Q, we have only there to apply a tangent to the curve and to draw hq parallel thereto ; hq represents the oblique strain at Q ; aq represents the whole weight from the crown A to Q, and therefore pq is proportional to the weight imposed upon the position PQ of the arc. Using the language of trigonometry, the horizontal thrust is to the oblique strain at any part of the curve as radius is to the secant of the angle of inclination to the horizon; also the same horizontal thrust is to the weight of the superstructure as radius is to the tangent of the same inclination. And thus, if the intrados be a known curve, such as a circle, an ellipse, or a parabola, we are able without much trouble to compute, on this hypothesis, the load to be placed over each part. If we use the method of rectangular co-ordinates placing x along OH and z vertically downwards, so that PTT may be the increment of x, -n-Q that of z, the tangent of the inclina- &amp;lt;ts tion at P is v- and therefore if h stand for the horizon- ox tal strain, and for the weight of the arch, we have I!} ft OX while the oblique strain is h / ( &quot; V Also the change of weight from P to a proximate point Q is Let RST be the outline which the mason-work would have if placed compactly over the arch-stones, in which case RST is called the extrados, then the weight supported at P is proportional to the surface ARSP, and the incre ment of the weight is proportional to PSTQ, hence if the weights and strains be measured in square units of the vertical section of the structure, and if y be put for PS, the thickness of the mason- work, we have whence y = h o-v When the curve APQ is given, the relations of s and of its differentials to x are known, and thus the configuration of the extrados can be traced, and we are able to arrange the load so as to keep all the strains in equilibrium. But when the form of the extrados is prescribed and that of the intrados is to be discovered, we encounter very great difficulties. Seeing that our hypothesis is not admissible in practice, it is hardly worth while to engage in this inquiry ; it may suffice to take a single, and that the most interesting case. If the whole space between the arch-stones and the road way be filled up, the extrados becomes a straight line, and when this is horizontal we have y = z, so that the form of the arch must be such as to satisfy the condition o-z ^ ox* that is to say, z must be a function of x such as to be pro portional to its own second derivative or differential coefficient. Now this character is distinctive of the cate narian functions, and therefore ultimately V/A where A is AO, the thickness at the crown of the arch, and e the basis of the Napierian system of logarithms. In this case, since 8w = z?&amp;gt;x, Fig. 5. so that the form of the arch and also its weight may readily be computed by help of a table of catenarian func tions. Let us now consider the case when the ends of the arch- stones are dressed continuously, while the imposed load is formed of stones having vertical faces. The weight of the column PSTQ resting on the oblique face PQ is prevented from sliding by a resist ance on the vertical surface QT, which resistance goes to partly oppose the horizontal strain transmitted by the preceding arch-stone; and thus the out-thrust of the arch, instead of being entirely resisted by the ultimate abutment, is spread over the whole depth of the structure. In this case the horizontal thrust against QT is to the weight of the column as QTT. the increment of z, is to Pvr, the increment of x ; wherefore, putting H for the hori zontal thrust at the crown of the arch, and // for that part of it which comes down to P, the decrement of h from P to Q is proportional to the rectangle under PS and QTT, that is to say, 8A=y82. Now, the whole decrement from the crown downwards is the sum or integral of all such partial decrements, and therefore the horizontal thrust transmitted to P is expressed by the symbol while the whole weight supported at P is the analogous integral _ w =fyox. But the resultant of these two pressures must be perpendi cular to the joint of the arch-stones, or parallel to the line of the curve ; wherefore ultimately we obtain, as the con dition of equilibrium in such a structure, the equation Since the vertical pressure at P is w, while the horizontal strain is h, the intensity of the oblique strain at P must be J{w- + h z }. Now, in passing to the proxi mate point Q, w becomes w + 8w, while h is reduced to h - Siv^-, so that the oblique strain at Q must be 02 or, neglecting the second power of the mfinitesimally small increment $w, J j w~ + p 2 + 2w8w - 2/t3^ j, but
 * -H-/y&&amp;gt;