e 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 - w=fy8x.
But the resultant of these two pressures must be perpendicular 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 equationSz(II - fy8z) = SxfySx Since the vertical pressure at P is so, while the horizontal strain is h, the intensity of the oblique strain at P must be ,/(w2 +M). Now, in passing to the proximate point Q, so becomes so + 8w, while it is reduced to it - Stoaz - so that the oblique strain at Q must be - J( (20 + 802 ÷ (it - Stutz)2 ), zu =1 - wherefore the strain at Q is „,/(w2 + h2), or exactly the same as that at P. This result might have been obtained from the consideration that the thrust upon the surface PQ is perpendicular to the oblique strain, and can tend neither to augment nor to diminish it. Hence, as a characteristic of this arrangement, we have the law that the tension across the joints of the arch-stones is the same all along, and therefore is equal to H, the horizontal tension at the crown of the arch.
From this it at once follows that if r be the radius of curvature at the point P, y being the vertical thickness of the mason-work there, H = ry, so that if R be the radius of curvature at the crown of the arch, and A the thickness there, the horizontal thrust there, or the strain transmitted along the arch-stones, is H = RAH, being measured in square units of surface ; hence also A : y :r :11, or the thickness at any place, is inversely proportional to the radius of curvature there.
When the form of the intrados is given, its curvature at any point is known, and from that the thickness of the stone-work and the shape of the extrados can be found. The most useful case of the converse problem is, again, that in which the extrados is a horizontal straight line.
Let OH, figure 6, be the horizontal extrados, and A the crown of the arch ; make also AB such that its square may represent the horizontal thrust there; then, having joined OB and drawn BC perpendicular to it, and meeting the continuation of OA in C, C is the centre of curvature for the crown of the arch. Or, if the radius of curvature and the thickness of the arch at the crown be pre • scribed, we may obtain the horizontal thrust by describing on CO a semicircle, cutting a horizontal line through A in the point B, then the horizontal thrust is equal to the weight of the quantity of the stone-work which would fill up the square on AB. The conditions of the problem require that the curve APQ be so shaped as that the radius of curvature at any point P shall be inversely proportional to the ordinate HP.
Resuming the general equation of conditionSz(H - fy8z)= 8.rfy8x , and observing that in this case y=z, we haveSz(H - .118z)=8*Sx Now the integral .11-8: is 1z2, but as it must be reckoned only from A where 2= A, the equation becomes Sz(11 + 1-A2 - iz2)= 8.rfiSx .
The coefficient of Sz becomes less when z increases, and when 1;z2= 112 + 2A2, this coefficient becomes zero, at which time az also becomes zero in proportion to Sz ; that is to say, the direction of the curve becomes vertical. Wherefore, if we make OD = D such that D2 = A2+ 2112, we shall obtain that depth at which the curve is upright, or at which the horizontal ordinate DQ is.the greatest, and then the equation takes the form-- Sz(D2 - z2) = 28:03.r , difficulty, and therefore it may be convenient to attempt a graphical solution. Since, for any vertical ordinate HP( = z), the horizontal thrust is 1(D2 - 7.2), while the oblique strain is 1(D2 - A2), the obliquity of the curve at P has for - its cosine the value D2 z2 wherefore the angle at which the curve crosses the horizontal line pP is knoWn. Let then a multitude of such lines be drawn in the space between BA and DQ, and let the narrow spaces thus marked be crossed in succession from A downwards by lines at the proper inclination, and we