ARCHES.] BRIDGES 313 increase or diminish the span of the rib .it the springinga, the rest of the rib being assumed free from strain. Then" catling I the moment of inertia of the cross section of the rib, and E the modulus of elasticity, we shall have- My.AL As IE aa appears from the following considerations :-~ In fig. 58, let Oo = Aa=B6=AL; conceive the surface at AB as fixed, and let the action of the couple M be such as to extend the Fig. 58. top and compress the bottom of the rib, moving the point a to a, and the points b to i l5 then, calling jj the intensity of the stress at a, we have (equation 4, 14) also we have ( 24)- and therefore d . . E MAL d join o and Q and draw the line oQj, making the angle QoQ 1 equal to aoa^, at Q draw QQ t perpendicular to oQ. Then the effect of the couple M on the length AL of the rib, the rest being unstrained, will be to move the point Q to Q lf and by similar triangles we have aa 1 : QQ[ = -^ : &> and therefore Then resolving the motion OOj into horizontal and vertical com ponents As and Av, we have by similar triangles, As : QQ^y : oQ, or, as above I.E But if, as above stated, the rib does not act as a spring in either direction, the span will remain constant, and the sum of all the changes in span produced by all the successive lengths AL will be nil, or 2As = 0. Hence, since E is constant, we have 1 as a necessary condition for the equilibrium of a loaded rib, hinged at the abutments when these do not yield. This condition must be satisfied whatever be the form or load of the rib, the reasoning by which it was obtained being independent of the form either of the rib or linear arch. When the cross section of the rib is constant, we have 3 ....... 2JIy.AL = 0. We shall now proceed to show how the linear arch satisfying this condition can be found for the case of a uniform rib. Let the line OOjO./)^, fig. 59, be the geometrical axis of the rib, and let OC 1 C. J C 3 4 be the linear arch required ; this arch will, as shown above, cut the geometrical axis at some point, as at 3 . Let C 4 C represent the resultant pressure on the rib at any point C 4 in direction and magnitude, iig. (JO. If this pressure be resolved in to its vertical and horizontal components, the latter C^H will be equal to the horizontal thrust A (constant throughout the linear arch smce the loads are vertical). Fig. 59. This force applied at C 4 will be equivalent to an equal and parallel force, l^H,,, applied at the point, 4 in the axis, added to a left handed couple, of which the moment is h 4 C 4 . This couple Fig. 60. is, for the point 4 , the couple M* required for equations 1 , 2, and 3, the magnitude of which, h being constant, is proportional to the vertical distance 00 between the curves at any point. Equation 3 requires that the sum of all the values of My shall be equal to zero, and we now see that this condition results when the sum of all the products of OC into y is equal to zero, or when when the cross section is not constant the value of y.OC must be substituted for the simple product y.OC. The problem of discovering the actual linear arch which will be called into play with a given rib is now reduced to that of finding the linear arch fulfilling the condition in equation 4. We might proceed tentatively, drawing numerous linear arches, and selecting jy trial that which most nearly fulfils the condition, as proposed jy Mr Bell, Proc. I.C.E., vol. xxxii., but Professor Fuller of Belfast
- ias shown, Proc. I.C.E., vol. xl., that the ordinates of the required
inear arch can at once be calculated from the values of the bending noment at the several sections of a beam of equal span and simi- .arly loaded. Let 0(/jgr 2 (/ 3 (fig. 61), be the curve of bending mo ments which, as was shown in 30, is one form_of linear arch cor- espouding to the given load, the lengths oo lt o^, o 2 o 3 &c., being equal and representing AL ; let y as bejore be any ordinate of the urve oo^o./ifr the axis of the rib ; let sg be any ordinate of the
- iven curve of bending moments; let _sc be any ordinate of the
equired linear arch. Then, since oc scj, we have for the case if uniform cross section and hinged abutments the equation 5 (sc
- Tliis couple is sufficient to shift the force from C 4 to 4 , but the
esultant of the force at <7 4 and the couple would not be tangent to rhe geometrical axis of the rib. To alter the direction of the force in this naimer a vertical component must be added, but this vertical com- xment may be looked upon as a shearing force, which, being vertical, ends neither to extend nor to diminish the span.
IV. 40
