292 SUPPLEMENT TO CHAP. XIV. [CHAP. IV 



y 



L ^ = 4- a = y, where, as in the Fig., e a is negative. In any case, 



ivr 



is given by e, = =- ; hence, for the intersection curve, 



Ve M 



y = g~~ ....... (33) 



(J) Direction curvet and segments. 



The direction of the resultants R and R' can be determined in two ways. 

 First, by the points of intersection <f> and fy with the verticals through the 

 centres of the end cross-sections ; second, by means of the curves enveloped 

 by these resultants for every position of P. We call the first distances 

 A <f> = Ci, B ^ = d, the direction segments, and the enveloped curves the 

 direction curves. 



Taking Ci and c as positive when laid off upwards above the ends, we 

 have MI = H Ci M a = H e, ; therefore 



(84) 



We may also easily determine the equation of the direction curves. Let 

 the co-ordinates with reference to the crown of any point be v and w (Fig. 

 92). If the load P is now moved through an indefinitely small distance, 

 the new resultant cuts the former in a point of the curve required. These 

 two resultants intersect the vertical through O in two points. Let the dis- 

 tances of these points from O be c and c + d c, and let y and y + d y be 

 the angles of the resultants with the vertical. Then v = (w + c) tan y, 

 c = (w + c + d c) tan (y + d y), 



Eliminating v, 



(c +d c) tan (y + d y) c tany _ d (c tan y) 

 tan (y + d y) tan y d tan y 



From the first of the equations above we have then 



dc 



d tally 



tan" 



H Mo Mo . 



But tan y = -, c = - -^-, c tan y = =-, hence 



. . (85) 



Thus we see that in any case we have only to determine H, V and M c , and 

 we can then from (83) and (34) or (85) determine at once the intersection 



