94 SIMPLE GIRDERS. LCHAP. VII. 



Max. M = _ -p P. 



The shearing forces are, then, given by a straight line inter- 

 secting the span in the middle, the ordinate at either end being 



\pL [PI. 14, Fig. 47.] 



The moments, as we have already seen [Art. 44, Fig. 30], are 

 given by a parabola whose vertex is in the centre of the span 



and whose middle ordinate is -3 p P. Since we have seen [Art. 



o 



70] that a load at any point causes at every point a negative 

 moment, the maximum moment at any point will be when the 

 whole span is loaded. 



78. method of Loading caning Maximum Shearing 

 Force. We have seen [Art. 70] that a single load causes at 

 any point a positive or negative shear, according as it lies upon 

 the right or left side of the cross-section at that point. Hence, 

 for a uniform load, 



The shearing force will be a positive or negative maximum 

 according as the load reaches from the right or left support to 

 the cross-section in question. For the positive maximum we 



have V x = p (I so) ' = p - = -. Therefore, max. -f S = 



'2. I i 



1 (l-xf 



For the graphical determination we can apply the method 

 given in Art. 72, Fig. 42, by which we have for max. + S and 

 max. S two parabolas whose vertices are at the ends of the 



span, and whose ordinates at these points are +~ and -t-~ 



Since, however, each point is found thus from the preceding, 

 the construction is not very exact. We may deduce a better 

 construction as follows. [PI. 14, Fig. 48.] Through any point 

 F of the curve drop a vertical intersecting A B in and the 

 line B K parallel to the tangent at F in G. Let the tangent 



at F intersect AB in H. Then CH = BH ; hence, CF = - C G. 



2 



We have, then, A E = 1 A D =-l^> Z. Since C F = - C G, 

 a ' 39 '2 



we have also AI =_ A K ; therefore, AI:AE;:AK:AD. 



2 



