CHAP. VH.] SIMPLE GIRDERS. 9ft 



Hence the following construction : 



Make A E = ]-p I 



a 



Divide A E and A B into an equal number of equal parts, and 

 draw lines from B to the points of division of A E, and verticals 

 through the points of division of A B. The curve passes 

 through the points of intersection of corresponding lines. 



79. Uve and Dead Loads. Let p be the load per unit of 

 length for dead, and m for live load. The maximum moment 

 for any point will be as before. 



M= -(j?+m) x (lx); that is, will be 



given by a parabola whose middle ordinate is -- (p+m) P. 



8 



For the shearing force, we have 



Max. + S = \p (Z-2 a)+i| fe^j 

 2 2 L 



1 1 or 5 



Max. S =- p (12 x)-m -j. 

 2i 2i L 



Indicate A C [Fig. 49, PI. 14] by a?!, for which max. S= 0. 

 then 



=p I (12 xjm a?! 8 , 



or 



m m, 



hence 



. 



I m m 2 m 



For the point D for which max. +S = 0, B D = a? t . The 

 shearing force within A C is positive, within B D negative, 

 while within C D it is both positive and negative. 



For 1 = 5, 10, 20, 50, 75, 100, 150 metres. 



^ = 0.12 0.19 0.31 0.64 1.05 1.55 3.12 

 in 



= 0.24 0.29 0.33 0.38 0.42 0.44 0.46Z 

 CD = 0.52 0.42 0.34 0.24 0.16 0.12 0.08Z; 

 that is, C D diminishes with increasing span. 



Recapitulation. For girders of a length of about 100 feet 

 or more, then, we may consider the live load as distributed 

 per unit of length. The maximum shearing force can then be 



