190 On the Flexure of Continuous Girders. 



Then, from (4), 



_ M«-M, p »225P 



M 3 -M 2 31P „M 3 -M 4 _9P 



,_ M 4 -M 3 _ 9P 



^ 3 ~ / "256' 



and from (5), 



225 P , 40P r 9P^ 



^2-^2- j> 56 > R 3 = S 2 + S 3 = 256 ^ R 4 =S 3 =- 256 - 



For the inflection-point in the unloaded span, we have from (6), 



M 3 -S 3 # 3 = or x 3 =^=-l; 



for the one between the abutment and P, 



*> = S7 = 225 /=(H73/i 



for the one between P and the pier, 



M 2 -P« 25 nurk£;7 

 ^=-S^p- = 31 / =°- 806 ^ 



7. To find the greatest deflections in both spans of the girder 

 in the last problem. 



The equation of the elastic curve between P and the pier is, 

 from (II.) (since the tangent at the fixed end is horizontal, £ = 0), 



6EIy = M^ 2 -S 2 ^ 3 + P(^-fl) 3 . 



Substituting the values of M q , S 2 , and a, and differentiating, we 



find y to be a maximum when # = 0-413/ ; substituting this, we 



have 



P/ 3 

 y =00034 ^r for tne value of the maximum deflection. 



The equation of the curve for the unloaded span is 



6EI?, = 6EI t 6 x + M 3 x - S 3 #, 



where t 3 is given by (III.) or (IIP.), or 



6EI* 8 = -2M 8 /-M 4 /= -9 PI 2 . 



Substituting t 3 , M 3 , and S 3 , we find that y becomes a negative 

 maximum for x=-\l) then from the equation of the curve we get 



P/ 3 

 y= — 0*0009 y, r - as the minimum deflection. 



