STRENGTH OF MATERIALS 



and 



At A, x = and y = ; therefore C 2 = 0. In order to determine d it is necessary 

 to find the equation of BC. 



Taking a section on the right of B, M = - j ' and consequently 

 nr ffiy Pd(l-x) 



Pdl' 2 

 At (7, x = I and y = ; therefore C 4 = -- C 8 i. 



o 



Now at B both branches of the elastic curve have the same ordinate and the 

 same slope. Therefore, putting x = d in the above integrals and equating the slopes 

 and ordinates of the two branches, 



Solving these two equations simultaneously for Ci and C, 



Substituting these values of Ci and C 8 in the above integrals, the equation 

 branch AB becomes, after reduction, 



and the equation of BC becomes 



Since the load is not at the center of the beam, the maximum deflection will 

 occur in the longer segment. Moreover, at the point of maximum defl*'-ti>ii tin- 

 tangent is horizontal, that is, = 0. Therefore, equating to zero the first differ- 

 ential coefficient of the branch AB, 



