RESTRAINED, OR BUILT-IN, BEAMS 



and, inserting this value in the first equation, we have 



(97) R, = P l - 



85 



3 Q 



I* 



Also, inserting this value of R^ in the expression for the moment, 

 it is found that 



(98) 



' 



Similarly, by forming the expressions for the total deflection of the 

 point E with respect to the tangent at A, we obtain the equations 



7? 7 3 P7 3 P7 2 7 

 zi 2 t **! ^Ma c\ 



(99) 



2 

 2 El 3 El 3Ji 



g/ PI? 



~EI^~ 2EI 2 j/ 



= 0; 



and, solving these equations simultaneously for M 2 and R Z , as above, 

 the results are 



(100) , = !>*: 



(101) 



If AC is the longer of the two segments, the maximum deflection 

 will occur somewhere between A and C. Also, since the tangent at 

 this point must be horizontal, the total angular deflection from one 

 end, say J, to the point of maximum deflection is zero. Let x 

 denote the distance of this point of maximum deflection from A. 

 Then, computing the total angular deflection up to this point and 

 equating it to zero, we have 



whence 



Inserting the values just obtained for M^ and 



2^ t 

 = *. + 3 1 ' 



this becomes 



(102) 



