CHAP. III.] SUPPLEMENT TO CHAP. VH. 121 



Now, for x = I, y' = ; hence V = P i -75 - -. If the load is in the 



<u If 



middle, V = P. 



V, or the reaction at the free end, is now known, and substituting it in 

 the value of y' above, we have the equation of the deflection curve between 

 the weight and the free end. 



Substituting it also in the value of above, and placing then -r-^- equal 



dx dx 



to zero, we find for the value of , which makes the deflection a maximum, 



/I a 



when x is greater than a, x I Zl/ = - . 



8 l> ~~ Ok 



Substituting this value of x in the value of y' above, we have for the 

 maximum deflection itself 



A = 



F Z 3 1 



When the weight is at the middle, this becomes A = x , or 



48 E I 1/5 



only : , as much as for a beam of same length fixed at end and with 



load at other end, and only -= as much as for same beam simply sup- 



r 5 

 ported at ends. 



Breaking weight. 



Having now V, we know M and M'. Rupture will occur where the 

 moment is greatest, that is, either at the fixed end or at the weight. Now 

 the moment at P is V (Z a) = V 1 + V a. The moment at the fixed 

 end is V Z + F a. Now, as V is always less than P, we see at once that 

 for any value of a less than I, the moment at the weight is greatest. 



We have for the moment at the weight from (8) 



rr n \ -r, * ( 3 Z-) /7 \ 2 T * A t, 



M = V (I a) = P 5^-jj : (I a) = ^ , and hence 

 2 I / 



4 TI Z 3 



for the breaking weight P = - ,= -. 



n a* (3 I a) (I a) 



64 T I 

 If the weight is in the middle, P = -T-T, 



or f ths as much as for the same beam supported at the ends. 



17. Beam as before fixed at one end and supported at 

 the other Uniform load. In this case the moment at any point is 



1 d 2 y 



M = V (Z a) + -p (Z )* = E I -=-^|. Integrating twice and determin- 



& d 2! 



ing the constants by the conditions that, for = 0, = and y = 0, we 

 easily obtain 



