122 SUPPLEMENT TO CHAP. VIL [CHAP. HI 



y= f4 V (3 lx) p (6 I* *-. 



o 



For a = Z, y = 0, and hence V = j> Z. 



8 



Substituting this value of V 



88 E I 



d y __ p as 



(6f-15Za;+8aj 9 )- 



_ 



Putting this last equal to zero, we find as - - I, or x 0.5785 I, 



16 



for the value of a*, which makes the deflection a maximum, and this in- 

 serted in the value of y gives for this maximum deflection itself, 



-^^=0.0054 I?. 



For the "breaking weight, we have, since the greatest moment is at the fixed 



1 .1 72 __1 72 2TI. 16 T I 



end and equal to p r, M = -p I * = hence p I = - . 

 8 8 n hi 



The strength is, then, f times as great as for the same load in the middle, 

 but no greater than for a beam of same length and load supported at both 

 ends. 



1. Beam fixed at botb ends Constant cross-section 

 Concentrated load. Taking our notation as before (Art. 12), we 

 have in this case not only a reaction at the right end, but also a positive 

 moment there as well, both of which must be found. If l\ be the distance 

 from left end to weight, and It from weight to right end, and if Vj and 

 V z are reactions, Mi and M a the moments at left and right ends respec- 

 tively, then for equilibrium we must have Vi + V s = P, M, + V a li = 

 M,-V S Z,. 



<Z 2 y 

 For <e less than Z, we have M = M, + Vi a = E I - ~ . Integrating 



Cl X 



d *z/ 

 once, since the constant is zero, because, for a = 0, , y = 0, we have 



d x 



^= 1 [~2M 1+ V ia! ~k 

 dx 2E I L J 



Integrating again, since, for x = 0, y = 0, and the constant is zero, 



For the distance from the right end to the weight we may obtain similar 

 expressions, if we take that end as the origin, only we should have V, 



and M a in place of Vi and ML At the weight itself -~ and y must in 



d z 



each case be equal, but -y-^ of opposite sign. Therefore we have the equa- 



d *c 



tions (2 M, + V, Z.) ^ = -(2 M 2 -V, l t ) I,, 



(3 M, + V, J,) Z, s = (3 M 4 ~V a I,) IS. 



