murphy: maximum bending moments. 151 



21 



Solving (34) for z we have z=7;o and z=— , . 



K 



The value z=o shows that the moment at the right hinge is zero for 



all value of x. The second value of z is a function of x whose limits 



1 1 X . 



from Table II are |1 when x— o, and — when x= — . As -— - varies 

 •^ 2 2 1 



from — to 1 there is only one point of zero moment on the right, viz., 

 2 



the hinge. Hence this second point of zero moment on the right moves 



from |1 to 41 as x increases from zero to — . 

 5 2 2 



From Mi=o in (27) we have 



( wx^ . , , ,x I w ( x^k' ) 



^ 1 ^" — ^^'+^ > s T ' ' — i^ r=° ^35) 



Solving (35) we have 



4x|x-^(.+k')[ 



z=.o andz=- , .a • (35) 



2~ - ^ '- k' 



" (IS' 



The value z=o shows that the moment is zero at the left hinge for 

 all values of x. The second value of z can be put in the form 



z ■ X ' 4-^(2 + k')l 



and from Table II we find that the ratio —r- is greater than the ratio 



1 "^ 



X X 



— r- until ——=1, that is, this point of zero moment is not under the 



load until x becomes equal to — . As x varies from — to 1 this point 



,2 2 



moves from — to |1. That is, this movable point of zero moment 



moves from |1 to |1 while x varies from o to 1. 



The maximum moment may be under the load or to the right of the 

 load. 



To find the maximum moment on the right of the load we have 



WX" ( z I X- 



M„= I ,- ■ — w--k'(zl— z3) 



dMj. wx2 wx2 



-~k (1 — 2Z) 



dz 2I 4I 



2 



]■ + ¥')■ (36) 



