murphy: maximum bending moments. 



145 



Table I gives values of k for values of 













TABLE I. 













X 



T 



.01 



. I 



. 2 



•3 



•4 



•5 



.6 



■7 



.8 



•9 



1 .0 



k 



.ugy 



.980 



.928 



.847 



■744 



.625 



.496 



•363 



.232 



. 109 



0.0 



Substituting for H its value in (i) and (2) we have 

 Mr=Px|i-^[-^(zl-z^)k. 



(5) 



Mi=P 



1 



5Px, , 



>)k 



(6) 



Differentiating H with respect to x to find the position of the load 

 when H is a maximum, we have 



dH ,p i .rxy3_^^x^ 



dx 



h-]*UJ "Htj +'f 



(7) . 



solving (7) we have -t = >^ and ^^±^1/3- The first of these roots 



is the only one which satisfies the conditions of the problem and since 

 it makes the second derivative of H with respect to x in (7) negative 

 it corresponds to a maximum value of H. H is a maximum theii, 

 when the load is at the middle of the span. 

 Equation (i) can be put in the form 



M. 



\- 



=Hf^ 



v^v- 



•y 



(8) 



k . ^ I 



_ IT J 



The second factor in (8) is the distance from the moment curve to 



the rib and the first term of this second factor is the distance from the 



' ax:is of x to the moment curve. This term being of the first degree in 



z the moment curve on the right of the load, for any position of the 



load is a straight line. 



In the same way (2) can be put in the form 



,^ ^^ ( 8h(l— x) ) , . , , , , 



Mi=H - — ^— — -z — y r which shows that the moment curve on the 



,.left, for any position of the load, is a straight line passing through the 

 origin. ' 



To find the locus of the intersection of these two moment curves: 

 > Let the ordinate of the moment curve be denoted by u, then 



These two ordi- 



u,=- =-(l-x), and u 



Mr 

 ll7 



8h 



