murphy: maximum bending moments. 



149 



From the equation of the "x displacement we have 



rMiydz+ I Mrydz=o (22) 

 O *' X 



Substituting for Mj and Mj. their values and simplifying we have 

 — — I z2ydz-|-Vj I zydz — wx I -. z — " ydz — H I y^dz^o 



2»/q ^' o •'^x' 2) •^o 



(23) 



Substituting for y its value in (23), integrating and simplifying we 

 have 



wx^P wx^l wx5 



=t'51^H13 (24) 



24 



24 



60 



From (24) we find 



H: 



—5 



l^.l 



i6h ( L 1 I ^11 

 Table II gives values of k' for values of 



TABLE II. 



+ 5 



X 



T 



i6h 



k' 



(25) 



1 



.01 

 4.999 



.1 

 4.952 



.2 

 4.816 



.3 

 4.601 



.4 

 4.328 



.5 



4.000 



.6 

 3.632 



.7 

 3 236 



.8 

 2.824 



.9 



2.408 



1.0 

 2.000 



Substituting for Vj and H their values in (20) and (21) we have 



M, 



M, 



:2 ( z I ^^ wx' 



-1 ^- -r r-Hy=— r 



41= 



-k'(zl— z2) 



(26) 



Hy-j 



X2 [ 



2I ) 2 ' { 2\ \ 2 4I2 



k'(zl— z2) (27) 



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

 when H is a maximum we have 



dH x^ 



10^- 



dx 



-20— +IOX=:0 



(28) 



From (28) we find x=o, x=l and xr=^±|| '5. 



The second of these, viz., x=l satisfies the conditions of the prob- 

 lem and makes the second derivative of H negative; hence, H is a 

 maximum when the load covers the whole span. 



Equation (26) can be put in the form 



Mr= 



wx^k' ( 8h 



i6h Ik' I 



z 1 \ „ i 8h 



(29) 



