Long Beams under Transverse Forces. 305 



+ B*/l- „„£„ „ + 



I (4E* 2 ) 7.6 T (4H/ 8 ) 2 . 13. 12. 7. 6 



« 6n "i 



± (±MY(6n+ 1)6/1. . . 7. 6 + ' ' * 'J =0 ' 



As in the case of the cantilever with single load make 



dd ; ' 



— equal to zero at the origin, and zero when x is equal 

 to Z; then B = 0, and the condition for critical stability, if R be 

 put for jg, is 



i-A+ R 



6.5 12.11.6.5 



± 6n(6n- 1) (6n-6) (6n - 7) . . . 6 . 5 + * ' * , = ' * * * 

 The first root of this equation is 



11 = 41-305.... 



VI. Another mode of loading, which is interesting as 

 leading to solutions in finite terms and as reducing to a well- 

 known result in a particular case, is that in which equal and 

 opposite forces parallel to the axis are applied at symmetrical 

 points in the plane of greatest rigidity. 



Let forces — P, + P be applied at the points (/, o, h) 

 ( — 1, o, h) respectively; then 



B a =-Py + PA0, I 



Thus 



and 



also 



G=-pA 



dx 



dd_ _?hdy 

 dx 7 dx* 



d*6_ _Vhdfy 

 dx" 2 <y dx 1 



J 



^/ = B_ 2 B, 

 Phil. Mag. S. 5. Vol. 48. No. 292. Sept. 1899. 



