Buckling of Deep Beams. 

 equal to zero. Then 



m s x 8 m 12 x V2 



305 



f ., ??l 4 A' 4 



}.(30) 



4.5.8.9 4.5.8.9.12.13 



T x = a \ l -^T + 17578" I" ' • (31) 



) zero when x = ^l. 

 Writing s for - 1 1 g??^ 4 / 4 the equation for s is 



Therefore 



This has to be zero when >v = U. 



1 _ s 



4 + 4.5.8 



4 . 5 . 8 . 9 . 12 



= 0. 



(32) 



It is worth while to show how this can be solved. 

 The equation can be written, after multiplying up by 

 4.5.8, 



-405= -160 + 



9 . 12 9.12.13.16 



that is 



(s-20) 2 = 400-160 + 



= 240 



9.12 



9.12 



Neglecting the cube and all higher powers of s we get 



s — 20= + s/ 240= + 15*5 approximately. 



Since we are seeking the smallest root of our equation, this 

 smallest root being the one that corresponds to the most stable 

 state, just as in the case of Euler's strut problems, we must 

 take the negative sign on the right. Then 



5 = 4*5 approximately. 



If we now use this approximate value of 5 in the terms 

 containing s 3 and s 4 , and add the correction for these terms 

 on to the 240, we get 



4.53 4.54 



(s-20) 2 = 240 + 



9 . 12 9 . 12 . 13 . 16 



240-83, 



5 = 20- V240-83 

 = 4-48. 



