COLUMNS FIXED AT THE BASE 89 



Now (38) equals H I times the deflection of the column from 

 y z= d to y = h f which equals zero by hypothesis. 

 Solving (38) we have 



3 hd* 



Zd * (39) 



z ' z 



In a beam fixed at one end there is a point of inflection at some 

 point, between y = o and y = d, where the bending moment equals 

 zero. Now if y equals the value of y for the point of inflection, we 

 have from (32) 



B (d y ) =C (h y ) and 



= 



Equating the second members of equations (39) and (40) and 

 solving for y , we have 



y*=r} 



To find the relations between y and d, we will substitute h in 

 terms of d in (41) and solve for y . 



For d= h y=\d 



& o 



1 = 1*. y. = \* 



Solving (31) and (39) foi C, we have 



c= H^ 3d* _ 



To find the moment M b at the base of the leeward column, we 

 have from (32) 



M b = B d C h 



