STRESSES IN A DERRICK. 



Art. 62. 



V=P+P- 



c 



b=l sina 



-P . 



From this equation it is evident that for a maximum stress 

 in the mast, c should be a minimum. This occurs when the boom 

 is swung so that the angle B'G'F' is 90. 



Also if the boom cannot drop below a horizontal position 

 it is evident that the maximum stress in the mast occurs when 

 a =90, or the boom is horizontal. (Sina is maximum.) 



If it is possible to drop the boom below a horizontal posi- 

 tion the position for a maximum stress in the mast may be 

 found by differentiating equation (c) and setting the first differ- 

 ential equal to zero. 



d(BA) PI , PI . 



= cosa H sina =0. 

 da c h 



cos a 

 c 



sin a sin a h 



or =tana = . 



h cos a. c 



This gives the value for a for a maximum stress in the mast. 



Maximum stress in a guy. The position of the boom in 

 elevation should be such as to produce a maximum vertical 

 component to the resultant of the stresses in the two guys in 

 action. 



Vert. comp. of restulant =P . 



From this it is seen that the boom should be horizontal. 



The position of the boom in plan may be obtained from 

 resolutions at B' normal to one of the guys, using horizontal 

 components. To obtain the stress in B'D' we will take resolu- 

 tions normal to B'F 1 ' . 



D'B' cos (0-90) =B'C' sin 0, 



D'B' =B'C' 



cos (0-90) 



This is evidently a maximum when sin is a maximum, or when 

 0=90. In words the stress in one guy is maximum when the 

 boom is rotated so that it stands perpendicular to the hori- 

 zontal projection of one of the other guys. 



