:.62. 



STRESSES IN A DERRICK. 



85 



Fig. 62 shows a horizontal and vertical projection of a 

 derrick with three guys. This is the fewest number that the 

 derrick could have and be stable in all positions of the boom 

 unless some of the guys were stiff, or capable of taking com- 

 pression as well as tension. Also the maximum angle between 

 any two adjacent guys must not equal 180 or the derrick will 

 not be stable for certain position of the boom. 



The stresses in the members of the derrick will vary for 

 different positions of the boom, both horizontally and vertically, 

 and the problem is to find the position of the boom which will 

 give maximum stresses in each member. 



Maximum stress in the boom. The boom may be raised and 

 lowered by means of the tie BC. Taking a section mn through 

 the tie and the boom, and a center of moments at B, 



ACXh sin a = PXl sin a, 



AC =Stress in boom=P (a) 



h 



This is independent of the angle a, therefore the stress in 

 the boom is constant for any given load. 



Maximum stress in the tie. With the same . section as 

 before, and a center of moments at A, 



BCxh sin @=PXt sin /?, 



BC= Stress in tie =P 



(6) 



From this it is seen that the stress in the tie is a maxi- 

 mum when the length t of the tie is a maximum. This would 

 occur when the boom is hanging vertically downward, if such a 

 position were possible. Usually the boom cannot drop below 

 a horizontal position. 



Maximum stress in the mast. By considering the joint 

 B' in the horizontal projection, Fig. 62 (6), it is evident that 

 the guy B'E' is not in service as long as the angle D'B'C' or 

 F'B'C' does not exceed 180, also the resultant of the stresses 

 in the guys B'D' and B'F' must lie in the plane of the two guys 

 and directly opposite to the tie B'C'. Again considering joint 

 A of the elevation, by vertical resolutions, 



B A = Stress in mast = V -AC cos a. 



