70 



STRESSES IN A CRANE. 



Art. 51. 



unknown without finding a function of an angle, it being as- 

 sumed that li is a known dimension of the truss. 



The angle Q should always be measured from the direction of 

 the loads, and its functions should always be expressed as ratios 

 of lengths and thus used on the slide rule 1 . This necessitates the 

 calculation of the lengths of the diagonal members which may 

 easily be done by means of a table of squares. 



The stress D 2 is equal to D L because the truss is symmetrical. 

 They may also be gotten at joint 2 from D^ sinO = D 2 sinOj 

 hence D 1 = Z>, and 2D cos = P; Z> = y 2 P secB which is the 

 same as was gotten above since R^ = %P. Using a moment equa- 

 tion we have, 



^ 3 = y 2 ppD 1 pcose = o ; D = y 2 Pseco 



All stress diagrams are simply extensions of the above simple 

 case excepting a few apparently statically indeterminate cases 

 (42). 



51. Stresses in a Crane. Fig. 49 shows a crane with sup- 

 ports at A and B and carrying a single load P. 



The reaction at A is shown divided into its vertical and 

 horizontal components R and H. The reaction at B is assumed 

 to be horizontal. 



The vertical reaction at A must be equal to P from S Vert. 

 Comp. =0 7 and H at A equals H at B from 2 Horiz. Comp. =0. 



By moments about A, 



Hh=Pb and 



-jo- 



The stresses in the members may be obtained by resolutions 

 at the joints. 



Joint 2. U sina=H 



U=H-(tension) 

 b 



U cos a = V =H-- =H- (tension) 



b l v b 



Joints. Dsin[l=H 



D =H (compression) 







It is to be noted that those equations have been chosen which 

 involve but two forces, and it thus becomes a very simple matter 

 to determine the character of the stress. Thus the horizontal 



x An exception to this rule is the case of = 45. sin 45 = cos 45 = 

 0.707; ton 45 = 1.000; sec. 45 = 1.414. 



