irt. 55. 



STRESSES IN A CANTILEVER TRUSS. 



75 



tion at A must be equal to the sum of the two loads. By taking 

 moments about A, 4# =8000X6 +6000X3 =66000 ft. Ibs., from 

 which H = 16500 Ibs. The H at A must equal the H at B, as 

 they are the only horizontal forces. 



For the algebraic solution the functions of the angles should 

 be expressed in terms of the lengths of the members of the 

 truss (50). The diagonal length here = \/3- f2 2 =Vl3=3.6 ft. 



sin 6 = cos 6 = 

 3.6 3.6 



By resolutions at the joints the following stresses are found- 

 Joint 1. L 1 cos 0=8000 1^=8000 = 14400 Ibs. (+) 



L 1 sinO = U l = SOOO- X - = 12000 Ibs. ( -) 

 & 0.6 



Joint 2. F!= 6000 Ibs. (+) 



U 2 = Ut = 12000 Ibs. (-) 



q fi 



Joints. L 2 sin 0=H = 16500 Ibs. L 2 = 16500 =19800 Ibs. (+) 



L 2 cos 0- 14000 = V 2 = 19800 -14000 =-3000 lbs.(+) 



3.6 



Joint 4. Z) j cos = F 2 =3000 1^=3000 =5400 Ibs. (-) 



2 



=H = 16500 Ibs. 



D, = (16500- 12000) =5400 Ibs. (-). Check. 



o 



Graphically, using Bow's notation, the force polygon of 

 external forces is abcdea. The stress diagram may be con- 

 structed, beginning either at joint 1, 4, or 5. The scaled stresses 

 are given on the truss diagram in 1000-lb. units. 



56. Stresses in a Compound Fink Roof Truss for 

 Vertical Loads. Fig. 53 shows the case mentioned in Art. 42 

 as being apparently statically indeterminate. The external force 

 diagram is laid out and the stress diagram started, beginning 

 at joint 1. There is no difficulty until joint 4 is reached, when 

 it is found that there are more than two unknowns at each of 



