Art. 52. STRESSES IN AN UNSYMMETRICAL TRUSS. 



71 



component of U at joint 2 must act away from the joint to 

 balance H. 



Fig. 49. 



The stresses may be checked by writing other equations of 

 equilibrium, as follows : 



Joint 1. U sin a =D sin /?. 



U cosa+P=Dcosp. 

 Joint3. Dcosp=P + V. 



M l =Pb-H(a+h)+Vb=Q. 



The above equations are but solutions of the triangles in the 

 stress diagram. The force polygon of external forces is abaca. 

 The force polygon for joint 1 is PDU; for joint 2, HUV; and 

 for joint 3, RHVD. 



52. Stresses in an Unsym metrical Truss Unsym- 

 metrically Loaded. Fig. 50 shows a truss of three equal panels 

 with parallel chords, and supports at joints 1 and 7. The 

 reactions are readily found as explained in Chapter IV. 



Graphically the force polygon for external forces is abcda. 

 The force polygon for each joint is easily followed and the signs 

 of the stresses determined. These are also readily found by 

 inspection because the truss is a beam supported at its ends; 

 it bends so that the upper chord is shortened and the lower 

 chord lengthened; hence the upper chord members are in com- 

 pression and the lower chord members in tension. The diagonals 

 are determined by the shears in the panels; thus, the shear in 

 panel 1-3 shortens D v the shear in panel 3-5 (2000 Ibs. up) 



