120 PRACTICAL STRUCTURAL DESIGN 



When the load is uniformly distributed | is carried by each strut, 



W 

 so in the above expressions substitute -5- for P. 



o 



When the horizontal member is in one piece and uniformly 

 loaded, each strut carries ^ of W. 



The above formulas are true for any depth of truss of the single or 

 double strut kind. The truss may be reversed so that the sloping 

 members are above and the long horizontal member is below. 

 Then the lower member carries tension and the upper members 

 are in compression. The methods of computation are not altered 



...L. .j 



3 1 



Fig. 77. Double Strut Belly Rod Beam 



except that the vertical pieces are ties instead of struts. With 

 the load applied at the upper end the ties carry no stress from the 

 load but are used merely to maintain the horizontality of the 

 lower chord. With the load applied at the lower end each vertical 

 carries the amount it would carry if the beam were inverted. With 

 a double-tied beam the tie serves to hold the frame together in 

 case of a rolling load, or a load applied other than vertically, in 

 which case it does carry stress. Diagonal counters set between 

 the ties will take care of such stresses and the ties merely serve 

 to hold the frame together. It is advisable to have ties many 

 times in trusses when an analysis shows they are not stressed, 

 in order to carry the weight of the lower chord. If the lower 

 chord must carry all of its own weight, or any load between sup- 

 ports, bending and shearing stresses will be set up in the lower 

 chord in addition to the direct tensile stress. This is one reason 

 for making all tension members of metal when possible. 



Dimensions are on center lines. The tension rods should go 

 through the ends of the compression member at the neutral axis. 

 The plates at the ends should be normal to the direction of the 

 tie. The area of each plate is obtained by dividing the tension 

 in the rod by the allowable safe unit compressive stress on the end 

 of the wood. 



The area of each strut is obtained by dividing the compression 



