Art. 127. 



STRESSES IN SIMPLE TRUSSES. 



237 



method of sections is simpler than the other method, for any 

 particular stress, it should be used. Involved numerical work 

 should be avoided in order to reduce the probability of error (40) . 

 Some general principles will be deduced which will determine 

 positions of live load for maximum stresses in certain simple 

 trusses; but it should be remembered that the simplest way of 

 finding maximum live load stresses, and one which is applicable 

 to any case, is to calculate all stresses for each panel load sepa- 

 rately and then to make combinations of stresses from such panel 

 loads as may reasonably be supposed to act at the same time. 

 If certain panel loads produce compressive stresses in a mem- 

 ber, and all of the others produce tensile stresses, the sum of 

 the former gives the maximum compressive stress and the 

 sum of the latter the maximum tensile stress. The following 



Fig. 175. 



example applied to the web members of the truss of Fig. 175 

 will illustrate this. 



The stresses are to be found for a dead load of 300 Ibs. and 

 a five load of 800 Ibs. per lineal foot of truss. 



Dead load panel load=300X20= 6000 Ibs. 

 Live load panel load=800X 20=16000 Ibs. 

 The dead load is, of course, a full load, there being five panel 

 loads of 6000 Ibs. each, which, for the sake of simplicity, are as- 

 sumed to act, all at the lower joints. 

 Dead load stresses. 



B 1 =B S r=2%X 6,000=15,000 Ibs. 



Sect. 1-1. D 1 =R 1 sec. 45. D 1 =15,OOOX 1.414=21,200 Ibs. 



Sect. 2-2. Z> 2 =(K 1 -P 5 ) sec. 45. Z) 2 = 9,000X1-414=12,700 Ibs. 

 Sect. 3-3. Z> 3 =(^ 1 -P 5 -P 4 )sec.45. 



D s = 3,000X1-414= 4,200 Ibs. 

 Since the shear acts upward on the left of each section, D 1 



