106 



OBLIQUE LOADING. 



Art. 73. 



parallel to the roof while the load, which it carries to the roof 

 trusses, acts vertically. This is illustrated for a timber purlin 

 in Fig. 79. 



In order to find the maximum fiber 

 stress, it is not necessary to find the posi- 

 tion of the neutral axis. The bending mo- 

 ment is M and its plane is vertical. If it 

 is resolved into two components whose 

 planes contains the principal axes (as 

 shown), the extreme fiber stresses for each 

 Fig. 79. component may be calculated, according to 



equation (10), and the results added algebraically. 



Mcosa , Msina 

 JT 



(19) 



Mcosa produces compression at a and &, and tension at d 

 and c. Msina produces compression at b and c, and tension at a 

 and d. The maximum resultant compression will evidently be 

 at b and the maximum tension at d. 



A single channel is frequently used as 

 a roof purlin, as shown in Fig. 80. For a 

 quarter-pitch roof (rise equals one-fourth 

 of the span), a =26 34'. Knowing the 

 span and the load, M is easily calculated. 

 Assuming M = 68000 in. Ibs., in this case, 

 Msin a -- = 68000 X 0.447 = 30400 in. Ibs., 

 and Mcos a == 68000 X 0.894 == 60900 in. 

 Ibs. The other quantities of equation (19) 

 are given in the Cambria handbook. 



8 at b = -^9- 3 + -?^- 1.4 = 74800 Ibs. per sq. in. 



Fig. 80. 



s at d = ^^- 3 + -^_ 0.52 = 36600 Ibs. per sq. in. 



lo.U U. i 



At c and <i, the stresses will be less, because the two terms 

 will have opposite signs. 



Since these stresses exceed 16000 Ibs. per square inch, the 

 usual working stress, the channel is too small, although as com- 

 monly designed, the maximum stress is found to be less than 

 16000 Ibs. per sq. in. This stress is calculated as if the plane of 

 M coincided with that of the principal axis, in which case 



M 



68000 



3 = 15700 Ibs. per sq. in. 



J ' 13 



The above discrepancy may be much reduced, as it often is. 



