252 



REPORT 1865. 



For the position of the neutral axis NO we 

 have by equation (1), art. 18, 



7l =£(«o a o +«!<*, + tf 2 aj, (1) 



where a , a v a 2 are the areas of the material in the 

 sides, in the top cells, and in the bottom plates re- 

 spectively, and so on. 



If D = AC, the depth of the beam, then 

 7 h =J)-h. 



From equation (2), art. 18, we get 



I =Q + Q 1 + Q 2 +«„(> t -O 2 



+ a 1 (h-* i y + a 2 (h-ccj, (2) 



where Q =the moment of inertia of the side plates 

 about their centre of gravity, Q 2 that of the top 

 cells, and Q, 2 that of the bottom plates. 



If the beam be loaded in the middle and sup- 

 ported at the ends, then M=|WZ, and 



iWZ=M=|l or^I ; 



Fig. 12. 



W= 



4SI 



-J or Sk 



hi ° \l ' 



which gives the load corresponding to any assumed value of S or S r 

 versely S or S x may be determined for any given load, W. 



(1) The following is a more simple and practical method of calculating 

 (approximately) the strength of these beams. 



Putting a, rtj for the areas of the top and bottom parts respectively, g, g i for 

 the distance of their respective centres of gravity from NO, and (i=g-]-g 1 , 

 neglecting the side plates, for the position of the neutral axis, we have 



a ff= a iffi 



And from General Theorem II., art. 19, we get 



Io =a 9*+ a i9i*= a 9(9+9i) or a i9,(9+9d 

 = agG or a^fi ; 



.-. M=t«5 , G or j^a^gfi 



(4) 

 (5) 



= SaD X |jj or S^D X j±q 



(6) 



=S«D or SjfitjD nearly, 



taking K~. and 'Apr as constants, each being nearly equal to unity. 



Here the strength varies as the top or bottom areas multiplied by the depth. 

 If K be put for the whole section, and taking ~rfr= I -^ — a constant, we 



get 



M=C(a+a 1 )D=CKD 



Here the strength varies as the whole section multiplied by the depth. 

 When M=5WZ, we get 



_ 4SaD 4S,«,D 

 W= — 7— or 



(7) 



I 



I 



(8) 



