224 



Kansas Academy of Science. 



from the top of the beam, and a similar rod at the same distance 

 from the bottom. Then b = 4, h = 16, di = dii = l|, As = A,i = .7854 

 sq. in. Assume E, = 29,000,000 and Ee = 2,900,000. Then e = 10. 

 Substituting these values in the author's formula (1), we have: 



A 16- 



7854(1. 5)2 +.7854(14. 5)2 



= 25.8 inches, 

 we have 



64 +(1.5708)9 

 Substituting these same values in our new formula, 



162 



4 + ^ + 



y = 



. 7854X 1.5 + . 7854X 14 . 5 9 



■4 



64 + (1 . 5708) X 9 



which is the correct result, since our beam was assumed symmet- 

 rical. Probably the reason that the author did not discover such 

 an obvious mistake is because he assumed in his numerical ex- 

 ample a beam having no reenforcement on the upper side. And it 

 is in the square of the large term h — di^ that the greatest error ap- 

 pears. Now, in the case of a built-in beam, reenforcement would be 

 put in the top of the beam for at least part of its length. 

 The next formula that requires comment is formula {8) 



12 



Ic.= ^-(a,V^ + A,H'^) 



This should be L 



f +bh(|-y) -(A,v-' + A,H'^ 



That is, in order to subtract one moment of inertia from another 

 we must have them about the same axis. From the moment of 

 inertia of the entire area about its center of gravity the author sub- 

 tracts the moment of inertia of the steel about the neutral axis of 

 the beam. This is correct only when these two coincide. And 

 this is not generally the case. The extra term may or may not be 

 a negligible quantity. 



If we omit the rod in the upper part of the beam of our preced- 

 ing example, we have b=4", h = 16", As=.7854 sq. in. 



A,i = o, di=l|''ande = 10. 



4X162 



+ 



.7854x1.5x9 



64 + .7854x9 

 v=y_di = 7.37-t.5=:5.87" 



7.37" 



I 



