192 WROUGHT IRON AND STEEL BEANS. 



load of about 16 or 17 tons. So that, taking the readings 

 up to that point, and taking the differences as before, the 

 following results are obtained : 



0102 G'096 G'090 0084 0-077 



0-052 0-046 0040 0'034 0'027 



0-050 0-050 0-050 0-Q50 0050 



0-071 0-065 0-058 



0-020 0-013 0-006 



0-051 0-052 0052 



Taking the mean of these 



0-050 



0-050 



0-050 



0-050 



0-050 



0-051 



0-052 , 



0-052 



8)0-405 



./I* 



8 tons increment of load. 



o-O'Ofi/ * ncne8 average deflection per 



For a section such as the one in question, which is not 

 perfectly uniform, the best way to obtain the Moment of 

 Inertia is by the graphic method, but here it will be 

 sufficient to make use of the following approximate 

 formula : 



_ B H 3 - b h* 



12 



Here B = 6 in. 

 H = 6 

 6 = 5,, 

 h = 5 

 the thickness being | in. 



So that, T (6) 4 - (5)* 



12 

 1296 - 625 



"S" 



The values of the symbols in the formula are therefore 

 W = 8 x 2,240 = 17,920 Ib. 



I = 60 in. 



8 = 0-0506 in. 

 and I = 55 '9. 



