180 



CAST-IRON BEAMS. 



To find the " equivalent breaking load " on the standard 

 2 in. x 1 in. section. We know that 



(width) x (depth) 2 

 span 



is proportional to the breaking load in the centre. Or, if 

 Wx is the actual breaking load and W is the similar load 

 for a 2 in. x 1 in. section and 36 in. span, then, 



Ix2 2 



I 



where b, d, and I are the width, depth, ^and span respec- 

 tively of the broken beam. 



w w *i_ 



I l-|* \\f VV 



L) CL oO 



As the span I can generally be varied at will, and is 

 usually set to 3 ft., I may be taken = 36, and the equation 



reduces to 



iir A. 



W = 



bd* 

 In the above example, 



W 1= 3125 Ib. 

 b = 1-04 in. 

 d = 2-03 in. 



So that, using these values, 



3125x4 



W = 



1-04 x (2-03) 2 

 = 2910 Ib. 



= 26 cwt. 



The results thus obtained may be put on a report form 

 somewhat as follows : 



REPORT OF TEST OF CAST-IRON BEAM. 



