CAMBRIA STEEL. 



Thus, to find, from the preceding table, the deflection in inches for 

 Cambria shapes used as Beams under their safe loads uniformly dis- 

 tributed including the weight of the beam : 

 Let D = deflection in inches. 



L = length between supports in feet. 



H = coefficient for deflection from table for fibre stress of 16 000 



pounds per square inch. 

 H' = coefficient for deflection from table for fibre stress of 12 500 



pounds per square inch. 



d = depth of beam in inches for symmetrical sections. 

 x x = distances in inches from neutral axis to most remote fibre 

 for unsymmetrical sections. 



FOR SYMMETRICAL SECTIONS. 



TT 



For fibre stress of 16 000 pounds per square inch D = 



TT I 



For fibre stress of 12 500 pounds per square inch D = 



d 



FOR UNSYMMETRICAL SECTIONS. 



TT 



For fibre stress of 16000 pounds per square inch D = 



2xj 



TT/ 



For fibre stress of 12 500 pounds per square inch D = 



2xj 

 EXAMPLES. 



Case I. To find the deflection of a 9" I-Beam weighing 30 pounds 

 per foot, for a span of 15 feet and a maximum fibre stress of 16 000 

 pounds per square inch, under its safe load uniformly distributed. 



From the above table the deflection coefficient for this case is found 

 to be 3.724 which divided by 9, the depth of the beam in inches, gives 

 .414, which is the required deflection in inches. 



The safe load for this beam under the conditions named is 16 100 

 pounds including the weight of the beam itself as stated in the Tables 

 of Safe Loads for Cambria I-Beams on page 109. 



Case II. To find the deflection of a 6" X 4" X 5" an gle, sup- 

 ported at the ends on its short leg as a horizontal base, for a span of 9 

 feet and a maximum fibre stress of 16 000 pounds per square inch under 

 its safe load uniformly distributed including its own weight. 



From the table of " Properties of Angles " on page 207 the distance 

 x' from the neutral axis to the back of the shorter leg is found to be 

 1.99 inches, which subtracted from the length of long leg, 6 inches, 

 gives 4.01 as the distance x x from the neutral axis to the most remote 

 fibre. From the above table the deflection coefficient for this case is 

 found to be 1.341, which divided by 8.02, twice x 1? gives .167, which 

 is the required deflection in inches. 



NOTE. For deflections of Beams and Channels due to any central or 

 uniform load see coefficients of deflection N and N' in the Tables of Properties 

 relating to these sections and the accompanying explanations. 



For deflections of any symmetrical beams due to various systems of loading, 

 see general formulae and diagrams on pages 160 to 165 inclusive. 



