162 _ CONSOLIDATED STEEL CORPORATION _ __ 



DEFLECTION OF STEEL BEAMS AND GIRDERS 

 UNDER TRANSVERSE LOADS 



Using the notation given on page 159, the deflection, in inches, of a steel 

 beam or other section under a uniformly distributed load of W, in pounds, 

 is found from the formula 



JL WP_ 5 



~ ~ 



_ 

 38 U EI~38U El 



When W is the safe uniformly distributed load corresponding to a coeffi 

 cient of strength C, the following relations exist between W and C and the prop 

 erties of the shape: 



W=^ and C-H/S-H/- 



Substituting these values in the above formula, then, 



nE 



When the fiber stress is I6,004) pounds per square inch and the modulus of 

 elasticity of steel taken as 29, 000, 000, then the deflection, in inches, is given 

 by the formula 



0:01 655 L 2 

 2n 



In the case of a beam, girder or other section symmetrical about its neutral 

 axis, 2n equals the depth of the beam. The deflection, in inches, of such a sec- 

 tion under its safe uniformly distributed load which produces a fiber stress of 

 16,000 pounds per square inch is given by the simple formula 



0.01655L* 1 U 



D= j- or very nearly = w~d' 



The table on the opposite page gives the value of the expression 0.01655L 2 

 for spans from 1 foot to 60 feet. 



The safe loads and corresponding deflections for other usual cases of load- 

 ing, as compared with the safe uniformly distributed loads given in the tables, 

 are as follows: 



Beam supported at both ends and loaded with a single load concentrated 

 at center of span. Safe load = ^ tabular load. Deflection = T V 



Cantilever beam, fixed at one end and unsupported at the other, uniformly 

 loaded. Safe load = \i tabular load. Deflection = 2 T V 



Cantilever beam, fixed at one end and unsupported at the other, single load 

 concentrated at free end. Safe load = Y* tabular load. Deflection = 3 T V 



