CAMBBIA STEEL. 161 



GENERAL FORMULA FOR FLEXURE OF BEAMS. 



(CONTINUED.) 



DEFLECTIONS. 



(1) Beam supported at both ends and Uniformly Loaded: 



5 W13 5 (Wi + Wz) 1 

 Deflection for Total Load = D = ^ -g j = ^ - g| - 



Deflection for Superimposed Load = Dw t = ^ -~ 



(2) Beam supported at both ends with load concentrated at the middle: 



PI* 5 W2l 3 

 Deflection for Total Load = D = 



P1 J 



Deflection for Superimposed Load = D p = 



4OE/1 



(3) Beam fixed at one end, unsupported at the other, and Uniformly Loaded: 



Defection for Tota, Load - D - | = < + 



Wil* 

 Deflection for Superimposed Load = Dw t = -ggj 



(4) Beam fixed at one end, and unsupported at the other, with load concen- 

 trated at the unsupported end: 



Deflection for Total Load = D = g^j + ggj 



PI* 



Deflection for Superimposed Load = D p = 5=7 



KE] 



. where W = (W^ + W^ = 1 000 pounds and 

 oo oo i/ 



1=12" 



N' - -!^L where P = 1 000 pounds and 1 = 12" 

 4or!/l 



Total Deflection, in inches, due to a Beam Uniformly Loaded for any span in 

 _ n - NWL3 N ( 



~ 1000 = 1000 



Total Deflection, in inches, due to a Su.^ 

 Beam W 2 for any span in feet = D = YQQQ + 



FOB SYMMETRICAL SECTIONS. 



Total Deflection, in inches, for a fibre stress of 16 000 Ibs. per square inch 



-f 



Total Deflection, in inches, for a fibre stress of 12 500 Ibs. per square inch 



D= f 



FOB UNSYMMETBICAL SECTIONS. 



Total Deflection, in inches, for a fibre stress of 16 000 pounds per square inch 



D=S 2| 



Total Deflection, in inches, for a fibre stress of 12 500 pounds per square inch 



