164 



CAMBKIA STEEL. 



BENDING MOMENTS AND DEFLECTIONS FOR 

 BEAMS OF UNIFORM SECTION. 



W = Total Load, in Ibs., uniformly 

 distributed, including the weight of 

 beam. 



Wi = Total Superimposed or Live 

 Load, in Ibs., uniformly distributed. 



Wj = Total Weight of Beam or 

 Dead Load, in Ibs., uniformly dis- 

 tributed. 



P, Pi, P 2 , Ps = Loads, in Ibs., con- 

 centrated at any points. 



M = Total Bending Moment.in inch-lbs. 



M w i,M p =BendingMoments,ininch-lbs., 

 due to Weights Wi and P respectively. 



I = Moment of Inertia, in inches 4 . 



1 = Length of Span, in inches. 



E = Modulus of Elasticity, in Ibs., per 

 square inch = 29 000 000 for steel. 



W a = Total Safe Load, in Ibs., uni- 

 formly distributed, including the weight 

 of beam = Total Safe Load of Tables. 



The ordinates in diagrams give the bending moments for corresponding points 

 on beam. For superimposed load only, make W2 in formulae equal to zero. 



(7) Beam Supported at both ends 

 with Loads Concentrated 

 at various Points. 



CD E 



@p QPI Q P * 



The total bending moment at any 

 point produced by all the weights is 

 equal to the sum of the moments at 

 that point produced by each of the 

 weights separately. 



Diagram for Dead Load similar to 

 Case (1). 



The Maximum Bending Moment occurs 

 at the point where the vertical shear 

 equals zero and will be at one of the 

 loads P, Pi, or P 2 depending upon their 

 amounts and spacing if W2 is neglected. 



Let R = Reaction at Left Support. 

 Bending Moment at P = 



Bending Moment at Pi = 

 p, - Rai - P^ + P (a, - a)] 

 Bending Moment at Pi = M p2 = Raj 



Shear or Reaction at Left Support 

 P 2 b 2 +Pibi + Pb , W 2 

 1 "*" 2 ' 



Shear or Reaction at Right Support = 



P 2 a 2 



+ Pi at + Pa W 2 



1 "*" 2 ' 



Diagram for Superimposed Load: 

 Draw as in Case (5) the Ordinates FC, 

 GD and HE representing the bending 

 moments due to loads P, Pi and P 2 re. 

 spectively. Produce FC to P, making PC 

 = FC + 1C + JC; GD to Q, making 

 QD - GD + KD + LD; and HE to R, 

 making RE = HE + ME + NE. Join 

 the points A, P, Q, R and B, then the 

 ordinates between A B and polygon A P 

 QRB will represent the bending moments 

 for corresponding points on beam. 



