EXTERNAL FORCES 25 



The formulas for the reactions are the same as though the load 

 was concentrated at a point. 



For two equidistant loads on a beam, as in Fig. 18, the formula 

 for the reactions reads 



#!-/,= P + 2jp 



The moment due to the loads only is a maximum under each 

 load and is constant at all points on the beam between the loads. 

 M ( for loads only) = Pa. 



This can be proved. The loads being placed on the beam at 

 the same distance from the end, call each load P. Then the total 

 load - 2P. One-half of the 



total load goes to each end, /^K /\ 

 so the reaction must be H ^^ ^-^ h 



2P 



= P. The moment is 



equal to the reaction multi- Fig. 18 Two Equidistant Concentrated 

 plied by the arm through Loads on Beam on Two Supports as Shown 

 which it acts, therefore M = Pa. 

 Adding the weight of the beam gives us under the load, 



wa* 



,, 



M 



m 



in which w = weight per lineal foot of beam 



a - length of beam between load and reaction. 

 To prove that the moment is the same under the middle of the 

 beam as it is under each load we must multiply the reaction by 

 half the span and subtract from it the load multiplied by half the 

 span minus the length of the arm from the load to the reaction. 

 The full written-out expression is as follows: 



NOTE. In the above expression algebra has been used for the 

 first time. The product of two positive (+) quantities is positive. 

 The product of two negative (-) quantities is positive. The 

 product of a positive (+) and a negative (-) quantity is negative. 

 In the above expression, where the subtraction is indicated, the 

 load P is negative and the first quantity within the parenthesis 

 is positive, for when no sign is written the positive (+) sign is 



