44 RESISTANCE OF MATERIALS 



For a uniform load the expressions for moment and shear, as 



shown in article 29, are 



,_ wl war 

 M=- X -, 



wl 



The first of these equations evidently represents a parabola, and the 

 second an inclined straight line of slopB = w. 



From these results it follows that for any combination of uniform 

 and concentrated loads the moment diagram is a connected series of 

 parabolic arcs, and the shear diagram is a succession of inclined lines 

 or sloping steps. 



Since Skx represents an elementary vertical strip of the shear 

 "diagram, the area subtended by the shear diagram between any two 

 given points is V$A#. Making use of the relation &M=S&x, and 

 summing between two points P l and Jf^, we have 



p 2 

 = ^!M=M^-M^ 



?! 



where M l and M 2 denote the moments at the two points in question. 

 Hence the difference between the moments at any two given points is 

 equal to the area of the shear diagram between these points. 



At the ends of a simple beam the moment is always zero. There- 

 fore, by the theorem just proved, for a simple beam the area of the 

 shear diagram from one end to any point is equal to the moment at 

 this point. 



31. General directions for sketching diagrams. To economize time 

 and effort it is important to follow a definite program in drawing 

 the diagrams and determining the expressions for shear and moment. 

 The following outline of procedure for either cantilever beams or 

 simple beams resting on two supports is therefore suggested. 



1. Find each reaction by summing the moments of all the ex- 

 ternal forces about a point on the opposite reaction as moment 

 center. Check this calculation by noting that the sum of the reac- 

 tions must equal the sum of the loads. 



2. Note that the expressions for moment and shear both change 

 whenever a concentrated load is passed. Consequently, there will 



