38 



RESISTANCE OF MATERIALS 



each point of the beam. This may be done by means of a bending- 

 moment diagram and a shear diagram, obtained by plotting the 

 general expressions for the moment and shear, such as those given in 

 the examples in the preceding paragraph. Thus, the shear diagram is 

 obtained by plotting the shear at any arbitrary section mn as ordinate 

 and the distance x of this section from a fixed origin as abscissa. 

 Similarly, the moment diagram is obtained by plotting the moment 

 at any arbitrary section mn as ordinate and the distance x as abscissa. 



The following simple applica- 

 tions illustrate the method of 

 drawing the diagrams. 



1. Simple beam bearing a single 

 concentrated load P at its center 

 (Fig. 39). From symmetry, the 



reactions R^ and R 2 are each equal 



P 

 to Let mn denote any section 



of the beam at a distance x from 

 the left support, and consider the 

 portion of the beam on the left of 

 this section. Then the moment at 



SHEAK 



FIG. 39 



mn 



Af-J 



= x ) and the shear 



For a section on the 



right of the center the bending moment is R Z (I x) and the shear 

 is R z . Consequently, the bending moment varies as the ordinates 



of a triangle, being zero at either support and attaining a maximum 



PI 

 value of at the center, while the shear is constant from A to B, 



and also constant, but of opposite sign, from B to C. 



The diagrams in Fig. 39 represent these variations in bending 

 moment and shear along the beam under the assumed loading.' 

 Consequently, if the ordinates vertically beneath B are laid off to 

 scale to represent the bending moment and shear at this point, the 

 bending moment and shear at any other point D of the beam are 

 found at once from the diagram by drawing the ordinates EF and 

 HK vertically beneath D. 



