MECHANICS 



107 



of the reaction, 19,4661 lb., is divided by the 3,000 lb., the 

 result will be the number of feet of the uniform load required 

 to counteract the remaining part of the reaction, and this 

 will give the distance of the section, beyond which the result- 

 ant of the forces at the left becomes negative, from the edge 

 of the uniform load at o; thus, 19,466f-J- 3,000 = 6.49 ft. 

 The distance from R\ to the edge of the uniform load is 8 ft. 

 The entire distance to the section of change of sign of the 

 shear is, therefore, 8 + 6.48 = 14.48 ft. from RI. 



Shear Diagram. It is sometimes necessary to plot the 

 shear of the forces acting on a beam in a diagram known as 

 the shear diagram. In order to illustrate how this may be 



done the beam shown in Fig. 12 (o), the span of which is 

 12 feet, will be considered. Since the beam is symmetrically 

 loaded, each reaction is equal to one-half the sum of the 



400 + 1,200 + 400 

 loads; that is, each reaction is equal to 



= 1,000 lb. The shear at RI is equal to the reaction, or 

 1,000 lb. To plot the diagram, proceed as follows: Draw 

 a line ab to any convenient scale to represent the length of 



