SIMPLE; BEAMS 



Load = w Ibs- per lin- ft. 



k- 



*-, 



(b) 



j Moment Diagram 



(a) 

 Force Polygon 



R 2 



i 

 -i 



(c) 



Shear Diagram 



FIG. 38. 



required parabola. It will be seen in Fig. 38 that points on the parabola 

 are also obtained. 



The shear at any point x, will be 



==#! wx = ~wL wx = w (~ x\ 



which is the equation of the inclined line shown in (c) Fig. 38. The 

 shear at any point is therefore represented by the ordinate to the shear 

 diagram at the given point. 



Property of the Shear Diagram. Integrating the equation for 

 shear between the limits, x = o and x x we have 



which is the equation for the bending moment at any point, x } in the 

 beam, and is also the area of the shear diagram between the limits 

 given. From this we see that the bending moment at any point in a 

 simple beam uniformly loaded is equal to the area of the shear dia- 

 gram to the left of the point in question. The bending moment is also 

 equal to the algebraic sum of the shear areas on either side of the point. 



