580 Prof. E. H. Barton on the Lateral 



integrating over the whole cross-section of the bar of area 

 co say. Then, writing k for the radius of gyration of the 

 cross-section about its intersection with the neutral surface, 

 we obtain for the bending moment 



M = f qrfdco/R = qcoKr/H = qo)fc 2 d 2 i//d.r- 



(i) 



Bending Moments and Forces on a Bar at Rest. — Let us take 

 the next step in the elementary treatment of the problem by 

 finding relations between the bending moment, the shearing- 

 forces, and the applied forces on a bar by the following method. 

 Consider the bar bent in the plane of xy and in equilibrium 

 under given forces. Let it have a free end at the origin of 

 coordinates and lie almost along the axis of x, the further end 

 being maintained at rest by clamps or any other convenient 

 arrangements (see fig. 1). Then suppose first an isolated 



Fis. 1. 



force /i to act at the origin along the axis of y. The shearing 

 force in the bar is accordingly J\ from its free end and at 

 every point up to where it is clamped. But the bending 

 moment at x is f x x. That is, the bending moment is zero at 

 the origin, and increases with x at the rate/i per unit length 

 along x. Thus if M denotes the bending moment at x we 

 have M=f x x and dM/<ii'==/ 1 . Or, if M be plotted as a graph 

 we obtain a straight line through the origin at an angle <j> l 

 with the axis of x, where tan0 1 =/ 1 (see fig. 2). 



Next at x 2 between the origin and x, let a second force f 2 

 act together with and parallel to the force f x at the origin. 

 Then the graph for M will consist of two straight lines, their 

 junction occurring at the point x 2 where the second force is 

 applied, The bending moment at x, where x'>x 9i ^is now 



