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fore reduces to a, transverse " shearing force " F, and a couple 

 or " bending moment " M. These will be functions of x, the 

 longitudinal coordinate. If rj denote the lateral displacement, 

 parallel to the plane of symmetry, then, resolving transversally 

 the forces acting on an element Sx of the length, we have 



Y) 



or P &) o^ = o~ ...................... (1) 



dt z dx 



Again, if K denote the radius of gyration of the area of the 

 cross-section G> about an axis through its centre of gravity, 

 normal to the plane of flexure, the element of mass is 

 ultimately a disk of area o>, thickness 8x, and moment of 

 inertia pwSx . K 2 *. Since the axis of this disk has been turned 

 through a small angle drj/dx from the position of equilibrium, 

 the equation of angular motion is 



whence pmK = + F . .................. (2) 



Tf we eliminate F between (1) and (2) we have 



provided the sectional area 6> be uniform. 



We have next to express M in terms of the deformation 

 of the bar. Consider in the first instance the case of a bar 

 uniformly bent, so that its axis becomes an arc of a circle. 

 It is evident from symmetry that the shearing force F now 

 vanishes, and it hardly needs calculation to shew that the 

 strain in any part of the cross-section will be proportional to 

 the curvature. Hence by Hooke's law the resultant couple M 

 will also vary as the curvature, or 



M=WIR, ........................ (4) 



where R is the radius of curvature, and 23 is a constant 

 depending on the shape and size of the cross-section, and on 

 the elastic properties of the material. 



* The symbol K is not required at present in its former sense as an elastic 

 constant. 



