260 Trans. Acad. Sci. of St. Louis. 



Tcott? 39 



Kx ' v 



in which D is the displacement, I the moment of inertia of 

 the section with reference to a horizontal line passing through 

 its center of gravity, the angle which an oblique force 

 makes with the axis of the beam before flexure, E the area of 

 section, and x the abscissa of the center of gravity of the 



section. 



In case of a compressive longitudinal force the neutral 

 axis will be moved toward the side of the beam which is in 

 tension and in case of a tensile force in the opposite direction. 

 With equal longitudinal forces in compression and in tension, 

 the displacement of the neutral axis in the two halves of a 

 telescope is opposite in direction and may be assumed to be 

 equal in amount. Now the minimum value of the moment 

 of inertia of a section with reference to a horizontal line 

 in the plane of the section is the moment of inertia with 

 reference to the horizontal line through the center of gravity 



* Professor DeVolson Wood in his Resistance of Materials (p. 91) gives 

 as the expression for the linear displacement of the neutral axis in case of a 

 beam subjected to an oblique force, P, 



PI 



h = -jtt^ COS0. 



cEK 

 In which h is the displacement and c = -,p being the radius of curvature of 



the flexure curve at the point of whose abscissa is x. 

 a. , 



Now o = — A + — 2 V and since in this ca8e t is a very sma11 quan " 

 ' d 2 y \ dx 2 / ox 



1 d 2 y 1 



tity we may put - = ^. By equation (1) this becomes p = — p 2 z + q i y 



EI 



nr -— = -pr- , and the above expression for h becomes 



or ' — Wxumo + Wycosd ' 



I 



h = ~ K (xtanfl + «/)' 

 Now y is a very small quantity and may be neglected in the case of a 

 telescope without appreciable error and we get, omitting the minus sign, 



Kx 

 From this formula the linear displacement of any point of the neutral axis 

 whose abscissa is x may be computed. 



