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



i, = ^V (39) 



in which D is the displacement, / 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, K 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 ttie expression for the linear displacement of the neutral axis in case of a 

 beam subjected to an oblique force, P, 



I 

 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. 



2. 



Now p = — — - II -[- y— ) and since in this case -? is a very small quan- 

 a^y \ ax^/ dx 



1 d^v 1 



tity we may put — = ^. By equation (1) this becomes p = — « 4. Q«y 



EI 



or p = — Wxiim0-\- WycosB ' *°^ *^® above expression for h becomes 



— ~~k (xi&wd -\- yY 

 Now J/ is a very small quantity and may be neglected in the case of & 

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



. Jcot^ 



fi = = D. 



Kx 



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

 whose abscissa is x may be computed. 



