﻿6 Prof. L. X. G. Filon : Investigation of Stresses 



case the solution is well-known *. If M is the bending 

 moment at any section ivhieh is at a distance from the points 

 of loading, S the total shear at that section, A the area of 

 the cross-section, k the radius of gyration of this area about 

 a horizontal through its centroid in its plane (the so-called 

 " neutral axis "), we have 



Now the inclinations (?) to the .I'-axis of the principal axes 

 of stress at the point are given by the equation 



~ M V tjb 2 J~M y ' * ' w 



since, the section being rectangular, P = b 2 /?>. 



It follows that if g = or M = (unless S = at the 

 same time) tan 2i=co or f = 45°, that is, when the nicols are 

 at 45° to the vertical, the isoclinic line will be along the axis 

 of the bar, and along the sections of zero bending moment. 



It z = 0, that is, when the nicols are horizontal and vertical, 

 the isoclinic line should be along the top and bottom edges 

 of the bar. To get the shape of the isoclinic lines for inter- 

 mediate values of i we note that over any part of the bar 

 which is free from loading 



M = M + S .r, (3) 



where M is a constant. 



If we measure x from a section of zero bending moment, 

 this takes the form 



M=S #, 



and we have for the isoclinic lines 



a^tan2;+(# 2 -& 2 )=0 (I) 



These are a set of hyperbolas whose asymptotes are 



y = Q, ?/4-^tan2i = 0, 

 and all of which pass through the points 

 lV = 0, i/=±b. 



* See Love, ' Theory of Elasticity.' 



