310 Profs. W. E. Ayrton and J. Perry. [Feb. 14, 



We will now calculate the value of f } the maximum stress at any 

 point in the rectangular section. 



The shear stress at any point, xy, in such a section of the prism 

 which has received a twist, t, per unit of length is, where the axes of 

 x and y are parallel to the edges of the section, and the origin the 

 centre of the section, z being at right angles to the section, 



N (tx + — ^ in the directions of y and z, 

 \ dyj 



and N"( — ry + ^1 ) in the directions of z and x, 



\ dxj 



where 7 equals — txij-\- an expression which vanishes if the thickness of 

 the strip is very small. 



Hence since — = — tx, 



dy 



and ^=-ry, 



dx 



it follows that the shear stress at a point xy is 

 in the direction of y, 

 and — ZNry in the direction of x, 



so that the whole shear stress at a point is — ZNry, and the shear 

 stress, q, at the middle of the longer edge of the rectangular section, 

 where we may suppose the shear stress is the greatest, is 2Nt5. 

 Now if C is the twisting couple, we know that 



G 



T ~ a' 



, 3Fr cos a. , o , 7 o\ 



hence T= T6W (a ~ +H ' 



and 2=^-^ (oS + P). 



Also, since Fr sin a is the bending couple, the greatest stress due 

 to bending, 



3 Fr sin a. 



Hence if / is the greatest total stress anywhere in a section, it may be 

 shown that 



