202 



Mr Clvree, On thin rotating isotropic disks. [May 4, 



importance, I have on the advice of Professor Pearson made the 

 proof complete in itself. 



The term disk is here restricted to mean a thin plate of which 

 a section parallel to the faces is bounded by a circle or two con- 

 centric circles. The disk is supposed of uniform density p and 

 of an isotropic material, for which m and n are the elastic con- 

 stants in the notation of Thomson and Tait's Natural Philosophy. 



Taking the axis of the disk for axis of z with the origin in 

 the central plane — or plane bisecting the thickness of the disk — 

 we see from the symmetry that as the disk rotates about its axis 

 with uniform angular velocity co the displacement at every point 

 is in the plane through that point and the axis, having for its 

 components w parallel to the axis and u along the perpendicular 

 r on the axis directed outwards. At any point r, z in the disk 

 the strains are as follows : 



Normal strains. 



Tangential or shearing strain. 



dw 



longitudinal -j- parallel Oz, 

 radial -=- along r, 



du dw , „ , „ 



T- + -p- m the plane ot zr. 

 dz dr 



transverse - perpendicular to r and z. 



It is obvious from the symmetry that the two other tangential 

 strains must vanish. 



The expression for the dilatation 8 is 



£ _ du u dw 

 dr r dz 



The stress system is as follows : 



_ s dw 



iz = (pi — ft) 8 + 2ft -=— parallel to oz, 



Normal J ^ ^ u 



stresses | rr = (m - n) 8 + 2n ^- along r, 



.(1). 



4>(j> = (m — n)8+ 2n u/r perpendicular to Oz and to r ; 



(~ (du dw\ . , 



r =7i U + ^J inplane ^ 



The other two shearing stresses must vanish from the symmetry. 



Shearing (~ 

 stress 



