272 P. W. Bridg-man — Crystalline Cylinders. 



is rotated through 180° about this digonal axis, the applied stress 

 system is unaltered. It follows that the displacements must have 

 digonal symmetry about this axis. A moment's consideration 

 si lows that u r must he an even function of and tt an odd 

 function. The two simplest trigonometric functions satisfying 

 all these conditions are 7c + cos 40, and sin 40, where 7c is any 

 constant. Symmetry relations impose no conditions on the kind 

 of function of r that the solution is, and we therefore try a solu- 

 tion of the form 



i 



{ u r = A,r + A t r~ l + |>,(r) + <£,(r)cos 40] 



( Ug = <£ 3 (r)sin 40 



where <£, , $ a , an d <K are a ^ ^'' st order terms. Direct substitu- 

 tion in the equilibrium equations, keeping only first order terms, 

 verifies that this is a possible solution and gives ordinary differ- 

 ential equations of the second order in <£, , <p 2 , and <£ 3 , which are 

 simple enough to be readily solved. The solutions are poly- 

 nomials, giving for the explicit form of the first approximation 



C 2 r~ 5 lcos40 



L«- 



sin 40. 



La i 3a + b a -\- 3b J 



There are just enough arbitrary constants in this solution to 

 enable one to make rr = when r = r , rr — — P when r — r t , 

 and r0 = when r = r and r = i\ . The condition rz = when 

 r = r a and r—-r 1 is already satisfied by putting w s =s 0. It is 

 not worth while writing out the explicit values of the constants ; 

 they may be determined numerically readily enough in any 

 special case. 



Better approximations may be obtained by adding terms in 

 cos 80, cos 120 and sin 80, sin 120, etc. Each of these higher 

 terms will involve new arbitrary constants, and the equations for 

 determining these constants will become more complicated. 



Tetragonal Crystals.— (A) Six Constant Group. The stress- 

 strain relations are given in Love for rectangular coordinates. 

 When transformed to cylindrical coordinates they are as follows : 



rr = (a — c cos 40)e,.,, + (b + c cos 40)c gg + c rt e zz + c sin 40 e,. g 



00 = (b + c cos 46)e rr + (a — c cos 40)^ + <\.,e zz - c sin 40 e /(? 



zz = e n e rr + c„e^ + c 88 e Z2 



