$7.4 Circular Plates and Diaphragms 199 in a similar manner to the equation dy M=E dx2 used in the Macaulay beam method,i.e.it may be successively integrated to determine 6, and hence y,in terms of constants of integration,and these can then be evaluated from known end conditions of the plate. It will be noted that the expressions have been derived using cartesian coordinates(X,Y and Z).For circular plates,however,it is convenient to replace the variable x with the general radius r when the equations derived above may be re-written as follows: dr rdr (7.13) radial stress =+ Eu (7.14) tangential stress Eu o二1-内'i+ (7.15) moments 「d0 Mr=D d (7.16) r [d0.8 M,=D+ (7.17) In the case of applied uniformly distributed loads,i.e.pressures q,the effective shear load O per unit length for use in eqn.(7.13)is found as follows. At any radius r,for equilibrium, 0×2nr=9×πr2 i.e. 0-号 Thus for applied pressures eqn.(7.13)may be re-written (7.18) 7.4.General case of a circular plate or diaphragm subjected to combined uniformly distributed load q(pressure)and central concentrated load F For this general case the equivalent shear O per unit length is given by Q×2πr=qXπr2+F Q=号+品$7.4 Circular Plates and Diaphragms 199 in a similar manner to the equation d2 Y M = EI￾dX2 used in the Macaulay beam method, i.e. it may be successively integrated to determine 8, and hence y, in terms of constants of integration, and these can then be evaluated from known end conditions of the plate. It will be noted that the expressions have been derived using Cartesian coordinates (X, Y and Z). For circular plates, however, it is convenient to replace the variable x with the general radius r when the equations derived above may be re-written as follows: radial stress tangential stress moments - [' Q __ (rz)] = dr r dr Eu Eu uz = ~ (7.13) (7.14) (7.15) (7.16) (7.17) In the case of applied uniformly distributed loads, i.e. pressures q, the effective shear load Q per unit length for use in eqn. (7.13) is found as follows. At any radius r, for equilibrium, 2 Q x 2nr = q x nr i.e. Q=- qr 2 Thus for applied pressures eqn. (7.13) may be re-written (7.18) 7.4. General case of a circular plate or diaphragm subjected to combined uniformly distributed load q (pressure) and central concentrated load F For this general case the equivalent shear Q per unit length is given by Q x 2nr = q x nr2 + F