204 Mechanics of Materials 2 87.7 Fr dr 2πD og。r+Cir dy F 「r2 Integrating, 4 2+C2 0= dy F 5+9 (7.28) Fr2 Integrating, y= ogr-刂+g+6gr+G (7.29) 8πD Again,taking the origin at the centre of the deflected plate as shown in Fig.7.5,the following conditions apply: For a non-infinite slope at the centre C2=0 and at r=0,y=0,.'.C3 =0. Also,at rR,slope 0 dy/dr =0. Therefore from eqn.(7.28), CR F [loge R 1 πD2 The maximum deflection will be at the centre and again equivalent to that obtained when rR,i.e.from eqn.(7.29). FR2 FR2[1og。R_11 maximum deflection = 3Dliog,R- FR2 167Dl-2log,R+2+2log,R-1] FR2 =16rD Substituting for D, FR212(1-2) ymax三 16π Et 3FR2 4xEr3(I-v) (7.30) Again substituting for de/dr and 0/r from eqn.(7.28)into eqns(7.14)and (7.15)yields 3F Crn三 (7.31) 2nt2 3vF (7.32) 2nt2204 Mechanics of Materials 2 $7.7 Integrating, (7.28) Fr2 c1r2 = - - [log, r - I] + __ + ~2 log, r + C3 (7.29) 8nD 4 Integrating, Again, taking the origin at the centre of the deflected plate as shown in Fig. 7.5, the following conditions apply: For a non-infinite slope at the centre C2 = 0 and at r = 0, y = 0, :. CS = 0. Also, at r = R, slope 8 = dy/dr = 0. Therefore from eqn. (7.28), log, R - E] = 4 F log, R I~D [ 2 :] =" - The maximum deflection will be at the centre and again equivalent to that obtained when r = R, i.e. from eqn. (7.29), FR~ FR2 log,R maximum deflection = --[log,R - 11 -t- - - - - 8nD 4nD [ 2 :] -- FR~ [-21og,R+2f21ogeR- 11 16nD FR~ -- 16nD Substituting for D, FR~ 12(1 - u2) Ymax = - 16n Et3 3FR2 4nEt3 -- (1 - "2) (7.30) Again substituting for dO/dr and O/r from eqn. (7.28) into eqns (7.14) and (7.15) yields (7.31) (7.32)