198 Mechanics of Materials 2 $7.3 It is now possible to write the stress equations in terms of the applied moments, 124 ie. Os =Mxy (7.10) 12w 0:Myz (7.11) 7.3.General equation for slope and deflection Consider now Fig.7.4 which shows the forces and moments per unit length acting on a small element of the plate subtending an angle 8 at the centre.Thus Mxy and Myz are the moments per unit length in the two planes as described above and o is the shearing force per unit length in the direction Oy. 8中/2 于☑M,sin Q+dQ Fig.7.4.Small element of circular plate showing applied moments and forces per unit length. For equilibrium of moments in the radial XY plane,taking moments about the outside edge, (MxY +8Mxy)(x +8x)8-Mxyx80-2Myz8x sin 38+Qx808x =0 which,neglecting squares of small quantities,reduces to MxYδr+δMxYx-Myz8r+Qx8x=0 In the limit,therefore, MxY+x dMxY-Myz =-Ox dx Substituting eqns.(7.8)and (7.9),and simplifying d281d00 dxi+xdx=- D This may be re-written in the form d「1dx dxx dx D (7.12) This is then the general equation for slopes and deflections of circular plates or diaphragms. Provided that the applied loading is known as a function of x the expression can be treated
198 Mechanics qf Materials 2 57.3 It is now possible to write the stress equations in terms of the applied moments, i.e. (7.10) (7.11) 12u U, =Mxyt3 12u U, = MyZ -- t3 73. General equation for slope and deflection Consider now Fig. 7.4 which shows the forces and moments per unit length acting on a small element of the plate subtending an angle 84 at the centre. Thus Mxy and MYZ are the moments per unit length in the two planes as described above and Q is the shearing force per unit length in the direction OY. Fig. 7.4. Small element of circular plate showing applied moments and forces per unit length. For equilibrium of moments in the radial XY plane, taking moments about the outside edge, (Mxy + GMxy)(x + Sx)S$ - MxyxG$ - 2My~Sx sin 484 + QxS4Sx = 0 which, neglecting squares of small quantities, reduces to MxyGx + GMxyx - MyzGx + QXSX = 0 Substituting eqns. (7.8) and (7.9), and simplifying d20 1 dO 0 Q -+ dx2 xdx x2 D This may be re-written in the form Q D (7.12) This is then the general equation for slopes and deflections of circular plates or diaphragms. Provided that the applied loading Q is known as a function of x the expression can be treated
$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 = EIdX2 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
200 Mechanics of Materials 2 $7.5 Substituting in eqn.(7.18) 品品（器-号- Integrating, ()=6∫+] D 4 Integrating, +C2 slope = qr3 Fr C2 dr -i6D -8mD [2l0g,r-11+C+ (7.19) Integrating again and simplifying, deflection y = qr4 Fr2 2 64D 8 Dlog。r-刂+Ci4+Calog,+C (7.20) The values of the constants of integration will be determined from known end conditions of the plate;slopes and deflections at any radius can then be evaluated.As an example of the procedure used it is now convenient to consider a number of standard loading cases and to determine the maximum deflections and stresses for each. 7.5.Uniformly loaded circular plate with edges clamped The relevant fundamental equation for this loading condition has been shown to be 品品()】 Integrating. 品() 9r2 4D +C1 品() 93 4D Integrating. 94 16D+C12+C slope =9r3 C2 dr -16D+C2+ (7.21)
200 Mechanics of Materials 2 Substituting in eqn. (7.18) $7.5 Integrating, --(rg)=-;/[y+--$]dr Id r dr 1 .. - d (rg) =-- 1 [$+Qlog,r +Clr dr D 4 2n Integrating, (7.19) qr3 Fr r c2 slope B= - = -- - ---[2log, r - 11 + CI - + - dY .. Integrating again and simplifying, dr 160 8x0 2r qr4 Fr2 r2 deflection y = -~ - -[log, r - 11 + CI- + Czlog, r + C3 640 8nD 4 (7.20) The values of the constants of integration will be determined from known end conditions of the plate; slopes and deflections at any radius can then be evaluated. As an example of the procedure used it is now convenient to consider a number of standard loading cases and to determine the maximum deflections and stresses for each. 7.5. Uniformly loaded circular plate with edges clamped The relevant fundamental equation for this loading condition has been shown to be Integrating, Integrating, - d [-- Id (91 dr rdr r dr dr dy r- dr d 1’ slope 6’ = - dr (7.21)
$7.5 Circular Plates and Diaphragms 201 Integrating, deflection y=rC 64D+ 4 -C2 loge r +C3 Now if the slope 6 is not to be infinite at the centre of the plate,C2=0. Taking the origin at the centre of the deflected plate,y=0 when r=0. Therefore,from eqn.(7.22),C3=0. At the outside,clamped edge where r=R,0=dy/dr=0. Therefore substituting in the slope eqn.(6.21), 9R3 CiR 16δ+ ,=0 2 C1=9R2 8D The maximum deflection of the plate will be at the centre,but since this has been used as the origin the deflection equation will yield y=0 at r=0;indeed,this was one of the conditions used to evaluate the constants.We must therefore determine the equivalent amount by which the end supports are assumed to move up relative to the"fixed"centre. Substituting r=R in the deffection eqn.(7.22)yields mmmo=g需+6-g8 The positive value indicates,as usual,upwards deflection of the ends relative to the centre, i.e.along the positive y direction.The central deflection of the plate is thus,as expected,in the same direction as the loading,along the negative y direction (downwards). Substituting for D, R4「12(1-2) 64 E13 3qR4 = 16E1- (7.23) Similarly,from eqn.(7.21). slope 9r3 qR2r i6D+60三-160P-R1 de 3gr2 qR2 dr 16D +=83- Now,from eqn.(7.14) Eu 0r= +] (1-v2)dr Eu 9r2 (1-2) 16D3++ 9 6D1+
$7.5 Circular Plates and Diaphragms Integrating, -qr4 Clr2 640 4 deflection y = - + ~ + C2 log, r + C3 Now if the slope 0 is not to be infinite at the centre of the plate, C2 = 0. Taking the origin at the centre of the deflected plate, y = 0 when r = 0. Therefore, from eqn. (7.22), C3 = 0. At the outside, clamped edge where r = R, 8 = dy/dr = 0. Therefore substituting in the slope eqn. (6.2 l), qR3 CfR ___ +--=o 160 2 20 1 (7.22) The maximum deflection of the plate will be at the centre, but since this has been used as the origin the deflection equation will yield y = 0 at r = 0; indeed, this was one of the conditions used to evaluate the constants. We must therefore determine the equivalent amount by which the end supports are assumed to move up relative to the "fixed" centre. Substituting r = R in the deflection eqn. (7.22) yields qR4 qR4 qR4 maximum deflection = -__ + __ = __ 640 320 640 The positive value indicates, as usual, upwards deflection of the ends relative to the centre, i.e. along the positive y direction. The central deflection of the plate is thus, as expected, in the same direction as the loading, along the negative y direction (downwards). Substituting for 0, qR4 12(1 - v2) = - 64 [ Et3 ] Similarly, from eqn. (7.21), d8 - - ___ 3qr2 + __ qR2 = --[3r- 4? - R2] - 160 160 160 .. dr Now, from eqn. (7.14) (7.23)