CHAPTER 7 CIRCULAR PLATES AND DIAPHRAGMS Summary The slope and deflection of circular plates under various loading and support conditions are given by the fundamental deflection equation 品品()】=号 ar rar where y is the deflection at radius r:dy/dr is the slope 6 at radius r;is the applied load or shear force per unit length,usually given as a function of r;D is a constant termed the “flexural stiffness'”or“flexural rigidity”=Et/12(1-v2】and t is the plate thickness. For applied uniformly distributed load (i.e.pressure g)the equation becomes d「1d.dy1 ar rar'dr For central concentrated load F F 0= and the right-hand-side becomes- 2πr 2πrD For axisymmetric non-uniform pressure (e.g.impacting gas or water jet) g K/r and the right-hand-side becomes -K/2D The bending moments per unit length at any point in the plate are: de Mr Mxy D 十- d. M:=My= D dr Similarly,the radial and tangential stresses at any radius r are given by: Eu de radial stress o,= (1-v2)dr Eu 「d tangential stress 124 124 Alternatively, 0二 M,and d:M: 193
CHAPTER 7 CIRCULAR PLATES AND DIAPHRAGMS Summary The slope and deflection of circular plates under various loading and support conditions are given by the fundamental deflection equation - [’ Q __ (.$)I - [’ _- (rz)] =-% dr rdr where y is the deflection at radius r; dyldr is the slope e at radius r; Q is the applied load or shear force per unit length, usually given as a function of r; D is a constant termed the “flexural stiffness” or “flexural rigidity” = Et3/[12(l - u2)] and t is the plate thickness. For applied uniformly distributed load (i.e. pressure q) the equation becomes qr dr rdr For central concentrated load F F F Q = - and the right-hand-side becomes - - 21rr 2nrD For axisymmetric non-uniform pressure (e.g. impacting gas or water jet) q = K/r and the right-hand-side becomes - K/2D The bending moments per unit length at any point in the plate are: Similarly, the radial and tangential stresses at any radius r are given by: radial stress a, = EU tangential stress a, = ___ Alternatively, 12u 12u t3 13 a, = -Mr and a: = -MZ 193
194 Mechanics of Materials 2 For a circular plate,radius R.freely supported at its edge and subjected to a load F distributed around a circle radius R F R ymax [3v)(R2)-Ri log.R] 8πD2(1+y) 3F 2(+)logR -)R + (R2-R) and 三0zm Table 7.1.Summary of maximum deflections and stresses. Loading Maximum Maximum stresses condition deflection (ymax) Urmas Ozmux Uniformly loaded, 3gR4 1661-2) 3qR2 3gR2 edges clamped 4r2 82I+v) Uniformly loaded. edges freely 3qR 3qR2 3g supported 16E735+1-) 823+ 82(3+) Central load F, 3FR2 3F 3vF edges clamped 4πE1-2) 2πt2 2n2 Central load F, From 3FR2 From edges freely 4玩E3+(1- 3 R 3F 2721+)log (I+v)log +(1-v) supported 2πt2 For an annular ring,freely supported at its outside edge,with total load F applied around the inside radius Ri,the maximum stress is tangential at the inside radius, 3F(1+v) R2 R i.e. πt2 (R-R) loge Ri If the outside edge is clamped the maximum stress becomes 3F (R2-R) Omax= 2πt2 R2 For thin membranes subjected to uniform pressure g the maximum deflection is given by ymax =0.662 R 9R7'3 Et For rectangular plates subjected to uniform loads the maximum deflection and bending moments are given by equations of the form 9b4 ymax a- Et3 M=Bgb2
194 Mechanics of Materials 2 .. Central load F, edges clamped Central load F, edges freely support e d For a circular plate, radius R, freely supported at its edge and subjected to a load F distributed around a circle radius RI 1 3FR2 3F 3uF 2nr2 2nr2 From From 3FR2 3F R 3F r 2nr2 4?rEt" (I - 2) ~ - 1 and Table 7.1. Summary of maximum deflections and stresses. I Loading I ~ condition I ~aflxri~~~ Maximum stresses I Uniformly loaded, edges freely supported 16Et" For an annular ring, freely supported at its outside edge, with total load F applied around the inside radius RI , the maximum stress is tangential at the inside radius, i.e. If the outside edge is clamped the maximum stress becomes 3F (R' - R:) = [ R2 ] For thin membranes subjected to uniform pressure q the maximum deflection is given by For rectangular plates subjected to uniform loads the maximum deflection and bending moments are given by equations of the form 9b4 ymax = aEt3 M = &h2
$7.1 Circular Plates and Diaphragms 195 the constants a and B depending on the method of support and plate dimensions.Typical values are listed later in Tables 7.3 and 7.4. A.CIRCULAR PLATES 7.1.Stresses Consider the portion of a thin plate or diaphragm shown in Fig.7.1 bent to a radius Rxy in the XY plane and Ryz in the YZ plane.The relationship between stresses and strains in a three-dimensional strain system is given by eqn.(7.2), 6=glo,-w 1 1 lo:-vox-voyl Fig.7.1. Now for thin plates,provided deflections are restricted to no greater than half the plate thickness,the direct stress in the y direction may be assumed to be zero and the above equations give E 0x≠ 0-2e:+e] (7.1) E 0-2吗8，+e (7.2) EJ.Hearn,Mechanics of Materials /Butterworth-Heinemann,1997. S.Timoshenko,Theory of Plates and Shells,2nd edn.McGraw-Hill.1959
$7.1 Circular Plates and Diaphragms 195 the constants a and ,~9 depending on the method of support and plate dimensions. Typical values are listed later in Tables 7.3 and 7.4. A. CIRCULAR PLATES 7.1. Stresses Consider the portion of a thin plate or diaphragm shown in Fig. 7.1 bent to a radius RxU in the XY plane and RYZ in the YZ plane. The relationship between stresses and strains in a three-dimensional strain system is given by eqn. (7.2): 1 E - -[a, - uay - ua,] ,-E 1 E E? = -[a, - ua, - !Jay] Y Z / Fig. 7.1. Now for thin plates, provided deflections are restricted to no greater than half the plate thickness3 the direct stress in the Y direction may be assumed to be zero and the above equations give E a, = [E, + UE,] ~ (1 - u2) E.J. Hearn, Mechanics of Materials I, Butterworth-Heinemann, 1997. S. Timoshenko, Theory of Plates and Shells, 2nd edn., McGraw-Hill, 1959
196 Mechanics of Materials 2 $7.1 If u is the distance of any fibre from the neutral axis,then,for pure bending in the XY and YZ planes, Mo E 7=y=R and E-R =8 8x- RxY and 8:= Ryz 1 d2y u Now R dx2 and,for small deflections, tan =0(radians). dx 1 d2y de RxY dxdx de and x=4-(=radial strain) (7.3) dx Consider now the diagram Fig.7.2 in which the radii of the concentric circles through C and D on the unloaded plate increase to [(x+dx)+(0+d)u]and [x +ue],respectively, when the plate is loaded. dx Unlogded plate E N.A. /de e Looded plate N.A.(Unstrained) u(8+d8) u8 Fig.7.2. Circumferential strain at D2 2π(x+u0)-2πx =6:= 2πx 0 -(circumferential strain) (7.4)
196 Mechanics of Materials 2 $7.1 If u is the distance of any fibre from the neutral axis, then, for pure bending in the XY and YZ planes, MaE (Tu - ---- and -E- =E IYR ER -- U U and E? = - Rxr RYZ E, = ~ 1 d2y du R dx2 dx Now - = ~ and, for small deflections, - = tan0 = 0 (radians). d0 dx and E, = u- (= radial strain) (7.3) Consider now the diagram Fig. 7.2 in which the radii of the concentric circles through CI and D1 on the unloaded plate increase to [(x + dx) + (6' + dO)u] and [x + uO], respectively, when the plate is loaded. Circumferential strain at 02 2n(x + ue) - 2rx 2TcX - - E; = Ue = - (= circumferential strain) X (7.4)
§7.2 Circular Plates and Diaphragms 197 Substituting egns.(7.3)and (7.4)in egns.(7.1)and (7.2)yields E de ox=(1)"dx Eu [de ie. 0x= 1-吗a+vg (7.5) Eu 「d81 Similarly, =0-x (7.6) dx Thus we have equations for the stresses in terms of the slope and rate of change of slope de/dx.We shall now proceed to evaluate the bending moments in the two planes in similar form and hence to the procedure for determination of 0 and de/dx from a knowledge of the applied loading. 7.2.Bending moments Consider the small section of plate shown in Fig.7.3,which is of unit length.Defining the moments M as moments per unit length and applying the simple bending theory, Unit length Fig.7.3. Substituting eqns.(7.5)and (7.6). Et3 8 MxY= 121-2)dc (7.7) Et3 Now 12(1-y2) =D is a constant and termed the flexural stiffness 「d81 so that Mxy =D (7.8) dx x 「6 de and,similarly, Myz =D +V (7.9) d
$7.2 Circular Plates and Diaphragms 197 Substituting eqns. (7.3) and (7.4) in eqns. (7.1) and (7.2) yields E d6 u6 a, = ~ i.e. Similarly, Eu Eu de u, = ~ + v- (1-9) [: dx] (7.5) (7.6) Thus we have equations for the stresses in terms of the slope 6 and rate of change of slope d6/dx. We shall now proceed to evaluate the bending moments in the two planes in similar form and hence to the procedure for determination of 6 and d@/dx from a knowledge of the applied loading. 7.2. Bending moments Consider the small section of plate shown in Fig. 7.3, which is of unit length. Defining the moments M as moments per unit length and applying the simple bending theory, az c 1 x t3 ct3 M=-=- - -- y u [ 12 1 - 12u m Fig. 7.3. Substituting eqns. (7.5) and (7.6), Et3 [ 9 + 21 12(1- 9) dx Mxr = Et3 Now = D is a constant and termed the j-lexural stiffness 12(1 - I?) so that and, similarly, (7.7)
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)
202 Mechanics of Materials 2 §7.6 The maximum stress for the clamped edge condition will thus be obtained at the edge where r=R and at the surface of the plate where u=t/2, E 12gR2 3gR2 ie. 0rm=Q-)216D42 (7.24) N.B.-It is not possible to determine the maximum stress by equating dor/dr to zero since this only gives the point where the slope of the o,curve is zero(see Fig.7.7).The value of the stress at this point is not as great as the value at the edge. Similarly, Eu de 0:=1 Eu (1-2)16D +1)+ 16D1+ Unlike o,,this has a maximum value when r=0,i.e.at the centre. E 1 gR2 OZm= Γ(1-u2)216D (1+) 821+ (7.25) 7.6.Uniformly loaded circular plate with edges freely supported Since the loading,and hence fundamental equation,is the same as for $7.4,the slope and deflection equations will be of the same form,i.e.eqns (7.21)and(7.22)will apply.Further, the constants C2 and C3 will again be zero for the same reasons as before and only one new condition to solve for the constant Ci is required. Here we must make use of the fact that the bending moment is always zero at any free support, i.e.at r=R. M,=0 Therefore from egn.(7.16), D+=0 「d0.0 de dr Substituting from eqn.(7.21)with r =R and C2=0. 16D CI-9 R2「(3+v) D(1+)
202 Mechanics of Materials 2 57.6 The maximum stress for the clamped edge condition will thus be obtained at the edge where r = R and at the surface of the plate where u = t/2, i.e. E t 2qR2 - 3qR2 f,, = - __ (1 - v2)2 160 4t2 (7.24) N.B.-It is not possible to determine the maximum stress by equating dar/dr to zero since this only gives the point where the slope of the a,. curve is zero (see Fig. 7.7). The value of the stress at this point is not as great as the value at the edge. Similarly, .;=---["..$I Eu (1 -v2) r Unlike a,, this has a maximum value when r = 0, i.e. at the centre. E t qR2 (1 - u2)2 160 .zmsx = (1 +VI 4 3qR2 - -(1+ tit2 (7.25) 7.6. Uniformly loaded circular plate with edges freely supported Since the loading, and hence fundamental equation, is the same as for $7.4, the slope and deflection equations will be of the same form, i.e. eqns (7.21) and (7.22) will apply. Further, the constants C2 and C3 will again be zero for the same reasons as before and only one new condition to solve for the constant Cl is required. Here we must make use of the fact that the bending moment is always zero at any free SUPPOfl, i.e. at r = R. M,=O Therefore from eqn. (7.16), .. d8 8 dr r _- -vSubstituting from eqn. (7.21) with r = R and C2 = 0,