CHAPTER 4 RINGS,DISCS AND CYLINDERS SUBJECTED TO ROTATION AND THERMAL GRADIENTS Summary For thin rotating rings and cylinders of mean radius R,the tensile hoop stress set up is given by OH pa2R2 The radial and hoop stresses at any radius r in a disc of uniform thickness rotating with an angular velocity rad/s are given by =A-月9-8+ 8 0H=A+2-(1+3w)Pg B where A and B are constants,p is the density of the disc material and v is Poisson's ratio. For a solid disc of radius R these equations give ,=3+o8R2-2 gga+-+3wr时 At the centre of the solid disc these equations yield the maximum stress values Ol =or=(3) 8 At the outside radius, 0r=0 0H=1-)w2R 4 For a disc with a central hole, w-答3+（因+后+）-+3or 117
CHAPTER 4 RINGS, DISCS AND CYLINDERS SUBJECTED TO ROTATION AND THERMAL GRADIENTS Summary For thin rotating rings and cylinders of mean radius R, the tensile hoop stress set up is given by 22 UH=WR The radial and hoop stresses at any radius r in a disc of uniform thickness rotating with an angular velocity w rads are given by B pw2 r2 a, =A- - r2 - (3+u)- 8 B pw2r2 CH =A + - r2 - (1 + 3~)- 8 where A and B are constants, p is the density of the disc material and u is Poisson's ratio. For a solid disc of radius R these equations give At the centre of the solid disc these equations yield the maximum stress values At the outside radius, a, = 0 For a disc with a central hole, 117
118 Mechanics of Materials 2 S4.1 the maximum stresses being CHmax= 0w [3+)R吃+(I-)R] at the centre and 0n=3+) 8®-RJP atr=√(R1R2) For thick cylinders or solid shafts the results can be obtained from those of the corre- sponding disc by replacing v by v/(1-v), e.g.hoop stress at the centre of a rotating solid shaft is OH 3+a- 0w2r2 8 Rotating thin disc of uniform strength For uniform strength,i.e.oH=or=a(constant over plane of disc),the disc thickness must vary according to the following equation: 1=t0e-w2r2/2a) 4.1.Thin rotating ring or cylinder Consider a thin ring or cylinder as shown in Fig.4.1 subjected to a radial pressure p caused by the centrifugal effect of its own mass when rotating.The centrifugal effect on a unit length of the circumference is p=mo'r Fig.4.1.Thin ring rotating with constant angular velocity @ Thus,considering the equilibrium of half the ring shown in the figure, 2F p x 2r (assuming unit length) F=pr where F is the hoop tension set up owing to rotation
118 Mechanics of Materials 2 $4.1 the maximum stresses being PO2 OH,,, = - [(3 + u)R; + (1 - u)R:] at the centre at r = J(RlR2) 4 PW2 and Ormx = (3 f u)- [R2 - R1I2 8 For thick cylinders or solid shafts the results can be obtained from those of the correby ~/(l - u), sponding disc by replacing e.g. hoop stress at the centre of a rotating solid shaft is Rotating thin disc of uniform strength For uniform strength, i.e. OH = or = o (constant over plane of disc), the disc thickness = toe(-~2r2)/(2c) must vary according to the following equation: 4.1. Thin rotating ring or cylinder Consider a thin ring or cylinder as shown in Fig. 4.1 subjected to a radial pressure p caused by the centrifugal effect of its own mass when rotating. The centrifugal effect on a unit length of the circumference is 2 p=mwr F F Fig. 4.1. Thin ring rotating with constant angular velocity o. Thus, considering the equilibrium of half the ring shown in the figure, 2F = p x 2r (assuming unit length) F = pr where F is the hoop tension set up owing to rotation
§4.2 Rings,Discs and Cylinders Subjected to Rotation and Thermal Gradients 119 The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed constant across the wall thickness. F=mass×acceleration=mw2r2×r This tension is transmitted through the complete circumference and therefore is resisted by the complete cross-sectional area. F mo2r2 hoop stress = A A where A is the cross-sectional area of the ring. Now with unit length assumed,m/A is the mass of the material per unit volume,i.e.the density p. hoop stress=pa2r2 4.2.Rotating solid disc (a)General equations (g,+8c,(r+8r)88 C.F.pr2uw28r8 CHx8rxI Fig.4.2.Forces acting on a general element in a rotating solid disc. Consider an element of a disc at radius r as shown in Fig.4.2.Assuming unit thickness: volume of element =r80 x 8r x I =r808r mass of element pr 806r Therefore centrifugal force acting on the element mo'r pr808ra2r pr2a280 8r
$4.2 Rings, Discs and Cylinders Subjected to Rotation and Thermal Gradients 119 The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed constant across the wall thickness. .. F = mass x acceleration = mw2r2 x r This tension is transmitted through the complete circumference and therefore is resisted by the complete cross-sectional area. .. F mw2r2 hoop stress = - = - A A where A is the cross-sectional area of the ring. density p. .. hoop stress = po2r2 Now with unit length assumed, m/A is the mass of the material per unit volume, i.e. the 4.2. Rotating solid disc (a) General equations ( u, t 8 u~( r + 8 r 8 6 4 XI Fig. 4.2. Forces acting on a general element in a rotating solid disc Consider an element of a disc at radius r as shown in Fig. 4.2. Assuming unit thickness: volume of element = r SO x Sr x 1 = r SO& mass of element = pr SO&- Therefore centrifugal force acting on the element = mw2r = pr~~rw'r = pr2w260Sr
120 Mechanics of Materials 2 §4.2 Now for equilibrium of the element radially 66 2awor sin+r(+8)(r+5r6-pr280 8r If 80 is small, 6868 sin radian Therefore in the limit,as r-0(and therefore 8o,-0)the above equation reduces to OH -O-r- rdor =pr2o (4.1) dr If there is a radial movement or"shift"of the element by an amount s as the disc rotates, the radial strain is given by ds I Er= -=(O,-voH) (4.2) dr E Now it has been shown in $9.1.3(a)that the diametral strain is equal to the circumferential strain. 5 EoH-or） (4.3) 5= E(OH-vo,) ds I Differentiating, dr=E(on -va,)+ (4.4) dr dr Equating eqns.(4.2)and(4.4)and simplifying, (oH-o,)1+v)+r -ur do.=0 (4.5) dr dr Substituting for (oH-o,)from eqn.(4.1), (0+n）1++r dr dr don dor =-pro?(1+v) dr dr Integrating, pr2w2 OH十O,三- 2(1+)+24 (4.6) where 24 is a convenient constant of integration. Subtracting egn.(4.1), 2o,+r dor pr202 dr -2(3+)+2A But 2+20=品× EJ.Hearn,Mechanics of Materials 1.Butterworth-Heinemann,1997
120 Mechanics of Materials 2 $4.2 Now for equilibrium of the element radially If SO is small, 68 68 22 sin - = - radian Therefore in the limit, as Sr + 0 (and therefore Sa, + 0) the above equation reduces to d ur 22 UH -ar - r- = pr o dr (4.1 ) If there is a radial movement or “shift” of the element by an amount s as the disc rotates, (4.2) Now it has been shown in $9.1.3(a)’ that the diametral strain is equal to the circumferential strain. 1 E Differentiating, dr E E [dr-dr] Equating eqns. (4.2) and (4.4) and simplifying, s = --(OH - war) ds 1 r dDH Vdo, - = -(OH - va,.) + - dOH do,. (CH - Gr)(l + V) + r- - vr- = 0 dr dr Substituting for (OH - a,.) from eqn. (4.1), dcTH do,. (I + v) + r- - vr- = o dr dr dCJH do,. 2 .. - + - = -prw (1 + v) dr dr Integrating, OH +ar = -- pr2w2 (1 + ”) + 2A 2 where 2A is a convenient constant of integration. Subtracting eqn. (4.1), But pr2w2 (3 + v) + 2A do, 2ar + r- = -- dr 2 dar d 2 I 20, +r- = - [(r a,)] x - dr dr r (4.3) (4.4) (4.5) (4.6) E.J. Hearn, Mechanics ofMatericrls 1. Butterworth-Heinemann, 1997
$4.2 Rings,Discs and Cylinders Subjected to Rotation and Thermal Gradients 121 品=++网 r2a=- pr4 2Ar2 83+)+ 2 B where -B is a second convenient constant of integration, ,=A-克-B+A B 8 (4.7) and from eqn.(4.5), %=A+,月-(1+3a2r2 、B 8 (4.8) For a solid disc the stress at the centre is given when r=0.With r equal to zero the above equations will yield infinite stresses whatever the speed of rotation unless B is also zero, i.e.B=0 and hence B/r2=0 gives the only finite solution. Now at the outside radius R the radial stress must be zero since there are no external forces to provide the necessary balance of equilibrium if o,were not zero. Therefore from egn.(4.7), 0r=0=A-(3+) Dw2R2 8 A=(3+)Pw2R2 8 Substituting in eqns.(4.7)and (4.8)the hoop and radial stresses at any radius r in a solid disc are given by o4=(3+)w 8-1+3w)0w2r2 8 83+R2-(1+3r2] (4.9) 0=(3+yw2 8-(3+v)eo 8 8+ 8R2-内 (4.10) (b)Maximum stresses At the centre of the disc,where r =0,the above equations yield equal values of hoop and radial stress which may also be seen to be the maximum stresses in the disc,i.e.maximum hoop and radial stress (at the centre) =3+吵w2R2 8 (4.11)
$4.2 Rings, Discs and Cylinders Subjected to Rotation and Thermal Gradients 2 pr4w2 2Ar2 r a, = -- (~+v)+--B 8 2 where -B is a second convenient constant of integration, B po2r2 r2 8 ~r = A - - -. (3 + u)- and from eqn. (4.3, 121 (4.7) (4.8) For a solid disc the stress at the centre is given when r = 0. With r equal to zero the above equations will yield infinite stresses whatever the speed of rotation unless B is also zero, i.e. B = 0 and hence B/r2 = 0 gives the only finite solution. forces to provide the necessary balance of equilibrium if a,. were not zero. Now at the outside radius R the radial stress must be zero since there are no external Therefore from eqn. (4.7), Substituting in eqns. (4.7) and (4.8) the hoop and radial stresses at any radius r in a solid disc are given by = .o‘ [(3 + v)R2 - (1 + 3u)r2] 8 pw2 R2 pw2r2 Or = (3 + u)- 8 - (3 + u)---- 8 (4.9) (4.10) (b) Maximum stresses At the centre of the disc, where r = 0, the above equations yield equal values of hoop and radial stress which may also be seen to be the maximum stresses in the disc, i.e. maximum hoop and radial stress (at the centre) (4.1 I)
122 Mechanics of Materials 2 $4.3 At the outside of the disc,at r =R,the equations give =0 and on (1-v)20R2 4 (4.12) The complete distributions of radial and hoop stress across the radius of the disc are shown in Fig.4.3. Hoop stress o Rodial stress a Fig.4.3.Hoop and radial stress distributions in a rotating solid disc. 4.3.Rotating disc with a central hole (a)General equations The general equations for the stresses in a rotating hollow disc may be obtained in precisely the same way as those for the solid disc of the previous section, ie. ,=A-治-g+心以 8 m=A+g-1+3列心 8 The only difference to the previous treatment is the conditions which are required to evaluate the constants A and B since,in this case,B is not zero. The above equations are similar in form to the Lame equations for pressurised thick rings or cylinders with modifying terms added.Indeed,should the condition arise in service where a rotating ring or cylinder is also pressurised,then the pressure and rotation boundary conditions may be substituted simultaneously to determine appropriate values of the constants A and B. However,returning to the rotation only case,the required boundary conditions are zero radial stress at both the inside and outside radius, i.e.at r=R1, 0,=0 -3+)w2R B 0=A- 8 and at r=R2, 0,=0 B 0=A- -3+)w2 8
122 Mechanics of Materials 2 $4.3 At the outside of the disc, at r = R, the equations give po2R2 a, = 0 and UH = (1 - u)- 4 (4.12) The complete distributions of radial and hoop stress across the radius of the disc are shown in Fig. 4.3. kootal streis u, Fig. 4.3. Hoop and radial stress distributions in a rotating solid disc 43. Rotating disc with a central hole (a) General equations The general equations for the stresses in a rotating hollow disc may be obtained in precisely the same way as those for the solid disc of the previous section, i.e. B pW2r2 r2 8 (T, = A - - - (3 + u)- The only difference to the previous treatment is the conditions which are required to evaluate the constants A and B since, in this case, B is not zero. The above equations are similar in form to the Lam6 equations for pressurised thick rings or cylinders with modifying terms added. Indeed, should the condition arise in service where a rotating ring or cylinder is also pressurised, then the pressure and rotation boundary conditions may be substituted simultaneously to determine appropriate values of the constants A and B. However, returning to the rotation only case, the required boundary conditions are zero radial stress at both the inside and outside radius, i.e. at r = RI, a, = 0 and at r = R2, a, = 0 .. B po2 R: R: 8 0 = A - - - (3 + u)-
§4.3 Rings,Discs and Cylinders Subjected to Rotation and Thermal Gradients 123 Subtracting and simplifying, B=(3+)Aw2R程 8 and A=(3+)w2R+R3) 8 Substituting in eqns.(4.7)and(4.8)yields the final equation for the stresses =(3+)8 R+殴- r2 -2 (4.13) g=+(+对+)-a+3r (4.14) (b)Maximum stresses The maximum hoop stress occurs at the inside radius where r =R1, ie. CHmat= 0w2 =8[【3+R++R)-(1+3DR 4[3+R经+1-R别 (4.15) As the value of the inside radius approaches zero the maximum hoop stress value approaches D 4(3+3 This is twice the value obtained at the centre of a solid disc rotating at the same speed.Thus the drilling of even a very small hole at the centre of a solid disc will double the maximum hoop stress set up owing to rotation. At the outside of the disc when r =R2 4[3+R+(1-vR】 0 Hmit- The maximum radial stress is found by consideration of the equation =a+g+-- (4.13)(bis) dor This will be a maximum when =0, dr i.e. when 0 0=R号-2r 4=RR7 r=V(RiR2) (4.16)
94.3 Rings, Discs and Cylinders Subjected to Rotation and Thermal Gradients 123 Subtracting and simplifying, pW2 R: R; 8 B = (3 + U) po2(R: + R;) 8 and A = (3 + U) Substituting in eqns. (4.7) and (4.8) yields the final equation for the stresses (b) Maximum stresses The maximum hoop stress occurs at the inside radius where r = R1, OH,,, = - [(3 + u)(R: + R: + RZ) - (1 + 3u)R:] PW2 i.e. 8 (4.13) (4.14) (4.15) As the value of the inside radius approaches zero the maximum hoop stress value approaches PO2 -(3 + v)RZ This is twice the value obtained at the centre of a solid disc rotating at the same speed. Thus the drilling of even a very small hole at the centre of a solid disc will double the maximum hoop stress set up owing to rotation. 4 At the outside of the disc when r = R2 The maximum radial stress is found by consideration of the equation dOr This will be a maximum when - = 0, dr i.e. dr (4.13)(bis) 0 = R,R27 2 2L - 2r (4.16)
124 Mechanics of Materials 2 $4.4 Substituting for r in eqn.(4.13). 0r=(3+y 8 Ri+-R1R:-RRa] =3+ 8 (R2-R12 (4.17) The complete radial and hoop stress distributions are indicated in Fig.4.4. Hoop折tsW OH Rodial stress Fig.4.4.Hoop and radial stress distribution in a rotating hollow disc. 4.4.Rotating thick cylinders or solid shafts In the case of rotating thick cylinders the longitudinal stress oL must be taken into account and the longitudinal strain is assumed to be constant.Thus,writing the equations for the strain in three mutually perpendicular directions(see $4.2), 1 eL=EoL-oH-o） (4.18) Er= E(,-WOH -voL)= 1 ds (4.19) dr 1 EH =E(OH-vO,-VOL)= (4.20) From eqn.(4.20) Es r[oH-v(a,+L)] Differentiating, +1 [OH-vo,-vOL] Substituting for E(ds/dr)in eqn.(4.19), [doH dor doL Or-VOH VOL r dr dr dr +OH VO,-VOL doL 0=(oH-o)1+)+r- dr dr dr
124 Mechanics of Materials 2 94.4 Substituting for r in eqn. (4.13). PO2 = (3 + u)- [R2 - R112 8 (4.17) The complete radial and hoop stress distributions are indicated in Fig. 4.4. Fig. 4.4. Hoop and radial stress distribution in a rotating hollow disc. 4.4. Rotating thick cylinders or solid shafts In the case of rotating thick cylinders the longitudinal stress OL must be taken into account and the longitudinal strain is assumed to be constant. Thus, writing the equations for the strain in three mutually perpendicular directions (see $4.2), (4.19) S (4.20) 1 E r &H = -(cH - war - u0L) = - From eqn. (4.20) Differentiating, ES = r[O,y - ~(0,. + OL)] + 1 [OH - ua, - UOL] Substituting for E(ds/dr) in eqn. (4.19), 0,. - vaH - v0L = r [% - + OH - u(T, - u0L d0H dor dot 0 = (ojl - 0,.)(1 + w) + r- - wr- - ur- dr dr dr
s4.5 Rings,Discs and Cylinders Subjected to Rotation and Thermal Gradients 125 Now,since &L is constant,differentiating eqn.(4.18), L=v [doH dor dr dr dr 0=(on dom dr dr Dividing through by (1+v), 0=(oH-or)+r(1- doH-vr- dor dr dr But the general equilibrium equation will be the same as that obtained in $4.2,eqn.(4.1), dor i.e. OH-Or-r dr =po'r Therefore substituting for (oH-o,), 0=pa2r2+r +r() dr dr 0=pw2r2+r(1-) aoH dr dou dr +do=- Pu2r dr 1-) Integrating, 0H+O,=- 0w2r2 +2A 2(1-) where 2A is a convenient constant of integration.This equation can now be compared with the equivalent equation of $4.2,when it is evident that similar results for o#and o,can be obtained if (1+v)is replaced by 1/(1-v)or,alternatively,if v is replaced by v/(1-v), see $8.14.2.Thus hoop and radial stresses in rotating thick cylinders can be obtained from the equations for rotating discs provided that Poisson's ratio v is replaced by v/(1-v),e.g.the stress at the centre of a rotating solid shaft will be given by eqn.(4.11) for a solid disc modified as stated above, ie. vpa2R2 3+a-8 (4.21) 4.5.Rotating disc of uniform strength In applications such as turbine blades rotating at high speeds it is often desirable to design for constant stress conditions under the action of the high centrifugal forces to which they are subjected. Consider,therefore,an element of a disc subjected to equal hoop and radial stresses, ie. 0H=0,=0(Fig.4.5)
$4.5 Rings, Discs and Cylinders Subjected to Rotation and Thermal Gradients 125 Now, since EL is constant, differentiating eqn. (4.18), daL dOH da, - dr = [r + dr] 2 daH dar 0 = (OH - ar)(l + u) + r(l - u )- - ur(1 + u)- dr .. dr Dividing through by (1 + u), dCJH do, 0 = (OH - or) + r(l - u)- - ur- dr dr But the general equilibrium equation will be the same as that obtained in $4.2, eqn. (4.1), i.e. dar 22 a~-O,-r-=p@ r dr Therefore substituting for (OH - a,), 2 2 dar dQH do, 0 = pw r + r- + r(1- u)- - ur- dr dr dr Integrating, where 2A is a convenient constant of integration. This equation can now be compared with the equivalent equation of $4.2, when it is evident that similar results for OH and a, can be obtained if (1 + u) is replaced by 1/(1 - u) or, alternatively, if u is replaced by u/(l - u), see $8.14.2. Thus hoop and radial stresses in rotating thick cylinders can be obtained from the equations for rotating discs provided that Poisson's ratio u is replaced by u/(l - u), e.g. the stress at the centre of a rotating solid shaft will be given by eqn. (4.1 1) for a solid disc modified as stated above, i.e. (4.21) 45. Rotating disc of uniform strength In applications such as turbine blades rotating at high speeds it is often desirable to design for constant stress conditions under the action of the high centrifugal forces to which they are subjected. Consider, therefore, an element of a disc subjected to equal hoop and radial stresses, i.e. OH = a, = a (Fig. 4.5)
126 Mechanics of Materials 2 $4.6 气0 Fig.4.5.Stress acting on an element in a rotating disc of uniform strength. The condition of equal stress can only be achieved,as in the case of uniform strength canti- levers,by varying the thickness.Let the thickness be t at radius r and (t+8t)at radius (r+8r). Then centrifugal force on the element =mass x acceleration =(ptr808r)wr pta2r2808r The equilibrium equation is then pta2r2808r+a(r +8r)80(t +8t)2ot8r sin 80+0,t80 i.e.in the limit otdr pa'r"idr atdr ordt ordt =-po2r2idr =- port dr Integrating, loge1 pw2 -log.A 20 where loge A is a convenient constant. 1=Ae(-pur)/(20) where r=0 t=A=10 i.e.for uniform strength the thickness of the disc must vary according to the following equation, t=toe(-Dukr)/(2o) (4.22) 4.6.Combined rotational and thermal stresses in uniform discs and thick cylinders If the temperature of any component is raised uniformly then,provided that the material is free to expand,expansion takes place without the introduction of any so-called thermal or temperature stresses.In cases where components,e.g.discs,are subjected to thermal
126 Mechanics of Materials 2 $4.6 Fig. 4.5. Stress acting on an element in a rotating disc of uniform strength. The condition of equal stress can only be achieved, as in the case of uniform strength cantilevers, by varying the thickness. Let the thickness be t at radius r and (t + st) at radius (r + 6r). Then centrifugal force on the element = mass x acceleration = (ptr606r)w2r = ptw2r2606r The equilibrium equation is then ptw2r2606r + a(r + 6r)60(t + st) = 2at6r sin + arttie i.e. in the limit atdr = pw2r2tdr + atdr + ordt ardt = -pw r tdr dt pw2rt dr a 22 - - -- .. Integrating, pw2 r2 log, t = -- + log, A 2a where log,A is a convenient constant. .. t = Ae(-pW2r2)/(2d where r = 0 t =A =to i.e. for uniform strength the thickness of the disc must vary according to the following equation, t = toe(-mzrz)/(~) (4.22) 4.6. Combined rotational and thermal stresses in uniform discs and thick cylinders If the temperature of any component is raised uniformly then, provided that the material is free to expand, expansion takes place without the introduction of any so-called thermal or temperature stresses. In cases where components, e.g. discs, are subjected to thermal
