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)