CHAPTER 10 THICK CYLINDERS Summary The hoop and radial stresses at any point in the wall cross-section of a thick cylinder at radius r are given by the Lame equations: 分 hoop stress aH=A+ B radial stress ,=A- With internal and external pressures P:and P2 and internal and external radii R:and R2 respectively,the longitudinal stress in a cylinder with closed ends is PR-P2R3 0L= (R-R) =Lame constant A Changes in dimensions of the cylinder may then be determined from the following strain formulae: circumferential or hoop strain diametral strain =OH_0, E ,0L E ongn-登- E-V E For compound tubes the resultant hoop stress is the algebraic sum of the hoop stresses resulting from shrinkage and the hoop stresses resulting from internal and external pressures. For force and shrink fits of cylinders made of different materials,the total interference or shrinkage allowance (on radius)is [eH。-eH]r where eand eare the hoop strains existing in the outer and inner cylinders respectively at the common radius r.For cylinders of the same material this equation reduces to Elon-on For a hub or sleeve shrunk on a solid shaft the shaft is subjected to constant hoop and radial stresses,each equal to the pressure set up at the junction.The hub or sleeve is then treated as a thick cylinder subjected to this internal pressure. 215
CHAPTER 10 THICK CYLINDERS Summary The hoop and radial stresses at any point in the wall cross-section of a thick cylinder at radius r are given by the Lam6 equations: B hoop stress OH = A + - r2 B radial stress cr, = A - - r2 With internal and external pressures P, and P, and internal and external radii R, and R, respectively, the longitudinal stress in a cylinder with closed ends is P1R: - P2R: aL = = Lame constant A (R: - R:) Changes in dimensions of the cylinder may then be determined from the following strain formulae: circumferential or hoop strain = diametral strain 'JH cr OL =-- v- - vEEE OL or OH longitudinal strain = - - v- - v- EEE For compound tubes the resultant hoop stress is the algebraic sum of the hoop stresses resulting from shrinkage and the hoop stresses resulting from internal and external pressures. For force and shrink fits of cylinders made of diferent materials, the total interference or shrinkage allowance (on radius) is CEH, - 'Hi 1 where E", and cH, are the hoop strains existing in the outer and inner cylinders respectively at the common radius r. For cylinders of the same material this equation reduces to For a hub or sleeve shrunk on a solid shaft the shaft is subjected to constant hoop and radial stresses, each equal to the pressure set up at the junction. The hub or sleeve is then treated as a thick cylinder subjected to this internal pressure. 21 5
216 Mechanics of Materials §10.1 Wire-wound thick cylinders If the internal and external radii of the cylinder are R,and R2 respectively and it is wound with wire until its external radius becomes R3,the radial and hoop stresses in the wire at any radius r between the radii R2 and R3 are found from: hoop stress r-(+)()} where T is the constant tension stress in the wire. The hoop and radial stresses in the cylinder can then be determined by considering the cylinder to be subjected to an external pressure equal to the value of the radial stress above when r=R2. When an additional internal pressure is applied the final stresses will be the algebraic sum of those resulting from the internal pressure and those resulting from the wire winding. Plastic yielding of thick cylinders For initial yield,the internal pressure P,is given by: R,=录[R好-R1 For yielding to a radius R, P-,[是，-购 and for complete collapse, -,[是] 10.1.Difference in treatment between thin and thick cylinders-basic assumptions The theoretical treatment of thin cylinders assumes that the hoop stress is constant across the thickness of the cylinder wall(Fig.10.1),and also that there is no pressure gradient across the wall.Neither of these assumptions can be used for thick cylinders for which the variation of hoop and radial stresses is shown in Fig.10.2,their values being given by the Lame equations: =M+,片4ndg,=A- B Γ2 Development of the theory for thick cylinders is concerned with sections remote from the
216 Mechanics of Materials $10.1 Wire-wound thick cylinders If the internal and external radii of the cylinder are R, and R, respectively and it is wound with wire until its external radius becomes R,, the radial and hoop stresses in the wire at any radius r between the radii R, and R3 are found from: radial stress = ( -27i-) r2 - R: Tlog, (-) Ri - R: r2 - Rt r2 + R: R; - R: hoop stress = T { 1 - ( - 2r2 )'Oge(r2-Rf)} where T is the constant tension stress in the wire. The hoop and radial stresses in the cylinder can then be determined by considering the cylinder to be subjected to an external pressure equal to the value of the radial stress above when r = R,. When an additional internal pressure is applied the final stresses will be the algebraic sum of those resulting from the internal pressure and those resulting from the wire winding. Plastic yielding of thick cylinders For initial yield, the internal pressure P, is given by: For yielding to a radius R,, and for complete collapse, 10.1. Difference io treatment between thio and thick cylinders - basic assumptions The theoretical treatment of thin cylinders assumes that the hoop stress is constant across the thickness of the cylinder wall (Fig. lO.l), and also that there is no pressure gradient across the wall. Neither of these assumptions can be used for thick cylinders for which the variation of hoop and radial stresses is shown in Fig. 10.2, their values being given by the Lame equations: B B an=A+- and q=A-- r2 r2 Development of the theory for thick cylinders is concerned with sections remote from the
§10.2 Thick Cylinders 217 2t Fig.10.1.Thin cylinder subjected to internal pressure. c,(tensile】 Stress distributions a.(compressive c-p A+B/r2 ,A-B/r2 Fig.10.2.Thick cylinder subjected to internal pressure. ends since distribution of the stresses around the joints makes analysis at the ends particularly complex.For central sections the applied pressure system which is normally applied to thick cylinders is symmetrical,and all points on an annular element of the cylinder wall will be displaced by the same amount,this amount depending on the radius of the element.Consequently there can be no shearing stress set up on transverse planes and stresses on such planes are therefore principal stresses(see page 331).Similarly,since the radial shape of the cylinder is maintained there are no shears on radial or tangential planes,and again stresses on such planes are principal stresses.Thus,consideration of any element in the wall of a thick cylinder involves,in general,consideration of a mutually prependicular,tri-axial, principal stress system,the three stresses being termed radial,hoop (tangential or circumferential)and longitudinal (axial)stresses. 10.2.Development of the Lame theory Consider the thick cylinder shown in Fig.10.3.The stresses acting on an element of unit length at radius r are as shown in Fig.10.4,the radial stress increasing from a,to o,+do,over the element thickness dr (all stresses are assumed tensile), For radial equilibrium of the element: o,+do,)r+dn)d0x1-o,×rd6×1=2oH×dr×1×sin2
0 10.2 Thick Cylinders 217 Fig. 10.1. Thin cylinder subjected to internal pressure. Stress distributions uH=A + B/r2 u,= A-B/r2 Fig. 10.2. Thick cylinder subjected to internal pressure. ends since distribution of the stresses around the joints makes analysis at the ends particularly complex. For central sections the applied pressure system which is normally applied to thick cylinders is symmetrical, and all points on an annular element of the cylinder wall will be displaced by the same amount, this amount depending on the radius of the element. Consequently there can be no shearing stress set up on transverse planes and stresses on such planes are therefore principal stresses (see page 331). Similarly, since the radial shape of the cylinder is maintained there are no shears on radial or tangential planes, and again stresses on such planes are principal stresses. Thus, consideration of any element in the wall of a thick cylinder involves, in general, consideration of a mutually prependicular, tri-axial, principal stress system, the three stresses being termed radial, hoop (tangential or circumferential) and longitudinal (axial) stresses. 10.2. Development of the Lam6 theory Consider the thick cylinder shown in Fig. 10.3. The stresses acting on an element of unit length at radius rare as shown in Fig. 10.4, the radial stress increasing from a, to a, + da, over the element thickness dr (all stresses are assumed tensile), For radial equilibrium of the element: de (a,+da,)(r+dr)de x 1-6, x rd0 x 1 = 2aH x dr x 1 x sin- 2
218 Mechanics of Materials §10.2 +do dr Fig.10.3. a,+do, Unit length de Fig10.4. For small angles: do do sin22 radian Therefore,neglecting second-order small quantities, rdo,+a,dr oHdr dor o,+dr =0H dor or GH-c,=r (10.1) Assuming now that plane sections remain plane,i.e.the longitudinal strain &L is constant across the wall of the cylinder, then 6L=E[aL-vo,-vOH] -E [GL-Y(o,+oH)]=constant It is also assumed that the longitudinal stress aL is constant across the cylinder walls at points remote from the ends. ,+H constant 2A (say) (10.2)
218 Mechanics of Materials 410.2 Fig. 10.3. q +do, length Fig. 10.4. For small angles: . d9 d9 22 sin - - radian Therefore, neglecting second-order small quantities, rda, + a,dr = aHdr .. or do, a, + r- = an dr (10.1) Assuming now that plane sections remain plane, Le. the longitudinal strain .zL is constant across the wall of the cylinder, 1 E 1 E then EL = - [aL - va, - VaH] = - [aL - v(a, + OH)] = constant It is also assumed that the longitudinal stress aL is constant across the cylinder walls at points remote from the ends. .. a, + aH = constant = 2A (say) (10.2)
§10.3 Thick Cylinders 219 Substituting in (10.1)for oH, dor 2A-0,-0,= 'dr Multiplying through by r and rearranging, 2a,7+r2dg-24=0 dr o,2-Ar内=0 d i.e. Therefore,integrating, a,r2-Ar2 constant =-B (say) 6,=A- B (10.3) and from eqn.(10.2) B GH=A+ (10.4) The above equations yield the radial and hoop stresses at any radius r in terms of constants A and B.For any pressure condition there will always be two known conditions of stress (usually radial stress)which enable the constants to be determined and the required stresses evaluated. 10.3.Thick cylinder-internal pressure only Consider now the thick cylinder shown in Fig.10.5 subjected to an internal pressure P,the external pressure being zero. Fig.10.5.Cylinder cross-section. The two known conditions of stress which enable the Lame constants A and B to be determined are: Atr=R1 ,=-P and at r=R2 G,=0 N.B.-The internal pressure is considered as a negative radial stress since it will produce a radial compression(i.e.thinning)of the cylinder walls and the normal stress convention takes compression as negative
$10.3 Thick Cylinders 219 Substituting in (10.1) for o~, dor 2A-ar-ar = r- dr Multiplying through by r and rearranging, dor 2orr + r2 - - 2Ar = 0 dr i.e. Therefore, integrating, .. d -(or? -A?) = 0 dr orrZ - Ar2 = constant = - B (say) and from eqn. (10.2) (10.4) B UH=A+- rz The above equations yield the radial and hoop stresses at any radius r in terms of constants A and B. For any pressure condition there will always be two known conditions of stress (usually radial stress) which enable the constants to be determined and the required stresses evaluated. 10.3. Thick cylinder - internal pressure only Consider now the thick cylinder shown in Fig. 10.5 subjected to an internal pressure P, the external pressure being zero. Fig. 10.5. Cylinder cross-section. The two known conditions of stress which enable the Lame constants A and B to be determined are: At r = R, or= -P and at r = R, or =O N.B. -The internal pressure is considered as a negative radial stress since it will produce a radial compression (i.e. thinning) of the cylinder walls and the normal stress convention takes compression as negative
220 Mechanics of Materials §10.4 Substituting the above conditions in eqn.(10.3), -P=A- B R 0=A B R PR i.e. A-7 R经-R) and B=1 PR2R3 R好-R) B radial stress a,=A- R-] PR2 []-P鬥 (10.5) where k is the diameter ratio D2/D=R2/R1 and bpms6“R[+] PR? R]-P[] (10.6) These equations yield the stress distributions indicated in Fig.10.2 with maximum values of both o,and oH at the inside radius. 10.4.Longitudinal stress Consider now the cross-section of a thick cylinder with closed ends subjected to an internal pressure P and an external pressure P2 (Fig.10.6). Closed ends Fig.10.6.Cylinder longitudinal section. For horizontal equilibrium: P,×πR子-P2×πR3=LXπ(R经-R)
220 Mechanics of Materials 0 10.4 Substituting the above conditions in eqn. (10.3), i.e. B radial stress 6, = A - - r2 (10.5) where k is the diameter ratio D2 /Dl = R , f R , and hoop stress o,, = (10.6) PR: rZ+Ri wzm2 + 1 (R;-R:) [TI='[ k2-1 ] - These equations yield the stress distributions indicated in Fig. 10.2 with maximum values of both a, and aH at the inside radius. 10.4. Longitudinal stress Consider now the cross-section of a thick cylinder with closed ends subjected to an internal pressure PI and an external pressure P, (Fig. 10.6). UL - t t\ Closed ends Fig. 10.6. Cylinder longitudinal section. For horizontal equilibrium: P, x ITR: - P, x IT R$ = aL x n(R; - R:)
§10.5 Thick Cylinders 221 where ot is the longitudinal stress set up in the cylinder walls, longitudinal stress=一】 PR-P2R3 R子-R1 (10.7) i.e.a constant. It can be shown that the constant has the same value as the constant A of the Lame equations. This can be verified for the "internal pressure only"case of $10.3 by substituting P2=0in egn.(10.7)above. For combined internal and external pressures,the relationship oL=A also applies. 10.5.Maximum shear stress It has been stated in $10.1 that the stresses on an element at any point in the cylinder wall are principal stresses. It follows,therefore,that the maximum shear stress at any point will be given by eqn.(13.12) as imax=01-03 2 i.e.half the difference between the greatest and least principal stresses. Therefore,in the case of the thick cylinder,normally, OH-O 2 since o is normally tensile,whilst o,is compressive and both exceed ot in magnitude. m-(a+)-(4-)] B (10.8) The greatest value oftthus normally occurs at the inside radius wherer=R 10.6.Change of cylinder dimensions (a)Change of diameter It has been shown in $9.3 that the diametral strain on a cylinder equals the hoop or circumferential strain. Therefore change of diameter diametral strain x original diameter circumferential strain x original diameter With the principal stress system of hoop,radial and longitudinal stresses,all assumed tensile, the circumferential strain is given by H ELon-va,-vaL]
8 10.5 Thick Cyli&rs 22 1 where bL is the longitudinal stress set up in the cylinder walls, PIR; - P, Ri R;-R: .. longitudinal stress nL = (10.7) i.e. a constant. It can be shown that the constant has the same value as the constant A of the Lame equations. This can be verified for the “internal pressure only” case of $10.3 by substituting P, = 0 in eqn. (10.7) above. For combined internal and external pressures, the relationship (TL = A also applies. 10.5. Maximum shear stress It has been stated in $10.1 that the stresses on an element at any point in the cylinder wall It follows, therefore, that the maximum shear stress at any point will be given by eqn. (13.12) are principal stresses. as bl-a3 7max= ___ 2 i.e. half the diference between the greatest and least principal stresses. Therefore, in the case of the thick cylinder, normally, OH - Qr 7ma7.= ~ 2 since on is normally tensile, whilst Q, is compressive and both exceed nL in magnitude. B nux = - r2 (10.8) The greatest value of 7,,thus normally occurs at the inside radius where r = R,. 10.6. Cbange of cylinder dimensions (a) Change of diameter It has been shown in $9.3 that the diametral strain on a cylinder equals the hoop or Therefore arcumferential strain. change of diameter = diametral strain x original diameter = circumferential strain x original diameter With the principal stress system of hoop, radial and longitudinal stresses, all assumed tensile, the circumferential strain is given by 1 E EH = - [QH - Vbr - VbL]
222 Mechanics of Materials §10.7 Thus the change of diameter at any radius r of the cylinder is given by aD=Eaa-vG,-vo】 2r (10.9) (b)Change of length Similarly,the change of length of the cylinder is given by △L= ELoL-av,-vou] (10.10) 10.7.Comparison with thin cylinder theory In order to determine the limits of D/t ratio within which it is safe to use the simple thin cylinder theory,it is necessary to compare the values of stress given by both thin and thick cylinder theory for given pressures and D/t values.Since the maximum hoop stress is normally the limiting factor,it is this stress which will be considered. From thin cylinder theory: D OH=P 2t i.e. OH K P=2 where K D/t For thick cylinders,from eqn.(10.6), PR? R 0H=(R经-R) 1+ i.e. -P at r=R (10.11) Now,substituting for R2 =R+t and D 2R1, T生D2+D+0]P t(D+t) D2 -22+D+1P i.e. 2水++ K2 (10.12) Thus for various D/t ratios the stress values from the two theories may be plotted and compared;this is shown in Fig.10.7. Also indicated in Fig.10.7 is the percentage error involved in using the thin cylinder theory. It will be seen that the error will be held within 5%if D/t ratios in excess of 15 are used
222 Mechanics of Materials Thus the change of diameter at any radius r of the cylinder is given by 2r E AD = -[uH- VU, - VUL] (b) Change of length Similarly, the change of length of the cylinder is given by L E AL = -[uL-uv,-vuH] 0 10.7 (10.9) (10.10) 10.7. Comparison with thin cylinder theory In order to determine the limits of D/t ratio within which it is safe to use the simple thin cylinder theory, it is necessary to compare the values of stress given by both thin and thick cylinder theory for given pressures and D/t values. Since the maximum hoop stress is normally the limiting factor, it is this stress which will be considered. From thin cylinder theory: i.e. where K = D/t _- OHP2 For thick cylinders, from eqn. (10.6), 1.e. Now, substituting for R, = R, + t and D = 2R,, aHmx t(D + t) D2 + 11, = [ 2t2 (D/t + 1) i.e. K2 -- P 2(K+l)+l (10.1 1) (10.12) Thus for various D/t ratios the stress values from the two theories may be plotted and compared; this is shown in Fig. 10.7. Also indicated in Fig. 10.7 is the percentage error involved in using the thin cylinder theory. It will be seen that the error will be held within 5 % if D/t ratios in excess of 15 are used
§10.8 Thick Cylinders 223 error Thick cylinder 60 theory 50 40 6 30 Thin cylinder ole theory 20 0 24古日4后 K=D/t Fig.10.7.Comparison of thin and thick cylinder theories for various diameter/thickness ratios. However,if D is taken as the mean diameter for calculation of the thin cylinder values instead of the inside diameter as used here,the percentage error reduces from 5%to approximately 0.25%at D/t =15. 10.8.Graphical treatment-Lame line The Lame equations when plotted on stress and I/r2 axes produce straight lines,as shown in Fig.10.8. Stress Hoop stress .A+8/r2 Slope B Radiol stress Slope-B .A-B/r2 Fig.10.8.Graphical representation of Lame equations-Lame line. Both lines have exactly the same intercept A and the same magnitude of slope B,the only difference being the sign of their slopes.The two are therefore combined by plotting hoop stress values to the left of the axis(again against 1/r2)instead of to the right to give the single line shown in Fig.10.9.In most questions one value of o,and one value of o#,or alternatively two values of o,,are given.In both cases the single line can then be drawn. When a thick cylinder is subjected to external pressure only,the radial stress at the inside radius is zero and the graph becomes the straight line shown in Fig.10.10
$10.8 Thick Cylinders 223 1 Thick cylinder theory a 60 40 K= D/1 Fig. 10.7. Comparison of thin and thick cylinder theories for various diarneter/thickness ratios. However, if D is taken as the mean diameter for calculation of the thin cylinder values instead of the inside diameter as used here, the percentage error reduces from 5% to approximately 0.25 % at D/t = 15. 10.8. Graphical treatment -Lame line The Lame equations when plotted on stress and 1 /rz axes produce straight lines, as shown in Fig. 10.8. Stress b Rodiol' stress yo' / =A- B/r' Fig. 10.8. Graphical representation of Lam6 equations- Lam6 line. Both lines have exactly the same intercept A and the same magnitude of slope B, the only difference being the sign of their slopes. The two are therefore combined by plotting hoop stress values to the left of the aaxis (again against l/rz) instead of to the right to give the single line shown in Fig. 10.9. In most questions one value of a, and one value of oH, or alternatively two values of c,, are given. In both cases the single line can then be drawn. When a thick cylinder is subjected to external pressure only, the radial stress at the inside radius is zero and the graph becomes the straight line shown in Fig. 10.10
224 Mechanics of Materials §10.9 Hoop stresses Rodial stresses Cylinder wall snipo A Externol r2 pressure Internal snipoJ Jaino (an-)ssaJs Cylinder Fig.10.9.Lame line solution for cylinder with internal and external pressures. snipDJ snipo Jeino snipo A External Dr色ssUr色 (negafive 乐0a1 inner rodius 6 Fig.10.10.Lame line solution for cylinder subjected to external pressure only. N.B.-From $10.4 the value of the longitudinal stress oL is given by the intercept A on the o axis. It is not sufficient simply to read off stress values from the axes since this can introduce appreciable errors.Accurate values must be obtained from proportions of the figure using similar triangles. 10.9.Compound cylinders From the sketch of the stress distributions in Fig.10.2 it is evident that there is a large variation in hoop stress across the wall of a cylinder subjected to internal pressure.The material of the cylinder is not therefore used to its best advantage.To obtain a more uniform hoop stress distribution,cylinders are often built up by shrinking one tube on to the outside of another.When the outer tube contracts on cooling the inner tube is brought into a state of
224 Mechanics of Materials $10.9 Fig. 10.9. Lam15 line solution for cylinder with internal and external pressures. Fig. 10.10. Lam6 line solution for cylinder subjected to external pressure only. N.B. -From $10.4 the value of the longitudinal stress CT is given by the intercept A on the u axis. It is not sufficient simply to read off stress values from the axes since this can introduce appreciable errors. Accurate values must be obtained from proportions of the figure using similar triangles. 10.9. Compound cylinders From the sketch of the stress distributions in Fig. 10.2 it is evident that there is a large variation in hoop stress across the wall of a cylinder subjected to internal pressure. The material of the cylinder is not therefore used to its best advantage. To obtain a more uniform hoop stress distribution, cylinders are often built up by shrinking one tube on to the outside of another. When the outer tube contracts on cooling the inner tube is brought into a state of
