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