§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]