【正文】 所以表達式σl能從靜力學中獲得：σl = (Piri^2Poro^2)]/(ro^2ri^2) ()然后σl已知，：w={ r^2(Piri^2Poro^2)(12μ )+(PiPo)Ri^2ro^2(1+μ )}/Er(ro^2ri^2) (1. 27)(選自Maan H. Jawad and James R. Farr，結(jié)構(gòu)分析和工藝設(shè)備設(shè)計，約翰威力父子出版社1984)Reading Material 5Static and Dynamic Balance of Rotating BodiesThe unbalance of a single disk can detected by allowing the disk to rotate on its axle betweentwo parallel knifeedges, as shown in Fig. 1, 22. The disk will rotate and e to rest with the heavy side on the bottom. This type of unbalance is called static unbalance, since it can be detected by static means.In general, the mass of a rotor is distributed along the shaft such as in a motor armature or an automobileengine crankshaft. A test similar to the one above may indicate that such parts are in static balance, but the system may show a considerable unbalance when rotated.As an illustration, consider a shaft with two disks, as shown in Fig. 1. 23. If the two unbalance weights are equal and 180 deg. apart, the system will be statically balanced about the axis of the shaft. However, when the system is rotated, each unbalanced disk would set up a rotating centrifugal force tending to rock the shaft on its bearings. Since this type of unbalance results only from rotation we refer to it as dynamic unbalance.Fig. 1. 24 shows a general case where the system is both statically and dynamically unbalanced. It will now be shown that the unbalanced forces P and Q can always be eliminated by the addition of two correction weights in any two parallel planes of rotation.Consider first the unbalance force P, which can be replaced by two parallel forces P*a/l and P *b/l In a similar manner Q can be replaced by two parallel forces Q *c/l and Q *d/l . The two forces in each plane can then be bined into a single resultant force that can be balanced by a single correction weight as shown. The two correction weights C1 and C2 introduced in the two parallel planes pletely balanced P and Q, and the system is statically and dynamically balanced. It should be further emphasized that a dynamically balanced system is also statically balanced. The converse, however, is not always true。一個平面的軸向壓力在施加壓力前保持著增加。 10時的精確結(jié)果。（選自R. K. Sinnott 化學動力第6卷第二版，帕加馬出版社，1996 選自Raymond F. Neathery，靜力學與應(yīng)用強度材料，約翰威力父子出版公司，1985）Reading Material 4Stresses in Cylindrical Shells due to Internal PressureThe classic equation for determining stress in a thin cylindrical shell subjected to pressure is obtained from Fig. 1. 16. Summation of forces perpendicular to plane ABCD givesPL * 2r =2σθLt or σθ=PL/t ()where P=pressure, L=length of cylinder,σθ=hoop stress ,r=radius ,t = thickness the strain εθ is defined as εθ=(final lengthoriginal length)/original length and from ., εθ=[2π(r + W)2πr r]/ 2πr r or εθ=W/r ()also εr=d W /d r (1. 19) The radial deflection of a cylindrical shell subjected to internal pressure is obtained by substituting the quantity into Eq. (1. 18). Hence for thin cylindersW=Pr^2/Et ()where W = radial deflection, E =modulus of elasticity.Equations (1. 17) and (1. 20) give accurate results when r/t10. As r/t decreases, however, a more accurate expression is needed because the stress distribution through the thickness is not uniform. Recourse is then made to the thick shell” theory first developed by Lame. The derived equations are based on the forces and stresses shown in Fig. 1. 18. The theory assumes that all shearing stresses are zero due to symmetry and a plane that is normal to the longitudinal axes before pressure is applied remains plane pressurization. In other words , εl is constant at any cross sectionA relationship between σr and σθcan be obtained by taking a freebody diagram of ring dr as shown in Fig. 1. 18b. Summing forces in the vertical direction and neglecting higherorder terms, we then haveσθσr =dσr /d r ()A second relationship is written asσθ=E[εθ(1μ)+ μ(εr+εl)]/[(1+μ)(12μ)]σr=E[εr(1μ)+ μ(εθ+εl)]/[(1+μ)(12μ)] (1. 22)σl=E[εl (1μ)+ μ(εθ+εr)]/[(1+μ)(12μ)]Substituting Eqs. (1. 18) and (1. 19) into the first two expressions of Eq. () and substituting the result into Eq. (1. 21) results in d^2w/dr^2 + dw/rdr – w/r^2=0A solution of this equation is w=A r + B /r ()where A and B are constants of integration and are determined by first substituting Eq. (1 23) into the first one of Eq. (1. 22) and then applying the boundary conditionsσr = pi at r = ri and σr= po at r=roExpression (1, 23) then bees w = μrε1+1[r^2(1μ2μ^2)(Piri^2Poro^2)+ri^2ro^2(1+μ)(PiPo))/Er(ro^2r1^2) (1. 24)Once w is obtained, the values of σθdetermined from Eqs. (1. 18), and (1. 19), and () and expressed for thick cylinders asσθ=[Piri^2Poro^2+(PiPo)(ri^2 ro^2/r^2)]/(ro^2ri^2) ()σr=[ Poro^2Piri^2 +(PiPo)(ri^2 ro^2/r^2)]/(ro^2ri^2)where σr = radial stress σθ = hoop stress Pi =internal pressure P0=external pressure ri=inside radius ro=outside radius r = radius at any pointThe longitudinal stress in a thick cylinder is obtained by substituting Eqs. (1. 18)，(1. 19) and (1. 24) into the last expression of Eqs. (1. 22) to give σl = Eε1+[ 2μ(Piri^2Poro^2)]/(ro^2ri^2)This equation indicates thatσl is constant throughout a cross section because εl is constant and r does not appear in the second term. Thus the expressionσl can be obtained from statics asσl = (Piri^2Poro^2)]/(ro^2ri^2) ()With σl known, Eq. (1. 24) for the deflection of a cylinder can be expressed asw={ r^2(Piri^2Poro^2)(12μ )+(PiPo)Ri^2ro^2(1+μ )}/Er(ro^2ri^2) (1. 27)(Selected from Maan H. Jawad and James R. Farr,Structural Analysis and Design of Process Equipment,John Wiley amp。(σ3σ1)/2在