【正文】 r(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。 Sons Inc. , 1984. )材料4內(nèi)部壓力引起的柱狀殼體應(yīng)力。ABCD垂直面商的合力如下：PL * 2r =2σθLt or σθ=PL/t ()P是壓力，L是圓柱體長度，σθ是環(huán)向應(yīng)力，r是半徑，t 是厚度，張力定義如下：εθ=（最終長度原始長度）/原始長度：εθ=[2π(r + W)2πr r]/ 2πr r or εθ=W/r ()同時(shí)也有：εr=d W /d r (1. 19)。因此，對(duì)于薄壁圓柱體有：W=Pr^2/Et ()W是徑向變形偏移，E是彈性模量。 10時(shí)的精確結(jié)果。當(dāng)r/t增加時(shí)，由于應(yīng)力在厚度層上的分布不均勻所以需要更精確的表達(dá)式。救生索被應(yīng)用于厚殼體理論。這個(gè)理論由于對(duì)稱性而假設(shè)所有的剪切應(yīng)力都為0。一個(gè)平面的軸向壓力在施加壓力前保持著增加。換句話說，εl在任何截面上是不變的。σr 。在垂直方向的合力忽略高階條件的話我們得：σθσr =dσr /d r ()第二種關(guān)系如下寫：σθ=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)]，：d^2w/dr^2 + dw/rdr – w/r^2=0方程的一個(gè)解是：w=A r + B /r ()A和B是綜合的常數(shù)，應(yīng)用邊界條件σr = pi at r = ri and σr= po at r=row = μrε1+1[r^2(1μ2μ^2)(Piri^2Poro^2)+ri^2ro^2(1+μ)(PiPo))/Er(ro^2r1^2) (1. 24)w一旦獲得，,：σθ=[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)σr=徑向應(yīng)力 σθ=環(huán)向應(yīng)力 Pi =內(nèi)部壓力 P0=外部壓力 ri=內(nèi)徑 ro=外徑 r = 任意半徑,，：σl = Eε1+[ 2μ(Piri^2Poro^2)]/(ro^2ri^2)這個(gè)方程表示通過截面的σl是常數(shù)，因?yàn)棣舕是常數(shù)，r在第二時(shí)期沒有出現(xiàn)。所以表達(dá)式σl能從靜力學(xué)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(jì)，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。 a statically balanced system may be dynamically unbalanced.Example A rotor 4 in. long has an unbalance of 3 oz. in. in a plane 1 in. from the left end, and 2 oz. in. in the middle plane. Its angular position is 90 deg. in the clockwise direction from the first unbalance when viewed from the left end. Determine the corrections in the two end planes, giving magnitude and angular positions.Solution. The 3oz. in. unbalance is equivalent to2*1/4 oz. in. at the left end and 3/4 oz. in. at the right end,as shown in Fig. 1. 25. The 2 oz. in. at the middle is obviously equal to 1 oz. in. at the ends.Combining the two unbalances at each end, the corrections are：Left end：oz. in. to be rem