力學專業(yè)外文翻譯

1、<p><b>　　中文2500字</b></p><p><b>　　附 錄：外文翻譯</b></p><p>　　5.1Introduction</p><p>　　Cylindrical shells are used innuclear,fossil and petrochemical industrie

2、s. They are also used in heat exchangers of the shell and tube type.Generally.These vessels are easy to fabricate and install and economical to maintain. The design procedures in pressure vessel codes for cylindrical she

3、lls are mostly based on linear elastic assumption,occasionally allowing for limited inelastic behavior over a localized region.The shell thickness is the major design parameter and is usually controlledby int</p>

4、<p>　　5.2 Thin-shell equations</p><p>　　A shell is a curved plate-type structure.We shall limit our discussion to Shells of revolutions.Referring to Figure5.1 this is denoted by anangle ,The meridional

5、radius r1 and the conical radius r2,from the center line.The horizontal radius when the axis is vertical is r. If the shell thickness is t,with z being the coordinate across the thickness,following the convention of Flug

6、ge, We have the following stress resultants: </p><p><b>　　(5.1)</b></p><p><b>　　(5.2)</b></p><p><b>　　(5.3)</b></p><p>　　Figure 5.1

7、Thin shell of revolution.</p><p><b>　　(5.4)</b></p><p>　　These stress resultants are assumed to be due only to an internal pressure, p,acting in the direction of r. For membrane she

8、lls where the Effects of bending can be ignored,all the moments are zero and further development leads to</p><p>　　The following equations result from considering force equilibrium along with the additional

9、 requirement of rotational symmetry:</p><p><b>　　(5.6)</b></p><p><b>　　(5.7)</b></p><p>　　Noting that,we have,by solving Eqs.(5.6)and(5.7),</p><p&

10、gt;<b>　　(5.8)</b></p><p><b>　　(5.9)</b></p><p>　　The above two equations are the results for a general shell of revolution. Two speci?c cases result:</p><p>

11、　　For a spherical shellof radius R, r1= r2 =R,which gives</p><p><b>　　(5.10)</b></p><p>　　For a cylindrical pressure vessel of radius R,we have r1 =; r2 = R,which gives</p>&l

12、t;p><b>　　(5.11)</b></p><p><b>　　(5.12)</b></p><p>　　This gives the hoop stress</p><p><b>　　(5.13)</b></p><p>　　and the longitudin

13、al stress</p><p><b>　　(5.14)</b></p><p>　　These results will be shown to be identical to the results that follow. Let us consider a long thin cylindrical shell of radius R and thickn

14、ess t, subject to an internal pressure p.By thin shell we mean the ones having the ratio R/t typically greater than about10.If the ends of the cylindrical shell are closed,there will be stresses in the hoop as well as th

15、e axial (longitudinal) directions.</p><p>　　A section of such a shell is shown in Figure5.2. The hoop(circumfer-ential)stress, and the longitudinal stress, are indicated in the ?gure.The shell is assumed to

16、 be long and thin resulting in andto be uniform through the thickness.Therefore in this case and are also referred to as membrane stress(there are no bending stresses associated with this type of loading).</p><

17、;p>　　Considering equilibrium across the cut section,we have,</p><p>　　Figure 5.2 Thin cylindrical shell.</p><p>　　which gives</p><p><b>　　(5.15)</b></p><p&

18、gt;　　Considering a cross-section of the shell perpendicular to its axis,we have</p><p>　　Which gives</p><p><b>　　（5.16</b></p><p>　　5.3 Thick-shell equations</p>

19、<p>　　For R/t ratios typically less than 10,Eqs.(5.15) and (5.16) tend not to be accurate,and thick-shell equations have to be used. Consider a thick cylindrical shell of inside radius Ri and outside radius Ro subje

20、cted to an internal pressure p as shown in Figure5.3 .The stress function for this case(refer to AppendixI)is given as a Function of radius r as</p><p><b>　　(5.17)</b></p><p>　　Figur

21、e 5.3 Thick cylindrical shell.</p><p>　　with A and B to be determined by the boundary conditions. If we indicate the radial stress as and the hoop and longitudinal Stress as indicated previously by and,we ha

22、ve</p><p><b>　　(5.18)</b></p><p><b>　　(5.19)</b></p><p>　　The constants A and B are determined from the following boundary conditions:</p><p>&l

23、t;b>　　at </b></p><p>　　at (5.20)</p><p>　　Substituting(5.20)into(5.18)and(5.19),we have</p><p><b>　　(5.21)</b></p><p>　　Denoting the

24、 ratio of the outside to inside radii as m,so that m =Ro/Ri, We obtain theradial and hoop stresses</p><p><b>　　(5.22)</b></p><p><b>　　(5.23) </b></p><p>　　F

25、igure5.4 shows the radial and hoop stress distributions.</p><p>　　The longitudinal stress, is determined by considering the Equilibrium of forces across a plane normal to the axis of the shell,which gives<

26、;/p><p><b>　　(5.24)</b></p><p>　　This is of course based on the assumption that the longitudinal stress is a Form of membrane stress in that there is no variation across the thickness o

27、f The shell.Thus we have</p><p>　　Figure 5.4 Hoop and radial stress distribution.</p><p><b>　　(5.25)</b></p><p>　　It should be noted however that the solutions indicated

28、 by Eqs.(5.22), (5.23),and(5.25) are valid for regions remote from discontinuities.</p><p>　　5.4 Approximate equations</p><p>　　For a moderately thick shell employing thin-shell theory and using

29、 the Mean radius Rm we get the expression of the hoop stress, ,as</p><p><b>　　(5.26)</b></p><p>　　Equating the hoop stress, ,to the code-allowable design stress, , We have</p>

30、<p><b>　　(5.27)</b></p><p>　　Rewriting Eq.(5.27) in terms of the outside radius, we have,</p><p><b>　　(5.28)</b></p><p>　　Once again equating the hoo

31、p stress to the code-allowable design stress, S,we have</p><p><b>　　(5.29)</b></p><p>　　The equations in the ASME Boiler and Pressure Vessel Code are based On equating the maximum me

32、mbrane stress to the allowable stress Corrected for weld joint ef?ciency.The allowable stress, ,is replaced by The term (to be explained later).In ASME Boiler and Pressure Vessel</p><p>　　Code,SectionVIII Di

33、vision1, The Eqs .(5.27) and (5.29) are modi?ed as:</p><p><b>　　(5.30)</b></p><p><b>　　(5.31)</b></p><p>　　In ASME Boiler and Pressure Vessel Code SectionIII

34、,Division1, the equationsusedare</p><p><b>　　(5.32)</b></p><p><b>　　(5.33)</b></p><p>　　In Eqs .(5.30)and (5.32), R stands for the inside radius, ,whereas in

35、 Eqs.(5.31) and (5.33) it stands for the outside radius, .In both of the above equations, S is the allowable stress and E is the joint ef?ciency.This joint ef?ciency is employed because cylindrical shells are often fabri

36、cated by welding.The values of E depend on the type of radio graphic examination performed at various welded seams of the shell.</p><p>　　5.5Buckling of cylindrical shells</p><p>　　Consider a lo

37、ng,thin cylindrical shell of mean diameter D and wall thickness t subjected to an external pressure P.The cylinder is in a stable con?guration as long as it remains circular in shape.If there is an initial ellipticity,th

38、e cylinder will be in an unstable condition and will eventually buckle.</p><p>　　If the cylinder is suf?ciently long,the end effects may be neglected and The problem may be considered as two-dimensional.<

39、/p><p>　　Summing up the forces in the radial direction,we have</p><p><b>　　or</b></p><p><b>　　or</b></p><p>　　Summation of forces in the tangential

40、 direction(see Figure5.5) gives</p><p><b>　　-</b></p><p>　　Figure 5.5 Equilibrium of a shell element.</p><p>　　With ,we have</p><p>　　Assuming deviation fro

41、m circular shape to be small,we have</p><p>　　Where C is a constant</p><p><b>　　When</b></p><p><b>　　Therefore</b></p><p>　　For a curved shell:&

42、lt;/p><p><b>　　with </b></p><p><b>　　With </b></p><p><b>　　at </b></p><p><b>　　at </b></p><p>　　With a minimum

43、 value of n=1,we get</p><p><b>　　(5.34)</b></p><p>　　For cylinder with shorter lengths,where the ends are free to expand Axially and rotate with the restriction of expanding radially

44、,thecritical Pressure is given by</p><p>　　When is large:</p><p><b>　　(5.35)</b></p><p>　　So that it becomes identical to the buckling pressure inEq.(5.34)for n= 2. In t

45、he ASME code,the critical pressure is calculated for two situations, Involving the ratio of the outside diameter to the thickness(Do/t)</p><p><b>　　1.</b></p><p><b>　　2.</b&

46、gt;</p><p>　　The basic Eq.(5.19) is modi?ed to include inelastic buckling.For the ?rst Case above,a factor of safety of 3.0 isused.Fo rthe second case,avariable Factor of safety isused starting with a factor

47、 of safety of 3.0 for Do/t= 10 to 2.0 for Do/t =4.As the cylinder becomes progressively thicker,the buckling Ceases t be a plausible mode of failure.The ASME procedure is an Involved one in which two sets of curves have

48、 to be used to investigate buckling.The procedure becomes complicated for large Do/t,w</p><p>　　5.6 Discontinuity stresses in pressure vessels </p><p>　　Let us take the special case of discontin

49、uity at a juncture between a cylindrical vessel and ahemispherical head subjected to internal pressure p. For simplicity let us assume the spherical head and the cylindrical shell are Of the same thickness.If the mean

50、radius and thethickness of the shell are Denoted by Rm and t respectively,then the hoop and the longitudinal Stresses in the cylindrica lshell are given by:</p><p><b>　　(5.36)</b></p><

51、p><b>　　(5.37)</b></p><p>　　The hoop and the longitudinal stresses in the spherical shell are given by:</p><p><b>　　(5.38)</b></p><p><b>　　(5.39)<

52、/b></p><p>　　The radial growth or dilation of the cylindrical shell under internal pressure p is given by</p><p><b>　　(5.40)</b></p><p>　　That of the spherical region

53、 is given by</p><p><b>　　(5.41)</b></p><p>　　Where is the Poisson’s ratio.</p><p>　　If the spherical and the cylindrical portions were separated,the Difference in the ra

54、dial growth would be</p><p><b>　　(5.42)</b></p><p>　　In the actual vessel the hemispherical head and the cylindrical shell are Kept in place by shear force, V and moment M per unit c

55、ircumference. These discontinuity forces produce local bending stresses in the adjacent Portions of the vessel.The de?ection and the slope induced at the edges of</p><p>　　The cylindrical and spherical porti

56、ons by the force V are equal.The Continuity at the juncture will bes atis?ed if M equals zero and V is such That it produces a de?ection of /2. Applying the results from semi-in?nite beam on an elastic foundation Due to

57、M and V, And substituting the spring rate of the foundation by </p><p><b>　　(5.43)</b></p><p>　　where is the attenuation factor,given by</p><p><b>　　(5.44)</b&

58、gt;</p><p>　　We have with and M=0.at x=0</p><p><b>　　(5.45)</b></p><p>　　Substituting the value of from Eq.(5.45)we have</p><p><b>　　(5.46)</b>

59、;</p><p>　　The longitudinal stress and the hoop stress distribution in the cylindrical region is then given by</p><p>　　(5.47) (5.48)</p><p>　　Using numerical values as, p= 2

60、MPa, Rm = 1m, t = 25mm,and Poisson’sratio, =0.3,we have from Eq.(5.44), = 0.008127/mm and the Longitudinal and the hoop stresses become</p><p>　　Mpa (5.49) </p><p>　　Mpa

61、 (5.50)</p><p>　　In Eq .(5.49) the ?rst quantity–the membrane longitudinal stress–is a constant (equal to 40MPa) along the length of the vessel,while the second quantity–the bending stress–varies alo

62、ng the length.In Eq.(5.50) the membrane hoop stress (equal to 80MPa) stays constant along the length, while the direct compression stress due to shortening of the radius and the bending stress varies along the length of

63、the vessel.</p><p>　　1.Find the thickness of a cylindrical shell 2m in diameter if it is required to contain an internal pressure of 7MPa.The allowable stress in the material is140MPa.</p><p>　　

64、2.A thick cylindrical shell of 1.2m inside diameter and 1.5m outside diameter is subjected to an internal pressure of 35MPa.Determine the following:</p><p>　　a.Magnitude and location of the maximum hoop stre

65、ss</p><p>　　b.Magnitude of the maximum radial stress and its location</p><p>　　c.Average hoop stress</p><p>　　3.A thick cylinder has an inside diameter of 300mm and an outside Diame

66、ter of 450mm.If the allowable stress is175MPa,what is the maximum internal pressure that can be applied?</p><p>　　4.A cylinder has an inside radius of 1.8m and is subjected to an Internal pressure of 0.35 MP

67、a.What is th required thickness if the Allowable stress is105 MPa?</p><p>　　5.For problem 4,what is the required thickness if thick cylinder equations were used?</p><p>　　6.Using ASME Boiler and

68、 Pressure Vessel Code equations,determine the thickness of a cast-iron pressure vessel subjected to an internal pressure of 0.5 Mpa using a joint ef?ciency of 85 percent and a corrosion allowance of 1.5mm.The allowable s

69、tress (Sm)of the material is14MPa.</p><p><b>　　第五章</b></p><p><b>　　5.1簡介</b></p><p>　　圓柱形容器主要用于核能，石油化工工業(yè)，它們也用于管殼式換熱器。通常，這些容器制造、安裝容易，維護經濟。壓力容器標準中圓柱形容器的設計程序主要是

70、基于線性彈性假設，有時允許局部區(qū)域有限的非彈性行為。外殼厚度是主要設計參數(shù)，通常是由內部壓力控制，有時是由產生屈曲的外部壓力控制。添加載荷、不連續(xù)應力、熱應力在控制厚度的過程中也很重要。圓柱形容器的基本厚度主要是基于簡化應力分析和建筑物料的許用應力。在不同的設計標準中基本方程有一些變化。有一些方程是基于厚殼方程。在這一章節(jié)，這些理論將被討論。另外，我們還將討論圓柱形容器外壓失穩(wěn)在情形。</p><p><b

71、>　　5.2薄殼方程</b></p><p>　　對于彎板型結構，我們只討論旋轉殼。參考圖5.1，從中心，記為角，經向半徑記為r1和錐形半徑記為r2當軸是垂直時水平半徑為r。當沿Z坐標方向的壁厚為t，根據(jù)下面的草圖，我們得到殘余應力。</p><p><b>　　(5.1)</b></p><p><b>　　(5.

72、2) </b></p><p><b>　　(5.3)</b></p><p><b>　　圖5.1：旋轉殼</b></p><p><b>　　(5.4)</b></p><p>　　徑向的殘余應力由內壓產生，對于膜殼，彎曲的影響可以忽略。所有的彎矩為零，進一步推

73、得：</p><p>　　由旋轉對稱產生的力的平衡導出下面的方程：</p><p><b>　　(5.6)</b></p><p><b>　　(5.7)</b></p><p>　　由于，我們解出(5.6)和(5.7)得：</p><p><b>　　(5.8)&

74、lt;/b></p><p><b>　　(5.9)</b></p><p>　　以上兩個方程是普通殼的旋轉的結果。兩個具體的例子：</p><p>　　1.對于球殼，半徑為R則 r1= r2 =R得出：</p><p><b>　　(5.10)</b></p><p>

75、;　　2.對于圓柱形壓力容器，半徑為R， 則r1 =; r2 = R得出：</p><p><b>　　(5.11)</b></p><p><b>　　(5.12)</b></p><p><b>　　環(huán)向應力：</b></p><p><b>　　(5.13)&l

76、t;/b></p><p><b>　　軸向應力</b></p><p><b>　　(5.14)</b></p><p>　　這些結果與下面的結果一致?？紤]一細長圓柱殼的半徑為R，厚度為t，受內壓為p。薄殼是指殼的半徑和厚度R/t之比大于10。如果圓柱殼的兩端是封閉的，還受環(huán)向應力和縱向應力。殼的局部圖如圖5.1所

77、示。環(huán)向應力和縱向應力如圖所示。外殼被假定為細長所以和沿厚度方向分布均勻。因此，這種情況下的和也稱薄膜應力（這種類型載荷不存在彎曲應力）</p><p>　　考慮切節(jié)的平衡，得出：</p><p><b>　　圖5.2：圓柱殼</b></p><p><b>　　得出：</b></p><p>&l

78、t;b>　　(5.15)</b></p><p>　　考慮橫截面的殼垂直于軸心，我們得出：</p><p><b>　?。?.16）</b></p><p><b>　　5.3 厚殼理論</b></p><p>　　殼的半徑和厚度R/t之比小于10，式(5.15)和 式(5.16)

79、將會不準確，這時將用到厚殼理論，厚殼圓柱形容器的內徑為Ri，外徑為Ro，受內壓為p，如圖5.3這種情況下的應力函數(shù)（參閱附錄I）由半徑r的函數(shù)給出：</p><p><b>　　(5.17)</b></p><p>　　圖5.3厚殼圓柱形容器</p><p>　　A 和 B由邊界條件決定。如果用表示徑向應力、用表示環(huán)向應力和用表示縱向應力，我們

80、得出：</p><p><b>　　(5.18)</b></p><p><b>　　(5.19)</b></p><p>　　常量A 和 B由下面的邊界條件決定：</p><p><b>　　at </b></p><p>　　at

81、 (5.20)</p><p>　　將式 (5.20) 代入式(5.18)和式(5.19)得出：</p><p><b>　　(5.21)</b></p><p>　　外徑和內徑之比記為m，則m =Ro/Ri我們得出徑向和周向應力</p><p><b>　　(5.22)</b>&l

82、t;/p><p><b>　　(5.23)</b></p><p>　　圖5.4顯示出徑向和周向應力的分布</p><p>　　縱向壓力垂直殼軸的面受力平衡得出：</p><p><b>　　(5.24)</b></p><p>　　假定縱向應力是薄膜應力的一種，而且殼的厚度沒有

83、變化。因此，得出：</p><p>　　圖5.4徑向和周向應力的分布</p><p><b>　　(5.25)</b></p><p>　　值得一提的是：由式(5.22)、 (5.23) 和(5.25)推出的結果在遠離不連續(xù)性區(qū)域處有效。</p><p>　　對于中等厚殼，運用薄殼方程和平均半徑Rm，我們得出環(huán)向應力的方

84、程。</p><p><b>　　(5.26)</b></p><p>　　等值的環(huán)向應力，允許設計應力Sm，得出：</p><p><b>　　(5.27)</b></p><p>　　將外徑代入(5.27)式得：</p><p><b>　　(5.28)<

85、/b></p><p>　　又等同的環(huán)向應力的允許設計應力S得：</p><p><b>　　(5.29)</b></p><p>　　這些方程在ASME鍋爐及壓力容器標準中主要是基于等同最大薄膜應力的容許應力，根據(jù)焊接接頭效率校正。許用應力Sm用SE（稍后解釋）替代，在ASME VIII -1鍋爐及壓力容器標準中，.(5.27) 式和(

86、5.29)式修改為：</p><p><b>　　(5.30)</b></p><p><b>　　(5.31)</b></p><p>　　在ASME VIII -1鍋爐及壓力容器標準中，方程變?yōu)椋?lt;/p><p><b>　　(5.32)</b></p>&l

87、t;p><b>　　(5.33)</b></p><p>　　在(5.30) 式和(5.32)式中，Ri代表內徑，然而在(5.31)式和 (5.33)式中Ro代表外徑。在上面兩個式子中，S為許用應力，E是接頭效率。用接頭效率是因為圓柱殼容器通常是用焊接制造的。E的值取決于殼上采用的焊縫的種類。</p><p><b>　　5.5圓柱殼的屈曲</b

88、></p><p>　　細長圓柱殼容器的中徑為D，厚度為t，受外壓為p，只要它仍然是圓形，圓柱體是一個穩(wěn)定的配置，如果最初的橢圓，柱體將工作在不穩(wěn)定工況，最終將屈曲。</p><p>　　如果圓柱體足夠長，端面的因素可以忽略，問題可以認為在二維平面內。</p><p>　　總結徑向方向上的受力，得出：</p><p>　　總結切向方向上

89、的受力（見圖5.5），得出：</p><p><b>　　-</b></p><p>　　圖5.5殼單元的平衡</p><p><b>　　代入 ,得</b></p><p>　　假定圓形變形偏差可以忽略不計，得出：</p><p><b>　　其中C 是一個常量&

90、lt;/b></p><p><b>　　因為，，</b></p><p><b>　　所以：</b></p><p><b>　　對于彎曲殼體:</b></p><p><b>　　其中 </b></p><p><b

91、>　　其中 </b></p><p><b>　　at </b></p><p><b>　　at </b></p><p>　　由n的最小值為1,得出：</p><p><b>　　(5.34)</b></p><p>　　對于短圓筒

92、，端面軸向自由拓展、徑向拓展受到限制，推導出臨界壓力：</p><p><b>　　當 很大時：</b></p><p><b>　　(5.35)</b></p><p>　　恩此，它和屈曲壓力一致，在(5.34)中n= 2，在ASME標準中，臨界壓力根據(jù)兩種情況計算，還涉及到外徑和壁厚之比。</p><

93、;p><b>　　1.</b></p><p><b>　　2.</b></p><p>　　(5.19)式的修改包括非線性屈曲。在上面的第一種情況，安全系數(shù)取3.0，第二種情況，可變的安全系數(shù)當= 10時取3.0，當 =4時取. 2.0。隨著圓筒的厚度增大，屈曲停止在一個合理的模式。在ASME規(guī)范中，用2組曲線來探究屈曲。當增大時，程序變

94、得復雜，需要檢查屈曲是在彈性屈曲或者是在塑性區(qū)域，</p><p>　　5.6壓力容器的不連續(xù)應力</p><p>　　舉一個特別的例子，在受內壓為p的圓柱形容器與半球形封頭的連接處受不連續(xù)應力。為了簡化，我們假設球形封頭和圓柱殼的厚度相同。如果平均半徑和厚度分別記為Rm和t，從而圓柱容器的環(huán)向和切向應力為：</p><p><b>　　(5.36)<

95、;/b></p><p><b>　　(5.37)</b></p><p>　　球殼的環(huán)向和切向應力為：</p><p><b>　　(5.38)</b></p><p><b>　　(5.39)</b></p><p>　　受內壓的圓柱殼體的徑向

96、增長</p><p><b>　　(5.40)</b></p><p>　　受內壓的球殼的徑向增長</p><p><b>　　(5.41)</b></p><p><b>　　是泊松比</b></p><p>　　如果圓柱殼和球殼的部分分離，徑向增長變

97、為：</p><p><b>　　(5.42)</b></p><p>　　在實際的容器中，半球形封頭和圓柱殼受剪切力和每單位周長上彎矩的約束。這些不連續(xù)應力在相鄰的容器區(qū)域產生局部彎曲應力。在半球形封頭和圓柱殼的邊緣處的力產生的變形和坡口是相等的。如果M等于零V能關生/2大小的變形，過渡處的連續(xù)性將得到滿足。地基上由于M 和 V受彈性的半無限長梁應用結果。代入彈性模

98、量得：</p><p><b>　　(5.43)</b></p><p><b>　　衰減系數(shù) </b></p><p><b>　　(5.44)</b></p><p>　　當x=0時 ， M=0.</p><p><b>　　(5.45)

99、</b></p><p>　　把值代入(5.45)式得：</p><p><b>　　(5.46)</b></p><p>　　分布在圓柱殼上的軸向應力和環(huán)向應力為：</p><p><b>　　(5.47)</b></p><p><b>　　(5.4

100、8)</b></p><p>　　當 p= 2MPa, Rm = 1m, t = 25mm,泊松比為0.3時,我們由式(5.44)得 = 0.008127軸向應力和環(huán)向應力為</p><p>　　Mpa (5.49)</p><p>　　Mpa (5.50)</p><p>　　在

101、(5.49)式中，第一數(shù)量膜的縱向應力沿容器的長度是常量（等于40MPa）。第二數(shù)量彎曲應力沿容器的長度改變。在(5.50)式中膜環(huán)向應力（等于80MPa）沿容器的長度不變。壓應力由于半徑和彎曲應力的減小沿容器的長度改變</p><p>　　當受的內壓為7MPa材料的許用應力為140MPa圓柱形容器的內徑為2m</p><p>　　厚圓柱殼容器的內徑為1.8m外徑為1.5m，承受內壓為35

102、MPa</p><p>　　最大環(huán)向應力的大小及位置</p><p>　　最大徑向應力的大小及位置</p><p><b>　　平均環(huán)向應力</b></p><p>　　厚圓柱殼容器的內徑為300mm外徑為450mm，如果許用應力為175MPa，可承受最大的內壓為多少。</p><p>　　厚圓柱