過程裝備與控制工程專業英語翻譯(部分)-資料下載頁

　　

【正文】 的組成部分通常是不統一的。那就是說，一塊區域可能比周圍的含有更多的碳。這種成分的不同會影響熱處理的方法。如果加熱到一個高的溫度，其中的碳能擴散到四周，結果理所當然的刀均勻的成分。這樣鋼的成分就會更均勻從而使熱處理方法更統一。由于鑄鐵的固有性質，對工作前的鑄鐵塊，硬化前的鋼的鑄件和鍛件，正火處理會運用得更頻繁。去應力當一塊金屬被加熱時就會發生膨脹，膨脹的多少跟溫度的上升成比例。當冷卻一金屬，就會發生相反的反應。那就是說，金屬的收縮可以被觀測到。當一根鋼棒或鋼板被加熱到一個點高于另一個點，就像焊接或鍛造一樣，內應力就會產生。在加熱過程中，被加熱部分不能自由膨脹，就會轉為變形。在冷卻時，熱部分周圍的冷金屬就會阻止收縮。打算使金屬收縮的力并沒有減少，當金屬再次被冷卻時，這個力與內應力一樣不變。體積的變化同樣能產生應力，這個應力隨著金屬轉變和沉淀。內應力和殘余應力是不好的因為它使鋼的加工部分產生彎曲。為了減少這種應力，鋼被加工到約595176。C，假設所有的部分被均勻加熱，然后緩慢冷卻到室溫。這步驟就叫做去應力退火，或僅僅叫去應力。第四篇：過程裝備與控制工程專業英語翻譯4Reading Material4Stresses in Cylindrical Shells due to Internal PressureThe classic equation for determining stress in a thin cylindrical shell subjected to pressure isobtained from of forces perpendicular to plane ABCD gives:PL ．2r =2σqLtorσqq=PrtWhereP=pressure,L=length of cylinder ,σThe strain εε=is defined as_2prr2prr=hoop stress ,r=radius, t=thicknessq2p(r+W)2prrr=qorεq=WrAlsoεdWdrThe radial deflection of a cylindrical shell subjected to internal pressure is obtained by substituting the quantity into Eq.().Hence for thin cylindersW=PrEt2Where W= radial deflection, E= modulus of elasticityEquations()and()give accurate results when r rt decreases,however , a more accurate expression is needed because the stress distribution through the thickness is not is then made to the “thick shell” theory first developed by derived equations are based on the forces and stresses shown in 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 other words ,?1 is constant at any cross sectionA relationship between σrand σqcan be obtained by taking a freebody diagram ofring dr as shown in forces in the vertical direction and neglecting higherorde terms ,we then haveσ—σr=dσdrrqA second relationship is written asσ σr=q=E（1+μ）（12μ）E[εq(1μ)+μ（εr+ε）]σ1=（1+μ）（12μ）E[εr(1μ)+μ（ε(1μ)+μ（εq+ε+ε）]）]（1+μ）（12μ）[ε1qrSubstituting Eqs.()and()into the first two expressions ofEq.()and substituting the result into Eq.()results in:dwdr+1dwrdr—wr=0A solution of this equation isW=Ar+BrWhere A and B are constants of integration and are determined by first substituting Eq.()into the first one ofEq.()and applying the boundary conditionsσr= —piatr=riatσr= —poatr=roExpression()then bees: w= — μrε1+1Er(rOri)［r2(1μ2μ)（PiriPOrO）+rirO(1+μ)（PiPO）］2222Once w is obtained, the values of σand expressed for thick cylinders asPiriPOri+(PiPO)（rOriqare determined from Eqs.(),and ,(),and()rirOσq=r)σr= —POriPiri+(PiPO)（rOririrOr)whpressureσr=radial stressσq=hoop stressPi=internal pressurePO=externalri=inside radiusrO=outside radiusr=radius at any pointThe longitudinal stress in a thick cylinder is obtained by substituting Eqs.(),(),and()into the last expression of Eqs.()to giveσ1=Eε+2μ(PiriPOrO）r2OriThis equation indicates that σ1 is constant throughout a cross section because εis constantand r does not appear in the second the expressionσ1 can be obtained from statics Asσ1=PiriPOrOr2OriWith σ1 known, Eq.()for the deflection of a cylinder can be expressed asw=r(PiriPOrO)(12μ)+(PiPO)rirO(1+μ)Er(rOri)閱讀材料4圓柱殼體的應力源于內部壓力。累積力通過正交于平面ABCD給出：PL ．2r =2σqLtorσqq=Prt說明： P=壓力，L=圓柱體的長度，σ正應變εε=由下面式子決定：_2prr=環繞切應力，r=半徑，t=殼體的壁厚q2p(r+W)2prrrq并且：ε=2prrdWdrorεq=Wr圓柱狀殼體的半徑偏差源于內部壓力。因此，對薄殼體有：W=PrEt說明：W=半徑的偏差值，E=彈性橫量當rt0時，方程式（）和（）給出精確的結果。隨著r的減少，然而，一個更精確的表述是非常必要的，因為厚度的壓力分配并不是均衡的。接著要求助于Lame首先提出來的“厚殼”理論。該理論假定所有的剪切力都為0是由于對稱性和一個正?？v軸的平面，在施壓前仍然是平面壓力。換句話說，?1 是恒在任意橫截面上。σr和σq的關系可以通過一個環dr的自由體示意圖獲得。綜合垂直dσdr方向上的力和不計高階的項，于是我們得到σEq—σr=r第二種關系可以表達為：σq=（1+μ）（12μ）[εq(1μ)+μ（εr+ε）]σr=Eσ1=（1+μ）（12μ）E[εr(1μ)+μ（ε(1μ)+μ（εq+ε+ε）]）]（1+μ）（12μ）[ε1qr.()與及把結果代入Eq.()得：dwdr+1dwrdr—wr=0Br方程的一個值為W=Ar+r=riatσ：w= — μrε1+= —pir= —poatr=roOEr(rri)r2(1μ2μ（）PiriPOrO）+rirO(1+2222μ)（PiPO）］ 當W被求得，σ、圓柱的厚度可以表達為：rirOσq=PiriPOri+(PiPO)（rOr)riσr= —POriPiri+(PiPO)（r2OrirOr)riσr=徑向應力σq=環向應力Pi=內部壓力PO=外部壓力ri=內半徑rO=外半徑r=任意點的半徑、。其為： σ1=Eε+2μ(PiriPOrO）r2O2ri這個表達式說明了σ1是一個橫截面的常數因為ε的表達式可以由靜態值表示:σ1=PiriPOrOr2O2是常數和r不會在第二項出現。因此σri,隨著σ1知道了，表示為：w=r(PiriPOrO)(12μ)+(PiPO)rirO(1+μ)Er(rri)2O22222第五篇：過程裝備與控制工程專業英語翻譯 17Reading Material 17Stress CategoriesThe various possible modes of failure which confront the pressure vessel designer are:(1)Excessive elastic deformation including elastic instability.(2)Excessive plastic deformation.(3)Brittle fracture.(4)Stress rupture/creep deformation(inelastic).(5)Plastic instabilityincremental collapse.(6)High strainlow cycle fatigue.(7)Stress corrosion.(8)Corrosion dealing with these various modes of failure, we assume that the designer has at his disposal a picture of the state of stress within the part in would be obtained either through calculation or measurements of the both mechanical and thermal stresses which could occur throughout the entire vessel during transient and steady state question one must ask is what do these numbers mean in relation to the adequacy of the design? Will they insure safe and satisfactory performance of a ponent? It is against these various failure modes that the pressure vessel designer must pare and interpret stress example, elastic deformation and elastic instability(buckling)cannot be controlled by imposing upper limits to the calculated stress must consider, in addition, the geometry and stiffness of a ponent as well as properties of the plastic deformation mode of failure can, on the other hand, be controlled by imposing limits on calculated stresses, but unlike the fatigue and stress corrosion modes of failur