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过程装备与控制工程专业英语翻译

过程装备与控制工程专业英语翻译
过程装备与控制工程专业英语翻译

过程装备与控制工程专业英语阅读材料翻译

Reading Material 1

Static Analysis of Beams

A bar that is subjected to forces acting transverse to its axis is called a beam. In this section we will consider only a few of simples types of beams, such as those shown in Fig.1.2. In every instance it is assumed that the beam has a plane of symmetry that is parallel to the plane of the figure itself. Thus ,the cross section of the beam has a vertical axis of symmetry. Asle ,it is assumed that the applied loads act in the plane of symmetry ,and hence bending of the beam occurs in that plan. Later we will consider a more general kind of bending in which the beam may have an unsymmetrical cross section .

The beam in Fig.1.2(a),with a pin support at one end and a roller support at the other ,is called a simply supported beam, or s simple beam. The essential feature of a simple beam is that both ends of beam may rotate freely during bending ,but they cannot translate in the lateral direction. Also ,one end of the beam can move freely in the axial direction (that is ,horizontally ). Supports of a simple beam may sustain vertical reactions acting either upward or downward.

The beam in Fig.1.2(b) which is built-in or fixed at one end and free at the other end ,is called a cantilever beam. At the fixed support the beam can neither rotate nor translate , while at the free end it may do both. The third example in the figure shows a beam with an overhang. This beam is simply supported at A and B and has a free end at C.

Loads on a beam may be concentrated forces , such as P1 and P2 in Fig.1.2(a) and (c) , or distributed loads ,such as the load q in Fig.1.2(b). distributed loads are characterized by their intensity ,which is expressed in units of force per unit distance along the axis of the beam. For a uniformly distributed load ,illustrated in Fig.1.2(b), the intensity is constant; a varying load ,on the other hand ,is one in which the intensity varies as a function of distance along the axis of the beam.

The beam shown in Fig.1.2 are statically determinate because all their reactions can be determined from equations of static equilibrium. For instance ,in the case of the simple beam supporting the load P1[Fig.1.2(a)], both reactions are vertical ,and their magnitudes can be found by summing moments about the ends ;thus, we find

R A=P1 (l-a)/L R B=P1a/L

The reactions for the beam with an overhang [Fig.1.2(c )] can be found in the same manner.

For the cantilever beam [Fig.1.2(b )],the action of the applied load q is equilibrated by a vertical force RA and a couple MA acting at the fixed support, as shown in the figure. From a summation of forces in the vertical direction , we conclude that

R A= qb

And, from a summation of moments about point A ,we find

M A= qb(a+b/2)

The reactive moment MA acts counterclockwise as shown in the figure.

The preceding examples illustrate how the reactions (forces and moments ) of statically determinate beams may be calculated by statics. The determination of the reactions for statically indeterminate beams requires a consideration of the bending of the beams , and hence this subject will be postponed.

The idealized support conditions shown in Fig.1.2 are encountered only occasionally in practice. As an example ,long –span beams in bridges sometimes are constructed with pin and roller supports at the ends. However, in beams of shorter span , there is usually some restraint against horizontal movement of the supports. Under most conditions this restraint has little effect on the action of the beam and can be neglected. However ,if the beam is very flexible ,if the horizontal restraints at the ends are very rigid ,it may be necessary to consider their effects

Example

Find the reactions at the supports for a simple beam loaded as shown in Fig.1.3(a).Neglect the weight of the beam. Solution

The loading of the beam is already given in diagrammatic form . The nature of the supports is examined next and the

unknown components of these reactions are boldly indicated on the diagram. The beam, with the unknown reaction components and all the applied forces ,is redrawn in Fig.1.3(b) to deliberately emphasize this important step in constructing a free-body diagram . At A ,two unknown reaction components may exist ,since the end is pinned. The reaction at B can only act in a vertical direction since the end is on a roller. The points of application of all forces are carefully noted. After a fee-body diagram of the beam is made ,the equations of statics are applied to obtain the solution.

ΣF x=0, R Ax=0

ΣM A=0+,2000+100(10)+160(15)-R B(20)=0,R B=+2700 lb↑

ΣM B=0+,R Ay(20)+2000-100(10)-160(5)=0,R Ay=-10 lb↓

Check:ΣF y=0↑+,-10-100-160+270=0

Note that ∑Fx =0 uses up one of the three independent equations of statics, thus only two additional reaction components may be determined from statics. If more unknown reaction components or moments exist at support, the problem becomes statically indeterminate.

Note that the concentrated moment applied at C enters only into the expressions for the summation of moments. The positive sign of R B indicates that the direction of R B has been correctly assumed in Fig.1.3(b). the inverse is the case of R Ay , and the vertical reaction at A is downward. Note that a check on the arithmetical work is available if the calculations are made as shown.

(Selected from Stephen P.Timoshenko and James M. Gere,Mechanics of Materials ,

Van Nostrand Reinhold Company Ltd.,1978.

* Selected from Egor P. Popov ,Introduction to Mechanics of Solids, Prentice-Hall Inc. , 1968)

横梁的静态分析

一条受力的作用横向坐标的轴被称为横梁。在本章中,我们只考虑横梁的一些简单的类型。在例1.2.的每种情况下,假定横梁水平对称,也就是说与横梁平整的外形平行。因此,横梁的横截面有一个垂直的对称轴,假设在平面上施加力,则在那个平面上横梁会因此发生弯曲。我们后面要考虑一种常见的弯曲,即横梁所具有的不对称的截面弯曲。

例1.2横梁的类型

在例题1.2(a)中,简单支撑梁或者叫简单支梁是一端有一固定支座另一端有一滚动支座的梁。简单支梁的基本作用是横梁在弯曲时其两端可以自由旋转,但它们不能转换成横向。同时,横梁的一端能在纵向自由移动。简单支座可以支撑向上或者是向下的任何一种垂直反力。

在例1.2(b)中,悬臂梁是一端安装或固定好,另一端不受约束的横梁。在固定支撑处的横梁既不能旋转也不能调动,然而不受约束端既能旋转又能调动。例3显示的是外伸梁。这根横梁在A和B处是简单梁,C处是自由端。

作用于横梁的负载可能是集中力,如例1.2(a)和(c)中的P1和P2,如1.2(b)中的负载q。分布负载具有集度的特性,力的单位表示方法是用沿着横梁轴向的单位长度表示。对于例1.2(b)中列举的均匀分布负载,强度是不变的,变动负载,另一方面,就是其强度沿着横梁的轴向根据距离的变化而变化。

例1.2所示的横梁是静定的,因为它们所有的反作用力都可以根据静力平衡方程得出。例如,简支梁例子1.2(a)中支持的负载P1,两个反作用都是垂直的,并且它们的大小可以通过分析受力完成瞬间来确定。这样,我们发现:

RA=P1(L--a)/ L RB =P1a/ L

作用在(例1.2(c))外伸梁上的反作用力可以用相同的方法来确定。

对于例1.2(b)中的悬臂梁,外加负载q的作用,是由固定支座的竖直方向的力R A和力偶M A

平衡的。如图所示。从垂直方向上的合力,我们可以推断出:

RA =qb

而且,从A点的瞬间合力,我们发现

MA =qb(a+b/2)

反力偶MA是逆时针的,如图所示。

前面的例子说明了确定静定梁的作用力可以通过静力学计算。分析静定梁的作用需要考虑横梁的弯曲,这个论述将会在后面的章节中学习。

例1.2所示的理想支持条件在实践中只是偶尔发生。例如,桥梁的大跨度横梁有时就是末端用铰链和滚动支座构建的。然而,在一些较短跨度的横梁中,常常会约束其在水平支撑面的运动。在多数情况下,这种约束对横梁的作用影响很小,可以忽略不计。可是,假如横梁是很柔韧的,而且末端的水平约束很坚硬,这就需要考虑它们的作用效果。

例:

找出支撑简支梁的作用反力,载荷如例1.3(a)所示,横梁自身的重量忽略不计。

解答:

横梁受到的载荷已在图中标出。支撑的性质接下来就被测出,这些未知的反作用力在图示中明确指出。有未知反作用力和所有已知作用力作用的横梁已经在图 1.3(b)中标出,用于刻意强调构建这个自由体图示的重要步骤。在A处,因为末端是铰链,有两个方向的大小未知的反作用力存在。由于末端是滚动支座,B处的反作用力仅仅是在垂直方向的。所有作用力的作用点都已经认真标出。作出横梁自由端的受力图后,要用静力学方程来解答。

注意到∑F x=0使用了静力学的三个独立方程中的一个。如此仅有两个额外的反作用力组成部分会从静力学中确定。如果在支撑中有更多的反力组成部分或力矩存在,问题就变成了静不定。

注意到作用于C中心的集中力仅是所有集中力的综合表现。RB的正号指示对RB的方向的假设是正确的。Ray的方向与实际方向相反,在A处垂直反力是向下的。注意如果运算如上所述,对算术过程的验证是可靠的。

Reading Material 2

Shear Force And Bending Moment In Beams

Let us now consider, as an example, a cantilever beam acted upon by an inclined load P at its free end [Fig.1.5 (a)]. If we cut through the beam at a cross section mn and isolate the left-hand part of the beam as free body [Fig.1.5 (b)], we see that the action of the removed part of the beam (that is, the right-hand part) upon the left-hand part must be such as to hold the left-hand part in equilibrium. The distribution of stresses over the cross section me is not known at this stage in our study, but we do know that the resultant of these stresses must be such as to equilibrate the load P.

It is convenient to resolve the resultant into an axial force N acting normal to the cross section and passing through the centric of the cross section, a shear force V acting parallel to the cross section, and a bending moment M acting in the plane of the beam.

The axial force, shear force, and bending moment acting a across section of a beam are known as stress resultants. For a statically determinate beam, the stress resultants can be determined from equations of equilibrium. Thus, for the cantilever beam pictured in Fig.1.5, we may write three equations of static’s for the free-body diagram shown in the second part of the figure. From summations of force in the horizontal and vertical directions we find, respectively,

N=P cos βV=P sinβ

(a)

(b) (c)

Fig.1.5 stress resultants N,V and M

And, from a summation of moments about an axis through the centric of cross section m n ,we obtain: M=Pox sinβ

M=P x sinβ

Where x is the distance from the free and to section menthes, through the use of a free-body diagram and equations of static equilibrium, we are able to calculate the stress resultants without difficulty. The stresses in the beam due to the axial force N acting alone have been discussed in the text of Unit.2;now we will see how to obtain the stresses associated with bending moment M and the shear force V.

The stress resultants N, V and M will be assumed to be positive when they act in the directions shown in Fig.1.5 (b).this sign convention is only useful, however, when we are discussing the equilibrium of the left-hand part of the beam. If the right-hand part of the beam is considered, we will find that the stress resultants have the same magnitudes but opposite directions [see Fig.1.5 (a)]. Therefore, we must recognize that the algebraic sign of a stress resultant does not depend upon its direction in space, such as to the left or to the right, but rather it depends upon its direction in space, such as to the left or to the right, but rather it depends upon its direction with respect to the material against which it acts. To illustrate this fact, the sign conventions for N, V and M are repeated in FIig1.6, where the stress resultants are shown acting on an element of the beam.

We see that a positive axial force is directed away from the surface upon which it acts (tension), a positive shear force acts clockwise about the surface upon which it acts ,and a positive bending moment is one that compresses the upper part of the beam.

(a) (b)

Fig.1.6

Example

A simple beam A

B carries two loads, a concentrated force P and a couple M O , acting as shown in Fig.1.7 (a). Find the

shear force and bending moment in the beam at cross sections located as follows; (a) a small distance to the left of the

middle of the beam and (b) a small distance to the right of the middle of the beam.

The first step in the analysis of this beam is to find the reactions R A and R B .taking moments about ends A and B

gives two equations of equilibrium, from which we find

R A =4P 3—L M O R B =4

P + L M O Next, the beam is cut at a cross section just to the middle, and a free-body diagram is drawn of either half of the beam.

In this example we choose the left -hand half of the beam, and the corresponding diagram is shown in Fig.1.7(b) .The force P and the reaction R A appear in this diagram as also do the unknown shear force V and bending moment M, both of which are shown in their positive directions ,The couple M O does not appear in the figure because the beam is cut to the left of

the point where M O is applied. A summation of forces in the vertical direction gives

V= R A —P= —4

P —L M O Which shows that the shear force is negative; hence, it acts in the opposite direction to that assumed in Fig.1.7 (b).

Taking moments about an axis through the cross section where the beam is cut [Fig.1.7 (b)] gives M =2L R A —4PL =8

PL —2M O Depending upon the relative magnitudes of the terms in this equation, we see that the bending moment M may be

either positive or negative.

To obtain the stress resultants at a cross section just to the right of the middle, we cut the beam at that section and again

draw an appropriate free-body diagram [Fig.1.7(c)]. The only difference between this diagram and the former one is that

the couple M O now acts on the part of the beam to the left of the cut section. Again summing forces in the vertical

direction, and also taking moments about an axis through the cut section, we obtain

V=—4P —L M O M = 8

PL +2M O

Fig.1.7

We see from these results that the shear force does not change when the cut section is shifted from left to right of the couple M O , but the bending moment increase algebraically by an amount equal to M O .

梁上的切变应力与弯矩

现在让我们假设,在一根悬臂梁的自由端有一倾斜的载荷P 作用在其上,如图1.5(a )所示。

如果我们在m-n 截面处截开并隔离梁的左端部分作为自由体,如图1.5(b )所示,我们看到被移走的那部分横梁的作用(也就是,梁的右端部分)必须是要保持梁的左端部分受力平衡。在我们研究的现阶段还无法了解在截面m-n 处的压力分布,但是我们能确定的是这些压力的结果一定是与载荷P 平衡。这方便解答这个结果,通过把其化解成作用于垂直截面的轴向力N ,该力同时经过截面的重心,以及作用于平行截面的剪切力V ,和作用于横梁所在平面的弯矩M 。

作用于横梁的截面上的轴向力,剪应力和弯矩称为内力。对于静定梁,内力可以由静力平衡方程

求出。因此,图示例1.5中的悬臂梁,对于第二个图形所表示的自由体图解,我们可以写出三个静力

学平衡方程。从水平方向和垂直方向对力求和,我们分别得到,

N=Pcosβ V=Psinβ

(a)

(b)

(c)

例1.5 内力N ,V 和M 从通过截面重心的轴线上的瞬间求和,我们得到,

βsin Px M = 其中x 是从自由端到截面m-n 的距离。因此,通过利用自由体的图解和静力学平衡方程,我们能

够很容易的算出内力。由于单独作用在横梁的轴向力N 已经在第二单元讨论过;现在我们将会看到如何得到与弯矩M 和剪切力V 有关的压力。

内力N ,V ,和M 作用于图示1.5(b )的方向假设为其正方向。然而,只有当我们讨论梁的左部

分的平衡时,这个符号规定才适用。如果梁的右部分也需要考虑,我们会发现内力有相同的大小但是方向却相反,【如图 1.5(c )所示】。因此,我们要意识到一个内力的代数符号,并不是取决于它在空间的方向,比如说向左或者向右,而是取决于力作用其材料上产生的反作用力的方向。为了说明这

个事实,内力N ,V ,M 的惯例符号被重复展示在例1.6中,其中内力作用在一个单元梁上。

我们发现,正向轴力的方向是远离它的作用表面的方向,它相当于张力,剪切力的正方向是使它作用的表面顺时针旋转的方向,而弯矩的正方向是横梁上部分被压缩的方向。

(a)

(b)

图1.6 例子:

一个简支梁AB 受到两个负载以及一个集中力P 和一个集中力偶Mo 作用,受力如图1.7(a )所示。发现在梁的截面上的剪切力和弯矩作用如下:(a )一小段在中间梁的左侧,(b )一小段在中间梁的右侧。

解答:

分析这根梁的第一步是找出支反力A R 和B R ,在A 和B 两端我们列出两个平衡方程,从中我们得到:

L Mo P R A -=

43 L Mo P R B +=4 然后,从中间将横梁截断成两个如图所示的自由体结构,画出梁的任意一半的自由体图解。在这个例子中,我们选取横梁的左段部分,相对应的图示1.7(b )。集中力P 和支反力A R 作用在这一段上,同时还有未知的剪切力V 和弯矩M ,假设它们方向均为正方向。集中力偶Mo 没有在图中出现,因为梁被切取的是力偶Mo 作用点的左边。在垂直方向对所有的力求和有

L Mo P P V R A --=-=4 其中剪切力是负的;因此,它与图 1.7(b )中假定的方向相反。对通过横梁被劫截面的轴线上的一

瞬间列方程式【如图1.7(b )】有:

2842Mo PL PL L

R A M -=-=

在这个等式中,根据有关大小的条件,我们可以看到弯矩可能是正的也可能是负的。

为了得到截面右段的内力,我们再次截开横梁在同一截面上,并画出相应的自由体受力图【如图

1.7(c )】。这个图和前面那个图唯一的区别是现在集中力偶Mo 作用在梁的截面的左侧的部分上。再次在垂直方向上求和力,在过截面的轴线上的一瞬间列方程,我们得到

L Mo P V --=4 28Mo PL M += 我们从以上结果可知,当截面从左端移动到有集中力偶Mo 的右端时,剪切力不变,但弯矩在代

数上逐渐增加至趋近于Mo 。

3 Theories of strength

1. Principal stresses

The state of the stress at a point in a structural member under a complex system of loading is described by the magnitude and direction of the principal stresses. The principal stresses are the maximum values of the normal stresses at the point; which act on the planes on which the shear stress is zero. In a two-dimensional stress system, Fig.1.11, the principal stresses at any point are related to the normal stress in the x and y d irections σx and σy and the shear stress τxy at the point by the following equation:

Principal stresses,22214)(2

1)(21xy x y x y τσσσσσσ+-±+=??? The maximum shear stress at the point is equal to half the algebraic difference between the principal stresses.

stresses: Maximum shear stress,)(2

121max σστ-= Compressive stresses are conventionally taken as negative; tensile as positive.

2. Classification of pressure vessels

For the purpose of design and analysis, pressure vessels are sub-divided into two classes depending on the ratio of the wall thickness to vessel diameter : thin-wall vessels, with a thickness ratio of less than 1/10, and thick-walled above this ratio.

The principal stresses acting at a point in the wall of a vessel, due to a pressure load, are shown in Fig.1.12. If the wall is thin, the radial stress σ3 will be small and can be neglected in comparison with the other stresses , and the longitudinal and circumferential stresses σ1 and σ2 can be taken as constant over the wall thickness. In a thick wall, the magnitude of the radial stress will be significant, and the circumferential stress will vary across the wall. The majority of the vessels used in the chemical and allied industries are classified as thin-walled vessels. Thick-walled vessels are used for high pressures.

3. Allowable stress

In the first two sections of this unit equations were developed for finding the normal stress and average shear stress in a structural member. These equations can also be used to select the size of a member if the member’s strength is known. The strength of a material can be defined in several ways, depending on the material and the environment in which it is to be used. One definition is the ultimate strength or stress. Ultimate strength of a material will rupture when subjected to a purely axial load. This property is determined from a tensile test of the material. This is a laboratory test of an accurately prepared specimen, which usually is conducted on a universal testing machine. The load is applied slowly and is continuously monitored. The ultimate stress or strength is the maximum load divided by the original cross-sectional area. The ultimate strength for most engineering materials has been accurately determined and is readily available

If a member is loaded beyond its ultimate strength it will fail----rupture. In the most engineering structures it is desirable that the structure not fail. Thus design is based on some lower value called allowable stress or design stress. If, for example, a certain steel is known to have an ultimate strength of 110000 psi, a lower allowable stress would be used for design, say 55000 psi. this allowable stress would allow only half the load the ultimate strength would allow. The ratio of

the ultimate strength to the allowable stress is known as the factor of safety :

Sa

Su n or stress allowable strength ultimate safety of Factor == We use S for strength or allowable and σ for the actual stress in material. In a design:

This so-called factor of safety covers a multitude of sins. It includes such factors as the uncertainty of the load, the uncertainty of the material properties and the inaccuracy of the stress analysis. It could more accurately be called a factor of ignorance! In general, the more accurate, extensive, and expensive the analysis, the lower the factor of safety necessary.

4. Theories of failure

The failure of a simple structural element under unidirectional stress (tensile or compressive) is easy to relate to the tensile strength of the material, as determined in a standard tensile test, but for components subjected to combined stresses (normal and shear stress) the position is not so simple, and several theories of failure have been proposed. The three theories most commonly used are described below:

Maximum principal stress theory: which postulates that a member will fail when one of the principal stresses reaches the failure value in simple tension, σ’e . The failure point in a simple tension is taken as the yield-point stress, or the tensile strength of the material divided by a suitable factor of safety.

Maximum shear stress theory: which postulates that failure will occur in a complex stress system when the maximum shear stresses reaches the value of the shear stress at failure in simple tension. For a system of combined stresses there are three shear stresses maxima:

22

11σστ-±=,23

22σστ-±=,21

33σστ-±= (1.10)

In the tensile test,

2

'e e στ= (1.11) The maximum shear stress will depend on the sign of the principal stresses as well as their magnitude, and in a two-dimensional stress system, such as that in the wall of a thin-walled pressure vessel, the maximum value of the shear stress may be given by putting σ 3 =0 in equations 1.10. T he maximum shear stresses theory is often called Tresca’s, or Guest’s theory 。

Maximum strain energy theory: which postulates the failure will occur in a complex stress system when the total strain energy per unit volume reaches the value at which failure occurs in simple tensile.

The maximum shear-stress theory has been found to be suitable for predicting the failure of ductile material under complex loading and is the criterion normally used in the pressure-vessel design.

强度理论

1、主应力

在一个负载复杂的系统中构件在某点处的应力状态是通过主应力的大小和方向来描述的。主应力是这点处正应力的最大值,作用处的平面切应力等于零。在二维力学体系中,如例1.1,在任何一处的主应力都与正应力在X 和Y 方向的应力σx 和σy 有关,在该处的切应力τxy ;可以用下面的方程式得到:

主应力: 22214)(21)(21xy x y x y τσσσσσσ+-±+=??? (1.7)

在该处,最大切应力等于主应力代数差的一半。

最大切应力,τmax=21

(σ1—σ2) (1.8)

挤压应力通常视为是负的,拉伸应力视为正的。

2、压力容器的分类

以设计和分析为目的,压力容器可以根据壁厚与管道直径的比值分为两类:薄壁比小于1/10的薄壁容器,厚度比大于1/10的厚壁容器。

由于压力负载作用,作用在容器壁点某处的主应力如例1.12所示。如果壁是薄的,径向应力σ3将很小,对比于其他应力可以忽略不计。而纵向和四周的应力σ1和σ2就可以认为是与壁厚无关的常数。在厚壁上,径向应力大小将是很重要的,应用于化学和联合工业上的被分为薄壁容器。厚壁容器被用于高压环境。

3、许用应力

在这个章节开始的两节的方程式是为了找到构件中的正应力和平均切应力。这些方程式同样能在结构强度已知的情况下,用于选择构件的尺寸。材料的强度可以从几个方面被定义,依据自身的材料和其使用环境。一种定义是其极度强度和应力。极度强度是指当其受到纯轴向负载作用将要破裂时的应力。这项属性将会从材料的拉伸试验中测出。这是在一个万能材料试验机对某一精准拟定的试样进行分析检测的实验试验。负载缓慢增加,并且被连续不断地检测。极度压力或强度是在原始横断面被分开的最大负载。对于大多数的工程材料,极度强度已经被准确测定并且容易得出。

假如某一部分承受的载荷超过了它的强度极限,它就将断裂。在大多数的设计结构中,通常想让结构不会断裂。所以,设计是根据一些下限值设计的,这些值通常被称作允许应力或设计应力。例如,某一钢坯已知有最大强度值110000每平方英寸磅,一个较低的许用应力将被用于设计,比如55000每平方英寸磅。这个许用应力将仅能允许载荷强度极限的一半。极限强度与许用应力的比值被称作安全系数

许可应力值强度极限值安全系数=或A U S S =n

我们用S 表示强度值或者许用应力值,用σ表示材料受到的实际负载,在设计中

σ≤SA

这里所谓的安全因子涵盖了诸多因素,这些因素包括负载的不确定性,材料性能的不确定性,以及应力分析的不准确性。它被称为忽略系数可能更准确!一般而言,分析得越准确、越广泛、越昂贵,安全系数的必要性就越小。

4、断裂理论

在单向应力(拉伸和压缩)的作用下,一个简单构件的断裂就容易与材料的抗拉强度有关,正如在标准拉伸性实验的检测中,但对于承受复合应力(正应力或切应力)的部分,作用时的位置就不那么简单了,几个断裂理论已经被提出。三个最常用的理论如下所述:

最大主应力理论:在简单拉伸中,假定当主应力中的某一个达到断裂值σ’e 时,某一部件将会断裂。在简单拉伸中的失效点被称为屈服点,或者材料的拉伸强度除以一个合适的安全系数。

最大切应力理论:在简单拉伸中,假定当最大切应力达到断裂切应力时,断裂将会在复合应力体系中发生。

对于一个联合应力系统,有三个切应力最大值:

τ1=±σ1-σ2/2, τ2=±σ2-σ3/2, τ3=±σ3-σ1/2 (1.10)

在拉伸性试验中,

τe=σ’e/2 (1.11)

最大切应力取决于主应力的方向和大小,并且在二维的应力体系中,比如在薄壁压力容器的薄壁上,切应力的最大值可以从方程式1.10中的σ3=0得到。最大切应力理论经常会被称为“特雷斯卡理论”或“盖斯特理论”。

最大应变能理论:在简单拉伸中,假定当单位体积的总应变能达到断裂发生的临界值时,断裂将会在复合应力体系中发生。

最大切应力理论已经被发现适用于预测塑性材料在复杂载荷下的屈服点,并且也是应用在压力容器设计中的普遍准则。

Reading Material 4

Stresses in Cylindrical Shells due to Internal Pressure

The 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 gives:

PL .2r =2σθLt or σθ=t r P (1.17)

Where P=pressure, L=length of cylinder ,σθ=hoop stress ,r=radius, t=thickness

(a)

(b)

The strain εθ is defined as

εθ=rr W r ππ2)(2+_rr rr ππ22 or εθ =r W (1.18)

Also εr =dr

dW (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 cylinders

W=Et r 2

P (1.20)

Where W= radial deflection, E= modulus of elasticity

Equations (1.17) and (1.20) give accurate results when t r >0. As t r 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 ,ε

1 is constant at any cross

section

(a) (b) A relationship between σr and σθ can be obtained by taking a free-body diagram of ring dr as shown in Fig.118b. Summing forces in the vertical direction and neglecting higher-order terms ,we then have

σθ—σr =dr d r σ (1.21)

A second relationship is written as

σθ=μ)

μ)((211E -+[εθ(1-μ) +μ(εr +ε1)] σr =μ)

μ)((211E -+ [εr (1-μ)+μ(εθ+ε1)] (1.22) σ1=μ)μ)((211E -+ [ε1(1-μ)+μ(εθ+εr )]

Substituting Eqs.(1.18) and (1.19) into the first two expressions of Eq.(1.22) and substituting the result into Eq.(1.21) results in:

22dr w d +r 1dr dw —2r

w =0 A solution of this equation is

W=Ar+r B (1.23)

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 applying the boundary conditions

σr = —p i at r=r i at σr = —p o at r=r o

Expression (1.23) then becomes:

ω= — μr ε

1+)(122i O r r Er -[r 2(1-μ-2μ2)(P i r 2i -P O r 2O )+r 2i r 2O (1+μ)(P i -P O )](1.24)

Once w is obtained, the values of σ

θ are determined from Eqs.(1.18) ,and , (1.19) ,and (1.22) and expressed for thick

cylinders as σθ=2

222222))(

(i

O O i O i i O i i r r r r r P P r P r P --+- (1.25) σr = —2222222))(

(P i O O i O i i i i O r r r r r P P r r P --+-

wh σr =radial stress σθ=hoop stress P i =internal pressure P O =external pressure

r i =inside radius r O =outside radius r=radius at any point

The 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

σ1=E ε1+2222i P (2i

O O O i r r r P r --)μ This equation indicates that σ1 is constant throughout a cross section because ε

1 is constant and r does not appear in

the second term. Thus the expression σ1 can be obtained from statics

as

σ1=2222P i O O O i i r r r P r -- (1.26) With σ1 known, Eq.(1.24) for the deflection of a cylinder can be expressed as

ω=)()1()()21)((2222222i O O i O i O O i i r r Er r r P P r P r P r -+-+--μμ (1.27)

作用于圆柱壳的内应力

当薄壁圆柱壳体承受压力时,确定其应变的等式可以由图1.16得到。对垂直于平面ABCD 的力求和可得

Lt r PL σθ22=? 或 t Pr =σθ (1.17)

其中P=压力,L=圆柱的高,θσ代表环向压力,r =半径,t=厚度。

(a) (b)

图1.16

应变θε定义为:

ε

θ=(最终长度-原始长度)/原始长度 而从图1.17得到

rr rr W r πππεθ22)(2-+= 或 r W =εθ (1.18)

同时 dr dW =εθ (1.19)

圆柱壳受内部压力从而得到径向变形,把数值带入方程式1.18。 因此,对于薄壁圆柱有:

Et P W r 2= (1.20) 其中W=径向变形,E=弹性系数。

当r/t>10时,方程式1.17和1.20能得到较精确的结果。但是,当r/t 减少时,需要更精确的表述,因为应力在厚度方向上的分布是不均匀的。通常所依赖的“厚壁”原理由Lame 首先提出。导出的等式是以图1.18中 的力与应力为基础的。这个原理假设所有的剪应力是零对称的,且有一平面在压力作用前垂直纵向轴,保持平面压力作用,换句话说,1ε在任何截面都是常量。

σr 和σθ的关系可以通过选取环形分离体进行图表分析,如图1.18b 中所示。。求垂直方向的所有力的和时,忽略高阶条件,我们有

dr d r r σσσθ=-, dr d r r σσσθ=-

(a) (b)

图1.18

其他的关系如下:

)]()1([)21)(1()]

()1([)21)(1()]

()1([)21)(1(1111εεεεεεεσεεεσθθθθμμμμμμμμμμμμr r r r E E E ++--+=++--+=++--+= (1.22)

将方程式1.18和1.19代入方程式1.22中的第一个,得到结果后再代入方程式1.21,结果得

01222=-+r r d

dr d r d ωωω 这个方程的解是 r B Ar +=ω (1.23)

其中A 和B 是积分常数,是将方程式1.23代入方程式1.22再应用临界条件确定的

p i r -=σ r i r = p r οσ-= r r ο=

由此表达式1.23变成 )])(1())(21([)(1222222221P P r r r P r P r r r i i i i i Er r οοοοομμμωμε-++----+-= (1.24)

如果知道ω,从等式1.18,1.19,1.13以及厚圆柱体的表达式我们可以得到σθ的值

r r r r r p p r p r p r r r r r p p r p r p i i i i

i r i i i i i 22222222222222))(())((--+--=--+-=οοοοοοοοοοθσσ

(1.25)

其中r σ=径向应力,θσ=环压力,i P =内部压力,o P =外部压力,i r =内径,o r =外径,r =任意半径。

将方程式1.18,1.19和1.24代入方程式1.24可以得到厚壁圆柱的纵向压力:

r r r p r p i

i i E 2

22211)(2--+

=οοομεσ 这个等式表明,在整个横截面上,1σ是常量,又因为1ε是常量,r 没有在第二部分中出现。因此,可以从静力学得到1σ的表达式,如下:

r r r p r p i i i 22221--=

οοοσ (1.26) 若1σ已知,则偏向圆柱的式子1.24可以表述为: )()1()()21)((2222222r r r r p p r p r p r i o o i o i o o i i Er -+-+--=

μμω (1.27)

Reading Material 5

Static and Dynamic Balance of Rotating Bodies

The unbalance of a single disk can detected by allowing the disk to rotate on its axle between two parallel knife-edges, as shown in Fig.1.22. The disk will rotate and come 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 automobile-engine crankshaft. A test similar to the one above many 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. l. 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 l a P and l

b P . In a similar manner Q can be replaced by two parallel forces l

c Q an

d l a Q . Th

e two forces in each plane can then be combined into a single resultant force that can be balanced by a single correction weight as shown. The two correction weights C 1 and C 2 introduced in the two parallel planes completely 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 3-oz. in. unbalance is equivalent to 412 oz. in. at the left end and 4

3 oz. in. at the right end, as shown in Fig. 1. 25. The 2 oz. in. at the middle is obviously equal to l oz. in. at the ends.

旋转体的静态和动态平衡

单个圆盘的失衡可以通过允许其在两边缘间通过转轴旋转检测,如图1.22所示。该圆盘将会旋转而且停止在底部的沉重一侧,这类失衡称为静不定,因为它可以用静态平衡的方法来检测。

一般而言,大多数的转子都会安装在前轴,比如电 动 机 电 枢 或汽 车 发 动 机 曲 轴。类似于上述的一个试验可以表明,这种零件处于静态平衡,但是当系统旋转时,就可以看出严重的失衡了。

举例说明,假想两个圆盘套在一根轴上,如图1.23所示,如果两个不等重的圆盘处于平衡,质心位于轴的两侧且成180°,系统会沿轴线方向处于静态平衡。然而,当系统旋转时,每个失衡的圆盘会受到一个旋转离心力作用而沿其轴承摇摆轴。由于这类不平衡仅受旋转而导致的,我们称为动态失衡。

图1.24所示的是一般情况下系统既有静态又有动态的失衡。由图现在可看到不平衡的力P 和Q 总会被处于任何两个旋转平行平面上的两个修正项消去。

例1.24中旋转体的平衡需要在两个平面内修正。先设想不平衡力P 可分解为两个平行力Pa/l 和Pb/l 。同理力Q 也可分解为两个平行的力Qc/l 和Qd/l 。每个平面上的两个力又可合成一个与单个校正重量平衡的力。 这两个引进于两个平行面的校正重量C1和C2分别与力P 和Q 晚完全平衡,这时系统处于静态和动态的平衡。需要再次强调的是一个处于动平衡的系统也可处于静态平衡。然而,相反的情况可能就会不成立,处于平衡的系统可能会处于动态的失衡中。

举例:

一个4英寸的转子在距离最左端一英寸的一个平面上有3盎司的不平衡重量,在中心面上有2盎司。从左边看去至第一个不平衡量沿顺时针方向两者成90°。确定在两个断面处的校正量,指出大小和角位置。

解答:

3盎司不平衡重量等于左端面2.24盎司的重量和右端面的0.75盎司的重量,如图1.25所示。中部2盎司的重量显然等于两端面处1盎司的重量之和。

在每个端面分别合成两个不平衡量,得校正量为:

左端面:

()47.202512

21=+=C '02425

.21tan 1

1o ==-θ 第一个不平衡点的顺时针方向 右端面: 25.1143222=+??

? ??=C o 5375.01tan 11==-θ 第一个不平衡点的逆时针方向

Reading Material 6

Stainless Steel

Stainless steels do not rust in the atmosphere as most other steels do. The term "stainless" implies a resistance to staining, rusting, and pitting in the air, moist and polluted as it is, and generally defines a chromium content in excess of 11 % but less than 30%. And the fact that the stuff is "steel" means that the base is iron.

Stainless steels have room-temperature yield strengths that range from 205 MPa (30 ksi) to more than 1725 MPa (250 ksi). Operating temperatures around 750 C (1400 F) are reached. At the other extreme of temperature some stainless steels maintain their toughness down to temperatures approaching absolute zero.

With specific restrictions in certain types, the stainless steels can be shaped and fabricated in conventional ways. They can be produced and used in the as-cast condition; shapes can be produced by powder-metallurgy techniques; cast ingots can be rolled or forged (and this accounts for the greatest tonnage by far). The rolled product can be drawn, bent, extruded, or spun. Stainless steel can be further shaped by machining, and it can be joined by soldering, brazing, and welding. It can be used as an integral cladding on plain carbon or low-alloy steels.

The generic term "stainless steel" covers scores of standard compositions as well as variations bearing company trade names and special alloys made for particular applications. Stainless steels vary in their composition from a fairly simple alloy of, essentially, iron with 11% chromium, to complex alloys that include 30% chromium, substantial quantities of nickel, and half a dozen other effective elements. At the high-chromium, high-nickel end of the range they merge into other groups of heat-resisting alloys, and one has to be arbitrary about a cutoff point. If the alloy content is so high that the iron content is about half, however, the alloy falls outside the stainless family. Even with these imposed restrictions on composition, the range is great, and naturally, the properties that affect fabrication and use vary enormously. It is obviously not enough to specify simply a "stainless steel.‖

Classification

The various specifying bodies categorize stainless steels according to chemical composition and other properties. However, all the stainless steels, whatever specifications they conform to, can be conveniently classified into six major classes that represent three distinct types of alloy constitution, or structure. These classes are ferritic, martensitic, austenitic, manganese-substituted austenitic, duplex austenitic ferritic, and precipitation-hardening. Each class is briefly described below.

(1) Ferrous stainless steels: This class is so named because the crystal structure of the steel is the same as that of iron at room temperature. The alloys in the class are magnetic at room temperature and up to their Curie temperature (about 750 C; 1400 F). Common alloys in the ferrous class contain between 11% and 29% chromium, no nickel, and very little carbon in the wrought condition. (2) Martensitic stainless steels: Stainless steels of this class, which necessarily contain more than 11 % chromium, have such a great hardenability that substantial thickness will harden during air cooling, and nothing more

drastic than oil quenching is ever required. The hardness of the as-quenched martensitic stainless steel depends on its carbon content. However, the development of mechanical properties through quenching and tempering is inevitably associated with increased susceptibility to corrosion. (3) Austenitic stainless steels: The traditional and familiar austenitic stainless steels have a composition that contains sufficient chromium to offer corrosion resistance, together with nickel to ensure austenite at room temperature and below. The basic austenitic composition is the familiar l8% chromium, 8% nickel alloy. Both chromium and nickel contents can be increased to improve corrosion resistance, and additional elements (most commonly molybdenum) can be added to further enhance corrosion resistance. (4) Manganese-substituted austenitic stainless steels: The austenitic structure can be encouraged by elements other than nickel, and the substitution of manganese and nitrogen produces a c1ass that we believe is sufficiently different in its properties to be separated from the chromium-nickel austenitic class just described. The most important difference lies in the higher strength of the manganese-substituted alloys.

(5) Duplex austenitic-ferrous stainless steels: The structure of these steels is a hybrid of the structures of ferrite and austenite; and the mechanical properties likewise combine qualities of each component steel type. The duplex steels combine desirable corrosion and mechanical properties, and their use is as a result increasing in both wrought and cast form. (6) Precipitation-hardening stainless steels: Stainless steels can be designed so that their composition is amenable to precipitation hardening. This class cuts across two of the other c1asses, to give us martensitic and austenitic precipitation-hardening stainless steels. In this class we find stainless steels with the greatest useful strength as well as the highest useful operating temperature.

(2)temperature.

Properties In selection of stainless steels, three kinds of properties have to be considered: (1) Physical properties: density, thermal conductivity, electrical resistivity, and so on; (2) Mechanical properties: strength, ductility, hardness, creep resistance, fatigue, and so on; and (3) corrosion-resistant properties. Note that properties of stainless steels are substantially influenced by chemica1 composition and microstructure. Hence specifications include chemical composition, or, more correctly, an analysis of the most important elements (traces of unreported elements also may be present) as well as a heat treatment that provides the optimum structure.

Applications

Since stainless steels were first used in cutlery industry, the number of applications has increased dramatically. The relative importance of the major fields of application for flat and long stainless steel products is shown in Table 1. Chemical and power engineering is the largest market for both long and flat products. It began in about 1920 with the nitric acid industry. Today, it includes an extremely diversified range of service conditions, including nuclear reactor vessels, heat exchangers, oil industry tubular, components for the chemical processing and pulp and paper industries, furnace parts, and boilers used in fossil fuel electric power plants.

不锈钢

不锈钢不像其他大多数钢在空气中容易生锈。术语“不锈钢”意味着具有抵抗染色、生锈、潮湿污染空气中被腐蚀的能力,并且一般规定铬的含量在11%—30%之间。实际上原料是“钢”而本质是铁。

常温下,不锈钢的屈服强度从205MPa(30ksi)到725MPa(250KSI)。工作温度达约750℃(1400F)。

在温度的另一个极端,一些不锈钢能在接近于绝对零度的条件下保持很好的韧性。

由于某些类型的特定限制,不锈钢也可以通过常规方法成型和制造。它们可以在铸造状态下生产和使用;通过粉冶金技术获得外形;铸锭可以轧制或者锻造(这是迄今占有最大吨位的)。轧制产品可以取出,弯曲,挤压或旋转。不锈钢可以进一步塑造加工,也可以加入锡焊,铜焊和焊接。还可以用于普通碳钢和低合金钢的完整电镀。

通用术语“不锈钢”不仅包括数十种标准成分,还有变更轴承公司的商标名称和特殊合金的特定应用的意思。不锈钢的组成成分各不相同,从一种相当简单的合金,如含11%铬的铁,到复合金,包括含30%铬,大多数的镍和五六个其他的有效成分。在高铬、高镍范围内最后组成其他的耐热合金,且分界点是任意的。但是,如果合金含量过高以至于是铁含量的一半,则该合金就不是不锈钢了。即使对成分强加限制,范围是很宽的,自然的,其性能对合金的制造和使用仍有很大影响。很明显这是不足以简单的规定为“不锈钢”的。

分类:

不锈钢是按照其化学成分和其它性能分类的。但是不管有怎样的规格,所有不锈钢均可以很容易的分成六大类,体现各种同类型合金的结构。这些分类分别是铁素体,马氏体,奥氏体锰代铬奥氏体,奥氏体-铁素体双相和沉淀硬化型。各种类型描述如下:

(1)铁素体不锈钢:它是在室温下钢的晶体结构和铁一样而得名。室温下这种合金是有磁性的,直到居里温度(约750℃,1400F)。在有色金属中,普通合金含11%—29%的铬,不含镍,在可锻条件下含有微量的碳。

马氏体不锈钢:这类不锈钢含有不低于11%的铬,有很好的淬透性,在空冷时也能变硬。对油淬没有硬性要求。对于淬火马氏体不锈钢,它的硬度取决于含碳量的多少。然而,通过淬火和回火生产的机械必然会极易被腐蚀。

(2)奥氏体不锈钢:正宗和不正宗的奥氏体不锈钢都有足够的铬耐蚀性,连同镍来确保室温下奥氏体紧缩。奥氏体的基本成分是约18%的铬和8%的镍的合金。铬和镍都可以提高耐蚀性,其他的元素(大部分是钼)可以进一步增加耐蚀性。

(3)锰代铬奥氏体不锈钢:这种奥氏体的成分能被镍以外的元素激发,而锰和氮生成的成分,我们认为其属性是不同的,是有别于刚刚描述的铬镍奥氏体的。最主要的区别是锰代铬合金的高强度。

(4)奥氏体—铁素体双相不锈钢:这种钢的成分是铁素体和奥氏体的混合物,并且机械性能也结合了每个组件的优良钢种。这种双相钢结合了腐蚀性和机械性能,结果增加了锻造和铸造的形式。

(5)沉淀硬化型不锈钢:这种不锈钢的设计是经得起沉淀硬化的。这类钢横跨其它两类钢,给我们马氏体和奥氏体沉淀硬化不锈钢。我们发现了具有最佳力以及有益的最佳工作温度的不锈钢。性能:

不锈钢的选择,要考虑到三中性能:(1)物理性能:密度、热传导性、电阻率等等;(2)机械性能:强度、韧性、硬度、抗蠕变、疲劳度等等;(3)耐蚀性。注意不锈钢受化学成分和微观结构影响。因此,说明书要包含化学成分,或者,更准确的重要元素的分析(未公布的元素也尽可能标明)以及提供优化结构的热处理。

应用:

自从不锈钢首次用于工业刀具以来,应用范围越来越广。平面的和长的不锈钢产品在主要应用领域的重要性如表1所示。化工和电力工程是最广的应用领域。开始于1920年的硝酸行业。今天,应用极为多样化,包括核反应容器,换热器,石油行业的管道,用于化学加工、纸浆和造纸工业,炉零件组件的范围,和使用化石燃料发电厂的锅炉。

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CQ螺纹球阀CQ Thread Ball Valves L形三通式L-pattern three way T形三通式T-pattern three way 安全阀Safety valve 暗杆闸阀Inside screw nonrising stem type gate valve 百叶窗; 闸板shutter 百叶窗式挡板louver damper 摆阀式活塞泵swing gate piston pump 保温式Steam jacket type 报警阀alarm valve 报警阀; 信号阀; 脉冲阀sentinel valve 背压调节阀back pressure regulating valve 背压率Rate of back pressure 本体阀杆密封body stem seal 波纹管阀Bellows valves 波纹管密封阀bellow sealed valve 波纹管密封式Bellows seal type 波纹管平衡式安全阀Bellows seal balance safety valve 波纹管式减压阀Bellows reducing valve

波纹管式减压阀Bellows weal reducing valve 薄膜thin film 薄膜; 隔膜diaphragm 薄膜式减压阀Diaphragm reducing valve 薄型闸阀Thin Gate Valves 不封闭式Unseal type 槽车球阀Tank Lorry Ball Valves 颤振Flutter 常闭式Normally closed type 常开式Normally open type 超低温阀门Cryogenic valve 超高压阀门Super high pressure valve 超过压力Overpressure of a safety valve 衬胶隔膜阀rubber lined diaphragm 衬胶截止阀rubber lined globe valve 垂直板式蝶阀Vertical disc type butterfly valve 磁耦合截止阀Magnetic Co-operate Globe Valves 带补充载荷的安全阀Supplementary loaded safety valve 带辅助装置的安全阀Assisted safety valve

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户外用品整理 个人装备Personal equipment ==============登山靴Climbing boots 防寒运动靴Snow training shoes 攀岩鞋Climbing shoes 毛衬衫Woolen Shirt 登山裤Climbing trousers 运动衣裤Training wear 毛内衣裤Woolen undershirts 毛袜Woolen socks 毛手套Woolen glove 丝手套Silk glove 棉手套Cotton glove 毛衣Sweater 冲锋衣Jaket(Windbreaker) 外裤Over-trousers 外手套Over-gloves 外鞋罩Long spats 防寒帽Bataclave 高处帽High altitude cap 太阳帽Glacier cap 太阳镜Sunglasses 睡垫Mattress 鸭绒睡垫Down sleeping bag 鸭绒衣Down jacket 鸭绒裤Down trousers 鸭绒背心Down vest 鸭绒袜Down tent shoes/slippers 睡袋套Sleeping bag cover 背包Duffel bag 整理袋Stuff bag 冰爪Grampons 冰爪带Grampons strap 冰爪袋Grampons case 外靴Over-shoes 安全帽Helmet 冰镐Ice axe(PIckel) 安全带Harness 铁锁Carabiner with safety ring 铁锁Carabmer 小绳套Sling 下降器Eight rings

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材料科学与工程专业英语-课文翻译-Unit one

United 1 材料科学与工程 材料在我们的文化中比我们认识到的还要根深蒂固。如交通、房子、衣物,通讯、娱乐和食物的生产,实际上,我们日常生活中的每一部分都或多或少地受到材料的影响。历史上社会的发展、先进与那些能满足社会需要的材料的生产及操作能力密切相关。实际上,早期的文明就以材料的发展程度来命名,如石器时代,铜器时代。 早期人们能得到的只有一些很有限的天然材料,如石头、木材、粘土等。渐渐地,他们通过技术来生产优于自然材料的新材料,这些新材料包括陶器和金属。进一步地,人们发现材料的性质可以通过加热或加入其他物质来改变。在这点上,材料的应用完全是一个选择的过程。也就是说,在一系列非常有限的材料中,根据材料的优点选择一种最适合某种应用的材料。直到最近,科学家才终于了解材料的结构要素与其特性之间的关系。这个大约是过去的60年中获得的认识使得材料的性质研究成为时髦。因此,成千上万的材料通过其特殊的性质得以发展来满足我们现代及复杂的社会需要。 很多使我们生活舒适的技术的发展与适宜材料的获得密切相关。一种材料的先进程度通常是一种技术进步的先兆。比如,没有便宜的钢制品或其他替代品就没有汽车。在现代,复杂的电子器件取决于所谓的半导体零件。 材料科学与工程 有时把材料科学与工程细分成材料科学和材料工程学科是有用的。严格地说,材料科学涉及材料到研究材料的结构和性质的关系。相反,材料工程是根据材料的结构和性质的关系来设计或操纵材料的结构以求制造出一系列可预定的性质。从功能方面来说,材料科学家的作用是发展或合成新的材料,而材料工程师是利用已有的材料创造新的产品或体系,和/或发展材料加工新技术。多数材料专业的本科毕业生被同时训练成材料科学家和材料工程师。“structure”一词是个模糊的术语值得解释。简单地说,材料的结构通常与其内在成分的排列有关。原子内的结构包括介于单个原子间的电子和原子核的相互作用。在原子水平上,结构包括原子或分子与其他相关的原子或分子的组织。在更大的结构领域上,其包括大的原子团,这些原子团通常聚集在一起,称为“微观”结构,意思是可以使用某种显微镜直接观察得到的结构。最后,结构单元可以通过肉眼看到的称为宏观结构。 “Property”一词的概念值得详细阐述。在使用中,所有材料对外部的刺激都表现出某种反应。比如,材料受到力作用会引起形变,或者抛光金属表面会反射光。材料的特征取决于其对外部刺激的反应程度。通常,材料的性质与其形状及大小无关。 实际上,所有固体材料的重要性质可以概括分为六类:机械、电学、热学、磁学、光学和腐蚀性。对于每一种性质,其都有一种对特定刺激引起反应的能力。如机械性能与施加压力引起的形变有关,包括弹性和强度。对于电性能,如电导性和介电系数,特定的刺激物是电场。固体的热学行为则可用热容和热导率来表示。磁学性质表示一种材料对施加的电场的感应能力。对于光学性质,刺激物是电磁或光照。用折射和反射来表示光学性质。最后,腐蚀性质表示材料的化学反应能力。 除了结构和性质,材料科学和工程还有其他两个重要的组成部分,即加工和性能。如果考虑这四个要素的关系,材料的结构取决于其如何加工。另外,材料的性能是其性质的功能。因此,材料的加工、结构、性质和功能的关系可以用以下线性关系来表示: 加工——结构——性质——性能。 为什么研究材料科学与工程? 为什么研究材料科学与工程?许多应用科学家或工程师,不管他们是机械的、民事的、化学的或电子的领域的,都将在某个时候面临材料的设计问题。如用具的运输、建筑的超级结构、油的精炼成分、或集成电路芯片。当然,材料科学家和工程师是从事材料研究和设计的专家。

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附录:外文翻译 5.1Introduction Cylindrical shells are used innuclear,fossil and petrochemical industries. 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 shells 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 internal pressure and sometimes by external pressure which can produce buckling.Applied loads are also important in controlling thickness and so are the disconti-nuity and thermal stresses.The basic thicknesses of cylindrical shells are Based on simpli?ed stress analysis and allowable stress for the material of construction.There are some variations of the basic equations in various design codes.Some of the equations are based on thick-wall Lame equations.In this chapter such equations will be discussed.Also we shall discuss the case of cylindrical shells under external pressure where there is a propensity of buckling or collapse. 5.2 Thin-shell equations 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 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 Flugge, We have the following stress resultants: ?-+ = 2 2 1 1) ( t t dz r z r N θ θ σ(5.1) ?-+ = 2 2 2 2) ( t t dz r z r N φ φ σ(5.2) ?-+ = 2 2 2 2) ( t t dz r z r N θφ θφ σ(5.3)

电子信息工程专业英语 课文翻译 Unit 12 译文

Unit 12 生物识别技术 Unit 12-1 第一部分:指纹识别 在所有的生物技术中,指纹识别是最早期的一种技术。我们知道,每个人都有自己独特的、不可变更的指纹。指纹是由手指表皮上的一系列峰谷组成的。指纹的独特性是由这些峰谷的形状以及指纹的细节点所决定的。指纹的细节点是指纹局部凸起处的一些特性,这些特性出现在凸起的分叉处或是凸起的截止处。 指纹匹配技术可以被分为两类:基于细节的指纹匹配技术和基于相关性的指纹匹配技术。基于细节的指纹匹配首先要找出细节点,然后在手指上对应出与它们相关的位置,如图12.1所示。但是,使用这种方法存在一些困难。要精确地提取指纹的细节点是很困难的。而且,这种方法不能很好地考虑指纹峰谷的整体形状。基于相关性的指纹匹配技术可以解决部分基于细节的指纹匹配方法存在的问题,但它也存在一些自身的缺陷。基于相关性的匹配技术需要给出已注册过的特征点的精确位置,并且该方法会受图像平移和旋转的影响。 图12.1 基于细节的指纹匹配 基于细节的指纹匹配技术在匹配不同大小的细节模型时(未注册过的)会存在一些问题。指纹上局部的凸起结构不能完全由指纹细节实现特征化。我们可以尝试另一种表达指纹的方法,它可以获得更多的指纹局部信息并且得到固定长度的指纹编码。于是,我们只需要计算两个指纹编码之间的欧几里得距离,匹配过程有望变得相对简单。 研发对于指纹图像中噪声更稳健并能实时提供更高精度的算法是重要的。商用指纹(身份)认证系统对给定的错误接受率要求具有很低的错误拒绝率。在这点上,任何一项简单的技术都很难实现。我们可以从不同的匹配技术中汇总多个证据从而提高系统的总体精确度。在实际应用中,传感器、采集系统、性能随时间的变化是关键因素。为了评价系统性能,我们有必要对少数使用者在一段时间内进行现场试验。 每天我们可以从法医鉴定、出入口控制、驾驶证登记等多个方面的应用中采集并保存大量的指纹。基于指纹的自动识别系统需要把输入的指纹与数据库中大量的指纹进行匹配验证。为了缩短搜索时间、降低计算复杂度,要以准确而一致的方式将这些指纹分类,从而使输入的指纹只需与数据库某一子集中的指纹进行比对。 指纹分类是将指纹划分入多个预定义指纹类型中的一类中的一项技术。这些预定义的指纹类型是由提供索引机制的文献建立的。图12.2表示了不同类型的指纹。它们有螺纹状的、右旋状的、左旋状的、弓形的、帐篷形的。输入指纹首先粗略地被匹配为预定义的类型中的一类,随后,输入指纹仅与这一大类指纹库中的子集作更精细的比较。 图12.2 不同类型的指纹 指纹自动匹配中关键的一步是自动并且可靠地从输入指纹图像中提取出细节。然而,细节提取算法的好坏很大程度上依赖于输入图像质量的好坏。为确保自动指纹识别或确认系统的性能对于不同指纹图像质量具有稳健性,必须在细节提取模块中加入指纹增强算法。快速指纹增强算法能根据对局部指纹脊方向和频率(密度)的估计自适应地改善输入指纹图像中脊和沟结构的清晰度,如图12.3所示。实验证明加入增强算法可以显著增强指纹认证的精度。 图12.3 指纹增强 Unit 12-2 第二部分:说话人辨认介绍 介绍 现代安全系统范围很广, 通常要通过多层系统才能完全被突破。除了标准的锁和警报系统以外,还有很复杂的方法来保护重要资料。其中多数的方法为可以允许或者不允许一个特定人员获取资料——计算机系统必须能够检测出指纹、读取个人眼孔图样、或者确定说话者的真实身份。最后一点是本文讨论的重点——说话人辨认。说话人辨认经常会与其他类似的术语混淆。以下对这些术语的精确定义做出解释。 说话人识别:确定是谁在说话。 说话人辨认:初始情况下不知道说话人是谁,必须在与模板比较后确定说话人的身份。通常会有很多相关的模板。 说话人确认:确定说话者是否就是他(她)自称的那个人。仅将说话者的话音与一个样板进行比对,即他(她)自称的那一个。 语音识别:识别出人们说话时的语句。换句话说,识别出一个人在说什么而不是谁在说话。这个术语经常与声音识别相混淆。声音识别是识别出说话人。

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