1、土木工程专业英语全部Lesson 1Compression MembersNew Words1. achieve achievement 2. eccentricity center, 中心; eccentric 偏心的;eccentricity 偏心,偏心距3. inevitable evitable 可避免的 avoidable; inevitable 不可避免的 unavoidable 4. truss 桁架 triangular truss, roof truss, truss bridge5. bracing brace 支柱,支撑;bracing, 支撑,撑杆6. slender
2、细长,苗条;stout; slenderness7. buckle 压曲,屈曲;buckling load8. stocky stout9. convincingly convince, convincing, convincingly10. stub 树桩, 短而粗的东西;stub column 短柱11. curvature 曲率;curve, curvature12. detractor detract draw or take away; divert; belittle, 贬低,诽谤;13. convince14. argument dispute, debate, quarrel,
3、 reason, 论据(理由)15. crookedness crook 钩状物,v弯曲,crooked 弯曲的16. provision 规定,条款Phrases and Expressions1. compression member2. bending moment shear force, axial force3. call upon (on) 要求,请求,需要4. critical buckling load 临界屈曲荷载 critical 关键的,临界的5. cross-sectional area6. radius of gyration 回转半径 gyration 7. sl
4、enderness ratio 长细比8. tangent modulus 切线模量9. stub column 短柱10. trial-and-error approach 试算法11. empirical formula 经验公式 empirical 经验的12. residual stress 残余应力 residual13. hot-rolled shape 热轧型钢 hot-rolled bar 14. lower bound 下限 upper bound 上限16. effective length 计算长度Definition (定义)Compression members ar
5、e those structural elements that are subjected only to axial compressive forces: that is, the loads are applied along a longitudinal axis through the centroid of the member cross section, and the stress can be taken as fa=P/A, where fa is considered to be uniform over the entire cross section. 受压构件是
6、仅受轴向压力作用的构件,即:荷载是沿纵轴加在其截面形心上的,其应力可表示为,式中,假定fa在整个截面上均匀分布。 This ideal state is never achieved in reality, however, and some eccentricity of the load is inevitable. 然而,现实中从来都不可能达到这种理想状态,因为荷载的一些偏心是不可避免的。This will result in bending, but it can usually be regarded as secondary and can be neglected if the
7、theoretical loading condition is closely approximated. 这将导致弯曲,但通常认为它是次要的,如果理论工况是足够近似的,就可将其忽略。This cannot always be done if there is a computed bending moment, and situation of this type will be considered in Beam-Columns. 但这并非总是可行的,如有计算出的弯矩存在时,这种情形将在梁柱理论中加以考虑。 The most common type of compression mem
8、ber occurring in buildings and bridges is the column, a vertical member whose primary function is to support vertical loads. 在建筑物和桥梁中最常见的受压构件就是柱,其主要功能就是支承竖向荷载。In many instances these members are also called upon to resist bending, and in these cases the member is a beam-column. Compression members c
9、an also be found in trusses and as components of bracing systems. 在许多情况下,它们也需要抵抗弯曲,在此情况下,将它们称为梁柱。受压构件也存在于桁架和支撑系统中。Column Theory (柱理论) Consider the long, slender compression member shown in Fig.1.1a. 考虑如图1.1.a所示的长柱If the axial load P is slowly applied, it will ultimately reach a value large enough to
10、 cause the member to become unstable and assume the shape indicated by the dashed line. 如果慢慢增加轴向荷载P,它最终将达到一个足够大的值使该柱变得不稳定(失稳),如图中虚线所示。 The member is said to have buckled, and the corresponding load is called the critical buckling load. 这时认为构件已经屈曲,相应的荷载称为临界屈曲荷载。If the member is more stocky, as the on
11、e in Fig.1.1b, a larger load will be required to bring the member to the point of instability. 如果该构件更粗短些,如图1.1b所示,则需要更大的荷载才能使其屈曲。For extremely stocky members, failure may be by compressive yielding rather than buckling. 对特别粗短的构件,破坏可能是由受压屈服引起而非由屈曲引起。For these stocky members and for more slender colum
12、ns before they buckle, the compressive stress P/A is uniform over the cross section at any point along the length. 对这些短柱以及更细长的柱,在其屈曲前,在其长度方向上任意点处横截面上的压应力P/A都是均匀的。As we shall see, the load at which buckling occurs is a function of slenderness, and for very slender members this load could be quite sma
13、ll. 我们将会看到,屈曲发生时的荷载是长细程度的函数,非常细长的构件的屈曲荷载将会很低。 If the member is so slender (a precise definition of slenderness will be given shortly) that the stress just before buckling is below the proportional limitthat is, the member is still elasticthe critical buckling load is given by 如果构件如此细长(随后将会给出细长程度的精确定
14、义)以致即将屈曲时的应力低于比例极限即,构件仍是弹性的,临界屈曲荷载如下式给出: (1.1)where E is the modulus of elasticity of the material, I is the moment of inertia of the cross-sectional area with respect to the minor principal axis, and L is the length of the member between points of support. 式中E为材料弹性模量,I为关于截面副主轴的惯性矩,L 为支座间的距离。For Eq1
15、.1 to be valid, the member must be elastic, and its ends must be free to rotate but not translate laterally. This end condition is satisfied by hinges or pins. 要使方程1.1成立,构件必须是弹性的,且其两端必须能自由转动,但不能侧向移动。 This remarkable relationship was first formulated by Swiss mathematician Leonhard Euler and publishe
16、d in 1975. 此著名公式是瑞士数学家欧拉于1975年提出的。The critical load is sometimes referred to as the Euler load or the Euler buckling load. The validity of Eq.1.1 has been demonstrated convincingly by numerous tests. 因此有时将临界荷载称为欧拉荷载或欧拉临界荷载。欧拉公式的有效性(正确性)已由许多试验充分证实。 It will be convenient to rewrite Eq.1.1 as follows:
17、方程1.1可方便地写为 (1.1a)where A is the cross-sectional area and r is the radius of gyration with respect to the axis of buckling. The ratio L/r is the slenderness ratio and is the measure of a compression members slenderness, with large values corresponding to slender members. 式中A 为截面面积,r为关于屈曲轴的回转半径,L/r为长
18、细比,它是对受压构件细长程度的一种度量,该值越大,构件越细长。 If the critical load is divided by the cross-sectional area, the critical buckling stress is obtained: 如果将屈曲荷载除以截面面积,便可得到以下屈曲应力: (1.2)This is the compressive stress at which buckling occur about the axis corresponding to r. 这便是绕相应于r的轴发生屈曲时的压应力。Since buckling will take
19、 place as soon as the load reaches the value by Eq.1.1, the column will become unstable about the principle axis corresponding to the largest slenderness ratio. This usually means the axis with the smaller moment of inertia. 由于一旦荷载达到式1.1之值,柱将在与最大长细比对应的主轴方向变得不稳定(失稳),通常该轴是惯性矩较小的轴。Thus, the minimum mom
20、ent of inertia and radius of gyration of the cross section should be used in Eq.1.1 and 1.2. 因此,应在方程1.1和1.2中采用截面的最小惯性矩和最小回转半径。 Early researchers soon found that Eulers equation did not give reliable results for stocky, or less slender, compression members. 早期的研究者很快发现对短柱或不太细长的受压构件,欧拉公式并不能给出可靠的结果,This
21、 is because of the small slenderness ratio for members of this type, which results in a large buckling stress (from Eq.1.2). 这是因为这种构件的长细比较小,从而产生较高的屈曲应力。If the stress at which buckling occurs is greater than the proportional limit of the material, the relation between stress and strain is not linear,
22、 and the modulus of elasticity E can no longer be used. 如果屈曲发生时的应力大于材料的比例极限,应力应变关系就不再是线性的,也不能再用弹性模量E。This difficulty was initially resolved by Friedrich Engesser, who proposed in 1889 the use of a variable tangent modulus Et in Eq.1.1. 这一困难最初由Friedrich Engesser 所克服,他在1889年将可变的切线模量用于方程1.1. For a mate
23、rial with a stress-strain curve like the one in Fig.1.2, E is not a constant for stress greater than the proportional limit Fpl. The tangent modulus Et is defined as the slope of the tangent to the stress-strain curve for values of f between Fpl and Fy. 对于如图1.2所示的应力应变曲线(的材料),当应力超过比例极限时,E并非常数,当应力处于Fp
24、l和Fy之间时,将切线模量定义为应力应变曲线的切线的斜率,If the compressive stress at buckling, Pcr/A, is in this region, it can be shown that 如果屈曲时的压应力在此范围时,可以证明 (1.3)This is identical to the Euler equation, except that Et is substituted for E. 除公式中将E代之以Et外,上式与欧拉公式完全相同。Effective Length(计算长度)Both the Euler and tangent modulus
25、equations are based on the following assumptions: 欧拉和切线模量方程都是基于如下假定:1. The column is perfectly straight, with no initial crookedness. 柱完全竖直,无初始弯曲。2. The load is axial, with no eccentricity. 荷载是轴向加载,无偏心。3. The column is pinned at both ends. 柱在两端铰结。The first two conditions mean that there is no bendin
26、g moment in the member before buckling. 前两(假定)条件意味着在屈曲前无弯矩存在。As mentioned previously, some accidental moment will be present, but in most cases it can be neglected. 如前所述,可能偶然会存在一些弯矩,但在大多数情况下都可被忽略。The requirement for pinned ends, however, is a serious limitation, and provisions must be made for other
27、 support conditions. 然而,铰结要求是一个严重的局限,必须对其它支撑条件作出规定。The pinned-end condition is one that requires that the member be restrained from lateral translation, but not rotation, at the ends. 铰结条件要求约束构件两端不发生侧移,但并不约束转动。Since it is virtually impossible to construct a frictionless pin connection, even this sup
28、port condition can only be closely approximated at best. 由于实际上不可能构造无摩擦铰连接,即使这种支撑条件最多也只能是非常近似。Obviously, all columns must be free to deform axially. 显然,所有柱必须在轴向自由变形。 In order to account for other end conditions, the equations for critical buckling load will be written as 为了考虑其它边界条件,将临界荷载写为如下形式 or (1.
29、4)where KL is the effective length, and K is called the effective length factor. Values of K for different cases can be determined with the aid of the Commentary to the AISC Specification. 式中KL为计算长度,K称为计算长度系数,各种情况下的K值可借助于AISC(美国钢结构学会American Institute of Steel Construction)规范的条文说明加以确定。Lesson 2 Intro
30、duction to structural designNew words1. framework frame+work=frame 构架,框架; frame structure, 框架结构2. constraint vt. constrain 约束,强迫;n. constraint 约束 3. collaborate vt. 合作,通敌;collaboration, collaborative 4. evaluation vt. evaluate, value; assess, assessment5. fixture vt. fix, fixture 固定设备,固定物,夹具6. parti
31、tion vt. n 分割,划分, make apart; partition wall7. overlook8. crane n. 超重机,鹤9. fatigue vt.vi.n 疲劳 fatigue strength, fatigue failure10. drift v.n 漂流,漂移,雪堆11. enumerate v. list 列举12. plumbing n. (卫生,自来水)管道,plumber 管道工13. ventilation n. 通风, ventilate, ventilate a room, a well-ventilated room, vent 通风口14. a
32、ccessibility n. 可达性, access, n. vt. 通道,接近;accessible 易接近的,可达到的 15. code n. vt 代码,编码,规范16. administer v. 管理,执行;administrate, 管理 17. metropolitan a. 大城市,of metropolis 18. consolidate v. 巩固,strengthen,reinforce; consolidation19. prescription n. 规定,命令,药方;prescribe 20. municipality n. 市政当局,直辖市, municipal government21. specification n. 详述,规格, 规范;specify 22. mandate n. 书面命令,委托,Phrases and expressions 1. functional de
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