土木工程专业英语课后翻译参考(二班)

Civil EngineeringCivil engineering, the oldest of the engineering specialties, is the planning, design, construction, and management of the built environment. This environment includes all structures built according to scientific principles, from irrigation and drainage systems to rocket-launching facilities.土木工程学作为最老的工程技术学科，是指规划，设计，施工及对建筑环境的管理。

此处的环境包括建筑吻合科学规范的所有结构，从灌溉和排水系统到火箭发射设施。

Civil engineers build roads, bridges, tunnels, dams, harbors, power plants, water and sewage systems, hospitals, schools, mass transit, and other public facilities essential to modern society and large population concentrations. They also build privately owned facilities such as airports, railroads, pipelines, skyscrapers, and other large structures designed for industrial, commercial, or residential use. In addition, civil engineers plan, design, and build complete cities and towns, and more recently have been planning and designing space platforms to house self-contained communities.土木工程师建筑道路，桥梁，管道，大坝，海港，发电厂，给排水系统，医院，学校，公共交通和其他现代社会和大量人口集中地区的基础公共设施。

土木工程专业英语翻译第一单元Fundamentally, engineering is an end-product-oriented discipline that is innovative, cost-conscious and mindful of human factors. It is concerned with the creation of new entities, devices or methods of solution: a new process, a new material, an improved power source, a more efficient arrangement of tasks to accomplish a desired goal or a new structure. Engineering is also more often than not concerned with obtaining economical solutions. And, finally, human safety is always a key consideration.从根本上，工程是一个以最终产品为导向的行业，它具有创新、成本意识，同时也注意到人为因素。

它与创建新的实体、设备或解决方案有关：新工艺、新材料、一个改进的动力来源、任务的一项更有效地安排，用以完成所需的目标或创建一个新的结构。

工程是也不仅仅关心获得经济的解决方案。

最终，人类安全才是一个最重要的考虑因素。

Engineering is concerned with the use of abstract scientific ways of thinking and of defining real world problems. The use of idealizations and development of procedures for establishing bounds within which behavior can be ascertained are part of the process.工程关心的是，使用抽象的科学方法思考和定义现实世界的问题。

土木工程专业英语翻译第一篇：土木工程专业英语翻译1.第一课土木工程，这个最古老的工程专业，是指对被建设环境的规划、设计、建筑和管理。

这个环境包括按科学原理所建的一切结构，从灌溉和排水系统到火箭发射设备。

土木工程师们修路、建桥、打隧道，木工程师始终（将介词throughout转译成副词）都要充分利用计算机。

用计算机来设计工程的各要素（计算机辅助设计CAD）且用计算机来管理这个工程项目。

对于现代土木工程师而言，计算机是必备的工具，因为它们允许工程师高效建水坝，海港，发电厂，供水排水系统，建医院，学校，公共交通设施和其他公共设施实质上就是要建设现代化社会和大量人口集中地。

他们也建设私有的设施，例如：机场，铁路，管线，摩天大楼，和其他大型建筑物，它们设计用于工业，商业和居住等用途。

此外，土木工程师规划，设计和建设完整的城市和乡镇，近年来，已经在规划和设计空间平台来构建自给自足型社区。

2.土木这个词来源于拉丁文，原意是市民。

1782年，英国人JohnSmeaton用这个术语将非军事工程工作从在当时占绝大多数的军事工程师所人事的工程工作中区别开来。

从那以后，土木工程这个词常常用于表示建设公用设施的工程师们所人事的工作，尽管这个领域要宽广得多。

3.范围：因为它的面太广，所以土木工程被分成许多技术专业。

各专业的土木工程专家所需要的技能取决于工程项目的类型。

当一个项目开始时，场地被土木工程师所测绘，他们给定给水排水设施和电力线路的实际位置。

岩土工程专家们完成土壤实验来确定地基是否能承受工程项目的自重。

环境专家们研究项目对当地的影响：潜在的空气和地下水资源的污染，工程项目对当地动植物的影响，为满足保护环境的管理要求，怎样才能把工程项目设计好。

4.对于任何一个给定的工程项目，土地处理大量的用来制定工程最佳施工方法的数据。

5.结构工程.在这个专业中，土木工程师计划和设计各种各样的结构，包括桥，水坝，发电厂，设备的支柱，海岸工程的特殊结构，美国空间项目，发射塔，巨大的天文射电望远镜，和许多种其它的工程项目。

专业英语课后翻译参考Unit12、词组翻译（1）受弯构件（2）临界屈曲荷载（3）长细比（4）短柱（5）折减模量（6）effective length （7）residual stress（8）trial and error approach （9）radius of gyration（10）tangent modulus3、英译汉(1)This ideal state is never achieved in reality, however, and some eccentricity of the load is inevitable.在现实中,这种理想状态从来没有达到，然而，一些荷载偏心是不可避免的(2)In many instances the members are aslo called upon to resist bending, and in these cases the member is a beam-column.在许多情况下，需要构件能够抵抗弯矩，并在这些情况下，构件是梁柱。

(3)If the member is so slender that the stress just before buckling is below the proportional limit---that is, the member is still elastic---the critical buckling load is given by Q.如果该构件是是细长，则在屈曲前应力就低于比例限制---也就是说，该构件仍然是弹性---由公式Q 给出该临界屈曲荷载。

(4)The ratio L/r is the slenderness ratio and is the measure of a compression member’s slenderness, with large values corresponding to slender members.L/r 是长细比，是衡量受压构件的长细的一种方法，对于细长的构件它具有较大值。

（完整版）土木工程专业英语翻译(1)Concrete and reinforced concrete are used as building materials in every country. In many, including Canada and the United States, reinforced concrete is a dominant structural material in engineered construction.(1)混凝土和钢筋混凝土在每个国家都被用作建筑材料。

在许多国家，包括加拿大和美国，钢筋混凝土是一种主要的工程结构材料。

(2)The universal nature of reinforced concrete construction stems from the wide availability of reinforcing bars and the constituents of concrete, gravel, sand, and cement, the relatively simple skills required in concrete construction.(2) 钢筋混凝土建筑的广泛存在是由于钢筋和制造混凝土的材料，包括石子，沙，水泥等，可以通过多种途径方便的得到，同时兴建混凝土建筑时所需要的技术也相对简单。

(3)Concrete and reinforced concrete are used in bridges, building of all sorts, underground structures, water tanks, television towers, offshore oil exploration and production structures, dams, and even in ships.(3)混凝土和钢筋混凝土被应用于桥梁，各种形式的建筑，地下结构，蓄水池，电视塔，海上石油平台，以及工业建筑，大坝，甚至船舶等。

土木工程专业英语课文翻译土木工程专业英语课文翻译土木工程专业，是大学的一种自然学科。

专门培养掌握各类土木工程学科的基本理论和基本知识，能在房屋建筑、地下建筑、道路、隧道、桥梁建筑、水电站、港口及近海结构与设施。

以下是小编整理土木工程专业英语课文翻译的资料，欢迎阅读参考。

weight of the project. Environmental specialists study the project’s impact on the local area: the potential for air and groundwater pollution, the project’s impact on local animal and plant life, and how the project can be designed to meet government requirements aimed at protecting the environment. Transportation specialists determine what kind of facilities are needed to ease the burden on local roads and other transportation networks that will result from the completed project. Meanwhile, structural specialists use preliminary data to make detailed designs, plans, and specifications for the project. Supervising and coordinating the work of these civil engineer specialists, from beginning to end of the project, are the construction management specialists. Based on information supplies by the other specialists, construction management civil engineers estimate quantities and costs of materials and labor, schedule all work, order materials and equipment for the job, hire contractors and subcontractors, and perform other supervisory work to ensure the project is completed on time and as specified.领域。

参考译文第一单元第一部分钢筋混凝土混凝土混凝土由水，砂，石子和水泥构成。

这些不同的，分散的材料混合在一起就构成了一种坚硬的大块状物体（形状各异），有着良好的性能。

混凝土被用作建筑材料已有150年的历史。

它的普遍应用主要由于以下几点：（1）恶劣环境下的耐久性（包括耐水）（2）极易被浇铸成不同的形状和尺寸（3）相对经济实惠，极易获得（4）有极强的抗压能力但众所周知，与其较强的抗压强度相比，混凝土抗拉和抗弯强度较低。

因此，每当荷载，限制收缩或是温度发生变化，产生的拉应力超过混凝土的拉伸强度时，就会有裂缝出现。

在结构应用方面，通常的做法是利用钢筋来抵抗拉力或者是给混凝土施加压力来抵消这些拉力。

预应力混凝土对混凝土构件加载之前，对其进行压缩的方法称为预应力。

把钢筋和混凝土使用很强的力结合在一起就被称为预应力混凝土。

预应力混凝土的优点如下：1.在预应力操作过程中，混凝土和钢筋经过严格测试，较低的安全系数也是正当的。

2.混凝土中可容许的工作压力通常是抗压强度的三分之一，从而使保证金来弥补劣质混凝土在临界区发生的风险。

3.预应力减少风险，是由于混凝土在预应力操作期间产生的应力可能是其抗压强度的50%到75%。

今天，预应力混凝土被应用于建筑物，地下结构，电视塔，浮动储藏器和海上结构，电站，核反应堆容器和包括拱形桥和斜拉桥在内的各种桥梁系统当中。

这说明了预应力概念的多方面适应性以及对它的广泛应用。

所有这些结构的发展和建造的成功都是由于材料技术的进步，尤其是预应力钢和在估计预应力长期和短期损失方面积累的知识。

钢筋钢筋是一种极好的建筑材料。

与其他材料相比，钢筋有着较高的抗拉强度。

尽管在体积上是木材的十倍以上。

钢筋有着较高的弹性模量，因此在荷载下容易发生小的变形。

到目前为止所描述的钢筋的特性只适用于温度保持在70F上下的情况，大约从30F到110F。

这个温度区间覆盖了大多数结构的运行状况，但搞清楚当温度远远超出正常水平时所发生的情况仍然非常重要。

一.1.受压构件是只承受轴向压力的结构构件。

Compression members are those structural elements that are subjected only to axial compressive forces.2.公式1.1要成立，构件必须是弹性的并且其两端能自由转动但不能横向移动。

For Eq.1.1 to be valid,the member must be elastic,and it's ends must be free to rotate but not translate laterally.3.临界荷载有时被称为欧拉荷载或欧拉屈服荷载。

The critical load is sometimes referred to as the Euler load or the Euler buckling load.4.图1.2中的应力-应变曲线不同于延性钢的应力-应变曲线，因为它有明显的非线性区域。

The stress-strain curve in Fig.1.2 is differet from the ones for ductile steel because it has a pronounced region of nonlinearity.5.其他因素，像焊接和冷弯，都能影响残余应力，但冷却过程是残余应力的主要来源。

Other factors,such as welding and cold bending to create curvature in a beam, can contribute to be the residual stress,but the cooling process is the chief source.二1。

作用在结构上的力被称为荷载The forces that act on a structure are called loads.2．恒载就是固定不变的荷载，包括结构自身的重量。

(2) Cement gel 水泥凝胶体(4) The stability of the structure 结构稳定 (6) Moisture content 含水量LESSON 4 2、 Translate the following phrases into Chinese /English ・(1) Sustained lotid 长期荷载 (3) Water-cement ratio 水灰比(5) The expansion joint 伸缩缝(7) Cement paste 水泥浆 (8) The coefficient of thermal expansion of concrete 混凝土热膨胀系(9) Pernkinent plastic strain 永久塑性应变(10) The fatigue strength of concrete 混凝土疲劳强度3、Translate the fol lowing sentenee into Chinese ・(l) The cause of the volume changes in concrete can be attributed to changes in moisture content, chemicalretiction of the cement with water, variation in temperature, and applied loads ・ 混凝土的体积变化的原因可以归结为水分含最，水泥和水的化学反应，温度以及施加荷载的变化。

(2) High 一early 一strength and low-heat cement show more shrinkage than nonnal Portland cement. 高早强和低热水泥显示超过普通硅酸盐水泥收缩。

(3) The greater the aggregate content, the snitiller is the shrinkage ・ 骨料含量越大，收缩越小。

Reading Material(1)地铁工程是一个涉及规划、总图布置、详细设计、建造和地铁操作的交通运输系统的一个分支学科。

在大规模交通运输中，这些地下系统是一个主要要素。

随着城市化规模和人口面积增加，车辆交通变得更加拥挤，在规划城镇大量交通系统时，地铁正在被深度考虑。

1863年，第一个地铁在伦敦开放。

用焦炭和煤做燃料的蒸汽机被用到电车上。

1896年，欧洲大陆的第一个地铁用在服务上。

1895年和1897年，在波士顿、麻萨诸塞州，地铁线路制定。

从那以后，地铁被建在美国的好几个城市中。

最现代化的地铁是为在舒服的空调汽车里旅行时最大安全与高速一起设计的。

火车运行由电子设备控制。

自从地铁建造被投放在特别的设计地下铁路中，在地铁建设中，需要大多数的有关铁路的工程师。

然而，在地铁施工中大量而又昂贵的建造是重点；因为伴随着隧道经常要穿过足够深的地下水体；需要考虑地下乘客的极限；详细的空气流通设备；冗长的地下电子能源分布系统；照明设施；把乘客运输到地面的电梯；噪音控制；还有安全和灵活的电子信号设施。

规划。

有关地铁的工程计划包含去决定是否一个地铁系统是经济可行的。

这项工作包含大量的对工程造价的预测，乘客量、乘客费用，和税收、运营费用、仪器的贬值和维修、乘客的安全的估计的综合分析。

施工。

往往很难安装，地铁的施工明挖法和暗挖法同时存在。

运用明挖法时，一个基坑需要被挖掘。

这经常需要在基坑的每一边搭建支撑杆，另外不同的层面需要大的交叉框架。

敞开的基坑程序限制了工作可能完成的深度。

暗挖法允许地铁安置在很深的地方。

但是由于土的类型或者说所建造的地基的物质，不稳定的状况，巨大的水流量可能会造成运营很昂贵。

地铁的最佳位置的确定需要更仔细研究和考虑的各种植物，包括基础。

在一些大城市地区把地铁建在中央商务区下面和地上的其它地方。

在芝加哥南部的地铁被放在一条州际公路在相对地面公路上中央分隔带的交通空间。

在另一个芝加哥线从中央商务区领导西部，地铁是建在国会街高速公路中央分隔带下面。

【例1.1】Civil engineering offers a particular challenge , because almost every structure orsystem that is designed and built by civil engineers is unique.One structure rarely duplicatesanother exactly.译文:土木工程提出了特殊的挑战,因为由土木工程师设计建造的每个结构或系统几乎都是唯一的。

一个结构几乎不能完全复制另一个【例1.5】If the structure is saved or returned to its original state , additional foundationsupport must be provided.译文:假如建筑物要加以补救或恢复原貌,对基础做支护加固则是非常必要的。

为了使句子简洁精炼,专业英语中大量使用不定式.动名词、分词。

【例1.6】The total weight being less , it is possible to build much taller buildings.译文:由于总重量减轻,就有可能建造更高的楼房。

【例1.7】All the material forming the crust of the earth likely to be affected by the pressureof structures is divided by engineers into two major groups : rocks and soils.译文:受建筑物压力作用的地壳材料被工程人员分成两组,即岩石和土。

【例1.8】Compared with structural materials , such as steel and timber , soil is difficult to investigate scientifically.译文:与钢材、木材等建筑材料相比较,土研究起来颇为困难。

Structural behavior of low- and normal-strength interface mortar of masonryThomas Zimmermann1 , Alfred Strauss1 and Konrad Bergmeister1(1) Institute for Structural Engineering, University of Natural Resources and Life Sciences, Peter-Jordan-Strasse 82, 1190 Vienna, AustriaThomasZimmermann(Correspondingauthor)Email:Zimmermann.Thomas@boku.ac.atAlfredStraussEmail:Alfred.Strauss@boku.ac.atKonradBergmeisterEmail:Konrad.Bergmeister@boku.ac.atReceived:12 April 2011 Accepted:29 August 2011 Published online:8 November 2011 Abstract Building with masonry is based on the experience of many centuries. Although this design is used worldwide, knowledge about the material behaviour of masonry is still subject to uncertainties. The determination of safety of these structures against earthquakes is a complex challenge. For instance it depends on the resistance of the structure, the seismic action and on many uncertain structural details. One of the key parameters regarding the resistance is the shear strength of the masonry. A series of tests on mortar prisms according to EN 1015-11 was performed in which the mortar properties were varied in order to measure bending and compressive strength. In a second test program, the shear strength of the masonry was tested according to EN 1052-3. Shear triplets were made to establish the shear strength variation due to deliberate variation of the mortar properties. In addition, for both tests on mortar prisms and tests on shear triplets, descriptive statistical parameters were calculated and an attempt was made to describe the datasets with probabilistic distributions for further dimensioning and stochastic assessments. Keywords Shear strength – Coefficient of friction – Old masonry1IntroductionMasonry is a typical construction material which can withstand compression, but has low shear and bending resistance. This makes unreinforced masonry buildings highly interesting: (a) to gather mechanical properties and their wide scatter, which is characteristic for old masonry, and, (b) to obtain appropriate tools for assessment, analysis and retrofit methods.General rules and design aspects are stated in specific Eurocodes (EC). For masonry structures, rules and design aspects are regulated in EC 6 [1]. The ultimate limit state distinguishes between three major conditions: (a) masonry under vertical loading, (b) masonry under shear, and, (c) masonry under bending. The most critical loading conditions are cases (b) and (c), especially in the case of unreinforced masonry. Thereby the inappropriate horizontal loading situation is caused by wind loads or by seismic actions.Regarding the material behaviour under horizontal loading, two types of material parameters could be distinguished. The first type directly affects the stress side e.g. energy dissipation and behaviour factor. The second type directly affects the resistant side e.g. shear resistance, tensile strength and shear modulus. According to EC 6, the design value of shear strength depends on initial shear strength and the coefficient of friction as well as on geometrical parameters. With a testing program according to EN 1052-3 [2] it is possible to characterize these two material parameters for masonry. Further it is possible to define the shear resistance if sliding shear failure takes place. Therefore an extensive study can be found in Tomazevic [3], but it is focused on new brick material and mortar respectively.The procedure described in EN 1052-3 is the state of the art testing method to evaluate masonry shear strength without distinguishing between old and new masonry. Thereby two specimen layouts can be used. The smallest practical specimen consists of two brick units and one mortar layer while the second layout consists of three brick units and two mortar layers. In the case of testing old masonry the second specimen layout is more appropriate because the normative requirements can be easier achieved and further a symmetric loading situation occurs.The testing program presented in this paper is focused on old masonry and was carried out with different mortar properties. The results of this testing program, as well as a stochastic approach to describe the material strength in combination with an extensive literature review, are presented in this paper.2Material properties2.1BricksThe shear specimens were made with only one type of old, solid masonry bricks, see Fig. 1. This type of brick is typical for houses from the nineteenth century in Vienna. The mean dimensions of bricks were L/B/H = 29.12/14.13/7.05 cm. The dimensions were measured according to EN 772-16 [4]. Based on the obtained minimum dimensions in length and width, the bricks were cut to provide a consistent interface between the bricks and the mortar layer. The final dimensions of bricks were L = 25 cm and B = 12 cm. The height of bricks remained unchanged. The mean value of the dry density of bricks was ρ = 1,467 kg/m3.Fig. 1 Old, solid bricks used for shear testsThe compressive strength f b was obtained according to EN 772-1 [5] whereby the mean value of compressive strength resulted in f b = 19.28 MPa.2.2MortarTo determine the initial and shear strength between brick and mortar, a mortar mixture was chosen which was of low strength and a simple composition. Two mortar compositions out of four mixtures were chosen such that (a) the mortar had almost the same characteristics as the mortar for shear tests on masonry walls to provide comparability and (b) it was a very low strength mortar. These shear tests on masonry walls have already been carried out and are documented in [6]. In a testing program, consisting of four different mixtures, mortar prisms with dimensions of 40 × 40 × 160 mm were tested to obtain the compressive strength, f m and flexural strength, f m,fl. Table 1 shows the composition of all four mortar mixtures.Table 1 Investigated mortar mixtures, units in gramMix. I Mix. II Mix. III Mix. IVCEM 32.5 1,000 1,500 2,000 0Lime 400 400 400 400Rock flour 1,200 1,200 1,200 0Fine sand 0–1 4,650 4,650 4,650 4,650Course sand 0–4 12,445 12,445 12,445 12,445Water 3,500 3,500 3,500 3,250Compressive strength and flexural strength were obtained according to EN 1015-11 [7] after a curing time of 28 days. Table 2 shows the results of the testing program.Table 2 Material parameters of investigated mortar mixtures, units in MPaMix. I Mix. II Mix. III Mix. IVFlexural strength 0.58 1.02 1.39 –Compressive strength 1.50 3.58 4.06 0.22Based on these results mortar mixture II was chosen for a first triplet shear test groupbecause its characteristics are closest to the mortar characteristics of the mortar which was used in the shear tests on masonry walls. Mortar mixture IV was chosen for a second triplet shear test group.2.3Masonry specimensSpecimens for the triplet shear tests were built which consisted of three brick units with two mortar joints. The cut bricks provide a smooth surface for the bearings as well as for the load application area. The upper and lower surfaces of the specimens were confined with a cement mortar. After the specimens were built, each one was loaded with a compression load of about 3.0 × 10−3 MPa until testing. Simultaneously, while building the specimens for the triplet shear tests, additional mortar specimens of both mixtures were built for further mortar tests.3Testing methodsThe general problem in testing the shear behaviour along mortar joints and brick units is in applying a uniform distribution of both shear stress and normal stress. To avoid additional moments, the shear load should be applied as close as possible to the mortar joints, see [8]. There should also be no tensile stresses along the joint because these stresses could affect the failure load. However, some stress concentrations occur around the load introduction area and also some moment is introduced at the joint, which means that it is nearly impossible to introduce a pure shear stress distribution. The shear strength of masonry is dependent on the shear bond properties of the mortar joints, the vertical compression level and the friction angle. To obtain these properties different types of specimens can be used. Figure 2 shows a variety of different testing methods.Fig. 2 Various test arrangements for shear tests, a triplet test according to EN 1052-3, b Hoffmann and Stoeckl [9], c Riddington et al. [10], d Van der Pluijm [11], e Hamid et al. [12], f Abdou et al. [13] and g Popal and Lissel [14]These test methods consist of either two, three or four bricks. A review can be found in Jukes et al. [15] and additional experimental investigations are presented by Abdou et al. [13]. Further, several test arrangements have been investigated via FEM. Results are proposed by Stoeckl et al. [16]. Hence, it could be shown that peaks of both shear and normal stresses occur in all arrangements. There are also some approaches to combine the advantages of different test methods, e.g. [14].However, all the mentioned methods have it in common that they require very complex equipment and they are not a standard test method, expect triplet shear tests according to EN 1052-3.4Investigation of the shear behaviorPreviously mentioned test methods are designed so that the bricks only partially overlap. It does not matter with new bricks with more or less even surfaces. In the case of old bricks, a complete overlap is more advantageous because possible influences from uneven surfaces and imprints are taken into account. Thus the triplet test method according to EN 1052-3 was used for the investigations presented in this paper.According to EN 1052-3, two different test procedures are possible. In procedure (a) specimens have to be tested under at least three different normal stress levels with at least three specimens for each level. Procedure (b) is performed without any pre-compression with at least six specimens. In order to avoid normal tensile stresses along the mortar bed joints, procedure (a) was chosen. This normal stress is undesirable since the results for the shear strength can be affected by the tensile strength of the mortar bed joints.Two groups of specimens were tested. Table 3shows the properties of shear specimens. The bricks for both groups are the same, but the mortar mixtures differ. Mortar mixture II was used for group A while mortar mixture IV was used for group B.Table 3 Characteristics of masonry specimens, units in MPaCompressive strength ofBricks f b Mortar f m Masonry f kGroup A 19.28 3.58 5.65Group B 19.28 0.22 2.81Based on the compressive strengths of both bricks f b, and mortar f m, the compressive strength of masonry f k was calculated according to EC 6, National Annex B 1996-1-1 [17].(1)Shear strength was measured using the set up shown in Fig. 3. The brick in the middle is sheared and the upper and lower bricks are supported. The horizontal shear load was applied with a hydraulic jack. The varying pre-compression load was applied perpendicular to the shear surface.Fig. 3 Triplet shear test set upIn the case of group A, five vertical stress levels (3, 7, 15, 25 and 40% of f k) were applied and five tests were performed at each level for statistical evaluation. This resulted in a total number of 25 specimens for group A. In the case of group B, three vertical stress levels (13, 28 and 48% of f k) were applied. Hence, three tests were performed at each level. This resulted in a total number of nine specimens for group B.Each test took about 5 min until shear failure occurred. When the specimen cracks and pure shearing starts, the pre-compression load fluctuates. This was adjusted manually in order to keep it constant. During testing, the shear load and the applied pre-compression load were measured simultaneously.The evaluation of shear tests was based on the maximum horizontal force H max obtained during testing. Since the middle brick was loaded, the horizontal force had to be divided by two times the corresponding shear area (250 × 120 mm = 30,000 mm2). Hence, i the shear strength for each specimen f v,i could be calculated as:(2)The applied normal stress level σd was calculated with the applied pre-compressionforce with respect to the corresponding shear area of the specimen i.(3)The shear strength of masonry depends on the applicable friction forces in the horizontal joints, the tensile strength of the bricks, the compressive strength of masonry and the bond strength between bricks and mortar. The shear strength is essentially determined by the normal stress level. According to EC 6 it can be calculated as:(4)where f vko is the initial shear strength without any vertical stresses; σd the normal stress level perpendicular to the shear force and μk the coefficient of friction (both characteristic values).The evaluation of the shear test results was done (a) based on mean values, (b) based on a statistical approach using 5% fractiles of a Lognormal distribution and (c) according to EN 1052-3. Finally both evaluations were compared to each other, see Table 4.Table 4 Mean values of initial shear strenght and coefficient of friction and comparison of characteristic valuesInitial shear strength (MPa) Coefficient of friction (–)Mean EC 6 5% fractile EN 1052-3 Mean EC 6 5% fractil EN 1052-3f vo f vko f vko,5%f vkoμμkμk,5%μkGroup A 0.210 0.200 0.174 0.168a0.709 0.400 0.624 0.566 Group B 0.027 0.100 0.014 0.010b0.643 0.400 0.623 0.514a Calculated from mean value by multiplying with 0.8;b smallest single value of testdata4.1Failure modesGenerally, four failure modes during shear tests can appear. Mode (a) is a fracture plane localised at one brick mortar interface. Mode (b) is a fracture plane at each brick mortar interface combined with a vertical crack in the mortar layer. Mode (c) is a pure shear failure in the mortar layer and mode (d) is a fracture plane through both mortar and bricks, see Fig. 4. For the shear tests presented in this paper, only the failure modes (a) and (b) were observed during testing.Fig. 4 Failure modes of masonry specimens during shear testing4.2Results group AFigure 5shows the shear strength with respect to the corresponding normal stress level for the tested specimens of group A. For each stress level the mean value, the 5% fractile based on a Lognormal distribution and characteristic value according toEN 1052-3 were calculated. Linear best fits through (a) and (b) values were carried out using the least square method. Thereby, for case (a) a Mohr–Coulomb relationship was obtained as:(5)and for case (b) as:(6)Fig. 5 Shear strength with respect to vertical stress, group ACase (c), the determination of characteristic value according to EN 1052-3, can be directly calculated from mean values by multiplying with 0.8 or it corresponds to the smallest single value of the testdata. The smaller value is decisive:(7)Further, the normative relationship (norm) is plotted in Fig. 5. Due to the mortar properties, the corresponding mortar class, according to EC 6, is M2.5–M9. Hence the initial shear strength f vko norm= 0.20 MPa and the coefficient of friction μk norm = 0.4. Table 5 summarizes the test results and descriptive statistical parameters.Table 5 Test results of shear strength f v, i (MPa) with respect to normal stress levelSymbol Normal force level (kN)5.0 11.0 24.0 40.5 65.0Group AMean 0.3040.47960.8104 1.15001.7436Standard deviation s0.04890.04920.0634 0.0890 0.1413Coefficient of variation cov0.16090.10250.0783 0.0774 0.08105% fractile x50.11250.39910.7057 1.0036 1.5116Group BMean –0.2610.5440 0.8933 –Standard deviation s–0.01250.0092 0.0302 –Coefficient of variation cov–0.04760.0169 0.0338 –5% fractile x5–0.2410.5291 0.8445 –4.3Results group BFigure 6shows the shear strength with respect to the corresponding normal stress level for the tested specimens of group B. Again, linear best fits through (a) the mean values and (b) the fractile values were carried out using the least square method.Fig. 6 Shear strength with respect to vertical stress, group BThereby for case (a), a Mohr–Coulomb relationship was obtained as:(8)and for case (c) as:(9)Again, case (c), the determination of characteristic value according to EN 1052-3, can be directly calculated from mean values by multiplying with 0.8 or it corresponds to the smallest single value of the testdata. The smaller value is decisive:(10)Also the normative relationship (norm) is plotted in Fig. 6. Due to the mortar properties, the corresponding mortar class according to EC 6 is M1 – M2. Hence theinitial shear strength f vko norm= 0.10 MPa and the coefficient of friction μk norm = 0.4. Table 5 summarizes the test results and descriptive statistical parameters.5Probabilistic modelsThis section provides an overview of the investigated probabilistic models which were considered here to describe test results and literature data of the coefficient of friction of masonry. Depending on the distribution function, different procedures were used for estimating the unknown parameters e.g. Method of Moments and Method of Maximum Likelihood. Detailed studies regarding parameter estimation can be found in [18–20] and other sources. The functions of the investigated distributions relate to the two and three parameter function respectively.The investigated probabilistic models are the usual distribution functions like Normal and Lognormal, and also common distribution functions to describe material strength, such as Gamma and Weibull. Different methods have been used for choosing the best fit model to a given data set. These methods are the Kolmogrorv Smirnov (KS), the χ2 and the Anderson Darling (AD) test. The last method was chosen for this study as it is more sensitive to the tail behaviour. The sensitivity to the tail behaviour is particularly useful in structural engineering applications, where the tail is important in computing the structural reliability.The KS procedure involves the comparison between the assumed hypothetical and the empirical cumulative distribution function. For computing models, it is natural to choose a particular model for a given sample whereby the discrepancy is low. Otherwise, if the discrepancy is large with respect to what is normally expected from a given sample, the hypothetical model is rejected.The χ2-test is used to determine if a sample comes from a population with a specificdistribution. It compares the observed frequencies in k intervals of thevariate with the corresponding frequencies from an assumed hypothetical distribution.Finally, the AD-procedure is a general test to compare the fit of an empirical cumulative distribution function to a hypothetical cumulative distribution function. This test gives more weight to the tails than the KS-test.The various probabilistic models were applied to a data set consisting of values from an extensive literature review as well as of values from laboratory tests, as described in Sect. 4. A total number of n = 2,028 values were used. Table 6 shows the mean and characteristic values of the coefficient of friction from literature.Table 6 Mean and characteristic values of coefficient of friction, form literatureName Ref. Coef. of frictionμkAbdou et al. [13] 0.886 0.709Amadio and Rajgelj [21] 0.700 0.560Benjamin and Williams [22] 1.100 0.880Chin [23] 0.750 0.600Ghazali and Riddington [24] 0.778 0.622Hegemioer et al. [25] 0.941 0.753Jukes [26] 0.797 0.638Khalaf [27] 0.793 0.635Page [28] 0.700 0.560Sinha and Hendry [29] 0.700 0.560Van der Pluijm [11, 30, 31] 0.850 0.680Vermeltfoort [32, 33] 0.747 0.598Min 0.700 0.560Max 1.100 0.880Figure 7shows proportion–proportion plots (PP-plots) for some investigated distribution functions. The empirical cumulative proportion is plotted against the hypothetical cumulative proportion. The straight line is added as a reference line. The further the points vary from this line, the greater the indication of departures from the designated distribution. Table 7shows the results of the goodness of fit tests for different distribution functions.Fig. 7 PP-plots of different distribution functionsTable 7 KS-distances, AD-values and χ2-values for different distribution functionsPDF KS AD χ2Normal 0.05884 1.6669 16.827Lognormal 0.07429 2.3188 26.802Gamma 0.06551 1.8732 22.899Weibull 0.07256 4.4672 10.128Gumbel max 0.11476 6.147 39.993All probabilistic models can represent the lower and upper tail behaviour of the observed data, except Weibull where the points of lower tail are above the reference line. This indicates shorter than Weibull tails, i.e. less variance than expected. Further, a comparison between the median area and the remaining distributions shows that the slightest deviations arise for Normal and Lognormal distributions. This is in correlation with the applied goodness of fit tests.6ConclusionsAs a part of the SEISMID research project, several tests on masonry were carried out. In this case, the focus was on testing the shear behaviour of masonry triplets under different conditions according to EN 1052-3. Additional tests on bricks and mortar were carried out to determine the basic material properties.To estimate possible influences on the shear behaviour of masonry, two different groups of shear triplets were built and tested under different normal stress levels. The two groups (A and B) differed in terms of compressive strength of mortar (f m,A = 3.58 MPa and f m,B= 0.22 MPa). The evaluation of the test results show that the shear behaviour can be described by the Mohr–Coulomb friction law. Hence, the initial shear strength f vko and the coefficient of friction μk were determined. When compared to the values according to EC 6, some agreement can be seen, but also some values which are not in agreement.In the case of specimen group A, the mean value of the initial shear strength from testing (0.210 MPa) is very consistent with the suggested normative value (0.200 MPa). In case of group B, there is no consistency between the values from testing (0.027 MPa) and EC 6 (0.100 MPa). This inconsistency is mainly due to the mortar mixture in that the mortar of group B contains no cement and just a small amount of lime (compare Table 1). Hence, no significant initial shear strength between mortar joints and bricks can be developed.The evaluation of the characteristic value of initial shear strength according to EN 1052-3 results in f vko= 0.168 MPa for mortar group A and f vko= 0.010 MPa for mortar group B. If the evaluation is based on 5% fractiles of a Lognormal distribution the values results in f vko = 0.174 MPa for mortar group A and f vko = 0.014 MPa for mortar group B. As can be seen there are no significant differences of the calculated values. This indicates that both evaluation procedures are suitable to derive characteristic values from experimental test results.The comparison of the coefficient of friction shows a gap between the test results andthe value according to EC 6. The normative value for the coefficient of friction is suggested to be 0.400. The experimental data show that the percentage of normal stress on the shear strength amounts μ = 0.709 in case of group A and μ = 0.643 in case of group B, based on mean values. The evaluation of the characteristic value of coefficient of friction according to EN 1052-3 results in μk = 0.566 for mortar group A and μk= 0.514 for mortar group B. If the evaluation is based on 5% fractiles of a Lognormal distribution the values results in μk= 0.624 for mortar group A and μk = 0.623 for mortar group B. As can be seen there are differences of the calculated values. This indicates that the evaluation procedure according to EN 1052-3 procedure is more conservative because mean values are multiplied by the factor 0.8 to derive characteristic values but any additional information of test results are neglected. These additional information are accounted by the statistical approach. In addition, the literature review shows that the normative value for the coefficient of friction is too low.The choice of a probabilistic model plays an important role for a probabilistic based design approach and reliability assessment. In this work different statistical distribution functions were considered in order to critically analyze the coefficient of friction of masonry. Hence, two- and three-parameter distributions were used. The data set for the statistical distribution fitting was collected from both literature and laboratory tests.Based on the set of strength data and using several statistical criteria, like KS-test, χ2-test and AD-procedure, the Normal and Lognormal distributions appear to be more appropriate than the others. A further result is that all distributions, except Weibull, show an accurate tail behaviour in the lower as well as the upper bound. It is also reflected in the PP-plots. This is important since the sensitivity to the tail behaviour is particularly useful in structural engineering approaches and reliability.The overall conclusion from these investigations it is that the friction property of bricks should be characterized using a Lognormal distribution. Since the coefficient of friction is a low value (close to 0), the Lognormal distribution should be preferred over the Normal because its domain is limited to zero or a certain bound (γ > 0) wh ile the domain of a Normal distribution is between andThe assessment of existing structures is becoming more and more important for social and economical reasons, while most codes deal explicitly only with design situations of new structures. The assessment of an existing structure may, however, differ much from the design of a new one. In general, the safety assessment of an existing structure differs from that of a new one in a number of aspects, see Diamantidis [34] and Vrouwenvelder [35]. The main differences are: (1) Increasing safety levels usually involves more costs for an existing structure than for structures that are still in the design phase. The safety provisions embodied in safety standards have also to be set off against the cost of providing them, and on this basis improvements are more difficult to justify for existing structures. For this reason and under certain circumstances, a lower safety level is acceptable. (2) The remaining lifetime of an existing building is often less than the standard reference period of 50 or 100 yearsthat applies to new structures. The reduction of the reference period may lead to reductions in the values of representative loads as for instance indicated in the Eurocode for Actions.Therefore the safety philosophy for existing structures must be discussed with respect to the reliability levels in terms of the β-values for (a) new structures, and (b) for existing structures and with respect to monitoring and inverse analysis concepts [36, 37].Required β-values must be derived for masonry structures and anchored in code specifications such as ISO 13822 ―Assessment of existing structures‖ [38] or EC 8 part 3 ―Assessment and retrofitting of buildings‖ [39].Acknowledgments Research results discussed in this paper were carried out within the European research project SEISMID, supported and financed in cooperation with the Centre for Innovation and Technology (ZIT). We also wish to thank Mr. Walter Brusatti (Brusatti GmbH) for providing bricks and further Mr. Johann Lang from the College of Civil Engineering (HTBL Krems) Austria, for his efficient help during testing in the laboratory.References1. EN-1996-1-1 (2006) Eurocode 6: Design of masonry structures—part 1-1: common rules for reinforced and unreinforced masonry structures2. EN-1052-3 (2007) Methods of test for masonry—part 3: determination of initial shear strength3. Tomazevic M (2008) Shear resistance of masonry walls and eurocode 6: shear versus tensile strength of masonry. Mater Struct 42:889–9074. EN-772-16 (2005) Methods of test for masonry units—part 1: determination of dimensions5. EN-772-1 (2000) Methods of test for masonry units—part 1: determination of compressive strength6. Zimmermann T, Strauss A, Bergmeister K (2010) Numerical investigations of historic masonry walls under normal and shear load. Constr Build Mater 24:1385–13917. EN-1015-11 (2007) Methods of test for mortar for masonry—part 11: determination of flexural and compressive strength of hardened mortar8. Edgell G (2005) Testing of ceramics in construction. Whittles Publishing Ltd.,。

1.1 许多天然物质，如粘土、砂子和岩石，甚至树枝和树叶都已经被用作建筑材料。

Many naturally occurring substances, such as clay, sand, wood and rocks, even twigs and leaves have been used to construct buildings.1.2 砖块是由窑中烧制材料作成的块体，通常由粘土或页岩制成，但也可由炉渣制成。

A brick is a block made of kiln-fired material, usually clay or shale, but also maybe of lower quality mud.1.3 与水混合后，水泥便发生水化反应，并最终形成像石头一样的材料。

After mixing, the cement hydrates and eventually hardens into a stone-like material.1.4 金属可用作大型结构的框架，也可用来装饰建筑物外表。

Metal is used as structural framework for larger buildings such as skyscrapers, or as an external surface covering.1.5 明亮的窗户不但能使光线进入建筑物，而且也能将恶劣气候隔绝于建筑物之外。

Clear windows provided humans with the ability to both let light into rooms while at the same time keeping inclement weather outside.2.1 材料的抗拉强度是一种广延性质，因此它并不因试件尺寸的不同而改变。

Tensile strength is an intensive property and, consequently, does not depend on the side of the test specimen.2.2 屈服强度是材料从弹性变形到塑性变形转化时的应力。

土木工程专业英语翻译Ch 2 Overview of Engineering MechanicsAs we look around us we see a world full of ‘things’: machines, devices, tools; things that we have designed, 当我们环顾四周，我们可以看到充满物质的世界：机器，仪器，工具：这些被我设计、建造和使用过built and used; things made of wood, metals, ceramics, and plastics. We know from experience that some 的东西：这些东西是由木头，金属，陶瓷和塑料做成。

我们从经验可以知道一些东西比别的东things are better than others; they last longer, cost less, are quieter, look better, or are easier to use.比别的东西好：他们比较耐用，低成本，安静，看起来好点，或者使用比较简单。

Ideally, however, every such item has been designed according to some set of ‘functional requirements; as 不管怎么样，最理想的是这样一个东西都是被设计者所认为的那样根据某些“使用要求”设计出来的。

perceived by the designers—that is, it has been designed so as to answer the question, ‘Exactly what function也就是说他的设计是为了回答这些问题，确切的说它应该执行怎么样的功should it perform?’ In the world of engineering, the major function frequently is to support some type of 能？在工程的领域里，最主要的功能是经常支持一些由于重力，惯性，压力等作用产loading due to weight, inertia, pressure etc. From the beams in our homes to the wings of an airplane, there 生的荷载。

专业英语课文翻译Lesson 4Phrases and Expressions1.moisture content 含水量，含湿度; water content 2.cement paste 水泥浆 mortar 3.capillary tension 毛细管张力，微张力 4.gradation of aggregate 骨料级配 coarse fine (crushed stone, gravel) 5.The British Code PC100 英国混凝土规范PC100; nowaday BS8110 6. coefficient of thermal expansion of concrete 混凝土热膨胀系数 7. The B.S Code 英国标准规范 8. sustained load 永久荷载，长期荷载 9. permanent plastic strain 永久的塑性应变 stress 10. crystal lattice 晶格, 晶格 11. cement gel 水泥凝胶体 12. water-cement ratio 水灰比 13. expansion joint 伸缩缝 14. stability of the structure 结构的稳定性 structural stability15. fatigue strength of concrete 混凝土的疲劳强度Volume Changes of ConcreteConcrete undergoes volume changes during hardening. 混凝土在硬结过程中会经历体积变化。

If it loses moisture by evaporation, it shrinks, but if the concrete hardens in water, it expands. 如果蒸发失去水分，混凝土会收缩；但如果在水中硬结，它便膨胀。

Civil EngineeringCivil engineering, the oldest of the engineering specialties, is the planning, design, construction, and management of the built environment. This environment includes all structures built according to scientific principles, from irrigation and drainage systems to rocket-launching facilities.土木工程学作为最老的工程技术学科，是指规划，设计，施工及对建筑环境的管理。

此处的环境包括建筑符合科学规范的所有结构，从灌溉和排水系统到火箭发射设施。

Civil engineers build roads, bridges, tunnels, dams, harbors, power plants, water and sewage systems, hospitals, schools, mass transit, and other public facilities essential to modern society and large population concentrations. They also build privately owned facilities such as airports, railroads, pipelines, skyscrapers, and other large structures designed for industrial, commercial, or residential use. In addition, civil engineers plan, design, and build complete cities and towns, and more recently have been planning and designing space platforms to house self-contained communities.土木工程师建造道路，桥梁，管道，大坝，海港，发电厂，给排水系统，医院，学校，公共交通和其他现代社会和大量人口集中地区的基础公共设施。