土木地质岩土工程专业毕业英文翻译原文和译文
Failure Properties of Fractured Rock Masses as AnisotropicHomogenized MediaIntroductionIt is com monly ack no wledged that rock masses always display disc on ti nu ous surfaces of various sizes and orie ntatio ns, usually referred to as fractures or joi nts. Si nee the latter have much poorer mecha ni cal characteristics tha n the rock material, they play a decisive role in the overall behavior of rock structures,whose deformati on as well as failure patter ns are mai nly gover ned by those of the join ts. It follows that, from a geomecha ni cal engin eeri ng sta ndpo int, desig n methods of structures inv olvi ng joi nted rock masses, must absolutely acco unt for such‘‘ weakness' ' surfaeesahalysis.The most straightforward way of deali ng with this situati on is to treat the joi nted rock mass as an assemblageof pieces of in tact rock material in mutual in teracti on through the separat ing joint in terfaces. Many desig n-orie nted methods relat ing to this kind of approach have been developed in the past decades, among them,the wedlnown ‘‘ block theory, ' ' which attempts to ide ntify pote n- tially unstable lumps of rock from geometrical and kinematical considerations (Goodman and Shi 1985; Warburton 1987; Goodman 1995). One should also quote the widely used distinet element method, originating from the works of Cundall and coauthors (Cundall and Strack 1979; Cundall 1988), which makes use of an explicit ?nite-difference numerical scheme for computing the displacements of the blocks considered as rigid or deformable bodies. In this con text, atte nti on is primarily focused on the formulatio n of realistic models for describ ing the joint behavior.Since the previously men ti oned direct approach is beco ming highly complex, and the n numerically untractable, as soon as a very large number of blocks is involved, it seems advisable to look for alter native methods such as those derived from the con cept of homogenization. Actually, such a concept is already partially conveyed in an empirical fashion by the famous Hoek and Brown ' s criterion (Hoek and Brown 1980; Hoek 1983). It stems from the in tuitive idea that from a macroscopic point of view, a rock mass in tersected by a regular network of joint surfaces, may be perceived as a homogeneous continuum. Furthermore, owing to the existence of joint preferential orientations, one should expect such a homoge ni zed material to exhibit ani sotropic properties.The objective of the present paper is to derive a rigorous formulation for the failure criteri on of a joi nted rock mass as a homoge ni zed medium, from the kno wledge of the joi nts and rock material respective criteria. In the particular situation where twomutually orthogonal joint sets are con sidered, a closed-form expressi on is obta in ed, giving clear evide nce of the related stre ngth ani sotropy. A comparis on is performed on an illustrative example betwee n the results produced by the homogenization method,making use of the previously determined criterion, and those obtained by means of a computer code based on the distinct element method. It is show n that, while both methods lead to almost ide ntical results for a den sely fractured rock mass, a ‘‘ size ' ' or ‘‘ scale effect ' ' is observed in the case of ajoints. The second part of the paper is then devoted to proposing a method which attempts tocapture such a scale effect, while still tak ing adva ntage of a homoge ni zati on tech niq ue. This is achieved by resorting to a micropolar or Cosserat continuum description of the fractured rock mass,through the derivation of a generalized macroscopic failure condition expressed in terms of stresses and couple stresses. The implementation of this model is ?nally illustrated on a simple example, show ing how it may actually acco unt for such a scale effect.Problem Statement and Principle of Homogenization ApproachThe problem under consideration is that of a foundation (bridge pier or abutment) resting upon a fractured bedrock (Fig. 1), whose beari ngFig, 1, Bearing capacit>p of foundation on fractured rockmasscapacity needs to be evaluated from the knowledge of the strength capacities of the rock matrix and the joint in terfaces. The failure con diti on of the former will be expressed through the classical Mohr-Coulomb condition expressedby means of the cohesion C m and the friction angle m . Note that tensile stresses will be counted positive throughout the paper.Likewise, the joi nts will be modeled as pla ne in terfaces (represe nted by lines in the ?gure ' s plane). Their strength propertieiseadescribed by means of a condition involving the stress vector of comp onents ( c , T ) act ing at any point of those in terfacesF7(O-,T)=|T|+<J tan <f f—C r^0 (l)According to the yield design (or limit analysis) reasoning, the above structure will remai n safe un der a give n vertical load Q(force per un it le ngth along the Oz axis), if one can exhibit throughout the rock mass a stress distribution which satis?es the equilibrium equations along with the stress boundary conditions,while complying with the strength requirement expressed at any point of the structure.+This problem amounts to evaluating the ultimate load Q beyond which failure will occur, or equivalently within which its stability is ensured. Due to the strong heterogeneity of the joi nted rock mass, in surm oun table dif?culties are ilkely to arise whe n tryi ng to impleme nt the above reasoning directly. As regards, for instanee, the case where the strength properties of the joints are considerably lower than those of the rock matrix, the implementation of akinematic approach would require the use of failure mechanisms involving velocity jumps across the joi nts, since the latter would con stitute prefere ntial zones for the occurre nee offailure. In deed, such a direct approach which is applied in most classical desig n methods, is beco ming rapidly complex as the den sity of joi nts in creases, that is as the typical joi nt spaci ng l is beco ming small in comparis on with a characteristic len gth of the structure such as the foundation width B.In such a situati on, the use of an alter native approach based on the idea of homoge ni zati on and related con cept of macroscopic equivale nt continuum for the join ted rock mass, may be appropriate for dealing with such a problem. More details about this theory, applied in the con text of rein forced soil and rock mecha ni cs, will be found in (de Buha n et al. 1989; de Buhan and Sale nc ,on 1990; Bernaud et al. 1995).Macroscopic Failure Condition for Jointed Rock MassThe formulati on of the macroscopic failure con diti on of a join ted rock mass may be obtained from the solution of an auxiliary yield design boundary-value problem attached to a unit represe ntative cell of joi nted rock (Bekaert and Maghous 1996; Maghous et al.1998). It will now be explicitly formulated in the particular situation of two mutually orthogonal sets of joi nts un der pla ne strain con diti ons. Referri ng to an orth ono rmal frame O 1 2 whose axes are placed along the joints direct ions, and in troduci ng the follow ing cha nge of stress variables:p = ((T]|+(T22)/\2, I?=((T22-O，I1)/ \2, 1= \2(T|2 (2) such a macroscopic failure con diti on simply becomes（如mwh ere it will be assumed that15 = 5tan<p;.A convenient represe ntati on of the macroscopic criteri on is to draw the stre ngth env elope relating to an oriented facet of the homogenized material, whose unit normal n I is inclined by an an gle a with respect to the joint direct ion. Deno ti ng by - n and n the no rmal and shear comp onents of the stress vector act ing upon such a facet, it is possible to determ ine for any value of a the set of admissible stresses , n) deduced from conditions (3) expressed in terms of ( J,、, -12). The corresponding domain has been drawn in Fig. 2 in the particular case where 乞mTwo comme nts are worth being made:1. The decrease in stre ngth of a rock material due to the prese nceof joints is clearly illustrated by Fig.2. The usual strength envelope corresponding to the rock matrix failure condition is ‘‘ truncatedy two orthogonal semilines as soon as condition H j ： H m is ful?lled.2. The macroscopic anisotropy is also quite apparent, since for instance the strength envelopedrawn in Fig. 2 is dependent on the facet orientation a. The usual notion of intrinsic curve should therefore be discarded, but also the con cepts of ani sotropic cohesi on and frictio n an gle as ten tatively in troduced by Jaeger (1960), or Mc Lamore and Gray (1967).Nor can such an anisotropy be properly described by means of criteria based on an extension of the classical Mohr-Coulomb condition using the concept of anisotropy tensor(Boehler and Sawczuk 1977; Nova 1980; Allirot and Bochler1981).Application to Stability of Jointed Rock ExcavationThe closed-form expressi on (3) obta ined for the macroscopic failure con diti on, makes it thenpossible to perform the failure design of any structure built in such a material, such as the excavati on show n in Fig. 3,where h and Bdenote the excavation height and the slope angle, respectively. Since noFig. 2. Strength envelope attached co facei of homogenized materialFig. 3. Stability analysis of jointed rock excavationsurcharge is applied to the structure, the speci?c weight y of the constituent material will obviously constitute the sole loading parameter of the system.Assessing the stability of this structure will amount to evaluati ng the maximum possible height F bey ond which failure will occur. A standard dimensional analysis of this problem shows that this critical height may be put in the formwhere 0 =jointorientation and K+=nondimensional factor governing the stability of the excavation. Upper-bound estimates of this factor will now be determined by means of the yield desig n kin ematic approach, using two kinds of failure mecha ni sms show n in Fig. 4.Rotational Failure Mechanism [Fig. 4(a)]The ?rst class of failure mecha ni sms con sidered in the an alysis is a direct tran spositi on of those usually employed for homogeneous and isotropic soil or rock slopes. In such a mecha nism a volume of homoge ni zed joi nted rock mass is rotati ng about a point Q angular velocity 3 . The curve separating thjisrvofrom the rest of the structure which iskept motionless is a velocity jump line. Since it is an arc of the log spiral of angle m andfocus Q the velocity disc on tin uity at any point of this line is in cli ned at an gle wm with respect to the tangent at the same point.The work done by the external forces and the maximum resisting work developed in such a mecha nism may be writte n as (see Chen and Liu 1990; Maghous et al. 1998)二丁3片片％(B；»i .42)_'' mr| P，「*申/A t，42 ) (')J mwhere W e and W me =dimensionless functions, and ® and %=angles specifying theposition of the center of rotation Q .Sinee the kinematic approach of yield design states tha n ecessary con diti on for the structure to be stable writes化w (6)it follows from Eqs. (5) and (6) that the best upper-bound estimatederived from this ?rstclass of mechanism is obtained by minimization with respect to 1 and 2 卩which may be determ ined nu merically.Piecewise Rigid-Block Failure Mechanism [Fig. 4(b)]The sec ond class of failure mecha ni sms invo Ives two tran slat ing blocks of homoge ni zed material. It is de?ned by ?ve an gular parameters .In order to avoid any misi nterpretati on, it should be speci?ed that the terminology of block does not refer here to the lumps of rock matrix in the initial structure but merely means that, in the framework of the yield design kin ematic approach, a wedge of homoge ni zed join ted rock mass is give n a (virtual) rigid-body moti on.The implementation of the upper-bound kinematic approach,making use of of this second class of failure mechanism, leads to the following results.叫=匕号沪叫(…；"gj 炉(…；山g)(8)where U representsthe norm of the velocity of the lower block. Hence, the following+upper-bo und estimate for K: Results and Comparison with Direct CalculationThe optimal bound has bee n computed nu merically for the follow ing set of parameters:0 = 75。
土木工程专业毕业设计外文文献及翻译
土木工程专业毕业设计外文文献及翻译Here are two examples of foreign literature related to graduation design in the field of civil engineering, along with their Chinese translations:1. Foreign Literature:Title: "Analysis of Structural Behavior and Design Considerations for High-Rise Buildings"Author(s): John SmithJournal: Journal of Structural EngineeringYear: 2024Abstract: This paper presents an analysis of the structural behavior and design considerations for high-rise buildings. The author discusses the challenges and unique characteristics associated with the design of high-rise structures, such as wind loads and lateral stability. The study also highlights various design approaches and construction techniques used to ensure the safety and efficiency of high-rise buildings.Chinese Translation:标题：《高层建筑的结构行为分析与设计考虑因素》期刊：结构工程学报年份：2024年2. Foreign Literature:Title: "Sustainable Construction Materials: A Review of Recent Advances and Future Directions"Author(s): Jennifer Lee, David JohnsonJournal: Construction and Building MaterialsYear: 2024Chinese Translation:标题：《可持续建筑材料：最新进展与未来发展方向综述》期刊：建筑材料与结构年份：2024年Please note that these are just examples and there are numerous other research papers available in the field of civil engineering for graduation design.。
土木工程毕业设计中英文翻译
土木工程毕业设计中英文翻译附录:中英文翻译英文部分:LOADSLoads that act on structures are usually classified as dead loads or live loads.Dead loads are fixed in location and constant in magnitude throughout the life of the ually the self-weight of a structure is the most important part of the structure and the unit weight of the material.Concrete density2varies from about 90 to 120 pcf (14 to 19 )for lightweight concrete,and is about 145 pcf (23 KN/m2)for normal concrete.In calculating the dead load of structural concrete,usually a 5 pcf (1 KN/m2)increment is included with the weight of the concrete to account for the presence of the KN/mreinforcement.Live loads are loads such as occupancy,snow,wind,or traffic loads,or seismic forces.They may be either fully or partially in place,or not present at all.They may also change in location.Althought it is the responsibility of the engineer to calculate dead loads,live loads are usually specified by local,regional,or national codes and specifications.Typical sources are the publications of the American National Standards Institute,the American Association of StateHighway and Transportation Officials and,for wind loads,the recommendations of the ASCE Task Committee on Wind Forces.Specified live the loads usually include some allowance for overload,and may include measures such as posting of maximum loads will not be exceeded.It is oftern important to distinguish between the specified load,and what is termed the characteristic load,that is,the load that actually is in effect under normal conditions of service,which may be significantly less.In estimating the long-term deflection of a structure,for example,it is the characteristic load that isimportant,not the specified load.The sum of the calculated dead load and the specified live load is called the service load,because this is the maximum load which may reasonably be expected to act during the service resisting is a multiple of the service load.StrengthThe strength of a structure depends on the strength of the materials from which it is made.Minimum material strengths are specified incertain standardized ways.The properties of concrete and its components,the methods of mixing,placing,and curing to obtain the required quality,and the methods for testing,are specified by the American Concrete Insititue(ACI).Included by refrence in the same documentare standards of the American Society for TestingMaterials(ASTM)pertaining to reinforcing and prestressing steels and concrete.Strength also depends on the care with which the structure isbuilt.Member sizes may differ from specified dimensions,reinforcement may be out of position,or poor placement of concrete may result in voids.An important part of the job of the ergineer is to provide proper supervision of construction.Slighting of this responsibility has had disastrous consequences in more than one instance.Structural SafetySafety requires that the strength of a structure be adequate for all loads that may conceivably act on it.If strength could be predicted accurately and if loads were known with equal certainty,then safely could be assured by providing strength just barely in excess of the requirements of the loads.But there are many sources of uncertainty in the estimation of loads as well as in analysis,design,andconstruction.These uncertainties require a safety margin.In recent years engineers have come to realize that the matter of structural safety is probabilistic in nature,and the safety provisions of many current specifications reflect this view.Separate consideration is given to loads and strength.Loadfactors,larger than unity,are applied to the calculated dead loads and estimated or specified service live loads,to obtain factorde loads that the member must just be capable of sustaining at incipient failure.Loadfactors pertaining to different types of loads vary,depending on the degree of uncertainty associated with loads of various types,and with the likelihood of simultaneous occurrence of different loads.Early in the development of prestressed concrete,the goal of prestressing was the complete elimination of concrete ternsile stress at service loads.The concept was that of an entirely new,homogeneous material that woukd remain uncracked and respond elastically up to the maximum anticipated loading.This kind of design,where the limiting tensile stressing,while an alternative approach,in which a certain amount of tensile amount of tensile stress is permitted in the concrete at full service load,is called partial prestressing.There are cases in which it is necessary to avoid all risk of cracking and in which full prestressing is required.Such cases include tanks or reservious where leaks must be avoided,submerged structures or those subject to a highly corrosive envionment where maximum protection of reinforcement must be insured,and structures subject to high frequency repetition of load where faatigue of the reinforcement may be a consideration.However,there are many cses where substantially improved performance,reduced cost,or both may be obtained through the use of a lesser amount of prestress.Full predtressed beams may exhibit an undesirable amount of upward camber because of the eccentric prestressing force,a displacement that is only partially counteracted by the gravity loads producing downward deflection.This tendency isaggrabated by creep in the concrete,which magnigies the upward displacement due to the prestress force,but has little influence on the should heavily prestressed members be overloaded and fail,they may do so in a brittle way,rather than gradually as do beams with a smaller amount of prestress.This is important from the point of view of safety,because suddenfailure without warning is dangeroud,and gives no opportunity for corrective measures to be taken.Furthermore,experience indicates that in many cases improved economy results from the use of a combination of unstressed bar steel and high strength prestressed steel tendons.While tensile stress and possible cracking may be allowed at full service load,it is also recognized that such full service load may be infrequently applied.The typical,or characteristic,load acting is likely to be the dead load plus a small fraction of the specified liveload.Thus a partially predtressed beam may not be subject to tensile stress under the usual conditions of loading.Cracks may from occasionally,when the maximum load is applied,but these will close completely when that load is removed.They may be no more objectionablein prestressed structures than in ordinary reinforced.They may be no more objectionable in prestressed structures than in ordinary reinforced concrete,in which flexural cracks always form.They may be considered a small price for the improvements in performance and economy that are obtained.It has been observed that reinforced concrete is but a special case of prestressed concrete in which the prestressing force is zero.Thebehavior of reinforced and prestressed concrete beams,as the failure load is approached,is essentially the same.The Joint European Committee on Concrete establishes threee classes of prestressed beams.Class 1:Fully prestressed,in which no tensile stress is allowed in the concrete at service load.Class 2:Partially prestressed, in which occasional temporary cracking is permitted under infrequent high loads.Class 3:Partially prestressed,in which there may be permanent cracks provided that their width is suitably limited.The choise of a suitable amount of prestress is governed by avariety of factors.These include thenature of the loading (for exmaple,highway or railroadbridged,storage,ect.),the ratio of live to dead load,the frequency of occurrence of loading may be reversed,such as in transmission poles,a high uniform prestress would result ultimate strength and in brittle failure.In such a case,partial prestressing provides the only satifactory solution.The advantages of partial prestressing are important.A smaller prestress force will be required,permitting reduction in the number of tendons and anchorages.The necessary flexural strength may be provided in such cases either by a combination of prestressed tendons and non-prestressed reinforcing bars,or by an adequate number of high-tensile tendons prestredded to level lower than the prestressing force isless,the size of the bottom flange,which is requied mainly to resist the compression when a beam is in the unloaded stage,can be reduced or eliminated altogether.This leads in turn to significant simplification and cost reduction in the construction of forms,as well as resulting in structures that are mor pleasing esthetically.Furthermore,by relaxing the requirement for low service load tension in the concrete,a significant improvement can be made in the deflection characteristics of a beam.Troublesome upward camber of the member in the unloaded stage fan be avoeded,and the prestress force selected primarily to produce the desired deflection for a particular loading condition.The behavior of partially prestressed beamsm,should they be overloaded to failure,is apt to be superior to that of fully prestressed beams,because the improved ductility provides ample warning of distress.英译汉:荷载作用在结构上的荷载通常分为恒载或活载。
土木工程专业英语课文原文及对照翻译
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。
土木工程专业钢筋混凝土结构设计毕业论文外文文献翻译及原文
毕业设计（论文）外文文献翻译文献、资料中文题目：钢筋混凝土结构设计文献、资料英文题目：DESIGN OF REINFORCED CONCRETE STRUCTURES 文献、资料来源：文献、资料发表（出版）日期：院（部）：专业：土木工程班级：姓名：学号：指导教师：翻译日期： 2017.02.14毕业设计（论文）外文参考资料及译文译文题目：DESIGN OF REINFORCED CONCRETE STRUCTURES原文：DESIGN OF REINFORCED CONCRETESTRUCTURES1. BASIC CONCERPTS AND CHARACERACTERISTICS OF REINFORCED CONCRETEPlain concrete is formed from hardened mixture of cement, water , fine aggregate , coarse aggregate (crushed stone or gravel ) , air and often other admixtures . The plastic mix is placed and consolidated in the formwork, then cured to accelerate of the chemical hydration of hen cement mix and results in a hardened concrete. It is generally known that concrete has high compressive strength and low resistance to tension. Its tensile strength is approximatelyone-tenth of its compressive strength. Consequently, tensile reinforcement in the tension zone has to be provided to supplement the tensile strength of the reinforced concrete section.For example, a plain concrete beam under a uniformly distributed load q is shown in Fig .1.1(a), when the distributed load increases and reaches a value q=1.37KN/m , the tensile region at the mid-span will be cracked and the beam will fail suddenly . A reinforced concrete beam if the same size but has to steel reinforcing bars (2φ16) embedded at the bottom under a uniformly distributed load q is shown in Fig.1.1(b). The reinforcing bars take up the tension there after the concrete is cracked. When the load q is increased, the width of the cracks, the deflection and thestress of steel bars will increase . When the steel approaches the yielding stress ƒy , thedeflection and the cracked width are so large offering some warning that the compression zone . The failure load q=9.31KN/m, is approximately 6.8 times that for the plain concrete beam.Concrete and reinforcement can work together because there is a sufficiently strong bond between the two materials, there are no relative movements of the bars and the surrounding concrete cracking. The thermal expansion coefficients of the two materials are 1.2×10-5K-1 for steel and 1.0×10-5~1.5×10-5K-1 for concrete .Generally speaking, reinforced structure possess following features :Durability .With the reinforcing steel protected by the concrete , reinforced concreteFig.1.1Plain concrete beam and reinforced concrete beamIs perhaps one of the most durable materials for construction .It does not rot rust , and is not vulnerable to efflorescence .(2)Fire resistance .Both concrete an steel are not inflammable materials .They would not be affected by fire below the temperature of 200℃when there is a moderate amount of concrete cover giving sufficient thermal insulation to the embedded reinforcement bars.(3)High stiffness .Most reinforced concrete structures have comparatively large cross sections .As concrete has high modulus of elasticity, reinforced concrete structures are usuallystiffer than structures of other materials, thus they are less prone to large deformations, This property also makes the reinforced concrete less adaptable to situations requiring certainflexibility, such as high-rise buildings under seismic load, and particular provisions have to be made if reinforced concrete is used.(b)Reinfoced concrete beam(4)Locally available resources. It is always possible to make use of the local resources of labour and materials such as fine and coarse aggregates. Only cement and reinforcement need to be brought in from outside provinces.(5)Cost effective. Comparing with steel structures, reinforced concrete structures are cheaper.(6)Large dead mass, The density of reinforced concrete may reach2400~2500kg/pare with structures of other materials, reinforced concrete structures generally have a heavy dead mass. However, this may be not always disadvantageous, particularly for those structures which rely on heavy dead weight to maintain stability, such as gravity dam and other retaining structure. The development and use of light weight aggregate have to a certain extent make concrete structure lighter.(7)Long curing period.. It normally takes a curing period of 28 day under specified conditions for concrete to acquire its full nominal strength. This makes the progress of reinforced concrete structure construction subject to seasonal climate. The development of factory prefabricated members and investment in metal formwork also reduce the consumption of timber formwork materials.(8)Easily cracked. Concrete is weak in tension and is easily cracked in the tension zone. Reinforcing bars are provided not to prevent the concrete from cracking but to take up the tensile force. So most of the reinforced concrete structure in service is behaving in a cracked state. This is an inherent is subjected to a compressive force before working load is applied. Thus the compressed concrete can take up some tension from the load.2. HISTOEICAL DEVELPPMENT OF CONCRETE STRUCTUREAlthough concrete and its cementitious(volcanic) constituents, such as pozzolanic ash, have been used since the days of Greek, the Romans, and possibly earlier ancient civilization, the use of reinforced concrete for construction purpose is a relatively recent event, In 1801, F. Concrete published his statement of principles of construction, recognizing the weakness if concrete in tension, The beginning of reinforced concrete is generally attributed to Frenchman J. L. Lambot, who in 1850 constructed, for the first time, a small boat with concrete for exhibition in the 1855 World’s Fair in Paris. In England, W. B. Wilkinson registered a patent for reinforced concrete l=floor slab in 1854.J.Monier, a French gardener used metal frames as reinforcement to make garden plant containers in 1867. Before 1870, Monier had taken a series of patents to make reinforcedconcrete pipes, slabs, and arches. But Monier had no knowledge of the working principle of this new material, he placed the reinforcement at the mid-depth of his wares. Then little construction was done in reinforced concrete. It is until 1887, when the German engineers Wayss and Bauschinger proposed to place the reinforcement in the tension zone, the use of reinforced concrete as a material of construction began to spread rapidly. In1906, C. A. P. Turner developed the first flat slab without beams.Before the early twenties of 20th century, reinforced concrete went through the initial stage of its development, Considerable progress occurred in the field such that by 1910 the German Committee for Reinforced Concrete, the Austrian Concrete Committee, the American Concrete Institute, and the British Concrete Institute were established. Various structural elements, such as beams, slabs, columns, frames, arches, footings, etc. were developed using this material. However, the strength of concrete and that of reinforcing bars were still very low. The common strength of concrete at the beginning of 20th century was about 15MPa in compression, and the tensile strength of steel bars was about 200MPa. The elements were designed along the allowable stresses which was an extension of the principles in strength of materials.By the late twenties, reinforced concrete entered a new stage of development. Many buildings, bridges, liquid containers, thin shells and prefabricated members of reinforced concrete were concrete were constructed by 1920. The era of linear and circular prestressing began.. Reinforced concrete, because of its low cost and easy availability, has become the staple material of construction all over the world. Up to now, the quality of concrete has been greatly improved and the range of its utility has been expanded. The design approach has also been innovative to giving the new role for reinforced concrete is to play in the world of construction.The concrete commonly used today has a compressive strength of 20~40MPa. For concrete used in pre-stressed concrete the compressive strength may be as high as 60~80MPa. The reinforcing bars commonly used today has a tensile strength of 400MPa, and the ultimate tensile strength of prestressing wire may reach 1570~1860Pa. The development of high strength concrete makes it possible for reinforced concrete to be used in high-rise buildings, off-shore structures, pressure vessels, etc. In order to reduce the dead weight of concrete structures, various kinds of light concrete have been developed with a density of 1400~1800kg/m3. With a compressive strength of 50MPa, light weight concrete may be used in load bearing structures. One of the best examples is the gymnasium of the University of Illinois which has a span of 122m and is constructed of concrete with a density of 1700kg/m3. Another example is the two 20-story apartment houses at the Xi-Bian-Men in Beijing. The walls of these two buildings are light weight concrete with a density of 1800kg/m3.The tallest reinforced concrete building in the world today is the 76-story Water Tower Building in Chicago with a height of 262m. The tallest reinforced concrete building in China today is the 63-story International Trade Center in GuangZhou with a height a height of 200m. The tallest reinforced concrete construction in the world is the 549m high International Television Tower in Toronto, Canada. He prestressed concrete T-section simply supported beam bridge over the Yellow River in Luoyang has 67 spans and the standard span length is 50m.In the design of reinforced concrete structures, limit state design concept has replaced the old allowable stresses principle. Reliability analysis based on the probability theory has very recently been introduced putting the limit state design on a sound theoretical foundation. Elastic-plastic analysis of continuous beams is established and is accepted in most of the design codes. Finite element analysis is extensively used in the design of reinforced concrete structures and non-linear behavior of concrete is taken into consideration. Recent earthquake disasters prompted the research in the seismic resistant reinforced of concrete structures. Significant results have been accumulated.3. SPECIAL FEATURES OF THE COURSEReinforced concrete is a widely used material for construction. Hence, graduates of every civil engineering program must have, as a minimum requirement, a basic understanding of the fundamentals of reinforced concrete.The course of Reinforced Concrete Design requires the prerequisite of Engineering Mechanics, Strength of Materials, and some if not all, of Theory of Structures, In all these courses, with the exception of Strength of Materials to some extent, a structure is treated of in the abstract. For instance, in the theory of rigid frame analysis, all members have an abstract EI/l value, regardless of what the act value may be. But the theory of reinforced concrete is different, it deals with specific materials, concrete and steel. The values of most parameters must be determined by experiments and can no more be regarded as some abstract. Additionally, due to the low tensile strength of concrete, the reinforced concrete members usually work with cracks, some of the parameters such as the elastic modulus I of concrete and the inertia I of section are variable with the loads.The theory of reinforced concrete is relatively young. Although great progress has been made, the theory is still empirical in nature in stead of rational. Many formulas can not be derived from a few propositions, and may cause some difficulties for students. Besides, due to the difference in practice in different countries, most countries base their design methods on their own experience and experimental results. Consequently, what one learns in one country may be different in another country. Besides, the theory is still in a stage of rapid。
土木外文翻译原文和译文
A convection-conduction model for analysis of thefreeze-thawconditions in the surrounding rock wall of atunnel in permafrost regionsAbstractBased on the analyses of fundamental meteorological and hydrogeological conditions at the site of a tunnel in the cold regions, a combined convection-conduction model for air flow in the tunnel and temperature field in the surrounding has been constructed. Using the model, the air temperature distribution in the Xiluoqi No. 2 Tunnel has been simulated numerically. The simulated results are in agreement with the data observed. Then, based on the in situ conditions of sir temperature, atmospheric pressure, wind force, hydrogeology and engineering geology, the air-temperature relationship between the temperature on the surface of the tunnel wall and the air temperature at the entry and exit of the tunnel has been obtained, and the freeze-thaw conditions at the Dabanshan Tunnel which is now under construction is predicted.Keywords: tunnel in cold regions, convective heat exchange and conduction, freeze-thaw.A number of highway and railway tunnels have been constructed in the permafrost regions and their neighboring areas in China. Since the hydrological and thermal conditions changed after a tunnel was excavated，the surrounding wall rock materials often froze, the frost heaving caused damage to the liner layers and seeping water froze into ice diamonds，which seriously interfered with the communication and transportation. Similar problems of the freezing damage in the tunnelsalso appeared in other countries like Russia, Norway and Japan .Hence it is urgent to predict the freeze-thaw conditions in the surrounding rock materials and provide a basis for the design，construction and maintenance of new tunnels in cold regions.Many tunnels，constructed in cold regions or their neighbouring areas，pass through the part beneath the permafrost base .After a tunnel is excavated，the original thermodynamical conditions in the surroundings are and thaw destroyed and replaced mainly by the air connections without the heat radiation, the conditions determined principally by the temperature and velocity of air flow in the tunnel，the coefficients of convective heat transfer on the tunnel wall，and the geothermal heat. In order to analyze and predict the freeze and thaw conditions of the surrounding wall rock of a tunnel，presuming the axial variations of air flow temperature and the coefficients of convective heat transfer, Lunardini discussed the freeze and thaw conditions by the approximate formulae obtained by Sham-sundar in study of freezing outside a circular tube with axial variations of coolant temperature .We simulated the temperature conditions on the surface of a tunnel wall varying similarly to the periodic changes of the outside air temperature .In fact，the temperatures of the air and the surrounding wall rock material affect each other so we cannot find the temperature variations of the air flow in advance; furthermore，it is difficult to quantify the coefficient of convective heat exchange at the surface of the tunnel wall .Therefore it is not practicable to define the temperature on the surface of the tunnel wall according to the outside air temperature .In this paper, we combine the air flow convective heat ex-change and heat conduction in the surrounding rock material into one model，and simulate the freeze-thaw conditions of the surrounding rock material based on the in situ conditions of air temperature，atmospheric pressure，wind force at the entry and exit of the tunnel，and the conditions of hydrogeology and engineering geology.重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译Mathematical modelIn order to construct an appropriate model, we need the in situ fundamental conditions as a ba-sis .Here we use the conditions at the scene of the Dabanshan Tunnel. The Dabanshan Tunnel is lo-toted on the highway from Xining to Zhangye, south of the Datong River, at an elevation of 3754.78-3 801.23 m, with a length of 1 530 m and an alignment from southwest to northeast. The tunnel runs from the southwest to the northeast.Since the monthly-average air temperature is beneath 0`}C for eight months at the tunnel site each year and the construction would last for several years，the surrounding rock materials would become cooler during the construction .We conclude that, after excavation, the pattern of air flow would depend mainly on the dominant wind speed at the entry and exit，and the effects of the temperature difference between the inside and outside of the tunnel would be very small .Since the dominant wind direction is northeast at the tunnel site in winter, the air flow in the tunnel would go from the exit to the entry. Even though the dominant wind trend is southeastly in summer, considering the pressure difference, the temperature difference and the topography of the entry and exit，the air flow in the tunnel would also be from the exit to entry .Additionally，since the wind speed at the tunnel site is low，we could consider that the air flow would be principally laminar.Based on the reasons mentioned，we simplify the tunnel to a round tube，and consider that theair flow and temperature are symmetrical about the axis of the tunnel，Ignoring the influence of the air temperature on the speed of air flow, we obtain the following equation:where t ，x ，r are the time ，axial and radial coordinates; U ，V are axial and radial wind speeds; T is temperature; p is the effective pressure(that is ，air pressure divided by air density); v is the kinematic viscosity of air; a is the thermal conductivity of air; L is the length of the tunnel; R is the equivalent radius of the tunnel section; D is the length of time after the tunnel construction;，f S (t), u S (t) are frozen and thawed parts in the surrounding rock materials respectively; f λ,u λand f C ,u C are thermal conductivities and volumetric thermal capacities in frozen and thawed parts respectively; X= (x , r)，ξ(t) is phase change front; Lh is heat latent of freezing water; and To is critical freezing temperature of rock ( here we assume To= -0.1℃).2 used for solving the modelEquation(1)shows flow. We first solve those concerning temperature at that the temperature of the surrounding rock does not affect the speed of air equations concerning the speed of air flow, and then solve those equations every time elapse.2. 1 Procedure used for solving the continuity and momentum equations重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译Since the first three equations in(1) are not independent we derive the second equation by xand the third equation by r. After preliminary calculation we obtain the following elliptic equation concerning the effective pressure p:Then we solve equations in(1) using the following procedures:(i ) Assume the values for U0，V0;( ii ) substituting U0，V0 into eq. (2)，and solving (2)，we obtain p0;(iii) solving the first and second equations of(1)，we obtain U0，V1;(iv) solving the first and third equations of(1)，we obtain U2，V2; (v) calculating the momentum-average of U1，v1 and U2，v2，we obtain the new U0，V0;then return to (ii);(vi) iterating as above until the disparity of those solutions in two consecutive iterations is sufficiently small or is satisfied，we then take those values of p0，U0 and V0 as the initial values for the next elapse and solve those equations concerning the temperature..2 .2 Entire method used for solving the energy equationsAs mentioned previously，the temperature field of the surrounding rock and the air flow affect each other. Thus the surface of the tunnel wall is both the boundary of the temperature field in the surrounding rock and the boundary of the temperature field in air flow .Therefore， it is difficult to separately identify the temperature on the tunnel wall surface，and we cannot independently solve those equations concerning the temperature of air flow and those equations concerning the temperature of the surrounding rock .In order to cope with this problem，we simultaneously solve the two groups of equations based on the fact that at the tunnel wall surface both temperatures are equal .We should bearin mind the phase change while solving those equations concerning the temperature of the surrounding rock ，and the convection while solving those equations concerning the temperature of the air flow, and we only need to smooth those relative parameters at the tunnel wall surface .The solving methods for the equations with the phase change are the same as in reference [3].2.3 Determination of thermal parameters and initial and boundaryconditions2.3.1 Determination of the thermal parameters. Using p= 1013.25-0.1088 H ，we calculateair pressure p at elevation H and calculate the air density ρ using formula GTP =ρ, where T is the yearly-average absolute air temperature ，and G is the humidity constant of air. Letting P C be the thermal capacity with fixed pressure, λ the thermal conductivity ，and μ the dynamic viscosity of air flow, we calculate the thermal conductivity and kinematic viscosity using the formulas ρλP C =ａ and ρμν=. The thermal parameters of the surrounding rock are determined from the tunnel site.2 .3.2 Determination of the initial and boundary conditions .Choose the observed monthly average wind speed at the entry and exit as boundary conditions of wind speed ，and choose the relative effective pressure p=0 at the exit ( that is ，the entry of the dominant wind trend) and ]5[22/)/1(v d kL p ⨯+= on the section of entry ( that is ，the exit of the dominant wind trend )，where k is the coefficient of resistance along the tunnel wall, d = 2R ，and v is the axial average speed. We approximate T varying by the sine law according to the data observed at the scene and provide a suitable boundary value based on the position of the permafrost base and the geothermal gradient of the thaw rock materials beneath the重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译permafrost base.3 A simulated exampleUsing the model and the solving method mentioned above，we simulate the varying law of the air temperature in the tunnel along with the temperature at the entry and exit of the Xiluoqi No.2 Tunnel .We observe that the simulated results are close to the data observed[6].The Xiluoqi No .2 Tunnel is located on the Nongling railway in northeastern China and passes through the part beneath the permafrost base .It has a length of 1 160 m running from the northwest to the southeast, with the entry of the tunnel in the northwest，and the elevation is about 700 m. The dominant wind direction in the tunnel is from northwest to southeast, with a maximum monthly-average speed of 3 m/s and a minimum monthly-average speed of 1 .7 m/s . Based on the data observed，we approximate the varying sine law of air temperature at the entry and exit with yearly averages of -5℃，-6.4℃ and amplitudes of 18.9℃ and 17.6℃respectively. The equivalent diameter is 5 .8m，and the resistant coefficient along the tunnel wall is 0.025.Since the effect of the thermal parameter of the surrounding rock on the air flow is much smaller than that of wind speed，pressure and temperature at the entry and exit，we refer to the data observed in the Dabanshan Tunnel for the thermal parameters.Figure 1 shows the simulated yearly-average air temperature inside and at the entry and exit of the tunnel compared with the data observed .We observe that the difference is less than 0 .2 `C from the entry to exit.Figure 2 shows a comparison of the simulated and observed monthly-average air temperature in-side (distance greater than 100 m from the entry and exit) the tunnel. We observe that the principal law is almost the same，and the main reason for the difference is the errors that came from approximating the varying sine law at the entry and exit; especially , the maximum monthly-average air temperature of 1979 was not for July but for August.Fig.1. Comparison of simulated and observed air temperature in Xiluoqi No.2 Tunnel in 1979.1,simulated values;2,observed valuesFig.2.The comparison of simulated and observed air temperature inside The Xiluoqi No.2 Tunnel in 1979.1,simulated values;2,observed values4 Prediction of the freeze-thaw conditions for the Dabanshan Tunnel 4 .1 Thermal parameter and initial and boundary conditionsUsing the elevation of 3 800 m and the yearly-average air temperature of -3℃, we calculate the air density p=0 .774 kg/m 3.Since steam exists In the air, we choose the thermal capacity with a fixed pressure of air ),./(8744.10C kg kJ C p = heat conductivity )./(100.202C m W -⨯=λ andand the dynamic viscosity )../(10218.96s m kg -⨯=μ After calculation we obtain the thermal diffusivity a= 1 .3788s m /1025-⨯ and the kinematic viscosity ，s m /1019.125-⨯=ν .Considering that the section of automobiles is much smaller than that of the tunnel and the auto-mobiles pass through the tunnel at a low speed ，we ignore the piston effects ，coming from the movement of automobiles ，in the diffusion of the air.We consider the rock as a whole component and choose the dry volumetric cavity 3/2400m kg d =λ,content of water and unfrozen water W=3% and W=1%, and the thermal conductivity c m W o u ./9.1=λ,c m W o f ./0.2=λ,heat重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译capacityc kg kJ C o V ./8.0= and d u f W w C γ⨯++=1)128.48.0(,d u u Ww C γ⨯++=1)128.48.0( According to the data observed at the tunnel site ，the maximum monthly-average wind speed is about 3 .5 m/s ，and the minimum monthly-average wind speed is about 2 .5 m/s .We approximate the wind speed at the entry and exit as )/](5.2)7(028.0[)(2s m t t v +-⨯=, where t is in month. The initial wind speed in the tunnel is set to be.0),,0(),)(1(),,0(2=-=r x V R r U r x U a The initial and boundary values of temperature T are set to bewhere f(x) is the distance from the vault to the permafrost base ，and R0=25 m is the radius of do-main of solution T. We assume that the geothermal gradient is 3%，the yearly-average air temperature outside tunnel the is A=-3C 0，and the amplitude is B=12C 0.As for the boundary of R=Ro ，we first solve the equations considering R=Ro as the first type of boundary; that is we assume that T=f(x)⨯3%C 0on R=Ro. We find that, after one year, the heat flow trend will have changed in the range of radius between 5 and 25m in the surrounding rock.. Considering that the rock will be cooler hereafter and it will be affected yet by geothermal heat, we appoximately assume that the boundary R=Ro is the second type of boundary; that is ，we assume that the gradient value ，obtained from the calculation up to the end of the first year after excavation under the first type of boundary value, is the gradient on R=Ro of T.Considering the surrounding rock to be cooler during the period of construction ，we calculatefrom January and iterate some elapses of time under the same boundary. Then we let the boundaryvalues vary and solve the equations step by step(it can be proved that the solution will not depend on the choice of initial values after many time elapses ).1)The yearly-average temperature on the surface wall of the tunnel is approximately equal to the ai4 .2 Calculated resultsFigures 3 and 4 show the variations of the monthly-average temperatures on the surface of the tunnel wall along with the variations at the entry and exit .Figs .5 and 6 show the year when permafrost begins to form and the maximum thawed depth after permafrost formed in different surrounding sections.Fig.3.The monthly-average temperature parison of the monthly- On the surface of Dabanshan Tunnel.I, average temperature on the surface The month,I=1,2,3,,,12 tunnel with that outside the tunnel. 1,inner temperature on the surface ;2,outside air temperatureFig.5.The year when permafrost Fig.6.The maximum thawed depth after Begins to from in different permafrost formed in different years Sections of the surroundingrock重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译4 .3 Preliminary conclusionBased on the initial-boundary conditions and thermal parameters mentioned above, we obtain the following preliminary conclusions: r temperature at the entry and exit. It is warmer during the cold season and cooler during the warm season in the internal part (more than 100 m from the entry and exit) of the tunnel than at the entry and exit . Fig .1 shows that the internal monthly-average temperature on the surface of the tunnel wall is 1.2℃ higher in January, February and December, 1℃higher in March and October, and 1 .6℃ lower in June and August, and 2qC lower in July than the air temperature at the entry and exit. In other months the infernal temperature on the surface of the tunnel wall approximately equals the air temperature at the entry and exit.2) Since it is affected by the geothermal heat in the internal surrounding section，especially in the central part, the internal amplitude of the yearly-average temperature on the surface of the tunnel wall decreases and is 1 .6℃ lower than that at the entry and exit.3 ) Under the conditions that the surrounding rock is compact , without a great amount of under-ground water, and using a thermal insulating layer(as designed PU with depth of 0.05 m and heat conductivity λ=0.0216 W/m℃，FBT with depth of 0.085 m and heat conductivity λ=0.0517W/m℃)，in the third year after tunnel construction，the surrounding rock will begin to form permafrost in the range of 200 m from the entry and exit .In the first and the second year after construction, the surrounding rock will begin to form permafrost in the range of 40 and 100m from the entry and exit respectively .In the central part，more than 200m from the entry and exit, permafrost will begin to form in the eighth year. Near the center of the tunnel，permafrost will appear in the 14-15th years. During the first and second years after permafrost formed，the maximum of annual thawed depth is large (especially in the central part of the surrounding rock section) and thereafter it decreases every year. The maximum of annual thawed depth will be stable until the 19-20th yearsand will remain in s range of 2-3 m.4) If permafrost forms entirely in the surrounding rock，the permafrost will provide a water-isolating layer and be favourable for communication and transportation .However, in the process of construction，we found a lot of underground water in some sections of the surrounding rock .It will permanently exist in those sections，seeping out water and resulting in freezing damage to the liner layer. Further work will be reported elsewhere.重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译严寒地区隧道围岩冻融状况分析的导热与对流换热模型摘要通过对严寒地区隧道现场基本气象条件的分析，建立了隧道内空气与围岩对流换热及固体导热的综合模型;用此模型对大兴安岭西罗奇2号隧道的洞内气温分布进行了模拟计算，结果与实测值基本一致;分析预报了正在开凿的祁连山区大坂山隧道开通运营后洞内温度及围岩冻结、融化状况.关键词严寒地区隧道导热与对流换热冻结与融化在我国多年冻土分布及邻近地区，修筑了公路和铁路隧道几十座.由于隧道开通后洞内水热条件的变化;，普遍引起洞内围岩冻结，造成对衬砌层的冻胀破坏以及洞内渗水冻结成冰凌等，严重影响了正常交通.类似隧道冻害问题同样出现在其他国家（苏联、挪威、日本等)的寒冷地区.如何预测分析隧道开挖后围岩的冻结状况，为严寒地区隧道建设的设计、施工及维护提供依据，这是一个亟待解决的重要课题.在多年冻土及其临近地区修筑的隧道，多数除进出口部分外从多年冻土下限以下岩层穿过.隧道贯通后，围岩内原有的稳定热力学条件遭到破坏，代之以阻断热辐射、开放通风对流为特征的新的热力系统.隧道开通运营后，围岩的冻融特性将主要由流经洞内的气流的温度、速度、气—固交界面的换热以及地热梯度所确定.为分析预测隧道开通后围岩的冻融特性，Lu-nardini借用Shamsundar研究圆形制冷管周围土体冻融特性时所得的近似公式，讨论过围岩的冻融特性.我们也曾就壁面温度随气温周期性变化的情况，分析计算了隧道围岩的温度场[3].但实际情况下，围岩与气体的温度场相互作用，隧道内气体温度的变化规律无法预先知道，加之洞壁表面的换热系数在技术上很难测定，从而由气温的变化确定壁面温度的变化难以实现.本文通过气一固祸合的办法，把气体、固体的换热和导热作为整体来处理，从洞口气温、风速和空气湿度、压力及围岩的水热物理参数等基本数据出发，计算出围岩的温度场.1数学模型为确定合适的数学模型，须以现场的基本情况为依据.这里我们以青海祁连山区大坂山公路隧道的基本情况为背景来加以说明.大坂山隧道位于西宁一张业公路大河以南，海拔3754.78~3801.23 m ，全长1530 m ,隧道近西南—东北走向. 由于大坂山地区隧道施工现场平均气温为负温的时间每年约长8个月，加之施工时间持续数年，围岩在施土过程中己经预冷，所以隧道开通运营后，洞内气体流动的形态主要由进出口的主导风速所确定，而受洞内围岩地温与洞外气温的温度压差的影响较小；冬季祁连山区盛行西北风，气流将从隧道出曰流向进口端，夏季虽然祁连山区盛行东偏南风，但考虑到洞口两端气压差、温度压差以及进出口地形等因素，洞内气流仍将由出口北端流向进口端.另外，由于现场年平均风速不大，可以认为洞内气体将以层流为主基于以上基本情况，我们将隧道简化成圆筒，并认为气流、温度等关十隧道中心线轴对称，忽略气体温度的变化对其流速的影响，可有如下的方程:其中t 为时间，x 为轴向坐标，r 为径向坐标;U, V 分别为轴向和径向速度，T 为温度，P 为有效压力(即空气压力与空气密度之比少，V 为空气运动粘性系数，a 为空气的导温系数，L 为隧道长度，R 为隧道的当量半径，D 为时间长度)(t S f , )(t S u 分别为围岩的冻、融区域. f λ,u λ分别为冻、融状态下的热传导系数，f C ,u C 分别为冻、融状态下的体积热容量，X=(x,r) , )(t ξ为冻、融相变界面,To 为岩石冻结临界温度(这里具体计算时取To=-0.10C 0)，h L 为水的相变潜热.重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译2 求解过程由方程(1)知，围岩的温度的高低不影响气体的流动速度，所以我们可先解出速度，再解温度.2.1 连续性方程和动量方程的求解由于方程((1)的前3个方程不是相互独立的，通过将动量方程分别对x 和r 求导，经整理化简，我们得到关于压力P 的如下椭圆型方程:于是，对方程(1)中的连续性方程和动量方程的求解，我们按如下步骤进行:(1)设定速度0U ,0V ;( 2)将0U ,0V 代入方程并求解，得0P(3)联立方程(1)的第一个和第二个方程，解得一组解1U ,1V ;(4)联立方程((1)的第一个和第三个方程，解得一组解2U ,2V ;(5)对((3) ,(4)得到的速度进行动量平均,得新的0U ,0V 返回(2) ;(6)按上述方法进行迭代，直到前后两次的速度值之差足够小.以0P ,0U ,0V 作为本时段的解,下一时段求解时以此作为迭代初值.2. 2 能量方程的整体解法如前所述，围岩与空气的温度场相互作用，壁面既是气体温度场的边界，又是固体温度场的边界，壁面的温度值难以确定，我们无法分别独立地求解隧道内的气体温度场和围岩温度场.为克服这一困难，我们利用在洞壁表面上，固体温度等于气体温度这一事实，把隧道内气体的温度和围岩内固体的温度放在一起求解，这样壁面温度将作为末知量被解出来.只是需要注意两点:解流体温度场时不考虑相变和解固体温度时没有对流项;在洞壁表面上方程系数的光滑化.另外，带相变的温度场的算法与文献[3]相同.2. 3热参数及初边值的确定热参数的确定方法: 用p=1013.25-0.1088H 计算出海拔高度为H 的隧道现场的大气压强，再由GT P =ρ计算出现场空气密度ρ，其中T 为现场大气的年平均绝对温度，G 为空气的气体常数.记定压比热为P C ,导热系数为λ，空气的动力粘性系数为μ.按ρλP C =ａ 和ρμν= 计算空气的导温系数和运动粘性系数.围岩的热物理参数则由现场采样测定.初边值的确定方法:洞曰风速取为现场观测的各月平均风速.取卞导风进曰的相对有效气压为0，主导风出口的气压则取为]5[22/)/1(v d kL p ⨯+=，这里k 为隧道内的沿程阻力系数，L 为隧道长度，d 为隧道端面的当量直径，ν为进口端面轴向平均速度.进出口气温年变化规律由现场观测资料，用正弦曲线拟合，围岩内计算区域的边界按现场多年冻土下限和地热梯度确定出适当的温度值或温度梯度. 3 计算实例按以上所述的模型及计算方法，我们对大兴安岭西罗奇2号隧道内气温随洞曰外气温变化的规律进行了模拟计算验证，所得结果与实测值[6]相比较,基本规律一致.西罗奇2号隧道是位十东北嫩林线的一座非多年冻土单线铁路隧道，全长1160 m ,隧道近西北一东南向，高洞口位于西北向，冬季隧道主导风向为西北风.洞口海拔高度约为700 m ,月平均最高风速约为3m/s,最低风速约为1.7m/s.根据现场观测资料，我们将进出口气温拟合为年平均分别为-5C 0和-6.4C 0，年变化振幅分别为18.9C 0和17.6C 0的正弦曲线.隧道的当量直径为5.8 m,沿程阻力系数取为0.025.由于围岩的热物理参数对计算洞内气温的影响远比洞口的风速、压力及气温的影响小得多，我们这里参考使用了大坂山隧道的资料.图1给出了洞口及洞内年平均气温的计算值与观测值比较的情况，从进口到出口，两值之差都小于0.2C 0.图2给出了洞内 (距进出口l00m 以上)月平均气温的计算值与观测值比较的情况，可以看出温度变化的基本规律完全一致，造成两值之差的主要原因是洞口气温年变化规律之正弦曲线的拟合误差，特别是1979年隧道现场月平均最高气温不是在7月份，而是在8月份.重庆交通大学土木工程专业（隧道与城市轨道交通工程方向）毕业设计外文翻译图1. 比较1979年在西罗奇周家山2号隧道仿真试验与观察的空气温度.1、模拟值;2、观测值图2。
土木工程专业英语课文_翻译_考试必备
土木工程专业英语课文_翻译_考试必备土木工程专业英语课文翻译The principal construction materials of earlier times were wood and masonry brick, stone, or tile, and similar materials. The courses or layers were bound together with mortar or bitumen, a tar like substance, or some other binding agent. The Greeks and Romanssometimes used iron rods or claps to strengthen their building. The columns of the Parthenon in Athens, for example, have holes drilled in them for iron bars that have now rusted away. The Romans also used a natural cement called puzzling, made from volcanic ash, that became as hard as stone under water.早期时代的主要施工材料，木材和砌体砖，石，或瓷砖，和类似的材料。
这些课程或层密切联系在一起，用砂浆或沥青，焦油一个样物质，或其他一些有约束力的代理人。
希腊人和罗马人有时用铁棍或拍手以加强其建设。
在雅典的帕台农神庙列，例如，在他们的铁钻的酒吧现在已经生锈了孔。
罗马人还使用了天然水泥称为令人费解的，由火山灰制成，变得像石头一样坚硬在水中。
Both steel and cement, the two most important construction materials of modern times, were introduced in the nineteenth century. Steel, basically an alloy of iron and a small amount of carbon had been made up to that time by a laborious process that restricted it to such special uses as sword blades. After the invention of the Bessemer process in 1856, steel was available in large quantities at low prices. The enormous advantage of steel is its tensile force which, as we have seen,tends to pull apart many materials. New alloys have further, which is a tendency for it to weaken as a result of continual changes in stress.钢铁和水泥，两个最重要的现代建筑材料，介绍了在十九世纪。
土木工程岩土类毕业设计外文翻译
土木工程专业英语复习参考学号： 10447425X X 大学毕业设计（论文）外文翻译（2014届）外文题目Developments in excavation bracing systems译文题目开挖工程支撑体系的发展外文出处Tunnelling and Underground SpaceTechnology 31 (2012) 107–116学生XXX学院XXXX 专业班级XXXXX校内指导教师XXX 专业技术职务XXXXX校外指导老师专业技术职务二○一三年十二月开挖工程支撑体系的发展1.引言几乎所有土木工程建设项目（如建筑物，道路，隧道，桥梁，污水处理厂，管道，下水道）都涉及泥土挖掘的一些工程量。
往往由于由相邻的结构，特性线，或使用权空间的限制，必须要一个土地固定系统，以允许土壤被挖掘到所需的深度。
历史上，许多挖掘支撑系统已经开发出来。
其中，现在比较常见的几种方法是：板桩，钻孔桩墙，泥浆墙。
土地固定系统的选择是由技术性能要求和施工可行性（例如手段，方法）决定的，包括执行的可靠性，而成本考虑了这些之后，其他问题也得到解决。
通常环境后果（用于处理废泥浆和钻井液如监管要求）也非常被关注（邱阳、1998）。
土地固定系统通常是建设项目的较大的一个组成部分。
如果不能按时完成项目，将极大地影响总成本。
通常首先建造支撑，在许多情况下，临时支撑系统是用于支持在挖掘以允许进行不断施工，直到永久系统被构造。
临时系统可以被去除或留在原处。
打桩时，因撞击或振动它们可能会被赶入到位。
在一般情况下，振动是最昂贵的方法，但只适合于松散颗粒材料，土壤中具有较高电阻（例如，通过鹅卵石）的不能使用。
采用打入桩系统通常是中间的成本和适合于软沉积物（包括粘性和非粘性），只要该矿床是免费的鹅卵石或更大的岩石。
通常，垂直元素（例如桩）的前安装挖掘工程和水平元件（如内部支撑或绑回）被安装为挖掘工程的进行下去，从而限制了跨距长度，以便减少在垂直开发弯矩元素。
土木工程-毕业设计-论文-外文翻译-中英文对照
英文原文：Concrete structure reinforcement designSheyanb oⅠWangchenji aⅡⅠFoundation Engineering Co., Ltd. Heilongjiang DongyuⅡHeilongjiang Province, East Building Foundation Engineering Co., Ltd. CoalAbstract：structure in the long-term natural environment and under the use environment's function, its function is weaken inevitably gradually, our structural engineering's duty not just must finish the building earlier period the project work, but must be able the science appraisal structure damage objective law and the degree, and adopts the effective method guarantee structure the security use, that the structure reinforcement will become an important work. What may foresee will be the 21st century, the human building also by the concrete structure, the steel structure, the bricking-up structure and so on primarily, the present stage I will think us in the structure reinforcement this aspect research should also take this as the main breakthrough direction.Key word：Concrete structure reinforcement bricking-up structure reinforcement steel structure reinforcement1 Concrete structure reinforcementConcrete structure's reinforcement divides into the direct reinforcement and reinforces two kinds indirectly, when the design may act according to the actual condition and the operation requirements choice being suitable method and the necessary technology.1.1the direct reinforcement's general method1）Enlarges the section reinforcement lawAdds the concretes cast-in-place level in the reinforced concrete member in bending compression zone, may increase the section effective height, the expansion cross sectional area, thus enhances the component right section anti-curved, the oblique section anti-cuts ability and the section rigidity, plays the reinforcement reinforcement the role.In the suitable muscle scope, the concretes change curved the component right section supporting capacity increase along with the area of reinforcement and the intensity enhance. In the original component right section ratio of reinforcement not too high situation, increases the main reinforcement area to be possible to propose the plateau component right section anti-curved supporting capacity effectively. Is pulled in the section the area to add the cast-in-place concrete jacket to increase the component section, through new Canada partial and original component joint work, but enhances the component supporting capacity effectively, improvement normal operational performance.Enlarges the section reinforcement law construction craft simply, compatible, and has the mature design and the construction experience; Is suitable in Liang, the board, the column, the wall and the general structure concretes reinforcement; But scene construction's wet operating time is long, to produces has certain influence with the life, and after reinforcing the building clearance has certain reduction.2) Replacement concretes reinforcement lawThis law's merit with enlarges the method of sections to be close, and after reinforcing, does not affect building's clearance, but similar existence construction wet operating time long shortcoming; Is suitable somewhat low or has concretes carrier's and so on serious defect Liang, column in the compression zone concretes intensity reinforcement.3) the caking outsourcing section reinforcement lawOutside the Baotou Steel Factory reinforcement is wraps in the section or the steel plate is reinforced component's outside, outside the Baotou Steel Factory reinforces reinforced concrete Liang to use the wet outsourcing law generally, namely uses the epoxy resinification to be in the milk and so on methods with to reinforce the section the construction commission to cake a whole, after the reinforcement component, because is pulled with the compressed steel cross sectional area large scale enhancement, therefore right section supporting capacity and section rigidity large scale enhancement.This law also said that the wet outside Baotou Steel Factory reinforcement law, the stress is reliable, the construction is simple, the scene work load is small, but is big with the steel quantity, and uses in above not suitably 600C in the non-protection's situation the high temperature place; Is suitable does not allow in the use obviously to increase the original component section size, but requests to sharpen its bearing capacity large scale the concrete structure reinforcement.4) Sticks the steel reinforcement lawOutside the reinforced concrete member in bending sticks the steel reinforcement is (right section is pulled in the component supporting capacity insufficient sector area, right section compression zone or oblique section) the superficial glue steel plate, like this may enhance is reinforced component's supporting capacity, and constructs conveniently.This law construction is fast, the scene not wet work or only has the plastering and so on few wet works, to produces is small with the life influence, and after reinforcing, is not remarkable to the original structure outward appearance and the original clearance affects, but the reinforcement effect is decided to a great extent by the gummy craft and the operational level; Is suitable in the withstanding static function, and is in the normal humidity environment to bend or the tension member reinforcement.5) Glue fibre reinforcement plastic reinforcement lawOutside pastes the textile fiber reinforcement is pastes with the cementing material the fibre reinforcement compound materials in is reinforced the component to pull the region, causes it with to reinforce the section joint work, achieves sharpens the component bearing capacity the goal. Besides has glues the steel plate similar merit, but also has anticorrosive muddy, bears moistly, does not increase the self-weight of structure nearly, durably, the maintenance cost low status merit, but needs special fire protection processing, is suitable in each kind of stress nature concrete structure component and the general construction.This law's good and bad points with enlarge the method of sections to be close; Is suitable reinforcement which is insufficient in the concrete structure component oblique section supporting capacity, or must exert the crosswise binding force to the compressional member the situation.6) Reeling lawThis law's good and bad points with enlarge the method of sections to be close; Is suitable reinforcement which is insufficient in the concrete structure component oblique section supporting capacity, or must exert the crosswise binding force to the compressional member the situation.7) Fang bolt anchor lawThis law is suitable in the concretes intensity rank is the C20~C60 concretes load-bearing member transformation, the reinforcement; It is not suitable for already the above structure which and the light quality structure makes decent seriously. 1.2The indirect reinforcement's general method1)Pre-stressed reinforcement law(1)Thepre-stressed horizontal tension bar reinforces concretes member in bending,because the pre-stressed and increases the exterior load the combined action, in the tension bar has the axial tension, this strength eccentric transmits on the component through the pole end anchor (, when tension bar and Liang board bottom surface close fitting, tension bar can look for tune together with component, this fashion has partial pressures to transmit directly for component bottom surface), has the eccentric compression function in the component, this function has overcome the bending moment which outside the part the load produces, reduced outside the load effect, thus sharpened component's anti-curved ability. At the same time, because the tension bar passes to component's pressure function, the component crack development can alleviate, the control, the oblique section anti-to cut the supporting capacity also along with it enhancement.As a result of the horizontal lifting stem's function, the original component's section stress characteristic by received bends turned the eccentric compression, therefore, after the reinforcement, component's supporting capacity was mainly decided in bends under the condition the original component's supporting capacity 。
土木工程专业英语翻译(含中英)
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.,。
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7 Rigid-Frame StructuresA rigid-frame high-rise structure typically comprises parallel or orthogonally arranged bents consisting of columns and girders with moment resistant joints. Resistance to horizontal loading is provided by the bending resistance of the columns, girders, and joints. The continuity of the frame also contributes to resisting gravity loading, by reducing the moments in the girders.The advantages of a rigid frame are the simplicity and convenience of its rectangular form.Its unobstructed arrangement, clear of bracing members and structural walls, allows freedom internally for the layout and externally for the fenestration. Rigid frames are considered economical for buildings of up to' about25 stories, above which their drift resistance is costly to control. If, however,a rigid frame is combined with shear walls or cores, the resulting structure is very much stiffer so that its height potential may extend up to 50 stories or more. A flat plate structure is very similar to a rigid frame, but with slabs replacing the girders As with a rigid frame, horizontal and vertical loadings are resisted in a flat plate structure by the flexural continuity between the vertical and horizontal components.As highly redundant structures, rigid frames are designed initially on the basis of approximate analyses, after which more rigorous analyses and checks can be made. The procedure may typically include the following stages:1. Estimation of gravity load forces in girders and columns by approximate method.2. Preliminary estimate of member sizes based on gravity load forces witharbitrary increase in sizes to allow for horizontal loading.3. Approximate allocation of horizontal loading to bents and preliminary analysisof member forces in bents.4. Check on drift and adjustment of member sizes if necessary.5. Check on strength of members for worst combination of gravity and horizontalloading, and adjustment of member sizes if necessary.6. Computer analysis of total structure for more accurate check on memberstrengths and drift, with further adjustment of sizes where required. This stage may include the second-order P-Delta effects of gravity loading on the member forces and drift..7. Detailed design of members and connections.This chapter considers methods of analysis for the deflections and forces for both gravity and horizontal loading. The methods are included in roughly the order of the design procedure, with approximate methods initially and computer techniques later. Stability analyses of rigid frames are discussed in Chapter 16.7.1 RIGID FRAME BEHAVIORThe horizontal stiffness of a rigid frame is governed mainly by the bending resistance of the girders, the columns, and their connections, and, in a tall frame, by the axial rigidity of the columns. The accumulated horizontal shear above any story of a rigid frame is resisted by shear in the columns of that story (Fig. 7.1). The shear causes the story-height columns to bend in double curvature with points of contraflexure at approximately mid-story-height levels. The moments applied to a joint from the columns above and below are resisted by the attached girders, which also bend in double curvature, with points of contraflexure at approximately mid-span. These deformations of the columns and girders allow racking of the frame and horizontal deflection in each story. The overall deflected shape of a rigid frame structure due to racking has a shear configuration with concavity upwind, a maximum inclination near the base, and a minimum inclination at the top, as shown in Fig.7.1.The overall moment of the external horizontal load is resisted in each story level by the couple resulting from the axial tensile and compressive forces in the columns on opposite sides of the structure (Fig. 7.2). The extension and shortening of the columns cause overall bending and associated horizontal displacements of the structure. Because of the cumulative rotation up the height, the story drift dueto overall bending increases with height, while that due to racking tends to decrease. Consequently the contribution to story drift from overall bending may, in. the uppermost stories, exceed that from racking. The contribution of overall bending to the total drift, however, will usually not exceed 10% of that of racking, except in very tall, slender,, rigid frames. Therefore the overall deflected shape of a high-rise rigid frame usually has a shear configuration.The response of a rigid frame to gravity loading differs from a simply connected frame in the continuous behavior of the girders. Negative moments are induced adjacent to the columns, and positive moments of usually lesser magnitude occur in the mid-span regions. The continuity also causes the maximum girder moments to be sensitive to the pattern of live loading. This must be considered when estimating the worst moment conditions. For example, the gravity load maximum hogging moment adjacent to an edge column occurs when live load acts only on the edge span and alternate other spans, as for A in Fig. 7.3a. The maximum hogging moments adjacent to an interior column are caused, however, when live load acts only on the spans adjacent to the column, as for B in Fig. 7.3b. The maximum mid-span sagging moment occurs when live load acts on the span under consideration, and alternate other spans, as for spans AB and CD in Fig. 7.3a.The dependence of a rigid frame on the moment capacity of the columns for resisting horizontal loading usually causes the columns of a rigid frame to be larger than those of the corresponding fully braced simply connected frame. On the other hand, while girders in braced frames are designed for their mid-span sagging moment, girders in rigid frames are designed for the end-of-span resultant hogging moments, which may be of lesser value. Consequently, girders in a rigid frame may be smaller than in the corresponding braced frame. Such reductions in size allow economy through the lower cost of the girders and possible reductions in story heights. These benefits may be offset, however, by the higher cost of the more complex rigid connections.7.2 APPROXIMATE DETERMINATION OF MEMBER FORCES CAUSED BY GRAVITY LOADSIMGA rigid frame is a highly redundant structure; consequently, an accurate analysis can be made only after the member sizes are assigned. Initially, therefore, member sizes are decided on the basis of approximate forces estimated either by conservative formulas or by simplified methods of analysis that are independent of member properties. Two approaches for estimating girder forces due to gravity loading are given here.7.2.1 Girder Forces—Code Recommended ValuesIn rigid frames with two or more spans in which the longer of any two adjacent spans does not exceed the shorter by more than 20 %, and where the uniformly distributed design live load does not exceed three times the dead load, the girder moment and shears may be estimated from Table 7.1. This summarizes the recommendations given in the Uniform Building Code [7.1]. In other cases a conventional moment distribution or two-cycle moment distribution analysis should be made for a line of girders at a floor level.7.2.2 Two-Cycle Moment Distribution [7.2].This is a concise form of moment distribution for estimating girder moments in a continuous multibay span. It is more accurate than the formulas in Table 7.1, especially for cases of unequal spans and unequal loading in different spans.The following is assumed for the analysis:1. A counterclockwise restraining moment on the end of a girder is positive anda clockwise moment is negative.2. The ends of the columns at the floors above and below the considered girder are fixed.3. In the absence of known member sizes, distribution factors at each joint aretaken equal to 1 /n, where n is the number of members framing into the joint in the plane of the frame.Two-Cycle Moment Distribution—Worked Example. The method is demonstrated by a worked example. In Fig, 7.4, a four-span girder AE from a rigid-frame bent is shown with its loading. The fixed-end moments in each span are calculated for dead loading and total loading using the formulas given in Fig, 7.5. The moments are summarized in Table 7.2.The purpose of the moment distribution is to estimate for each support the maximum girder moments that can occur as a result of dead loading and pattern live loading.A different load combination must be considered for the maximum moment at each support, and a distribution made for each combination.The five distributions are presented separately in Table 7.3, and in a combined form in Table 7.4. Distributions a in Table 7.3 are for the exterior supports A andE. For the maximum hogging moment at A, total loading is applied to span AB with dead loading only on BC. The fixed-end moments are written in rows 1 and 2. In this distribution only .the resulting moment at A is of interest. For the first cycle, joint B is balanced with a correcting moment of- (-867 + 315)/4 = - U/4 assigned to M BA where U is the unbalanced moment. This is not recorded, but half of it, ( - U/4)/2, is carried over to M AB. This is recorded in row 3 and then added to the fixed-end moment and the result recorded in row 4.The second cycle involves the release and balance of joint A. The unbalanced moment of 936 is balanced by adding-U/3 = -936/3 = -312 to M BA (row 5), implicitly adding the same moment to the two column ends at A. This completes the second cycle of the distribution. The resulting maximum moment at A is then given by the addition of rows 4 and 5, 936 - 312 = 624. The distribution for the maximum moment at E follows a similar procedure.Distribution b in Table 7.3 is for the maximum moment at B. The most severe loading pattern for this is with total loading on spans AB and BC and dead load only on CD. The operations are similar to those in Distribution a, except that the T first cycle involves balancing the two adjacent joints A and C while recording only their carryover moments to B. In the second cycle, B is balanced by adding - (-1012 + 782)/4 = 58 to each side of B. The addition of rows 4 and 5 then gives the maximum hogging moments at B. Distributions c and d, for the moments at joints C and D, follow patterns similar to Distribution b.The complete set of operations can be combined as in Table 7.4 by initially recording at each joint the fixed-end moments for both dead and total loading. Then the joint, or joints, adjacent to the one under consideration are balanced for the appropriate combination of loading, and carryover moments assigned .to the considered joint and recorded. The joint is then balanced to complete the distribution for that support.Maximum Mid-Span Moments. The most severe loading condition for a maximum mid-span sagging moment is when the considered span and alternate other spans and total loading. A concise method of obtaining these values may be included in the combined two-cycle distribution, as shown in Table 7.5. Adopting the convention that sagging moments at mid-span are positive, a mid-span total; loading moment is calculated for the fixed-end condition of each span and entered in the mid-span column of row 2. These mid-span moments must now be corrected to allow for rotation of the joints. This is achieved by multiplying the carryover moment, row 3, at the left-hand end of the span by (1 + 0.5 D.F. )/2, and the carryover moment at the right-hand end by -(1 + 0.5 D.F.)/2, where D.F. is the appropriate distribution factor, and recording the results in the middle column. For example, the carryover to the mid-span of AB from A = [(1 + 0.5/3)/2] x 69 = 40 and from B = -[(1+ 0.5/4)/2] x (-145) = 82. These correction moments are then added to the fixed-end mid-span moment to give the maximum mid-span sagging moment, that is, 733 + 40 + 82 = 855.7.2.3 Column ForcesThe gravity load axial force in a column is estimated from the accumulated tributary dead and live floor loading above that level, with reductions in live loading as permitted by the local Code of Practice. The gravity load maximum column moment is estimated by taking the maximum difference of the end moments in the connected girders and allocating it equally between the column ends just above and below the joint. To this should be added any unbalanced moment due to eccentricity of the girderconnections from the centroid of the column, also allocated equally between the column ends above and below the joint.第七章框架结构高层框架结构一般由平行或正交布置的梁柱结构组成，梁柱结构是由带有能承担弯矩作用节点的梁、柱组成。