土木地质岩土工程专业毕业英文翻译原文和译文
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.,。
土木工程毕业设计外文翻译最终中英文
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。
土木工程专业毕业设计外文文献翻译2篇
XXXXXXXXX学院学士学位毕业设计(论文)英语翻译课题名称英语翻译学号学生专业、年级所在院系指导教师选题时间Fundamental Assumptions for Reinforced ConcreteBehaviorThe chief task of the structural engineer is the design of structures. Design is the determination of the general shape and all specific dimensions of a particular structure so that it will perform the function for which it is created and will safely withstand the influences that will act on it throughout useful life. These influences are primarily the loads and other forces to which it will be subjected, as well as other detrimental agents, such as temperature fluctuations, foundation settlements, and corrosive influences, Structural mechanics is one of the main tools in this process of design. As here understood, it is the body of scientific knowledge that permits one to predict with a good degree of certainly how a structure of give shape and dimensions will behave when acted upon by known forces or other mechanical influences. The chief items of behavior that are of practical interest are (1) the strength of the structure, i. e. , that magnitude of loads of a give distribution which will cause the structure to fail, and (2) the deformations, such as deflections and extent of cracking, that the structure will undergo when loaded under service condition.The fundamental propositions on which the mechanics of reinforced concrete is based are as follows:1.The internal forces, such as bending moments, shear forces, and normal andshear stresses, at any section of a member are in equilibrium with the effect of the external loads at that section. This proposition is not an assumption but a fact, because any body or any portion thereof can be at rest only if all forces acting on it are in equilibrium.2.The strain in an embedded reinforcing bar is the same as that of thesurrounding concrete. Expressed differently, it is assumed that perfect bonding exists between concrete and steel at the interface, so that no slip can occur between the two materials. Hence, as the one deforms, so must the other. With modern deformed bars, a high degree of mechanical interlocking is provided in addition to the natural surface adhesion, so this assumption is very close to correct.3.Cross sections that were plane prior to loading continue to be plan in themember under load. Accurate measurements have shown that when a reinforced concrete member is loaded close to failure, this assumption is not absolutely accurate. However, the deviations are usually minor.4.In view of the fact the tensile strength of concrete is only a small fraction ofits compressive strength; the concrete in that part of a member which is in tension is usually cracked. While these cracks, in well-designed members, are generally so sorrow as to be hardly visible, they evidently render the cracked concrete incapable of resisting tension stress whatever. This assumption is evidently a simplification of the actual situation because, in fact, concrete prior to cracking, as well as the concrete located between cracks, does resist tension stresses of small magnitude. Later in discussions of the resistance of reinforced concrete beams to shear, it will become apparent that under certain conditions this particular assumption is dispensed with and advantage is taken of the modest tensile strength that concrete can develop.5.The theory is based on the actual stress-strain relation ships and strengthproperties of the two constituent materials or some reasonable equivalent simplifications thereof. The fact that novelistic behavior is reflected in modern theory, that concrete is assumed to be ineffective in tension, and that the joint action of the two materials is taken into consideration results in analytical methods which are considerably more complex and also more challenging, than those that are adequate for members made of a single, substantially elastic material.These five assumptions permit one to predict by calculation the performance of reinforced concrete members only for some simple situations. Actually, the joint action of two materials as dissimilar and complicated as concrete and steel is so complex that it has not yet lent itself to purely analytical treatment. For this reason, methods of design and analysis, while using these assumptions, are very largely based on the results of extensive and continuing experimental research. They are modified and improved as additional test evidence becomes available.钢筋混凝土的基本假设作为结构工程师的主要任务是结构设计。
土木工程毕业论文中英文翻译
外文翻译班级:xxx 学号:xxx 姓名:xxx一、外文原文:Structural Systems to resist lateral loadsCommonly Used structural SystemsWith loads measured in tens of thousands kips, there is little room in the design of high-rise buildings for excessively complex thoughts. Indeed, the better high-rise buildings carry the universal traits of simplicity of thought and clarity of expression.It does not follow that there is no room for grand thoughts. Indeed, it is with such grand thoughts that the new family of high-rise buildings has evolved. Perhaps more important, the new concepts of but a few years ago have become commonplace in today’ s technology.Omitting some concepts that are related strictly to the materials of construction, the most commonly used structural systems used in high-rise buildings can be categorized as follows:1.Moment-resisting frames.2.Braced frames, including eccentrically braced frames.3.Shear walls, including steel plate shear walls.4.Tube-in-tube structures.5.Core-interactive structures.6.Cellular or bundled-tube systems.Particularly with the recent trend toward more complex forms, but in response also to the need for increased stiffness to resist the forces from wind and earthquake, most high-rise buildings have structural systems built up of combinations of frames, braced bents, shear walls, and related systems. Further, for the taller buildings, the majorities are composed of interactive elements in three-dimensional arrays.The method of combining these elements is the very essence of the design process for high-rise buildings. These combinations need evolve in response to environmental, functional, and cost considerations so as to provide efficient structures that provoke the architectural development to new heights. This is not to say that imaginative structural design can create great architecture. To the contrary, many examples of fine architecture have been created with only moderate support from the structural engineer, while only fine structure, not great architecture, can be developedwithout the genius and the leadership of a talented architect. In any event, the best of both is needed to formulate a truly extraordinary design of a high-rise building.While comprehensive discussions of these seven systems are generally available in the literature, further discussion is warranted here .The essence of the design process is distributed throughout the discussion.Moment-Resisting FramesPerhaps the most commonly used system in low-to medium-rise buildings, the moment-resisting frame, is characterized by linear horizontal and vertical members connected essentially rigidly at their joints. Such frames are used as a stand-alone system or in combination with other systems so as to provide the needed resistance to horizontal loads. In the taller of high-rise buildings, the system is likely to be found inappropriate for a stand-alone system, this because of the difficulty in mobilizing sufficient stiffness under lateral forces.Analysis can be accomplished by STRESS, STRUDL, or a host of other appropriate computer programs; analysis by the so-called portal method of the cantilever method has no place in today’s technology.Because of the intrinsic flexibility of the column/girder intersection, and because preliminary designs should aim to highlight weaknesses of systems, it is not unusual to use center-to-center dimensions for the frame in the preliminary analysis. Of course, in the latter phases of design, a realistic appraisal in-joint deformation is essential.Braced Frame sThe braced frame, intrinsically stiffer than the moment –resisting frame, finds also greater application to higher-rise buildings. The system is characterized by linear horizontal, vertical, and diagonal members, connected simply or rigidly at their joints. It is used commonly in conjunction with other systems for taller buildings and as a stand-alone system in low-to medium-rise buildings.While the use of structural steel in braced frames is common, concrete frames are more likely to be of the larger-scale variety.Of special interest in areas of high seismicity is the use of the eccentric braced frame.Again, analysis can be by STRESS, STRUDL, or any one of a series of two –or three dimensional analysis computer programs. And again, center-to-center dimensions are used commonly in the preliminary analysis.Shear wallsThe shear wall is yet another step forward along a progression of ever-stiffer structural systems. The system is characterized by relatively thin, generally (but not always) concrete elements that provide both structural strength and separation between building functions.In high-rise buildings, shear wall systems tend to have a relatively high aspect ratio, that is, their height tends to be large compared to their width. Lacking tension in the foundation system, any structural element is limited in its ability to resist overturning moment by the width of the system and by the gravity load supported by the element. Limited to a narrow overturning, One obvious use of the system, which does have the needed width, is in the exterior walls of building, where the requirement for windows is kept small.Structural steel shear walls, generally stiffened against buckling by a concrete overlay, have found application where shear loads are high. The system, intrinsically more economical than steel bracing, is particularly effective in carrying shear loads down through the taller floors in the areas immediately above grade. The system has the further advantage of having high ductility a feature of particular importance in areas of high seismicity.The analysis of shear wall systems is made complex because of the inevitable presence of large openings through these walls. Preliminary analysis can be by truss-analogy, by the finite element method, or by making use of a proprietary computer program designed to consider the interaction, or coupling, of shear walls.Framed or Braced TubesThe concept of the framed or braced or braced tube erupted into the technology with the IBM Building in Pittsburgh, but was followed immediately with the twin 110-story towers of the World Trade Center, New York and a number of other buildings .The system is characterized by three –dimensional frames, braced frames, or shear walls, forming a closed surface more or less cylindrical in nature, but of nearly any plan configuration. Because those columns that resistlateral forces are placed as far as possible from the cancroids of the system, the overall moment of inertia is increased and stiffness is very high.The analysis of tubular structures is done using three-dimensional concepts, or by two- dimensional analogy, where possible, whichever method is used, it must be capable of accounting for the effects of shear lag.The presence of shear lag, detected first in aircraft structures, is a serious limitation in the stiffness of framed tubes. The concept has limited recent applications of framed tubes to the shear of 60 stories. Designers have developed various techniques for reducing the effects of shear lag, most noticeably the use of belt trusses. This system finds application in buildings perhaps 40stories and higher. However, except for possible aesthetic considerations, belt trusses interfere with nearly every building function associated with the outside wall; the trusses are placed often at mechanical floors, mush to the disapproval of the designers of the mechanical systems. Nevertheless, as a cost-effective structural system, the belt truss works well and will likely find continued approval from designers. Numerous studies have sought to optimize the location of these trusses, with the optimum location very dependent on the number of trusses provided. Experience would indicate, however, that the location of these trusses is provided by the optimization of mechanical systems and by aesthetic considerations, as the economics of the structural system is not highly sensitive to belt truss location.Tube-in-Tube StructuresThe tubular framing system mobilizes every column in the exterior wall in resisting over-turning and shearing forces. The term‘tube-in-tube’is largely self-explanatory in that a second ring of columns, the ring surrounding the central service core of the building, is used as an inner framed or braced tube. The purpose of the second tube is to increase resistance to over turning and to increase lateral stiffness. The tubes need not be of the same character; that is, one tube could be framed, while the other could be braced.In considering this system, is important to understand clearly the difference between the shear and the flexural components of deflection, the terms being taken from beam analogy. In a framed tube, the shear component of deflection is associated with the bending deformation of columns and girders (i.e, the webs of the framed tube) while the flexural component is associated with the axial shortening and lengthening of columns (i.e, the flanges of the framed tube). In abraced tube, the shear component of deflection is associated with the axial deformation of diagonals while the flexural component of deflection is associated with the axial shortening and lengthening of columns.Following beam analogy, if plane surfaces remain plane (i.e, the floor slabs),then axial stresses in the columns of the outer tube, being farther form the neutral axis, will be substantially larger than the axial stresses in the inner tube. However, in the tube-in-tube design, when optimized, the axial stresses in the inner ring of columns may be as high, or even higher, than the axial stresses in the outer ring. This seeming anomaly is associated with differences in the shearing component of stiffness between the two systems. This is easiest to under-stand where the inner tube is conceived as a braced (i.e, shear-stiff) tube while the outer tube is conceived as a framed (i.e, shear-flexible) tube.Core Interactive StructuresCore interactive structures are a special case of a tube-in-tube wherein the two tubes are coupled together with some form of three-dimensional space frame. Indeed, the system is used often wherein the shear stiffness of the outer tube is zero. The United States Steel Building, Pittsburgh, illustrates the system very well. Here, the inner tube is a braced frame, the outer tube has no shear stiffness, and the two systems are coupled if they were considered as systems passing in a straight line from the “hat” structure. Note that the exterior columns would be improperly modeled if they were considered as systems passing in a straight line from the “hat” to the foundations; these columns are perhaps 15% stiffer as they follow the elastic curve of the braced core. Note also that the axial forces associated with the lateral forces in the inner columns change from tension to compression over the height of the tube, with the inflection point at about 5/8 of the height of the tube. The outer columns, of course, carry the same axial force under lateral load for the full height of the columns because the columns because the shear stiffness of the system is close to zero.The space structures of outrigger girders or trusses, that connect the inner tube to the outer tube, are located often at several levels in the building. The AT&T headquarters is an example of an astonishing array of interactive elements:1.The structural system is 94 ft (28.6m) wide, 196ft(59.7m) long, and 601ft (183.3m) high.2.Two inner tubes are provided, each 31ft(9.4m) by 40 ft (12.2m), centered 90 ft (27.4m)apart in the long direction of the building.3.The inner tubes are braced in the short direction, but with zero shear stiffness in the longdirection.4. A single outer tube is supplied, which encircles the building perimeter.5.The outer tube is a moment-resisting frame, but with zero shear stiffness for the center50ft(15.2m) of each of the long sides.6. A space-truss hat structure is provided at the top of the building.7. A similar space truss is located near the bottom of the building8.The entire assembly is laterally supported at the base on twin steel-plate tubes, because theshear stiffness of the outer tube goes to zero at the base of the building.Cellular structuresA classic example of a cellular structure is the Sears Tower, Chicago, a bundled tube structure of nine separate tubes. While the Sears Tower contains nine nearly identical tubes, the basic structural system has special application for buildings of irregular shape, as the several tubes need not be similar in plan shape, It is not uncommon that some of the individual tubes one of the strengths and one of the weaknesses of the system.This special weakness of this system, particularly in framed tubes, has to do with the concept of differential column shortening. The shortening of a column under load is given by the expression△=ΣfL/EFor buildings of 12 ft (3.66m) floor-to-floor distances and an average compressive stress of 15 ksi (138MPa), the shortening of a column under load is 15 (12)(12)/29,000 or 0.074in (1.9mm) per story. At 50 stories, the column will have shortened to 3.7 in. (94mm) less than its unstressed length. Where one cell of a bundled tube system is, say, 50stories high and an adjacent cell is, say, 100stories high, those columns near the boundary between .the two systems need to have this differential deflection reconciled.Major structural work has been found to be needed at such locations. In at least one building, the Rialto Project, Melbourne, the structural engineer found it necessary to vertically pre-stressthe lower height columns so as to reconcile the differential deflections of columns in close proximity with the post-tensioning of the shorter column simulating the weight to be added on to adjacent, higher columns.二、原文翻译:抗侧向荷载的结构体系常用的结构体系若已测出荷载量达数千万磅重,那么在高层建筑设计中就没有多少可以进行极其复杂的构思余地了。
土木工程毕业论文中英文翻译
外文翻译班级:xxx学号:xxx姓名:xxx一、外文原文:Structural Systems to resist lateral loads Commonly Used structural SystemsWith loads measured in tens of thousands kips, there is little room in the design of high-rise buildings for excessively complex thoughts. Indeed, the better high-rise buildings carry the universal traits of simplicity of thought and clarity of expression.It does not follow that there is no room for grand thoughts. Indeed, it is with such grand thoughts that the new family of high-rise buildings has evolved. Perhaps more important, the new concepts of but a few years ago have become commonplace in today’ s technology.Omitting some concepts that are related strictly to the materials of construction, the most commonly used structural systems used in high-rise buildings can be categorized as follows:1.Moment-resisting frames.2.Braced frames, including eccentrically braced frames.3.Shear walls, including steel plate shear walls.4.Tube-in-tube structures.5.Core-interactive structures.6.Cellular or bundled-tube systems.Particularly with the recent trend toward more complex forms, but in response also to the need for increased stiffness to resist the forces from wind and earthquake, most high-rise buildings have structural systems built up of combinations of frames, braced bents, shear walls, and related systems. Further, for the taller buildings, the majorities are composed of interactive elements in three-dimensional arrays.The method of combining these elements is the very essence of the design process for high-rise buildings. These combinations need evolve in response to environmental, functional, and cost considerations so as to provide efficient structures that provoke the architectural development to new heights. This is not to say that imaginative structural design can create great architecture. To the contrary, many examples of fine architecture have been created with only moderate support from the structural engineer, while only fine structure, not great architecture, can be developed without the genius and the leadership of a talented architect. In any event, the best of both is needed to formulate a truly extraordinary design of a high-rise building.While comprehensive discussions of these seven systems are generally available in the literature, further discussion is warranted here .The essence of the design process is distributed throughout the discussion.Moment-Resisting FramesPerhaps the most commonly used system in low-to medium-rise buildings, the moment-resisting frame, is characterized by linear horizontal and vertical members connected essentially rigidly at their joints. Such frames are used as a stand-alone system or in combination with other systems so as to provide the needed resistance to horizontal loads. In the taller of high-rise buildings, the system is likely to be found inappropriate for a stand-alone system, this because of the difficulty in mobilizing sufficient stiffness under lateral forces.Analysis can be accomplished by STRESS, STRUDL, or a host of other appropriate computer programs; analysis by the so-called portal method of the cantilever method has no place in today’s technology.Because of the intrinsic flexibility of the column/girder intersection, and because preliminary designs should aim to highlight weaknesses of systems, it is not unusual to use center-to-center dimensions for the frame in the preliminary analysis. Of course, in the latter phases of design, a realistic appraisal in-joint deformation is essential.Braced Frame sThe braced frame, intrinsically stiffer than the moment –resisting frame, finds also greater application to higher-rise buildings. The system is characterized by linear horizontal, vertical, and diagonal members, connected simply or rigidly at their joints. It is used commonly inconjunction with other systems for taller buildings and as a stand-alone system in low-to medium-rise buildings.While the use of structural steel in braced frames is common, concrete frames are more likely to be of the larger-scale variety.Of special interest in areas of high seismicity is the use of the eccentric braced frame.Again, analysis can be by STRESS, STRUDL, or any one of a series of two –or three dimensional analysis computer programs. And again, center-to-center dimensions are used commonly in the preliminary analysis. Shear wallsThe shear wall is yet another step forward along a progression of ever-stiffer structural systems. The system is characterized by relatively thin, generally but not always concrete elements that provide both structural strength and separation between building functions.In high-rise buildings, shear wall systems tend to have a relatively high aspect ratio, that is, their height tends to be large compared to their width. Lacking tension in the foundation system, any structural element is limited in its ability to resist overturning moment by the width of the system and by the gravity load supported by the element. Limited to a narrow overturning, One obvious use of the system, which does have the needed width, is in the exterior walls of building, where the requirement for windows is kept small.Structural steel shear walls, generally stiffened against buckling by a concrete overlay, have found application where shear loads are high. The system, intrinsically more economical than steel bracing, is particularly effective in carrying shear loads down through the taller floors in the areas immediately above grade. The system has the further advantage of having high ductility a feature of particular importance in areas of high seismicity.The analysis of shear wall systems is made complex because of the inevitable presence of large openings through these walls. Preliminary analysis can be by truss-analogy, by the finite element method, or by making use of a proprietary computer program designed to consider the interaction, or coupling, of shear walls.Framed or Braced TubesThe concept of the framed or braced or braced tube erupted into the technology with the IBM Building in Pittsburgh, but was followed immediately with the twin 110-story towers of the World Trade Center, New York and a number of other buildings .The system is characterized by three –dimensional frames, braced frames, or shear walls, forming a closed surface more or less cylindrical in nature, but of nearly any plan configuration. Because those columns that resist lateral forces are placed as far as possible from the cancroids of the system, the overall moment of inertia is increased and stiffness is very high.The analysis of tubular structures is done using three-dimensional concepts, or by two- dimensional analogy, where possible, whichever method is used, it must be capable of accounting for the effects of shear lag.The presence of shear lag, detected first in aircraft structures, is a serious limitation in the stiffness of framed tubes. The concept has limited recent applications of framed tubes to the shear of 60 stories. Designers have developed various techniques for reducing the effects of shear lag, most noticeably the use of belt trusses. This system finds application in buildings perhaps 40stories and higher. However, except for possible aesthetic considerations, belt trusses interfere with nearly every building function associated with the outside wall; the trusses are placed often at mechanical floors, mush to the disapproval of the designers of the mechanical systems. Nevertheless, as a cost-effective structural system, the belt truss works well and will likely find continued approval from designers. Numerous studies have sought to optimize the location of these trusses, with the optimum location very dependent on the number of trusses provided. Experience would indicate, however, that the location of these trusses is provided by the optimization of mechanical systems and by aesthetic considerations, as the economics of the structural system is not highly sensitive to belt truss location.Tube-in-Tube StructuresThe tubular framing system mobilizes every column in the exterior wallin resisting over-turning and shearing forces. The term‘tube-in-tube’is largely self-explanatory in that a second ring of columns, the ring surrounding the central service core of the building, is used as an inner framed or braced tube. The purpose of the second tube is to increase resistance to over turning and to increase lateral stiffness. The tubes need not be of the same character; that is, one tube could be framed, while the other could be braced.In considering this system, is important to understand clearly the difference between the shear and the flexural components of deflection, the terms being taken from beam analogy. In a framed tube, the shear component of deflection is associated with the bending deformation of columns and girders , the webs of the framed tube while the flexural component is associated with the axial shortening and lengthening of columns , the flanges of the framed tube. In a braced tube, the shear component of deflection is associated with the axial deformation of diagonals while the flexural component of deflection is associated with the axial shortening and lengthening of columns.Following beam analogy, if plane surfaces remain plane , the floor slabs,then axial stresses in the columns of the outer tube, being farther form the neutral axis, will be substantially larger than the axial stresses in the inner tube. However, in the tube-in-tube design, when optimized, the axial stresses in the inner ring of columns may be as high, or evenhigher, than the axial stresses in the outer ring. This seeming anomaly is associated with differences in the shearing component of stiffness between the two systems. This is easiest to under-stand where the inner tube is conceived as a braced , shear-stiff tube while the outer tube is conceived as a framed , shear-flexible tube.Core Interactive StructuresCore interactive structures are a special case of a tube-in-tube wherein the two tubes are coupled together with some form of three-dimensional space frame. Indeed, the system is used often wherein the shear stiffness of the outer tube is zero. The United States Steel Building, Pittsburgh, illustrates the system very well. Here, the inner tube is a braced frame, the outer tube has no shear stiffness, and the two systems are coupled if they were considered as systems passing in a straight line from the “hat” structure. Note that the exterior columns would be improperly modeled if they were considered as systems passing in a straight line from the “hat” to the foundations; these columns are perhaps 15% stiffer as they follow the elastic curve of the braced core. Note also that the axial forces associated with the lateral forces in the inner columns change from tension to compression over the height of the tube, with the inflection point at about 5/8 of the height of the tube. The outer columns, of course, carry the same axial force under lateral load for the full height of the columns because the columns because the shearstiffness of the system is close to zero.The space structures of outrigger girders or trusses, that connect the inner tube to the outer tube, are located often at several levels in the building. The AT&T headquarters is an example of an astonishing array of interactive elements:1.The structural system is 94 ft wide, 196ft long, and 601ft high.2.Two inner tubes are provided, each 31ft by 40 ft , centered 90 ft apartin the long direction of the building.3.The inner tubes are braced in the short direction, but with zero shearstiffness in the long direction.4.A single outer tube is supplied, which encircles the buildingperimeter.5.The outer tube is a moment-resisting frame, but with zero shearstiffness for the center50ft of each of the long sides.6.A space-truss hat structure is provided at the top of the building.7.A similar space truss is located near the bottom of the building8.The entire assembly is laterally supported at the base on twinsteel-plate tubes, because the shear stiffness of the outer tube goes to zero at the base of the building.Cellular structuresA classic example of a cellular structure is the Sears Tower, Chicago,a bundled tube structure of nine separate tubes. While the Sears Towercontains nine nearly identical tubes, the basic structural system has special application for buildings of irregular shape, as the several tubes need not be similar in plan shape, It is not uncommon that some of the individual tubes one of the strengths and one of the weaknesses of the system.This special weakness of this system, particularly in framed tubes, has to do with the concept of differential column shortening. The shortening of a column under load is given by the expression△=ΣfL/EFor buildings of 12 ft floor-to-floor distances and an average compressive stress of 15 ksi 138MPa, the shortening of a column under load is 15 1212/29,000 or per story. At 50 stories, the column will have shortened to in. 94mm less than its unstressed length. Where one cell of a bundled tube system is, say, 50stories high and an adjacent cell is, say, 100stories high, those columns near the boundary between .the two systems need to have this differential deflection reconciled.Major structural work has been found to be needed at such locations. In at least one building, the Rialto Project, Melbourne, the structural engineer found it necessary to vertically pre-stress the lower height columns so as to reconcile the differential deflections of columns in close proximity with the post-tensioning of the shorter column simulatingthe weight to be added on to adjacent, higher columns.二、原文翻译:抗侧向荷载的结构体系常用的结构体系若已测出荷载量达数千万磅重,那么在高层建筑设计中就没有多少可以进行极其复杂的构思余地了;确实,较好的高层建筑普遍具有构思简单、表现明晰的特点;这并不是说没有进行宏观构思的余地;实际上,正是因为有了这种宏观的构思,新奇的高层建筑体系才得以发展,可能更重要的是:几年以前才出现的一些新概念在今天的技术中已经变得平常了;如果忽略一些与建筑材料密切相关的概念不谈,高层建筑里最为常用的结构体系便可分为如下几类:1.抗弯矩框架;2.支撑框架,包括偏心支撑框架;3.剪力墙,包括钢板剪力墙;4.筒中框架;5.筒中筒结构;6.核心交互结构;7.框格体系或束筒体系;特别是由于最近趋向于更复杂的建筑形式,同时也需要增加刚度以抵抗几力和地震力,大多数高层建筑都具有由框架、支撑构架、剪力墙和相关体系相结合而构成的体系;而且,就较高的建筑物而言,大多数都是由交互式构件组成三维陈列;将这些构件结合起来的方法正是高层建筑设计方法的本质;其结合方式需要在考虑环境、功能和费用后再发展,以便提供促使建筑发展达到新高度的有效结构;这并不是说富于想象力的结构设计就能够创造出伟大建筑;正相反,有许多例优美的建筑仅得到结构工程师适当的支持就被创造出来了,然而,如果没有天赋甚厚的建筑师的创造力的指导,那么,得以发展的就只能是好的结构,并非是伟大的建筑;无论如何,要想创造出高层建筑真正非凡的设计,两者都需要最好的;虽然在文献中通常可以见到有关这七种体系的全面性讨论,但是在这里还值得进一步讨论;设计方法的本质贯穿于整个讨论;设计方法的本质贯穿于整个讨论中;抗弯矩框架抗弯矩框架也许是低,中高度的建筑中常用的体系,它具有线性水平构件和垂直构件在接头处基本刚接之特点;这种框架用作独立的体系,或者和其他体系结合起来使用,以便提供所需要水平荷载抵抗力;对于较高的高层建筑,可能会发现该本系不宜作为独立体系,这是因为在侧向力的作用下难以调动足够的刚度;我们可以利用STRESS,STRUDL 或者其他大量合适的计算机程序进行结构分析;所谓的门架法分析或悬臂法分析在当今的技术中无一席之地,由于柱梁节点固有柔性,并且由于初步设计应该力求突出体系的弱点,所以在初析中使用框架的中心距尺寸设计是司空惯的;当然,在设计的后期阶段,实际地评价结点的变形很有必要;支撑框架支撑框架实际上刚度比抗弯矩框架强,在高层建筑中也得到更广泛的应用;这种体系以其结点处铰接或则接的线性水平构件、垂直构件和斜撑构件而具特色,它通常与其他体系共同用于较高的建筑,并且作为一种独立的体系用在低、中高度的建筑中;尤其引人关注的是,在强震区使用偏心支撑框架;此外,可以利用STRESS,STRUDL,或一系列二维或三维计算机分析程序中的任何一种进行结构分析;另外,初步分析中常用中心距尺寸;剪力墙剪力墙在加强结构体系刚性的发展过程中又前进了一步;该体系的特点是具有相当薄的,通常是而不总是混凝土的构件,这种构件既可提供结构强度,又可提供建筑物功能上的分隔;在高层建筑中,剪力墙体系趋向于具有相对大的高宽经,即与宽度相比,其高度偏大;由于基础体系缺少应力,任何一种结构构件抗倾覆弯矩的能力都受到体系的宽度和构件承受的重力荷载的限制;由于剪力墙宽度狭狭窄受限,所以需要以某种方式加以扩大,以便提从所需的抗倾覆能力;在窗户需要量小的建筑物外墙中明显地使用了这种确有所需要宽度的体系;钢结构剪力墙通常由混凝土覆盖层来加强以抵抗失稳,这在剪切荷载大的地方已得到应用;这种体系实际上比钢支撑经济,对于使剪切荷载由位于地面正上方区域内比较高的楼层向下移特别有效;这种体系还具有高延性之优点,这种特性在强震区特别重要;由于这些墙内必然出同一些大孔,使得剪力墙体系分析变得错综复杂;可以通过桁架模似法、有限元法,或者通过利用为考虑剪力墙的交互作用或扭转功能设计的专门计处机程序进行初步分析框架或支撑式筒体结构:框架或支撑式筒体最先应用于IBM公司在Pittsburgh的一幢办公楼,随后立即被应用于纽约双子座的110层世界贸易中心摩天大楼和其他的建筑中;这种系统有以下几个显着的特征:三维结构、支撑式结构、或由剪力墙形成的一个性质上差不多是圆柱体的闭合曲面,但又有任意的平面构成;由于这些抵抗侧向荷载的柱子差不多都被设置在整个系统的中心,所以整体的惯性得到提高,刚度也是很大的;在可能的情况下,通过三维概念的应用、二维的类比,我们可以进行筒体结构的分析;不管应用那种方法,都必须考虑剪力滞后的影响;这种最先在航天器结构中研究的剪力滞后出现后,对筒体结构的刚度是一个很大的限制;这种观念已经影响了筒体结构在60层以上建筑中的应用;设计者已经开发出了很多的技术,用以减小剪力滞后的影响,这其中最有名的是桁架的应用;框架或支撑式筒体在40层或稍高的建筑中找到了自己的用武之地;除了一些美观的考虑外,桁架几乎很少涉及与外墙联系的每个建筑功能,而悬索一般设置在机械的地板上,这就令机械体系设计师们很不赞成;但是,作为一个性价比较好的结构体系,桁架能充分发挥它的性能,所以它会得到设计师们持续的支持;由于其最佳位置正取决于所提供的桁架的数量,因此很多研究已经试图完善这些构件的位置;实验表明:由于这种结构体系的经济性并不十分受桁架位置的影响,所以这些桁架的位置主要取决于机械系统的完善,审美的要求,筒中筒结构:筒体结构系统能使外墙中的柱具有灵活性,用以抵抗颠覆和剪切力;“筒中筒”这个名字顾名思义就是在建筑物的核心承重部分又被包围了第二层的一系列柱子,它们被当作是框架和支撑筒来使用;配置第二层柱的目的是增强抗颠覆能力和增大侧移刚度;这些筒体不是同样的功能,也就是说,有些筒体是结构的,而有些筒体是用来支撑的;在考虑这种筒体时,清楚的认识和区别变形的剪切和弯曲分量是很重要的,这源于对梁的对比分析;在结构筒中,剪切构件的偏角和柱、纵梁例如:结构筒中的网等的弯曲有关,同时,弯曲构件的偏角取决于柱子的轴心压缩和延伸例如:结构筒的边缘等;在支撑筒中,剪切构件的偏角和对角线的轴心变形有关,而弯曲构件的偏角则与柱子的轴心压缩和延伸有关;根据梁的对比分析,如果平面保持原形例如:厚楼板,那么外层筒中柱的轴心压力就会与中心筒柱的轴心压力相差甚远,而且稳定的大于中心筒;但是在筒中筒结构的设计中,当发展到极限时,内部轴心压力会很高的,甚至远远大于外部的柱子;这种反常的现象是由于两种体系中的剪切构件的刚度不同;这很容易去理解,内筒可以看成是一个支撑或者说是剪切刚性的筒,而外筒可以看成是一个结构或者说是剪切弹性的筒;核心交互式结构:核心交互式结构属于两个筒与某些形式的三维空间框架相配合的筒中筒特殊情况;事实上,这种体系常用于那种外筒剪切刚度为零的结构;位于Pittsburgh的美国钢铁大楼证实了这种体系是能很好的工作的;在核心交互式结构中,内筒是一个支撑结构,外筒没有任何剪切刚度,而且两种结构体系能通过一个空间结构或“帽”式结构共同起作用;需要指出的是,如果把外部的柱子看成是一种从“帽”到基础的直线体系,这将是不合适的;根据支撑核心的弹性曲线,这些柱子只发挥了刚度的15%;同样需要指出的是,内柱中与侧向力有关的轴向力沿筒高度由拉力变为压力,同时变化点位于筒高度的约5/8处;当然,外柱也传递相同的轴向力,这种轴向力低于作用在整个柱子高度的侧向荷载,因为这个体系的剪切刚度接近于零;把内外筒相连接的空间结构、悬臂梁或桁架经常遵照一些规范来布置;美国电话电报总局就是一个布置交互式构件的生动例子;1、结构体系长米,宽米,高米;2、布置了两个筒,每个筒的尺寸是米×米,在长方向上有米的间隔;3、在短方向上内筒被支撑起来,但是在长方向上没有剪切刚度;4、环绕着建筑物布置了一个外筒;5、外筒是一个瞬时抵抗结构,但是在每个长方向的中心米都没有剪切刚度;6、在建筑的顶部布置了一个空间桁架构成的“帽式”结构;7、在建筑的底部布置了一个相似的空间桁架结构;8、由于外筒的剪切刚度在建筑的底部接近零,整个建筑基本上由两个钢板筒来支持;框格体系或束筒体系结构:位于美国芝加哥的西尔斯大厦是箱式结构的经典之作,它由九个相互独立的筒组成的一个集中筒;由于西尔斯大厦包括九个几乎垂直的筒,而且筒在平面上无须相似,基本的结构体系在不规则形状的建筑中得到特别的应用;一些单个的筒高于建筑一点或很多是很常见的;事实上,这种体系的重要特征就在于它既有坚固的一面,也有脆弱的一面;这种体系的脆弱,特别是在结构筒中,与柱子的压缩变形有很大的关系,柱子的压缩变形有下式计算:△=ΣfL/E对于那些层高为米左右和平均压力为138MPa的建筑,在荷载作用下每层柱子的压缩变形为1512/29000或毫米;在第50层柱子会压缩94毫米,小于它未受压的长度;这些柱子在50层的时候和100层的时候的变形是不一样的,位于这两种体系之间接近于边缘的那些柱需要使这种不均匀的变形得以调解;主要的结构工作都集中在布置中;在Melbourne的Rialto项目中,结构工程师发现至少有一幢建筑,很有必要垂直预压低高度的柱子,以便使柱不均匀的变形差得以调解,调解的方法近似于后拉伸法,即较短的柱转移重量到较高的邻柱上;。
(完整版)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。
土木工程毕业设计外文翻译原文+翻译
The bridge crack produced the reason to simply analyseIn recent years, the traffic capital construction of our province gets swift and violent development, all parts have built a large number of concrete bridges. In the course of building and using in the bridge, relevant to influence project quality lead of common occurrence report that bridge collapse even because the crack appears The concrete can be said to " often have illness coming on " while fracturing and " frequently-occurring disease ", often perplex bridge engineers and technicians. In fact , if take certain design and construction measure, a lot of cracks can be overcome and controlled. For strengthen understanding of concrete bridge crack further, is it prevent project from endanger larger crack to try one's best, this text make an more overall analysis , summary to concrete kind and reason of production , bridge of crack as much as possible, in order to design , construct and find out the feasible method which control the crack , get the result of taking precautions against Yu WeiRan.Concrete bridge crack kind, origin cause of formation In fact, the origin cause of formation of the concrete structure crack is complicated and various, even many kinds of factors influence each other , but every crack has its one or several kinds of main reasons produced . The kind of the concrete bridge crack, on its reason to produce, can roughly divide several kinds as follows :(1) load the crack caused Concrete in routine quiet .Is it load to move and crack that produce claim to load the crack under the times of stress bridge, summing up has direct stress cracks , two kinds stress crack onces mainly. Direct stress crack refer to outside load direct crack that stress produce that cause. The reason why the crack produces is as follows, 1, Design the stage of calculating , does not calculate or leaks and calculates partly while calculating in structure; Calculate the model is unreasonable; The structure is supposed and accorded with by strength actually by strength ; Load and calculate or leak and calculate few; Internal force and matching the mistake in computation of muscle; Safety coefficient ofstructure is not enough. Do not consider the possibility that construct at the time of the structural design; It is insufficient to design the section; It is simply little and assigning the mistake for reinforcing bar to set up; Structure rigidity is insufficient; Construct and deal with improperly; The design drawing can not be explained clearly etc.. 2, Construction stage, does not pile up and construct the machines , material limiting ; Is it prefabricate structure structure receive strength characteristic , stand up , is it hang , transport , install to get up at will to understand; Construct not according to the design drawing, alter the construction order of the structure without authorization , change the structure and receive the strength mode; Do not do the tired intensity checking computations under machine vibration and wait to the structure . 3, Using stage, the heavy-duty vehicle which goes beyond the design load passes the bridge; Receive the contact , striking of the vehicle , shipping; Strong wind , heavy snow , earthquake happen , explode etc.. Stress crack once means the stress of secondary caused by loading outside produces the crack. The reason why the crack produces is as follows, 1, In design outside load function , because actual working state and routine , structure of thing calculate have discrepancy or is it consider to calculate, thus cause stress once to cause the structure to fracture in some position. Two is it join bridge arch foot is it is it assign " X " shape reinforcing bar , cut down this place way , section of size design and cut with scissors at the same time to adopt often to design to cut with scissors, theory calculate place this can store curved square in , but reality should is it can resist curved still to cut with scissors, so that present the crack and cause the reinforcing bar corrosion. 2, Bridge structure is it dig trough , turn on hole , set up ox leg ,etc. to need often, difficult to use a accurate one diagrammatic to is it is it calculate to imitate to go on in calculating in routine, set up and receive the strength reinforcing bar in general foundation experience. Studies have shown , after being dug the hole by the strength component , it will produce the diffraction phenomenon that strength flows, intensive near the hole in a utensil, produced the enormous stress to concentrate. In long to step prestressing force of the continuous roof beam , often block the steel bunch accordingto the needs of section internal force in stepping, set up the anchor head, but can often see the crack in the anchor firm section adjacent place. So if deal with improper, in corner or component form sudden change office , block place to be easy to appear crack strength reinforcing bar of structure the. In the actual project, stress crack once produced the most common reason which loads the crack. Stress crack once belong to one more piece of nature of drawing , splitting off , shearing. Stress crack once is loaded and caused, only seldom calculate according to the routine too, but with modern to calculate constant perfection of means, times of stress crack to can accomplish reasonable checking computations too. For example to such stresses 2 times of producing as prestressing force , creeping ,etc., department's finite element procedure calculates levels pole correctly now, but more difficult 40 years ago. In the design, should pay attention to avoiding structure sudden change (or section sudden change), when it is unable to avoid , should do part deal with , corner for instance, make round horn , sudden change office make into the gradation zone transition, is it is it mix muscle to construct to strengthen at the same time, corner mix again oblique to reinforcing bar , as to large hole in a utensil can set up protecting in the perimeter at the terms of having angle steel. Load the crack characteristic in accordance with loading differently and presenting different characteristics differently. The crack appear person who draw more, the cutting area or the serious position of vibration. Must point out , is it get up cover or have along keep into short crack of direction to appear person who press, often the structure reaches the sign of bearing the weight of strength limit, it is an omen that the structure is destroyed, its reason is often that sectional size is partial and small. Receive the strength way differently according to the structure, the crack characteristic produced is as follows: 1, The centre is drawn. The crack runs through the component cross section , the interval is equal on the whole , and is perpendicular to receiving the strength direction. While adopting the whorl reinforcing bar , lie in the second-class crack near the reinforcing bar between the cracks. 2, The centre is pressed. It is parallel on the short and dense parallel crack which receive the strength direction toappear along the component. 3, Receive curved. Most near the large section from border is it appear and draw into direction vertical crack to begin person who draw curved square, and develop toward neutralization axle gradually. While adopting the whorl reinforcing bar , can see shorter second-class crack among the cracks. When the structure matches muscles less, there are few but wide cracks, fragility destruction may take place in the structure 4, Pressed big and partial. Heavy to press and mix person who draw muscle a less one light to pigeonhole into the component while being partial while being partial, similar to receiving the curved component. 5, Pressed small and partial. Small to press and mix person who draw muscle a more one heavy to pigeonhole into the component while being partial while being partial, similar to the centre and pressed the component. 6, Cut. Press obliquly when the hoop muscle is too dense and destroy, the oblique crack which is greater than 45?? direction appears along the belly of roof beam end; Is it is it is it destroy to press to cut to happen when the hoop muscle is proper, underpart is it invite 45?? direction parallel oblique crack each other to appear along roof beam end. 7, Sprained. Component one side belly appear many direction oblique crack, 45?? of treaty, first, and to launch with spiral direction being adjoint. 8, Washed and cut. 4 side is it invite 45?? direction inclined plane draw and split to take place along column cap board, form the tangent plane of washing. 9, Some and is pressed. Some to appear person who press direction roughly parallel large short cracks with pressure.(2) crack caused in temperature changeThe concrete has nature of expanding with heat and contract with cold, look on as the external environment condition or the structure temperature changes, concrete take place out of shape, if out of shape to restrain from, produce the stress in the structure, produce the temperature crack promptly when exceeding concrete tensile strength in stress. In some being heavy to step foot-path among the bridge , temperature stress can is it go beyond living year stress even to reach. The temperature crack distinguishes the main characteristic of other cracks will be varied with temperature and expanded orclosed up. The main factor is as follows, to cause temperature and change 1, Annual difference in temperature. Temperature is changing constantly in four seasons in one year, but change relatively slowly, the impact on structure of the bridge is mainly the vertical displacement which causes the bridge, can prop up seat move or set up flexible mound ,etc. not to construct measure coordinate , through bridge floor expansion joint generally, can cause temperature crack only when the displacement of the structure is limited, for example arched bridge , just bridge etc. The annual difference in temperature of our country generally changes the range with the conduct of the average temperature in the moon of January and July. Considering the creep characteristic of the concrete, the elastic mould amount of concrete should be considered rolling over and reducing when the internal force of the annual difference in temperature is calculated. 2, Rizhao. After being tanned by the sun by the sun to the side of bridge panel , the girder or the pier, temperature is obviously higher than other position, the temperature gradient is presented and distributed by the line shape . Because of restrain oneself function, cause part draw stress to be relatively heavy, the crack appears. Rizhao and following to is it cause structure common reason most , temperature of crack to lower the temperature suddenly 3, Lower the temperature suddenly. Fall heavy rain , cold air attack , sunset ,etc. can cause structure surface temperature suddenly dropped suddenly, but because inside temperature change relatively slow producing temperature gradient. Rizhao and lower the temperature internal force can adopt design specification or consult real bridge materials go on when calculating suddenly, concrete elastic mould amount does not consider converting into and reducing 4, Heat of hydration. Appear in the course of constructing, the large volume concrete (thickness exceeds 2. 0), after building because cement water send out heat, cause inside very much high temperature, the internal and external difference in temperature is too large, cause the surface to appear in the crack. Should according to actual conditions in constructing, is it choose heat of hydration low cement variety to try one's best, limit cement unit's consumption, reduce the aggregate and enter the temperature of the mould , reduce the internal andexternal difference in temperature, and lower the temperature slowly , can adopt the circulation cooling system to carry on the inside to dispel the heat in case of necessity, or adopt the thin layer and build it in succession in order to accelerate dispelling the heat. 5, The construction measure is improper at the time of steam maintenance or the winter construction , the concrete is sudden and cold and sudden and hot, internal and external temperature is uneven , apt to appear in the crack. 6, Prefabricate T roof beam horizontal baffle when the installation , prop up seat bury stencil plate with transfer flat stencil plate when welding in advance, if weld measure to be improper, iron pieces of nearby concrete easy to is it fracture to burn. Adopt electric heat piece draw law piece draw prestressing force at the component , prestressing force steel temperature can rise to 350 degrees Centigrade , the concrete component is apt to fracture. Experimental study indicates , are caused the intensity of concrete that the high temperature burns to obviously reduce with rising of temperature by such reasons as the fire ,etc., glueing forming the decline thereupon of strength of reinforcing bar and concrete, tensile strength drop by 50% after concrete temperature reaches 300 degrees Centigrade, compression strength drops by 60%, glueing the strength of forming to drop by 80% of only round reinforcing bar and concrete; Because heat, concrete body dissociate ink evaporate and can produce and shrink sharply in a large amount(3) shrink the crack causedIn the actual project, it is the most common because concrete shrinks the crack caused. Shrink kind in concrete, plasticity shrink is it it shrinks (is it contract to do ) to be the main reason that the volume of concrete out of shape happens to shrink, shrink spontaneously in addition and the char shrink. Plasticity shrink. About 4 hours after it is built that in the course of constructing , concrete happens, the cement water response is fierce at this moment, the strand takes shape gradually, secrete water and moisture to evaporate sharply, the concrete desiccates and shrinks, it is at the same time conduct oneself with dignity not sinking because aggregate,so when harden concrete yet,it call plasticity shrink. The plasticity shrink producing amount grade is very big, can be up toabout 1%. If stopped by the reinforcing bar while the aggregate sinks, form the crack along the reinforcing bar direction. If web , roof beam of T and roof beam of case and carry baseplate hand over office in component vertical to become sectional place, because sink too really to superficial obeying the web direction crack will happen evenly before hardenning. For reducing concrete plasticity shrink,it should control by water dust when being construct than,last long-time mixing, unloading should not too quick, is it is it take closely knit to smash to shake, vertical to become sectional place should divide layer build. Shrink and shrink (do and contract). After the concrete is formed hard , as the top layer moisture is evaporated progressively , the humidity is reduced progressively , the volume of concrete is reduced, is called and shrunk to shrink (do and contract). Because concrete top layer moisture loss soon, it is slow for inside to lose, produce surface shrink heavy , inside shrink a light one even to shrink, it is out of shape to restrain from by the inside concrete for surface to shrink, cause the surface concrete to bear pulling force, when the surface concrete bears pulling force to exceed its tensile strength, produce and shrink the crack. The concrete hardens after-contraction to just shrink and shrink mainly .Such as mix muscle rate heavy component (exceed 3% ), between reinforcing bar and more obvious restraints relatively that concrete shrink, the concrete surface is apt to appear in the full of cracks crackle. Shrink spontaneously. Spontaneous to it shrinks to be concrete in the course of hardenning , cement and water take place ink react, the shrink with have nothing to do by external humidity, and can positive (whether shrink, such as ordinary portland cement concrete), can negative too (whether expand, such as concrete, concrete of slag cement and cement of fly ash). The char shrinks. Between carbon dioxide and hyrate of cement of atmosphere take place out of shape shrink that chemical reaction cause. The char shrinks and could happen only about 50% of humidity, and accelerate with increase of the density of the carbon dioxide. The char shrinks and seldom calculates . The characteristic that the concrete shrinks the crack is that the majority belongs to the surface crack, the crack is relatively detailed in width , and criss-cross, become the fullof cracks form , the form does not have any law . Studies have shown , influence concrete shrink main factor of crack as follows, 1, Variety of cement , grade and consumption. Slag cement , quick-hardening cement , low-heat cement concrete contractivity are relatively high, ordinary cement , volcanic ash cement , alumina cement concrete contractivity are relatively low. Cement grade low in addition, unit volume consumption heavy rubing detailed degree heavy, then the concrete shrinks the more greatly, and shrink time is the longer. For example, in order to improve the intensity of the concrete , often adopt and increase the cement consumption method by force while constructing, the result shrinks the stress to obviously strengthen . 2, Variety of aggregate. Such absorbing water rates as the quartz , limestone , cloud rock , granite , feldspar ,etc. are smaller, contractivity is relatively low in the aggregate; And such absorbing water rates as the sandstone , slate , angle amphibolite ,etc. are greater, contractivity is relatively high. Aggregate grains of foot-path heavy to shrink light in addition, water content big to shrink the larger. 3, Water gray than. The heavier water consumption is, the higher water and dust are, the concrete shrinks the more greatly. 4, Mix the pharmaceutical outside. It is the better to mix pharmaceutical water-retaining property outside, then the concrete shrinks the smaller. 5, Maintain the method . Water that good maintenance can accelerate the concrete reacts, obtain the intensity of higher concrete. Keep humidity high , low maintaining time to be the longer temperature when maintaining, then the concrete shrinks the smaller. Steam maintain way than maintain way concrete is it take light to shrink naturall. 6, External environment. The humidity is little, the air drying , temperature are high, the wind speed is large in the atmosphere, then the concrete moisture is evaporated fast, the concrete shrinks the faster. 7, Shake and smash the way and time. Machinery shake way of smashing than make firm by ramming or tamping way concrete contractivity take little by hand. Shaking should determine according to mechanical performance to smash time , are generally suitable for 55s / time. It is too short, shake and can not smash closely knit , it is insufficient or not even in intensity to form the concrete; It is too long, cause and divide storey, thickaggregate sinks to the ground floor, the upper strata that the detailed aggregate stays, the intensity is not even , the upper strata incident shrink the crack. And shrink the crack caused to temperature, worthy of constructing the reinforcing bar againing can obviously improve the resisting the splitting of concrete , structure of especially thin wall (thick 200cm of wall ). Mix muscle should is it adopt light diameter reinforcing bar (8 |? construct 14 |? ) to have priority , little interval assign (whether @ 10 construct @ 15cm ) on constructing, the whole section is it mix muscle to be rate unsuitable to be lower than 0 to construct. 3%, can generally adopt 0 . 3%~0. 5%.(4), crack that causes out of shape of plinth of the groundBecause foundation vertical to even to subside or horizontal direction displacement, make the structure produce the additional stress, go beyond resisting the ability of drawing of concrete structure, cause the structure to fracture. The even main reason that subside of the foundation is as follows, 1, Reconnoitres the precision and is not enough for , test the materials inaccuratly in geology. Designing, constructing without fully grasping the geological situation, this is the main reason that cause the ground not to subside evenly . Such as hills area or bridge, district of mountain ridge,, hole interval to be too far when reconnoitring, and ground rise and fall big the rock, reconnoitring the report can't fully reflect the real geological situation . 2, The geological difference of the ground is too large. Building it in the bridge of the valley of the ditch of mountain area, geology of the stream place and place on the hillside change larger, even there are weak grounds in the stream, because the soil of the ground does not causes and does not subside evenly with the compressing. 3, The structure loads the difference too big. Under the unanimous terms, when every foundation too heavy to load difference in geological situation, may cause evenly to subside, for example high to fill out soil case shape in the middle part of the culvert than to is it take heavy to load both sides, to subside soon heavy than both sides middle part, case is it might fracture to contain 4, The difference of basic type of structure is great. Unite it in the bridge the samly , mix and use and does not expand the foundation and a foundation with thefoundation, or adopt a foundation when a foot-path or a long difference is great at the same time , or adopt the foundation of expanding when basis elevation is widely different at the same time , may cause the ground not to subside evenly too 5, Foundation built by stages. In the newly-built bridge near the foundation of original bridge, if the half a bridge about expressway built by stages, the newly-built bridge loads or the foundation causes the soil of the ground to consolidate again while dealing with, may cause and subside the foundation of original bridge greatly 6, The ground is frozen bloatedly. The ground soil of higher moisture content on terms that lower than zero degree expands because of being icy; Once temperature goes up , the frozen soil is melted, the setting of ground. So the ground is icy or melts causes and does not subside evenly . 7, Bridge foundation put on body, cave with stalactites and stalagmites, activity fault,etc. of coming down at the bad geology, may cause and does not subside evenly . 8, After the bridge is built up , the condition change of original ground . After most natural grounds and artificial grounds are soaked with water, especially usually fill out such soil of special ground as the soil , loess , expanding in the land ,etc., soil body intensity meet water drop, compress out of shape to strengthen. In the soft soil ground , season causes the water table to drop to draw water or arid artificially, the ground soil layer consolidates and sinks again, reduce the buoyancy on the foundation at the same time , shouldering the obstruction of rubing to increase, the foundation is carried on one's shoulder or back and strengthened .Some bridge foundation is it put too shallow to bury, erode , is it dig to wash flood, the foundation might be moved. Ground load change of terms, bridge nearby is it is it abolish square , grit ,etc. in a large amount to put to pile with cave in , landslide ,etc. reason for instance, it is out of shape that the bridge location range soil layer may be compressed again. So, the condition of original ground change while using may cause and does not subside evenly Produce the structure thing of horizontal thrust to arched bridge ,etc., it is the main reason that horizontal displacement crack emerges to destroy the original geological condition when to that it is unreasonable to grasp incompletely , design and construct in thegeological situation.桥梁裂缝产生原因浅析近年来,我省交通基础建设得到迅猛发展,各地建立了大量的混凝土桥梁。
土木工程专业英语课文翻译
土木工程专业英语课文翻译土木工程专业英语课文翻译土木工程专业,是大学的一种自然学科。
专门培养掌握各类土木工程学科的基本理论和基本知识,能在房屋建筑、地下建筑、道路、隧道、桥梁建筑、水电站、港口及近海结构与设施。
以下是小编整理土木工程专业英语课文翻译的资料,欢迎阅读参考。
weight of the project. Environmental specialists study the project’s impact on the local area: the potential for air and groundwater pollution, the project’s impact on local animal and plant life, and how the project can be designed to meet government requirements aimed at protecting the environment. Transportation specialists determine what kind of facilities are needed to ease the burden on local roads and other transportation networks that will result from the completed project. Meanwhile, structural specialists use preliminary data to make detailed designs, plans, and specifications for the project. Supervising and coordinating the work of these civil engineer specialists, from beginning to end of the project, are the construction management specialists. Based on information supplies by the other specialists, construction management civil engineers estimate quantities and costs of materials and labor, schedule all work, order materials and equipment for the job, hire contractors and subcontractors, and perform other supervisory work to ensure the project is completed on time and as specified.领域。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
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。