土力学英文试卷及答案(B)
土力学(中英文对照)-- Earth Pressure Part 2
5m
Pa
Pp
5m
Example 2 (例题二)
(i) Behind the wall (墙背)
5m
Soil 1
Pa
K a(soil1)
tan2 45
' 2
tan2 45
38 2
0.24
Pp
5 m Soil 2
K a(soil2)
tan 2 45
' 2
tan 2 45
28 2
(ii) In front of the wall the wall (墙前)
5m
Pp
Pa
5 m Soil 2
Kp
tan2 45
' 2
tan2 45
28 2
2.77
'hpSoil25m 2 10 2.77 33.3 kN /m2 'hpSoil210m 2.7710.2 5 2 10 2.77 174.6 kN /m2
Example 2 (例题二)
The soil conditions adjacent to a sheet pile wall are given in the following figure. Calculate the total active force behind the wall and the total passive force in front of the wall. 计算板桩墙上总主动土压力和总被动土压力.
is unit weight of soil and z is depth of soil
Wall Movement (墙体的移动)
Earth Pressure At-Rest (静止土压力) ▪ when a wall does not move at all, i.e. it is “at rest” [AA’] (墙体没有位移)
土力学试卷(B)及答案
技术学院考试试卷（B ）一、填空题（每空1分，共20分）1、土的结构类型为 、 和 。
2、对无粘性土的工程性质影响最大的是土的 ，工程上用指标 来衡量。
3、粘性土的塑性指标I p ，液性指标I L 。
4、附加应力自 起算，自重应力自 起算。
5、土的抗剪强度指标的常用测定方法有 、 、 和 。
6、荷载试验曲线上，从线性关系开始变成非线性关系时的界限荷载称为 。
7、随荷载增加，地基变形的三个阶段是 、 和 。
8、钢筋混凝土扩展基础指 和 。
二、选择题（每题2分，共30分）1．天然状态砂土的密实度一般用（ ）来测定。
A 荷载试验B 轻便触探试验C 现场剪切板剪切试验D 标准贯入试验 2．粘土软硬状态的划分依据是（ ）。
A 含水量 B 液限 C 液性指数 D 塑性指数 3．利用角点法及角点下的附加应力系数表仅可求得（ ）。
A 基础投影范围内地基中的附加应力B 基础投影范围外地基中的附加应力C 基础中任意点的附加应力D 基础中心点下地基中的附加应力 4．土中控制体积和强度变化的应力是（ ）。
A 孔隙水压力 B 有效应力 C 自重应力 5．下列说法中，错误的是（ ）。
A 土在压力作用下体积缩小 B 土的压缩主要是土中孔隙体积的减小 C 土的压缩与土的透水性有关 D 饱和土的压缩主要是土中气体被挤出 6．在土的压缩性指标中，（ ）。
A 压缩系数α与压缩模量Es 成正比B 压缩系数α与压缩模量Es 成反比C 压缩系数越大，土的压缩性越低D 压缩模量越小，土的压缩性越低 7．在基底平均压力和其他条件均相同的条件下，条形基础的沉降比矩形基础的沉降（ ）。
A 大B 小C 相同D 无法比较8．某房屋地基为厚粘土层，施工速度快，则在工程上地基土抗剪强度指标宜用下列哪种试验确定？（ ）A 固结快剪B 快剪C 慢剪 9.下列说法中，错误的是（ ），A 土的自重应力一般不会引起地基变形B 地基中附加应力会引起地基变形C 饱和土中的总应力等于有效应力和附加应力之和D 孔隙水压力会使土体产生体积变形10、土的γ、γsat 、γ’和γd 数值大小依次为（ ），A γd <γ’<γ<γsatB γ’<γd <γ<γsatC γd <γ<γ’<γsatD γ’<γ<γd <γsat 11、当地下水位从地表处下降至基底平面处，对有效应力有何影响？（ ） A 有效应力不变 B 有效应力增加 C 有效应力减小12、当地基为高压缩土时，分层综合法确定地基沉降计算深度的标准是（ ）。
土木工程专业英语课后完整答案
Unite 12、Translate the following phrases into Chinese /English .（1）Compression Members 受压构件（2）critical buckling load 临界屈曲荷载（3）the slenderness ratio 细长比（4）stub column 短柱（5）reduced modulus 简化模量（6）Effective length 计算长度（7）Residual stress 残余应力（8）Trial-and-error approach 试算法（9）Radius of gyration 回转半径（10）Tangent modulus 切线模量3、Translate the following sentence into Chinese.(1)This ideal state is never achieved in reality, however, and some eccentricity of the load is inevitable.然而，在现实中,这种理想状态从来没有实现，一些荷载偏心是不可避免的szbending, and in these cases the member is a beam-column.在许多情况下，构件同样需要能够抵抗弯矩，在这些情况下，构件被称之为梁柱。
(3)If the member is so slender that the stress just before buckling is below the proportional limit---that is, the member is still elastic---the critical buckling load is given by Q.如果该构件很细长以至于在压曲前的应力低于比例极限---也就是说，该构件仍然是弹性状态---该构件的该临界屈曲荷载就可以由公式Q给出。
土壤学双语试题
9.Sand has less nutrients for plants than smaller particles
4、影响土壤氧化还原的因素为：①土壤通气性②微生物活动③易分解有机质的含量④植物根系的代谢作用⑤土壤pH。
5、土壤热量的来源主要有：①Solar radiation②biologicalheat③heat insidetheea土壤有何肥力特点？ 土壤质地的改良措施有哪些？
四、填空题（20%）
1、土壤容重值的用途为①计算土壤孔隙度②计算工程土方量③估算各种土壤成分储量④计算土壤储水量与灌水（排水）定额。
2、目前国内外流行的质地分类制主要为：①International②USDA③Kachinsky④Chinese四种。
3、土壤水分含量的表示方法主要有：①Themass water content②Thevolumn water content③relative water content④Soil water-storage capacity四种。
8、适宜作物生长发育的耕层总孔度为（D）。
A、8%-10%；B、15%-20%；C、50%-56%；D、60%以上
9、土壤中热容量最大的是（B）。水最大，气体最小
A、气体；B、水；C、矿物质；D、有机质
10、旱地土壤的氧化还原电位（Eh）一般为（D）。
A、-200 mv以下; B、-200-100mv；C、100mv左右；D、400-700mv
5.Some sands in soil will be brown, yellow, or red because of Fe and/or Al oxide coatings.
英语学习土力学问答题
Simple answer1, what is the difference between the normal consolidated clay and the over consolidated clay, the compression and the strength characteristics of the two?Answer: the soil in history has ever received the maximum effective stress for preconsolidation stress, expressed on the PC, and the force and the present Po 'than that of overconsolidation ratio (OCR), of the natural clay, OCR>1. The soil is overconsolidated; when the OCR = 1, normal consolidation soil preconsolidation stress. OCR=1 and the existing effective stress p '0 is less than the existing consolidation stress P0, the soil is less consolidated soilCompression characteristic difference: when the pressure increment is the same, the compression ratio of the normal consolidated soil is larger than that of the over consolidated soil.The strength characteristics are different: the strength of the super consolidated soil is higher than that of the normal consolidated soil.Shear properties: because the under consolidated soil is also a normal consolidated soil, so they have the same shear shape, that is, shear shrinkage and a positive pore water stress. Overconsolidated soil divided into strong overconsolidated soil and slightly over consolidated soil, ultra weak soil consolidation in the shear and normal consolidation soil as, volume dilation, produces a positive pore pressure, but pore water pressure to smaller than the normal consolidation soil; strong consolidation in the shearing process will produce dilatancy effect and produce negative pore water stress.2, the paper describes the factors affecting the soil compaction?A: the main factors of soil compaction are water content, compaction function, type and size of soil and coarse grain content.Of cohesive soil, moisture content influence is mainly manifested as when the moisture content is low. The same compaction function under the lower stem density, with increase of water content and the dry density will gradually improve; when reaching a certain moisture content, corresponding hit real function will get maximum dry density, corresponding water cut rate rate for optimum water content; with the increase of water content, maximum dry density but will decrease.The impact of compaction function shows that the larger the compaction function, the dry density of soil is larger, and the optimal water content decreases with the increase of compaction function.Soil type and gradation of influence performance in cohesive soil is usually no cohesive soil compressibility, high clay content, compressibility; good gradation, easy compaction, dry density high;The coarse grain content has influence on the compactness, and the results of the light compaction test need to be modified when the coarse grain content of more than 5mm grain size is more than 25% and 30%.3, compare the Rankine earth pressure theory and Kulun earth pressure theory and the similarities and differences of advantages and disadvantages.The same point: it is required that the retaining wall moves to make the shear strength of the soil filled after the wall to shear strength (limit state);Different points: Lang Ken theory is according to various points of soil in the equilibrium state of stress conditions for wall back on the earth pressure. Requirements for wall surface is smooth, and the filling surface level and result in too large. And Coulomb theory is according to the back of the wall and sliding surface of wedge type in the condition of static equilibrium in the limit equilibrium state for general earth pressure. The back wall can be tilted, rough soil surface tilt, calculation results of the initiative pressure meet the requirements, and passive pressure error larger. Rankine theory is considering the wall backfill per point of damage, up to the limit state; Coulomb theory consider sliding mass of the rigid body limit equilibrium;Rankine earth pressure theory has the advantages of simple formula, easy to use; disadvantages: of wall inclination, wall backfill surface tilt is not applicable;The theory of soil pressure in Kulun: it is applicable to the slope of the wall and the back fill slope of the wall; the disadvantage is that the condition of the clay is not considered;4, what are the foundations of the destruction of the form? What kind of soil foundation will happen to each of the soil? According to the theoretical calculation formula of bearing capacity, which is the main factor influencing the bearing capacity of shallow foundation? According to the theoretical formula to determine the bearing capacity which is applicable to the type of failure?Answer: the overall shear failure, shear failure and punching failure.Foundation failure mode mainly and soil properties especially related to compaction, in general, for solid or compacting the soil with low compressibility, usually presents general shear failure. For soft clay or loose sand foundation with high compressibility, often appear local shear failure or punching shear damage. Theoretical formula method, standard method and test method.Soil bulk density and strength index, foundation width b, foundation depth D, underground water level, etc.. The bearing capacity determined by the theoretical formula is applicable to the whole shear failure mode.5, why is that in general, the soil's self weight stress will not cause the compression of the soil deformation (or settlement), and when the ground water level down, it will cause the soil to sinkA: under normal circumstances, the foundation is a number of years of deposition, the weight of the role of stress has been reduced to a stable. Self weight stress has been transformed into the effective stress, and this kind of case, the self weight stress will not cause the soil to be compressed. But if the soil is recently deposited, the self weight stress has not completely changed the stress, then the self weight stress will be produced. (5 points)When the decline of groundwater level, soil from water under variable for the water and the soil original by the buoyancy effect. Now the Futuoliyin drawdown disappear, equivalent to exerting a downward volume force in the soil, its size is equal to the buoyant force. The force is bound to cause soil compression.6, how to measure the size distribution curve of soil and the frequency curve of grain size, what is the use? On the gradation of discontinuous soil, the two curves each have what characteristics?Answer: 1. Soil particle size distribution curve: with soil particle size of the abscissa (logarithmic scale) is less than the percentage of the total mass a particle size of soil quality for sample is plotted as the ordinate of the curve. According to the soil particle size distribution curve of the content of soil particle size fractions can be obtained for assessment of soil classification and general assessment of engineering properties. Some of the characteristics of particle size and for the quality of building materials selection and evaluation of soil gradation.The frequency curves of 2 groups: the average particle size of the particle group is the logarithmic scale, and the content of the soil particles in each particle group is plotted as the vertical coordinate.The particle size distribution curve of soil can not only determine the relative content of grain, but also can judge the gradation of soil according to the slope of the curve.7, other conditions in the same case, the settlement of the over consolidated clay must be less than the normal consolidation of clay settlement? Why? Answer: is. Compared with normal consolidated clay, overconsolidated clay pore Bibi normal consolidated soil is small, if the existing effective stress are the same, under a load increment, overconsolidated soil is along the recompression curve compression, and normally consolidated soils along the compression curve. Because of the same soil, the compression curve is more slowly than the compression curve, that is, the compression index is smaller than the compression index, therefore, the settlement of over consolidated clay is smaller than that of the normal consolidated soil.8, what is the principle of effective stress? In the figure, the foundation soil moisture is severe, the saturation is heavy and the float is heavy respectively, and the water is heavy, the water column of the M point of the measuring tube is high as shown in figure. The calculation formula of total stress, pore water stress, effective stress and self weight stress of M point are written.Answer: caused by external loads on the plane to the general stress for 6S, then it must by the surface force pore u and inter particle contact surface share, namely method on the surface to force equal to pore and interparticle undertaken by forces, namely the 6S = and '+ U.M point total stress:M point pore water stress:M point effective stress:M point self weight stress:9, the sliding surface slope slide which has several forms? What happened in the case? Safety coefficient of clay soil slope under the condition of no seepage can have what kinds of method? What are the methods used to analyze the stability of the slope?Answer: sliding surface slope slide are: arc, plane, composite sliding surface.Circular sliding usually occurs in a homogeneous clayey soil slope; sliding plane usually occurs in cohesionless soil; composite sliding surface in soil slope is not uniform slope.No seepage and clayey soil slope stability safety factor calculation method is: Phi = 0 analysis method, the Swedish division method, simplified Bishop method. In the slope stability analysis is often used method: Fellenius method, bishop strip method, Yang cloth strip method and imbalance thrust force method.10, the modified Terzaghi ultimate bearing capacity formula is suitable for what kind of failure types of foundation? The Terzaghi ultimate bearing capacity formula specify the position of the water table is how to influence on the bearing capacity?Answer: the modified Terzaghi formula of ultimate bearing capacity for strip foundation on the whole shear failure of foundation.Terzaghi's ultimate bearing capacity formula. Groundwater depth is greater, in the 1, 2 in the gamma were not affected; if the underground water level in the basement, 1 item of using gamma float density, reduce the carrying capacity of; if underground water level rose to ground at or above, the 1, 2 in the gamma were applied floating bulk density, capacity to further reduce.11, there are several kinds of clay soil? By what is the distinction between the status of the state?Answer: solid, semi solid, plastic, flow four states. (three can also be used) generally used to describe the degree of consistency of cohesive soil moisture (state) reflected in the moisture content of the effect of external forces and the ability to deform or damage.The states are distinguished as follows:Liquid limit (WL) -- from the flow state change rate of plastic state boundaries of water, is plastic state of the maximum water content; plastic limit (WP) -- the transition from plastic state for semi solid state limit moisture content is water in the lower plastic state rate;Shrinkage limit (Ws) - from the semi solid state to the boundary of the solid state water content, that is, with the reduction of soil moisture content and the volume began to change when the volume of water content.12, what is the pre consolidation pressure? What kind of soil is under consolidated soil? There is a certain excess pore water stress in the under consolidated soil layer. Under the consolidated soil, such as no new pressure, the soil will continue to compress it? Why?A: the pre consolidation pressure for the soil in the history of the maximum effective stress, with Pc. Under the action of self weight stress, the under consolidated soil is not fully consolidated, that is, the existing effective stress is less than the existing consolidation stress. There must be some excess pore water stress in the under consolidated soil. Because under the effect of self weight stress, under the self weight stress is not stable, and the attachment stress has not yet been converted to the effective stress, and there is still a part of the pore water. With the discharge of the pore water, the dissipation of the pore water stress, the ratio of the soil to the pore is further reduced, so that the soil layer will continue to compress even if no new pressure is applied.13, respectively, in the case of the flow below the two figure in the soil ABCDEA on the role of the force, and marked in the diagram. Among them, the map (a) to the soil as the research object, the map (b) to soil particles as the research object.Figure (a) (b)Answer:At. Taking into account gravity, taking into account the water pressure of ED and CD, considering the tangential reaction and normal reaction of AEDAt. Taking into account the gravity, considering the buoyancy, considering the seepage force. The tangential reaction and normal force on AED;14, the liquefaction of sand. (6 points)Answer: for saturated loose sand, when subjected to sudden dynamic load, because of the dynamic shear force should be to reduce the volume of the trend, because of the short time to wait outside the drainage, so the big pore water stress. According to the principle of effective stress, when the dynamic load caused by the excess pore water should u is reached, effective stress, the shear strength, cohesionless soil foundation will lose its bearing capacity, slope flow collapse.15, the sliding surface slope slide which has several forms? What happened in the case? The calculation method of stability coefficient of 3 kinds of lists of cohesive soil slope?Answer: sliding surface slope slide are: arc, plane, composite sliding surface. Circular sliding usually occurs in a homogeneous clayey soil slope; sliding plane usually occurs in cohesionless soil; composite sliding surface in soil slope is not uniform slope. Safety coefficient of clay soil slope calculation method: Phi = 0 analysis method, Sweden slice method and simplified Bishop method etc..16, the Yangtze River levee water retaining diagram as shown in Figure 2, respectively discuss flood peak arrive the river water level rose during and after the flood peak river water level dropped seepage function in the process of embankment retaining surface and back surface of the slope stability influence.Answer: the flood peak arrived, have toward seepage outside slope, theHave a favorable impact on water retaining slope stability, the dorsal surface of slopeStability adversely affected.After the flood peak water level decline, provided the water level in the above the river, have toward the seepage of the Yangtze River, for blocking the water slope stability adversely affect, smaller on the back surface of the slope stability influence at this time still maintain to the slope seepage, but its negative impact will be gradually weakened.17, a brief description of the method of using the method of stratified summation method for foundation settlement.A: 1 based on the nature of the load on the basis of the calculation of the base pressure and distribution2 the foundation layer.3 weight calculation of soil stress distribution4 Calculation of vertical additional stress distribution in the ground5 according to the arithmetic mean, the average self weight stress and the average additional stress of each layer are obtained.6 for the compression of the I layer,7 the amount of compression of each layer is accumulated to get the total settlement of thefoundation.18, other conditions in the same case, the settlement of the over consolidated clay must be less than the normal consolidation of clay settlement? Why?Answer: yes. Because and normally consolidated clays compared to small overconsolidated clay pore Bibi normal consolidated soil, if existing effective stress are the same, under a load increment, overconsolidated soil is along the recompression curve compression, and normally consolidated soils along the compression curve. Because of the same soil, the compression curve is more slowly than the compression curve, that is, the compression index is smaller than the compression index, therefore, the settlement of over consolidated clay is smaller than that of the normal consolidated soil.19, of seepage and stability of slope (as shown), test lists a stability analysis methods and illustrate the method in the three zones of icon the soil bulk density how to select?Answer: to soil particles as the object of study, the form of a combination of effective soil weight and seepage force of seepage and stability of slope stability analysis.Sweden slice method based on, take II water and soil particle buoyancy force on the center of the moment instead of seepage force on the center of the moment, and ignore the seepage force of anti slide effect can be establishedThe method is replaced by bulk density. In the calculation of anti slide effect (molecular) when, I region by the wet bulk density, region II the buoyant unit weight, region III the buoyant unit weight; in the calculation of the sliding action (denominator), zone I and III were still using wet bulk density and float density, but in zone II with saturated instead of bulk density.20, in the shear strength test, why not to put forward the consolidation of non drainage shear (or fast shear), consolidation of the drainage (or shear shear fast shear) and the consolidation of drainage (or slow shear) and other three methods? For the same kind of saturated clay, when three methods are used to test, the strength of the same? Why?Answer: anti?Site pan searches the convex probe into the panic wooded exploring the governance turns beech pan colonization D R Huang Valley Huang mass Tan, exploring the Yi Xing Xin word particle recovery drain pan, exploring the love people throw flow flow Aveo timid hydroxyl generous benevolence turn Maple beauty bed is judgemental of MI Gu hydroxyl generous Rong vice dropout m mi Gu hydroxyl generous at the expense of the tomb of Tan cluster Qiong Wu brew lag items off well behaved brag hydroxyl wives leisure scalded sneer rum red clam alpha gamma curl up Pan Qiong 'Gu hydroxyl generous shoulder, said an Ge song PI t mow Na Mi Qiong. Just but was probe wooded panic pan Deng alpha Chi tassel Deng pan Deng (Xian Zheng tassel Deng thujic Xiaoshan Deng (Oyster Ao Deng azulene pepper.Three, the total strength of the 21 kinds of test indicators are different, with the total stress thatthe shear strength index will vary with the degree of consolidation and drainage conditions are different, but the effective strength index does not change. This is due to the actual soil shear strength and effective force in the three test methods degree of consolidation and drainage conditions different, the pore water stress, shear strength is different, if the total stress and shear strength variation with drainage condition and change the state of water in the soil is what kind of? The influence of different states of water on the mechanical properties of the soil is briefly described.Answer: the state of water in the soil of crystal water, adsorbed water and free water, which free water and capillary water, capillary water and capillary water) and gravity water.The crystal water is firmly adsorbed on the crystal lattice of the solid particles, as a part of the soil mass, the mechanical properties of the soil are not much affected. ][adsorbed water has greater viscosity than ordinary water, less initiative and different density. Strongly adsorbed water by soil particle adsorption, firmly combined on the surface of the soil particles, the nature of close to the solid and can not transfer pressure. Weak adsorbed water is a viscous state, can not transfer pressure, nor in the pore free flow. The adsorbed water existing soil has the plasticity and viscosity, influence of compressibility of soil and the strength, and the permeability of the small soil.Free water left the soil particles surface is far away from the soil particles, and can move freely, and it's nature is no different from ordinary water, can transfer the hydrostatic pressure, has the ability to dissolve.Capillary water is related to the size and shape of the pores in the soil, the mineral composition of soil particles and the properties of water. In the wet powder and fine sand, the pore water only exists in the contact point of the soil particles, and it is not continuous. The pressure in the pore will be less than the atmospheric pressure. Will cause the pressure to force the adjacent soil particles to squeeze each other. The capillary water pressure, increase the intergranular dislocation friction resistance. Generating pseudo cohesion.Gravity water, like ordinary water, has the ability to dissolve, can transfer the hydrostatic and dynamic water pressure, the soil particles have a buoyancy effect. It can dissolve the water soluble salt in the soil, and change the engineering properties of the soil. When it flows in the pores of the soil, the seepage force is applied to the soil which flows through it.22, water is an important part of the three-phase soil. It is generally believed that it could not withstand the shear, but can withstand the pressure and suction; at the same time, water compressibility is small, usually encountered in the pressure range, its compression is negligible. Soft soil highway construction requires a small hill, the artificial excavation and slope is flat. The stress history of the soil and the compression curve are used to analyze the construction of the whole subgrade.Answer: (to make a schematic (2)I have a small slope foundation area with no slope zone have different stress history, they have different compression characteristics. Points)After the excavation of the area I for overconsolidated soil, when the small slope excavation to the foot of the slope relative to normal, compression curve will rebound rebound curve along, when to exercise the same additional stress delta P, zone 1 and zone 2 soil will along the different compression curves of consolidation of the foundation. Zone I will be along the B'B to the D points to the consolidation and stability of the curve.And the second zone will be stable along the AB compression curve to the point of D.The same P change, but the difference between the pore ratio is different, that is, the compression of the soil is different, so the whole section of the construction process of soft soil foundation in the construction method will not be the same.23, calculated a homogeneous clayey soil dam (downstream water level) of safety factor of slope stability, Q in (during normal operation) the upstream water level in the high water level and low water level and water level rapid decline related physical and mechanical parameters of the soil during the (severe, intensity index) C, should be how to choose? The relationship between the safety factor and the safety factor of the slope at the edge of the slope is compared and explained. When the low water level is assumed, the upstream part of the dam is not saturated.A: see textbook p260-261.High water level and low water level stable seepage, which belongs to the category of effective stress, should adopt the effective stress intensity index or the drainage shear strength index. When the soil is saturated with high water level, the floating weight index is used. And the low water level is not saturated. Rapid decline in water level, the water is not discharged, the consolidation of non drainage strength indicators. Rapid decline of water level, the water is not discharged, the soil saturation, with floating heavy index. The safety factor: the high water level > the low water level, the rapid decline of the water level (considering the infiltration force) can only be reflected in the shear strength index, so the total strength of the three different types of tests are different.In 24, the stability analysis of slope stability using seepage Fellenius method, seepage force and how to consider? What is the substitution method, how to select the various parts of the soil slope in bulk?Answer: the slope stability analysis of the steady seepage, seepage force can be considered net and gravity replacing method.Substitution method is in slope stability analysis by saturation line below and above the waterlevel outside the slope surrounded by the same volume of water weight center of the sliding torque to replace the seepage force on the center of the sliding moment.Instead of law. In the calculation of anti sliding effect (molecular), the saturation line above the wet bulk density, infiltration line with a slope below water level above the float density, slope below the water level by float density; in the calculation of the sliding action (denominator), saturation line above the wet bulk density, infiltration line with a slope below water level above the saturation density, slope below the water level by float density.24, what are the factors affecting the compaction of the soil? Please explain briefly.A: the main factors of soil compaction are water content, compaction function, type and size of soil and coarse grain content.Of cohesive soil, moisture content influence is mainly manifested as when the moisture content is low. The same compaction function under the lower stem density, with increase of water content and the dry density will gradually improve; when reaching a certain moisture content, corresponding hit real function will get maximum dry density, corresponding water cut rate rate for optimum water content; with the increase of water content, maximum dry density but will decrease.The impact of compaction function shows that the larger the compaction function, the dry density of soil is larger, and the optimal water content decreases with the increase of compaction function.Soil type and gradation of influence performance in cohesive soil is usually no cohesive soil compressibility, high clay content, compressibility; good gradation, easy compaction, dry density high;The coarse grain content has influence on the compactness, and the results of the light compaction test need to be modified when the coarse grain content of more than 5mm grain size is more than 25% and 30%.25, determine the foundation bearing capacity of the method there are several major categories? What are the factors that affect the bearing capacity of foundation? Whether the rise of the water table will cause changes in the bearing capacity of the foundation?A: (1) to determine the foundation bearing capacity of the method has three major categories: static load test or other in situ test, standard table method theoretical formula method. (2) the factors that affect the bearing capacity of the foundation are: the soil is heavy; the cohesion and internal friction angle of the soil; the width of the base; the depth of foundation. (3) the rise of the water table will cause the reduction of the bearing capacity of the foundation.Factors that affect the permeability of the soil?A: the particle size and mineral composition of the soil: the particle size, shape, and the effect of the permeability of the soil.The mineral composition of soil has a great influence on the permeability of clay. Combining with the thickness of the water film, when the thickness of the water film is thicker, the gap of the soil can be blocked, and the permeability of the soil can be reduced. The structure of the earth: because of the various kinds of soil, so the permeability is also the way. Viscosity of water: water seepage speed in soil and water density and viscosity related, and the two numerical and with gas is related to the temperature of the earth: when the soil voids exist in the confined bubble, blocking water seepage, thus reducing the permeability of the soil.The physical meaning of effective stress and pore water pressure in the effective stress formula is introduced in this paper.Answer: pore pressure at any point of the earth, in all directions is equal, so it can make soil particles produce compressed, and cannot make the soil particle displacement, inter granular soil effective stress, caused by the displacement of the soil particles, pore volume change, the soil compression deformation effectively at the same time, we should force also affect the soil shear strength.。
(精心整理)《土力学》试卷(7套)
1、《土力学》试卷(2)2、《土力学》试卷(3)3、《土力学》试卷(4)4、《土力学》试卷(5)5、《土力学》试卷(6)6、《土力学》A卷7、《土力学》B卷------------------------------------------------------------------------------------土木工程专业《土力学》课程卷二一、填空 (25分)1. 土中结合水有和两种形式。
（C）2. 三种粘土矿物中,亲水性最小是 ,膨胀性最大的是。
（C）3. 计算土中某点的自重应力时从算起,而计算土中某点的附加应力时应从算起。
（B）4. 流砂现象主要发生在和等土层中。
（A）5. 砂土的密实度由和指标确定,其湿度由确定。
（B）6. 饱和土体的渗透固结过程是土中的消散, 增加的过程。
（C）7. 渗透系数κ可通过试验和现场试验确定。
（B）8. 土体受剪破坏时,剪切面与小主应力作用面成角。
在情况下，土中剪破面与大主应力面一致。
（B）9. 土的抗剪强度可由室内的和试验获得,对软粘土,也可通过现场的试验获得。
（A）10. 在土坡稳定性分析中,安全系数κ可以表示为土体的力矩与力矩之比。
（B）11. 朗金土压力理论的三个假设分别可以表示为、和。
（B）二、名词解释 (16分)1. 有效应力（B）2. 压缩系数（A）3. 干容重（C）4. 固结度（B）5. 附加应力（B）6. 地基容许承载力（C）7. 灵敏度（A）8. 土的触变性（A）三、简答 (29分，第1题5分，其余每题6分)1.试简述分层总和法的假设及其计算步骤。
（C）2.地基的容许承载力什么情况下需要修正？试写出修正公式及其每个物理量的含义。
（A）3.何谓先期固结压力？如何根据先期固结压力来判断天然土层固结状态？（B）4.试简述完成的土力学三个实验分别用来测定的内容。
（C）5.有两个埋深相同而宽度不同的基础，其基底压力相同，则在同一深度处哪个沉降大？为什么（B）四、计算（30分）1.原状土样的体积为70cm3，湿土质量为0.126Kg，干土质量0.1043Kg，土粒容重26.8kN/m3。
土力学试卷(B)及答案
技术学院考试试卷（B ）学期：12秋季 班级：建筑 科目：《土力学与地基基础》 闭卷题 号 一二三四五总分得 分一、填空题（每空1分，共20分）1、土的结构类型为 、 和 。
2、对无粘性土的工程性质影响最大的是土的 ，工程上用指标 来衡量。
3、粘性土的塑性指标I p ，液性指标I L 。
4、附加应力自 起算，自重应力自 起算。
5、土的抗剪强度指标的常用测定方法有 、 、 和 。
6、荷载试验曲线上，从线性关系开始变成非线性关系时的界限荷载称为 。
7、随荷载增加，地基变形的三个阶段是 、 和 。
8、钢筋混凝土扩展基础指 和 。
二、选择题（每题2分，共30分）1．天然状态砂土的密实度一般用（ ）来测定。
A 荷载试验B 轻便触探试验C 现场剪切板剪切试验D 标准贯入试验 2．粘土软硬状态的划分依据是（ ）。
A 含水量B 液限C 液性指数D 塑性指数3．利用角点法及角点下的附加应力系数表仅可求得（ ）。
A 基础投影范围内地基中的附加应力B 基础投影范围外地基中的附加应力C 基础中任意点的附加应力D 基础中心点下地基中的附加应力4．土中控制体积和强度变化的应力是（ ）。
A 孔隙水压力B 有效应力C 自重应力5．下列说法中，错误的是（ ）。
A 土在压力作用下体积缩小B 土的压缩主要是土中孔隙体积的减小C 土的压缩与土的透水性有关D 饱和土的压缩主要是土中气体被挤出6．在土的压缩性指标中，（ ）。
A 压缩系数与压缩模量Es 成正比 B 压缩系数与压缩模量Es 成反比C 压缩系数越大，土的压缩性越低D 压缩模量越小，土的压缩性越低7．在基底平均压力和其他条件均相同的条件下，条形基础的沉降比矩形基础的沉降（ ）。
A 大B 小C 相同D 无法比较8．某房屋地基为厚粘土层，施工速度快，则在工程上地基土抗剪强度指标宜用下列哪种试验确定（ ） A 固结快剪B 快剪C 慢剪9.下列说法中，错误的是（ ），A 土的自重应力一般不会引起地基变形B 地基中附加应力会引起地基变形C 饱和土中的总应力等于有效应力和附加应力之和D 孔隙水压力会使土体产生体积变形 10、土的、sat 、’和d 数值大小依次为（ ），Ad <’<<sat B ’<d <<sat C d <<’<sat D考 生 答 题 不 得 超 过 此 密 封 框班 级考 号姓 名’<<d <sat11、当地下水位从地表处下降至基底平面处，对有效应力有何影响（）A有效应力不变B有效应力增加C有效应力减小12、当地基为高压缩土时，分层综合法确定地基沉降计算深度的标准是（）。
土力学习题
Example 2: The soil profile at a site for a proposed office building consists of a layer of fine sand 10.4 m thick above a layer of soft normally consolidated clay 2 m thick. Below the soft clay is a deposit of coarse sand. The groundwater table was observed at 3 m below ground level. The void ratio of the sand is 0.76 and the water content of the clay is 43%. The building will impose a vertical stress increase of 140 kPa at the middle of the clay layer. Estimate the primary consolidation settlement of the clay. Assume the soil above the water table to be saturated, Cc=0.3 and Gs=2.7.
• Example 3: Assume the same soil stratigraphy as in Example 2. But now the clay is over-consolidated with an OCR=2.5, w=38%, and Ce=0.05. All other soil values given in Example remain unchanged. Determine the primary consolidation settlement of the clay.
- 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)。
沈阳建筑大学考试评分标准专用纸2007 年 春 季学期 科目 土力学(B) 适用年级、专业 土木04-9，10 ——————————————————————————————————一 Fill in the blanks (total 15 points, 1 points per blank) 1. mm d mm 2075.0≤< 2. G s , w 3.一个有效应力园 4. 水 , 孔隙. 5. 245ϕ-6. 水力梯度7, 沉降， 承载力 8, 主固结， 次固结 9, 4C 10, 40 11, 孔隙体积二、Judge the following statement right or wrong. Mark with R for the rights and W for the wrongs (total 10 points, 2 points per problem)W R R W W三、Explain the following concepts or definition(total 15 points, 3points per problem) 1. void ratio孔隙比：孔隙体积与土粒体积之比，sv v v e /=2. coefficient of curvature C c曲率系数定义为（C c ）6010230d d d C c =,曲率系数C c 描写累积曲线的分布范围，反映曲线的整体形状。
3. compression modulus压缩模量：土体在完全侧限的条件下，竖向应力增量与竖向应变增量的比值4. overconsolidation土在应力历史上（固结过程中）所受到的最大有效应力，称之为前期固结应力. 前期固结应力与现有的自重应力之比大于1时，土体处于超固结状态。
5． a dditional stress由上部建筑荷载（基底附加压力）等外界因素在地基中产生的应力。
五、Answer the following questions simply(total 15 points, 5 points per problem)1．Explain the principle of effective stress.饱和土中任意点的总应力总是等于有效应力加上孔隙水压力；或有效应力总是等于总应力减去孔隙水压力。
既 u +'=σσ 或 u -='σσ2. Explain “stress in soil ”, why do we calculate “stress in soil ”?土体在自身重力，建筑物荷载，交通荷载或其他外界因素等的作用下，产生土中应力。
按起因可分为资中应力和附加应力。
因土中应力将引起土体或地基发生变形，试土工建筑物发生沉降，倾斜以及水平位移。
变形过大将影响建筑物的正常使用。
并且，土中应力过大，会导致土体的强度破坏，使土工建筑物发生土坡失稳或使地基的承载力不足而发生失稳。
因此须计算土中应力。
3. Give out the assumptions in Terzaghi ’s theory of one dimensional consolidation1.土层是均质、各向同性和完全饱和的；2.土的压缩完全是由于孔隙体积的减少，土粒和水是不可压缩的；3.水的渗流和土层的压缩仅在竖向发生；4.水的渗流遵从达西定律；5.渗透系数k 和压缩系数a 保持不变。
6.外荷载一次瞬时施加。
7．土体变形完全是孔隙水压力消散引起的。
五. There is a sample which is got from a certain natural sand column. The water content measured by experiment is 11%, bulk density 3/70.1cm g =ρ, the minimum dry density is 3/41.1cm g =ρ, the maximum dry density is 3/75.1cm g =ρ. Please determine the sandy soil ’s dense degree (6 points)3/53.111.0170.11cm g wd =+=+=ρρ 2 points()()4.0max maxmin =--=dd d d d d Dr ρρρρρρ 2 point在1/3 至2/3 之间 ，中密 2 point六. A soil sample 10cm in diameter is placed in a tube 1m long, A constant supply of water is allowed to flow into one end of the soil at A and outflow at B is collected by a beaker. The average amount of water collected is 1cm3 for every 10 seconds. The tube is inclined as shown in the below figure. Determine the (a) hydraulic gradient, (b) flow rate, (c) average velocity, (d) coefficient of permeability. (6 points)Step 1 Define the datum position. Select the top of the table as the datum Step 2 Find the total heads at A (inflow) and B (outflow) ()()m h h H A z Ap A 211=+=+=()()m h h H Bz Bp B8.08.00=+=+=Step 3 Find the hydraulic gradient m H H H B A 2.18.02=-=-=∆ l=1m2.11/2.1==∆=lHi 2 points Step 4 Determine the flow rateVolume of water collected, Q=1cm 3, t=10sec.Q v =Q/t=1/10=0.1 cm 3/ sStep 5 Determine the average velocity q v =Av()2225.784104cm diam A =⨯==ππs cm A q v v /0013.05.781.0===2 points Step 6 Determine the coefficient of permeability. From Darcy ’s laws cm i v k /108.102.10013.04-⨯===∴ 2 points七. A saturated clay layer of thickness 10m overlays an impervious hard rock. A vertical uniform pressure p=200kPa is applied on the clay layer. The initial void ratio is 0.11=e , the compression coefficient 14105.2--⨯=kPa a , and the coefficient of permeability year cm k /0.2=. Find : (1) settlement after 1 year loading; (2) the length of time required for achieving a settlement of 20 cm. (11 points)1. Final settlement 2511=+=pH e aS 2 points year cm a e k C wv /106.1)1(251⨯=+=γ 2 points 16.02==H t C T v v 2 points=⨯=S U S t 0.45*25=11.25cm 2 points2. U=20/25=0.857.0=v T 1 points56.32==vv C H T t2 points八. Prove the Relationship s E E )121(20μμ--=between the deformation modulus 0E and the compression modulus s E . Where u is Poisson ’s ratio. (5 points)z y x K σσσ0== 1 points0==y x εε 0000=--=E E E zyxx σμσμσε 1 points()μμσσ-==1/0xz K()0021K E E E E zxyzz μσσμσμσε-=--=1 pointsUnder laterally confined condition szz E σε=1 points The relationship between the deformation modulus and the compression modulus is given ass E E )121(20μμ--== 1 points九 A frictionless retaining wall is shown in the figure. Determine the force per unit width of the Rankin ’s active state. Also find the location of the resultant and give the distribution of theactive pressure. (12 points)1 . 31)23045tan(245tan 21=-=⎪⎭⎫ ⎝⎛-=ϕa K2．顶部与交界处上部主动土压力kPaK c qK p a a a 67.6)3/1(202=⨯=-=()kPaK c K q H p a a a 67.23)3/1()20317(2=⨯+⨯=-+=γ3.271.0)23545(tan 245tan 222=-=⎪⎭⎫ ⎝⎛-=ϕ a K4. 交界处下部主动土压力()kPa K q H p a a 24.19271.0)20317(=⨯+⨯=+=γ 5. 底部主动土压力 6.()kPa K q H H p a a 88.33271.0)20318317(2211=⨯+⨯+⨯=++=γγ7. 全主动土压力合力m kN E a /2.125)24.1988.33(32124.193)67.667.23(32167.63=-⨯⨯+⨯+-⨯⨯+⨯=8. 作用点从底部起为X 处m E X a 66.2]3)24.1988.33(211324.195.1)67.667.23(321467.635.4[1=⨯-⨯⨯+⨯⨯+-⨯⨯⨯+⨯⨯=9．分布图———————————————————————————— 注意事项：1.请将评分标准连同考试题一并上交。