土力学英文试卷及答案(A)
2005级土木工程专业英语课程考试试卷A参考答案
2、All soils beneath the surface are in a state of compaction; that is, they are under a pressure that is equal to the weight of the soil column above it.
答：从废水排入江河处的下游取用饮水时,不仅要特别注意从废水中清除有机物,而且要注意进行纯化微生物的消毒处理.也要关注从工厂排出的有机化学污染物,还要关注,沿杀虫剂和除草剂所污染的河流所灌溉的农田流出的排出物。
得
分
五、判断以下各句下画线部分何处有错，并改错。（每句只有一处有错）
（每小题2分，共10分）
答：所有地表以下的土都处于受压状态，说得更精确些，这些土承受与作用在其上的土柱重量相等的压力。
3、Architecture calls for good proportions—a pleasing relationship of voids to solids, of height to width, of length to breadth.
3
Where wastewater discharges into rivers from which drinking water is takendownstream(下游), special attention is given not only to the removal of organic matter from the wastes but also to the disinfection processes that inactivatemicroorganisms(微生物). Also of concern are discharges of organic chemical contaminants from industry or runoff from farm lands along the river contaminated bypesticides(杀虫剂) andherbicides(除草剂).
土力学试卷(A卷及其答案)
《土力学》试卷（A卷）学号姓名成绩一、填空题：（每空2分，共20分）1．确定各粒组相对含量的方法称为颗粒分析试验，分为法和比重计法。
2、当砾类土或砂类土同时满足Cu≥5，Cc = 两个条件时，视为良好级配。
3、渗透系数的测定方法：常水头试验（砂土），（粘土）。
4、土坡分为和人工土坡。
5、地基破坏的形式有整体破坏、冲剪切破坏。
6、土压力的类型有：静止土压力、、被动土压力。
7、土的物理性质指标中可描述土体中孔隙被水充满的程度。
8、土渗透变形的基本形式有管涌和。
9、土体的压缩系数被认为是由于土体中减小的结果。
10、根据粘性土地基在荷载作用下的变形特征，可将地基最终沉降量分成三部分：瞬时沉降，沉降和次固结沉降。
二、选择题（单选）：（每题2分，共20分）1．下列哪个物理性质指标可直接通过土工试验测定（）。
A．孔隙比e B．孔隙率n C．饱和度S r D．土粒比重G s2．在土工试验室中，通常用（）测定土的密度A．联合测定法B．环刀法C．比重计法D．击实仪3．土的强度是特指土的（）A．抗剪强度B．抗压强度C．抗拉强度4．由建筑物荷载或其它外载在地基内产生的应力称为（）A．自重应力B．附加应力C．基底压力D．基底附加压力5．地基中，地下水位的变化，会引起地基中的自重应力（）A．增大B．减小C．不变D．可能增大，也可能减小6．下列说法正确的是（）A．土体的压缩是由于土体中孔隙体积减小B．土体的压缩是由于土体中土颗粒体积减小C．土体的压缩是由于土体中水和气体体积减小D．土体的压缩是由于动水压力的作用7浅基础的判断标准为（）A．d≤b B．d＝b C．d＞b D．d≥b8.土坡的稳定计算中其中瑞典圆弧法、瑞典条分法、毕肖普法计算其滑裂面的形状为( )A．直线 B．折线 C．圆弧9．粘性土的状态有：（）①固态②半固态③可塑态④流态⑤气态A．①②③④ B．①⑤②④ C．①③④D．①②③④⑤10.在土体应力计算中，当条形基础的长度和宽度之比L/b ∞时，地基中的应力状态属于平面问题，但是实际工程中不存在这样的条形基础，根据研究当L/b为（）是条形基础。
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)
土木工程地质双语试题答案A
土木工程地质（双语）试题A(标准答案)一．填空（每空2分，共50分）1.Engineering geology is about applying all branches ofthe geosciences to assure the （safety）,（ efficiency）, and （economy of engineering ）and（ environ mental projects.）2.Physical properties of minerals include：（ Crystalform）、（ Hardness）、（color）、（ Luster）、（Streak ）、（ cleavage）。
3.The classification of rock according to the cause offormation are (igneous rock）、（sedimentary rock）、（metamorphic rock）。
4.The physical properties of groundwater（temperature）、（color）、（taste）、（transparent）、（smell）、（radioactivity）5.Geological environment (geological condition) ：（thestudy about topography and landforms）、（the study about the fundamental engineering ）、（ properties of soils）、（ rock）、（geological structure）、（hydrogeology）。
二．名词解释（每个2分，共计20分））1.Geological processes:continually modify the Earth’s surface,destroy old rocks, create new rocks and add to the complexity of ground conditions.2.Weathering:the Rock is destroyed at the place where it was formed because of the outer dynamics.3. Metamorphism:Because of the high pressure and or temperature, the rock is changed to a new kind of rock4.The eluvial soils are the direct products of weathering,they are formed in the place where their mother rocks are located.5.The slope deposits are formed from erosion of surfaceflow along the mountain slope surface .6. Absolute dating methods are those that generate an agein years.7.Relative dating methods use geological principles to place events in chronological order.8.Faults are planar discontinuity surfaces along which there has been significant displacement in shear.9.Groundwater:is that part of water which exits in the space of rock’s, soil’pore below ground surface. 10.Phreatic water : is below the ground surface, above thefirst impermeable layer, and has free water surface. 三．连线题（每题1.5分，共计15分）relative unit weight(G) 土的结构类型unit weight 空隙度dry unit weight 饱和度saturated unit weight 土的相对密度effective unit weight 土的有效重度water content (w) 土的含水量saturation degree 饱和重度porosity(n) 孔隙比relative porosity(void ratio) 土的重度soils structure type 土的干重度四．What are factors that contribute to increased shear stress(15分)1.)Removal of lateral Supporta.Erosion by stream etc...b.Previous rock fall, subsidance, large scale faulting etc...c.Work of human agencies in which cuts, quarries, pits,canals etc...2.)Surchargea.Rain, snow, water from springb.Accumulation of talusc.Collapse of accumulated volcanic materiald.Vegetatione.Seepage pressures of percolating water。
浙大土力学习题有答案
第一章绪论1、土力学的英语是：(A)Soil Mechanics (B)Solid Mechanics ( C) Soil Foundation2、岩土工程的英语是：(A)Rock and Soil Mechanics( B)Geotechnical Engineering(C) Rock and Soil Engineering3、下列哪位被誉为土力学之父？(A)库仑(Coulomb) (B)朗肯(Rankine) (C)太沙基仃erzaghi)4、土力学学科正式形成是哪一年？( A) 1890 ( B) 1925 ( C) 19605、土力学主要研究地基那两方面的问题？(A)变形与渗流(B)变形和稳定(C)渗流与稳定6、浙江大学曾国熙教授倡导的岩土工程学科治学方法是？(A)理论研究与工程实践相结合(B)试验研究与理论研究相结合(C)基本理论、试验研究和工程实践相结合第二章土的物理性质与工程分类1、土颗粒的大小及其级配，通常是用颗粒累积级配曲线来表示的。
级配曲线越平缓表示：(A)土粒大小较不均匀，级配良好(B)土粒大小均匀，级配良好(C 土粒大小不均匀，级配不良2、土的不均匀系数Cu越大，表示土的级配：(A)土粒大小均匀，级配良好(B)土粒大小不均匀，级配良好(C 土粒大小不均匀，级配不良3、土的三相指标包括：土粒比重、含水量、重度、孔隙比、孔隙率和饱和度，其中哪些为直接试验指标？(A)孔隙比、含水量、土粒比重(B) 土粒比重、含水量、重度(C)含水量、重度、孔隙比4、测定土的液限的标准是把具有30 度锥角、质量76 克的平衡锥自由沉入土体，沉入多少深度时的含水量为液限？( A) 18mm ( B)2mm ( C)10mm5、压实能量越小，则(A)最优含水量越大 (B) 土越容易压实 (C) 土的最大干密度越大6、土的液限和塑限的差值(省去%符号)称为(A)液性系数 (B)塑性系数 (C)液性指数 (D)塑性指数7、土的含水量一般用什么测定：(A)比重瓶法 (B)烘干法 (C)环刀法 (D)搓条法8、某土的天然含水量为42%，液限35%，塑性指数17，孔隙比1.58 ，则该土应定名为：(A)淤泥 (B)粉质粘土 (C淤泥质粘土9、土的密度一般用什么方法测定：(A)比重瓶法 (B)烘干法 (C)环刀法 (D)搓条法10、关于土中的结合水，下列说法正确的是：(A)强结合水能传递静水压力(B)弱结合水能传递静水压力(C强结合水和弱结合水能传递静水压力( D 强结合水和弱结合水都不能传递静水压力11、一般来说，粗大土粒往往是岩石经过什么作用形成？A 物理和化学风化作用B 物理风化作用C 化学风化作用12、粘性土的塑限一般用什么方法测定？( A 比重瓶法 ( B 烘干法 ( C 环刀法 ( D 搓条法13、土的液性指数越大，则：(A 土的渗透性越大 ( B 土的塑性指数越小 ( C 土质越软14、土的塑性指数越小，则：（A）土的粘性越差（B）土的渗透性越好（C）土的变形越大15、土粒比重一般用什么方法测定：（A）比重瓶法（B）烘干法（C）环刀法（D）搓条法第三章土的渗透性与渗流1、不透水岩基上有水平分布的三层土，厚度均为1m渗透系数分别为0.5m/d、0.001m/d、0.02m/d，则土层等效水平渗透系数为多少？（A）0.426m/d （B）0.174m/d （C）0.028m/d2、某饱和土的含水量为40%，土粒比重为2.75 ，则该土发生流土的临界水力坡降为：（A） 1.02 （B）0.83 （C） 1.203、管涌形成的条件中，除具有一定的水力条件外，还与土粒几何条件有关，下列叙述正确的是：（A）不均匀系数大于10的土比不均匀系数小于10的土更容易发生管涌（B）不均匀系数小于10的土比不均匀系数大于10的土更容易发生管涌（C与不均匀系数没有关系4、在防治渗透变形措施中，哪些措施是在控制水力坡降？（A）溢出部位铺设反滤层（B）上游做垂直防渗帷幕或设水平铺盖（C下游挖减压沟5、不透水岩基上有水平分布的三层土，厚度均为1m,渗透系数分别为0.4m/d、0.05m/d、0.003m/d，则土层等效竖向渗透系数为多少？（A）0.367m/d （B）0.063m/d （C）0.008m/d6、变水头试验常用来测定哪种土的渗透系数（A）粘性土（B）砂土（C）任何土7、计算成层地基中等效渗透系数时，水平等效渗透系数的大小取决于渗透系数________ 的土层，而垂直等效渗透系数的大小取决于渗透系数_____ 的土层；（A）大小（B）大大（C）小大（D）小小8、下列哪种土样更容易发生流砂：（A）粗砂（B）粉土（C）粉砂9、下列土体中渗透系数最小的是：（A）砂（B）粘土（C）粉土10、常水头试验常用来测定哪种土的渗透系数？（A）粘性土（B）砂土（C）任何土第四章地基中的应力计算1 、地下水位下降将引起：（A）自重应力减小（B）地面沉降（C）超静孔压减小2、已知宽为3m长为6m和另一宽为6m长为12m的矩形基础，若两基础的基底附加应力相等，则两基础角点下竖向附加应力之间有何关系?（A）小尺寸基础角点下z深度处应力与大尺寸基础角点下2z深度处应力相等（B）大尺寸基础角点下z深度处应力与小尺寸基础角点下2z深度处应力相等（C 两基础角点下下z深度处竖向应力分布相同3、当地下水位从地表处下降至基底平面处，对基底附加应力有何影响？（A）不变（B）减小（C）增加4、条形均布荷载中心线下，附加应力随深度减小，其衰减速度与基础宽度有何关系？（A）基础宽度越大，衰减越慢（B）基础宽度越大，衰减越快（C与基础宽度无关5、双层地基中上层土与下层土的变形模量之比越小，则下层土中的附加应力：（A）越小（B）不变（C）越大6、某方形基础中心点下1m处竖向附加应力为70kPa,则该基础中心点下2m处的竖向附加应力为：（A）95.2kPa （B）80.4kPa （C）56.3kPa7、在荷载作用范围外，竖向附加应力随深度的变化是：（A）竖向附加应力随深度的增加而减少（B）竖向附加应力随深度的增加，先增大，然后减小（C竖向附加应力随深度的增加，先减小，然后增大8、对于天然地基，若土体竖向变形模量大于水平向变形模量，则与均质地基相比地基中竖向附加应力：（A）产生应力集中（B）产生应力扩散（C没有变化9、对于均布条形荷载p 作用下的附加应力为0.1p 的等值线深度大约为基础宽度的几倍？（A）6；（B）3.5 ；（C）2；（D）1.510、在均布条形荷载作用下，最大剪应力发生在基础的什么部位？（A）基础中心（B）基础边缘（C）不确定第五章土的压缩性和固结理论1、土体压缩性e—p 曲线是在何种条件下试验得到的？（A）无侧限条件（B）部分侧限条件（C）完全侧限2、从现场载荷试验p—s 曲线上求得的土的模量为：（A）变形模量（B）弹性模量（C）压缩模量3、在室内压缩试验中，土样的应力状态与实际中哪一种荷载作用下土的应力状态相一致：（A）条形均布荷载（B）矩形均布荷载（C）无限均布荷载4、某地基土，已测得压缩指数为0.5 ，则回弹指数可能为：（A） 1.0 （B）0.01 （C）0.15、某地基土，已测得压缩模量6MPa则变形模量可能为：（A）7MPa （B）6MPa （C）5MPa6、双面排水，在土层厚度相同、性质相同的两个粘土层顶面，分别瞬时施加无限均布荷载p仁100kPa, p2=200kPa,试问经过相同时间t ，两种情况的固结度有何不同？（A）p1=200kPa的固结度大（B）p2=400kPa的固结度大（C）两种情况的固结度相同7、某地基粘土层的厚度均为4m情况1是双面排水，情况2是单面排水，当地面瞬时施加一无限均布荷载，达到同一固结度时，所对应的时间有何关系？（A）t1=4t2 （B）t1=2t2 （C）t1=t2 （D）t2=4t18、土的渗透性越好，则：（A）变形稳定越快（B）强度越小（C）有效应力增长越慢9、土的压缩性越大，则：（A）固结越快（B）固结越慢（C）与固结无关10、地基中的超静孔隙水压力产生的原因一般为？（A）地下水；（B）土自重与外加荷载；（C地下水与外加荷载；（D）外加荷载11、某土样经压缩试验测得其在100kPa和200kPa压力作用下压缩24小时后的孔隙比分别为0.89和0.85，则该土样的：（A）压缩指数为0.25 （B压缩模量为4.625MPa （C）压缩系数为0.4MPa-112、在固结过程中，饱和土体中的有效应力和孔隙水压力按如下规律变化：（A）有效应力减小，孔隙水压力减小（B）有效应力不变，孔隙水压力增大（C）有效应力增大，孔隙水压力减小第六章地基沉降计算1、在疏浚河道形成的新冲填土上建造建筑物，其沉降产生的原因：）（A）原地基的自重应力（B）冲填土自重（C）冲填土自重及建筑物荷载（D建筑物荷载2、地基沉降计算深度是指：（A）主要压缩层厚度（B）主要持力层厚度（C）20倍地基宽度3、工后沉降一般是指：(A)全部的次固结沉降(B)工程完工后使用期内地基所发生的沉降(C大部分的次固结沉降4、关于地基沉降，下列说法错误的是：(A)初始沉降的产生也需要一定的时间(B)土体固结沉降也有短时间完成的(C)将地基沉降分为三部分是从时间角度考虑的5、浙江沿海饱和软粘土地基一般情况下次固结沉降约占总沉降的：( A) 40% ( B) 10% ( C) 15%6、在半无限体表面，瞬时施加一局部荷载，这时按弹性理论计算得到的应力是：(A)有效应力 (B)总应力 (C孔隙压力7、初始沉降产生的原因：(A)土体体积不变，形状发生改变(B)土体体积发生变化(C上述两种情况都有8、在常规的沉降计算方法中，地基中的应力状态是根据什么理论计算得到的？(A 线弹性理论(B 弹塑性理论(C 塑性理论9、考虑应力历史的影响时，沉降计算应采用哪种压缩曲线？( A e—p 曲线( B e—logp 曲线( C 上述两种均可10、规范法计算地基沉降时，采用的是哪种附加应力系数？(A 平均附加应力系数(B 附加应力系数 (C 上述两种均可第七章土的抗剪强度1、饱和软粘土在不同竖向荷载p1>p2>p3作用下在直剪仪中进行快剪，所得的强度有何不同？(A p 越大，竖向荷载越大，所以强度越大( B 三者强度相同( C p 越大，孔隙水压力越大，所以强度越低2．十字板试验测得的抗剪强度相当于实验室用什么方法测得的抗剪强度？(A 不固结不排水试验(B 应力路径试验 (C 固结排水试验3、直剪试验土样的破坏面在上下剪切盒之间，三轴试验土样的破坏面在什么位置上？( A 与试样顶面夹角为45(B)与试样顶面夹角为45- © /2(C)与试样顶面夹角为45+© /24、有一个砂样，在三轴试验时，在围压100kPa应力下，增加轴向应力使砂样破坏，已知砂样的内摩擦角为30度，则破坏时破坏面上的正应力为多少？(A 200kPa ( B 180kPa (C 150kPa5、下列那个试验测试的不是土的天然抗剪强度：( A 十字板剪切试验(B 无侧限抗压强度试验(C 直接剪切试验(D 三轴不固结不排水试验6、现场原位测试土的抗剪强度的试验有：(A 十字板剪切试验(B 无侧限抗压试验( C 直接剪切试验( D 三轴压缩试验7、在做三轴试验时，加荷速率越快，则测得的：(A 抗剪强度越大(B 抗剪强度越小( C 对抗剪强度没影响8、对于蠕变，下列说法错误的是：(A 蠕变可能导致土体在没有达到抗剪强度的条件下破坏(B)随着土体蠕变，土体的粘聚力逐渐减小，直到某个限值( C) 随着土体蠕变，土体的内摩擦力逐渐减小，直到某个限值9、固结不排水试验剪切阶段的总应力路径轨迹为：(A)曲线 (B)直线 (C)不确定10、考虑土体的各向异性，下列说法正确的是：(A)正常固结粘土的水平向土样的强度常常小于竖直向土样的强度(B)正常固结粘土的水平向土样的强度常常大于竖直向土样的强度(C不确定第八章土压力与支档结构1、挡土墙后填土发生朗肯被动破坏，滑动面的方向如何确定？(A)与水平面夹角45+© 12(B)与水平面夹角45- © /2(C与水平面夹角452、库仑土压力理论通常适用于哪种土类？(A)粘性土 (B)砂性土 ( C)各类土3、设被动土压力和主动土压力充分发挥时所需的挡土墙水平位移分别为△1和厶2U:)(A)^〔＞△ 2 (B) △仁△ 2 (d〔＜△ 24、关于静止土压力、主动土压力、被动土压力下列说法正确的是:(A)静止土压力＞主动土压力＞被动土压力(B)被动土压力＞静止土压力＞主动土压力(C主动土压力＞静止土压力＞被动土压力5、对于无粘性土，在填土面水平、墙背竖直、墙背光滑的条件下，对朗肯土压力理论和库仑土压力计算的结果比较，则:( A 库仑土压力＞朗肯土压力(B 库仑土压力＜朗肯土压力(C 库仑土压力=朗肯土压力6、由库仑土压力理论计算出的主动土压力，其作用方向:( A 必垂直墙背( B 不一定垂直墙背( C 一般应为水平的7、刚性挡土墙墙背铅垂光滑，墙后土体之内摩擦角为©，挡土墙后移使墙后土体产生与水平面夹角为9的破裂面，土楔的重量为G,则按库仑土压力理论，作用在墙上的被动土压力为：( A)Gtan ( 9 +© )( B)G /tan ( 9 + © )( C)Gtan ( 9 -© )8、在下列公式的建立过程中，未涉及摩尔—库仑强度理论的是：(A)朗肯被动土压力(B)库仑主动土压力(C)静止土压力9、一般情况下，下列哪种土静止土压力系数最大？(A)粘性土 (B)砂性土 ( C)粉土10、在其他条件相同的情况下，当档土结构背离填土方向移动相同的位移时，下列哪种土最可能先达到主动土压力？(A)密砂 (B)松砂 (C)软粘土第九章地基承载力1、地基的极限承载力是指：(A)地基中形成连续滑动面时的承载力(B)地基中开始出现塑性区时的承载力(C地基的变形达到上部结构极限状态时的承载力2、地基的极限承载力公式时根据下列何种假设推导得到的？( A 根据建筑物的变形要求推导得到的(B 根据地基中滑动面的形状推导得到的( C 根据塑性区发展的大小得到的3、地基荷载达到某一数值后，基础的一侧到另一侧形成连续滑动面，基础四周地面隆起，基础倾斜，甚至倒塌，这种破坏模式称为：(A)整体剪切破坏(B)刺入破坏(C局部剪切破坏4、通常，加大浅基础埋深，会使得：(A)承载力降低，沉降减小(B)承载力提高，沉降增大(C)承载力提高，沉降减小(D承载力降低，沉降增大5、设地基为理想弹塑性体，基础为浅埋条形基础，地基的临塑荷载与下列哪项无关？(A)基础的埋深(B)基础的宽度(C地基土的内摩擦角(D 地基土的粘聚力6、下列说法正确的是：(A 地基破坏时，土中都将出现延伸至地表的滑面(B 荷载—沉降曲线上由直线转为曲线时所对应的荷载为极限荷载(C 以临塑荷载作为地基的承载力通常是偏于保守的第十章土坡稳定性分析1、堤坝填筑过程中，若施工速度很快，且土体和地基的渗透系数很小，则在分析堤坝竣工时的稳定性时，应采用下述哪种分析方法：( A 总应力分析法(B 有效应力分析法( C 总应力分析法或有效应力分析法皆可2、分析土坡稳定性时，对挖方土坡最小安全系数的说法正确的是：( A 挖方结束时，安全系数最小(B 最小安全系数出现在挖方过程中某个时间(C 稳定安全系数存在一个逐渐下降的过程，直至土坡形成后很长时间才趋于常数3、等厚度无限长砂性土坡下为基岩，坡面与水平面的夹角为25.7 度，砂性土的内摩擦角为30 度，则土体的安全系数为： ( A 1.0 ( B 1.2( C 1.44、无粘性土坡，内摩擦角为©，坡角为其稳定安全系数为：(A ) tan ©/tan B(B) sin ©/sin B( C tan B /tan ©。
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 anddifferences 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 sink it?A: 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 originalby 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 methodsused 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 porewater 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 the foundation.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 floatdensity, 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 that the 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 water level 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 thebearing 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.。
土力学试卷三
Seat No.: _____ Enrolment No.______GUJARAT TECHNOLOGICAL UNIVERSITYB. E. Sem. - V - Examination – June- 2011Subject code: 150604Subject Name: Geotechnical Engineering - IDate:27/06/2011Time: 10:30 am – 01:00 pmTotal Marks: 70Instructions:1. Attempt all questions.2. Make suitable assumptions wherever necessary.3. Figures to the right indicate full marks.Q.1 (a) 1) Differentiate Physical weathering and Chemical weathering. 2) Differentiate Flocculated structure and Honeycombed structure. 3) Define phase diagram and draw phase diagrams in terms of void ratio ‘e’ and porosity ‘n’.0202 03(b) An undisturbed soil sample has total weight of 2060 gm, volume of 1200cc, water content = 11 % and specific gravity G = 2.68. Compute (i) void ratio (ii) porosity (iii) degree of saturation (iv) water content to make sample fully saturated & (v) effective unit weight of the soil sample.07Q.2 (a) 1) Classify the given soil sample:- G = 20 %│% N : 10 20 30 60 90 100 S = 78 %│ f = 02 %│(Dmm): 1.28 2.98 3.07 4.80 4.92 5.25 2) Differentiate between coarse grained soils (CGS) and fine grained soils (FGS). A FGS has liquid limit of 65 % and plastic limit of 23 %. Classify the same.0304(b) 1) Define the term ‘Soil Structure’ and briefly explain the commonly observed soil structures. 2) Describe any one field identification test. 0502OR (b) 1) Define Toughness Index, Activity, Sensitivity and Thixotropy. 2) In a 10m thick sand deposit, ground water table (GWT) lies at 4.0m depth below the GL. Sand deposit has γt = 18 kN/m 3 & γsat = 20 kN/m 3. Compute effective stress values at the depths of 2.0m, 4.0m and 6.0m below the GL.0403Q.3 (a) 1) Explain the factors affecting permeability of soils. 2) Define ‘Critical Hydraulic Gradient’ and briefly explain ‘Quick Sand’ condition.0403(b) The following data were recorded while performing the compaction test:- Water content (%): 05 10 14 20 25 Bulk density (kN/m 3): 17.7 19.8 21.0 21.8 21.6Plot the MDD-OMC curve and obtain the optimum water content and maximum dry density. Calculate the water content necessary to completely saturate the sample at its maximum dry density, assuming no change in the volume. Also plot zero air voids curve. Take G = 2.6807OR Q.3 (a) A Falling Head permeameter accommodates a soil sample 6cm high and 50cm 2 in cross sectional area. The permeability of the sample is expected to be 1 x 10-4 cm/sec. If it is desired that the head in the Stand pipe should fall from 30 cm to 10 cm in 40 minutes, determine the size of the standpipe which should be used. If on the same soil sample a constant head of 200cm is maintained for 2 hours then how much quantity of water will flow?07(b) 1)Briefly explain the factors affecting compaction.2)During field compaction process, how the compacted density and themoisture content can be checked? 04 03Q.4 (a) 1)Briefly explain Coulomb’s, Mohr’s and Mohr-Coulomb’s failure theories.2)State merits and demerits of ‘Direct Shear Test’ and ‘TriaxialCompression Test’. 0304(b) Determine the shearing strength parameters from the Direct Shear Testresults given below. The proving ring constant is 0.5 kg/Div.Sr. No. Normal Stress (kg/cm2 ) Shear Force (kg)1. 1.0 1002. 2.0 1503. 3.0 220What would be shearing strength at the normal stress of 15 kg/cm2?07ORQ.4 (a) 1)Explain importance of ‘Unconfined Compression Test’ & ‘Laboratory Vane Shear Test’.2)Name and briefly explain the shear tests which may be performed basedon the different drainage conditions. 0304Q.4 (b) From the Undrained Triaxial test results given below, determine the total shear strength parameters c & φ by plotting conventional failure envelope ormodified failure envelope. Also state that at normal stress of 500 kPa, whatwould be the shear strength?Sr. No. Cell Pressure(kPa) Deviator Stress (kPa)1 100 1802 200 3203 300 50007Q.5 (a) Define the term ‘Consolidation’ and explain the same with the help of Terzaghi’s Spring Analogy concept. State the assumptions used inTerzaghi’s one dimensional consolidation theory.07(b) A 2.0 m x 2.0 m size footing placed at 2.0 m depth below the ground level(GL) is transmitting net pressure intensity of 200 kPa. The ground watertable lies at 2.0 m depth below the GL. Using the data given below, dividethe clay stratum in three parts and compute the settlement due toconsolidation:-(i)Top layer : 2.0 m thick sand, γt = 18 kN/m3(ii)Middle layer : 3.0 m thick NC clay, γsat = 20.1 kN/m3,γd = 16.0 kN/m3, Liquid Limit = 80 % & G = 2.7(iii)Bottom layer : sand07ORQ.5 (a) Define the terms coefficient of compressibility, coefficient of volume compressibility, compression index and coefficient of consolidation. Explainthe ‘square-root time fitting method’ for determination of coefficient ofconsolidation. 07(b) 1)Define the term ‘pre-consolidation pressure’ and briefly explain themethod for determination of the same.2)During consolidation test, the void ratio is determined to decrease from0.95 to 0.55 under the stress increment of 1.0 kg/cm2to 2.5 kg/cm2.Compute coefficient of compressibility, coefficient of volumecompressibility & compression index. 04 03*************。
三峡大学2013土力学双语试卷2
一、explain the following items(4point/word,total 20 point)1、compression coefficient2、Coulomb formula3、Hydraulic gradient4、Ultimate bearing capacity5、Stress increase二、fill vacancy1、Settlement can be divided into three component .initial immediate settlement ,primary consolidation settlement and secondary consolidation settlement .2、According to the overconsolidated raio ,soil can can be divided to 3 type;overconsolidated soil ,normally consolidated soil and underconsolidated soil3、Factor of safety is the ratio of the shear resistance to driving force4、Compare with Terzaghi’s theory . Hanse’s formulas including more factors such as shape depth inclination and tilt三、answer following questions1、how many types can be divided in the triaxial test according to consolidation and drainage conditions ?explain the consolidated-drained shear strength2、How many modes of shear failure may be defined ?what are the features of general shear failure?四、Calculation1、the volume of a soil specimen is 22.72cm2，and its mass is39.76g 。
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.
土建英语A卷答案
土建英语试卷A【闭卷】——参考答案及评分标准——Part I （15分，每题1分）1~5 BCABC 6~10 DCAAB11. last month 12.reliable 13.too long 14.friendly 15.Within one week.Part II （15分, 每题0.5 分）16~20 DABBD 21~25 CDCCB 26~29 CBBASection B (8分, 每题0.5 分)30~33 CDAB 34~37 DABC 38~41 BCDA 42~45CDBAPart III （30分，46—55 每题1分；56—65每题2分）46~50 BCDDC 51~55 CCADD56~60 ADACC 61~65 BBACDPart IV （25分，每题5分)66. 可行性报告是对项目的所有成本和总收入所做的财务分析，用于判断一个项目是否可以盈利。
完成一份可行性报告一般需要经过几个阶段。
67. 直接成本是指原材料成本、劳动力成本和设备成本，而间接成本包括办公费用、场地费用、工人的保险和交通费用等。
68. 每个建筑项目的设计都有其具体的特点，必要的设计步骤大致相同。
69. 施工就是将建筑设计转化为现实中的建筑物。
施工必须严格遵照设计图纸，从而建筑在要求的时间内完成施工。
70. 为一座建筑选择不同的建筑材料是设计师必须做出的选择。
天然材料，如木材、石头和泥砖，是最古老的建筑材料。
Part V （15分）作文评分标准1-5分格式不正确。
基本切题。
表达思想不清楚，连贯性差。
有较多的严重语言错误。
6-10分格式基本正确。
基本切题。
有些地方表达思想不够清楚，文字勉强连贯；有一定的语言错误。
11—15分格式正确。
切题。
表达思想清楚，文字通顺，连贯性好。
基本上无语言错误，或者仅有个别小错误。
公共外语II (土建英语) 试题A答案及评分标准第1 页（共 1 页）。
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.。
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.
长沙理工大学土力学模拟试卷及答案英文一
长沙理工大学拟题纸课程编号 080100285 拟题教研室（或老师）签名 教研室主任签名 ----------------------------------密------------------------------封--------------------------线------------------------------------ 课程名称（含档次） 土 力 学A （试卷2） 专业层次（本、 专 ） 本 科 专 业 土 木 类 考试方式（开、闭卷） 闭 卷一． 请对如下题目的说明给出对应的名词解释（要求中文和英文）：（每题4分，共20分） 1. A soil in which the current effective stress is lower than the past maximum stress.2. The water content at which a soil changes from a solid to a semisolid state without further change in volume.3. The stress carried by the soil particles and the liquids and gases in the voids.4. The pressure of the water held in the soil pores.5. The corresponding lateral pressure if the lateral strain in the soil is zero.二． 在下面题目的空格填写合适的英语单词 (每题4分,共20分)1. The effective stress represents the average stress carried by the soil solids and is the between the total stress and the .2. The upper and lower limits of the range of over which the soil exhibits plastic behaviour are defined as the liquid limit and the plastic limit, respectively. The water content range itself is defined as the .3. In the (LEM), the soil is considered to be on the verge of along an assumed or a known sliding surface.4. The of water flows through the soil is related to the (i) by Darcy’s law5. There are two methods to determine coefficient of (k) in the laboratory called test and Falling head test三．计算题: (第1题8分，其它每题9分，共44分)1. Derive the following formula from the definitions of the parameters:ω=eS r γwγsWhere ω，γs ，S r ，γw and e are the water content, the unit weight of soil particles, the degree of saturation , the unit weight of water and the void ratio respectively.共 2 页第 1 页2.某饱和土体积为97cm3，土的重力为1.98N，土烘干后重力为1.64N，求ω、γs、e及γd。
《土木工程专业英语》期末考试重点及参考答案
土木工程civil engineering结构工程structural engineering 土力学soil mechanics现场勘察site investigation项目经理project manager土木工程师civil engineer岩土工程Geotechnical Engineer 施工检查construction inspection 施工监管construction supervision助理工程师assistant engineer多孔材料porous material平衡条件equilibrium condition弹性模量elastic modulus应力张量stress tensor孔隙压力pore pressure有效应力principle of effectivestress专业技术人员professional andtechnical personnel压缩模块compressive modulus各向同性法向应力isotropicnormal stress应力分量stress components施工监管construction supervise1.Despite the essential尽管真正的工程师们在以上进步和人类幸福中所起的作用，但对他们所起作用的理解仍是不完全的。
munication lines道路、铁路、桥梁等交通网线是土木工程师的劳动果实，没有这样的交通网线，社会将不可能得到发展。
3.The principle of 流体力学原理可以用到日常生活中，如：飞机的飞行，水中鱼的游动，以及血管中的血液循环等。
4.Water supply engineering供水工程涉及水的定位和收集、水处理方法、标准极限试验和水的有效供给。
5.The role of an environmental环境工程师的作用是通过将各种技术应用到清理垃圾的工作中，从而在生物与技术之间建造一座桥梁。
土力学英文题(1)
Figure 1 Soil element with stresses acting on it
6.The plan of a uniformly loaded rectangular area is shown in Figure 2. Determine the vertical stress increase z below point A at a depth of z 4 m .
Soil Mechanics在此处键入公式。
1.The mass of a moist soil sample having a volume of 0.0057 m3 is 10.5kg. The moisture content(w) and the specific gravity of soil soilds(Gs) were determined to be 13% and 2.68 ,respectively. Determine a. Moist density,ρ (kg/ m3 ) b. Dry density,ρ d (kg/ m3 ) c. Void ratio, e d. Porosity, n e. Degree of saturation , s(%)
Impermeable layer
d
Figure 1 Soil profile with capillary rise
8. For the retaining wall shown in Figure 2, determine the force per unit width of the wall for Rankine’s active state. Also find the location of the resultant.
土力学试卷1(A)卷(后附答案)
2015-2016学年第1学期期末考试试卷（坏。（）
4、墙背光滑是朗肯土压力理论的基本假设。（）
5、在同样的地基上，基底附加应力相同的两个建筑物，其沉降值也相同。（）
四、计算题(共10分)
1、高度为5m的挡土墙，墙背直立和光滑，墙后填土面水平，试用朗肯理论计算墙背主动土压力分布及Ea值。
2015-2016学年第1学期期末考试试卷（A）答案及评分标准
8、某一土层排水固结，达到90%固结度时（即固结度 ）完成的沉降量 为20.0cm，
则该土层完全固结后，最终沉降量 将为：
A.16.2cm B.18.0cm C.20.0cm D.22.2cm
9、对同一种土，五个重度指标的大小顺序是（）
A.sat>>d>B>sat>>dC.sat>d>>D.sat>d>>。
4、提高挡土墙墙后的填土质量，使土体的抗剪强度提高，将使作用于墙背的（ ）
A．主动土压力增加B．主动土压力减小C．静止土压力增加D．被动土压力减小
5、饱和土的渗透固结过程应该是（）
A．孔隙水压力不断增加的过程B．有效应力的增加而孔隙水压力减小的过程
C．有效应力不断减小的过程D．有效应力的减小而孔隙水压力增加的过程
A.Δ1＞Δ2B.Δ1≥Δ2C.Δ1＜Δ2D.Δ1≤Δ2
13、评价粘性土软硬程度的指标常用______________。
《土壤学》双语-试卷(附整理答案).doc
读书破万卷下笔如有神一、土壤名词英汉互译（ 15% ）1、英译汉：(1) Pedosphere 土壤圈 , 土界 (2) Soil respiration 土壤呼吸 (3) Matric potential 衬质势 (4) Soil profile 土壤剖面(5) Plastic index 可塑指数2、汉译英：(1) 粘粒 clay particle （2）土壤容重 bulk density of soil （ 3）土壤质地soil texture （ 4）矿物 mineral （ 5）土壤肥力与土壤质量 soil fertility and soil quality二、简答题（ 4 题中任选 2 题用英语回答，另外 2 题用中文回答）（ 15%）1、 What are the three phases of a soil?土壤固相、液相、气相。
“固相”主要指土壤中各种岩石碎屑、矿物颗粒以及动、植物和微生物的残体；“液相”主要指土壤溶液或土壤水；“气相”指土壤空气或土壤中未被水占据的空隙部分。
2、 Where does soil organic matter come from?主要来源于动植物、微生物残体(trash) is employed)，排泄物(excretion/faeces) ，所施用有机肥料(the organic fertilizer和分泌物 (secretion)。
which3、 What are the major differences in composition between soil air and atmosphere?①． The level of CO2 in soil is about eight times higher than the normal atmospheric level.②． The O2 level in soil air is slightly lower than it is in atmospheric air.③． Soil air usually is much higher in water vapor than is the atmosphere.④． Also, under waterlogged conditions, the concentrations of gases such as methan （ CH4） and hydrogen sulfide (H2S) are notably higher in soil air.4、 What are basic( 碱性 /基性 ) cations(阳离子 ) and acidic cations in soils?三、选择题（10% ）（请您将正确答案的编号填入括号内）1、国际制土壤质地分类将粘粒定为（）。
浙江大学土力学题库及答案教学提纲
浙江大学土力学题库及答案第一章绪论1、土力学的英语是：（A）Soil Mechanics （B）Solid Mechanics （C）Soil Foundation2、岩土工程的英语是：（A）Rock and Soil Mechanics（B）Geotechnical Engineering（C）Rock and Soil Engineering3、下列哪位被誉为土力学之父？（A）库仑(Coulomb) （B）朗肯(Rankine) （C）太沙基(Terzaghi)4、土力学学科正式形成是哪一年？（A）1890 （B）1925 （C）19605、土力学主要研究地基那两方面的问题？（A）变形与渗流（B）变形和稳定（C）渗流与稳定6、浙江大学曾国熙教授倡导的岩土工程学科治学方法是？（A）理论研究与工程实践相结合（B）试验研究与理论研究相结合（C）基本理论、试验研究和工程实践相结合第二章土的物理性质与工程分类1、土颗粒的大小及其级配，通常是用颗粒累积级配曲线来表示的。
级配曲线越平缓表示：（A）土粒大小较不均匀，级配良好（B）土粒大小均匀，级配良好（C）土粒大小不均匀，级配不良2、土的不均匀系数Cu越大，表示土的级配：（A）土粒大小均匀，级配良好（B）土粒大小不均匀，级配良好（C）土粒大小不均匀，级配不良3、土的三相指标包括：土粒比重、含水量、重度、孔隙比、孔隙率和饱和度，其中哪些为直接试验指标？（A）孔隙比、含水量、土粒比重（B）土粒比重、含水量、重度（C）含水量、重度、孔隙比4、测定土的液限的标准是把具有30度锥角、质量76克的平衡锥自由沉入土体，沉入多少深度时的含水量为液限？（A）18mm （B）2mm （C）10mm5、压实能量越小，则（A）最优含水量越大（B）土越容易压实（C）土的最大干密度越大6、土的液限和塑限的差值（省去%符号）称为（A）液性系数（B）塑性系数（C）液性指数（D）塑性指数7、土的含水量一般用什么测定：（A）比重瓶法（B）烘干法（C）环刀法（D）搓条法8、某土的天然含水量为42%，液限35%，塑性指数17，孔隙比1.58，则该土应定名为：（A）淤泥（B）粉质粘土（C）淤泥质粘土9、土的密度一般用什么方法测定：（A）比重瓶法（B）烘干法（C）环刀法（D）搓条法10、关于土中的结合水，下列说法正确的是：（A）强结合水能传递静水压力（B）弱结合水能传递静水压力（C）强结合水和弱结合水能传递静水压力（D）强结合水和弱结合水都不能传递静水压力11、一般来说，粗大土粒往往是岩石经过什么作用形成？（A）物理和化学风化作用（B）物理风化作用（C）化学风化作用12、粘性土的塑限一般用什么方法测定？（A）比重瓶法（B）烘干法（C）环刀法（D）搓条法13、土的液性指数越大，则：（A）土的渗透性越大（B）土的塑性指数越小（C）土质越软14、土的塑性指数越小，则：（A）土的粘性越差（B）土的渗透性越好（C）土的变形越大15、土粒比重一般用什么方法测定：（A）比重瓶法（B）烘干法（C）环刀法（D）搓条法第三章土的渗透性与渗流1、不透水岩基上有水平分布的三层土，厚度均为1m，渗透系数分别为0.5m/d、 0.001m/d、 0.02m/d，则土层等效水平渗透系数为多少？（A）0.426m/d （B）0.174m/d （C）0.028m/d2、某饱和土的含水量为40%，土粒比重为2.75，则该土发生流土的临界水力坡降为：（A）1.02 （B）0.83（C）1.203、管涌形成的条件中，除具有一定的水力条件外，还与土粒几何条件有关，下列叙述正确的是：（A）不均匀系数大于10的土比不均匀系数小于10的土更容易发生管涌（B）不均匀系数小于10的土比不均匀系数大于10的土更容易发生管涌（C）与不均匀系数没有关系4、在防治渗透变形措施中，哪些措施是在控制水力坡降？（A）溢出部位铺设反滤层（B）上游做垂直防渗帷幕或设水平铺盖（C）下游挖减压沟5、不透水岩基上有水平分布的三层土，厚度均为1m，渗透系数分别为0.4m/d、0.05m/d、 0.003m/d，则土层等效竖向渗透系数为多少？（A）0.367m/d （B）0.063m/d (C)0.008m/d6、变水头试验常用来测定哪种土的渗透系数（A）粘性土（B）砂土（C）任何土7、计算成层地基中等效渗透系数时，水平等效渗透系数的大小取决于渗透系数____的土层，而垂直等效渗透系数的大小取决于渗透系数_____的土层；（A）大小（B）大大（C）小大（D）小小8、下列哪种土样更容易发生流砂：（A）粗砂（B）粉土（C）粉砂9、下列土体中渗透系数最小的是：（A）砂（B）粘土（C）粉土10、常水头试验常用来测定哪种土的渗透系数？（A）粘性土（B）砂土（C）任何土第四章地基中的应力计算1、地下水位下降将引起：（A）自重应力减小（B）地面沉降（C）超静孔压减小2、已知宽为3m，长为6m和另一宽为6m，长为12m的矩形基础，若两基础的基底附加应力相等，则两基础角点下竖向附加应力之间有何关系?（A）小尺寸基础角点下z深度处应力与大尺寸基础角点下2z深度处应力相等（B）大尺寸基础角点下z深度处应力与小尺寸基础角点下2z深度处应力相等（C）两基础角点下下z深度处竖向应力分布相同3、当地下水位从地表处下降至基底平面处，对基底附加应力有何影响？（A）不变（B）减小（C）增加4、条形均布荷载中心线下，附加应力随深度减小，其衰减速度与基础宽度有何关系？（A）基础宽度越大，衰减越慢（B）基础宽度越大，衰减越快（C）与基础宽度无关5、双层地基中上层土与下层土的变形模量之比越小，则下层土中的附加应力：（A）越小（B）不变（C）越大6、某方形基础中心点下1m处竖向附加应力为70kPa，则该基础中心点下2m处的竖向附加应力为：（A）95.2kPa （B）80.4kPa （C）56.3kPa7、在荷载作用范围外，竖向附加应力随深度的变化是：（A）竖向附加应力随深度的增加而减少（B）竖向附加应力随深度的增加，先增大，然后减小（C）竖向附加应力随深度的增加，先减小，然后增大8、对于天然地基，若土体竖向变形模量大于水平向变形模量，则与均质地基相比地基中竖向附加应力：（A）产生应力集中（B）产生应力扩散（C）没有变化9、对于均布条形荷载p作用下的附加应力为0.1p的等值线深度大约为基础宽度的几倍？（A）6；（B）3.5；（C）2；（D）1.510、在均布条形荷载作用下，最大剪应力发生在基础的什么部位？（A）基础中心（B）基础边缘（C）不确定第五章土的压缩性和固结理论1、土体压缩性e—p曲线是在何种条件下试验得到的？（A）无侧限条件（B）部分侧限条件（C）完全侧限2、从现场载荷试验p—s曲线上求得的土的模量为：（A）变形模量（B）弹性模量（C）压缩模量3、在室内压缩试验中，土样的应力状态与实际中哪一种荷载作用下土的应力状态相一致：（A）条形均布荷载（B）矩形均布荷载（C）无限均布荷载4、某地基土，已测得压缩指数为0.5，则回弹指数可能为：（A）1.0 （B）0.01 （C）0.15、某地基土，已测得压缩模量6MPa，则变形模量可能为：（A）7MPa （B）6MPa （C）5MPa6、双面排水，在土层厚度相同、性质相同的两个粘土层顶面，分别瞬时施加无限均布荷载p1=100kPa，p2=200kPa，试问经过相同时间t，两种情况的固结度有何不同？（A）p1=200kPa的固结度大（B）p2=400kPa的固结度大（C）两种情况的固结度相同7、某地基粘土层的厚度均为4m，情况1是双面排水，情况2是单面排水，当地面瞬时施加一无限均布荷载，达到同一固结度时，所对应的时间有何关系？（A）t1=4t2 （B）t1=2t2 （C）t1=t2 （D）t2=4t18、土的渗透性越好，则：（A）变形稳定越快（B）强度越小（C）有效应力增长越慢9、土的压缩性越大，则：（A）固结越快（B）固结越慢（C）与固结无关10、地基中的超静孔隙水压力产生的原因一般为？（A）地下水；（B）土自重与外加荷载；（C）地下水与外加荷载；（D）外加荷载11、某土样经压缩试验测得其在100kPa和200kPa压力作用下压缩24小时后的孔隙比分别为0.89和0.85，则该土样的：（A）压缩指数为0.25 （B）压缩模量为4.625MPa （C）压缩系数为0.4MPa-112、在固结过程中，饱和土体中的有效应力和孔隙水压力按如下规律变化：（A）有效应力减小，孔隙水压力减小（B）有效应力不变，孔隙水压力增大（C）有效应力增大，孔隙水压力减小第六章地基沉降计算1、在疏浚河道形成的新冲填土上建造建筑物，其沉降产生的原因：）（A）原地基的自重应力（B）冲填土自重（C）冲填土自重及建筑物荷载（D）建筑物荷载2、地基沉降计算深度是指：（A）主要压缩层厚度（B）主要持力层厚度（C）20倍地基宽度3、工后沉降一般是指：（A）全部的次固结沉降（B）工程完工后使用期内地基所发生的沉降（C）大部分的次固结沉降4、关于地基沉降，下列说法错误的是：（A）初始沉降的产生也需要一定的时间（B）土体固结沉降也有短时间完成的（C）将地基沉降分为三部分是从时间角度考虑的5、浙江沿海饱和软粘土地基一般情况下次固结沉降约占总沉降的：（A）40% （B）10% （C）15%6、在半无限体表面，瞬时施加一局部荷载，这时按弹性理论计算得到的应力是：（A）有效应力（B）总应力（C）孔隙压力7、初始沉降产生的原因：（A）土体体积不变，形状发生改变（B）土体体积发生变化（C）上述两种情况都有8、在常规的沉降计算方法中，地基中的应力状态是根据什么理论计算得到的？（A）线弹性理论（B）弹塑性理论（C）塑性理论9、考虑应力历史的影响时，沉降计算应采用哪种压缩曲线？（A）e—p曲线（B）e—logp曲线（C）上述两种均可10、规范法计算地基沉降时，采用的是哪种附加应力系数？（A）平均附加应力系数（B）附加应力系数（C）上述两种均可第七章土的抗剪强度1、饱和软粘土在不同竖向荷载p1>p2>p3作用下在直剪仪中进行快剪，所得的强度有何不同？（A）p越大，竖向荷载越大，所以强度越大（B）三者强度相同（C） p越大，孔隙水压力越大，所以强度越低2．十字板试验测得的抗剪强度相当于实验室用什么方法测得的抗剪强度？（A）不固结不排水试验（B）应力路径试验（C）固结排水试验3、直剪试验土样的破坏面在上下剪切盒之间，三轴试验土样的破坏面在什么位置上？（A）与试样顶面夹角为45（B）与试样顶面夹角为45-φ/2（C）与试样顶面夹角为45+φ/24、有一个砂样，在三轴试验时，在围压100kPa应力下，增加轴向应力使砂样破坏，已知砂样的内摩擦角为30度，则破坏时破坏面上的正应力为多少？（A）200kPa （B）180kPa （C）150kPa5、下列那个试验测试的不是土的天然抗剪强度：（A）十字板剪切试验（B）无侧限抗压强度试验（C）直接剪切试验（D）三轴不固结不排水试验6、现场原位测试土的抗剪强度的试验有：（A）十字板剪切试验（B）无侧限抗压试验（C）直接剪切试验（D）三轴压缩试验7、在做三轴试验时，加荷速率越快，则测得的：（A）抗剪强度越大（B）抗剪强度越小（C）对抗剪强度没影响8、对于蠕变，下列说法错误的是：（A）蠕变可能导致土体在没有达到抗剪强度的条件下破坏（B）随着土体蠕变，土体的粘聚力逐渐减小，直到某个限值（C）随着土体蠕变，土体的内摩擦力逐渐减小，直到某个限值9、固结不排水试验剪切阶段的总应力路径轨迹为：（A）曲线（B）直线（C）不确定10、考虑土体的各向异性，下列说法正确的是：（A）正常固结粘土的水平向土样的强度常常小于竖直向土样的强度（B）正常固结粘土的水平向土样的强度常常大于竖直向土样的强度（C）不确定第八章土压力与支档结构1、挡土墙后填土发生朗肯被动破坏，滑动面的方向如何确定？（A）与水平面夹角45+φ/2（B）与水平面夹角45-φ/2（C）与水平面夹角452、库仑土压力理论通常适用于哪种土类？（A）粘性土（B）砂性土（C）各类土3、设被动土压力和主动土压力充分发挥时所需的挡土墙水平位移分别为△1和△2，则：）（A）△1>△2 (B) △1=△2 （C）△1<△24、关于静止土压力、主动土压力、被动土压力下列说法正确的是：（A）静止土压力>主动土压力>被动土压力（B）被动土压力>静止土压力>主动土压力（C）主动土压力>静止土压力>被动土压力5、对于无粘性土，在填土面水平、墙背竖直、墙背光滑的条件下，对朗肯土压力理论和库仑土压力计算的结果比较，则：（A）库仑土压力>朗肯土压力（B）库仑土压力<朗肯土压力（C）库仑土压力=朗肯土压力6、由库仑土压力理论计算出的主动土压力，其作用方向：（A）必垂直墙背（B）不一定垂直墙背（C）一般应为水平的7、刚性挡土墙墙背铅垂光滑，墙后土体之内摩擦角为φ，挡土墙后移使墙后土体产生与水平面夹角为θ的破裂面，土楔的重量为G，则按库仑土压力理论，作用在墙上的被动土压力为：（A）Gtan（θ+φ）（B）G /tan（θ+φ）（C）Gtan（θ-φ）8、在下列公式的建立过程中，未涉及摩尔—库仑强度理论的是：（A）朗肯被动土压力（B）库仑主动土压力（C）静止土压力9、一般情况下，下列哪种土静止土压力系数最大？（A）粘性土（B）砂性土（C）粉土10、在其他条件相同的情况下，当档土结构背离填土方向移动相同的位移时，下列哪种土最可能先达到主动土压力？（A）密砂（B）松砂（C）软粘土第九章地基承载力1、地基的极限承载力是指：（A）地基中形成连续滑动面时的承载力（B）地基中开始出现塑性区时的承载力（C）地基的变形达到上部结构极限状态时的承载力2、地基的极限承载力公式时根据下列何种假设推导得到的？（A）根据建筑物的变形要求推导得到的（B）根据地基中滑动面的形状推导得到的（C）根据塑性区发展的大小得到的3、地基荷载达到某一数值后，基础的一侧到另一侧形成连续滑动面，基础四周地面隆起，基础倾斜，甚至倒塌，这种破坏模式称为：（A）整体剪切破坏（B）刺入破坏（C）局部剪切破坏4、通常，加大浅基础埋深，会使得：（A）承载力降低，沉降减小（B）承载力提高，沉降增大（C）承载力提高，沉降减小（D）承载力降低，沉降增大5、设地基为理想弹塑性体，基础为浅埋条形基础，地基的临塑荷载与下列哪项无关？（A）基础的埋深（B）基础的宽度（C）地基土的内摩擦角（D）地基土的粘聚力6、下列说法正确的是：（A）地基破坏时，土中都将出现延伸至地表的滑面（B）荷载—沉降曲线上由直线转为曲线时所对应的荷载为极限荷载（C）以临塑荷载作为地基的承载力通常是偏于保守的第十章土坡稳定性分析1、堤坝填筑过程中，若施工速度很快，且土体和地基的渗透系数很小，则在分析堤坝竣工时的稳定性时，应采用下述哪种分析方法：（A）总应力分析法（B）有效应力分析法（C）总应力分析法或有效应力分析法皆可2、分析土坡稳定性时，对挖方土坡最小安全系数的说法正确的是：（A）挖方结束时，安全系数最小（B）最小安全系数出现在挖方过程中某个时间（C）稳定安全系数存在一个逐渐下降的过程，直至土坡形成后很长时间才趋于常数3、等厚度无限长砂性土坡下为基岩，坡面与水平面的夹角为25.7度，砂性土的内摩擦角为30度，则土体的安全系数为：（A）1.0（B）1.2（C）1.44、无粘性土坡，内摩擦角为φ，坡角为β，其稳定安全系数为：（A）tanφ/tanβ（B）sinφ/sinβ（C）tanβ/tanφ。
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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(A) 适用年级、专业土木04-9，10 ——————————————————————————————————一Fill in the blanks (total 10 points, 1 points per blank)1. The soil fabric (结构) usually contains three kinds of basic types. (1) 单粒,(2) honeycomb fabrics, (3) 絮状.2. Basic physical indexes of soil include, bulk density of soil , G s and w3. In the triaxial test under unconsolidated-undrained (非固结非排水)，because only 一个有效应力园can be obtained, so that the internal friction angle could not be determined.4. The degree of saturation, S r, is defined as the ratio of the volume of 水to thevolume of 孔隙.5. If the soil mass pushes against a retaining wall such as to push it away, the soil becomes theactuating element and the pressure resulting thereby is known as 主动pressure. On the other hand, If the wall pushes against the soil the resultant pressure is known as 被动pressure.6. Based on Darcy’s law, for saturated soils, the average water flow velocity through soils is proportional to 水力梯度.二、Choose the right one from the four answers(total 20 points, 2 points per problem)A C CB B A A D A A三、Judge the following statement right or wrong. Mark with R for the rights and W forthe wrongs (total 10 points, 2 points per problem)R W W R W四、Explain the following concepts or definition(total 15 points, 3points per problem) 1. Relative Density of cohesionless soil (Dr)定义为：minmax max e e e e D r --=可用公式()()min max min max d d d d d d r D ρρρρρρ--=2. liquidity index I L液性指数是指粘性土的天然含水量和塑限的差值与塑性指数之比。
PPPL P L I w w w w w w I -=--=3. degree of consolidation饱和土在附加应力的作用下，孔隙中相应的一些自由水将随时间而逐渐排除，同时空隙体积也逐渐减少。
这个过程称为饱和土的渗透固结。
经历t 时间的固结变形量与最终变形量的比值或孔隙水压力的消散程度称为固结度。
4. net contact pressure基底压力与基底处建造前土中应力之差。
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 10 points, 5 points per problem)1． Give out the meanings of the concepts e+11 andee +1 in 3-phase diagram (3 相草图)e +11： 土粒子占总体积的比率 ee +1：孔隙所占总体积的比率2. 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。
六. A soil specimen has a water content w 1 of 10% and unit weight1γ of 18.0kN/m 3. Keepvoid ratio constant and increase water content to w 2 =20%. How much water will be infused into 1 m 3 soil . (8 points)()111111-+=γγw G e ws ()112222-+=γγw G e ws 2 points21s s G G = 21e e = 2 pointsSo that31212/64.1911m kN w w =++=γγ 2 pointskN w 64.1=∆ 2 points七. For a soil, the friction angle30=φ, cohesion c=10kPa, would failure occur on a planewithin a mass of this soil at point A where the principal stresses are 100kPa and 400kPa respectively, and point B where the shear stress is 100kPa and the normal stress 200kPa?(8 points)1.kPac f 6.334245tan 2245tan 231=⎪⎭⎫ ⎝⎛++⎪⎭⎫⎝⎛+=φφσσ 2 points11σσ<fffailure at A 2 points2.5.125tan =+=φστc f2 pointsττ<fnot failure at B 2 points八. 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 , thecompression coefficient14105.2--⨯=kPaa , and the coefficient of permeabilityyearcm k /0.2=.Find : (1) settlement after 1 year loading; (2) the length of time required for achieving asettlement of 20 cm. (12 points)1. Final settlement 2511=+=pH e a S 2 pointsy e a r cma e k C wv /106.1)1(251⨯=+=γ2 points16.02==Ht C T v v 2 points2 points=⨯=S U S t 0.45*25=11.25cm 2 points2. U=20/25=0.857.0=v T 2 points56.32==vv C H T t 2 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 the active 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 qKp a aa 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()kPaK q H p a a 24.19271.0)20317(=⨯+⨯=+=γ5. 底部主动土压力6.()kPaK q H H p a a 88.33271.0)20318317(2211=⨯+⨯+⨯=++=γγ7. 全主动土压力合力mkN 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mE X a66.2]3)24.1988.33(211324.195.1)67.667.23(321467.635.4[1=⨯-⨯⨯+⨯⨯+-⨯⨯⨯+⨯⨯=9．分布图2 points2 points2 points2 points———————————————————————————— 注意事项：1.请将评分标准连同考试题一并上交。
2.客观题注明每题的总分，每小题的分数。
3.主观性试题必须注明得分点及分值。
2 points3mEa ： 125.2kN/m。