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1、沈阳建筑大学考试评分标准专用纸2007年 春 季学期 科目土力学(B)适用年级、专业土木04-9,10一 Fill in the blanks (total 15 points, 1 points per blank)1- 0.075mm d 2mm2. Gs,ws3.一个有效应力园4. 水 ,孔隙.5. 45。-业26. 水力梯度7, 沉降， 承载力8, 主固结， 次固结9, 4。C10, 40。11, 孔隙体积二、Judge the following statement right or wrong. Mark with R for the rights and W forthe wro
2、ngs (total 10 points, 2 points per problem)W R R W W三、 Explain the following concepts or definition(total 15 points, 3points per problem)1. void ratio孔隙比：孔隙体积与土粒体积之比，e = v / vvs2. coefficient of curvature Cc曲率系数定义为(Cc) C汁，曲率系数Cc描写累积曲线的分布范围，反映曲线的整体形c d d10 60 状。3. compression modulus压缩模量：土体在完全侧限的条件下，
3、竖向应力增量与竖向应变增量的比值4. overconsolidation土在应力历史上(固结过程中)所受到的最大有效应力，称之为前期固结应力. 前期固结应力与现有 的自重应力之比大于1 时，土体处于超固结状态。5additional stress由上部建筑荷载(基底附加压力)等外界因素在地基中产生的应力。五、Answer the following questions simply(total 15 points, 5 points per problem)1Explain the principle of effective stress.饱和土中任意点的总应力总是等于有效应力加上孔隙水压力
4、；或有效应力总是等于总应力减去孔隙水压力。既 c = + u 或 = u2. Explain “stress in soil”, why do we calculate “stress in soil”?土体在自身重力，建筑物荷载，交通荷载或其他外界因素等的作用下，产生土中应力。按起因可分为 资中应力和附加应力。因土中应力将引起土体或地基发生变形，试土工建筑物发生沉降，倾斜以及水 平位移。变形过大将影响建筑物的正常使用。并且，土中应力过大，会导致土体的强度破坏，使土工 建筑物发生土坡失稳或使地基的承载力不足而发生失稳。因此须计算土中应力。3. Give out the assumptions
5、in Terzaghis theory of one dimensional consolidation1. 土层是均质、各向同性和完全饱和的；2. 土的压缩完全是由于孔隙体积的减少，土粒和水是不可压缩的；3. 水的渗流和土层的压缩仅在竖向发生；4. 水的渗流遵从达西定律；5. 渗透系数k和压缩系数a保持不变。6. 外荷载一次瞬时施加。7土体变形完全是孔隙水压力消散引起的。五. There is a sample which is got from a certain natural sand column. The water contentmeasured by experiment is
6、 11%, bulk density p 二 1.70g / cm3, the minimum dry densityis p 二 1.41g / cm 3, the maximum dry density is p 二 1.75 g / cm 3. Please determine the sandysoils dense degree (6 points)pdp _ 1.701+W _ 1 + 0.11_1.53g /cm32 points(p p)pDr _ dd min d max _ 0.42 point(p p )pd maxd d在1/3至2/3之间，中密 2 point六. A
7、 soil sample 10cm in diameter is placed in a tube 1m long, A constant supply of water is allowed to flow into one end of the soil at A and outflow at B is collected by a beaker. The average amount of water collected is 1cm3 for every 10 seconds. The tube is inclined as shown in the below figure. Det
8、ermine the (a) hydraulic gradient, (b) flow rate, (c) average velocity, (d) coefficient of permeability. (6 points)Step 1 Define the datum position. Select the top of the table as the datumStep 2/Find tpe gotal heads at A (inflow) and B (outflow)H =h 丿 +(h ) = 1 +1 = 2mA ( p z AH = n + (h 丿=0 + 0.8
9、= 0.8mB pBz BStep 3Find the hydraulic gradientAH = H - H = 2 - 0.8 = 1.2mABl=1mi=Ar=12/1=122 pointsStep 4 Determine the flow rateVolume of water collected, Q=1cm3, t=10sec.Q =Q/t=1/10=0.1 cm3/ svStep 5 Determine the average velocityq=Av,兀(diam)2兀 x IO2A = =44= 78.5cm 2q 0.1V = v =A 78.5= 0.0013cm /
10、s2 pointsStep 6 Determine the coefficient of permeability. From Darcys law.k = V =皿i 1.2=10.8x10-4cm/ s2 points七. A saturated clay layer of thickness 10m overlays an impervious hard rock. A vertical uniform pressure p=200kPa is applied on the clay layer. The initial void ratio ise1 = 1.0 , thecompre
11、ssion coefficient a =2.5x10-4kPa-1 , and the coefficient of permeabilityk = 2.0 cm / year . Find : (1) settlement after 1 year loading; (2) the length of time required for achieving a settlement of 20 cm. (11 points)Tv0.140.160.180.50.570.6U0.430.450.480.760.800.821. Final settlement s = a pH = 252
12、points1 + e1C =丄=1.6 x105 cm2 / year2 pointsv awCtt = Cvt = 0.162 pointsv H 2S = U x s = 045*25=1125cm2 pointst2 U=20/25=08t = 0.571 pointsvt = TvH2 = 3.562 pointsCv2u 2八. Prove the Relationship E 二(1 -) E between the deformation modulus e and the01 - p s0compression modulus E Where u is Poissons ra
13、tio (5 points) sG = G=KGxy0z=0xyGG&=x-p -JxEE00KGz=p / (1 - p0GxGG 二z=z-p p E00)Under laterallyG& 二 00二告G - 2必0 )0GxE01 points1 pointspointsconfinedcondition ezG zEs1 pointsThe relationship betweenthe deformation modulus and the compression modulus is given as2p2E0 = (1-总)Es1 points九 A frictionless
14、retaining wall is shown in the figure Determine the force per unit width ofthe Rankins active state Also find the location of the resultant and give the distribution of theactive pressure (12 points)Ka1=tan 2 45。一I 2丿=tan(45 -卫)=1232顶部与交界处上部主动土压力p = qK 一 2c ： K = 20 x (1/3)aaa= 6.67kPa3.Ka2=tan2 454.a=23.67kPa=tan2(45 35)= 0.271交界处下部主动土压力2 points2 pointsp =H + q)K = (17 X 3 + 20) x 0.271 = 19.24kPa aa5.底部主动土压力6. p =G H +Y H + q)K = (17x3 +18x3 + 20)x0.271 = 33.88kPaa 1 12 2 a7. 全主动土压力合力Ea = 3 x 6.67 + 2 x 3 x (23.67 6.67) + 3 x
