高等土力学(李广信)-课后习题解答 可直接打印版

13208.9kPa，q50kPap100kPa3()tg'yx2zxyx,y,z分别代三个方向上的主应力，以σ1＝σz,σx=σy=σ3为0，计算完成下表。关于的解释zzˊ＝60yxxy3()tg'yx2zxybyxzx2b(1)3ctg1z(zx)计y()算yx公22q1bb(zx)1bbz(2)式q的z(3)2推1bb导ybz(4)3pzy3x1bpz(5)x3bzxz(6)b0yxy(7)567天，U＝94％；n=100，U＝99％－时间？81Uet294%0.0046ctTvvH2n100,10000t80.01e10000t,t2.3h2已知砂土试样的1＝800kPa，2＝500kPa，3＝200kPa，计算I1、I2、I3、J2、J3、p、q和各是多少；如果1＝800kPa，2＝3＝200kPa，上述各值为多少？1800kPa,2500kPa,3200kPa1800kPa,2200kPa,3200kPad(13)d1dd1t(dd)d1d3123EEtEtt不排水试验d1•dididuduB[d3A(d1d3)ddE13td11A(12t)z＝k1.0x0.33,z/xkqφ’=32.1；φcu=11.9RTCCTCuB[3A(13)]1/34c1210kPa,RTC：φ’=20；φcu=13答案G=2.672me=0.89H=2mH=2m0.54t/m22t/m21.76t/m2H=2m3.76t/m2H=0.31mH=1.69mH=2m2t/m20.44t/m21896.519170.015225.1z4m1m4m1m1020304050(m)2.34.33.83.34.77.5扬压力(t/m2)，答案8820.36kPa78.48kPa答案1.8m10.8m9m21.18kPa12.83kPa105.95kPa6m6m51.21kPa111.84kPa，370，c0；3sat18.7KN/mi=9/12水头损失各4.5m90kPapaKazKa(j)zppKpzKp(j)z105kPa49.92kPa6m6m56.07kPa111.84kPa答案(a)0.91m10.8m9.89m62.63kPa0kPa0kPa2.81m1.8m0kPa17.66kPa1.99m3m11.37kPa23.25kPa2.5m2.5m27.85kPa21.29kPa3.5m3.5m57.11kPa0kPa17m5m72.65kPa88.04kPaθ=36.5H2h2Q1.366k5050Rr0hlHhr028.2mlglg(10.2)r0lr03.03m11.97m237.15kPa6m15m9m90kPa147.15kPa查表得μ0=0.8，μ1=0.70，Si=0.112m(未修正)qB2002S0.80.711.2cmi01E2000，讨论：从图可见：如果粘土在上部，它阻挡了第二层土的排水路线，可以近似用8m厚的Ⅰ层土计算；如果粘性土在下部，它实际上成为不排水层，可以近似用6m后的第Ⅱ层土计算。所以上述近似计算方法不合理。ⅠⅡⅡⅠ