清华版《土力学》陈希哲教材答案.ppt

习题,土木工程专业（建筑工程方向）,3.1（P134）,当第四层为强风化岩时,,,sckPa,Z,o,a,b,c,,,27.0,,60.84,,78.48,,132.48,,,,,,sckPa,Z,o,a,b,c,,,27.0,,60.84,,78.48,,,,3.2P135,解,3.3P135,解,,,,,,,,,z,,,,,,,,,,3.4P135,,,,,,,,,b/2,b/2,,,,,,L5b,,解取一半L/b10，按均布条形荷载边点下考虑,,,z/b0,x/b0.5,α0.5,σz1.0*0.5*10050.0kPa,z/b0.50,x/b0.5,α0.481,σz1.0*0.496*10048.1kPa,z/b1.0,x/b0.5,α0.410,σz1.0*0.410*10041.0kPa,z/b2.0,x/b0.5,α0.275,σz1.0*0.275*10027.5kPa,z/b4.0,x/b0.5,α0.153,σz1.0*0.153*10015.3kPa,z/b6.0,x/b0.5,α0.104,σz1.0*0.104*10010.4kPa,,,,,,,,,z,,,,,,,,,3.5P135,e～p曲线,中压缩性土,3.6P135,,,,,,,,,,,,,,14.0m,5.0m,5.0m,,6.0m,基础中心点下,A,M点下,,,,3.7P135,,,,,,,6.0m,,,,,,0.25m,P2400kN,,,,3.0m,,,,,,9.0m,,,,,,,,,,300,500,,,,,,,,,,,100,po300kPa,b9.0m,z9.0m的均布荷载,,,x/b0.5,z/b1.0,α10.410,σz10.41*300123.0kPa,po400kPa,b9.0m,z9.0m的三角形荷载,x/b0.5,z/b1.0,α20.25,σz20.25*30075.0kPa,po500kPa,b3.0m,z9.0m的均布荷载,x/b0.5,z/b3.0,α30.198,σz30.198*50099.0kPa,po100kPa,b3.0m,z9.0m的三角形荷载,x/b0.5,z/b3.0,α40.10,σz40.10*10010.0kPa,,σzσz1σz2–σz3-σz4123.075.0-99-10.089.0kPa,,,po200kPa,b9.0m,z9.0m的均布荷载,x/b0.5,z/b1.0,α10.410,σz10.41*20082.0kPa,po300kPa,b9.0m,z9.0m的三角形荷载,x/b-0.5,z/b1.0,α20.16,σz20.16*30048.0kPa,,po200kPa,b3.0m,z9.0m的均布荷载,x/b0.5,z/b3.0,α30.198,σz30.198*20039.6kPa,po100kPa,b3.0m,z9.0m的三角形荷载,x/b-0.5,z/b3.0,α40.10,σz40.10*10010.0kPa,σzσz1σz2–σz3-σz482.048.0-39.6-10.080.4kPa,,,,,po300kPa,b6.0m,z9.0m的均布荷载,x/b6.0/6.01.0,z/b9.0/6.01.5,α10.211,σz10.211*30063.3kPa,po200kPa,b6.0m,z9.0m的三角形荷载,x/b6.0/6.01.0,z/b9.0/6.01.5,α20.0.13,σz20.13*20026.0kPa,直接按条形荷载计算（计算点位于大边下）,σzσz1σz263.326.089.30kPa,（计算点位于小边下）,x/b-6.0/6.0-1.0,z/b9.0/6.01.5,α20.09,σz20.09*20018.0kPa,σzσz1σz263.318.081.30kPa,3.8P135,,,,,,,d1,,N1,,,,,,b1,,,,,b1,粉土,,,6b1,,,,2N1,,,,,,,2b1,粘土,问两基础的沉降量是否相同何故通过整d和b，能否使两基础沉降量接近说明有几种方案，并给出评介。,在第1层土内中心点下的沉降差,在第2层土内中心点下的沉降差,条形平均附加应力系数查不到，借用矩形面积上均布荷载角点下的平均附加应力系数,设,2、1基础间的沉降差,最有效的方法调整d，加大基础2的埋深。,调整基底宽度b，加大基础1的b，,亦可使,3.9P135,,3.10P136,,,,,,,2.0m,,F,,,,,,4.0m,,,,,4.0,人工填土,,,1.6m,,粘土,卵石,,求粘土层沉降量。,,用中心点下的系数,,3.11P136,,,,,,,2.0m,,N4720kN,,,,,,4.0m,,,,,4.0,细砂,,,3.0m,,粉质粘土,碎石,用分层总和法计算粉质粘土层沉降量。,,,4.5m,分层，取层厚1.0m，,,,各层面处的附加应力如下,L/b2.0/2.01.0,附加应力计算表,沉降量计算表h1.0m,,,3.12P136,,,,,,,1.0m,,N900kN,,,,,,2.0m,,,,粉质粘土,用规范法计算基础中心点的最终沉降量。,取,返算厚度,取,设,4.07.01.00.7,3.13P136,,,,,,,1.5m,,N576kN,,,,,,2.0m,,,,,4.4m,杂填土,,,6.5m,,卵石,粉土,用规范法计算基础中心点的最终沉降量。,,,天然地面,返算厚度,取,设,2.54.01.101.0,3.14P136,,,,,,,1.5m,,N706kN,,,,,,2.0m,,,,,2.5m,粉质粘土,,,6.6m,,卵石,粘土,用规范法计算基础中心点的最终沉降量。,,,天然地面,粉土,,,5.8m,返算厚度,取,设,2.54.01.101.0,3.15P136,,,,,,,,4.0m,,,8.0m,粘土,,,砂层,,,,砂层,,,计算地基沉降与时间的关系需补充a、e及t后才能计算。,基底尺寸42.5m*13.5m,4.1P185,,未坏。,4.2P185,,,1,2,3,4.3P185,1,,2,,4.4P186,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,300,,1800,,,,,,,,,,,4.5P186,,,,170kPa,,不会破坏。,4.6P186,,4.7P186,,4.8P186,K线表示法,,,,总应力,,,,总应力,,,,总应力,,,,有效应力,总应力,a13kPa,由图知σ125，τ49,,,有效应力,a3kPa,由图知σ100，τ59,4.9P186,未坏；,当,时，,破坏；,4.10P186,4.11P187,1,2,3,地基的临界荷载随基底宽度与埋深的增加而增加，相比之下，随埋深增大，临界荷载增加的更显著。,4.12P187,不满足,按太沙基极限承载力公式计算,根据,查P173图4.30得,按汉森极限承载力公式计算,根据,查表4.5得,4.13P187,按太沙基极限承载力公式计算,根据,查P173图4.30得,设,4.14P187,按汉森极限承载力公式计算,根据,查P179表4.5得,（1）,（2）,当地下水上升至基础底面时,4.15,1极限荷载,按斯肯普顿极限承载力公式计算,（2）按浅基础地基的临界荷载公式,查P165表面4.4得,4.16P187,按汉森极限承载力公式计算,根据,查P179表4.5得,,,,,,干砂,5.1P236,,,填土面,（1）求静止土压力值,（2）求产生主动土压力所需的位移,（3）求主动土压力值,（1）,密实砂土，产生主动土压力所需的位移约为墙高的0.5，,即,（2）,3,5.2P237,,,,,,砂土,,,填土面,,,,,地下水位面,,（1）求静止土压力值,（2）求主动土压力值,（3）求墙后水压力值,（1）,根据,得,1,2,3,,,,,,,,,,,,2,根据,得,,,,,,,,,,,,3,5.3P237,根据,查表5.1得,5.4P237,根据,查表得,5.5P237,,,,,,,,1.5m,,,,,,2.5m,,,,,,,,,5.0m,,,,,,,5/3,,可以,,不可以,5.6P237,,,,,,,,3.0m,,,,,,,,,,,,,,,,,,,,中砂,粗砂,3.0m,,,,地下水,,,4.0m,,,求主动土压力和水压力,1,2,3,4,,,,,,,,3.0m,,,3.0m,,,,,,4.0m,1,2,3,4,,,,,,,,,,,,,,,,,,,3.0m,3.0m,,4.0m,,,,,,,,,,,,,,,,,,,,,,5.7P237,,,,,,1.0m,,,,,,5.0m,,,,,,7.0m,,,,1.5m,,,,,中砂,细砂,（1）求作用在墙背上的土压力；（2）作用在墙前趾上的土压力，（3）验算挡土墙抗滑稳定性。,,,,（1）作用在墙背上的土压力；,根据,查表得,（2）作用在墙前趾上的土压力,,,1.5m,,,,,,根据,按式5.16计算得,,（3）验算挡土墙抗滑稳定性。,安全,5.8P237,根据,查图5.48得,,5.9P237,根据,查图5.48得,5.10P237,,,,,,,,,,3.0m,3.0m,,,,7.0m,,,,,,,,,,,,,,,,,,2,4,6,8,,,,,,,,,,,,,2,4,6,,第1个小土条,b1.0m,