最小支护抗力计算

兰州交通大学隧道及地下工程（最小支护抗力计算）班级:姓名:学号:编程序计算并绘图：按塑性应力平衡条件求解最小支护抗力Pamin值oC=0.294MPa,q）=36.9°,H=156.Om，Y=19.6kN/m3,a=10mo解：参数求解如下表:埋深容重洞径粘聚力内摩擦角RbZio15619.6100.29436.91.1767734.0052583.057615.20024按塑性应力平衡条件求解由卡柯公式，求得作用在M护结构上的围岩松散压力R值:c•ctg（p[l-（普）支护抗力值:Rb2RbaRb2Rbat——）（一）i尸-1八拓Pa=— F（ F° £-1圣+1由于a=10m在坑道不变的情 10/15.20024^-^1尸0绘出Pa 及P[-3的关系曲线图（见下方excel表格）找出两曲线的交点A,交点A处的七值即为Famin值。从图形中得出竺在0.672^0.673之间尸0根据Pi=Pa当-=0.67225时Pa值为Pamin<r0此时心=14.8754m=0.02698MPa埋深容重洞径粘聚力内摩擦角R>£ro15619.6100.29436.91.1767734.0052583.057615.20024Px0.0281370.0271750.027096a/r00.6578840.670.6710.0270160.6720.0269960.672250.0268560.6740.0267770.6750.0266970.6760.0266170.6770.0265370.6780.0264570.6790.0263770.680.0247680.70.020690.750.0165530.80.0157220.810.0148890.820.0140560.830.0132220.840.0123870.850.0115530.860.0107190.870.0098860.880.0090530.890.0082210.90.007390.910.006560.920.0057320.930.0049060.940.0040820.950.003260.960.0024410.970.0016240.980.0015430.9810.0014610.9820.001380.983P.0.0006680.0227790.02464a/r00.6578840.670.6710.0265070.6720.0269740.672250.0302570.6740.032140.6750.0340290.6760.0359240.6770.0378240.6780.039730.6790.0416410.680.0810630.70.1899060.750.3142740.80.3411010.810.3685990.820.3967760.830.4256410.840.45520.850.4854640.860.516440.870.5481360.880.580560.890.6137210.90.6476280.910.6822880.920.717710.930.7539020.940.7908720.950.8286280.960.867180.970.9065340.980.9105140.9810.9145020.9820.9184990.9830.0012980.9840.9225030.9840.0012170.9850.9265150.9850.0011360.9860.9305360.9860.0010540.9870.9345650.9870.0009730.9880.9386020.9880.0009650.98810.9390060.98810.0009570.98820.939410.98820.0009490.98830.9398140.98830.000940.98840.9402190.98840.000940.9884050.9402390.9884050.0009240.98860.9410280.98860.0009160.98870.9414330.98870.0009080.98880.9418370.98880.00090.98890.9422420.98890.0008920.9890.9426470.9890.0008840.98910.9430520.98910.0008750.98920.9434570.98920.0008670.98930.9438620.98930.0008590.98940.9442670.98940.0008510.98950.9446730.98950.0008430.98960.9450780.98960.0008350.98970.