隧道结构力学计算课程设计-二次衬砌内力计算书

= - 1..7×107×4450.844×10-6 =-219.2947×10-6Р闭合差Δ=0Р4.墙底(弹性地基上的刚性梁)位移Р单位弯矩作用下的转角:Р?βa= 1KI8 = 10.4×106×23.4375=58.5938×10-6Р主动荷载作用下的转角:Р βap0 = M8P0βa =-2137.479×58.5938× 10-6 = -.9102× 10-6Р单位抗力及相应摩擦力作用下的转角:Р?βaσ0 = M8σ0βa = -13.9667×58.5938× 10-6 = -818.3613× 10-6Р五解力法方程Р衬砌矢高Рf = y8 = 7.16712mР计算力法方程的系数为:Рa11 = δ11 + βa = (28.7659+58.5938)×10-6=87.3597×10-6Р?a12 = δ12 + f βa = (40.62215+7.16712×58.5938)×10-6=460.5709×10-6Рa22 = δ22 + f2 βa = (140.47023+7.×58.5938)×10-6 =3150.2936×10-6Рa10 = ∆1p+ βap0 + (∆1σ+ βaσ0)×σh Р?= -(22105.2397+.9102+34.6258σh+818.3613σh)×10-6Р?= -(.1499+852.9871σh)×10-6Рa20 = ∆2p+ fβap0 + (∆2σ+ fβaσ0)×σhР =-(69667.4704+7.16712×.9102+184.6689σh+7.16712×818.3613σh)×10-6Р= -(.43695+6049.9625σh)×10-6Р以上将单位抗力及相应的摩擦力产生的位移乘以,即为被动荷载的载位移。Р求解方程为:РX1 = a22a10-a12a20a122-a11a22