圆形桩基强度裂缝计算

一.已知条件输入R=0.75截面半径（m)c=80中心保护层厚度(mm)Rs=0.67钢筋所在圆周的半径（m）砼采用C30（砼强度等级）钢筋采用HRB335（钢筋级别）n=28配筋数量d=28钢筋直径（mm）Md=4188承载力弯矩（KN·m）Nd=8658承载力轴力（KN）Lo=10构件计算长度（m)ro=1结构重要性系数fsd=fsd'=280钢筋设计强度（Mpa）Es=200000钢筋弹性模量（Mpa）g=0.893Rs/Rρ=0.0098纵向钢筋配筋率As/(πr^2)εcu=0.0033混凝土极限压应变fcd=13.8混凝土设计强度二.强度验算1.计算偏心距增大系数ηi=0.375截面回转半径Lo/i=26.666667考虑偏心距影响eo=Md/Nd=0.484(m)ho=R+Rs=1.42(m)1.120应≤1故ζ1=11.083应≤1故ζ2=1计算η=1.093η=1.093偏心距增大系数eoη=0.529(m)2.迭代计算（D62，P112)ξ=Xo/2/r假定ξ=0.555手动调制值β=0.8截面受压区矩形应力分布高度x与实际受压区高度Xo的比值θc=1.459与矩形应力分布高度x相应的截面受压面积所对圆心角一半θsc=1.155由周边均匀配置的纵向钢筋变换的薄壁钢环，在压塑区起点所对的圆心角一半θst=2.279由周边均匀配置的纵向钢筋变换的薄壁钢环，在拉塑区起点所对的圆心角一半A=1.3473B=0.6542C=0.2604D=1.8851eo=0.529m偏心距误差0.1%小于2%，即可。3.强度验算公式ζ1=0.2+2.7e0/h0=ζ2=1.15-0.01L0/h=圆形截面偏心受压构件'3300'220sdcddsdcddfgrDfBreNfrCfArNroNd=8658KNAr^2fcd+Cρr^2fsd'=10858.3KN满足要求roNdeo=4578.30KN·mBr^3fcd+Dρgr^3fsd'=5749.25KN·m满足要求三.裂缝计算Ns=6714短期组合轴力（KN）Ms=3136Ns对应弯矩（KN·m）Nl=5926长期组合轴力（KN)eo=0.4670837轴力Ns对应偏心距（m）ηs=1.0000使用阶段偏心距增大系数（6.4.4-8）fcu,k=30强度等级σss=86.45MpaC1=1C2=1.44C=66钢筋保护层Wfk=0.11(mm)编制：中南市政设计院安徽分院赵国强QQ：115469346欢迎批评指正。'3300'220sdcddsdcddfgrDfBreNfrCfArN)(65.118.242.59320,2MParefrNskcusss)(52.1004.003.021mmCdECCWsssfk由周边均匀配置的纵向钢筋变换的薄壁钢环，在压塑区起点所对的圆心角一半由周边均匀配置的纵向钢筋变换的薄壁钢环，在拉塑区起点所对的圆心角一半强度等级fckftkfcdftdEcC15101.276.90.882.20E+04C2013.41.549.21.062.55E+04C2516.71.7811.51.232.80E+04C3020.12.0113.81.393.00E+04C3523.42.216.11.523.15E+04C4026.82.418.41.653.25E+04C4529.62.5120.51.743.35E+04C5032.42.6522.41.833.45E+04C5535.52.7424.41.893.55E+04C6038.52.8526.51.963.60E+04C6541.52.9328.52.023.65E+04C7044.5330.52.073.70E+04C7547.43.0532.42.13.75E+04C8050.23.134.62.143.80E+04钢筋种类fskfsdESR2352351952.10E+05HRB3353352802.00E+05HRB4004003302.00E+05KL4004003302.00E+05