地铁的站钢支撑轴力计算新

91,1.2N/φA=(1.2×2456×10³)/(0.791×222.5×10²)=165.67N/mm²<f=205N/mm²,满足要求。二、钢支撑拼接管处受力计算:钢支撑受力最小截面图查表得:f取215N/mm²,截面属于a类截面。(一)、受力截面几何特性截面积A=π(D²-d²)/4=3.14(60.9²-57.7²)/4=297.9cm²截面惯性矩Ix=π(D^4-d^4)/64=3.14(60.9^4-57.7^4)/64=131050cm^4Iy=π(D^4-d^4)/64=3.14(60.9^4-57.7^4)/64=131050cm^4回转半径ix=√Ix/A=√131050/297.9=21cmiy=√Iy/A=√131050/297.9=21cm(二)、截面验算1.强度σ=1.2N/A=(1.2×2456×10³)/(297.9×10²)=98.9N/mm²<f=205N/mm²,满足要求。2.刚度和整体稳定性λx=lox/ix=14.5×10²/21=69?<[λ]=150,满足λy=loy/iy=60.9/21=2.9查表,构件对x轴y轴屈曲均属a类截面,因此由λmax?λx,λy=69,查附表得φ=0.844,1.2N/φA=(1.2×2456×10³)/(0.844×297.9×10²)=117.2N/mm²<f=215N/mm²,满足要求。(2).刚度和整体稳定性(考虑单根最长钢支撑23.5米处)λx=lox/ix=23.5×10²/21=112?<[λ]=150,满足λy=loy/iy=60.9/21=2.9查表,构件对x轴y轴屈曲均属a类截面,因此由λmax?λx,λy=112,查附表得φ=0.548,1.2N/φA=(1.2×1737×10³)/(0.548×297.9×10²)=127.7N/mm²<f=215N/mm²,满足要求。