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1.本曲线按弹性变形及容许应力的原理编制。按受拉区不参与工作的大偏心受压构件验算公式进行强度计算。Fa’=Fa,当偏心距e=M/N>K时为大偏心受压。式中:M—弯矩,N—垂直力,K=W/F,F—断面积,W—断面模量。中性轴至力N的距离Y按下式计算:y~3+3y[2nFa’/b(c+c’)-g~2] +2[-3nFa’/b(c~2+c’~2)+g~3]=0式中:g=e=-b/2,c=e+(b/2)-a c’=g+a’。令q=2nFa’/b(c+c’)-g~2 q=-3nFa’/b(c~2+c’~2)+g~3则
1. The curve according to the principle of elastic deformation and allowable stress preparation. According to the tension zone does not participate in the work of large eccentric compression member checking formula for strength calculation. Fa ’= Fa, when the eccentricity e = M / N> K, it is big eccentric compression. Where: M-moment, N-vertical force, K = W / F, F-section area, W-section modulus. The distance Y from the neutral axis to the force N is calculated as follows: y ~ 3 + 3y [2nFa ’/ b (c + c’) - g ~ 2] +2 [-3nFa ’/ b (c ~ 2 + c’ ~ 2) + g ~ 3] = 0 where g = e = -b / 2, c = e + (b / 2) -a c ’= g + a’. Let q = 2nFa ’/ b (c + c’) - g ~ 2 q = -3nFa ’/ b (c ~ 2 + c’ ~ 2) + g ~ 3