根据题意计算如下:
①基础底面处自重应力为:pc1=∑yihi=5×20+3×10 = 130kPa;
②基础底面中心点以下深度18m处的自重应力为: pcz =∑yihi=5×20 +3×10 + 18×10 =310kPa;
③基础底面附加应力为:Po=pk-pc1=425-130=295kPa;
④m =l/6 = (42/2)/(30/2) =1.4,n =z/6 = 18/(30/2) =1.2;
查《建筑地基基础设计规范》(GB 50007 — 2002)表K. 0.1-1得,附加应力系数α= 0. 171,则计算点处的附加应力为:px =4αPo =4 ×0.171×295 =201. 8kPa。
![](https://assets.51tk.com/images/57cd23e248b37a19_img/f992d7fe9932e876.jpg)