题目内容
在求解土中一点的应力状态时，已知最大主应力σ1=410kPa，最小主应力σ3=150kPa，内摩擦角φ=25°，内粘聚力c=18kPa，则由σ1和σ3决定的莫尔应力圆（状态圆）的半径为kPa，圆心坐标为kPa，。以最小主应力σ3=150kPa不变，绘制的极限应力圆σ1f=kPa，半径为kPa；以最大主应力σ1=410kPa不变，绘制的极限应力圆σ3f=kPa，半径为kPa；土体处于半径为状态。（保留一位小数）
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