求助！~普通混凝土试件大于等于10组的强度评定计算实例！~最好详... [复制链接]

自己去看评定标准附录D,里面有公式 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã

这样不对,做为管理员,这样对待朋友,不合理 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã

    设计值    平均值    方差    最小值    判定系数    　 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
n    fcu,k    mfcu    sfcu    fcu,min    λ1    λ2    0.9fcu,k    　0.95fcu,k    1.15fcu,k    mfcu-λ1.sfcu    λ2.fcu,k    　 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
50    40    52.4     2.7     45.2     1.6    0.85    36.0     38.0     46.0     48.2     34.0     　 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
每组强度值（MPa） ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
1    2    3    4    5    6    7    8    9    10    11    12    13    14    15    16    17    18    19    20 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
50.3     48.6     50.3     48.5     54.1     50.2     51.5     50.9     51.9     56.6     51.9     52.2     55.3     53.4     54.8     54.1     53.5     56.2     55.2     53.6  ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
53.6     52.0     53.0     52.1     48.1     48.8     46.8     50.0     48.4     50.3     54.5     53.3     45.2     48.5     55.9     53.4     55.1     55.6     55.1     54.2  ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
50.9     50.6     53.9     54.2     54.3     54.5     54.2     55.0     52.3     52.7     　    　    　    　    　    　    　    　    　    　 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
　    　    　    　    　    　    　    　    　    　    　    　    　    　    　    　    　    　    　    　 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
因为：mfcu-λ1.sfcu=48.2>0.9fcu,k=36.0 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
       fcu,min=45.2>λ2.fcu,k=34.0 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã
所以，评定合格 ¾Y¼wbbs.3c3t.comvº¬<¬Ÿ;ȇã