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某场地进行岩土工程勘察,对某一层饱和黏性土取原状样,进行无限性抗压强度试验,测得10个士样的试验数据如下表所示,根据《土工试验方法标准》GB/T 50123 2019则该原状士层不排水抗剪强度cuc_u标准值为多少?

土样编号12345678910
无侧限抗压强度quq_u46.515.366.254.414.828.5100.854.368.540.2
解答
注意

按照三倍标准差作为舍弃标准

提示

cu=qu/2c_u = q_u / 2

《土工试验标准》附录A

qˉu=46.5+15.3+66.2+54.4+14.8+28.5+100.8+54.3+68.5+40.210=48.95KPa s=46.52+15.32+66.22+54.42+14.82+28.52+100.82+68.52+40.2210×48.952101=26.30 qˉu3s=48.953×26.3=20.95KPa qˉu+3s=48.95+3×26.3=127.85KPa \bar{q}_u = \cfrac{46.5 + 15.3 + 66.2 + 54.4 + 14.8 + 28.5 + 100.8 + 54.3 + 68.5 + 40.2}{10}= 48.95 KPa \\~\\ s = \sqrt{\cfrac{46.5^2 + 15.3^2 + 66.2^2 + 54.4^2 + 14.8^2 + 28.5^2 + 100.8^2 + 68.5^2 + 40.2^2 - 10 \times 48.95^2}{10 - 1}} = 26.30 \\~\\ \bar{q}_u - 3s = 48.95 - 3 × 26.3 = -20.95 KPa \\~\\ \bar{q}_u + 3s = 48.95 + 3 × 26.3 = 127.85 KPa \\~\\

无需舍弃数据

cv=26.348.95=0.537 γs=1(1.70410+4.678102)×0.537=0.686 quk=0.686×48.95=33.56KPa cuk=33.56/2=16.78KPac_v = \cfrac{26.3}{48.95} = 0.537 \\~\\ \gamma_s = 1 - (\cfrac{1.704}{\sqrt{10}} + \cfrac{4.678}{10^2}) × 0.537 = 0.686 \\~\\ q_{uk} = 0.686 × 48.95 = 33.56 KPa \\~\\ c_{uk} = 33.56 / 2= 16.78 KPa