3⭐

某建筑桩基础采用散口 PHS ，桩型为PHS-AB400(250)型承台埋深 2.0m，地下水埋深 4.5m，桩总长 24.0m，桩顶嵌入承台 0.5m， 桩端开口，桩尖长 50cm.持力层为5层，其地层单桥静力触探结果如下表，根据《建筑桩基技术规范》JGJ 94-2008，试计算单桩竖向承载力特征值为多少?

土层名称	层底深度(m)	静力触探 p_s(MPa)
粉质黏土	4.0	1.2
淤泥质土	12.5	0.62
细砂	17.5	5.0
粉质黏土	23.0	3.5
中砂	26.5	22.2

解答

注意

4d, 8d 的d使用400mm

$$p_s \in (20, 30) \Rarr p_{sk2} = 5/6 × 22.2 = 18.5MPa, p_{sk1} = \cfrac{3.5 × 1.2 + 18.5 × 2 }{3.2} = 12.875MPa \\~\\ \cfrac{p_{sk2}}{p_{sk1}} = 1.437 < 5 \Rarr \beta = 1, p_{sk} = \cfrac{12.875 + 18.5}{2} = 15.69MPa \\~\\ 6m q_{sk1} = 15kPa \\~\\ 6~12.5m q_{sk2} = 0.06 × 620 = 31KPa \\~\\ 12.5 ~ 17.5m q_{sk3} = 100kPa \\~\\ 17.5 ~ 23m q_{sk4} = 0.016 × 3500 +20.45 = 76.45kPa \\~\\ 23~25 q_{sk5} = 100kPa \\~\\ A_{p1} = \pi × 0.25^2 / 4 = 0.049 m^2 \\~\\ A_j = 0.4^2 - 0.049 = 0.111 m^2 \\~\\ h_b / d_1 = 2 / 0.25 = 8 \geq 5 \Rarr \lambda_p = 0.8 \\~\\ \alpha = 0.75 + \cfrac{0.9 - 0.75}{30 - 15} × (23 -15) = 0.83 \\~\\ Q_{uk} = 0.4 × 4 ×(15 × 4 + 6.5 × 31 + 5 × 100 + 5.5 × 76.45 + 2 × 100) + 0.83 × 15690 × (0.111 + 0.8 × 0.049) = 4167.2 \\~\\ R_a = 4167.2/2 = 2083.6kN$$