问答题
某一滑坡体，其参数见表，滑坡推力安全系数γ
t
=1.05，试计算推力F
3
。
| 块号 | 滑体重力G(kN) | 滑带长度l(m) | 倾角β(°) | 黏聚力c(kPa) | 摩擦角φ(°) |
|
| ① | 11000 | 50 | 40 | 20 | 20.3 | 0.733 |
| ② | 53760 | 100 | 18 | 18 | 19 | 1.0 |
| ③ | 5220 | 20 | 18 | 18 | 19 |
【正确答案】
【答案解析】解 根据《建筑边坡工程技术规范》(GB 50330—2002)第5.2.3条，块体抗滑力
R 1 =G 1 cosβ 1 f 1 +c 1 l 1 =11000cos40°×tan20.3°+50×20=4118kN
R 2 =53760×cos18°×tan19°+18×100=19388kN
R 3 =5220×cos18°×tan19°+20×18=2066kN
块体下滑力
T 1 =G 1 sinβ 1 =11000sin40°=11000×0.643=7071kN
T 2 =53760×0.309=16613kN
T 3 =5220×0.309=1613kN
推力
F 1 =1.05T 1 -R 1 =1.05×7071-4118=3307kN
F2=
+1.05T
2
-R
2
=3307×0.733+1.05×16613-19388=480kN
F 3 =
R 1 =G 1 cosβ 1 f 1 +c 1 l 1 =11000cos40°×tan20.3°+50×20=4118kN
R 2 =53760×cos18°×tan19°+18×100=19388kN
R 3 =5220×cos18°×tan19°+20×18=2066kN
块体下滑力
T 1 =G 1 sinβ 1 =11000sin40°=11000×0.643=7071kN
T 2 =53760×0.309=16613kN
T 3 =5220×0.309=1613kN
推力
F 1 =1.05T 1 -R 1 =1.05×7071-4118=3307kN
F2=
+1.05T
2
-R
2
=3307×0.733+1.05×16613-19388=480kN
F 3 =