WORKED EXAMPLES 115
Solutions
- Core depth 40 6 46 m. Mean ∆cfor core 20 46/2460 kN m^2.
Therefore
(^) 1c0.5 460 46/10^4 1.06 m (equation (2.30)).
Settlement under core, clay B: at mid-height,
∆B0.95 20 46 874 kN m^2 (equation (2.32)),
(^1) B0.52 m (equation (2.31));
hence the nominal crest level41.6 m AOD.
- Point 1:∆21.5 20 430 kN m^2. Therefore at mid-height in clay
A,
∆A0.98 430 421 kN m^2 (equation (2.32)),
(^1) A0.20 m (equation (2.31)),
and in clay B,
∆B0.95 430 408.5 kN m^2 ,
(^1) B0.25 m.
Therefore (^) total0.45 m at point 1.
Point 2: ∆21.5 40 860 kN m^2. Therefore (^) total0.90 m at
point 2.
Point 3: ∆ 20 40 800 kN m^2. Therefore ∆ in cut-off
0.98 800 784 kN m^2 ,
(^) cin cut-off0.24 m
and∆B0.95 800 760 kN m^2. (6 m depth cut-off balances 6 m
excavated in clay A.) Therefore,
(^) B0.46 m.
0.6 760 10
104
0.6784.6
104
0.6408.5 10
109
0.8 421 6
104
0.6 874 10
104