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Volume 6 – Geotechnical Manual, Site Investigation and Engineering ...

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Chapter 6 SLOPE STABILITY<br />

ii.<br />

The height of each slice is set off below the centre of the base, <strong>and</strong> the<br />

normal <strong>and</strong> tangential components, h cos α <strong>and</strong> h sin α respectively are<br />

determined graphically as shown in Figure 3.3. Thus:<br />

W cos α<br />

W sin α<br />

=<br />

=<br />

30h cos α<br />

30h sin α<br />

iii.<br />

The pore water pressure at the centre of the base of each slice is taken to be<br />

γwzw, where zw is the vertical distance of the centre point below the water<br />

table (Fig 3.3 refers). [Note: This procedure slightly overestimates the pore<br />

water pressure, which strictly should be γwze, where ze is the vertical<br />

distance below the point of intersection of the water table <strong>and</strong> the<br />

equipotential through the centre of the slice base. The error involved is<br />

however, on the safe side].<br />

iv. From Figure 6.13, the overall arc length, La is calculated as 14.35m. v.<br />

The results are summarised in Table 6.2 below.<br />

Table 6.5 Summary of Results<br />

Slice No.<br />

1<br />

2<br />

3<br />

4<br />

5<br />

6<br />

7<br />

8<br />

h cos α<br />

(m)<br />

0.75<br />

1.80<br />

2.70<br />

3.25<br />

3.45<br />

3.10<br />

1.90<br />

0.55<br />

h sin α<br />

(m)<br />

- 0.15<br />

- 0.10<br />

0.40<br />

1.00<br />

1.75<br />

2.35<br />

2.25<br />

0.95<br />

u<br />

(kN/m 2 )<br />

5.9<br />

11.8<br />

11.2<br />

18.1<br />

17.1<br />

11.3<br />

0<br />

0<br />

l<br />

(m)<br />

1.55<br />

1.50<br />

1.55<br />

1.10<br />

1.70<br />

1.95<br />

2.35<br />

2.15<br />

u.l<br />

(kN/m)<br />

9.1<br />

17.7<br />

25.1<br />

29.0<br />

29.1<br />

22.0<br />

0<br />

0<br />

17.50 8.45 14.35 132.0<br />

vi.<br />

Hence:<br />

Σ W cos α = 30 x 17.50<br />

Σ W sin α = 30 x 8.45<br />

Σ (W cos α - ul) = 525 <strong>–</strong> 132<br />

= 525 kN/m<br />

= 254 kN/m<br />

= 393 kN/m<br />

F = c' L a +tan' ∑Wcosα-ul<br />

∑ Wsinα<br />

= 10x14.35+(0.554x393)<br />

254<br />

= 143.5+218<br />

254<br />

= 1.42<br />

6A-2 March 2009

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