© 2012 Taylor & Francis

Chapter 3

Effective stress

3.1

Figure Q3.1

Total vertical stress at 5 m depth:

kPa

Pore water pressure:

kPa

Therefore, effective vertical stress:

kPa

Alternative method:Buoyant unit weight:

kPa

Effective vertical stress:

kPa σ′v = 51 kPa

The complete stress profile within the ground may be obtained using the spreadsheet tool Stress_CSM8 on the Companion Website. The complete solution is shown on the follow-ing page. The spreadsheet may then be used to query the stresses at 5 m depth, giving the values shown above.

Effective stress 23

© 2012 Taylor & Francis

3.2

Figure Q3.2

Total vertical stress at 5 m depth:

kPakPa

kPa

Alternative method:Buoyant unit weight:

kPakPa σ′v = 51 kPa

The complete stress profile within the ground may be obtained using the spreadsheet tool Stress_CSM8 on the Companion Website. The complete solution is shown on the follow-ing page. The spreadsheet may then be used to query the stresses at 5 m depth, giving the values shown above. Comparing the answer with Problem 3.1, it is clear that when the water table is above the ground surface, the depth of the overlying water has no effect on the effective stresses within the ground (though σv and u do change).

3.3

Figure Q3.3

24 Effective stress

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At the top of the clay:

kPakPa

kPa σ′v = 51.4 kPa

At the bottom of the clay:

kPa

The elevation head (due to the location of the WT) is 6 m. Artesian pressure is additional to this, 4 m above ground level (AGL), i.e. 6 m above the WT:

kPakPa σ′v = 33.3 kPa

(N.B. The alternative method of calculation shown in Problems 3.1 and 3.2 is not applica-ble under artesian ground water conditions.) The complete stress profile within the ground may be obtained using the spreadsheet tool Stress_CSM8 on the Companion Website. The complete solution is shown on the following page. The spreadsheet may then be used to query the stresses at 4 and 8 m depth, giving the values shown above.

3.4

Figure Q3.4

Total vertical stress at 8 m depth:

kPakPa

kPa σ′v = 106 kPa

The complete stress profile within the ground may be obtained using the spreadsheet tool Stress_CSM8 on the Companion Website. The complete solution is shown on the follow-ing page. The spreadsheet may then be used to query the stresses at 8 m depth, giving the values shown above.

Effective stress 25

© 2012 Taylor & Francis

3.5

Figure Q3.5

(a) Immediately after WT rise:At 8 m depth, pore water pressure is governed by the new WT level because the permeability of the sand is high.

kPakPa

kPa σ′v = 94 kPa

At 12 m depth, pore water pressure is governed by the old WT level because the permea-bility of the clay is very low. (However, there will be an increase in total stress of 9 kPa due to the increase in unit weight from 16 to 19 kN/m3 between 3 and 6 m depth: this is accompanied by an immediate increase of 9 kPa in pore water pressure.)

kPakPakPa σ′v = 154 kPa

(b) Several years after WT rise:At both depths, pore water pressure is governed by the new WT level, it being assumed that swelling of the clay is complete. Therefore, at 8 m depth:

kPa (as above) σ′v = 94 kPa

At 12 m depth:

kPakPa

kPa σ′v = 134 kPa

26 Effective stress

© 2012 Taylor & Francis

The complete stress profile within the ground may be obtained using the spreadsheet tool Stress_CSM8 on the Companion Website. The complete solution is shown on the follow-ing page. The spreadsheet may then be used to query the stresses at 8 m depth, giving the values shown above.

3.6

Figure Q3.6

Total weight of the element = vector ab:

Boundary water force on CD = vector bd. Referring to the figure above:

Boundary water force on BC = vector de. Referring to the figure above:

Resultant body force = vector ea. Measuring from the force diagram (drawn to scale) gives |ea| = 9.9 kN. This acts in a direction of 17° to the vertical. Resultant body force = 9.9 kN (@ 17° to vert.)

Effective stress 27

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3.7

Figure Q3.7

(a) For case (1), h = 2 m at the top of the element and h = 0 m at the bottom. Seepage is therefore occurring from top to bottom and at the centre of the sample, h = 1 m:

kPa

kPa

kPa σ′v = 30.2 kPa

For case (2), h = 2 m at the bottom of the element, h = 0 at the top and seepage is occurring from bottom to top. As before, h = 1 m at the centre of the sample:

kPa

kPa

kPa σ′v = 10.6 kPa

(b) In both cases, Δh = 2 m across the sample, which has a length of Δs = 4 m. The hydraulic gradient is therefore i = Δh/Δs = 0.5 from top to bottom in (1) and from bottom to top in (2). This gives a seepage pressure j = iγw = 4.9 kN/m3 downwards in (1) and upwards in (2). For case (1):

kPa σ′v = 30.2 kPa

For case (2):

kPa σ′v = 10.6 kPa

28 Effective stress

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3.8

The flownet for this problem was found in Problem 2.3, and is shown below:

Figure Q3.8

The average exit hydraulic gradient may be found directly using the results of Flownet_CSM8, or reading from the flownet sketch. In the latter case:

Loss in total head between adjacent equipotentials:

Δs ≈ 0.85 m, so exit hydraulic gradient:

The critical hydraulic gradient is given by Equation 3.12:

Therefore, factor of safety against ‘boiling’ (Equation 3.14):

F = 1.76

The pore pressures at C and D may also be found directly using the results of Flownet_CSM8 to get the total head, or reading from the flownet sketch.

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Total head at C:

m

Elevation head at C:

m

Pore water pressure at C:

kPa

Therefore, effective vertical stress at C:

kPa σ′vC = 14.5 kPa

For point D:

mm

kPakPa σ′vD = 93.4 kPa

3.9

The flow net for this problem was found in Problem 2.2, and is shown below:

Figure Q3.9

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For a soil prism 1.50 × 3.00 m adjacent to the piling (shown shaded in the above figure):

m (from Flownet_CSM8)

Factor of safety against ‘heaving’ (Equation 3.13):

F = 1.88

With a filter:

Depth of filter = 17.4/21 = 0.83 m (if above water level). Filter height = 0.83 m