CE 468 – Example: Stone Columns

CE 468 – Example: Stone Columns
CE 468 – Geotechnical Design
Example: Stone Columns
Given: 8m height and 25 m width (at base) fill (γ=18 kN/m3) will be constructed for the bridge
approach embankments in Gebze-Orhangazi Motorway Project. Site investigation show that there is
12 m thick soft-medium stiff high plasticity clay deposit overlies weathered rock with SPT-N>R. By
using SPT correlations and considering the laboratory test results soil properties are: γn = 20 kN/m3,
c’=5 kPa, Φ’=27°, cu = 18 kPa, E’=7 MPa. Ground water level is at ground surface. For 50 mm
allowable settlement, and FS=1.5, check the design criteria and design a proper stone column pattern.
Assume n=14 and D=0.7 m in present site conditions. Crushed stone used for the stone column has an
internal friction angle of 43° Assume Δcu / Δσ =~ 0.22
Solution:
1. Bearing Capacity Check
𝑞𝑠𝑢𝑟𝑐ℎ𝑎𝑟𝑔𝑒 = 𝛾 ∗ 𝐻 = 18 ∗ 8 = 144 𝑘𝑃𝑎
𝑞𝑛𝑒𝑡,𝑢𝑙𝑡 = 𝑐𝑢 ∗ 𝑁𝑐 = 18 ∗ 5.14 = 92.5 𝑘𝑃𝑎 𝑤ℎ𝑒𝑟𝑒 𝑁𝑐 𝑐𝑜𝑚𝑒𝑠 𝑓𝑟𝑜𝑚 𝑆𝑘𝑒𝑚𝑝𝑡𝑜𝑛 (1951)
𝐹𝑆 =
𝑞𝑛𝑒𝑡,𝑢𝑙𝑡
𝑞𝑠𝑢𝑟𝑐ℎ𝑎𝑟𝑔𝑒
=
92.5
= 0.642 < 1.5 𝑁𝑜𝑡 𝐴𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒
144
2. Settlement Check
𝑆𝑐 = 𝑚𝑣 ∗ ∆𝐻 ∗ ∆𝜎
1
Where: 𝑚𝑣 = 𝐸 ′ = 0.14 ∗ 10−3 𝑚2 ⁄𝑘𝑁
𝐹𝑑 =
𝑎𝑛𝑑 ∆𝜎 = 𝐹𝑑 ∗ 𝑞𝑠𝑢𝑟𝑐ℎ𝑎𝑟𝑔𝑒
𝐵∗𝐿
𝐵
=
𝑓𝑜𝑟 𝑠𝑡𝑟𝑖𝑝
(𝐵 + 𝑧) ∗ (𝐿 + 𝑧) 𝐵 + 𝑧
After dividing the soil into 4 sublayers, following table is obtained:
Layer ID
ΔH (m)
Zmid-depth (m)
Fd
1
3
1.5
0.943
2
3
4.5
0.847
3
3
7.5
0.769
4
3
10.5
0.704
Sci
S=201 mm >> 50mm
Not Accepted
Since both criteria are not satisfied, site can be improved by stone columns (vibro-replacement). For
D=0.7m stone columns:
3. Design of Stone column
3.1. Settlement Criteria
Sall = 50 mm= Streated (St)
𝑆𝑅𝑅 =
𝑆𝑡
𝑆𝑢𝑛𝑡𝑟𝑒𝑎𝑡𝑒𝑑
Dr. Zeynep ÇEKİNMEZ
=
50
1
= 0.25 =
−→ 𝑎𝑟 = 0.23 = 23% 𝑓𝑜𝑟 𝑛 = 14
201
1 + (𝑛 − 1) ∗ 𝑎𝑟
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CE 468 – Example: Stone Columns
For triangular pattern:
0.23 =
𝜋 ∗ 0.72 ⁄4
−→ 𝐷𝑒 = 1.46𝑚−→ 𝐷𝑒 = 1.05𝑠−→ 𝑠 ≅ 1.4𝑚
𝜋 ∗ 𝐷𝑒2 ⁄4
3.2. Bearing Capacity Check
For n=14:
𝜎𝑐 = 𝜎 ∗
𝑛
14
= 144 ∗
= 504 𝑘𝑃𝑎
1 + (𝑛 − 1) ∗ 𝑎𝑟
4
𝑞𝑢𝑙𝑡,𝑠𝑡𝑜𝑛𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = 𝑡𝑎𝑛2 (45 +
𝐹𝑆 =
43
) ∗ (4 ∗ 𝑐𝑢 + 𝜎𝑟′ ) = 5.29 ∗ 6 ∗ 𝑐𝑢 = 571.3 𝑘𝑃𝑎
2
𝑞𝑢𝑙𝑡,𝑠𝑡𝑜𝑛𝑒 𝑐𝑜𝑙𝑢𝑚𝑛
= 1.13 < 1.5 𝑤ℎ𝑒𝑟𝑒 1.5 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑚𝑖𝑛. 𝐹𝑆 𝑓𝑜𝑟 𝐵. 𝐶. 𝑠𝑡𝑜𝑛𝑒 𝑐𝑜𝑙𝑢𝑚𝑛
𝜎𝑐
Above results show that when the full height of the fill is constructed, bearing capacity criterion is not
satisfied. Hence, fill should be constructed in two stages (Stage1: 5 m + Stage2: 3 m)
For Stage 1
𝜎 = 5 ∗ 18 = 90 𝑘𝑃𝑎
Stone Column: For n=14  σc = 90*14/4 = 315 kPa  FS=571.3 / 315= 1.8 > 1.5 OK
Soil: σs = σc / n = 315 / 14 =22.5 kPa and qult = cu*Nc =18*5.14=92.5 kPaFS=92.5/22.5=4.1>1.5 OK
Where Nc is Skempton’s Nc for strip (CE 366 LN)
At the end of stage 1, after waiting sufficient time:
𝑐𝑢 = 𝑐𝑢,𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + ∆𝑐𝑢 𝑤ℎ𝑒𝑟𝑒 ∆𝑐𝑢 = 0.22 ∗ ∆𝜎
For σ=90 kPa, at the mid-depth of clay layer (z=12/2=6m);
∆𝜎 = 90 ∗
25
= 72.6 𝑘𝑃𝑎
25 + 6
𝑐𝑢,𝑓𝑖𝑛𝑎𝑙 = 18 + 0.22 ∗ 72.6 = 34 𝑘𝑃𝑎 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒 1
For Stage 2
𝜎 = 8 ∗ 18 = 144 𝑘𝑃𝑎
Stone Column: for n=14  σc = 144*14/4 = 504 kPa
For cu, final = 34 kPa qult, stone column = 5.29*6*34=1077.8 kPaFS=1077.8 / 504= 2.1 >1.5 OK
Soil: σs = σc / n = 504 / 14 =36 kPa and for cu, final = 34 kPa cu*Nc =34*5.14=174.8 kPa 
FS=174.8 / 36= 4.9 > 1.5 OK
4. Conclusion
0.7 m diameter vibro-replacement columns (stone columns) will be constructed in triangular pattern
with s= 1.4 m (ar = 23%) where fill should be constructed in two stages (as 5m+3m).
Dr. Zeynep ÇEKİNMEZ
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