A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 As one layer of clay of 5m thick: Divide the clay layer into (5) sub-layers each of 1m thick:Calculation of increase of stress below the center of each sub-layer Δσ ⃘(i): Weighted average pressure increase (Simpson's rule): "note in the picture solve the question but please solve in method 2:1"
A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 As one layer of clay of 5m thick: Divide the clay layer into (5) sub-layers each of 1m thick:Calculation of increase of stress below the center of each sub-layer Δσ ⃘(i): Weighted average pressure increase (Simpson's rule): "note in the picture solve the question but please solve in method 2:1"
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter10: Shear Strength Of Soil
Section: Chapter Questions
Problem 10.20P
Related questions
Question
A circular foundation 2m in diameter is shown in the figure below. A normally
consolidated clay layer 5m thick is located below the foundation. Determine the consolidation
settlement of the clay. method 2:1
- As one layer of clay of 5m thick:
- Divide the clay layer into (5) sub-layers each of 1m thick:Calculation of increase of stress below the center of each sub-layer Δσ ⃘(i):
- Weighted average pressure increase (Simpson's rule):
"note in the picture solve the question but please solve in method 2:1"
![A circular foundation 2m in diameter is shown in the figure below. A normally
consolidated clay layer 5m thick is located below the foundation. Determine the consolidation
settlement of the clay. method 2:1
(1) As one layer of clay of 5m thick:
- Divide the clay layer into (5) sub-layers each of 1m thick:
(2) Calculation of increase of stress below the center of each sub-layer Ao (1):
(3) Weighted average pressure increase (Simpson's rule):
Circular foundation
diameter B - 2m
G.S.
1.0m
Sand
q = 150 kN/m²
y = 17 kNhm²
0.5m
W.T.
Sand
0.5m
Normally consolidated clay
Y sat. = 18.5 kN/m²
C =0.16, e, =0.85
5.0m
Solution :
(1) As one layer of clay of 5m thick:
At the center of clay: 0,=1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²
For circular loaded area, the increase of stress below the center is given by:
1
Ao = q{1-
where: b=the radius of the circular foundation,
[(b/z)² +1]3/2
1
At mid-depth of the clay layer: z = 3.5m; Ao =150{1-
=16.66 kN/m²
(1/3.5)² +1]3/2
Ae =C, log10
o, + Ao
- = 0.16.log10
51.82 +16.66
= 0.0194
51.82
Де
-H; =
1+ 0.85
0.0194
-(5)(1000) =52.4 mm
1+e,
(2) Divide the clay layer into (5) sub-layers each of 1m thick:
• Calculation of effective stress at the middle of each sub-layer o'oi):
For 1*. Layer: oom=1.5(17) +0.5(19-9.81) + 0.5(18.5-9.81) = 35.44 kN/m²
For 2nd. Layer: oo(2)= 35.44 +1.0(18.5-9.81) = 35.44 + 8.69 = 43.13 kN/m²
For 3rd. Layer: oo(3) = 43.13 + 8.69 = 51.81 kN/m²
For 4. Layer: (4)= 51.81 + 8.69 = 60.51 kN/m²
For 5". Layer: (5)= 60.51 + 8.69 = 69.20 kN/m?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9724cf14-7cc2-4dfe-93b9-2d48372bf682%2F8ff15595-7970-451c-b355-156b27f3ab68%2Fsvb33mi_processed.png&w=3840&q=75)
Transcribed Image Text:A circular foundation 2m in diameter is shown in the figure below. A normally
consolidated clay layer 5m thick is located below the foundation. Determine the consolidation
settlement of the clay. method 2:1
(1) As one layer of clay of 5m thick:
- Divide the clay layer into (5) sub-layers each of 1m thick:
(2) Calculation of increase of stress below the center of each sub-layer Ao (1):
(3) Weighted average pressure increase (Simpson's rule):
Circular foundation
diameter B - 2m
G.S.
1.0m
Sand
q = 150 kN/m²
y = 17 kNhm²
0.5m
W.T.
Sand
0.5m
Normally consolidated clay
Y sat. = 18.5 kN/m²
C =0.16, e, =0.85
5.0m
Solution :
(1) As one layer of clay of 5m thick:
At the center of clay: 0,=1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²
For circular loaded area, the increase of stress below the center is given by:
1
Ao = q{1-
where: b=the radius of the circular foundation,
[(b/z)² +1]3/2
1
At mid-depth of the clay layer: z = 3.5m; Ao =150{1-
=16.66 kN/m²
(1/3.5)² +1]3/2
Ae =C, log10
o, + Ao
- = 0.16.log10
51.82 +16.66
= 0.0194
51.82
Де
-H; =
1+ 0.85
0.0194
-(5)(1000) =52.4 mm
1+e,
(2) Divide the clay layer into (5) sub-layers each of 1m thick:
• Calculation of effective stress at the middle of each sub-layer o'oi):
For 1*. Layer: oom=1.5(17) +0.5(19-9.81) + 0.5(18.5-9.81) = 35.44 kN/m²
For 2nd. Layer: oo(2)= 35.44 +1.0(18.5-9.81) = 35.44 + 8.69 = 43.13 kN/m²
For 3rd. Layer: oo(3) = 43.13 + 8.69 = 51.81 kN/m²
For 4. Layer: (4)= 51.81 + 8.69 = 60.51 kN/m²
For 5". Layer: (5)= 60.51 + 8.69 = 69.20 kN/m?
![• Calculation of increase of stress below the center of each sub-layer Aoi):
1
For 1“. Layer. Aoq) =150'1--
|
= 63.59 kN/m²
[(1/1.5)² + 13/2
1
For 2nd. Layer: Ao 2) =1
= 29.93 kN/m²
[(1/2.5)² + 1]3/2
1
For 3rd Layer: Ao) =150{1.
=16.66 kN/m²
[(1/3.5)² +1]3/2
1
For 4. Layer: Ao(4) =150
- =10.46 kN/m²
[(1/4.5)² +1]3/2
1
For 5th. Layer: Ao(5) =150
= 7.14 kN/m²
[(1/5.5)² +1]3/2
Ae 6)
AH;
Ae (1)
- AH;
1+e,
Layer no.
kN/m2
kN/m²
m
35.44
43.13
51.82
63.59
29.93
0.0727
0.0366
0.0393
0.0198
0.0105
0.0060
0.0037
1
2
1
3.
1
16.66
0.0194
4
1
60.51
10.46
0.0111
1
69.20
7.14
0.00682
Σ -00793
oʻoi) + Ao) .
Ae 6) = C, log10
Cc =0.16, e,=0.85, Sc= 0.0793 m= 79.3 mm.
(3) Weighted average pressure increase (Simpson's rule):
At the center of clay: o% =1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²
1
At z = 1.0m from the base of foundation: Ao =150
= 75 kN/m²
[(1/1)² +1}3/2]
1
At z = 3.5m from the base of foundation: Ao =150{1-
=16.66 kN/m?
[(1/3.5)² + 1}3/2 J
1
At z = 6.0m from the base of foundation: Ao =150{1-
= 6.04 kN/m²
[(1/6)? +1]3/2 ]
(Ao; + 4Aom + Ao) ==[75 + 4(16.66) + 6.04] = 24.61 kN/m²
o, +Ao -0.16.log10
.. Aoavg.
51.82 + 24.61
Ae =C, log10
= 0.027
51.82
Де
-Hi
1+e.
0.027
-(5)(1000) =72.9 mm
1+ 0.85](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9724cf14-7cc2-4dfe-93b9-2d48372bf682%2F8ff15595-7970-451c-b355-156b27f3ab68%2Fhiq212a_processed.png&w=3840&q=75)
Transcribed Image Text:• Calculation of increase of stress below the center of each sub-layer Aoi):
1
For 1“. Layer. Aoq) =150'1--
|
= 63.59 kN/m²
[(1/1.5)² + 13/2
1
For 2nd. Layer: Ao 2) =1
= 29.93 kN/m²
[(1/2.5)² + 1]3/2
1
For 3rd Layer: Ao) =150{1.
=16.66 kN/m²
[(1/3.5)² +1]3/2
1
For 4. Layer: Ao(4) =150
- =10.46 kN/m²
[(1/4.5)² +1]3/2
1
For 5th. Layer: Ao(5) =150
= 7.14 kN/m²
[(1/5.5)² +1]3/2
Ae 6)
AH;
Ae (1)
- AH;
1+e,
Layer no.
kN/m2
kN/m²
m
35.44
43.13
51.82
63.59
29.93
0.0727
0.0366
0.0393
0.0198
0.0105
0.0060
0.0037
1
2
1
3.
1
16.66
0.0194
4
1
60.51
10.46
0.0111
1
69.20
7.14
0.00682
Σ -00793
oʻoi) + Ao) .
Ae 6) = C, log10
Cc =0.16, e,=0.85, Sc= 0.0793 m= 79.3 mm.
(3) Weighted average pressure increase (Simpson's rule):
At the center of clay: o% =1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²
1
At z = 1.0m from the base of foundation: Ao =150
= 75 kN/m²
[(1/1)² +1}3/2]
1
At z = 3.5m from the base of foundation: Ao =150{1-
=16.66 kN/m?
[(1/3.5)² + 1}3/2 J
1
At z = 6.0m from the base of foundation: Ao =150{1-
= 6.04 kN/m²
[(1/6)? +1]3/2 ]
(Ao; + 4Aom + Ao) ==[75 + 4(16.66) + 6.04] = 24.61 kN/m²
o, +Ao -0.16.log10
.. Aoavg.
51.82 + 24.61
Ae =C, log10
= 0.027
51.82
Де
-Hi
1+e.
0.027
-(5)(1000) =72.9 mm
1+ 0.85
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