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 Aoo(): (3) Weighted average pressure increase (Simpson's rule): Circular foundation diameter B = 2m G.S. 1.0m Sand q = 150 kN/m y = 17 kNhm² 0.5m W.T. Sand 19 kNiw! 0.5m Normally consolidated clay Y sat. = 18.5 kN/m² C_ =0.16, e, =0.85 5.0m %3D Solution :

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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"please resolve the question but using method 2:1"

note / the value load (q) is KN/mis stree not load

method 2:1 q/(B+z) why the q * area

b is circular

"note in the picture solve the question but please solve in method 2:1"

• 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
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
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?
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?
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