HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY

FACULTY OF CIVIL ENGINEERING

_____________

PROJECT OF FOUNDATION ENGINEERING

DESIGN A STRIP FOUNDATION AND PILE FOUNDATION
(Subject code: CI3239)

Name & Student code

Le Hoang Phuc - 1952927
Advisor:
Assoc. Prof. Le Ba Vinh

Sem 212, May, 2022

1
 TABLE CONTENT

PART 1: DESIGN A STRIP FOUNDATION

I. The preliminary structural model and calculating data 6

II. Foundation materials 6

III. Preliminary size of foundation 7

IV. Check ground stability 8

V. Check of deformation 11

VI. The preliminary size of footing 12

VII. Check of differential settlement 13

VIII. Check of punching shear 14

IX. Calculation of reinforcement 14

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 PART 1: DESIGN A STRIP FOUNDATION
A: STATISTIC OF SOIL PROPERTIES 1A
Based on results survey at two bore hole (HK1, HK2), each bore hole has depth 30m,
including 4 soil layers:
• Water table has depth 1.2m
• Soil layer A
Concrete floor, debris, sand, soil, we do not take this layer to survey.
o Dry unit weight : γ = 22 kN/m3
• Soil layer B
clay, ash-gray, plastic state, we do not take this layer to survey.
• Soil layer 1
clay mixed with gravel gravel laterite, gray white - sepia – yellow brown, hard
plasticity:
Bore hole Depth, m Thickness, m SPT, Hammer
HK1 1.9 0.9 12-14
HK2 1.7 1.3 13-15
o Moisture content W : W = 23.035%
o Bulk unit weight : γI = 19.6 ± 0.584 kN/m3
γII = 19.6 ± 0.268 kN/m3
o Internal friction angle : φII = 12.075 ± 1.0885
o Cohesion coefficient : cII = 25.525 ± 5.2066 kN/m2
• Soil layer 2
sand (y) clay, grey yellow – sepia, hard plasticity.
Bore hole Depth, m Thickness, m SPT, Hammer
HK1 2.8 8.9 5-11
HK2 3.0 5.4 6-13

o Moisture content W : w = 20.7929 %

o Saturated unit weight : γI = 19.9286 ± 0.0876 kN/m3
: γII = 19.9286 ± 0.051 kN/m2
o Internal friction angle : φII = 13.4092 ± 0.7849
o Cohesion coefficient : cII = 23.6214 ± 3.7439 kN/m3
• Soil layer 3
sand (y), grey white, yellow-brown, sepia, hard plasticity:

Bore hole Depth, m Thickness, m SPT, Hammer

3
 HK1 11.7 13.6 6-11
HK2 8.4 18.5 5-13

o Moisture content W : w = 19.2156 %

o Bulk unit weight : γI = 20.25 ± 0.0437 kN/m3
: γII = 20.25 ± 0.0267 kN/m3
o Internal friction angle : φII = 25.5292 ± 0.3667
o Cohesion coefficient : cII = 9.5688 ± 1.7485 kN/m3
• Soil layer 4
coarse sand, yellow gray - reddish brown, medium compaction.

Bore hole Depth, m Thickness, m SPT, Hammer

HK1 25.3 4.7 11-15
HK2 26.9 3.1 11-15

o Moisture content W : w = 18.24 %

o Bulk unit weight : γI = 18.98 ± 0.1558 kN/m3
: γII = 18.98 ± 0.087kN/m3
o Internal friction angle : φII = 25.14º
o Cohesion coefficient : cII = 6.81 kN/m3
o Overall statistic :

Soil γII
W (%) e γtc (kPa) γI (kPa) Ctc (kPa) CI (kPa) CII (kPa) tanφtc tanφI tanφII
layer (kPa)

25.525 25.525 0.2139

19.6 19.6 ± 0.2139
1 23.035 0.7 19.6 25.525 ± ± 0.2139 ±
±0.584 0.268 ± 0.019
8.9387 5.2066 0.0326
19.9286 23.6214 23.6214 13.4092
19.9286 13.4092
2 20.793 0.6394 19.9286 ± 23.6214 ± ± 0.2389 ±
± 0.051 ± 1.266
0.0876 6.0397 3.7439 0.7849
9.5688 9.5688 0.4476 0.4776
20.25 ± 20.25 ±
3 19.216 0.5691 20.25 9.5688 ± ± 0.4776 ± ±
0.0437 0.0267
2.7809 1.7485 0.0102 0.0064
0.5504 0.5504
18.98 ± 18.98 ± 2.92 ± 2.92 ±
4 18.24 0.65 18.98 2.92 0.5504 ± ±
0.1558 0.087 6.4231 3.9727
0.0235 0.0146

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B: DESIGN A STRIP FOUNDATION

Figure 1.1: Foundation Plan View

5
 I. THE PRELIMINARY STRUCTURAL MODEL AND
CALCULATING DATA :

Figure 1.2: Structural model of strip foundation

• Applied loads on columns
With load factor n = 1.15
o Calculating value:
Table 1.1: Applied calculating loads on foundation

Vertical load Ntt Moment Mtt Horizontal load Htt

Column
(kN) (kNm) (kN)
B 430 37 44
C 506 -31 55
D 506 -31 -55
E 455 -31 -39

II. FOUNDATION MATERIALS :

o Concrete C16/20
fck = 200 MPa
E = 2.1×107

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o Steel CB200

Fyk = 300 MPa

Bulk unit weight of concrete and soil

γaverage = 22 kN/m3

o Load factor n = 1.15

III. PRELIMINARY SIZE OF FOUNDATION :

o Depth of foundation

Df = 2 m
o The extra span for both side of strip foundation
1 1 1 1
c1 = ( ÷ ) 𝑎1 = ( ÷ ) 4000 = (2000 ÷ 1000)
2 4 2 4

Choose c1 = 2 m
1 1 1 1
c2 = ( ÷ ) 𝑎3 = ( ÷ ) 5000 = (2500 ÷ 1250)
2 4 2 4

Choose c2 = 2 m
o Preliminary height hs
1 1 1 1
hs = ( ÷ )amax = ( ÷ )×6000 = (1000÷600)
6 10 6 10
amax: maximum span

Choose hs = 0.75 m
o The preliminary width of the foundation
b = 1.5 m
o Total length of the foundation
L = c1 + a1 + a 2+ a3 + c2 = 2 + 5 + 5 + 4 + 2 = 18 m

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 IV. CHECK GROUND SATBILITY

• Combination of column loads at the center of footing :

o Distance from set-point to centroid of each foundations :
𝐿 18
x1 = − 𝑙𝑎 = −2=7𝑚
2 2
𝐿 18
x2 = − 𝑙𝑎 − 𝑙𝑏 = −2−5=2𝑚
2 2
𝐿 18
x3 = − 𝑙𝑑 − 𝑙𝑒 = −4−2=3𝑚
2 2
𝐿 18
x4 = − 𝑙𝑒 = −2=7𝑚
2 2

o Total vertical load :

Ntt = 𝑁𝐵𝑡𝑡 + 𝑁𝐶𝑡𝑡 + 𝑁𝐷𝑡𝑡 + 𝑁𝐸𝑡𝑡 = 430 + 506 + 506 + 455 = 1897 kN
Htt = 𝐻𝐵𝑡𝑡 + 𝐻𝐶𝑡𝑡 + 𝐻𝐷𝑡𝑡 − 𝐻𝐸𝑡𝑡 = 44 + 55 − 55 − 39 = 5 kN
Mtt = ∑ 𝑀𝑖𝑡𝑡 + ∑ 𝑁𝑖𝑡𝑡 × 𝑥𝑖 + ∑ 𝐻𝑖𝑡𝑡 × ℎ𝑠 =(37 + 31 − 31 − 31) +
(−430 × 7 − 506 × 2 + 506 × 3 + 455 × 7) + 5 × 0.6 = 690 kN/m

o Total standard vertical load :

Load factor n = 1.15

𝑁𝑡𝑡
Ntc = = 1897 kN/m2
𝑛
o The primary capacity of shallow foundation (TCVN) :

𝑚1 𝑚2
RII = (𝐴𝑏𝛾𝐼𝐼 + 𝐵𝐷𝑓 𝛾𝐼𝐼′ + 𝐷𝑐𝐼𝐼 )
𝑘𝑡𝑐
Where:
▪ m1, m2 are coefficients of working condition of soil and
construction
(Section 4.6.10 TCVN 9362:2012)
Choose m1 = m2 = 1
▪ A, B, D are valuable coefficients bearing force depending on
internal friction angle φII
(Table 14 TCVN 9362:2012)
Because Df = 2 m, bottom of foundation places at soil layer 1
𝐴 = 0.2369
φII = 0.2107 ➔ : {𝐵 = 1.9477
𝐷 = 4.4305
▪ ktc – coefficient of reliability is chosen as follows:

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 ktc = 1.2 when calculations are taken directly from
experiments
Choose ktc = 1.2
1×1
 RII = × (0.2369 × 1.6 × 19.7971 + 1.9477 × 2 × 22 + 4.4305 × 24.2838)
1
= 196.0477 kN/m2

o Calculate bottom area :

𝑁𝑡𝑐 1649.5652
𝐹𝑠𝑏 ≥ = = 10.849 𝑚2
𝑅𝐼𝐼 −𝛾𝑡𝑏 ×𝐷𝑓 196.0477−22×2

10.849
▪ 𝑏𝑚𝑖𝑛 = = 0.6027 𝑚
18

So, choose b = 1.2 m

o Calculate the effective length and width :

▪ Condition of ground’s stability

𝜙
N≤
𝑘𝑡𝑐
Where
𝜙 = 𝑏̅𝑙(𝐴
̅ 𝐼 𝑏̅𝛾𝐼 + 𝐵𝐼 𝐷𝑓 𝛾𝐼′ + 𝐷𝐼 𝑐𝐼 )
Determinate non-dimensional coefficients AI, BI, DI
𝐴𝐼 = 𝜆𝛾 × 𝑖𝛾 × 𝑛𝛾
𝐵𝐼 = 𝜆𝑞 × 𝑖𝑞 × 𝑛𝑞
𝐷𝐼 = 𝜆𝑐 × 𝑖𝑐 × 𝑛𝑐
Determinate , q, c are the coefficient of bearing capacity
depending on the internal friction angle φI = 12.0736

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 Figure 3.1: Coefficients of affective angle friction

γ = 0.7, q= 3.1, c= 9.1 where tan(12.0736) = 0.2107

∑ 𝐻 𝑡𝑡 5
Determine i , iq , ic because tg(δ)/tg(φI)= ∑ = = 0.00725
𝑁 𝑡𝑡 690

 i , iq , ic = 1

Figure 3.2: Coefficients affective gradient of loading

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 In which :
𝑀𝑥𝑡𝑡 0
𝑒𝑏 = 𝑡𝑡 = =0𝑚
𝑁 1897

𝑀𝑦𝑡𝑡 690
𝑒𝑙 = 𝑡𝑡 = = 0.4183 𝑚
𝑁 1897
 𝑏̅ = 𝑏 − 2𝑒𝑏 = 1.2 m
 𝑙 ̅ = 𝑙 − 2𝑒𝑙 = 18 - 2×0.4183 = 17.1634 m
𝑙̅
 𝑛= = 14.3028 m
𝑏̅
0.25
𝑛𝛾 = 1 + = 1.0175
𝑛

1.5
𝑛𝑞 = 1 + = 1.1049
𝑛

0.3
𝑛𝑐 = 1 + = 1.021
𝑛

 𝐴𝐼 = 𝜆𝛾 × 𝑖𝛾 × 𝑛𝛾 = 0.7 × 1 × 1.0175 = 0.7122

 𝐵𝐼 = 𝜆𝑞 × 𝑖𝑞 × 𝑛𝑞 = 3.1 × 1 × 1.1049 = 3.4251
 𝐷𝐼 = 𝜆𝑐 × 𝑖𝑐 × 𝑛𝑐 = 9.1 × 1 × 1.021 = 9.2909
o Calculate:
𝜙 = 𝑏̅𝑙 (𝐴
̅ 𝐼 𝑏̅𝛾𝐼 + 𝐵𝐼 𝐷𝑓 𝛾𝐼′ + 𝐷𝐼 𝑐𝐼 )

Where: 𝛾𝐼 = 16.28 𝑘𝑁/𝑚2

𝐷𝑓 = 2 𝑚
cI = 14.3828
 𝜙 = 1.2×17.1634 × (0.7122×1.2×16.28 + 3.4251×2×16.28 + 9.2909 × 14.3828)
= 7249.2839 kN
𝑏̅𝑙 𝑅
̅ 𝐼 7249.2839
Check: 𝑁 𝑡𝑡 + 𝛾𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝐷𝑓 𝑏𝑙 ≤ =
𝑘𝑡𝑐 1.2

1897 + 22×2×1.2×18 ≤ 6041.0699

➔ 2847.4 < 6041.0699 (𝑆𝑎𝑡𝑖𝑓𝑖𝑒𝑑)
➔ Stable

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 V. Check of deformation
o Condition of the elastic ground

𝑡𝑐
𝑝𝑚𝑎𝑥 ≤ 1.2𝑅𝐼𝐼
𝑡𝑐
{ 𝑝𝑚𝑖𝑛 ≥ 0
𝑡𝑐
𝑝𝑡𝑏 ≤ 𝑅𝐼𝐼

𝑡𝑐 𝑁𝑡𝑐 6𝑒𝑙 1649.5652 6×0.418

𝑝𝑚𝑎𝑥 = (1 + ) + 𝛾𝑎𝑣𝑔 𝐷𝑓 = (1 + ) + 22 × 2
𝐹 𝑙 1.2×18 18
= 131.0169 kN/m2

𝑡𝑐 𝑁𝑡𝑐 6𝑒𝑙 1649.5652 6×0.418

𝑝𝑚𝑖𝑛 = (1 − ) + 𝛾𝑎𝑣𝑔 𝐷𝑓 = (1 − ) + 22 × 2
𝐹 𝑙 1.2×18 18
= 109.7206 kN/m2

𝑡𝑐 +𝑝𝑡𝑐
𝑝𝑚𝑎𝑥 131.0169+109.7206
𝑡𝑐 𝑚𝑖𝑛
𝑝𝑡𝑏 = = = 120.3688 kN/m2
2 2

RII = 196.0477 kN/m2

𝑡𝑐
𝑝𝑚𝑎𝑥 = 131 ≤ 1.2𝑅𝐼𝐼 = 235.2572
𝑡𝑐
=> { 𝑝𝑚𝑖𝑛 = 109.7206 ≥ 0 (Satisfied) => Stable
𝑡𝑐
𝑝𝑡𝑏 = 126.45 ≤ 𝑅𝐼 = 196.0477

o Divide soil layer into small soil layers

Choose ℎ𝑖 = 0.6 𝑚

o Pressure at middle of element soil layer 𝒊 before foundation

construction
′ ′
𝜎𝑧𝑖 +𝜎𝑧(𝑖+1)
𝑝1𝑖 = 𝜎𝑧′ 𝑡𝑏 = interpolate ➔ 𝑒1𝑖
2

o Pressure at middle of element soil layer 𝒊 after foundation construction

∆𝜎𝑧𝑖 +∆𝜎𝑧(𝑖+1)
𝑝2𝑖 = 𝑝1𝑖 + ∆𝜎𝑧𝑡𝑏 = 𝑝1𝑖 + interpolate ➔ 𝑒2𝑖
2

12
 o Calculate void ratio of each layer :
Soil layer 1 :
P (kN/m2) 0 100 200 400 800
HK 1-1 0.650 0.603 0.571 0.534 0.5103
HK 2-1 0.750 0.695 0.665 0.642 0.613
Average 0.700 0.649 0.618 0.588 0.588

Soil layer 2 :

P (kN/m2) 0 100 200 400 800

HK 1-2 0.692 0.64 0.606 0.573 0.534
HK 1-3 0.598 0.554 0.526 0.497 0.461
HK 2-2 0.709 0.658 0.627 0.595 0.56
HK 2-3 0.604 0.561 0.534 0.505 0.468
Average 0.651 0.603 0.573 0.543 0.506

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m = L/B = 18/1.2 = 15
n = z/b = 2z/B
From m and n we can find I4 due to table 10.9 :

Table 5.1:
Pressure at
element soil
layer i
before and

14
 after construction

𝜸
Soil hi
Sublayer z (kN/m3) 𝐦𝟏 𝐧𝟏 I4 𝝈′𝒛 ∆𝝈𝒛 ∆𝝈𝒛 ,ave P1 P2 e1 e2 Si (m)
layer (m)
0 9.6 0 1 19.2 84.3688 0.6496 0.0139
1 1 0.6 76.6912 22.08 98.7712 0.6887
0.6 9.6 1 0.818 24.96 69.0136
0.6 9.6 1 0.818 24.96 69.0136 0.6101 0.01
2 0.6 57.666 27.8525 75.5186 0.6375
1.2 9.6209 2 0.549 30.745 46.3184
1.2 9.9286 2 0.549 31.1143 46.3184 0.6156 0.0069
3 0.6 39.8221 34.0929 73.9149 0.6346
1.8 9.9286 3 0.395 37.0714 33.3257
18 9.9286 3 0.395 37.0714 33.3257 0.6177 0.0051
4 0.6 29.447 40.05 69.4947 0.6317
2.4 9.9286 4 0.303 43.0286 25.5637
15
2.4 9.9286 4 0.303 43.0286 25.5637 0.6179 0.004
2 5 0.6 23.0749 46.0071 69.082 0.6289
3.0 9.9286 5 0.244 48.9857 20.586
3.0 9.9286 5 0.244 48.9857 20.586 0.6171 0.0033
6 0.6 18.8142 51.9463 70.7785 0.6261
3.6 9.9286 6 0.202 54.9429 17.0425
3.6 9.9286 6 0.202 54.9426 17.0425 0.6158 0.0028
7 0.6 15.7348 57.9214 73.6562 0.6232
4.2 9.9286 7 0.171 60.9 14.4271
4.2 9.9286 7 0.171 60.9 14.4271 0.6140 0.0024
8 0.6 13.4146 63.8786 77.2932 0.6204
4.8 9.9286 8 0.147 66.8571 12.4022

Table 5.2: Settlement calculation

o Total settlement

𝑆0 = ∑ 𝑆𝑖 = 0.0139 + 0.01 + 0.0069 + 0.0051 + 0.004 + 0.0033 + 0.0028 + 0.0024

= 0.0461 𝑚 = 4.61 𝑐𝑚 ≤ 𝑆𝑔ℎ = 8 𝑐𝑚 => Stable

 Settlement is satisfi

15
 VI. THE PRELIMINARY SIZE OF FOOTING :

o Determine the preliminary area of the column

𝑡𝑡
𝑁𝑚𝑎𝑥 506
𝐹𝑐 = = = 0.0253 𝑚2 = 25300 𝑚𝑚2
𝑓𝑐𝑘 20000
Choose bc × lc = 200mm × 200mm (Fc = 40000 mm2)

o Determine the preliminary width of beam foundation

𝑏𝑠 ≥ 𝑏𝑐 + 100𝑚𝑚 ≥ 400 𝑚𝑚

Choose 𝑏𝑠 = 400 𝑚𝑚

o The height of the strip foundation

1 1 1 1
ℎ=( ÷ ) . 𝑎𝑚𝑎𝑥 = ( ÷ ) . 5000 = (500 ÷ 833) = 600 𝑚m
10 6 10 6

o The width of strip foundation

𝑏 = 1.2 𝑚

o The height of pad footing

ℎ𝑎 ≥ 200 𝑚𝑚

Choose ha = 200 mm

o The height of flange footing

Choose ℎ𝑏 = 400𝑚𝑚 (Slope < 1/3)

16
 Figure 6.1: Strip footing size x – direction

VII. CHECK OF DIFFERENTIAL SETTLEMENT :

o ETABS modeling :

Table 7.1: Under column displacement from Etabs

Column Settlement Differential Settlement

B 21.58 B&C 0.4616
C 23.888 C&D 0.4818
D 26.297 D&E 0.1155
E 25.835

From the Etabs analysis results, we have the differential settlement

between every column:
𝑆𝐵 − 𝑆𝐴 23.888 − 21.58
= = 0.0004616 ≤ 0.002 𝑚𝑚
5000 5000

𝑆𝐷 − 𝑆𝐶 26.297 − 23.888
= = 0.0004818 ≤ 0.002 𝑚𝑚
5000 5000

𝑆𝐷 − 𝑆𝐸 26.297 − 25.835
= = 0.0001155 ≤ 0.002 𝑚𝑚
4000 4000

17
 Therefore, the differential settlements between every two columns
are satisfied condition
VIII. CHECK OF PUNCHING SHEAR :
o Punching shear :
1 𝑡𝑡
𝑝𝑥𝑡 = 𝑝 (𝑏 − 𝑏𝑠 − 2. ℎ0 )
2 𝑚𝑎𝑥

𝑡𝑡 𝑁𝑡𝑡 6𝑒𝑙 1897 6×0.4183

Where 𝑝𝑚𝑎𝑥 = (1 + )= (1 + ) = 100.0694 kN/m2
𝐹 𝑙 1.2×18 18

Choose thickness of protection a = 0.07 m :

ℎ0 = ℎ𝑏 − 𝑎 = 0.4 − 0.07 = 0.23 𝑚

1
➔𝑝𝑥𝑡 = × 100.0694(1.2 − 0.4 − 2 × 0.43) = 7.0049 kN/m2
2

o Punching shear resistance :

𝑝𝑐𝑥 = 0.75𝑅𝑘 ℎ0 = 0.75 × 1000 × 0.33 = 247.5 kN/m2

 𝑝𝑥𝑡 ≤ 𝑝𝑐𝑥

Therefore, the thickness of footing is satisfied the punching shear condition

IX. Calculation of reinforcement :
o Determine moment and shear force of beam footing by Etabs :

Figure 9.1: Moment diagram from SAP2000

18
 Figure 9.2: Shear diagram from ETABS
Take cut at each column and mid span between column respectively from A to F
we have
Table 9.1: Summary of moments from ETABS
Moments max
Cut
(kNm)
1-1 217.198
2-2 -102.831
3-3 219.7487
4-4 -104.3343
5-5 248.2028
6-6 245.8552

o Calculate steel:
o Concrete CB16/20 : 𝑅𝑘 = 1000 kN/m2

𝑓𝑐𝑘 = 200𝑀𝑃𝑎
o Steel CB300 : 𝑓𝑦𝑘 = 200𝑀𝑃𝑎
𝑀
Area of reinforcement is calculated by : 𝐹𝑎 =
0.9𝑅𝑎 ℎ0

Determine position of neutral axis :

ℎ𝑓
𝑀𝑓 = 𝑛. 𝑓𝑐𝑑. 𝐵𝑒𝑓𝑓. ℎ𝑓 (ℎ𝑠 − 𝑎 − )
2
20000
=1× × 1.2 × 0.2 × (0.6 − 0.07 − 0.2/2)
1.5
= 1376 𝑘𝑁𝑚
We have 𝑀𝑓 = 1376 > 𝑀𝑚𝑎𝑥 = 248.2028 𝑘𝑁𝑚 => so the neutral axis goes
through the flange.

19
 ❖ Design for negative moment section (1200x600) :

Cut 2-2 :
M = -102.831 kNm
𝑀 102.831
k= = |− | = 0.0153 < 0.196 => OK
𝑏.𝑑2 .𝑓𝑐𝑘 1200×6002 ×20000
0.0153
𝑧 = 0.5𝑑(1 + √1 − 3𝑘/𝑛) = 0.5 × 600 (1 + √1 − 3 × ) = 0.5239 𝑚
1

𝑀 102.831
𝐴𝑠 = = |− | = 752.455 (𝑚𝑚2 )
𝑧. 𝑓𝑦𝑑 0.5239 × 261 × 106

Select rebar 3d22 : As, provided = 1140 mm2

𝐴𝑠 1140
𝑝= .100% = × 100% = 0.1583 %
𝑏.𝐿 600×1200

𝑝: 0.14% < 0.1583 % < 4% (𝐸𝑢𝑟𝑜𝑐𝑜𝑑𝑒)

 𝑂𝐾
𝐴𝑠. 𝑓𝑦𝑑
𝜆. 𝑥 𝑏. 𝑛. 𝑓𝑐𝑑
𝑀 𝑟𝑒𝑠 = 𝐴𝑠. 𝑓𝑦𝑑. 𝑧 = 𝐴𝑠. 𝑓𝑦𝑑. (𝑑 − ) = 𝐴𝑠. 𝑓𝑦𝑑. (𝑑 − )
2 2

= 1140× 261 × 106 × (530 – 2(×1140× 261 × 106 )/(600× 1 × 200 × 106 ))
= 152.0898 kNm
Check M res =152.0898 > |M| = 102.831 => Adequate
Similar with other cuts, result in the table:

Cut M k (<0.196) z As,req Select As,pro  M res Check

rebar v
4-4 -104.3343 0.0155 0.5238 763.5876 3d22 1140 0.1583 152.0898 Adequate

❖ Design for negative moment section (400x600) :

Similar with calculation about, result in the table:

20
Cut M k z As,req Select As,prov  M res Check
(<0.196) rebar
1-1 217.198 0.0967 0.04883 1705.0852 4d22+2d12 1747 0.7279 228.5606 Adequate
3-3 219.7487 0.0978 0.4878 1727.0052 4d22+2d12 1747 0.7279 228.5626 Adequate
5-5 248.2028 0.1104 0.4817 1975.2072 5d22+1d12 2014 0.8392 261.2051 Adequate
6-6 245.8552 0.0365 0.5151 1829.6916 5d22 1901 0.7921 247.4633 Adequate

❖ Stirrup :
Based in EN – 1992 – 1 – 1, section 6.2.3
The max shear force from ETABS Ved = 276.337 kN
Concrete C16/20 :
fck = 20 => fcd = fck/1.5 = 13.3333 MPa
Steel CB300 :
fyk = 300 => fyd = 300/1.12 = 261 MPa
h = 600 mm ; bw= 400 mm ; d = 530 mm
Asw, max = 2500 mm
Choose 𝜙 = 10 mm
200 200
𝑘1 = 1 + √ =1+√ = 1.6143
𝑑 530

Crd,c = 0.12
𝐴𝑠,𝑚𝑎𝑥
𝑝1 = .100% = 2500/(400*530) = 0.0118
𝑏.𝐿

𝑣𝑚𝑖𝑛 = 0.035. 𝑘 1.5 . √𝑓𝑐𝑘 = 0.035 x 1.61431.5 x √20 = 0.321

𝜎𝑐𝑝 = 0
1/3
𝑉𝑅𝑑, 𝑐 = (𝐶𝑅𝑑, 𝑐. 𝑘 (100𝑝1. 𝑓𝑐𝑘 ) + 𝑘1. 𝜎𝑐𝑝). 𝑏𝑤. 𝑑
= (0.12 x 1.6143(100 x 0.0118 x 20) + 1.6143 x 0) x 400 x 530
= 117.7727 < Ved = 276.337 => Shear rebar is require

𝑎𝑐𝑤 = 1
𝑧 = 0.9𝑑 = 0.9 x 530 = 477 mm

21
 v1 = 0.6
𝑉𝑒𝑑 = 𝑉𝑅𝑑, 𝑚𝑎𝑥
𝑎𝑐𝑤.𝑏𝑤.𝑧.𝑣1.𝑓𝑐𝑑
=> 276.337 = 1 => cot 𝛳 = 7.12 > 2.5
𝑐𝑜𝑡𝛳+
𝑐𝑜𝑡𝛳

=> choose cot ϴ = 2.5

𝑎𝑐𝑤.𝑏𝑤.𝑧.𝑣1.𝑓𝑐𝑑
=> 𝑉𝑟𝑑, 𝑚𝑎𝑥 = 1 = 526.3448 kN > Ved = 276.337 => Adequate
𝑐𝑜𝑡𝛳+
𝑐𝑜𝑡𝛳

𝜃𝑏𝑎𝑟 2 102
Asw = 2𝜋. = 2𝜋. = 157.0796 mm2
4 4
𝑓𝑦𝑘
𝑓𝑦𝑤𝑑 = = 300/1.24 = 240 MPa
1.25
(𝑧.𝑐𝑜𝑡𝜃.𝐴𝑠𝑤.𝑓𝑦𝑤𝑑)
𝑠𝑚𝑎𝑥 = = (447 × 2.5 × 157.0796 × 10)/276.337
𝑉𝑒𝑑

= 162.6861 mm
𝑓𝑦𝑘 300
𝑠𝑚𝑎𝑥 = 𝐴𝑠𝑤. = 157.0796 ×
𝑏. 0.08. √𝑓𝑐𝑘 400 × 0.08 × √20

= 329.2882 mm
𝑠𝑚𝑎𝑥 = 0.75𝑑 = 0.75 x 430 = 397.5 mm
o Design spacing of stirrup
Provide s ≤ min (s1,s2,s3) = 162.6831 → Choose S = 150 mm
Spacing at mid-span ≤ Smax = S3 = 450 mm
Provide Smid-span = 200 mm

22
23
 PART 2: DESIGN PILE FOUNDATION

PART 1: DETERMINE LOADS FOR PILE FOUNDATION .................................. 25

PART 2 : DERTERMINE MATERIALS FOR DESIGN .......................................... 25
PART 3 : GEOPHYSICAL STATISTIC ..................................................................... 25
PART 4 : DESIGN PILE FOUNDATION ................................................................... 26
4.1 Preliminary Section ............................................................................................ 26
4.2 Determine Bearing capacity .............................................................................. 27
4.2.1 Design Base Resistance capacity .............................................................. 27
4.2.2 Design Shaft Resistance............................................................................. 28
4.2.3 Design Compression resistance ................................................................ 33
4.3 Determine force on foundation for preliminary number of piles .................. 33
4.4 Design Sizing of Cap ........................................................................................... 34
4.5 Recalculate Force on foundation and Number of piles .................................. 34
4.6 Determine Load on Pile ...................................................................................... 35
4.7 Check Stability of Pile group ............................................................................. 35
4.8 Settlement of Pile group ..................................................................................... 36
4.9 Design rebar for Pile Cap .................................................................................. 38
4.10 Design shear resistance for Pile Cap ............................................................. 40
4.11 Design shear rebar for Column ...................................................................... 42
4.12 Design Compression and Shear rebar for Pile............................................... 42
4.13 Modeling in Etabs ............................................................................................. 43

24
 PART 1: DETERMINE LOADS FOR PILE FOUNDATION
Check the data sheet, we have properties of load:

LOAD Permanent load Gk Variable load Qk

Vertical (T) 2810 1350
Moment (T.m) 310 149
Horizontal (T.kN) 355 170

PART 2 : DERTERMINE MATERIALS FOR DESIGN

For Concrete choose: C25/30
For Steel choose: CB300
Properties of material:
- fcd = fck*0.85/(1.5*1.1)
- fyd = fyk/1.15
Material (kPa)
C25/30 fck 25
fcd 12.88
CB300 fyk 300
fyd 260.87

PART 3 : GEOPHYSICAL STATISTIC

The survey volume includes 5 boreholes, the ground consists of 11 layers of soil. With
layer A cement base, and debris in top layer.
Select HK3 which is the most dangerous case and has 8 layers to calculate, below is the
statistical table of the physical and mechanical properties of the soil layers in HK3 :

25
 PART 4: DESIGN PILE FOUNDATION
4.1 Preliminary section:
+ For Cap, choose data:
- D = dept of cap = 3 m
- h = height of cap = 2 m
- a1 = excess pile head height = 0.1 m
- a2 = dept of pile in cap = 0.6 m
For Pile:
- d = dept of pile = 55 m
- L = effective length = d – dcap + a1 + a2 = 58.7 m
- dnom = 1.2 m
- d = dnom – 0.05 = 1.15 m
- Ac = d =  m
- As,min =  mm
− Rcd = (fyd*As,min/10^6 +Ac*(1-0.0025)*fcd)*1000 = 14021 kN

26
+ For column of cap:
- Choose h = 0.8 m
- Choose b = 0.8 m

4.2 Determine Bearing capacity:

There are 3 approach with factors (for driven piles = precast pile) in table due to
EN 1997-1:
Approach 1 Approach Approach
Factors Combination Combination 2 3
1 2
Permanent actions G 1.35 1 1.35 1.35
Variable actions Q 1.5 1.3 1.5 1.5
Base resistance b 1 1.3 1.1 1
Sharft resistance s 1 1.3 1.1 1
Total resistance c 1 1.3 1.1 1
Effective cohensive c’ 1 1 1 1.25
Undrained strength cu 1 1 1 1.4
Soil friction angle φ 1 1 1 1.25

4.2.1 Design Base Resistance capacity:

Data:
- Df = ground water level = 0.75 m
- φ = soil friction angle = φi/φ
- c = effective cohesion = ci/c’
- 'z = effective stress cause by the soil’s weight at the pile’s tip = ∑i*hi
- N*q = end-bearing factor designed by Meyerhoff’s chart or table
- qbk = q*N*q ≤ 50*N*q*tan(φ)
- Rbk = end bearing resistance = qp*Ap
- Rbd = end bearing resistance design = Rbk/b

Example for Approach 1 – Combination 1:

- Df = 0.75 m
- φ = 21.067/1 = 21.067 (decimal)
- c = 8.6/1 = 8.6
- 'z = 22*0.75 + (22-10)*(3-0.75) + 5.1*8.9 + 9.7*1.6 + 9.5*4.1 + 10.2*4.1 +

27
10.3*3.9 + 10.2*10.2 + 9.7*3.6 + 10.6*14.1 + 10.3*(55-51.4) = 550.85 (kN)
- N*q = 13.914 kN
- qbk = q*N*q ≤ 50*N*q*tan(φ)
= 550.85 *13.914 = 7665 kN > 50*13.914*tan(21.067) = 268 kN
=> qbk = 268 kN
- Rbk = 268**1.42/4/1.25 = 242 kN
- Rbd = 242/1.5 = 162 kN
Similar with Approach 1-Combination 2, Approach 2, and Approach 3, get the
result in the table below:

4.2.2 Shaft resistance Rsd:

- zc = affected dept of pile = 15d = 15*1.42= 18 m
+ For layer is cohesion soils, calculate by follow step:
- Determine cu = effective cohesion of layer
- Determine  using Table Variation of  base on Terzaghi, Peck, and Mesri, 1996:

𝐜𝐮
𝐩𝐚 
≤ 0.1 1.00
0.2 0.92
0.3 0.82

28
 0.4 0.74
0.6 0.62
0.8 0.54
1.0 0.48
1.2 0.42
1.4 0.40
1.6 0.38
1.8 0.36
2.0 0.35
2.4 0.34
2.8 0.34

Note: pa = atmospheric pressure = 100kN/m2 or 2000lb/ft2

- fs,i = skin pile friction = *cu,i
- Rsk,i = shaft resistance = fs,i*As,i = fs,i**hi/s (with hi = Dept of layer)
Example for Layer 1 - Approach 1 - Combination 1:
- cu = 13.61
-=1
- fs,1 = 1*13.61 = 13.61
- Rsk,i = 13.61*  *8.9/1 = 457 kN
+ For layer is sand, calculate by follow step:
- Determine thickness of layer h
- Determine soil friction angle φi
- Determine pile skin friction angle φa = 2/3 φi for concrete pile
- Determine lateral earth pressure coefficient ks,i
For bored piles, group piles of close spacing:
ks,i= 1 – sinφi
For groups piles of far spacing:
ks,i = 1.4*(1 – sinφi)
- Determine effective stress at middle of layer of soil 'v,i = ∑*h
- fs,i = skin pile friction = ks,i*'v,i*tanφa
- Rsk,i = shaft resistance = fs,i*As,i = fs,i**hi/s (with hi = Dept of layer)
Example for Layer 4- Approach 1 - Combination 1:
- h = 4.1 m
- φ = 21.067
- φa = 2/3 φ = 14.045

29
- For bored piles, group piles of close spacing:
ks,I = 1 – sinφi = 0.64
- 'v,i = 152.27 kN
- fs,i = 24.4
- Rsk,i = 377.1 kN
Similar with other layer and other Approach, result in table below:

30
31
32
 Shaft resistance of pile Rsk = ∑Rsk,i :
+ Approach 1 - Combination 1: Rsk = 54774 kN
+ Approach 1 - Combination 2: Rsk = 5474 kN
+ Approach 2: Rsk = 5474 kN
+ Approach 3: Rsk = 5225 kN
4.2.3 Design compressive resistance:
- Rsd = design shaft resistance of pile = Rsk/s
+ Approach 1 - Combination 1: Rsd = 3650 kN
+ Approach 1 - Combination 2: Rsd = 2807 kN
+ Approach 2: Rsd = 3318 kN
+ Approach 3: Rsd = 3483 kN
- Rcd = design resistance = Rsd + Rbd
+ Approach 1 - Combination 1: Rcd = 3811 kN
+ Approach 1 - Combination 2: Rsd = 2969 kN
+ Approach 2: Rsd = 3479 kN
+ Approach 3: Rsd = 3610 kN

4.3 Determine force on foundation for preliminary number of piles:

Properties of Vertical load:
- Permanent load: Gk = 2810 kN
- Variable load: Qk = 1350 kN
- Selfweight of pile: Sw,pile = *dnom2/4*L*’concrete
= *1.22/4*53.7*(25-10) = 911 kN
- Design compressive action Fcd (without Sw,pile and Sw,cap)
= Q*Qk+G*Gk
- Determine factor  = 1.2 ~ 1.6
∑𝑁𝑒𝑑 𝑁+𝑊𝑐𝑎𝑝
- Design number of piles : n,req =  = 
𝑅𝑑 𝑅𝑑
- Choose number of piles n

33
Table of result:

4.4 Design Sizing of Cap:

- h = 2m
- Spacing between 2 adjecent piles: s = (3~6)d (precast piles)
Choose s = 3d = 3*1.2 = 3.6 m
- Distance from outer pile to pile cap’s border:
≥ 150mm and from d/2 ~ d/3
Choose distance to border = d/3 = 0.4 m
- Design length and width of cap:
B = s + distance to border*2 + dnom = 5.6 m
L = 2S + dnom + 2*distance to border = 9.2 m
- Selfweight of cap: Sw,cap = B*L*(distance to border*concrete) = 911 kN

4.5 Recalculate force on foundation, also number of piles:

Similar with 4.3 but Fcd including Selfweight of pile and cap, have the result:

34
4.6 Determine load on piles:
- Calculate moment of piles: ∑M = G*(MGk+HGk*Hcap)+Q*(MQk + HQk*Hcap)
- Determine x = coordinate of pile in x axis (take neutral axis at central of
foundation)
∑𝑁𝑡𝑡 ∑𝑀𝑡𝑡𝑦∗𝑥𝑖
- Calculate Pmax/min = ±
𝑛 ∑𝑥𝑖 2
- Check if Pmin > 0; Pmax < Rcd so design is acceptable.
Example for Approach 1 - Combination 1:
- M = 2111 kNm
- x = ±1.8
- Pmax = 2994 kN
- Pmin = 2323 kN

4.7 Check stability of pile group:

- n1 = number of rows of pile group = 2

- n2 = number of piles in a row = 3
𝑑
-  = tan( )-1 = 18.435
𝑠
- ng = 1- (((𝑛 − 1)𝑚 + (𝑚 − 1)𝑛)/(𝑚 ∗ 𝑛)) = 
- Total Rcd = ng*n*Rcd
+ Approach 1 - Combination 1: Total Rcd = 17403 kN, check acceptable

35
+ Approach 1 - Combination 2: Total Rsd = 13557 kN, check acceptable
+ Approach 2: Total Rsd = 15888 kN, check acceptable
+ Approach 3: Total Rsd = 16486 kN, check acceptable

4.8 Settlement of pile group:

Layer 10: sand-clay, grayish-white, grayish-blue, plastic

Self-weight of removed soil above the foundation:
Wsoil = L*B*∑’i*hi = 9.2*5.6*537 = 27684 kN
Dept affected under pile:
h = 28.6 m
Unit pressure at the bottom of equivalent pile group:
Lg = h + L = 28.6 +9.2 = 37.8
Bg = h + B = 28.6 +5.6 = 34.2
∆σ = (N+SW,cap)/(Bg*Lg) = (3215+941)/(37.8*34.2) = 3.2147
q = (N+Wsoil)/(B*L) = (3215+27684)/(9.2*5.6) = 599.75 kN/m

36
Number of SPT corrected:
Ncorrect = 1.25 Ntb = 1.25*19.25 = 24.062

The immediate settlement of foundation:

I = 1 – B/8N = 0.998
0.96𝑞 √𝐵𝑔𝐼 0.96∗599.75∗√5.6∗0.998
Si = = = 56.61 𝑚𝑚
𝑁𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 24.0625

Check Si = 56.61 mm < 80 mm => Satisfied

Consolidation settlement:
The settlement of pile groups in granular soils can be calculated in a manner
similar to that employed for pile groups in clay. In this case a virtual raft is
assumed at the base of the pile. The settlement is calculated using the results of
Dutch cone or Standard penetration tests.

In this thesis, the consolidation settlement of a group pile can be

estimated by using the 2:1 stress distribution method; the calculation
involves the following steps:
Step 1: Let the depth of embedment of the piles be 𝐿. The group is
subjected to a total load
of 𝑄𝑔 . If the pile cap is below the original ground surface, 𝑄𝑔 equals the
total load of the
superstructure on the piles, minus the effective weight of soil above the
group piles
removed by excavation.
Step 2: Assume that the load 𝑄𝑔 is transmitted to the soil beginning at a
depth of 2L/3 from the top of the pile, as shown in figure. The load 𝑄𝑔
spreads out along two vertical to one horizontal line from this depth.
Lines aa’ and bb’ are the two 2:1lines.
Step 3: Calculate the increase in effective stress caused at the middle of
each soil layer by
the load 𝑄𝑔 . The formula is:
𝑄𝑔
∆𝜎𝑖′ = 𝐻 𝐻
(𝐵𝑔 + )(𝐿𝑔 + )
2 2
Where: ∆𝜎𝑖′ is the increase in effective stress at the middle of layer 𝑖𝑡ℎ

37
 𝐻
is the depth of the middle of layer 𝑖𝑡ℎ from the bottom of
2
equivalent foundation
Step 4: Calculate the consolidation settlement of each layer caused by
the increased stress:
∆𝑒(𝑖)
∆𝑆𝑐(𝑖) = [ ]𝐻
1+𝑒0(𝑖)
Where: ∆𝑆𝑐(𝑖) is the consolidation settlement of layer 𝑖𝑡ℎ
∆𝑒(𝑖) is the change of void ratio caused by the increase in stress
in layer 𝑖𝑡ℎ
𝑒0(𝑖) is the initial void ratio of layer 𝑖𝑡ℎ (before the
construction)
𝐻 is the thickness of layer 𝑖𝑡ℎ
According to Braja M. Das (2014), the change of void ratio in normal
consolidation soil will be determined by the relationship:
𝜎0′ +∆𝜎𝑖′
∆𝑒(𝑖) = 𝐶𝑐(𝑖) 𝑙𝑜𝑔 [ ]
𝜎0′
Where: 𝜎0′ is the average initial effective vertical stress of layer 𝑖𝑡ℎ
𝐶𝑐(𝑖) is the compressive index of layer 𝑖𝑡ℎ
Step 5: The total consolidation settlement of the group piles is:
∆𝑆𝑐(𝑔) = ∑ ∆𝑆𝑐(𝑖)

4.9 Design rebar for pile cap:

At section 1-1:
P = Pmax = 1680kN
M = Pmax*(x-bcolumn/2) = 1680*(1.8-0.8/2) = 4703 kNm

38
Calculate k
- k = 𝑀/(𝑏 ∗ 𝑑 2 ∗ 𝑓𝑐𝑘)
=> k = 4703*106/(1000*1752*20) = 0.146

Calculate z
1
-z= 𝑑(1 + √1 − 3𝑘)
2

𝑘
26
=> z = (𝐵 ∗ 1000 − 50 − ) ∗ (0.5 + √0.25 − 1.134 ) = 965 mm
2

Calculate the area of tensile reinforcement require:

- fyd = fyk / 1.15 = 347.83 MPa
𝑀
- As,require =
𝑓𝑦𝑑.𝑧
𝑀 4703∗106
=> As,require = = = 18676 𝑚𝑚2
𝑓𝑦𝑑.𝑧 347.83∗167.25

Calculate the area of tensile reinforcement provide:

Choose 3726:
- As,prov = n*d2/4 = 37262/4 = 19644 mm2 > As,req
=> Reinforcement provide ratio condition has been satisfied.

39
Similar with section 2-2, we have:

4.10 Design Shear rebar for Cap:

Condition of no shear reinforcement: 𝑉𝑒𝑑 < 𝑉𝑅𝑑,𝑐
Critical section taken to be 20%×pile diameter (or ϕp/5) inside the face of the pile

40
 Shear force along critical section:
𝑉𝑒𝑑 = ∑(𝑝𝑖 𝑜𝑓 𝑝𝑖𝑙𝑒𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛)

Shear enhancement may be considered such that the shear force, 𝑉𝑒𝑑 , may be
𝑎
decreased by 𝑣
2𝑑

Where: 𝑎𝑣 = distance from the face of column to the critical section

𝑎𝑣 = 1128 𝑚𝑚
𝑎𝑣 1128
= = 0.47
2𝑑 2×1200

Shear force along critical section

𝑉𝑒𝑑 = 𝑃 ∗ 0.47 = 2 ∗ 1680 × 0.47 = 1579 𝑘𝑁
The shear capacity of the concrete, 𝑉𝑅𝑑,𝑐 , in such situations is given by an empirical
expression:
1
𝑉𝑅𝑑,𝑐 = [0.12𝑘 (100 × 𝜌𝑓𝑐𝑘 )3 ] 𝑏𝑑

With a minimum value of:

3 1
2
𝑉𝑅𝑑,𝑐 = [0.035𝑘 2 𝑓𝑐𝑘 ] 𝑏𝑑

Where: 𝑉𝑅𝑑,𝑐 = the design shear resistance of the section without s hear
reinforcement

200
𝑘 = (1 + √ ) ≤ 2 with 𝑑 expressed in mm
𝑑

𝐴𝑠
𝜌= ≤ 0.02
𝑏𝑑

𝐴𝑠 = the area of tensile reinforcement that extends beyond the section

being considered by at least a full anchorage length plus one effective depth (d)
𝑏 = the smallest width of the section in the tensile area (mm)
𝐴𝑠 14137
𝜌= = = 0.01
𝑏𝑑 5600×1885

200
𝑘 = (1 + √ ) = 1.2303
𝑑.ℎ

1
𝑉𝑅𝑑,𝑐 = [0.12 × 1.2303(100 × 0.01 × 25)3 ] 5600 × 1885 × 10−3

41
 = 21140 𝑘𝑁
𝑉𝑅𝑑,𝑐,𝑚𝑖𝑛 > 𝑉𝑒𝑑 (𝑂𝐾)

 Minimum shear reinforcement is require: 6@200

4.11 Design Shear rebar for Column:

Total vertical force act on column:
N = 1.35*Gk +1.5Qk = 1.35*2810 + 1.5*1350 = 5818.5 kN
Shear resitance max :
Vrd,max = 0.5*(0.6*(1-fck/250))*(fck/1.5)*n*bcolumn*d
= 0.5*(0.6*(1-25/250))*(25*/1.5)*6*800*1885 = 40716 kN > N
=> Satisfied

4.12 Design Compression and Shear resistance for pile:

For compression:
Vertiacal force act on pile:
N = Fcd/n + SW,pile = 8318/6 + 911 = 2297 kN
Moment act on pile:
M = 2265 kNm
d = 0.8*dpile = 0.8*1150 = 920 mm
Nd = N/(d2*fck) = 0.0695
Md = M/(d2*fck) = 0.0596
k = 0.05
As,req = k*d2**fck/(fyk*n) = 4400 mm2
As.prov = 1024 = 4524 mm2 > As,req => Satisfied

For Shear resistance :

k = 1 +(200/d)0.5 = 1.4663
p1 = 6**d2/4/d2 = 0.0032
Vrd,c = (0.12*k*(p1*fck)1/3)*d2 = 298.1 kN
Vrd,c min = (0.0035*k*(p1*fck)1/3)*d2 = 26.3 kN < Vrd,c => Satisfied
=> Minimun shear reinforece require, choose 6@200

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4.13 Modeling Pile in Etabs :

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