pile foundation design
TRANSCRIPT
1.0 COMPANY PROFILE
1.1 INTRODUCTION
Since its establishment in 1988, Reka Bangunan Construction has undertaken many
notable projects in Malaysia and overseas. Our corporate roots coincided with Malaysia’s
development as a modern nation and we have continuously met the demands of a
changing era and people's aspirations for the future through sound construction
operations.
As the world enters the second century of modern urban construction, there is a
strong need to preserve historic scenery, protect the environment and create attractive
public and private spaces. Aiming to provide customers with high-quality service at
reasonable prices, Reka Bangunan Construction is pursuing initiatives that apply expert
knowledge and expanded services to achieve even greater customer satisfaction.
Due to this reason we have been invited to submit a foundation design proposal
for a Proposed Elevated Interchange to Built Coastal Highway from Johor Bahru to
Nusajaya, Johor Darul Takzim which is located at the southern region of Peninsular
Malaysia. This project will involve the construction of several piers and abutments to
support the elevated structure.
This project is supervised under a very expert project manager, Mr Yek Nai
Chuang. Before the construction of this project, a site investigation has been carry out by
member of a professional design team. This is one of the crucial thing need to be done in
order to get the soil profile of the proposed site. Data obtained from the site investigation
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will be use to built the foundation of a pier.
Thus, in order to determine the type of foundation that is suitable for this project
several meetings have been done to choose the best approaches for this project. After a
very detailed analysis we finally managed to come out with a very suitable foundation.
The dedicated group of engineers includes Abdul Mu’iz B. Abdul Mubin, Cheong Chee
Hoe, Hanis Binti Omar, Mardhiah Binti Zainal Abidin, Mohd Shahrul Amin B. Yahya,
Noraini Binti Said, Pon Chew Leng, and Syafiqah Binti Syafruddin.
1.2 ORGANIZATION CHAR
1.3 SITE INVESTIGATION
Site investigation program consist of four stages which is
1) Desk study and reconnaissance
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2) Preliminary ground investigation
3) Detailed ground investigation
4) Monitoring
1.3.1 DESK STUDY
In order to gain as much information as possible, both geological and
historical of the proposed site is needed. Thus geological maps are use to
gives us the excellence indication of the sort of ground conditions. On the
other hand aerial photography is another source of information on
topography and ground conditions. This record are obtained from local
authorities and government agencies.
1.3.2 PRELIMINARY INVESTIGATION
In predicting the geological structures, soil profile and the position of the
ground water table by making a few boreholes. A borehole data we obtain
from Pakatan Geo Services Sdn Bhd. Gives us the description of soil. Soil
borings are the most common method of subsurface exploration in the
field. A borehole is used to determine the nature of the ground in a
qualitative manner and then recover disturb and undisturbed sample for
quantitative examination.
1.3.3 DETAILED INVESTIGATION
After obtaining all possible preliminary information as indicated in the
preceeding section, we proceed to the next stage. At this stage, the extent
of the test, number and depth of boreholes, selection of the appropriate
equipment for field testing and the choice of the laboratory testing are
made. All the data we get from this investigation is aimed to gives a
detailed which is to be use in designing the foundation of piers and
abutments.
Using data from BH15, the proposed foundations of a pier with the
loadings of
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1) Maximum total compression load = 20,000 kN
2) Maximum moment due to horizontal force = 1,500kNm
1.3.4 MONITORING
No one can ensure that the soil parameters used for the design is the most
representative of the soil condition at the site. Thus field observation is
crucial as it can give early diagnosis and redemption of any problems that
might counter during construction. Among the measurement that is taken
during the monitoring stage are settlement, displacement, deformations,
inclination and pore water pressure. This is because review of the design
can be made during construction based on the information gained from
monitoring program.
1.4 SOIL PROFILE
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Soil Profile of BH 15
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Hard, grey with red spots, SILT WITH PARTS OF CLAY, traces of sand and gravel
Hard,-ditto-
Hard, grey, SILT with parts of clay, and some sand
Hard,-ditto-
Hard, grey, SILT with parts of clay, little sand and traces of gravel
Hard,-ditto-
Hard,-ditto-
Top Soil
Sand
Very Stiff
Very Stiff
Very Stiff
Hard
Hard
Hard
Hard
Hard
Hard
Hard
0m
0.5m
1.3m
3.0m
4.95m
9.45m
10.95m
12.41m
13.84m
15.26m
16.76m
18.2m
Brownish grey, CLAY with some fine sand and roots
Very loose, grey, silty SAND with traces of gravel
Firm, reddish brown with yellow spots, SILT with parts of clay, traces of sand and a little gravel
Stiff, -ditto-
Very stiff, brown with grey mottled, -ditto-
Very stiff, brown with grey mottled, SILT with parts of clay, traces of sand and gravel
Very stiff, -ditto-
Firm
3.45mStiff
6.45m
7.95m
SHALLOW
FOUNDATION
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19.61m
2.0 SHALLOW FOUNDATION DESIGN
Given:a) Maximum total compression load, Qmax = Qall = 20000kNb) Maximum moment, Mmax = 1500 kNm
Assumption:a) Depth of foundation, Df = 3mb) Factor of safety, FS = 2.5c) Cohesion, c = 0 kN/m2
d) Inclination, β = 0°e) Internal friction angle of sand, φsand = 30˚f) Saturated unit weight of sand,Ysat (sand) = 19 kN/m3
g) Unit weight of water, Yw (water) = 9.81kN/m3
h) Unit weight of sand Yb (sand) = 17 kN/m3
i) Use square footing, so, width, Area = BXB
Case 1 Assume Df = 1.3 m, FS = 2.5
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γb = 17 kN/m3
c = 0φ = 30o
SAND1.3 m
Q = 20000kN
M = 1500kNm
Since the ground water level is below the footing, therefore the q and γ that we calculate are as below.
q = γb x Df
= 17(1.3) = 22.1 KN/m2
γ = ( 19 – 9.81) = 9.19
By using Meyerhoff equation :
qu = C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B.Nγ.Fγs.Fγd.Fγi
Nq = 18.40
Nγ = 22.40
Shape factor
Fqs =
Fγs =
Depth factor ( Assume Df /B ≤ 1)
Fqd =
Fγd = 1
inclination factor
Fci = Fqi =
Fγi =
qu= C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B.Nγ.Fγs.Fγd.Fγi
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= 0 + 22.1(18.4)(1.577)(1+0.375/B)(1) + 0.5(9.19)(B)(22.4)(0.6)(1)(1)
= 641.27 + 240.48/B + 61.76B
qnet(all) = (qu – q) / FS= (641.27 + 240.48/B + 61.76B – 22.1) / 2.5= 247.67 + 96.19/B + 24.7B
Qall = qnet(all) . Area
20000 = (247.67 + 96.19/B + 24.7B)(B2) 20000 = 247.67 B2 + 96.19B + 24.7B3
B = 6.82 m
Check the Df /B
Df /B = 1.3/6.82 = 0.191 < 1 ok!
Use B = 6.82 m
But, there is moment due to horizontal force, then
e = M/Q
= 1500kNm / 20000kN = 0.075 m < B/6= 1.137 m ok!
So the effective value
B’ = 6.82 – 2e L’ = L = 6.82m = 6.82 – 0.15 = 6.67 m
shape factor
Fqs =
Fγs =
9
Depth factor
Fqd =
Fγd = 1
inclination factor
Fci = Fqi =
Fγi =
Therefore,
qu’ = C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B’.Nγ.Fγs.Fγd.Fγi
= 0 + 22.1(18.4)(1.565)(1.055)(1) + 0.5(9.19)(6.67)(22.4)(0.609)(1)(1)
= 1089.49 KN/m2
q’u(net) = (qu
’ – q) = 1089.49 – 22.1= 1067.39 KN/m2
q’net(all) = 1067.39 /2.5 = 426.96 KN/m2
Q’ net(all) = q’net(all) . Area = 426.96 (6.82)( 6.67) = 19422.15 kN < Q = 20000 kN Not ok!
qmax = (Q / BL) (1 + 6e/B) = 20000/(6.82x6.82) (1 + (6x0.075)/6.82)) = 431.06 kN/m2 <448.54 KN/m2 ok!qmin = (Q / BL) (1 - 6e/B) = 20000/(6.82x6.82) (1 - (6x0.075)/6.82))
= 430.93 kN/m2
Therefore, square footing of 6.82m x 6.82m with depth of foundation 1.3m is not adequate.
Immediate settlement
Se = Bqo /Es (1 – μs 2 )αr
Take μs = 0.2
Es = 10.35 MN /m2
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αr = 0.89
Se = ((6.82x1121.35) / (10.35 x103))x(1 – 0.22)(0.89) = 0.631m = 631 mm > 50mm Not ok!
Case 2Assume Df = 3 m, FS = 2.5
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γb = 17 kN/m3
γsat = 19 kN/m3
c = 0φ = 30o
SAND
1.3 m
1.7 m
Q = 20000kN
M = 1500kNm
CALCULATIONS:
Use Meyerhoff’s Method
qu = C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B.Nγ.Fγs.Fγd.Fγi
q = (17x1.3) + (19-9.81)(1.7) = 22.1 + 15.623 = 37.723 KN/m2
γ' = 19 – 9.81 =9.19 KN/m3
Nq = 18.40
Nγ = 22.40
Shape factor
Fqs =
Fγs =
Depth factor ( Assume Df /B ≤ 1)
Fqd =
Fγd = 1
inclination factor
Fci = Fqi =
Fγi =
qu= C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B.Nγ.Fγs.Fγd.Fγi
= 0 + 37.723(18.4)(1.58)(1+0.87/B)(1) + 0.5(9.19)(B)(22.4)(0.6)(1)(1)
= 1096.68 + 954.1/B + 61.76B
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Qall = qnet(all) . Area
20000/B2 = ((qu – q) / FS) = (1096.68 + 954.1/B + 61.76B -37.723)/2.5 50000/B2 = (1096.68 + 954.1/B + 61.76B -37.723) 50000 = 1096.68B2 + 954.1B + 61.76B3 -37.723B2
61.76B3 + 1058.96B2 + 954.1B – 50000 = 0 B = 5.63 m
Check the Df /B
Df /B = 3/5.63 = 0.533 < 1 ok!
But, there is moment due to horizontal force, then
e = M/Q
= 1500kNm / 20000kN = 0.075 m < B/6= 0.938 m ok!
So the effective value
B’ = 5.63 – 2e L’ = L = 5.63m = 5.63 – 0.15 = 5.48 m
Since the ground water table is at the soil surface, so we will use Ysat in the calculation.
Bearing Capacity Factors
We ignore the top soil and consider soil at layer 2 for taking the φ = 30 kN/m2
Nq = 18.40, Nγ = 22.40
shape factor
Fqs =
Fγs =
13
Depth factor
Fqd =
Fγd = 1
inclination factor
Fci = Fqi =
Fγi =
Therefore,
qu’ = C.Nc.Fcs.Fcd.Fci + q.Nq.Fqs.Fqd.Fqi + ½.γ.B’.Nγ.Fγs.Fγd.Fγi
= 0 + 37.723(18.4)(1.565)(1.155)(1) + 0.5(9.19)(5.48)(22.4)(0.611)(1)(1)
= 1599.28 KN/m2
q’u(net) = (qu
’ – q) = 1599.28 – 37.723= 1561.56 KN/m2
q’net(all) = 1561.56 /2.5 = 624.62 KN/m2
Q’ net(all) = q’net(all) . Area = 624.62 (5.48)( 5.63) = 19271 kN < Q = 20000 kN Not ok!
qmax = (Q / BL) (1 + 6e/B) = 20000/(5.63x5.63) (1 + (6x0.075)/5.63)) = 681.41 kN/m2 >624.62 KN/m2 Not ok!qmin = (Q / BL) (1 + 6e/B) = 20000/(5.63x5.63) (1 - (6x0.075)/5.63))
= 580.54kN/m2
Therefore, square footing of 5.63m x 5.63m with depth of foundation 3m is not adequate.
Se = Bqo /Es (1 – μs 2 )αr
Take μs = 0.2
Es = 10.35 MN /m2
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αr = 0.89
Se = ((5.63x1561.56) / (10.35 x103))x(1 – 0.22)(0.89) = 0.726m = 726 mm > 50mm Not ok!
Pile foundation is required.
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PILE FOUNDATION
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3.0 PILE FOUNDATION DESIGN
BOREHOLE 15:
Using 1200 mm circular pre-cast pile (bored) with unit weight of 24 kN/
Assume safety factor, Fs = 2.5
Length of pile = 9.45m (until the hard layer)
Use Meyerhof’s static method:
= + ∑
End Bearing
Qb = qbAb
The pile tip is in the silt layer with
Cu = 121.5 kN/m2 ; = 3o ; γ sat = 19 kN/
From Figure 8.13
Nc* = 10 ; Nq* = 1.5
q = γ’ Df (γ’= γsat- γw)
= 17(1.3) + (19-9.81)(1.7) + (19-9.81)(9.45-3)
= 97 kN/m2
cNc* = 121.5(10)
= 1215 kN/m2
qNq* = 97(1.5)
= 145.5 kN/m2
qb is in the range of 145.5 kN/m2 and 1215 kN/m2.
Since, there is only a little sand in the layer, therefore,
we take qb = 1100 kN/m2
Thus, Qb = qbAb
= 1100 x π(1.2)2/4
= 1244.07 kN
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Skin Friction
Qs = Asf
= ∑P (∆L)f
15D = 15 x 1.2 = 18 m
For 0 < d ≤ 1.3 m
Sand ( = 300)
f = K0 σ'v tan
K0 = 1- sin = 1- sin 30° = 0.5
= 0.5 = 15o
σ'v = 17 x 1.3 = 22.1 kN/m2
since d=1.3m < 18 m, thus
f = 0.5(22.1/2)(tan 15o) = 1.480
QS1 = π (1.2) x 1.3 x 1.48 = 7.25 kN
For 1.3 m < d < 3m
Sand ( = 300)
f = K0 σ'v tan
K0 = 1- sin = 1- sin 30° = 0.5
= 0.5 = 15o
σ'v = 22.1 + (19-9.81)(1.7) = 37.723 kN/m2
since d=3m < 18 m, thus
f = 0.5(22.1+37.723)/2 x (tan 15o) = 4.007
QS2 = π(1.2) x 1.7 x 4.007 = 25.68kN
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For 3 m < d < 3.45m
Silt ( = 30)
f = αCu + K0 σ'v tan
since Cu = 31.25 kN/m2 < 50 kN/m2, thus
α = 1.0
K0 = 1- sin = 1- sin 3° = 0.95
= 0.5 = 1.5o
σ'v = 37.723 + (19-9.81)(3.45-3) = 41.859 kN/m2
since d=3.45m < 18 m, thus
f = (1.0 x 31.25) + [0.95(37.723 + 41.859)/2 x (tan 1.5o)] = 32.240
QS3 = π(1.2) x 0.45 x 32.24 = 54.69 kN
For 3.45 m < d < 4.95 m
Silt ( = 30)
f = αCu + K0 σ'v tan
From Figure 8.19, when Cu = 78.5 kN/m2
α = 0.56
K0 = 1- sin = 1- sin 3° = 0.95
= 0.5 = 1.5o
σ'v = 41.859 + (19-9.81)(4.95-3.45) = 55.644 kN/m2
since d=4.95m < 18 m, thus
f = (0.56 x 78.5) + [0.95(55.644 + 41.859)/2 x (tan 1.5o)] = 45.170
QS4 = π(1.2) x (4.95-3.45) x 45.170 = 255.429 kN
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For 4.95 m < d < 9.45 m
Silt ( = 30)
f = αCu + K0 σ'v tan
Cu = (113.5 + 120 + 121.5)/3 = 118.33 kN/m2
From Figure 8.19,
α = 0.4
K0 = 1- sin = 1- sin 3° = 0.95
= 0.5 = 1.5o
σ'v = 55.644 + (19-9.81)(9.45-4.95) = 96.999 kN/m2
since d=9.45m < 18 m, thus
f = (0.4 x 118.33) + [0.95(55.644 + 96.999)/2 x (tan 1.5o)] = 49.231
QS5 = π(1.2) x (9.45-4.95) x 49.231 = 835.18 kN
ΣQs = QS1 + QS2 + QS3 + QS4 + QS5
= 7.25 + 25.68 + 54.69 + 255.429 + 835.18
= 1178.18kN
Qu = Qb + ΣQs
= 1244.07 + 1178.18
= 2422.25kN
Qall = Qu / FS
= 2422.25
2.5
= 968.9kN
Hence, the number of piles needed, N = Q / Qall
= 20000 / 968.9
= 20.6
≈ 25 piles
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Take spacing = 2.5 dpile = 2.5 x 1.2
= 3.0 m
4 @ 3.0 m
Y
Take size of pile cap = 14m x 14m x 0.75m
= 147 m3
Thus,
Npile = (Q + Wpile cap) / Qall
= (20000 + 147 x 24) / 968.9
= 23528 / 968.9
= 24.3piles
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X
1.0m
14.0
0 m 4 @
3.0 m
14.00m
≈ 25 piles
Distribution of loads in a pile group:
4 @ 3.0m
Y
Maximum load apply on each pile
Qi = Q/N ± (Myx/x2) ± (Mxy/y2) (Mxy/y2) = 0
= Q/N ± (Myx/x2)
For pile #1 to #5 :
Qm =
= 941.12 -
= 921.12kN
22
Bg 4 @
3.0 m
Lg
9 19
14
24
5 10
20
15
25
1 6 16
11
21
4
3 8 18
13
23
2 7 17
12
22
For pile #6 to #10 :
Qm =
= 941.12 -
= 931.12 kN
For pile #11 to #15 :
Qm =
= 941.12 -
= 941.12 kN
For pile #16 to #20 :
Qm =
= 941.12 +
= 951.12 kN
For pile #21 to #25 :
Qm =
= 941.12 +
= 961.12 kN
Qi(max) = 961.12kN < Qall = 968.90 kN (OK!)
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Uplift Resistance of Pile:
Since Qm is always positive, therefore, we no need to check the uplift resistance of the piles.
Uplift resistance check is needed whenever Qm is a negative value.
Check settlement of a pile group:
Immediate settlement:
In sand and gravel:
Seg (mm) = Qg (0.92) √(BgI)
Lg Bg N’
Since
Qg = 20000 kN
Bg = 13.2 m
Lg = 13.2 m
N’ = 31
L = 9.45 m
Is= 2+0.35 √(L/B)___
= 2+0.35√ (9.45/1.2)
= 2.98
I =Is –L/Bg
=2.98 – 9.45/(13.2)
=2.27 ≥ 0.5 OK!
Seg (mm) = Qg (0.92) √ (BgI)
Lg Bg N
=20000 (0.92)√(13.2)(2.27)
(13.2)(13.2)(31)
= 18.65 mm < 50 mm OK!
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In clay:
Seg (mm) = Qg BgI
Lg Bg 2qc
Qg = 20000 kN
Bg = 13.2 m
Lg = 13.2 m
I = 2.27
σ'o = 96.999
qc = CuNk + σ'o take Nk = 16
= 121.5 x 16 + 96.999
= 2041 kN/m2
Seg (mm) = Qg BgI
Lg Bg 2qc
= 20000 x 13.2 x 1.9
13.2 x 13.2 2 x 2041
= 0.843 mm < 50 mm OK!
In silt:
The layer of silt is the combination of both sand and clay. Therefore, interpolation must be made
in order to get the actual immediate settlement of silt.
Take Seg = (18.65 + 0.843) /2
= 9.75 mm < 50mm, it is OK!
Consolidation settlement:
No need to consider consolidation settlement since the pile underlain on the hard layer. Thus, no
consolidation will occur.
25
PILE CAP
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4.0 REINFORCED CONCRETE DESIGN (PILE CAP)
4 @ 3.00 m
Result From Previous,Pile size, diameter = 1200mmDesign working load per pile = 961.12kN
Given Value,Maximum compression load, N = 20 000kNMaximum moment, M = 1500 kN.m
Assumption,Column size = 6000mm x 6000mmSpacing between Piles = 2.5 x dpile = 2.5 x 1200mm = 3000mmPile cap size = 14000mm×14000mmBearing capacity soil = 200kN/m2
fcu = 35 N/mm2
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1.0m
14.0
0 m 4 @
3.00 m
6.00 m
fy = 500 N/mm2
cover, c = 75mmsize bar = 32mm
Try pile cap 14000×14000×750Characteristic load, Qc = 20 000/1.45 = 13 793.10 kNCharacteristic moment, Mc = 1500/1.45 = 1034.48 kN.mUltimate load , P
= Ok!
Column Ultimate load
1
2
3
4
5
Steel Bar in Vertical Direction
28
My-y = (5×820) × (6-3) = 12300 kNmd = 750 – 75 – (32/2) = 659 mmb = 14000 mm
K =
z =
Use z = 0.93d = 0.93× 657 = 611.01 mm
Asmin = 0.13% bh = 0.13% bh = (0.0013) (14000) (1000) = 18200 mm2 Asmax = 4% bh = 4% bh = (0.04) (14000) (1000) = 560000 mm2
From Table of Properties,Use steel bar 60T32 ( As = 48300mm²)
Steel Bar in Horizontal Direction
Mx-x = (5×820) × (6-3) = 12300 kNmd = 750 – 75 – 75 – (32/2) = 584 mmb = 14000 mm
Use z = 0.91d = 0.91×584 = 531.44 mm
Asmin = 0.13% bh = 0.13% bh = (0.0013) (14000) (1000) = 18200mm2 Asmax = 4% bh = 4% bh = (0.04) (14000) (1000) = 560000mm2
From Table of Properties,Use steel bar 68T32 (As = 54740 mm²)
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Shear1. Check for Maximum Shear Stress
Shear Force,
Shear stress,
Critical shear stress,
Ok!
Shear Force maximum, V max =20 000kN
Shear Stress Maximum, V =
Shear capacity = Ok!
Cracking
Checking the outer most steel bar, 40T20.
BS 8110, Cl 3.12.11.2.7: fy = 500 N/mm2 and h = 750 > 200 mm
30
Table 3.8 :
True Allowable free distance between steel bars = < 750mm. OK!
Steel Ties
By using,
Asmin = 0.13% bh = 0.13% bh = (0.0013) (14000) (750) = 13650mm2 25T32 (As = 20215 mm2) is used in the distance 2/3 d from bottom, 2/3(909) = 606 mm from bottom with spacing 200 mm .
31
Reinforcement Detailing
32
CONCLUSION
In order to determine the types of foundation which is suitable for this project, we had
undergone a very detailed analysis. Analyses that have been made are to determining the
thickness and shape of the footing, amount and location of reinforcing steel and performing other
details of the actual structural design. Firstly, the calculation of shallow foundation is made for
two trials. Based on the calculation, the net allowable load, Q net and immediate settlement, Se of
the footing cannot sustain the load above it. From these two trials proves that we cannot use
shallow foundation for this proposed project as it gives inadequate result.
Thus we to proceed our calculation to pile foundation. From calculation of pile
foundation we found that the proposed elevated interchange needs 25 piles to sustain load above
it. Finally, we have decided to use the pile foundation rather than shallow foundation for the
Proposed Elevated Interchange to Built Coastal Highway from Johor Bahru to Nusajaya, Johor
Darul Takzim.
33