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

1

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

2

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

3

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

4

Soil Profile of BH 15

5

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

6

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

7

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

8

= 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

10

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

11

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

12

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

14

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.

15

PILE FOUNDATION

16

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

17

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

18

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

19

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

20

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

21

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!)

23

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!

24

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

26

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

27

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)

29

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