Perancangan Turap Kantilever | Deep Foundation | Components

Deep Foundation References:  Coduto, D.P. (1994): Foundation design: principles and practices  Day, R.W. (2010): Foundation engineering handbook  Hardiyatmo, H.C. (2011): Analisis dan Perancangan Fondasi, Bagian II

 Teng , Wayne C. (1992): Foundation Design  Tomlinson, M.J. (2001): Foundation design and construction

Topics (from SAP): Kapasitas dukung tiang terhadap gaya lateral dalam tanah kohesif

a. Ujung tiang bebas (tiang pendek dan tiang panjang) b. Ujung tiang terjepit (tiang pendek dan tiang panjang)

Defleksi tiang

a. Ujung tiang bebas (tiang pendek dan tiang panjang) b. Ujung tiang terjepit (tiang pendek dan tiang panjang)

Analisis stabilitas fondasi tiang

a. Beban tiang b. Kapasitas dukung tiang c. Jumlah tiang d. Susunan tiang e. Kontrol

Turap

a. Pengertian b. Tipe struktur turap c. Tipe turap dari segi bahan

Perancangan turap jenis kantilever

a. Gaya-gaya yang bekerja b. Panjang turap yang dipancang c. Dimensi turap dan pemilihan profil turap

Perancangan turap dengan angkur

a. Letak tumpuan angkur b. Dimensi batang angkur c. Konstruksi angkur

Fondasi caisson

a. Pengertian dan jenis fondasi caisson b. Bentuk tampang fondasi sumuran c. Analisis fondasi sumuran

Design of Sheet Pile • Determination of sheet pile stability a. Calculate the depth of embedded sheet pile into the ground b. Determinate the sheet pile dimension.

• Steel sheet pile  use the table of section profile • Concrete sheet pile  use the table of section profile for corrugated sheet pile or determine the thickness and arrange the steel reinforcement • Timber sheet pile  determine the dimension

a. Cantilever-type sheet pile 1. For non-cohesive soil Analyze the sheet pile width of 1 m  drawing area Determine d (depth of embedded part) (SF = 1,50 - 2,00)

Sheet pile dimension is determined based on Mmaks

• Assuming: The sheet pile is a fixed-solid structure, so that the elastic line appears as an inclined straight line rotating at Point Do

• Acting forces: – Right side – Left side

: ADo – active lateral earth pressure DDo – Passive lateral earth pressure : BDo – Passive lateral earth pressure DoD – Active lateral earth pressure

• Forces Diagram A

A

H

Ea B do d

d’

dog.KP

B

Do (H+do)g.Ka

d’g.Ka dog.Ka

(H+do)g.KP

Do’ D

EP1 -Ea d’g.KP

Do EP2 -Ea2 Do’ D

• Sheet pile stability at every point:

F

H

0

M  0

• Use SF = 1,50 – 2,00  B-do can be defined. Then the length of embedded sheet pile (B-D) can be calculated • In the analysis: 1. (SMP/SMA) > 1,50 - 2,00 at point Do , or 2. SF is used to divide EP 3. Length of embedded sheet pile is do = BDo and the implemented (real) sheet pile embedment d = (1,20 - 1,40)do

Stage of analysis: Assuming Do  same location as Do' A K force is acting on Do and remain unknown if : SMP & SMA id calculated on Do

SMDo= 0  calculate do and then decide  d = 1,2do

B d

x

do Do D

K

Sheet pile dimension is determined from the actual Mmaks. Mmaks at the point with distance of x from B, so (dMx/dx) = 0 or SD = 0

Example: A timber sheet pile with 2.00 m high, supporting the backfill with j = 30o, cohesion is neglected, g = 18 kN/m3 and s all timber = 10 MPa. Calculate the length and dimension of sheet pile to be used.

Solution: Calculation is being done on the sheet pile with the width of 1 m  drawing area: K a  tan 2  45  j   1 2 3  K p  tan 2  45  j   3 2 

A

A

H=2m

H=2m

B

B d

x do

do

Ea X

EP

Do Do

D

D

K



2 Ea  1 H1  g  K a 2

Ea  1



and

H1  H  d o  2  d o

  2  d   18  1  2 3 2

o

Ea  32  d o  kN 2

With the distance to Do:

  1 d 2

ea  1 2  d o  3

  18  3

2 E p  1 do  g  K p 2

Ep

2  o

E p  27d o  kN 2

With the distance to Do:

e p  1 d o  3

a. Determination of embedded sheet pile do

M or

Do

0

so

 E a  ea  E p  e p  0

2 2 32  d o   1 2  d o   27  d o  1 d o  3 3

2  d o 3  9  d o 3 2  d o   2,08  d o d o  1,85 m

The actual embedded sheet pile: d  1,2  d o  2,23 m Use d = 2,30 m Total length of sheet pile = 4,30 m

b. Determination of sheet pile dimension (thickness of timber sheet pile) Mmaks occured at the location between B & Do Analyzing point X with the distance of x m from B, so: 2 Eax  1 2  x  g  K a 2 2 Distance from x: Eax  32  x  kN

E px

1

eax  1 2  x  3

x  g  K p 2

2

E px  27x  kN 2

Distance from x:

e px  1 2  x  3

There are two methods to solve this problem

1. SD = 0 (the sum of lateral forces = 0)

Eax  E px 32  x   27x  2

2

x = 1 , from point B to downward direction 2.

 M E x



ax

 eax   E px  e px 

 



2 2 1       M  3 2  x  2  x  27 x  1 2  x   x 3 3 3 1     M   2  x  2  x  9 x  x 3 3

d  M  x

 0 So :

 32  x   27 x 2  0 2

dx x = 1 m, from point B to downward direction Mmaks = Mx=1 = -18 kNm

Assuming:

thickness = t m,

So:





2 2 3 t 1 W 1 t m  m3 6 6

Meanwhile

s all timber  10 MPa s all

M 18 kNm timber   2 t W 6

So:

t = 0,104 m = 10,4 cm

Use:

t = 11 or 12 cm

width = 1 m

a. Cantilever-type sheet pile 2. Cohesive soil For cohesive soil, the internal friction angle is 0 or nearly 0.

Based on Japan Port & Harbour Association-Design Standard for Port and Harbour Structures  coefficient of active lateral earth pressure for cohesive soil (Kac)= 0,50, meanwhile coefficient of passive lateral earth pressure for cohesive soil (KPc)= 2,00

Sheet pile supported by pile foundation A combination of pile and horizontal sheet or horizontal and vertical sheet pile

The solution of this type of sheet pile may be divided into two parts : a. The pile supports horizontal forces b. Horizontal sheet pile construction or the combination of horizontal and vertical sheet pile

The pile supports horizontal forces

Pile, with a certain distance

b Horizontal sheet pile

⅓H

B H

b B

A B

B B

30 cm

B

B Cross-section A-A

Layout

A

Cohesive Soil

Non-cohesive soil

A

A

⅓H

⅓H

B

B

1,5b

1,5b do 9 cu.b

Do D

do d 3b.do.g.KP

Do D

d

Example: A

1,25

H=1,2

⅓H

1,25

B

1,25

do 3b.do.g.KP

Do D

d

1,25

Sheet pile supported by piles (as shown in the picture) is installed on noncohesive soil with j= 30o, c 1,25 = 0, unit volume weight (g) = 17 kN/m3 Calculate: 1,25 a. The length of embedded sheet pile, 1,25 b. Dimension and thickness of the pile sall timber = 10 MPa Width of the pile = 10 cm

Solution: A 1,25 H=1,2

1,25

Ea

0,4 m

1,25

B

1,25

1,25 do

1,25 d

Do

3b.do.g.KP

D

1,25

a. Determination of the pile length Analyzing 1 m  drawing area



  3 2  1 H  g  K   1 1,2  17  1   4,08 kN (1 m) 2 2 3

K a  tan 2 45  30 Ea

2  13

2

K p  tan 2 45  30

2

1

a

Moment arm from Do = (0,40 + do) m





E p  1 3b  d o  g  K p  d o  1 3  0,1 d o  17  1  d o 2 2 3

E p  7,65  d o

2

Moment arm from Do = ⅓do m Equilibrium condition: SMDO = 0 Note: active lateral earth pressure should be calculated with the width of 1,25 m

1,25  Ea  0,4  d o   E p  1  d o 3 2 1,25  4,08  0,4  d o   7,65  d o  1  d o 3 3 2,55d o  5,1d o  2,04  0



Trial:

Use:



do = 1  -0,510 do = 1,1  -0,176 do = 1,12  -0,089  0 do = 1,12 m d = 1,2  do = 1,344 m or d = 1,40 m

Total length of the pile: t = H + d = 1,20 + 1,40 = 2,60 m

b. Determination of pile dimension A

 M E x

M

0,4 B

x

⅓x X

D

 eax   E px e px 

1,25   Ea 0,4  x   E p  1 x 3

3 M   2 , 04  5 , 1 x  2 , 55 x  x

Ea

EPX

x

ax

Maximum moment occurred if:

d  M  x

dx

0

We get x = 0,816 m Mmax = -4,8165 kNm

b

Assuming: thickness = t m, width = b m

t So:





W  1 b  t 2 m3 6 M s s all timber  10 MPa W

If b = 0,10 m, we get t = 16,9997 cm Use  t = 20 cm

c. Determination of sheet pile thickness A

1,25

1,25 B

p = H.g.Ka kN/m’ EP

D

1,25

p = H.g.Ka t Mmax B= 1,25 m

b

Use the width of sheet pile = 1 unit length and thickness = t The load acting of sheet pile  p = HgKa

M max  1 p.B 2  1,328 kNm 8 M s and W  1 b  t 2  6 W t = 0,0282 m or t = 2,82 cm Use t = 3 cm

use unit length of 1m