REKAYASAPONDASI

Pp D

b W2

W3

W1

W4

W5

W6

1

2

3

4

5

6

Pa1

Pa2

PaH

i

a

c

d e f g h

qmaxqmin

q = 1,03 t/m2

Pp diabaikan

1. Gravity Wall

q = 1,03 t/m2 (beban merata)

b = 0,12. H = 0,12. 7,03 = 0,844 m D = 1,3 ma = H-b = 7,03-0,844 = 6,186 m = 1,603 t/m3

d = b = 0,12.H = 0,844 m C = 2,303 t/m2

c = 0,7.H = 0,7. 7,03 = 4,921 m = 20,3°e = 0,05. 7,03 = 0,532 mg = e = 0,532 m Tanah Urugf = i = 1 m = 1,703 t/m3

h = c-2e-f-d = 2,013 m C = 0 = 30,3°

c = 1,703 t/m3

Pemeriksaan Terhadap Moment

No Luas (m2) Berat (t/m) Jarak (m)

1 4,921.0,844 = 4,153 4,153. 2,4 = 9,967 ½ . 4,921 = 2,461

2 ½ .0,532.6,168. = 1,645

1,645. 2,4 = 3,948 0,844+2/3.0,532 = 1,199

3 1. 6,168 = 6,186 6,186. 2,4 = 14,846 0,844+0,532+½ .1 = 2,461

4 ½ . 0,532. 6,186 = 1,645

1,645. 2,4 = 3,948 0,844+0,532+1+1/3.0,532 = 2,553

5 ½ . 0,532. 6,186 = 1,645

1,645. 1,603 = 2,637 0,844+0,532+1+2/3. 0,532 = 2,751

6 2,013. 6,186 = 12,452 12,452. 1,603 = 19,961 0,844+0,532. 2+1+½.2,013 = 3,915

307,55V

No Momen (berat x jarak)

1 9,967 . 2,461 = 24,5292 3,948 . 1,199 = 4,734

3 14,846. 2,461= 36,536

4 3,948 . 2,553 = 10,079

5 2,637 . 2,751 = 7,202

6 19,961 . 3,915= 78,147

H = H’ q = 1,03 t/m2

Ka = tan2 (45-

ϕ1

2 ) = tan2(45-20 ,3

2 ) = 0,485

Ph = PaPa1 = (Ka.q) H

= (0,485.1,03). 7,03 = 3,512 t/m

Pa2 = (1/2. Ka.γ 1 .H).H

307,55V=(1/2.1,603.7,032.0,485)= 19,211 t/m

Ph = Pa1 + Pa2 = 22,723 t/m

Ya = Ph

YPaYPa 1211 ..

=

(3 ,512. 1/2. 7 ,03 )+(19 ,211.1/3 . 7 ,03)22 ,723

= 2,524 m

∴Fs = ∑

M B

MO

=161 ,22757 ,353 = 2,811 > 2 OK !!!

Pemeriksaan Terhdap Sliding

K1 = K2 = 2/3

Pp = ½ .Kp.γ 2 .D2 + 2.C2.√K p .D Kp= tan2(45+ϕ

2 )

= ½ .3,037. 1,703. 1,32 + 2.0.√3 ,037 .1,3 = tan2(45+30,2

2 )= 4,370 + 0 = 4,370 t/m = 3,037

∴F(sliding) =

(∑V ) tan (2/3 .ϕ1 )+B .K2 .C2+Pp

Pa .Cosα

=

55 ,307 tan(23)+4 ,921 .(23). 0+4 ,370

22 ,723.Cos0

=

13 ,312+0+4 ,37022 ,723

=

17 ,68222 ,723

= 0,778 < 1,5 Tidak Aman !!!

Cek Bearing Capasity

e =

B2−∑ B−∑Mo

∑V= 4 ,921

2−161 ,227−57 ,353

55 ,307

= 0,582 m <

B6=4 ,921

6=0 ,820m

q max =

∑ V

B (1+6 eB )=55 ,307

4 ,921 (1+ 6 . 0 ,5824 ,921 )=19 ,214

tm2

q max =

∑ V

B (1−6eB )=55 ,307

4 ,921 (1−6 .0 ,5824 ,921 )=3 ,264

tm2

Pondasi Squere (tanah asli)ϕ2 = 30,3 Nc = 35

Nq = 23Nγ = 20

q = γ 2.D = 1,703.1,3 = 2,214 t/m2

qu = 1,3.C2.Nc + q.Nq + 0,4. γ 2 .B.N γ

= 1,3.0,35 + 2,214.23 + 0,4.1,703.4,921.20= 0 + 50,922 + 67,044= 117,966 t/m2

∴Fs (bearing capasity) =

quqmax

=

117 ,96619 ,214

=6 ,139> 3 OK !!!

Pp D

b W1

W2

W3

W43

1

24

5

a

c

d e f g

qmaxqmin

Pp diabaikan

W5

10,3°

Pa

Y

2. Cantilever Wall

H = 9,03 mD = 2,03 m

c = 0,7. H = 0,7. 9,03 = 6,321 mb = 0,1. H = 0,1. 9,03 = 0,903 m Tanah Dasara = H-b = 9,03-0,903 = 8,127 m = 1,703 t/m3

d = 0,1 H = 0,1.9,03 = 0,903 m C = 3,43 t/m2

e = 0,05.H = 0,05. 9,03 = 0,452 m = 23,0°e = 0,05. 7,03 = 0,532 mf = = 2 m Tanah Urugg = C-D-e-F = 2,966 m = 1,803 t/m3

C = 0

Y = 9 tan 10,3 = 2,966 tan 10,3 = 33,0°

= 0,539

c = 2,4 t/m3

Pemeriksaan Terhadap Moment

No Luas (m2) Berat (t/m) Jarak (m)

1 ½.0,452.8,127 = 1,837

1,837.2,4 = 4,409 0,903 + 2/3.0,452 = 1,204

2 2.8,127 = 16,254 16,254.2,4 = 39,009 0,903 + 0,452 + ½.2 = 2,355

3 6,321.0,903 = 5,708 5,708.2,4 = 13,699 ½.6,321 = 3,161

4 2,966.8,127 = 24,105 24,105.1,703 = 41,051 0,903+0,452+2+ ½.2,966 = 4,838

5 ½.0,539.2,966 = 0,799

0,799.1,703 = 1,361 0,903+0,452+2+2/3.2,966 = 5,332

6 Pv = 6,106 6,321

V = 105,635Pa = ½ 1.H12.ka H1 = H + y = 9,03 + 0,539 = 9,569 m

= ½ 1,703(9,569)2.0,438 Ka = tan2(45−φ2 )

= 34,15 t/m = tan2(45−232 )=0 ,438

Pv = Pa sin 10,3 = 34,15 sin 10,3 = 6,106 t/m

No. Momen (Jarak berat)1. 1,204.4,409 = 5,3082. 2,355.39,109 = 91,8663. 3,161.13,699 = 43,3024. 4,838.41,051 = 198,6055. 5,332.1,361 = 7,2576. 6,321.6,106 = 38,596

MB = 384,934

Ph = Pa. Cos.10,3 = 34,15.cos 10,3 = 33,6

Mo = Pa. (1/3H1) = 34,15 (1/3.9,569) = 108,927

Fs (overtuning) =

ΣM B

M o

=384 ,934108 ,927

= 3,534 > 2 OK!!!

Pemeriksaan terhadap slidingk1 = k2 = 2/3

Pp = ½ kp. 2. D2 + 2.C2√k p .D kp = tan2(45−φ

2 )= ½.3,392.1,803.2,032 + 0 = tan2(45−33

2 )=3 ,392

= 12,601

Fs (sliding) =

(ΣV ) tan(2/3Q 1)+Bk2 .c2+P p

Pa . cos10 ,3°

=

105 ,635 tan(2 /3 . 23)+0+12 ,60133 ,6

=1 ,237< 1,5

tidak aman

Cek Bearing capacity

e =

B2−ΣM B−ΣMO

ΣV=6 ,321

2−384 ,934−108 ,927

105 ,635

= 0,548 m <

B6=6 ,321

6=1 ,054

m

qmax =

ΣVB (1+ 6e

B )=105 ,6356 ,321 (1+ 6 .0 ,548

6 ,321 )=25 ,405 t/m2

qmin =

ΣVB (1−6e

B )=105 ,6356 ,321 (1−6 .0 ,548

6 ,321 )=8 ,019 t/m2

q= 2. D = 1,803.2,03 = 3,66 t/m2

B1 = B – 2e = 6,321 – 2.0,548 = 5,225 m

2 = 33 Nc = 45 Nq = 35 N = 33(Tanah asli)

Fcd = 1 + 0,4 ( DB1 )

=1 + 0,4 ( 2 ,03

5 ,225 )=1 ,155

Fqd = 1 + 2 tan 2 (1 – sin 2)2( DB1 )

= 1 + 2 tan 33 (1 – sin 33)2( 2 ,035 ,225 )=1 ,105

Fd = 1

Fcd = Fq = (1− ψ°

90 ° )2

= tan-1

( Pacos L

ΣV )= (1−180 ,132 °

90 ° )= tan-1

(34 ,15 cos10 ,3 °105 ,635 )

= 1,003 = 180,132

Fq = (1− ψ

φ2 )2

=(1−108 ,13233 )

2

=5 ,184 °

qu = c.Nc.Fcs.Fcd.Fct + q.Nq.Fqs.Fqd.Fqt +1/2..B1.N.Fs.Fd.Ft Cz = 0= 0 + 3,66.35.1,105.1,003 + ½.1,803.5,225.33.1.5,184= 947,782 t/m2

Fs (bearing capacity) =

quqmax

=947 ,78225 , 405

=37 ,307> 3 sangat OK !!!

TUGAS IV

REKAYASAPONDASI

1. The cantilever sheet pile wall penetrating a granular soil, given: L1 = 2,03 m and L2 = 3,3 m. = 30,3c = 0y = 1,503 t/m3

ysat = 1,83 t/m3

Determine: Depth of sheet pile penetrartion (d) and minimum size of sheet pile!

Answer:1) Calculate ka.and kp

ka = tan2(45−φ2 )

= tan2(45−30 ,32 )

= 0,329

kp = tan2(45−φ2 )

= tan2(45−3032 )

= 3,037

2) Calculate P1 dan P2

P1 = .L1ka = (1,503).(2,03).(0,329) = 1,004 kN/m2

P2 = (.L1 + 1L2) ka = ((1,503.2,03) + (1,83.3,3)) 0,329 = 2,991 kN/m2

3) Calculate L3

L3 =

P2

γ 1(kp−ka )= 2,991

1,83(3 ,037−0 ,329)=0 ,604

m

4) Calculate P and ZP = ½ P1L1 + P1L2 + ½ (P2 – P1)L2 + ½ P2L3.

= (½1,004.2,03)+(1,004.3,3)+(½(2,991–1,004)3,3)+(½ 2,991.3,3)= 1,019 + 3,313 + 3,279 + 4,935= 12,546 kN/m2

5) Calculate 2 (that is the center of pressure for the are ACDE) by taking the moment about E

Z =

112 ,546

[ 1,019(0 ,604+3,3+ 2 ,033 )+3 ,313(0 ,604+3,3

3 )+3,279

(0 ,604+3 ,333 )+4 ,935(0 ,604×2

3 )]=

112 ,546

[(4 ,668 )+(7 ,468)+(5 ,587 )+(1 ,987 )] = 1,571 m

6) Calculate P5

P5 = (L1 + 1L2) kp + 1L3 (kp – ka)

= (1,503.2,03 + 1,83.3,3) 3,037 + 1,83.0,604 (3,037 – 0,329)= 30,599 kN/m2

7) Calculate A1, A2, A3, and A4

A1 =

P5

γ 1(kp−ka )=30 ,599

1,83(3 ,037−0 ,329)=6 ,175

A2 =

8 Pγ 1(kp−ka )

=8 (12,546 )

1,83(3 ,037−0 ,329)=20 ,253

A3 =

6 P [2Σy1(kp−ka )+P5 ]γ

12( kp−ka)2

=

6 P(12,546 ) [2(1 ,571)(1 ,83)(3 ,037−0 ,329)+30 ,599 ](1 ,83 )2 (3 ,037−0329 )2

=

3472 ,45324 ,558

= 141,398

A4 =

P (6ΣP5+4 P )

γ12(kp−ka )2

=

12 ,546 [6 (1 ,571 )(30 ,599 )+4 (12,546 )](1 ,83)2 (3 ,037−0 ,329 )2

= 172,984

8) Determine L4 by trial and error of the equationL4

4+6 ,175 L43−20 ,253 L4

2−141 ,398L4−172 ,984=0The following table shows the stations of the preceding equation by trial and error

Assumed L4 (m)5

4,84,9

4,995

Left side of Fq 10049

+ 7,485

So, L4 = 4,995 m

9) Calculate P4

P4 = P5 + 1L4 (kp – ka)= 30,599 + (1,83)(4,995)(3,037 – 0,329) = 55,312 kN/m2

10) Calculate P3

P3 = 1 (kp – ka) L4

= 1,83 (3,037 – 0,329) 4,995= 24,753 kN/m2

11) Optain L5

L5 =

P3 .L4−2P

P3+P4

=

24 ,753 . 4 ,995−2 .12 ,54624 ,753+55 ,312 = 1,231 m

12) Draw the pressure distribution diagram

13) Obtain the theoretical depth and the actual depth of penetration by increasing by about 20 – 30%The actual depth of penetration = 1,3 (L3 + L4)

= 1,3 (0,604 + 4,995) = 7,279 mThe theoritical depth of penetration = 0,604 + 4,995

= 5,599 m

14) Calculate maximum building moment (Mmax)

21 = √ 2 Pγ1 (kp−ka )

= √ 2(12 ,546)1 ,83(3 ,037−0 ,329)

= 2,252 m

Mmax = P (Z + Z1) - [ 12γ1Z1

2(kp−ka )](Z1

3 )= 12,546 (1,571 + 2,25) -

[ 12(1 ,83 )(2 ,25)2 (3 ,037−0 ,329 )]−( 2 ,25

3 )= 38,53 kN/M2

15) Calculate required section modulus (s) of the sheet pileThe required section modulus of the sheet pile

S =

Mmax

σallWith tan = 17,25 t/m2

S =

22 ,306

17 ,25 .103

= 1,293.103 m3/m of wall

TUGAS V

REKAYASAPONDASI

1. The cantilever sheet pile wall penetrating a day soil. Given L1 = 3,3 m and L2 = 2,03 m. The granular soil (above the dradge line) has following properties. = 3,3c = 0 = 1,63 t/m3

sat = 1,903 t/m3

1) Calculate ka. For the granular soil

ka = tan2(45−φ2 )

= tan2(45−3,32 )

= 0,891

2) Calculate P1 dan P2 1 = 1= 1,63 1 = 0,63

P1 = .L1ka = (1,63).(3,3).(0,891) = 4,793 kN/m2

P2 = (.L1 + 1L2) ka= ((1,63.3,3) + (1,903 – 1) 2,03) 0,891 = 6,426 kN/m2

3) Calculate P3 and

P1 =

12 P1L1 + P1L2 +

12 (P2 – P1)L2 = 7,90845 + 9,72979 + (1,657)

= ( 1

24 ,793. 3,3)+(4 ,793 . 2 ,03 )+( 1

2(6 ,426−4 ,793 )2 ,03)

= 19,294 t/m

Z =

119 ,294 [7 ,908 (2 ,03+3,3

3 )+9 ,729( 2 ,032 )+1 ,657 ( 2,03

3 )]= 1,853 m

4) Calculate theoritical value of D

D2 [4c – (L1 + 1L2)] – 2DP1

P1(P1+12cΣ )

(γL1+γ1 L2 )+2c = 0

Substituting proper valuesD2 = {4(4,3) – [(1,63)(3,3) + (0,903)(2,03)]} – 2D(19,294)

4 ,793[ 4 ,793+(12)(4,3 )(1 ,853)[(1 ,63 )(3,3 )+(0 ,903 )(2,03 ) ]+2(4,3 )

=0

or 13,654 D2 – 38,588 D – 30,436 = 0Solving the preceding equation, D = 3,47 mCoba-coba13,654 (3,47)2 – 38,588 (3,47) – 30,436 = 0,0700

5) Calculate L4

L4 =

D [ 4c−(γL1+γ1 L2) ]−P1

γc4c – (L1 + 1L2) = 4(4,3) – [(1,63)(3,3) + (0,903)(2,03)

= 9,988 t/m2

so L4 =

3 ,47(9 ,988 )−4 ,7934( 4,3)

= 1,736 m

6) Calculate P6 and P7

P6 = 4c – (L1 + 1L2) = 4(4,3) – (1,63.3,3 + 0,903.2,03) = 9,989 t/m2

P7 = 4c + (L1 + 1L2) = 4(4,3) + (1,63.3,3 + 0,903.2,03) = 24,412 t/m2

7) Draw the pressure distribution diagram (hal sendiri)8) Obtain the theoritical depth and the actual depth of penetration by

increasing by about 40 – 60%D actual = 1,5 D theoritical = 1,5 (3,47) = 5,205 m

9) Calculate maximum building moment (Mmax)

Z1 =

P1

P6

=4 ,7939 ,989

=0 ,479 m

Mmax = P1(Z

1+Z1)−P

6¿

2So,

Mmax = 4 ,793(0 ,479+1 ,853)−

9 ,989 (0 ,479 )2

Z= 7,965 t/m2

10) S = 1 (Kp – Ka) L4

Kp = tan2(45+3,32 )

= 1,122= 0,903 (1,122 – 0,891).1,736 = 0,362

TUGAS VI

REKAYASAPONDASI

1. Dik : L = 20,3 m ; ukuran pile = 33,0 cm ; x = 33,0 cm = 0,33 M = 1,703 t/m3 ; Q = 33,0

Hit P?Jawab:QP = Ap. q1.Nq* = Ap..L.Nq* Nq* = 80

= (0,33 x 0,33). 1,703 (20,6) 80 = 76,408 tQP = Ap. q1 = Ap.Nq* tan

= (0,33)2 . 80. Tan 33 = 5,658

2. Hitung QSK = 1,3 = 0,8 , Hitung QS , L =20,3L1 = 15.D = 15.33,0 = 495 cm = 4,95 mZ = 0. t

1 = 0 f = 0Z = L1 = 4,95 mτ t

1= .L1 = 1,703 . 4,95 = 8,429 t/m2

f = K. τ t1 tan = 1,3 . 8,429 tan (0,8.33,0) = 5,439 t/m2

QS =

( fz=0+ fz=4 ,95 )2 PL1 + fz = P (L – L1)

= [ 0+5 ,439

2 ](4 . 0 ,33)4 ,95+5 ,439(20 ,3−4 ,95 )

= 17,769 + 83,489 = 101,258

3. Gambar

Ap =

π4 D2 =

π4 (0,33)2 = 0,085 m2

Qp = Ap. 9p = Ap. Nc* Cu(2) = 0,085.9.10,3 = 78,795 kN/mQs = 1.Cu(1) P.DLev = 30,3 kN/m2 ; = 1,0 ; Cu2 = 103 kN/m ; 2 = 0,491) Qs = 1.Cu(1) [()(0,303)7,66 + 0,49 (106) [()(0,303)] 20,3

= 1.30 [()(0,303)7,66 + (0,49) (106) [ (0,303)] 20,3= 220,91 + 1028,58 = 1249,49 kN/m

2) Fav = (1v + 2Cu)

=

Cu(1)(7 ,66 )+Cu(2)(20 ,3 )28 ,26

=30 .(7 ,66)+106 .(20 ,3 )28 ,26

= 85,39 kN/m2

τ t1

=

A1+A2+A3

L

1 = .3,03 = 1,73.3,03 = 53,852 = .3,03 + (sat - w) . 3,03 = 110,833 = 110,83 + (19,60 – 9,81) .20,3 = 309,567

A1 = ½ . 53,85.3,03 = 82,39

A2 =

53 ,85+110 ,832

. 3 ,03=251 ,96

A3 =

82 ,39+251 ,96+4360 ,3428 ,26

=166 ,12 kN/m2

SehinggaFav = 0,13 x [166,12 + (2).(85,30)] = 43,77 kN/m2

QS = PL.fav = (0,306)(28,26)43,77 = 1189,10 kNFor Z = 0 – 3,03

Fav1 = (1 – sin QR) tan QR τ t1(av)

= (1 – sin 33) tan 33

(0+53 ,85 )2 = 8,06 kN/m2

For Z = 3,03 – 43,0

Fav2 = (1 – sin 33) tan 33

(110 ,83+309 ,569 )2 = 47,521 kN/m2

Sehingga: QS = P [Fav1 (3,03) + Fav2 (4,3) + Fav3 (20,3)= (.0,303)[8,06.3,03+24,345.4,3+47,521.20,3] = 240,057

3) Estimasi the net allowable pile capaciti Fs = 4

Qs =

1249 ,49+1189 ,102

=1218 ,79

Qu = Qp + Qs = 14,198 + 1218,79 = 1236,467 t

Qall =

QuFs

=

1236 ,4674

=309 ,117kN.

4. GambarHitung Pmax!N = 28 buahXmax = 3,48m Ymax = 1,74 m

nx = 7 buahny = 4 buahJumlah kuadrat absis-absis tiang pancang:x2 = 4.2.3,032 + 4.2.22 + 4.2.0,842 = 111,092 m2

y2 = 7.2.1,742 + 7.2.0,582 = 47,096 m2

Pmax =

ΣVn

±My . Xmax

ny . Σx2±Mx .Y max

nx . Σy2

=

50328

±430 .3 ,484 . 111,092

±230 .1 ,747 . 47 ,096

= 17,964 3,367 1,215Pmax = 17,964 + 3,367 + 1,215 = 22,546 t

L1

L2L

P1

P2

5. Gambara. Daya dukung tiang pancang individu

A tiang = Fb + Fy = 30,6. 30,6 + (8,15 . 3,3) =

Gaya yang diperoleh pada tiangPtiang = Fb A tiang = 50,6 =

Berdasarkan kekuatan tanah

Q tiang =

Atiang×P2

+ 0×C5

A tiang = 30,6.30,6 = 936,36 cm2

Harga konus P = 20 kg/cm2

Harga cloet C = 480 kg/cm2

Keliling tiang O = 4.30,6 = 122,4 cm

Q tiang =

936 ,36 .202

+122 ,4 .4805

=21114 kg = 21,114 t

Jadi daya dukung tanah individu = 21,1114 t

b. Daya dukung kelompok tiang

Eff: = 1− φ

90° {(n−1 )m+(m−1)nm×n }

= Arc. Tg

D5 = Arc tg

30 ,65 = 80,7199 = 80 43’ 11,8”

Eff . <1−80° 43 ' 11 ,8} over { 90} } left lbrace { { \( 6 - 1 \) . 4+ \( 4 - 1 \) 6} over {4 . 6} } right rbrace } { ¿¿¿

= 0,163Daya dukung tiang0,163.21,114 = 3,447 t