tugas rekayasa pondasi ii
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J.jpgTUGAS REKAYASA PONDASI II
DISUSUN OLEH :
ERICHA MARTHINA 21010111120002
RAHAYU C RATRI 21010111120047
NUR FAHRIA R.D 21010111120049
ITA PUJI LESTARI 21010111120050
JURUSAN TEKNIK SIPIL FAKULTAS TEKNIK
UNIVERSITAS DIPONEGORO
SEMARANG
2013
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LEMBAR PENGESAHAN
TUGAS REKAYASA PONDASI II
Disusun Oleh :
ERICHA MARTHINA 21010111120002
RAHAYU C RATRI 21010111120047
NUR FAHRIA R.D 21010111120049
ITA PUJI LESTARI 21010111120050
Telah menyelesaikan laporan Tugas Rekayasa Pondasi II dan telah diperiksa sertadisahkan pada :
Hari : Jumat
Tanggal : 20 Desember 2013
Mengetahui
Asisten Dosen
Rekayasa Pondasi II
Kresno WS, ST. MEng
198207162012121004
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1. Diketahui :
Spread footing foundation
B = 2.5 m
L = 2.5 m
Df = 2 m
c = 20 kN/m2
= 5
= 16 kN/m3 (above GWT)
= 18 kN/m3 (saturated)
= 15 kN/m3
Ditanya :
a. Daya dukung (ultimate bearing capacity) dengan Terzaghi dan Mayerhof ?b. Daya dukung ijin (allowable bearing capacity) jika SF = 3 dan Qmaks ?
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c. Penurunan yang terjadi jika dibawah pondasi 4m terdapat lempung 5m ?
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Jawab :
a. 1. Terzaghi
= 5
Nc = 7.34
Nq = 1.64
N = 0.14
q = (2 x 16 ) = 32 kN/m2
= +( )
= (18 kN/m3 9.8 kN/ m3) +(18 (18-9.8))
= 16.04 kN/m3
qu = 1.3 cNc + qNq + 0.4 B N
= (1.3 x 20 x 7.34) + (32 x 1.64) + (0.4 x 16.04 x 2.5 x 0.14)
= 245.566 kN/m3
2. Mayerhof
Nc = 7.34
Nq = 1.64
N = 0.14
. Shape Factors :
Fcs = 1 + (
= 1 + (
= 1.242
Fqs = 1 + (
= 1 + (
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= 1.087
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F s = 1 0.4 (
= 1 0.4 (
= 0.6
. Depth Factors
Fcd = 1 + 0.4 (
= 1 + 0.4 (
= 1.32
Fqd = 1 + 2 tan (1 sin )2 (
= 1 + 2 tan (1 sin )2 (
= 1.116
F d = 1
. Inclination Factors
Fci = Fqi =
= 1
F q =
= 1 0
= 1
q = x Df
= 16 x 2
= 32 kN/m2
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qult = cNc Fcs Fcd Fci + q Nq + Fqs Fqd + B N F s F d F i
=1.1161+ 1211.042.50.450.611
= 70.97 + 6.335 + 5.4
= 82.705 kN/m2
b. 1. Terzaghi
q =kN/m2
2. Mayerhof
q =kN/m2
c. S =
Cc = 0.5
C0 = 0.8
H = 5m
. Sebelum ada pondasi
= (4 x +
= (4 x +
= 76.975 kN/m2
= x Ic
= 27.57 x Ic
m1 = L/B
= 2.5/2.5
= 1
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. Pada kedalaman 2m dari 4m
n1 =
=
= 1.6
Maka Ic = 0.449 (tabel Braja M das principles of foundationengineeringHal 21)
at depth 2m
= x Ic
= 27.57 x 0.449
= 12.378 kN/m2
. Pada kedalaman 4.5m dari 9m
n1 =
=
= 3.6
Maka Ic = 0.136
at depth 4.5m
= x Ic
= 27.57 x 0.136
= 3.750 kN/m2
. Pada kedalaman 7m dari 9m
n1 =
=
= 5.6
Maka Ic = 0.06
at depth 7m
= x Ic
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= 27.57 x 0.06
= 1.654 kN/m2
Jadi :
=
= 4.839 kN/m2
. Penurunan :
S =
Cc = 0.5
C0 = 0.8
H = 5m
Maka :
S =
= 0.037 m
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2. Diketahui :
Jawab :
Dengan menggunakan Metode Mayerhof
qu = cNc Fcs Fcd Fci + q Nq + Fqs Fqd + B N F s F d F i
. Shape Factors :
Fcs = 1 + (
= 1 + (
= 1.223
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Fqs = 1 + (
= 1 + (
= 1.08
F s = 1 0.4 (
= 1 0.4 (
= 0.632
. Depth Factors
Fcd = 1 + 0.4 (
= 1 + 0.4 (
= 1.32
Fqd = 1 + 2 tan (1 sin )2 (
= 1 + 2 tan (1 sin )2 (
= 1.116
F d = 1
. Inclination Factors
Fci = Fqi =
= 1
F i =
= 1 0
= 1
q = x Df
= 16 x 2
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= 32 kN/m2
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. Maka qult Metode Mayerhof
q= cNc Fcs Fcd Fci + q Nq + Fqs Fqd + B N F s F d F i
=1.116+ 12162.30.450.63211
= 209.544 + 50.24 + 1.205 + 5.233
= 266.222 kN/m3
Qult = qBL
= 266.222 x 2.3 x 2.5
= 1530.777 kN
. Dengan FS = 3, maka Q yang diijinkan
Qall =
=
= 510.259 kN
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3. Diketahui :
Pile dikonstruksi dengan cara
Ditanya:
a. Nilai Qu dan Qall ?
b. Efesiensi group pile?
c. Nilai Qu dan Qall dengan beban vertikal ?
d. Penurunan kelompok tiang?
e. Nilai Qu dan Qall dengan beban lateral pada single pile?
f. Nilai Qu dan Qall dengan beban lateral pada group pile?
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Jawab:
a) Calculate of the bearing capacity1. Mayerhoff Methode
The area of cross section of the pile, including the soil inside the pile, is:
Ap =2
=2
= 0,196 m2
Qp1 = Ap . Nc . Cu
= Ap . 9 . Cu
= 0,196 . 9 .125
= 220,5 KN
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..
..
..
..
.. .
.2. Vesic Methode
Depht(m)
AverageDepht(m)
Av. Verticalaff.stress(s0)(KN/m)
Cu
(KN/m)
Cu/s0(KN/m)
a
0-2
1
(0+32)/2 = 16
0
0
0
2-
2,5
(32+50)/2 =41
0
0
0
3-
4,5
(50+101)/2 = 75,5
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0
0
0
6-
13
(101+187,4)/2 =144,2
129,286
0,896
0,5
0 =
= (32+41+377,5+1730,4)/20
= 109,025 KN/m2
. Irr dari jenis tanah lempung clay = 100 (sumber:Braja M DAS hal 516)
. Karena Ir = Irr = 100 dan clay = 0, maka Nc=10,04 dan N. = 1,00
(sumber : Braja M DAS hal 517 tabel 11.5)
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Qp2 = Ap . ((Cu . Nc) + (s0 . N.))
= 0,196 (( 125.10,04) + (109,025.1))
= 267,349 KN
3. Janbu Methode
= 0 , . = 75
(sumber : tabel 11.6 Braja M DAS)aa
Nc = 5,74
Nq = 1,00
Qp3 = Ap . ((Cu . Nc) + (q . Nq))
q = (.1 .H 1 ) + (.2 . H2 ) + (.3 .H3 ) + (.4 + H4)
= (16 .2 ) + ( 18 . 1) + ( (20-9,8) . 5) + ((17-98).14)
= 197,4 KN/m2
Qp3 = 0,196 . ((125 .5,74)+(187,4 . 1)
= 177,36 KN
. Jadi Point Bearing Capacity :
Qp = ( Qp1 + Qp2 + Qp3)/3
= (220,5 + 267,34 + 177,36)/3
= 221,736 KN
b) Calculate of The Friction
1. a Methode
Qs1 = (a1 . Cu1 . L1 + a2 . Cu2 . L2 + a3 . Cu3 . L3+ a4 . Cu4 . L4) . P
= ( 0 + 0 + 0 + 0,5 . 129,286 . 14) . (3,14 . 0,5)
= 1421,574 KN
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2. . Methode
(Cu1 . L1 + Cu4 . L4)/Lclay = (0 . 2)+(129,286 . 14) /(2+14)
= 113,125 KN/m3
Fav = 0,173 ( 109,025 + (2) (113,125))
= 58,003 KN/m2
Qs2 = p . L .fav
= (3,14 . 0,5 ) . 20 . 58,003
= 1822,218 KN
. Jadi Qs = (Qs1 + Qs2)/2
= ( 1421,574 + 1822,218) / 2
= 1621,896 KN
. Qu = Qp + Qs
= 221,796 + 1621,896
= 1843,632 KN
. Qall = Qu / Fs
= 1843,632 / 3
= 614,54 KN
c) Group Efesiensi
. = 1 ) . .
n1 = 4
n2 = 2
d = 1
D = 0,5
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. = tan-1 ( D/d)
= tan-1 ( 0,5/1)
= 26,5650
. = 1 ) 26,5650
= 0,631
d) Bearing Capacity dan beban yang diijikan
Qult (1 pile) = 1843,632 . 0,631
= 1163,332 KN
Qult (8 pile) = Qult (1 pile) . 8
= 1163,332 . 8
= 9306,654 KN
Qall = 9306,654 / 3
= 3102,218 KN
e) Penurunan Kelompok Tiang
D = 0,5 m
d = 1 m
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. Settlement Tiang Tunggal.1) Semi Empirical Methode.
Diasumsikan distribusi tahanan kecil seragam, maka c menurut Vesic adalah 0,5.
Anggaplah Qall adalah beban maksimum yang terjadi pada pile tersebut.
= 614,544 kN73,912 kN
= 540,632 kN
E = 25743 N/mm2
= 25743.102 ton/mm2
= 25743.108 ton/m2
= 25743.103 kN/m2
Persamaan settlement St = Ss + Sp + Sps , maka:
.
.; Cp untuk sand 0,02 0,04, diambil 0,03
.
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a. Sps =
Cs = 0,93 + 0,16 . Cp
= 0,93 + 0,16 . 0,03
= 0,9603
Sps == 0,0229 m
= 22,9 mm
Sehingga penurunan tiang tunggal yang terjadi adalah
St = Ss + Sp + Sps
= 1,36 + 3,9 + 22,9
= 28,16 mm
2) Empirical Methode
St =+
=+
= 7,43 . 10-3 m
= 7,43 mm
b. Settelmen kelompok tiang. Vesic
SB = St
= 7,43
= 16,614 mm
. Mayerhofs
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SG = 2P ,dimana :
P === 206,815 KN/m = 2,15 ton/ft
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I = ( 1-)
= (1 -)
= 0,5 m
SG = 2 . 2,1597 . 0,07
= 0,04 in
= 0,109 m
= 10,9 mm
f) Daya Dukung ( Bearing Capacity ) dan Beban yang Diijinkan
( Allowable Load )
. Fixed Head Pile
E = 25743.103 kN/m2
I =
=
= 0,0023 m4
EI = E . I
= 25743. 103 x 2,33 . 10-3
= 60015,789 kNm2
Z =
Mu = Z . fb
. fb = 0,33 . fc ( asumsi fc = 30 MPa = 30 N/mm2 )
= 0,33 . 30000 kN/m2
= 9900 kN/m2
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Mu = 9,2 . 10-3 x 9900
= 91,08 kNm
R =K = 67 Cu
== 67 (113,125) = 7579,375 kN/m3
= 1,677 m
Long Piles
1) Fixed Head Pile
From figure 6.28b, for
. Qu = 9,8 x 73,912 ( 0,5 )2
= 181,084 kN
Fs = 3
Qall =
g) Daya Dukung ( Bearing Capacity ) dan Beban yang Diijinkan ( Allowable Load )pada Pembebanan Lateral untuk Group Pile.
From table 6.14, forhave a value of Ge = 0,3 , so :
. For 1 Pile
Qult eff = Qfixed . 0,3
= 181,084 . 0,3
= 54,325 kN
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. For 8 Pile
. Qult eff = 8 x 54,325 kN
= 434,602 kN
Qall =
= 144,867 kN
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