kuliah 7 (Daya Dukung Pondasi)

Powerpoint Templates Page 1Powerpoint Templates

PONDASI DANGKAL

MEKANIKA TANAH IIYULVI ZAIKA

Powerpoint Templates Page 2

SYARAT- SYARAT PEMILIHAN PONDASI1. KEDALAMAN TANAH KERAS2. BEBAN YANG DITAHAN3. BIAYA YANG TERSEDIA4. PENGARUH-PENGARUH LAIN

Powerpoint Templates Page 3

PONDASI TELAPAK

Lapisan Tanah Keras

Powerpoint Templates Page 4

Combined footing

2 footings close

to each other

P1 close to property

line and P2 > P1

property line

P1 P2

If P2/P1 < 1/2, use strap combined footing

property line

If 1/2 < P2/P1 < 1

use trapezoidal footing

property line

P1 P2

Powerpoint Templates Page 5

BENTUK PONDASI TELAPAK

Powerpoint Templates Page 6

Powerpoint Templates Page 7

PONDASI TELAPAK MENERUS/ LAJUR

Powerpoint Templates Page 8

Powerpoint Templates Page 9

Powerpoint Templates Page 10

PONDASI RAKIT

Powerpoint Templates Page 11

PONDASI RAKIT

Powerpoint Templates Page 12

Powerpoint Templates Page 13

PONDASI RAKIT

Powerpoint Templates Page 14

Powerpoint Templates Page 15

SYARAT PONDASI DANGKAL

Kedalaman tanah pondasi kurang atau sama dengan lebarnya atau kedalaman (Terzaghi, 1943).

Teori lain, kedalaman pondasi dangkal 3-4 kali lebar pondasi.

Pondasi setempat harus memenuhi syarat-syarat:1. Stabilitas, aman terhadap keruntuhan geser 2. deformasi harus lebih kecil dari yang diizinkan.

Lapisan Tanah Keras

Df

B

Powerpoint Templates Page 16

METODA PELAKSANAAN

Powerpoint Templates Page 17

KERUNTUHAN PADA PONDASI DANGKALPADA TANAH PASIR PADAT DAN LEMPUNG KAKU

B

Powerpoint Templates Page 18

MODEL KERUNTUHAN GESER

» keruntuhan umum geser» pasir padat Dr>67%

» lempung kaku NSPT >12

» Koruntuhan lokal» pasir atau

» pasir kelempungan» kurang padat (medium)

» 30%<Dr<67%

» Keruntuhan penetrasi » Pasir lepas

» Dr< 30%

Beban/luas

Penurunan

qu

Beban/luas

Penurunan

qu1

qu

Beban/luas

Penurunan

qu1

qu qu

Permukaan runtuh

Permukaan runtuh

Permukaan runtuh

(a)

(b)

(c)

Powerpoint Templates Page 19

MODEL KERUNTUHAN BERDASARKAN RELATIF DENSITY

Powerpoint Templates Page 20

Terzaghi Assumptions

• 1.    Soil under footing is homogeneous and isotropic  • 2. Soil surface is horizontally• 3. The base of footing is rough, to prevent the shear displacement.• 4      The foot is shallow foundation, i.e. the depth of foundation is less than

the width of foot…• Df      ≤        B

• 3.     Shear strength above the level of the base of footing is negligible. c = 0 above ( F.L ).

• 4.     Consider only the surcharge which produced as uniform pressure     •    q = DF at foundation level.• 5.      The load on foundation is vertical and uniform.• 6.     The foot is long strip footing (pondasi lajur, B/L 0). • 7. =

Powerpoint Templates Page 21

Shear failure happened on many stages

• I) Stage I: The soil in the elastic case and behave as the part of foundation it still that, and by increasing the load performed the region I which called active zone.

• II)     Stage II: At this stage the foundation load effect on the active zone and neighboring soil so perform the region which called arc of logarithmic spiral zone.

• III)    Stage III: By increasing the load performed the third part curve in which part the soil became in the passive case it make to resist the failure.

Powerpoint Templates Page 22

PROSES TERJADINYA KERUNTUHAN

Powerpoint Templates Page 23

Three components produced to resist the failure of soil.

I)  (Pp)γ   = Component produced by the weight of shear zone II, III.

II)(Pp)c = Component produced by the cohesive stress.

III) (Pp)q = Component produced by the surcharge q.

Pp     =       (Pp)γ +   (Pp)c  +   (Pp)q

Powerpoint Templates Page 24

DASAR PENURUNAN RUMUS TERZAGHI

• Kesetimbangan Gaya qu(2b)= -W+ 2Pp+f sin • W=(1/2)2b btan =b2tan• f=c DC= c. b/cos• Pp=(1/2) (b tan )2 K + c (b tan )Kc +q(b tan ) Kq• qu =c{tan (Kc +1)} +q(tan Kq) + B/2{(1/2) tan (K. tan -

1)}

Zona aktif

Zona pasif

Zona geser radial

quA B

J

PpPp

W

ff

B=2b

Powerpoint Templates Page 25

PERSAMAAN DAYA DUKUNG TERZAGHIUNTUK PONDASI LAJUR/ MENERUS

12u c qq cN qN BN=

245cos2 2

2

a

Nq =

= e

( ) cot1-qNNc =

Powerpoint Templates Page 26

TEORI DAYA DUKUNG TERZAGHI (1943)

Daya dukung:Pondasi lajur

Pondasi bujursangkar

Pondasi lingkaranUntuk keruntuhan geser lokal:

c’ = 2/3 c dan ’ = 2/3

qu

q=Df

B

Df

45-/2 45-/2 45-/2 45-/2 A C

D

EF

H G

J I

12u c qq cN qN BN=

1.3 0.4u c qq cN qN BN=

1.3 0.3u c qq cN qN BN=

Powerpoint Templates Page 27

GRAFIK FAKTOR DAYA DUKUNG

General shear failureLocal shear failure

Powerpoint Templates Page 28

PERBANDINGAN BIDANG RUNTUH

Powerpoint Templates Page 29

RUMUS UMUM DAYA DUKUNG PONDASIMEYERHOF (1963)

• Fcs , Fqs , Fs = faktor bentuk• Fcd , Fqd , Fd = faktor kedalaman• Fci , Fqi , Fi = faktor inklinasi

12u c cs cd ci q qs qd qi s d iq cN F F F qN F F F BN F F F =

Powerpoint Templates Page 30

FAKTOR PENGARUH MEYERHOF

• F. BENTUK F. KEDALAMAN F. INKLINASI

=

=

=

=

245tan

1.01

1.01

2.01

2

p

ps

pqs

pcs

K

LBKF

LBKF

LBKF

BDKF

BDKF

BDKF

pd

pqd

pcd

1.01

1.01

2.01

=

=

=

2

2

190

1

o

ci qi

i

F F

F

= = -

= -

Q

T

R

Powerpoint Templates Page 31

Powerpoint Templates Page 32

DISTRIBUSI TEGANGAN TANAH AKIBAT TEG. VERTIKAL DAN MOMEN

P

M

Powerpoint Templates Page 33

Tegangan kontak akibat beban vertikal dan momen

P

M

PMe=

e = B/6qmaxqmin

e < B/6qmaxqmin

e > B/6qmaxqmin

B

eR

Powerpoint Templates Page 34

P

e = B/6qmaxqmin

e < B/6qmaxqmin

e > B/6qmaxqmin

e

Powerpoint Templates Page 35

Pondasi Lajur

e

B

B-2e

Powerpoint Templates Page 36

Stabilitas Daya Dukung

C

Tegangan kontak : tegangan akibat beban

Menentukan eksentrisitas DPT

XBeV

MX net -=

=2

;