Mektan-9 · Title: Microsoft PowerPoint - Mektan-9 Author: TG Created Date: 5/23/2005 9:15:56 PM

TEGANGAN LATERAL TANAHTEGANGAN LATERAL TANAH

PERTEMUAN PERTEMUAN 19 19 –– 2020

TOPIK BAHASAN 9


merupakanmerupakan tegangantegangan tanahtanah padapada araharahhorisontalhorisontal dandan fungsifungsi daridari teganganteganganvertikalvertikaldapatdapat disebabkandisebabkan oleholeh massamassa tanahtanah dandanatauatau bebanbeban luarluarAdaAda 3 3 kondisikondisi

TeganganTegangan lateral lateral saatsaat diamdiam (at rest)(at rest)TeganganTegangan lateral lateral aktifaktifTeganganTegangan lateral lateral pasifpasif


Tegangan lateral saat diam (at rest)


Tegangan lateral aktif


Tegangan lateral pasif


q

σv

σh

z σv = γ . z + q

v

hKσσ

=

At rest, K = Ko

Jaky, Broker dan Ireland Ko = M – sin φ’Pasir, lempung terkonsolidasi normal M = 1

Lempung dengan OCR > 2 M = 0,95

Sherif dan Ishibashi Ko = λ + α (OCR – 1)λ = 0,54 + 0,00444 (LL – 20)

α = 0,09 + 0,00111 (LL – 20)

LL > 110% λ = 1,0 ; α = 0,19

Broker dan IrelandKo = 0,40 + 0,007 PI , 0 ≤ PI ≤ 40Ko = 0,64 + 0,001 PI , 40 ≤ PI ≤ 80

TEGANGAN LATERAL TANAH AKTIFTEGANGAN LATERAL TANAH AKTIF

Ka = tan2 (45 - φ/2)

σ1 = σ3 . tan2 (45+φ/2)+2c.tan (45+φ/2)

σa = σv . tan2(45-φ/2) – 2c . tan (45-φ/2)σa = σv . Ka – 2c√Ka

TEGANGAN LATERAL TANAH PASIFTEGANGAN LATERAL TANAH PASIF


σp= σv . tan2(45+φ/2) + 2c . tan (45+φ/2)


Kp = tan2 (45 + φ/2)

σh = σv . Kp + 2c√Kp

CONTOH SOALCONTOH SOAL

h1 = 2 m

h2 = 8 m

h3 = 4 m

γ1 = 15 kN/m3

φ1 = 10 o

c1 = 10 kN/m2

γ2 = 15 kN/m3

φ2 = 15 o

c2 = 10 kN/m2

q = 20 kN/m2

Pertanyaan :

1. Hitung tegangan dan tegangan tanah aktif dan pasif yang dialamioleh sheet pile

2. Hitung kestabilan guling dan geser dari sheet pile

Sheet Pile

PENYELESAIANPENYELESAIAN

2 m

8 m4 m

q = 20 kN/m2

Koefisien Tekanan Tanah :

Aktif ; ka = tan2(45-φ1/2) = 0,704

Pasif ; kp = tan2(45+φ2/2) = 1,698

Pa2

Pa1

Pp1Pw1Pw2

Pa1 = ka . γ1 . h1 – 2 . c . √ka = 0,704 . 15 . 2 – 2 . 10 . √0,704 = 4,34 kN/m2

Pa2 = ka . (γ1 . h1 + γ1’ . h2) – 2 . c . √ka = 32,5 kN/m2

Tekanan Tanah AktifPq1


2 m

8 m4 m

q = 20 kN/m2


Aktif ; ka = tan2(45-φ1/2) = 0,704

Pasif ; kp = tan2(45+φ2/2) = 1,698

Pa2

Pa1

Pp1Pw1Pw2

Pq1 = ka . q = 0,704 . 20 = 14,08 kN/m2

Pw1 = kw . γw . h2 = 1 . 10 . 8 = 80 kN/m2

Tekanan Tanah AktifPq1


2 m

8 m4 m

q = 20 kN/m2


Aktif ; ka = tan2(45-φ1/2) = 0,704

Pasif ; kp = tan2(45+φ2/2) = 1,698

Pa2

Pa1

Pp1Pw1Pw2

Pp1 = kp . γ2’ . h3 + 2 . c . √kp = 1,698 . 5 . 4 + 2 . 10 . √1,698 = 60,02 kN/m2

Pw2 = kw . γw . h3 = 1 . 10 . 4 = 40 kN/m2

Tekanan Tanah PasifPq1

2 m

8 m4 m

q = 20 kN/m2


Aktif ; ka = tan2(45-φ1/2) = 0,704

Pasif ; kp = tan2(45+φ2/2) = 1,698

Pa2

Pa1

Pp1Pw1Pw2

Gaya Tanah AktifPq1

Pp

Pa

Pa = 0,5 . Pa1 . h1 + (Pa1+Pa2)/2 . H2 + Pq1 . (h1+h2) + 0,5 . Pw1 . h2 = 612,5 kN/m

za = 3,32 m

za

zp



4 m

q = 20 kN/m2


Aktif ; ka = tan2(45-φ1/2) = 0,704

Pasif ; kp = tan2(45+φ2/2) = 1,698

Pa2

Pa1

Pp1Pw1Pw2

Gaya Tanah PasifPq1

Pp

Pa

Pp = 0,5 . Pp1 . h3 + 0,5 . Pw2 . h3 = 200,04 kN/m

zp = 4/3 m

za

zp

2 m

8 m


4 m

q = 20 kN/m2

KESTABILAN TERHADAP GULING

Pp

Pa

za

zp

2 m

8 m

oFaktor Keamanan

FK = Pp . zp / Pa . za = (200,04 . 4/3) / (612,5 . 3,32) = 0,13

4 m

q = 20 kN/m2

Pp

Pa

za

zp

2 m

8 m

KESTABILAN TERHADAP GESER


Faktor Keamanan

FK = Pp / Pa = 200,04 / 612,5 = 0,33


Asumsi :

-Tanah timbunanberupa tanah granular

- Gesekan antardinding dan timbunandiperhitungkan

- Selubung keruntuhanberbentuk bidang datar(BC1, BC2 …)

Pa = ½ Ka . γ . H2

( )2

2

2

)sin().sin()sin().sin(1sin.sin

)(sinKa

⎥⎦

⎤⎢⎣

⎡

α+βδ−βα−φδ+φ

+δ−ββ

φ+β=


( )2

2

2

)sin().sin()sin().sin(1sin.sin

)(sinKp

⎥⎥⎦

⎤

⎢⎢⎣

⎡

α+βδ+βα+φδ+φ

+δ+ββ

φ−β=

Pp = ½ Kp . γ . H2

TEGANGAN LATERAL MATERIAL GRANULARTEGANGAN LATERAL MATERIAL GRANULAR

φ−α+α

φ−α−αα=

22

22

coscoscos

coscoscoscosKa

φ−α−α

φ−α+αα=

22

22

coscoscos

coscoscoscosKp

Ka.H..Pa 221 γ=

Kp.H..Pp 221 γ=

TEGANGAN LATERAL AKIBAT BEBANTEGANGAN LATERAL AKIBAT BEBAN

( )222

2

baba.

nHq2

+=σ

( )222

2

baba.

nHq4

+=σ

( )22b16,0b203,0.

Hq

+=σ

a > 0,4

a ≤ 0,4

TEGANGAN LATERAL AKIBAT BEBANTEGANGAN LATERAL AKIBAT BEBAN

( )αβ−β=σ 2cos.sinHq

( )( )12H90qP θ−θ=

( ) ( )( )12

122

H2H'a30,57QRHz

θ−θ+−−θ−θ

=

⎟⎠⎞

⎜⎝⎛=θ −

H'btan 1

1

⎟⎠⎞

⎜⎝⎛ +

=θ −

H'b'atan 1

2

( ) ( )22 90'b'aR θ−+=

( )12 90'bQ θ−=

TEGANGAN LATERAL AKIBAT GEMPATEGANGAN LATERAL AKIBAT GEMPA

( ) aev2

21

ae Kk1H..P −γ=

( )

( ) ( ) ( )( ) ( )

2

2

2

ae

sin'sin'sinsin1'sin.sin'.cos

'sinK

⎥⎦

⎤⎢⎣

⎡β+αθ−δ−βα−θ−φδ+φ

+δ−θ−ββθ

θ−β+φ=

⎥⎦

⎤⎢⎣

⎡−

=θ −

v

h1

k1ktan'


( )( )

ae

ae

P

Pa3HPH6,0

z⎟⎠⎞

⎜⎝⎛+∆

=

PaPP aeae −=∆


( ) pev2

21

pe Kk1H..P −γ=

( )

( ) ( ) ( )( ) ( )

2

2

2

pe

sin'sin'sinsin190'sin.sin'.cos

'sinK

⎥⎦

⎤⎢⎣

⎡β+αθ+δ+βθ−α+φδ+φ

−−θ+β+δβθ

φ−θ+β=

APLIKASI TEGANGAN LATERALAPLIKASI TEGANGAN LATERAL

ANALISA STABILITAS DINDING PENAHAN ANALISA STABILITAS DINDING PENAHAN TANAHTANAH

GESERGESERGULINGGULING

APLIKASI TEGANGAN LATERALAPLIKASI TEGANGAN LATERAL

Mektan-9 · Title: Microsoft PowerPoint - Mektan-9 Author: TG Created Date: 5/23/2005 9:15:56 PM

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Transcript of Mektan-9 · Title: Microsoft PowerPoint - Mektan-9 Author: TG Created Date: 5/23/2005 9:15:56 PM