soal 1
TRANSCRIPT
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 1/12
SOAL 1
Diketahui suatu struktur dinding penahan dan batu kali ( gravity wall ) dengan pembebanan dan profl
lapisan tanah seperti pada gambar di bawah ini sebagai salah satu solusi untu keadaan sebenarnya di
lapangan di bawah ini.
KETENTUAN :
H1 = 3,00 m 1 = !,"0 m #anah $ ( urug ) #anah $$ ( asli)
H! = %,00 m ! = 0,"0 m &1 = 0 k'm &! = 10 k'm
H3 = 1,"0 m 3 = 0,"0 m 1 = 30* ! = 30*
H% = 3,00 m % = 1,"0 m +1 = !0 k'm3 +! = 1 k'm3
- = 10 k'm!
DIMINTA :
nalisis konstruksi tersebut terhadap /
1. tabilitas eser
!. tabilitas uling, dan
3. tabilitas daya dukung tanah
%. ambarkan konstruksi tersebut ( skala 1 / "0 ) beserta sistem drainase pada dinding.
PENYELESAIAN :
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 2/12
Berat Dinding Penahan Tanah dan Beton di atasnya
idang 1
Diambil berat 2enis beton = !" k'm3
1 = 4 . a . t . +
= 4 . 0,"0 . 5,00 . !"
= %3,5" k'm
idang !Diambil berat 2enis beton = !" k'm3
! = p . l . +
= 5,00 . 0,"0 . !"
= 5," k'm
idang 3
Diambil berat 2enis beton = !" k'm3
3 = p . l . +
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 3/12
= ",00 . 1,"0 . !"
= 15," k'm
idang %
% = p . l . +
= 3,00 . !,"0 . !0
= 1"0 k'm
idang "
" = p . l . ( +1 6 +w )
= %,00 . !,"0 . ( !0 6 10 )
= 100 k'm
eban kibat eban 7erata
= - . 8
= 10 k'm! 9 !,"0 m
= !" :'m
Jara Be!an Terhada" U#$ng Dinding Penahan % di titi O &
1. 91 = ( ; . 0,"0 ) < 1,"0 = 1,33 m
!. 9! = ( 4 . 0,"0 ) < 0,"0 < 1,"0 = !,!" m
3. 93 = ( 4 . ",00 ) = !,"0 m
%. 9% = ( 4 . !,"0 ) < 0,"0 < 0,"0 < 1,"0 = 3,5" m
". 9" = ( 4 . !,"0 ) < 0,"0 < 0,"0 < 1,"0 = 3,5" m
. 9 = ( 4 . !,"0 ) < 0,"0 < 0,"0 < 1,"0 = 3,5" m
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 4/12
Mo'en Terhada" U#$ng Dinding Penahan % Titi O &
71 = 1 . 91
= %3,5" . 1,33
= 0,1>35" k'
7! = ! . 9!
= 5," . !,!"
= 1>,5" k'
73 = 3 . 93
= 15," . !,"0
= %,5" k'
7% = % . 9%
= 1"0 . 3,5"
= "!," k'
7" = " . 9"
= 100 . 3,5"
= 35" k'
7 = . 9
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 5/12
= !" . 3,5"
= >3,5" k'
Ta!e( 1)1 *asi( Perhit$ngan Mo'en Ai!at +aya ,ertia(
Koe-sien Teanan Ati. % Ka &
Koe-sien Teanan Tanah Pasi. % K" &
Teanan Tanah Ati. % Pa &
?a1 = :a . - . H
= ⅓ . 10 ,"0
= !,333 k'
?a! = :a . +1 . H1 . ( H! < H3 )
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 6/12
= ⅓ . !0 . 3,00 . ( %,00 < 1,"0 )
= 1!0 k'
?a3 = 4 . :a . +@ . ( H! < H3 )!
= 4 . ⅓ . ( !0 6 10 ) . ( %,00 < 1,"0 ) !
= "0,%15 k'
?a% = 4 . +w . ( H! < H3 )!
= 4 . 10 . ( %,00 < 1,"0 )!
= 1"1,!" k'
?a" = 4 . :a . +1 . ( H1 )!
= 4 . ⅓ . !0 . ( 3,00 )!
= 30 k'
A ?a = ?a1 < ?a! < ?a3 < ?a% < ?a"
= !,333 < 1!0 < "0,%15 < 1"1,!" < 30
= 35>,>>>5 k'
Teanan Tanah Pasi. % P" &
?p = 4 . :p . + . ( H% )!
= 4 . 3. !0 . ( 3,00 ) !
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 7/12
= !50 k'
Jara ( Lengan Terhada" Titi O
l1 = 4 . H = 4 . ,"0 = %,!" m
l! = 4 . ( H! < H3 ) = 4 . %,00 . 1,"0 = 3,00 m
l3 = ⅓ . (H! < H3 ) = ⅓ . %,00 . 1,"0 = !,00 m
l% = ⅓ . (H! < H3 ) = ⅓ . %,00 . 1,"0 = !,00 m
l" = ( ⅓ . H1 ) < H! < H3 = ( ⅓ . 3,00 ) < %,00 < 1,"0 = ,"0 m
l = ⅓ . H% = ⅓ . 3,00 = 1,00 m
Ta!e( 1)/ +aya 0 +aya *orionta( 2 Perhit$ngan Mo'en
Ta!e( 1)3 +aya *orionta( Ai!at Teanan Pasi.
J$'(ah +aya 0 +aya *orionta(
A ?h = A ?a 6 A ?p
= 35>,>>>5 6 !50,0
= 10>,>>>5 k'
Mo'en yang Mengai!atan Pengg$(ingan
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 8/12
A 7g = A 7a 6 A 7p
= 105,5%> 6 !50,0
= 0,5%> k'
Menghit$ng Sta!i(itas Terhada" Penggeseran
#ahanan geser pada dinding sepan2ang = ",00 m, dihitung dengan menganggap dasar dinding
sangat kasar. ehingga sudut geser Bb = C! dan adhesi &d = &!.
ntuk tanah & 6 C ( C E 0 , dan & E 0 )
A Fh = &d . < tan Bb
Dengan A Fh = tahanan dinding penahan tanah terhadap penggeseran
&d = adhesi antara tanah dan dasar dinding
= lebar pondasi ( m )
= berat total dinding penahan dan tanah diatas plat pondasi
Bb = sudut geser antara tanah dan dasar pondasi
A Fh = &d . < tan Bb
= ( 10 k'm . ",00 m ) < ">3,5" k'm . tan 30*
= "0 k'm < 3%!,015 k'm
= 3>!,015 k'm
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 9/12
= 3,"50> G 1," 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 % di'ensi tida "er($ di"er!esar &
Dimana /
gs = Iaktor aman terhadap penggeseran
A ?h = 2umlah gaya 6 gaya horiJontal
Menghit$ng Sta!i(itas Terhada" Pengg$(ingan
#ekanan tanah lateral yang diakibatkan oleh tanah dibelakang dinding penahan, &enderung
menggulingkan dinding, dengan pusat rotasi terletak pada u2ung kaki depan dinding penahan tanah.
= 1,%5 G 1," 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 % di'ensi tida "er($ di"er!esar &
Dimana / gl = aktor aman terhadap penggulingan
A 7w = Kumlah momen yang melawan penggulingan
A 7a = Kumlah momen yang menyebabkan penggulingan
:arena Iaktor aman konstruksi dinding penahan tanah terhadap geser dan guling lebih dari 1,"
( G 1," ), maka dimensi konstruksi sudah aman dan tidak perlu diperbesar.
Sta!i(itas Terhada" Ker$nt$han Ka"asitas Daya D$$ng Tanah
Dalam hal ini akan digunakan persamaan Hansen pada perhitungan, dengan menganggap pondasi
terletak di permukaan.
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 10/12
Esentrisitas % e &
Le!ar E.eti. % B5 & 6 B 0 /e
= ",00 6 ( ! 9 1,3!% ) m
= !,3"! m
@ = @ 9 1
= !,3"! 9 1
= !,3"! m!
+aya 0 +aya yang ada "ada dinding
• aya horiJontal = 105,5%> k'm
• aya Lertikal = ">3,5" k'm
7ator Ke'iringan Be!an
= 0,505
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 11/12
Berdasaran ta!e( : % $nt$ 8 6 39 &
'& = 30,1%
'- = 1,%0
'+ = 1",05
= 0,>0
= 0,51
Ka"asitas D$$ng U(ti'it $nt$ Pondasi di "er'$aan 'en$r$t *ansen :
DI = 0
d& = d- = d+
& = - = +
Didapat /
-u = i- . M . '& < iy . 0," . @ . +! . '+
= 0,505 . 10 . 30,1% < 0,51 . 0," . !,3"! . 1 . 1",05
= !13,0> < !!>,0%3
7/18/2019 SOAL 1
http://slidepdf.com/reader/full/soal-1-56d4c7a5ca1a9 12/12
= %%!,13! k'm!
ila dihitung berdasarkan lebar pondasi eIektiI, yaitu tekanan pondasi ke tanah dasar terbagi rata
se&ara sama, maka
aktor aman terhadap keruntuhan kapasitas daya dukung tanah /
tau dapat pula dihitung dengan kapasitas berdasar distribusi tekanan kontak antara tanah dasar
pondasi dianggap linear.