tugas v mektan i redha rm
TRANSCRIPT
REDHA ARIMA RM 1310015211092
TUGAS V MEKANIKA TANAH I
1. Permukaan air dalamendapanpasirsetebal 8 m terletakpadakedalaman 3
dibawahpermukaantanah. Diataspermukaan air pasirdalamkeadaanjenuhdengan air kapiler.
Beratisipasir = 2,00 gram/cm3 . Hitungtekananefektifpadakedalaman: 1 m, 3 m, 8 m
dibawahpermukaantanah. Kemudiangambarkan diagram tekanan total, tekanannetral,
tekananefektifsampaikedalaman 8 m tersebut.
Diketahui :
Ditanya :
Tekanan total, tekanannetral, tekananevektif
Jawab :
2 gram/cm3 = 2T/m3
γ sat=γ d+γ w
γ sat=2+1
γ sat=3T /m2
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
Pasirγ sat=3T /m2
Pasir
γ sat=3T /m2
Pasir γ sat=3T /m2
A
B
C
D
Zona air capiler
2 m
1 m
5 m
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PadakedalamanA
σ=0
U = - (2 .γ w)
U = - (2 . 1)
U = - 2
σ '=σ−u
σ '=0+2
σ '=2T /m2
PadakedalamanB
σ=2 . γ sat
σ=2 .3
σ=6T /¿m2
U = - (1 .γ w ¿
U = - (1 .1¿
U = -1
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
σ '=σ−u
σ '=6−(−1)
σ '=7 T /¿m2
PadakedalamanC
σ=2 . γ sat+1 . γ sat
σ=2 .3 + 1 . 3
σ=9T /¿m2
U = (1 .γ w ¿
U = (1 .0¿
U = 0
σ '=σ−u
σ '=9−0
σ '=9 T /¿m2
PadakedalamanD
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M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
σ=2 . γ sat+1 . γ sat+5 . γ sat
σ=2 .3 + 1 . 3 + 5 . 3
σ=24 T /¿m2
U = (5 .γ w ¿
U = (5 .1¿
U = 5 T/m2
σ '=σ−u
σ '=24−5
σ '=19T /¿m2
Gambar Diagram
Tekanan total σ
2. Diketahuipenampangtanahsepertigambardibawahini
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
Tekanannetral U
0
6 T/m2
9 T/m2
24 T/m2
-2 T/m2
1 T/m2
0
5 T/m2
2 T/m2
7 T/m2
9 T/m2
19 T/m2
Tekananefektifσ '
2 m
1 m
5 m
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Jawab :
Padabidang A-B
γ sat=Ws+Ww
V=
Gs . γ w+e . Sr . γ w
1+e
γ sat=Gs . γ w+( n
1−n). Sr . γ w
1+(n
1−n)
γ sat=2,65 . 9,81+( 0,40
1−0,40 ) .0,3 .9,81
1+( 0,401−0,40
)
γ sat=16,746 KN /M 3
Padabidang B-C
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
3 m
1 m
2 m
4 m
4 m
PasirHalus n = 0,40 G = 2,65Sr = 30 %
Sr = 0 %e = 0,60 G = 2,68Lanau
Gambut e = 3 G = 2,10
LempungKelanauan
Wsat = 35 % G = 2,70
A
B
D
E
F
C
2 mE’
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γ¿ Ws+WwV
=Gs. γw+e . Sr . γw
1+e
γ¿2,68 .9,81+0,60 .0 . 9,81
1+0,60
γ = 16,43 KN/M3
Padabidang C-D
γ sat=Ws+Ww
V=
Gs . γ w+e . Sr . γ w
1+e
γ sat=2,68 . 9,81+0,60 . 1. 9,81
1+0,60
γ sat = 20,11 KN/M3
Padabidang D-E
γ sat=Ws+Ww
V=
Gs . γ w+e . Sr . γ w
1+e
γ sat=2,10 . 9,81+3 .1 .9,81
1+3
γ sat = 12,51 KN/M3
Padabidang E-F
γ sat=Ws+Ww
V=
Gs . γ w+w sat . Gs . γ w
1+w sat . Gs
S
γ sat=2,70 . 9,81+0,35 .2,70 . 9,81
1+0,35 .2,70
1
γ sat = 26,49 KN/M3
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
Menentukanteganganefektif
Padakedalaman A
σ=0
U = 0
σ '=σ−u
σ '=0−0
σ '=0 KN /M 2
PadaKedalaman B
σ=3 x γsat
σ=3 x16,746
σ=50,24 KN / M 2
U = 3 x γ w
U = 3 x 9,81
U = 29,43 KN/M2
σ '=σ−u
σ '=50,24−29,43
σ '=20,81 KN / M 2
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
Padakedalaman C
σ=1xγ
σ=1x 16,43
σ=16,43 KN /M 2
U = 1 x γ w
U = 1 x 0
U = 0 KN/M2
σ '=σ−u
σ '=16,43−0
σ '=16,43 KN / M 2
Padakedalaman D
σ=2 xγ sat
σ=2 x20,11
σ=40,22 KN / M 2
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
U = 2 x γ w
U = 2 x 9,81
U = 19,62 KN/M2
σ '=σ−u
σ '=40,22−19,62
σ '=20,6 KN /M 2
PadaKedalaman E
σ=4 xγ sat
σ=4 x15,51
σ=62,04 KN / M 2
U = 4 x γ w
U = 4 x 9,81
U = 39,24 KN/M2
σ '=σ−u
σ '=62,04−39,24
σ '=22,8 KN /M 2
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
REDHA ARIMA RM 1310015211092
Padakedalaman E’
σ=2 xγ sat
σ=2 x26,49
σ=52,98 KN /M 2
U = 2 x γ w
U = 2 x 9,81
U = 19,62 KN/M2
σ '=σ−u
σ '=52,98−19,62
σ '=33,36 KN /M 2
Gambar
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY
4 m
1 m
2 m
40,22KN¿m2
2 m
A
BC
D
E
E’
50,24KN¿m2
16,43 KN /m2
62,04 KN /m2
3 m
52,98KN¿m2
Tegangan total σ
0
29,43 KN¿m2
19,62 KN¿m2
0
39,24 KN¿m2
19,62 KN¿m2
0 0
20,81 KN¿m2
16,43 KN¿m2
20,6 KN¿m2
22,28 KN¿m2
33,36 KN¿m2
Tegangannetral U Teganganefektifσ’
12 m
REDHA ARIMA RM 1310015211092
CIVIL ENGINEERING MEKANIKA TANAH I
M,MEK
BUNG HATTA UNIVERSITY