1. CIVIL ENGINEERING MEKANIKA TANAH I
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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
𝛾𝑠𝑎𝑡 = 𝛾 𝑑 + 𝛾 𝑤
𝛾𝑠𝑎𝑡 = 2 + 1
𝛾𝑠𝑎𝑡 = 3 𝑇/𝑚2
Pasir
𝛾𝑠𝑎𝑡 = 3 𝑇/𝑚2
Pasir
𝛾𝑠𝑎𝑡 = 3 𝑇/𝑚2
Pasir 𝛾𝑠𝑎𝑡 = 3 𝑇/𝑚2
A
B
C
D
Zona air
capiler
2 m
1 m
5 m

2. CIVIL ENGINEERING MEKANIKA TANAH I
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PadakedalamanA
𝜎 = 0
U = - (2 .𝛾 𝑤)
U = - (2 . 1)
U = - 2
𝜎′
= 𝜎 − 𝑢
𝜎′
= 0 + 2
𝜎′
= 2 𝑇/𝑚2
PadakedalamanB
𝜎 = 2 . 𝛾𝑠𝑎𝑡
𝜎 = 2 . 3
𝜎 = 6 𝑇/m2
U = - (1 .𝛾 𝑤)
U = - (1 .1)
U = -1
𝜎′
= 𝜎 − 𝑢
𝜎′
= 6 − (−1)
𝜎′
= 7 𝑇/m2

3. CIVIL ENGINEERING MEKANIKA TANAH I
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PadakedalamanC
𝜎 = 2 . 𝛾𝑠𝑎𝑡 + 1 . 𝛾𝑠𝑎𝑡
𝜎 = 2 . 3 + 1 . 3
𝜎 = 9 𝑇/m2
U = (1 .𝛾 𝑤)
U = (1 .0)
U = 0
𝜎′
= 𝜎 − 𝑢
𝜎′
= 9 − 0
𝜎′
= 9 𝑇/m2
PadakedalamanD
𝜎 = 2 . 𝛾𝑠𝑎𝑡 + 1 . 𝛾𝑠𝑎𝑡 + 5 . 𝛾𝑠𝑎𝑡
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′
= 19 𝑇/m2

4. CIVIL ENGINEERING MEKANIKA TANAH I
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Gambar Diagram
Tekanan total 𝜎
2. Diketahuipenampangtanahsepertigambardibawahini
Jawab :
Padabidang A-B
𝛾𝑠𝑎𝑡 =
𝑊𝑠 + 𝑊𝑤
𝑉
=
𝐺𝑠 . 𝛾 𝑤 + 𝑒 . 𝑆𝑟 . 𝛾 𝑤
1 + 𝑒
𝛾𝑠𝑎𝑡 =
𝐺𝑠 . 𝛾 𝑤 + (
𝑛
1−𝑛
) . 𝑆𝑟 . 𝛾 𝑤
1 + (
𝑛
1−𝑛
)
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′
3 m
1 m
2 m
4 m
4 m
PasirHalus n = 0,40 G = 2,65
Sr = 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 m
1 m
5 m
2 m
E’

5. CIVIL ENGINEERING MEKANIKA TANAH I
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𝛾𝑠𝑎𝑡 =
2,65 . 9,81 + (
0,40
1−0,40
) . 0,3 . 9,81
1 + (
0,40
1−0,40
)
𝛾𝑠𝑎𝑡 = 16,746 𝐾𝑁/𝑀3
Padabidang B-C
γ=
𝑊𝑠+𝑊𝑤
𝑉
=
𝐺𝑠 .𝛾 𝑤+𝑒 .𝑆𝑟 .𝛾 𝑤
1+𝑒
γ=
2,68 .9,81+0,60 .0 .9,81
1+0,60
γ = 16,43 KN/M3
Padabidang C-D
𝛾𝑠𝑎𝑡 =
𝑊𝑠 + 𝑊𝑤
𝑉
=
𝐺𝑠 . 𝛾 𝑤 + 𝑒 . 𝑆𝑟 . 𝛾 𝑤
1 + 𝑒
𝛾𝑠𝑎𝑡 =
2,68 . 9,81 + 0,60 . 1 . 9,81
1 + 0,60
𝛾𝑠𝑎𝑡 = 20,11 KN/M3
Padabidang D-E
𝛾𝑠𝑎𝑡 =
𝑊𝑠 + 𝑊𝑤
𝑉
=
𝐺𝑠 . 𝛾 𝑤 + 𝑒 . 𝑆𝑟 . 𝛾 𝑤
1 + 𝑒
𝛾𝑠𝑎𝑡 =
2,10 . 9,81 + 3 .1 . 9,81
1 + 3
𝛾𝑠𝑎𝑡 = 12,51 KN/M3
Padabidang E-F
𝛾𝑠𝑎𝑡 =
𝑊𝑠 + 𝑊𝑤
𝑉
=
𝐺𝑠 . 𝛾 𝑤 + 𝑤𝑠𝑎𝑡 . 𝐺𝑠 . 𝛾 𝑤
1 +
𝑤 𝑠𝑎𝑡 .𝐺𝑠
𝑆
𝛾𝑠𝑎𝑡 =
2,70 . 9,81 + 0,35 .2,70 . 9,81
1 +
0,35 .2,70
1
𝛾𝑠𝑎𝑡 = 26,49 KN/M3

6. CIVIL ENGINEERING MEKANIKA TANAH I
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Menentukanteganganefektif
Padakedalaman A
𝜎 = 0
U = 0
𝜎′
= 𝜎 − 𝑢
𝜎′
= 0 − 0
𝜎′
= 0 𝐾𝑁/𝑀2
PadaKedalaman B
𝜎 = 3 𝑥 𝛾𝑠𝑎𝑡
𝜎 = 3 𝑥 16,746
𝜎 = 50,24 𝐾𝑁/𝑀2
U = 3 x 𝛾 𝑤
U = 3 x 9,81
U = 29,43 KN/M2
𝜎′
= 𝜎 − 𝑢
𝜎′
= 50,24 − 29,43
𝜎′
= 20,81𝐾𝑁/𝑀2

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Padakedalaman C
𝜎 = 1 𝑥 γ
𝜎 = 1 𝑥 16,43
𝜎 = 16,43 𝐾𝑁/𝑀2
U = 1 x 𝛾 𝑤
U = 1 x 0
U = 0 KN/M2
𝜎′
= 𝜎 − 𝑢
𝜎′
= 16,43 − 0
𝜎′
= 16,43 𝐾𝑁/𝑀2
Padakedalaman D
𝜎 = 2 𝑥 𝛾𝑠𝑎𝑡
𝜎 = 2 𝑥 20,11
𝜎 = 40,22 𝐾𝑁/𝑀2
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/𝑀2

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PadaKedalaman E
𝜎 = 4 𝑥 𝛾𝑠𝑎𝑡
𝜎 = 4 𝑥 15,51
𝜎 = 62,04 𝐾𝑁/𝑀2
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/𝑀2
Padakedalaman E’
𝜎 = 2 𝑥 𝛾𝑠𝑎𝑡
𝜎 = 2 𝑥 26,49
𝜎 = 52,98 𝐾𝑁/𝑀2
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/𝑀2

9. CIVIL ENGINEERING MEKANIKA TANAH I
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Gambar
4 m
1 m
2 m
40,22KN/m2
2 m
A
B
C
D
E
E’
50,24 KN/m2
16,43 KN/m2
62,04 KN/m2
3 m
52,98 KN/m2
52,98
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