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Sifat Tanah

This document provides solutions to example problems involving calculations related to soil mechanics. It first solves a multi-part problem involving computing dry density, void ratio, porosity, degree of saturation, and saturated density given soil properties. It then solves another multi-part problem involving similar calculations for a soil sample where total volume, wet mass, dry mass, and particle density are given. Formulas and step-by-step workings are shown for each calculation.

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dani
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740 views9 pages

Sifat Tanah

This document provides solutions to example problems involving calculations related to soil mechanics. It first solves a multi-part problem involving computing dry density, void ratio, porosity, degree of saturation, and saturated density given soil properties. It then solves another multi-part problem involving similar calculations for a soil sample where total volume, wet mass, dry mass, and particle density are given. Formulas and step-by-step workings are shown for each calculation.

Uploaded by

dani
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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MEKANIKA TANAH

NAMA : Dani Hendrawan


NIM : 41118110145
Soal :

1. Given : - density = 1,76 t/m3, density of solid = 2,7 t/m3 - water content = 10%
Required : Compute : dry density, void ratio, porosity, degree of saturation, saturated
density

2. Volume total suatu spesimen tanah adalah 80.000 mm3 dan beratnya 145 g, sedang
berat keringnya adalah 128 g. Kepadatan butir tanah tanah adalah 2,68. Berdasarkan
informasi tersebut, hitunglah :
a) kadar air d) derajat kejenuhan
b) void ratio e) kepadatan kering
c). porositas f) kepadatan jenuh

JAWAB :
1. Diketahui
P=1.76 t/m3
Ps=2.7 t/m3
W= 10%
Vt = 1 t/m3
Ditanya : A. Dry Density
B. Void Ratio
C. Porosity
D. Degree of Saturation
E. Saturated Density

Dijawab :
𝑀𝑡 𝑀𝑠+𝑀𝑡
P= =
𝑉𝑡 𝑉𝑡
1.76 = Mw + Mt
1.76 = 0.1. Ms + Mt
17.6 = 1.1 Ms
Ms= 1.6 t

𝑀𝑤
W=
𝑀𝑠
Mw= W.Ms
=0.1. Ms
Mw = 0.16t
𝑀𝑠
Ps = 2.7  Ps = =
𝑉𝑠
1.6
2.7 =  Vs = 0.59
𝑉𝑠
𝑀𝑤
Pw = 1  Pw = =
𝑉𝑤
0.16
1=  Vw = 0.16
𝑉𝑤

A). Dry Density


𝑀𝑠 1.6
Pd = = = 1.6
𝑉𝑡 1
B). Void Ratio
𝑉𝑣 0.25
e= = = 0.42
𝑉𝑠 0.59
C).Porosity
𝑉𝑣 0.25
n= x100% = x100% = 25%
𝑉𝑡 1
D). Degree of Saturation
𝑉𝑤 0.16
S= x100% = x100% = 64%
𝑉𝑣 0.25
E. Saturated Density
𝑀𝑤+𝑀𝑠 0.16+1.6
P= = =7.04
𝑉𝑡 0.75

2. Diketahui
Vt =80000 mm3 = 80 cm3
Mt =145 gr
Ms= 128 gr
Ps = 2.68 gr/ cm3
Ditanya : A. Kadar Air
B. Void Ratio
C. Porosity
D. Degree of Saturation
E. Kepadatan Kering
F. Kepadatan Jenuh
Dijawab :
𝑀𝑠
Ps = 2.68  Ps = =
𝑉𝑠
128
2.68 =  Vs = 47.76 cm3
𝑉𝑠

Vv = Vt-Vs = 80 – 47.76 = 32.24 cm3


𝑀𝑤
Pw = 1  Pw = =
𝑉𝑤
20
1=  Vw = 20
𝑉𝑤
A). Kadar Air
𝑀𝑤 20
W= x100% = x100% = 15.625%
𝑀𝑠 128

B). Void Ratio


𝑉𝑣 32.24
e= = = 0.675
𝑉𝑠 47.75
C).Porosity
𝑉𝑣 32.24
n= x100% = x100% = 40.3%
𝑉𝑡 80
D). Degree of Saturation
𝑉𝑤 20
S= x100% = x100% = 62.03%
𝑉𝑣 32.24
E). Kepadatan Kering
𝑀𝑠 128
Pd= = = 1.6
𝑉𝑡 80
F). Kepadatan Jenuh
𝑀𝑤 20
Pw= = = 0.62
𝑉𝑤 32.24
Contoh Soal :
A sample of wet soil in a drying dish has a mass of 462 g. After drying in an oven at
110 C overnigth, the sample and dish have a mass of 364 g. The mass of the dish alone
is 39 g
Required Determine the water content of the soil.
Perhitungan :
Berat cawan + tanah basah = (W1) = 462 gram
Berat cawan + tanah kering = (W2) = 364 gram
Berat cawan kosong = (W3) = 39 gram
Berat air = (W1-W2)
= (462-364) gram = 98 gram
Berat tanah kering = (W2-W3)
=(364-39) gram = 325 gram
Kadar air = (W1-W2) / (W2-W3) x 100%
=( 462-364) / (364-39) x 100%
= 98/325 x100 %= 30.15 %
Soal
Volume Pasir Lembab adalah 588 cm3 dengan berat 1010 gr. Berat Keringnya adalah
918 gr dan kepadatan butirannya 2.67 gr/cm3.
Hitung : Void Ratio, Porositas, Kadar Air, Derajat Jenuh, Kepadatan Kering,
Kepadatang Jenuh

Jawab :
Diketahui
Vt =588 cm3
Mt =1010 gr
Ms= 918 gr
Gs = 2.67 gr/ cm3
Ditanya : A. Void Ratio
B. Posositas
C. Kadar Air
D. Derajat Jenuh
E. Kepadatan Kering
F. Kepadatan Jenuh

Jawaban :
A. Void Ratio(e)
𝑀𝑠
Gs =
𝑉𝑠
918
2.67 = Vs = 343.82 cm3
𝑉𝑠
Vv = Vs-Vt = 588-343.82= 244.18 cm3
𝑉𝑣 244.18
e= = = 0.71
𝑉𝑠 343.82
B).Porositas
𝑉𝑣 244.18
n= x100% = x100% = 41.53% = 0.415
𝑉𝑡 58
C). Kadar Air (W)

Mw = Mt-Ms
= 1010 – 918 = 92 gr
𝑀𝑤 92
W= x100% = x100% = 10.02%
𝑀𝑠 918

D). Derajat Jenuh (S)


𝑀𝑤
Pw = 1  Pw = =
𝑉𝑤
92
1=  Vw = 92
𝑉𝑤
𝑉𝑤 92
S= x100% = x100% = 37.68%
𝑉𝑣 244.18

E). Kepadatan Kering


𝑀𝑠 98
Pd= = = 1.56 gr/ cm3
𝑉𝑡 588

F). Kepadatan Jenuh


𝑀𝑠+𝑀𝑡 918+92
Pw= = = 1.72 gr/ cm3
𝑉𝑤 588

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