Atk One

  • May 2020
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  • Words: 277
  • Pages: 4
Diketahui

:

sebuah lapisan tanah seperti tergambar.

Ditanya

:

hitung dan gambarkan

Jawab

:

Se = w . Gs → S = 1 1+e = w . Gs e = w . Gs 1 = 0,28 . 2,66 1 = 0,75

γ sat = γ w ( Gs + e) 1+e = 9,81 (2,66 + 0,75) 1 + 0,75 = 19,11 Kpa Titik A :

UA = 0 . 9,81 = 0

σA = 0 . 19,11 = 0 σ′ = 0 Titik B :

UB = Hx . γw = 2 . 9,81 = 19,62 KPa.

σ

B

= Hx .

γ sat

= 2. 19,11

σ, σ′, U.

= 38,22 KPa.

σ′B = σ

B

– UB

= 38,22 – 19,62 = 18,6 KPa. Titik C :

UC = Hx . γw = 4 . 9,81 = 39,24 KPa.

σ

C

= Hx

. γ sat

= 4. 19,11 = 76,4 KPa.

σ’ = σ C

C

– UC

= 76,4 – 39,24 = 37,16 KPa. Titik D :

UD = Hx . γw = 7 . 9,81 = 68,67 KPa.

σ

D

= Hx .

γ sat

= 7 . 19,11 = 133,8 KPa.

σ′ = σ – U D

D

= 133,8 – 68,67

D

= 65,13 KPa. Titik E :

UE = Hx . γw = 10 . 9,81 = 98,1 KPa.

σE = Hx . γ

sat

= 10 . 19,11 = 191,1 KPa.

σ′ = σ – U E

E

= 191,1 – 98,1 = 93 KPa.

E

TUGAS MEKANIKA TANAH I

OLEH NAMA

:

MARSEL NENOTEK

NIM

:

2190 / TS – ATK / 2008

AKADEMI TEKNIK KUPANG 2009

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