Geotechnical Engineering

Presentasi berjudul: "Geotechnical Engineering"— Transcript presentasi:

1 Geotechnical Engineering
Settlement Building 10 at MIT Geotechnical Engineering

2 Geotechnical Engineering

3 The building underwent differential settlement and was slightly

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5 Settlement of building in downtown Adapazari

6 Geotechnical Engineering

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11 TEORI KONSOLIDASI ANALOG KONSOLIDASI
Proses mengecilnya volume tanah secara perlahan-lahan pada tanah jenuh dengan permebilitas rendah akibat pengaliran sebagian air pori. Proses ini berlangsung terus sampai dengan kelebihan tekanan air pori yang disebabkan oleh kenaikan tegangan total telah hilang p ANALOG KONSOLIDASI U0 = 0 U1 = p p p p p = U2 + p1 p = U3 + p2 U4 = 0

12 Laju penurunan akibat keluarnya air pori

13 POROUS DOUBLE DRAINED Ht H POROUS

14 SINGLE DRAINED POROUS H Ht TIDAK POROUS
Casagrande (1938), Taylor (1948)

15 penurunan akibat konsolidasi

16 Casagrande (1938), Taylor (1948)
U Tv 10 0.008 20 0.031 30 0.071 40 0.126 50 0.197 60 0.287 70 0.403 80 0.567 90 0.848 100

17 Log-time Fitting Method U=50%
Casagrande dan Fadum (1940) Log-time Fitting Method U=50% 18.53 t H 19.202 0.25 19.074 1 18.712 2.25 18.541 4 18.44 9 18.306 16 18.237 25 18.196 36 18.167 49 18.13 64 18.107 81 18.096 100 18.093 121 18.082 144 18.075 169 196 18.06 400 18.051 1444 18.031 18,97 R0 R50 18.09 R100 2,3

18 U Tv 10 0.008 20 0.031 30 0.071 40 0.126 50 0.197 60 0.287 70 0.403 80 0.567 90 0.848 100 50 0.197 0.197

19 Square Root of Time Method U=90%
Taylor (1948) Square Root of Time Method U=90% t H 19.202 0.25 0.5 19.074 1 18.712 2.25 1.5 18.541 4 2 18.44 9 3 18.306 16 18.237 25 5 18.196 36 6 18.167 49 7 18.13 64 8 18.107 81 18.096 100 10 18.093 121 11 18.082 144 12 18.075 169 13 196 14 18.06 400 20 18.051 1444 38 18.031 P O 3,35 OR = 1,15 OP R

20 U Tv 10 0.008 20 0.031 30 0.071 40 0.126 50 0.197 60 0.287 70 0.403 80 0.567 90 0.848 100 90 0.848 0.848

21 Uji Konsolidasi Di Lab Z1, 1 Pasir Z2, 2 Lapisan lempung Cc p0=Z11 + Z22 p akibat bangunan H Pasir

22 P (kN/m2) Initial Dial Reading Final Dial Reading DH, mm H, mm e= (H-Hs)/Hs  10 20 25 10.215 0.215 19.79 50 10.351 0.351 19.65 100 10.481 0.481 19.52 200 10.652 0.652 19.35 400 10.849 0.849 19.15 800 11.05 1.05 18.95 10.75 0.75 19.25

23 P (kN/m2) Initial Dial Reading Final Dial Reading DH, mm H, mm e=
(H-Hs)/Hs  10 20 25 10.215 0.215 19.79 50 10.351 0.351 19.65 100 10.481 0.481 19.52 200 10.652 0.652 19.35 400 10.849 0.849 19.15 800 11.05 1.05 18.95 10.75 0.75 19.25 0.727 0.709 0.697 0.685 0.671 0.653 0.636 0.662

24 q = 400 kN/m2 Cc = 0.731 Cv = 5E-10 m2/det e0 = 0.727 B = 2 m Z = 7
g = 18,6 kN/m3 H1 = 5m H2 = 4m g = 18,4 kN/m3 Cc = 0,731 Cv = 5 X m2/det q = 400 kN/m2 B = 2m p0 = Dp = pasir lempung Z = 7 m B + Z = 9 m2 (B+Z)2 = 81 p0 = Sgz = 5*18,6+2*18,4 p0 = 129.8 kN/m2 Dp =A0/Az*q = S = 0.1042 m cm U = 0.45 Su = 4.6873 Tv = 0.159 Ht = 2 t = det jam Hr th

25 NORMALLY CONSOLIDATED
Log p Z  p = Z

26 OVER CONSOLIDATED  e Z0 Z1 p0 = Z0 p1 = Z1 Log p

27 NORMALLY CONSOLIDATED
OVER CONSOLIDATED e Log p e Log p e  Log p e0 e  Log p e  Log p 0,42e 0,42e P0

28 OVER CONSOLIDATED e Log p e0 0,42e P0 Pc

29 OVER CONSOLIDATED e Log p e  Log p e0 e  Log p 0,42e P0 Pc

30 p (kN/m2) H (mm) DH (mm) De = e 20 50 19,649 100 19,519 200 19,348 400 19,151 800 18,906 19,25 0,08635xH e = w Gs = 0,245 x 2,7 = 0,662 0,727 0,351 0,0303 0,696 0,481 0,0415 0,685 0,652 0,0563 0,670 0,849 0,0733 0,653 1,094 0,0945 0,632 0,75 0,0648 0,662

31 e1 = 0,663 e2 = 0,641 200 600 Cc = Indeks kompresi Compresion index

32 av = koefisien pemampatan mv =koefisen perubahan volume
p (kN/m2) H (mm) DH (mm) De= 0,08635xH e 20 0,727 50 19,649 0,351 0,0303 0,696 100 19,519 0,481 0,0415 0,685 200 19,348 0,652 0,0563 0,670 400 19,151 0,849 0,0733 0,653 800 18,906 1,094 0,0945 0,632 19,25 0,75 0,0648 0,662 av = koefisien pemampatan mv =koefisen perubahan volume