TUGAS2TEKNIK PONDASI 1KELASA

AGUS SETIAWANNIM121910301052

Semester52014

Problem 5.10 :Given:q0= 144 kN/m2B= 1,52 mL= 3,05 m= 0,5= 1,22 m

q0H= 12,2 mDetermine : Se ?Solution :

Df = 1, 22 m

1,52 m x 3,05 m

H = 12,2 m

A rigid rock layer

Se = A1 A2

Value of A1 : = = 8 = = 2 So, from curve, the value of = 0,88Value of A2 : = = 0,8 So, from curve, the value of = 0,93

So,= A1 A2= 0,88 . 0,93 . = 13,85 m

Problem 5.20 :Given :

Determine : The consolidation settlement?Solution : = = 268,745 0= 1,52 m . 15,7 + 1,22 m . (19,24 9,81) + 1,525 m . (19,24 9,81) = 49,75

= z (m) = = .

11,221,330,551= 148,08

12,74530,179= 48,11

14,274,670,084= 22,57

Now,= = ( 148,08 + 4 . 48,11 + 22,57 )= 60,515 So,= = = 0,102 log 2,216= 0,035 m= 35 mm

Problem 5.21 :Given :B= 1,83 mL= 1,83 m= 268,745 z = 1,22 + 3,05 = 4,27 m

Determine : the consolidation settlement?Solution := = = 24,187

With OCR (Over Consolidation Ratio) step :OCR= = = 1,486Base on table 5.14, we get the value of (with interpolation)So, = 0,972 . 35 = 34,02 mm