A rigid footing with B = 3m and L = 4m carries a total load of P = 890kN as shown. A.) Determine the elastic settlement of the footing if the Poisson’s ratio (µs) of soil = 0,32, modulus of elasticity of soil Es = 15900kPa, Influence factor Ip = 0.88. B.) Compute the primary consolidation settlement of clay layer if it is normally consolidated. C.) Compute the total consolidation of the clay 6 years after the completion of primary consolidation settlement. Time for completion of primary settlement is 2 years.
A rigid footing with B = 3m and L = 4m carries a total load of P = 890kN as shown. A.) Determine the elastic settlement of the footing if the Poisson’s ratio (µs) of soil = 0,32, modulus of elasticity of soil Es = 15900kPa, Influence factor Ip = 0.88. B.) Compute the primary consolidation settlement of clay layer if it is normally consolidated. C.) Compute the total consolidation of the clay 6 years after the completion of primary consolidation settlement. Time for completion of primary settlement is 2 years.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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- A rigid footing with B = 3m and L = 4m carries a total load of P = 890kN as shown.
A.) Determine the elastic settlement of the footing if the Poisson’s ratio (µs) of soil = 0,32, modulus of elasticity of soil Es = 15900kPa, Influence factor Ip = 0.88.
B.) Compute the primary consolidation settlement of clay layer if it is normally consolidated.
C.) Compute the total consolidation of the clay 6 years after the completion of primary consolidation settlement. Time for completion of primary settlement is 2 years.
![P= 890kN
LOOSE
1.5m
SAND
Teny= 16.2 kN/m
3m
1.5m
V GWT
Yan = 18 kN/m
2m
T= 19.4 kN/m
CLAY
3m
e, = 0.6
LL = 45
ROCK](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7d6b4a3-4711-4396-87e2-7d29596e0fe0%2F7fa9decf-aa73-46e1-bd0f-571189118de1%2Ftxdtimc_processed.png&w=3840&q=75)
Transcribed Image Text:P= 890kN
LOOSE
1.5m
SAND
Teny= 16.2 kN/m
3m
1.5m
V GWT
Yan = 18 kN/m
2m
T= 19.4 kN/m
CLAY
3m
e, = 0.6
LL = 45
ROCK
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