Question

In: Civil Engineering

a circular footing 10m in diameter is proposed to carry a total load of 9000 KN....

a circular footing 10m in diameter is proposed to carry a total load of 9000 KN. The footing is to be placed at a depth of 2m below ground surface. The soil is normally consolidated silty clay with a unit weight of 20KN/m^3 and undrained shear strength of 50 KPa.

1) determine the short-term factor of safety using the General bearing capacity equation.

2) Calculate the expected consolidation settlement of the footing if the compression index is 0.2 and the initial void ratio is 1.0.

3) if the calculated settlement is not acceptable, indicate the necessary steps to be taken before construction to reduce the consolidation settlement.

Solutions

Expert Solution

Load = P = 9000 kN

A = πx102/4 = 78.54 m2

Net Bearing pressure = qp = P/A = 9000/78.54

= 114.6 kN/m2

Ultimate Bearing Capacity of soil given by General Bearing Capacity equation is given by

qu =cNcscdcic + qNqsqdqiq + 0.5YBNysydyiy

where

c = cu = 50 kN/m2

For, φu = 0

Nc = 5.14 , Nq = 1 and Ny = 0

Shape factors

sc = 1.2

sq = 1

Depth factors

dc = 1.2

dq = 1

Inclination factors

For zero inclination with ground

ic = iq = 1

q = overburden pressure = YDf

where Y = unit wt of soil = 20 kN/m3

Df = depth of foundation = 2 m

=> q = 20x2 = 40 kN/m2

Putting values we get

qu = 50x5.14 x1.2x1.2 x1 + 40x1x1x1x1 + 0

qu = 370.08 + 40

= 410.08 kN/m2

Net ultimate Bearing Capacity is given by

qnu = qu - YDf

= 410.08 - 40

= 370.08 kN/m2

Factor of Safety against short - term Bearing Capacity is given by

F = qnu/qna

= 370.08/114.6

= 3.23

b) Taking clay depth H = 10 m

Initial effective overburden at mid-depth of clay(at 5 m from suface)= q' = 20x5 = 100 kN/m2

Footing is placed at a depth of 2 m from surface

Depth of soil below footing = 8 m

Increase in stress at mid-depth of clay layer from bottom of footing i.e at depth at 3 m from footing bottom by assuming 2: 1 load distribution is given by

dq' = 9000/(10 + 3)2

= 53.25 kN/m2

Final effective stress = q'f = 100 + 53.25 = 153.25 kN/m2

Settlement for normally Consolidated soils is given by

S = (CcH/1+e0)xlog(q'f/q')

= (0.2x10/1+1)xlog(153.25/100)

= 0.1854 m

= 185.4 mm

c) If the settlement is not acceptable we can do following methods to reduce settlement

1. By Consolidating the soil with a placement of fill to allow pore pressure to diisipate

2. By placing sand drains in the soil to shorten the drainage path and allow pore pressure tro dissipate

3. By compacting the soil with rollers

4. By reinforcing the soil with geotextiles to improve soil properties

5. By using Compaction piles to reinforce the soil


Related Solutions

Q3. The total load of a chimney and its circular footing is 35 MN, as shown...
Q3. The total load of a chimney and its circular footing is 35 MN, as shown in the figure. Assume the average unit weight of soil below and above the water table is the same. Presume: B (diameter) = 8 m, H = 6 m, and D = 2 m. According to SPT results, the average N-value for sandy soil below the footing is 20. Estimate the values of the friction angle and the average unit weight of soil based...
square footing 4×4 m carrying a load of 400 KN and the moment of 200 KN.m...
square footing 4×4 m carrying a load of 400 KN and the moment of 200 KN.m compute the maximum and minimum bearing pressure
A simply supported beam 10m long carries a uniform load of 24 kN/m. Using E =...
A simply supported beam 10m long carries a uniform load of 24 kN/m. Using E = 200 GPa, I = 240x10^6 mm4 Using CBM (conjugate beam method): A. Determine the rotation (in degrees) of the beam at a point 4m from the left support. B. Determine the deflection at a point 4m from the left support.
A rectangular footing (0.75m∙0.75m), embedded 0.5 m, carries a downward column load of 200 kN. The...
A rectangular footing (0.75m∙0.75m), embedded 0.5 m, carries a downward column load of 200 kN. The footing is supported on an overconsolidated clay with the following properties: Cr = 0.05         e = 0.7             OCR = 2.0 PI = 55            cu = 200 kPa ϒsat = 16.2 kN/m3 The ground water table is at a depth of 1.0 m below the ground surface. a) Calculate the primary consolidation settlement for this footing. b) Is the calculated Sc within...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT