worked example 5.8 - 5.10

Exercise 5.8:

HES2155 Geomechanics

A 10m thick normally consolidated saturated clay layer has been calculated to consolidate 400mm at 100 percent consolidation due to the layer of fill place on top. Calculate the time (in years, months and days) it will take for 160mm and 320mm consolidation settlement to occur if the coefficient of consolidation is equal to 1.34 m2/year.

10m

Normally consolidated clay(clay is fully saturated)

cv = 1.34 m2/year

Impervious Bedrock

Surcharge created by layer of Fill

%40100400160

U ionConsolidat of Degree c = 160 mm

v

drainage

cHT

t2

4040 years 034.9

34.1

10126.0 2

Therefore, days 25 and mths 4 yrs, 9

T40 = 0.126 as the surcharge is likely to create a uniform increase in pore pressure throughout the consolidating layer.

Exercise 5.8:


A 10m thick normally consolidated saturated clay layer has been calculated to consolidate 400mm at 100 percent consolidation due to the layer of fill place on top. Calculate the time (in years, months and days) it will take for 160mm and 320mm consolidation settlement to occur if the coefficient of consolidation is equal to 1.34 m2/year.

10m


cv = 1.34 m2/year

Impervious Bedrock


%80100400320


v

drainage

cHT

t2

8080 years 313.42

34.1

10567.0 2


T80 = 0.567 as the surcharge is likely to create a uniform increase in pore pressure throughout the consolidating layer.

Exercise 5.9:


If a 1m thick layer of sand existed between the saturated clay and impervious bedrock, recalculate the time it would take for 160mm (t40) and 320mm (t80) of consolidation to occur .

%40100400160


v

drainage

cHT

t2

4040 years 351.2

34.1

5126.0 2


10m


cv = 1.34 m2/year

Impervious Bedrock


Sand Layer 1m

T40 is still 0.126 but the length of the drainage path is now 5m

Exercise 5.9:


If a 1m thick layer of sand existed between the saturated clay and impervious bedrock, recalculate the time it would take for 160mm (t40) and 320mm (t80) of consolidation to occur .

%80100400320


v

drainage

cHT

t2

8080 years 578.10

34.1

5567.0 2


10m


cv = 1.34 m2/year

Impervious Bedrock


Sand Layer 1m

T40 is still 0.567 but the length of the drainage path is now 5m

Exercise 5.10:


Using the same normally consolidated saturated clay layer as in Exercise 5.8, a small isolated square pad footing is causing the clay layer to consolidate.

Calculate the time it will take for 40% and 80% of the consolidation to be complete. Also, determine the degree of consolidation complete after 6 months.

Degree of Consolidation: U = 40%

v

drainage

cHT

t2

4040 years 582.3

34.1

10048.0 2

days 29 and mths 6 yrs, 3

T40 is now 0.048, as the loading is likely to create a uniform decrease in pore pressure from beneath the footing and drainage is up.

10m


cv = 1.34 m2/year

Impervious Bedrock

Point Load

1m x 1m

Case Type 1

u

Pervious Layer

Impervious Layer

One-way drainage

Exercise 5.10:


Using the same normally consolidated saturated clay layer as in Exercise 5.8, a small isolated square pad footing is causing the clay layer to consolidate.

Calculate the time it will take for 40% and 80% of the consolidation to be complete. Also, determine the degree of consolidation complete after 6 months.

Degree of Consolidation: U = 80%

v

drainage

cHT

t2

8080 years 835.32

34.1

10440.0 2

day 1 and mths 10 yrs, 32

T80 is now 0.440, as the loading is likely to create a uniform decrease in pore pressure from beneath the footing and drainage is up.

10m


cv = 1.34 m2/year

Impervious Bedrock

Point load

1m x 1m

Case Type 1

u

Pervious Layer

Impervious Layer

One-way drainage

worked example 5.8 - 5.10

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