Question

Question 1 5 pts A clay layer is shown in the figure below. The OCR at point Ais 1. If the water table rises to the ground surface, then the OCR at point A becomes ym Ground surface Clay 2m Clay Figure: The clay layer profile. Answer (Round to 1 decimal place): Question 2 5 pts In a Casagrande's liquid limit test, the moisture content for a soil sample was 80% at 10 blows and 30% at 100 blows. The liquid limit of the soil sample is estimated as __%. Answer (Round to the nearest integer):

Answer 1

Question (1) .

Given data:-

Over consolidation ratio (OCR) = 1

Bluk unit weight upto 2 m depth of clay = 18 kN/m³,

Saturated unit weight of clay from 2 m to Point A (5 m depth ) = 20 kN/m³,

The Clay profile are given in question. So that,

We know that O.C.R. is given by as,

0.C.R. Maximum Applied EffectiveStressi n Past Present Applied EffectiveStress

Then,

we considering 1st condition of clay profile when water table is located at depth 2 m .

So that,

Present applied effective stress is given by as,

n-0=0

Where,

$\bar{\sigma }$ = Present effective Stress

$\sigma$ = Total Stress on soil

= Pore Water pressure

Then,

Total stress on soil is given by as,

$\sigma = 2X18+20X3$

2 0 = 96kN/m

Pore water pressure is

( Because water unit weight is 9.81 kN/m³)

So that, Present effective Stress is

n-0=0

o = 96 – 29.43

$\bar{\sigma } = 66.57 kN/m^{2}$

Then,

From Over Consolidation ratio, we determine Maximum applied effective stress is past,

0.C.R. Maximum Applied EffectiveStressi n Past Present Applied EffectiveStress

Put all data,

$1 = \frac{MaximumApplied Effective Stress In Past}{66.57}$

Then,

we considering 2nd  condition of clay profile when water table rises to ground surface.

We know that Effective stress which is applied in past time on Soil profile will be consatnt or its not change after sime time and its cant up & down in past applied stress. But we can increase Stress in Present time.

So that maximum applied effective stress in past is 66.57 kN/m².

For Determine OCR at point A when water table rises ground surface.

Calculating Present applied effective stress, when water table rises ground surface.

n-0=0

Where,

$\bar{\sigma }$ = Present effective Stress

$\sigma$ = Total Stress on soil

= Pore Water pressure

Then,

Total stress on soil is

$\sigma = 5X20$

$\sigma = 100kN/m^{2}$

Pore water pressure is,

Present effective stress is

n-0=0

$\bar{\sigma } = 100-49.05$

$\bar{\sigma } = 50.95 kN/m^{2}$

Then,

Over Consolidation Ratio is

0.C.R. Maximum Applied EffectiveStressi n Past Present Applied EffectiveStress

$O.C.R. = \frac{p}{\bar{\sigma }}$

Where,

p = Past applied effective stress = 66.57 kN/m².

$\bar{\sigma }$ = Present applied effective stress

$\bar{\sigma } = 50.95 kN/m^{2}$

put all known data,

$O.C.R. = \frac{66.57}{50.95}$

So that ratio is grater then 1, then Soil is over consolidated.

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Question (2).

Given data:-

Casagrande liquid limit test,

Sample A.

Moisture content (w) = 80%

No. of blows (N) = 10

Sample B

Moisture content (w ) = 30%

No. of blows (N) = 100

Liquid limi is determined by plotting a 'flow curve' on grapth between No.of bolws and water / moisture content. determine liquid limit corresponding to 25 blows with moisture content.The plot is approximately straight line .

The liquid limit is determine  from the plot corrsponding to 25 blows(N).

Flow curve is given below.

100 وه 80+ CON TENT (W1.) W = 70% N=25 70 - 1 60 - WATER 40 1 mt 30+- 20 - - 10 - دما *** 50 60 70 N 20 ta 100 10 20 25 30 Number of blows (N)

So that,

Liquid limit correspont to 25 blows by Flow curve is 70%.

Liquid limit = 70% = 0.7 Ans.