# Geotechnical Engineering and Underground Structures

This Blog is created to keep up the notes or articles for the courses related to Geotechnical Engineering and Underground Structures. If you need some article on any specific topic, please leave a comment in the comment box, or write in to my Google+ or other profiles(links given below).

## Tuesday, November 28, 2017

## Saturday, April 29, 2017

## Thursday, December 22, 2016

### Consolidation Settlement of Clay - Time Factor and coefficient of Consolidation- Single and Double Drainage

Hi,

One of the main factor of rate of consolidation and the time of the consolidation is the single or the double drainage. When there is a double drainage, i.e. drainage from both of the opposite faces, the drainage path becomes half of the total thickness of the clay layer. In such cases the time of settlement decreases significantly. The following example illustrates the same.

If the same clay has a thickness of 10mm, find out the time it would take to reach 90% consolidation with (b) single drainage (c) Double drainage. Draw the settlement curve/consolidation curve in each case.

Thank You!

One of the main factor of rate of consolidation and the time of the consolidation is the single or the double drainage. When there is a double drainage, i.e. drainage from both of the opposite faces, the drainage path becomes half of the total thickness of the clay layer. In such cases the time of settlement decreases significantly. The following example illustrates the same.

**Example:**(a) A 20 mm clay layer with double drainage reaches the 90% of total consolidation in 30 minutes. Find out the coefficient of consolidation.If the same clay has a thickness of 10mm, find out the time it would take to reach 90% consolidation with (b) single drainage (c) Double drainage. Draw the settlement curve/consolidation curve in each case.

**Solution:**For solution please zoom into the picture below.

Thank You!

## Tuesday, October 18, 2016

### Expression for permeability of an Unconfined Aquifer.

Hi,

In pumping out test, the water is pumped out through a tube well, until a steady state is reached. The water table around the well turns into a cone of depression. The maximum depression along the centre line of the well is known as the drawdown(d).

Following assumptions are made:

Further, permeability can be calculated using the influence radius as given in the formula in the image below.

In pumping out test, the water is pumped out through a tube well, until a steady state is reached. The water table around the well turns into a cone of depression. The maximum depression along the centre line of the well is known as the drawdown(d).

Following assumptions are made:

- Flow follows the Darcy's law.
- Flow is laminar.
- Soil mass is isotropic and homogeneous.
- Well penetrates the aquifer to reach the impervious bed.
- Steady flow along the soil.
- Co-efficient of permeability is constant throughout.
- Natural water ground regime doesn't change.

Also it is assumed that the slope of the hydraulic line is small and so it can be taken as the tangent of the angle, in place of sine of the angle.

or i = dz/dr

Consider flow through a cylindrical surface at a distance of 'r' and of depth 'z'. Further equation are given in the image below.

Flow can be estimated to a rough value using the influence radius of the well. Influence radius is the radius of the circle along which the effect of the pumping are observed. It varies from 150m to 300m. According the Sichardt, it can be found as follows.

R = 3000*d*(k)^0.5 , here d = Drawdown, k = permeability.

Further, permeability can be calculated using the influence radius as given in the formula in the image below.

Thanks!

Reference : Buy the book online from the link given below. Book Title: "Soil Mechanics and Foundation Engineering" by Dr. K R Arora.

## Friday, October 7, 2016

### Lateral Thrust on a Wall at rest against a Soil Backfill

Hi,

A lateral thrust is applied by the soil backfilled behind a wall at rest, this thrust is known as the lateral thrust at rest. Walls which are restricted against any movements or moment come under this category. Example can be a basement wall, which is topped with a RCC slab, and therefore restricts its movement. The amount of thrust on the wall depends upon the properties of the soil.

Density of the soil, angle of friction, degree of the consolidation and the level of the water table are the major factors which govern the magnitude of the lateral thrust.

OCR= 2, π = 35°, πΎπ = 19.7 kN/m3. Find the lateral thrust and its location from base.

πΎπ = 1 - sinπ = 1 - sin35° = 0.426

πΎπ(overconsolidated) = πΎπ(normally consolidated )√OCR

πΎπ(overconsolidated) = 0.426√2 = 0.602

At any depth z

π′π£ = z.πΎπ, [pore water pressure u= 0)

π′β = πΎπ*π′π£

Depth, z= 0

π′π£ = 0, π′β = 0

Depth, z= H=10 m

π′π£ = 10*19.7 = 197 kN/m2,

π′β = 0.602*197 = 118.594 kN/m2

Lateral Thrust = Area of the lateral stress diagram. = 1/2 (10*118.59) = 592.97 kN

Lateral thrust acts at the centroid of the lateral stress diagram, i.e. at a height of H/3 from base.

H/3 = 10/3 = 3.333 m from base.

Thanks!

A lateral thrust is applied by the soil backfilled behind a wall at rest, this thrust is known as the lateral thrust at rest. Walls which are restricted against any movements or moment come under this category. Example can be a basement wall, which is topped with a RCC slab, and therefore restricts its movement. The amount of thrust on the wall depends upon the properties of the soil.

Density of the soil, angle of friction, degree of the consolidation and the level of the water table are the major factors which govern the magnitude of the lateral thrust.

**Example:**A 10 m high basement wall, behind which is filled a soil with density 19.7 kN/m3, c=0,OCR= 2, π = 35°, πΎπ = 19.7 kN/m3. Find the lateral thrust and its location from base.

**Solution.**Fig. Lateral Thrust diagram for wall at rest. |

πΎπ(overconsolidated) = πΎπ(normally consolidated )√OCR

πΎπ(overconsolidated) = 0.426√2 = 0.602

At any depth z

π′π£ = z.πΎπ, [pore water pressure u= 0)

π′β = πΎπ*π′π£

Depth, z= 0

π′π£ = 0, π′β = 0

Depth, z= H=10 m

π′π£ = 10*19.7 = 197 kN/m2,

π′β = 0.602*197 = 118.594 kN/m2

Lateral Thrust = Area of the lateral stress diagram. = 1/2 (10*118.59) = 592.97 kN

Lateral thrust acts at the centroid of the lateral stress diagram, i.e. at a height of H/3 from base.

H/3 = 10/3 = 3.333 m from base.

Thanks!

## Friday, September 2, 2016

### When and why the mat or raft foundations are used?

Q. When and why the mat or raft foundations are used?

A.

Mat or raft foundations are type of the shallow foundations, which covers the entire area beneath a structure and provides support to all the columns and walls. It is also a type of the combined footing where all the footings are combined in the following cases:

A.

Mat or raft foundations are type of the shallow foundations, which covers the entire area beneath a structure and provides support to all the columns and walls. It is also a type of the combined footing where all the footings are combined in the following cases:

- When the columns are so close such that their footings nearly touch each other.
- When the soil is weak and having the low bearing capacity.
- Where chances of differential settlement exist due to either the existence of the different soils or variation of the moisture content.
- or where there is a large variation in the loading in the adjacent columns.

reference: 'Analysis and design of substructures' by Swami Saran.

thanks!

## Saturday, May 2, 2015

### Failure Criteria for Rockmasses

Hi there!

As we know, rock-masses are the complex structures. Failure of the rock-mass occur by the development of the fractures or slip surfaces, when the stresses get increased from the strength of the rock mass.

The failure process in rock-mass is complex, and mathematically difficult to quantify. Four failure criteria are:

As we know, rock-masses are the complex structures. Failure of the rock-mass occur by the development of the fractures or slip surfaces, when the stresses get increased from the strength of the rock mass.

The failure process in rock-mass is complex, and mathematically difficult to quantify. Four failure criteria are:

- Coulomb's Criteria
- Mohr's Criteria
- Hoek and Bray Criteria
- Griffith's Criteria.

First three uses empirical approach, and have considerable value for the design of excavation, while the last one has value for understanding the fracture initiation in the rock.

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