51 - NPTEL · 2017. 8. 4. · A Pole is a unique point located on the circumference of Mohr’s circle. The point of intersection of Mohr’sstress circle and line drawn through the

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

51


Module 4: Lecture 2 on Stress-strain relationship

and Shear strength of soils


Stress state, Mohr’s circle analysis and Pole, Principalstress space, Stress paths in p-q space;

Mohr-coulomb failure criteria and its limitations,correlation with p-q space;

Stress-strain behavior; Isotropic compression andpressure dependency, confined compression, large stresscompression, Definition of failure, Interlocking conceptand its interpretations, Drainage conditions;

Triaxial behaviour, stress state and analysis of UC, UU, CU,CD, and other special tests, Stress paths in triaxial andoctahedral plane; Elastic modulus from triaxial tests.

Contents


Mohr's circle is a geometric representation of the two-dimensional stress state and is very useful to performquick and efficient estimations.

It is also popularly used in geotechnical fields such assoil strength, stress path, earth pressure and bearingcapacity. It is often used to interpret the test data, toanalyze complex geotechnical problems, and topredict soil behaviours.

The pole point on Mohr's circle is a point so specialthat it can help to readily find stresses on anyspecified plane by using diagram instead ofcomplicated computation.

Mohr’s stress circle


Mohr’s stress circle: Pole points

A Pole is a unique point located on the circumference of Mohr’scircle.The point of intersection of Mohr’s stress circle and line drawn

through the pole parallel to a given plane, gives the stresses onthat plane.

Two pole points can be established,a) Relating to the direction of action of stresses, andb) Relating to the direction of planes on which stresses are

acting. Usually the pole point relating to the direction of the planes are

in use.


Mohr’s stress circle: Pole points for stresses- Procedure

Step : Project the line from the point (σz, τzx) in the line of action of σz (Vertical)OR

Project the line from the point (σx, τxz) in the line of action of σx(Horizontal) till it intersects the circumference of the circle.

The intersection point gives the POLE point Ps for stresses.

(σz, τzx)

Ps

τ

σ

(σz, τzx)

Ps

τ

σ

(σx, τxz) (σx, τxz)Pole points for stresses

τxzσx

σz

τzx


Mohr’s stress circle: Pole points for planes- ProcedureStep : Project the line from the point (σz, τzx) OR (σx, τxz) in the direction of plane on

which these stresses are acting till it intersect circumference of the circle.

The intersection point gives the POLE point Pp for planes.

(σz, τzx)

Ps

τ

σ

(σz, τzx)

Ps

τ

σ

(σx, τxz) (σx, τxz)Pole point for stresses

τxzσx

σz

τzx

Pp Pp

Pole point for planes


Mohr’s stress circle: Pole points for planes

The pole of a Mohr’s circle is defined thus:If a line is drawn from the pole to a point on the circlewhere the stresses are τi, σi then, in the (x, z) plane theline is parallel to the plane on which τI and σi act.


Mohr’s stress circle: Use of pole point

Pp E(σz, τzx)

(σx, τxz)

τxzσx

σzτzx

σc

τca

σc

τca

σc

τcaσa

τac

σa

τac

Objective: To find out stresses σc, σa, τac, and τca on theplane which is inclined at θ to the plane onwhich σz acts.

Step 1: Locate the pole point for planesextending point E horizontally. i.e. inthe direction of plane on which σz acts.

σ

τ


Mohr’s stress circle: Use of pole point

Pp E(σz, τzx)

(σx, τxz)

τxzσx

σzτzx

σc

τca

σc

τca

σc

τcaσa

τac

σa

τac

Step 2: Draw a line parallel to the plane onwhich σc acts.

Step 3: Extend a line from point D throughcentre of the circle till it intersects thecircle.

Parallel planes

D(σc, τca)

Stresses on σc plane

(σa, τac)

Stresses on σa plane

τ

σ


Use of pole point P to locate stresses (σc , τca) at angle θ to the reference stress direction

The stress point on theMohr circle is found bysimply projecting a linefrom Pp parallel to theplane on which (σc, τca)acts until it intersectsthe circle at point D.


Principal shearing stresses

Fundamental relationships by Mohr’s stress circleThe maximum shearing stress, often called the principalshearing stress, has a magnitude of (σ1-σ3)/2, whichequals the radius of the Mohr circle.

The principal shearingstress occurs onplanes inclined at 45°.


Conjugate shearing stresses

Fundamental relationships by Mohr’s stress circleShearing stress on planes at right angles to each otherare numerically equal but are of opposite sign. Thesestresses are called conjugate shearing stresses.


Obliquity and resultant stress

Fundamental relationships by Mohr’s stress circle

The resultant stress on any plane has a magnitudeexpressed by and has an obliquity which isequal to


Maximum Obliquity

Fundamental relationships by Mohr’s stress circleThe maximum of all the possible obliquity angles on the variousplanes is called the maximum angle of obliquity αm.The coordinates of thepoint of tangency are thestresses on the plane ofmaximum obliquity andis less than the plane ofprincipal shear (i.e.maximum shear stress).

Since a limiting obliquityis the criterion of slip andwhere as on the plane ofprincipal shear failure isliable to happen.


Mohr’s stress circle: Pole point- example

Draw the Mohr stress circle at failure on a cylindrical specimen ofstiff clay with a shear strength of 100 kPa, if the radial stress ismaintained constant at 80 kPa. By using pole point method find theinclination θ to the radial direction of the planes on which the shearstress is one-half the maximum shear stress, and determine thenormal stresses acting on these planes.


Mohr’s stress circle: Pole point- Solution

σ3=80

τmax =100

100

-100

τmax = radius of circleσ1 = σ3+2τmax = 280

Step 1: Plot Mohr’s circle based on above information, i.e. radius and two points oncircle.

σ1=280

(All units are in kPa)



80 280

τmax =100

100

-100

50

Step 2: Draw a line at τ = 50 kPa which is half to τmax




80 280

τmax =100

100

-100

50

Step 3: As the principal stresses are acting on edges the σ3 point will act as a pole.

Pp




80 280

τmax =100

100

-100

50

Step 3: Draw line from pole to the intersection of 50 kPa line and Mohr’s circle.

Pp75°

15°

Inclination of the plane on which τ = τmax/2 is, θ = 15° or 75°

σn15° = 267 kPa,σn75° = 93 kPa



In the figure below, the normal loads applied to the faces of asoil cube are F1 = 0.45 kN and F2 = 0.30 kN and the shearloads are F3 = F4 = 0.1 kN. The sides of the soil cube are each40 mm. Construct the Mohr’s circle of total stress and find themagnitudes of the principal total stresses and the direction ofthe principal planes in the soil .

Example problem: Mohr’s circle of total stress

F4

F1

F1F4

F2F2

F3

F3

xz


Solution: Mohr’s circle of total stress

σz = 0.45/(40x40x10-6) = 281.25 kN/m2

σx = 0.30/(40x40x10-6) = 187.5 kN/m2

τxz = τzx = 0.10/(40x40x10-6) = 62.5 kN/m2

(Minor differencein magnitudescan be ignored)


Example problem

Pole P


Solution:1)

Establish pole point P or origin of planes. A line throughthe pole inclined at an angle α = 35° from the horizontalplane would be parallel to the plane on the element.

2)

3) The intersection is at point C and we find thatσα = 39 kPa and τα = 18.6 kPa.


Example problem

Note: Resultingσα = 39 kPa and τα = 18.6 kPa are same becausenothing has changed except orientation in space of the element.


Example problem

The stress shown on the element in the figure below:

Required:

a) Evaluate σα , ταwhen α = 30°

b) Evaluate σ1 and σ3when α = 30°

a


Solution

1) Plot the state of stress on the horizontal plane (6, 2) atpoint a. Note that the shear stress makes a clockwisemoment about a and therefore is positive.

2) Plot point B (-4,-2). The shear stress on the vertical plane isnegative since it makes a counterclockwise moment.

3) Points A and B are two points on a circle; Construct theMohr Circle with center at (1,0).

4) Find the pole, by drawing horizontal line through A orvertical line through B.

5) Find the state of stress on the plane inclined at angle α =30°, draw the line PC. C (1.8, 5.3 ) MPa.

6) Lines drawn from P to σ1 and σ3 establish the orientation ofmajor and minor principal planes.


Solution


Mohr circles of total and effective stresses

σ1σ3σ′1σ′3

u

τ

σ

Total stress circleEffective stress circle

σ′1 = σ1 - uσ′3 = σ3 - u

σ′θ , τ′θ σθ , τθ

θ θ


Mohr circles of total and effective stresses The effective stress circle has the same diameter as

the total stress circle and is separated from it by thepore water pressure.

The stresses τ′θ and σ′θ are the effective stressesacting on plane inclined at an angle θ

By examining the circles we note that τ′θ = τθ

σ′θ = σθ - u

Thus, for a given state of total stress, changes in porepressure have no effect on the effective shear stresses,they alter only the effective normal stresses.


Mohr circles of total and effective stresses Use the pole construction on the effective stress

Mohr’s circle to calculate the effective stresses onany plane is exactly same way as we used the poleconstruction to calculate total stresses.

The position of the pole in the Mohr’s circle ofeffective stress is the same as in the Mohr’s circle oftotal stress and the locations of the principal planesof total and effective stresses in the soil areidentical.

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51 - NPTEL · 2017. 8. 4. · A Pole is a unique point located on the circumference of Mohr’s circle. The point of intersection of Mohr’sstress circle and line drawn through the