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

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2 Module 4: Lecture 2 on Stress-strain relationship and Shear strength of soils

3 Contents Stress state, Mohr s circle analysis and Pole, Principal stressspace, Stress pathsin p-q space; Mohr-coulomb failure criteria and its limitations, correlation with p-q space; Stress-strain behavior; Isotropic compression and pressure dependency, confined compression, large stress compression, Definition of failure, Interlocking concept and its interpretations, Drainage conditions; Triaxial behaviour, stress state and analysis of UC, UU, CU, CD, and other special tests, Stress paths in triaxial and octahedralplane; Elastic modulus from triaxial tests.

4 Mohr s stress circle Mohr's circle is a geometric representation of the twodimensional stress state and is very useful to perform quick and efficient estimations. It is also popularly used in geotechnical fields such as soil strength, stress path, earth pressure and bearing capacity. It is often used to interpret the test data, to analyze complex geotechnical problems, and to predict soil behaviours. The pole point on Mohr's circle is a point so special that it can help to readily find stresses on any specified plane by using diagram instead of complicated computation.

5 Mohr s stress circle: Pole points A Pole is a unique point located on the circumference of Mohr s circle. The point of intersection of Mohr s stress circle and line drawn through the pole parallel to a given plane, gives the stresses on that plane. Two pole points can be established, a) Relating to the direction of action of stresses, and b) 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.

6 Mohr s stress circle: Pole points for stresses- Procedure Step : Project the line from the point (σ z, τ zx ) in the line of action of σ z (Vertical) OR Project the line from the point (σ x, τ xz ) in the line of action of σ x (Horizontal) till it intersects the circumference of the circle. The intersection point gives the POLE point P s for stresses. τ (σ z, τ zx ) τ xz σ x τ zx τ (σ z, τ zx ) σ z σ σ (σ x, τ xz ) P (σ s Pole points x, τ xz ) for stresses P s

7 Mohr s stress circle: Pole points for planes- Procedure Step : Project the line from the point (σ z, τ zx ) OR (σ x, τ xz ) 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 P p for planes. Pole point for planes τ P p (σ z, τ zx ) σ x τ xz τ P p (σ z, τ zx ) τ zx σ σ z σ (σ x, τ xz ) P s Pole point (σ x, τ xz ) for stresses P s

8 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 circle where the stresses are τ i, σ i then, in the (x, z) plane the line is parallel to the plane on which τ I and σ i act.

9 Mohr s stress circle: Use of pole point τ σ x τ xz σ z σ c τca τ zx P p E(σ z, τ zx ) Objective: To find out stresses σ c, σ a, τ ac, and τ ca on the plane which is inclined at θ to the plane on which σ z acts. Step 1: Locate the pole point for planes extending point E horizontally. i.e. in the direction of plane on which σ z acts. σ a σ c τca σ τ ac τ ac (σ x, τ xz ) σ c τ ca σ a

10 Mohr s stress circle: Use of pole point τ σ x τ xz σ z σ c τca τ zx Stresses on σ a plane P p E(σ z, τ zx ) Step 2: Draw a line parallel to the plane on which σ c acts. Step 3: Extend a line from point D through centre of the circle till it intersects the circle. Parallel planes (σ a, τ ac ) (σ x, τ xz ) D(σ c, τ ca ) σ Stresses on σ c plane σ c τca σ a τ ac τ ac σ c τ ca σ a

11 Use of pole point P to locate stresses (σc, τca) at angle θ to the reference stress direction The stress point on the Mohr circle is found by simply projecting a line from Pp parallel to the plane on which (σ c, τ ca ) acts until it intersects the circle at point D.

12 Fundamental relationships by Mohr s stress circle The maximum shearing stress, often called the principal shearing stress, has a magnitude of (σ 1 -σ 3 )/2, which equals the radius of the Mohr circle. The principal shearing stress occurs on planes inclined at 45. Principal shearing stresses

13 Fundamental relationships by Mohr s stress circle Shearing stress on planes at right angles to each other are numerically equal but are of opposite sign. These stresses are called conjugate shearing stresses. Conjugate shearing stresses

14 Fundamental relationships by Mohr s stress circle The resultant stress on any plane has a magnitude expressed by and has an obliquity which is equal to Obliquity and resultant stress

15 Fundamental relationships by Mohr s stress circle The maximum of all the possible obliquity angles on the various planes is called the maximum angle of obliquity α m. The coordinates of the point of tangency are the stresses on the plane of maximum obliquity and is less than the plane of principal shear (i.e. maximum shear stress). Since a limiting obliquity is the criterion of slip and where as on the plane of principal shear failure is liable to happen. Maximum Obliquity

16 Mohr s stress circle: Pole point- example Draw the Mohr stress circle at failure on a cylindrical specimen of stiff clay with a shear strength of 100 kpa, if the radial stress is maintained constant at 80 kpa. By using pole point method find the inclination θ to the radial direction of the planes on which the shear stress is one-half the maximum shear stress, and determine the normalstresses acting on these planes.

17 Mohr s stress circle: Pole point- Solution 100 (All units are in kpa) τ max = radius of circle σ 1 = σ 3 +2τ max = 280 τ max =100 σ 3 =80 σ 1 = Step 1: Plot Mohr s circle based on above information, i.e. radius and two points on circle.

18 Mohr s stress circle: Pole point- Solution 100 (All units are in kpa) 50 τ max = Step 2: Draw a line at τ = 50 kpa which is half to τ max

19 Mohr s stress circle: Pole point- Solution 100 (All units are in kpa) 50 τ max =100 P p Step 3: As the principal stresses are acting on edges the σ 3 point will act as a pole.

20 Mohr s stress circle: Pole point- Solution (All units are in kpa) 100 Inclination of the plane on which τ = τ max /2 is, θ = 15 or P p τ max =100 σ n15 = 267 kpa, σ n75 = 93 kpa -100 Step 3: Draw line from pole to the intersection of 50 kpa line and Mohr s circle.

21 Example problem: Mohr s circle of total stress In the figure below, the normal loads applied to the faces of a soil cube are F 1 = 0.45 kn and F 2 = 0.30 kn and the shear loads are F 3 = F 4 = 0.1 kn. The sides of the soil cube are each 40 mm. Construct the Mohr s circle of total stress and find the magnitudes of the principal total stresses and the direction of the principal planes in the soil. F 1 F 3 F 4 F 2 z x F 2 F 4 F 1 F 3

22 Solution: Mohr s circle of total stress σ z = 0.45/(40x40x10-6 ) = kn/m 2 σ x = 0.30/(40x40x10-6 ) = kn/m 2 τ xz = τ zx = 0.10/(40x40x10-6 ) = 62.5 kn/m 2 (Minor difference in magnitudes can be ignored)

23 Example problem Pole P

24 1) Solution: 2) Establish pole point P or origin of planes. A line through the pole inclined at an angle α = 35 from the horizontal plane would be parallel to the plane on the element. 3) The intersection is at point C and we find that σ α = 39 kpa and τ α = 18.6 kpa.

25 Example problem Note: Resulting σ α = 39 kpa and τ α = 18.6 kpa are same because nothing has changed except orientation in space of the element.

26 Example problem The stress shown on the element in the figure below: Required: a) Evaluate σ α, τ α when α = 30 b) Evaluate σ 1 and σ 3 when α = 30 a

27 Solution 1) Plot the state of stress on the horizontal plane (6, 2) at point a. Note that the shear stress makes a clockwise moment about a and therefore is positive. 2) Plot point B (-4,-2). The shear stress on the vertical plane is negative since it makes a counterclockwise moment. 3) Points A and B are two points on a circle; Construct the Mohr Circle with center at (1,0). 4) Find the pole, by drawing horizontal line through A or vertical 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 σ 1 and σ 3 establish the orientation of major and minor principal planes.

28 Solution

29 Mohr circles of total and effective stresses τ Effective stress circle Total stress circle σ θ, τ θ σ θ, τ θ θ θ σ σ 3 σ 1 σ 3 σ 1 σ 1 = σ 1 - u σ 3 = σ 3 - u u

30 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 the pore water pressure. The stresses τ θ and σ θ are the effective stresses acting on plane inclined at an angle θ By examining the circles we note that τ θ = τ θ σ θ = σ θ - u Thus, for a given state of total stress, changes in pore pressure have no effect on the effective shear stresses, they alter only the effective normal stresses.

31 Mohr circles of total and effective stresses Use the pole construction on the effective stress Mohr s circle to calculate the effective stresses on any plane is exactly same way as we used the pole construction to calculate total stresses. The position of the pole in the Mohr s circle of effective stress is the same as in the Mohr s circle of total stress and the locations of the principal planes of total and effective stresses in the soil are identical.

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

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