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Module 4: Lecture 1 on Stress-strain relationship and Shear strength of soils
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.
Stress state The concept of stress: Defined as the force (internal resistance) per unit area Can not be measured directly Gives no indications of how forces are transmitted through stressed material The manner of transfer of forces in solid crystalline material is different from point to point contact transfer in materials like soil. F F Stress, = F/A
Stress state Simple axial stress: T = F sinθ P F F A Q P N F T θ F R Q N θ = n A/ cosθ θ F 2 = cos θ A T cosθ = A = F sin 2θ 2A
Stress state Simple axial stress: Derivation A A =, θ N = F cosθ, T = F sinθ, cosθ Therefore: 2 N F cos θ n θ = = A/ cosθ A And, θ = θ = T = A/ cosθ F sin 2θ 2A T A cosθ = F sinθ cosθ A P N F T θ F R Q
Stress state Simple axial stress: max plane d θ 0 For maximum value of θ dθ = d θ F = cos2θ dθ A Therefore: F cos 2θ = 0 A cos 2θ = 0 θ = 45 or(135 ) F and, = θ max 2A θ max occurs on a plane with 45 inclination with plane nθ max
Stress state Variation of normal stress nθ and shear stress θ with angle of plane θ in cylindrical test specimen θmax occurs on a plane with θ = 45 and nθmax on a plane with θ = 0.
Stress state Simple axial stress: Example A cylindrical specimen of rock, 75 mm in diameter and 150 mm height is subjected to an axial compressive force of 10 kn. Find: a) the normal stress nθ and shear stress θ on a plane inclined at 30 to the radial direction; b) the maximum value of shear stress; c) the inclination of planes on which the shear stress θ is equal to one-half θ max.
Stress state: Simple axial stress: Solution a) Area = A = π r = 4.42 10 m 2 3 2 2 2 F cos θ 10cos 30 nθ = = = 1696.83kPa 3 A 4.42 10 F 10 θ = sin 2θ = sin 60 = 979.67kPa 3 2A 2 4.42 10 b) F θ max = = 1131.22kPa 2A 1 θ max c) = sin 2θ, sin 2θ =, θ max 2 2 θ =15 or75
Stress at a point The point of application of a force within a soil mass could be on a particle or in a void. As void can not support any stress, stress is F/A, where A is gross cross-sectional area (both grain-tograin contacts as well as voids) F6 Consider F1, F2,--- F6 forces acting on a body in 2D plane F5 F1 O F4 F2 F3 α
Stress at a point The resolution of forces (F1, F2 F6) into normal and shear components acting on a plane passing through point O at an angle α Compressive stresses are +ve; Expanded view of a soil element at O +ve shear stress produce counterclockwise couple on element (i.e. clockwise moments about a point outside element)
Stress at a point At equilibrium, the sum of forces in any direction must be zero. F h = H - Tcosα - Nsinα = 0 F v = V - Tsinα - Ncosα = 0 Stresses on the α - plane are the normal stresses α and α x sinα - α cosα - α sinα = 0 (a) Solving (a) y cosα + α sinα - α cosα = 0
Stress at a point When this circle is plotted in - space, it is known as the Mohr circle of stress. o It represents the state of stress at a point at equilibrium o It applies to any material, not just soil (Note: Scales for and have to be same to obtain a circle)
Mohr s circle analysis and Pole Mohr s stress circle: Two dimensional Graphical representation of stress relationship Discovered by Culmann (1866) and developed in detail by Mohr (1882) Stresses are represented in the form of a circle + 2 s r P (x,y) 1 Considering any point P(x,y) on the circle, equation of the circle can be written as: 2 2 2 ( x s) + y = r r = radius of the circle (x, y) = co-ordinates of a point on the circle S = Horizontal distance of the center of the circle from origin
Mohr s stress circle: Graphical derivation Normal stress acting on any plane at an angle θ : 2 2 n θ = cos θ + sin 2θ + z zy y sin θ 2 sin θ = (1 cos2θ ) 2 1 1 2 n ( z + y ) = ( z y )cos2θ + zy sin 2θ...(1) cos θ = (1 + cos2θ ) 2 2 2 Shear stress acting on any plane at an angle θ : 1 θ = ( y z )sin 2θ + zy cos2θ...(2) 2 Squaring and adding equation (1) and (2) 1 2 2 1 2 2 [ ( )] [ ( )] n z y z y θ + + θ = + zy 2 2 θ 2 2 2 [ n s] + = r θ θ s r 2 Which represents an equation of a circle
Mohr s stress circle: Graphical representation 2 2 2 [ s] + = r n θ θ + yz 2 s z y S =( y + z )/2 =( 1 + 2 )/2 nθ 2θ 2α a circle with radius r and with a centre on a plot, at the point = s, = 0 θ 1 zy 1 and 2 are principal stresses as =0 on x-axis. z, y, zy, yz are boundary stresses which helps to plot the circle. nθ and θ are the normal and shear stresses on a plane at angle θ to the z plane. nθ and θ can be found on Mohr s circle by travelling clockwise around the circle from stress point ( z, zy ) to a distance 2θ at the centre of the circle 1 is at angle α to the plane of z
Mohr circles: Simple two-dimensional stress systems Biaxial Compression Biaxial Compression/tension Biaxial Pure shear 1 1 zy 2 2 yz + 2 + + 2θ + + zy + 1 2 1 1 = 2 2 1 yz
Mohr circles: Simple two-dimensional stress systems
Mohr circles: Simple two-dimensional stress systems
Mohr circles: Example The stress on a soil mass are shown in the following figure. Determine, a) the magnitude of principal stresses using Mohr s circle b) the magnitude of normal and shear stresses on plane AC shown in figure. B 100 kn/m 2 C y 35 50 kn/m 2 A 25 kn/m 2 z
Mohr circles: Solution Available information:- y = 50kN/m 2, z = 100 kn/m 2, yz = 25 kn/m 2 Step 1: Mark a point P( y, yz ) and point Q( z, zy ) on co-ordinate system. Q(100, 25) P(50, 25)
Step 2: Join point P and Q with a straight line Q(100, 25) P(50, 25)
Step 3: Draw a circle considering intersecting point of s axis and line PQ (Point O) as the centre and distance OP as the radius Mohr s stress circle Q(100, 25) o P(50, 25)
Step 3: Draw a circle considering intersecting point of s axis and line PQ (Point O) as the centre and distance OP as the radius 2 o Q(100, 25) 1 Principal stresses on the plane where shear stress is zero. i.e. where Mohr s circle cross axis P(50, -25) Major Principal stress = 110.36 kpa Minor Principal stress = 39.64 kpa
2 P(50, -25) o 225 45 Q(100, 25) 1 Angle 2α, i.e. two times the angle between z plane and major principal plane Major principal plane is inclined at 22.5 to the z plane. Minor principal plane is inclined at 112.5 to the z plane z plane is the plane on which z acts
Stresses on the plane inclined at 35 to z plane Q(100, 25) 2 70 o 1 Plane inclined at 35 to z plane P(50, -25)
Stresses on the plane inclined at 35 to z plane Q(100, 25) nθ θ 2 70 o 1 nθ = 42.96 kpa θ = 14.94 kpa P(50, -25)
Three-dimensional stresses on a cubical element
Mohr s stress circle: Three dimensional stress Three dimensional stress at a point can be represented as, T = z yz xz zy y xy zx yx x terms are the normal stresses and the terms are the shear stresses Total 6 terms are independent : x, y, z, xy, yz, zx. T = 1 0 0 0 0 2 0 0 3 If the reference axis are in the direction of 1, 2, and 3. Which is nothing but the direction of principal stresses
Mohr s stress circle: Three dimensional stress No simple method exists to draw a Mohr s circle to represent the general case, i.e. all normal and shear stresses acting on all the six faces of a cube. Two simple cases can be represented by using three Mohr s circles as below, a) A cubical element having only normal stresses on its faces. b) A cubical element which has only normal stresses acting on one pair of opposite faces and both normal and shear stresses on remaining two pair of faces.
Mohr s stress circle: Three dimensional stress It can be proved that stress conditions on any plane within the element must fall within the shaded area, but it is usually sufficient to be able to determine stresses on planes which are perpendicular to at least one opposite pair of element boundary faces. Stresses on these planes lie on the circles bounding the shaded areas.
Mohr s stress circle: Three dimensional stress Case (a) Case (b)
Mohr s stress circle: Three dimensional stress Case b, which depicts a cubical element with compressive normal stresses acting on all six faces and shear stresses on two pairs of opposite faces. Again, in this case, stresses on all planes within the element lie within the shaded area, with stresses on all planes which are perpendicular to at least one pair of element faces lying on one of the boundary circles. The sequence of drawing these circles consists firstly of locating stress points z, zy and y, yz, then drawing circle (i) through these with its centre on the = 0 axis. This locates the principal stresses 1 and 2. As the third principal stress is known, circles (ii) and (iii) can now be drawn. In the case shown 1 > 2 > 3.
Mohr s stress circle: Three dimensional stress - Example A piece of sandstone is cut into the shape of a cube with 100 mm long edges. Forces of 5 kn, 10 kn and 20 kn, respectively, act uniformly on, and normal to, the three pairs of faces of the cube. Evaluate the major, intermediate and minor principal stresses in the rock and draw the Mohr circles of stress. What is the maximum shear stress in the rock?
Mohr s stress circle: Three dimensional stress - Solution Area of each face, A = 0.01 m 2 So, three principal stresses are 1. Major principal stress = 1 = 20 x10-3 /0.01 = 2 MPa, 2. intermediate principal stress = 2 = 10 x10-3 /0.01 = 1 MPa, 3. Minor principal stress = 3 = 5 x10-3 /0.01 = 0.5 MPa. Plotting a Mohr s circle, 1 (MPa) 0 0.5 1 2 max (+) max (-) (MPa) Maximum shear stress is radius of largest Mohr s circle, max = ( 1-3 )/2 = 0.75 MPa -1