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


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2 Module 4: Lecture 1 on Stressstrain relationship and Shear strength of soils
3 Contents Stress state, Mohr s circle analysis and Pole, Principal stressspace, Stress pathsin pq space; MohrCoulomb failure criteria and its limitations, correlation with pq space; Stressstrain 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 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
5 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
6 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
7 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
8 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.
9 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 onehalf θ max.
10 Stress state: Simple axial stress: Solution a) Area = A = π r = m F cos θ 10cos 30 nθ = = = kPa 3 A F 10 θ = sin 2θ = sin 60 = kPa 3 2A b) F θ max = = kPa 2A 1 θ max c) = sin 2θ, sin 2θ =, θ max 2 2 θ =15 or75
11 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 crosssectional area (both graintograin 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
12 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)
13 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
14 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)
15 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: ( x s) + y = r r = radius of the circle (x, y) = coordinates of a point on the circle S = Horizontal distance of the center of the circle from origin
16 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 ) n ( z + y ) = ( z y )cos2θ + zy sin 2θ...(1) cos θ = (1 + cos2θ ) 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) [ ( )] [ ( )] n z y z y θ + + θ = + zy 2 2 θ [ n s] + = r θ θ s r 2 Which represents an equation of a circle
17 Mohr s stress circle: Graphical representation [ s] + = r n θ θ + yz 2 s z y S =( y + z )/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 xaxis. 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
18 Mohr circles: Simple twodimensional stress systems Biaxial Compression Biaxial Compression/tension Biaxial Pure shear 1 1 zy 2 2 yz θ + + zy = yz
19 Mohr circles: Simple twodimensional stress systems
20 Mohr circles: Simple twodimensional stress systems
21 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 kn/m 2 A 25 kn/m 2 z
22 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 coordinate system. Q(100, 25) P(50, 25)
23 Step 2: Join point P and Q with a straight line Q(100, 25) P(50, 25)
24 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)
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 = kpa Minor Principal stress = kpa
26 2 P(50, 25) o 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 to the z plane z plane is the plane on which z acts
27 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)
28 Stresses on the plane inclined at 35 to z plane Q(100, 25) nθ θ 2 70 o 1 nθ = kpa θ = kpa P(50, 25)
29 Threedimensional stresses on a cubical element
30 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 = If the reference axis are in the direction of 1, 2, and 3. Which is nothing but the direction of principal stresses
31 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.
32 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.
33 Mohr s stress circle: Three dimensional stress Case (a) Case (b)
34 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.
35 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?
36 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 x103 /0.01 = 2 MPa, 2. intermediate principal stress = 2 = 10 x103 /0.01 = 1 MPa, 3. Minor principal stress = 3 = 5 x103 /0.01 = 0.5 MPa. Plotting a Mohr s circle, 1 (MPa) max (+) max () (MPa) Maximum shear stress is radius of largest Mohr s circle, max = ( 13 )/2 = 0.75 MPa 1
Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay
51 Module 4: Lecture 2 on Stressstrain relationship and Shear strength of soils Contents Stress state, Mohr s circle analysis and Pole, Principal stressspace, Stress pathsin pq space; Mohrcoulomb failure
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