12.005 Lecture Notes 5

12.005 Lecture Notes 5 Quantities in Different Coordinate Systems How to express quantities in different coordinate systems? x3' x3 Direction Cosines P Axis φ11 x1 x2' φ23 x2 φ12 x1' x1' x1 x2 x3 α11 α12 α13 x2' α21 α22 α23 x3' α31 α32 α33 Figure 5.1 Figure 5.1 Figure by MIT OCW. Direction cosine ij is cosine of angle φ ij between primed axis i and unprimed axis j. If xˆi ' and x̂ j represent unit vectors that are the axes of two coordinate systems with the same origin, they are related by the equation 3 3 j=1 j=1 xˆ i ' = ∑ α ij xˆ j , x i ' = ∑ α ij x j → α ij x j where αij is the cosine of the angle between the primed axis xˆi ' and the unprimed axis x̂ j . For example, α12 is the cosine of the angle between xˆ1 ' and x̂ 2 . αij represents a 9component matrix called the transformation matrix. Unlike the stress tensor, it is not symmetric (αij ≠ αji). ⎡α 11 α 12 α 13 ⎤ α ij = ⎢⎢α 21 α 22 α 23 ⎥⎥ ⎢⎣α 31 α 32 α 33 ⎥⎦ In matix equations, the transformation law is written ⎡ x1 ' ⎤ ⎡α 11 α 12 α 13 ⎤ ⎡ x1 ⎤ ⎥⎢ ⎥ ⎢ x '⎥ = ⎢α ⎢ 2 ⎥ ⎢ 21 α 22 α 23 ⎥ ⎢ x 2 ⎥ ⎢⎣ x3 '⎥⎦ ⎢⎣α 31 α 32 α 33 ⎥⎦ ⎢⎣ x3 ⎥⎦ The inverse transformation law is written xˆ i = α ji xˆ j ' Consider the following transformation of coordinates: x2 x1' x2' 30o x3, x3' Figure 5.1a Figure 5.1a Figure by MIT OCW. x1 The transformation matrix is ⎡ 30° 60° 0⎤ α ij = ⎢⎢120° 30° 0⎥⎥ ⎢⎣ 0 0 0⎥⎦ The explicit transformation equations are xˆ1 ' = cos 30° xˆ1 + cos 60° xˆ 2 xˆ 2 ' = cos120° xˆ1 + cos 30° xˆ 2 Since xˆi ' and x̂ j are both unit length, these equations are easy to verify from the picture. First-order tensors First-order tensors or vectors have two components in 2D coordinates and three components in 3D coordinates. They transform according to the same laws as coordinate axes because coordinate axes are themselves vectors. If uj is a vector in the x̂ j coordinate system and ui’ is a vector in the xˆi ' coordinate system, then the following equations describe their transformation: u i ' = α ij u j u i = α ji u j ' Note that αij is positive if the angle is measured counterclockwise from xˆi ' to x̂ j . It is negative if the angle is measured clockwise. Second-order tensors The transformation law for second-order tensors like stress and strain is more complicated than the transformation law for first-order tensors. It may be derived as follows: 1. Begin with the vector transformation of traction Ti to Tk ' : Tk ' = α ki Ti 2. Rewrite Tk ' and Ti using Cauchy’s formulas: Tk ' = σ ' kl nl ' and Ti = σ ij n j Substitute Cauchy’s formulas into the original transformation equation: σ ' kl nl ' = α kiσ ij n j 3. Transform the normal vector nj to nl’ and substitute into the previous equation: n j = α lj nl ' σ ' kl nl ' = α kiσ ij α lj nl ' 4. Cancel the nl’ term on each side and group the αs: σ ' kl = α kiα jl σ ij Note that changing the position of the last α term changes the order of its subscripts. In vector notation, the equation is σ ' = ασα T where the double under ~ denote second-rank tensors and the superscript T denotes the transpose of matrix α. Mohr’s Circle We explained before that an object resting on a slope will slide down when the shear traction on the slope is greater than or equal to the product of the normal traction and the coefficient of friction. τ = f sσ n On a shallow slope, σn is large and the object will not slide. On a steep slow, τ is large and the object will slide. For any plane with normal n̂ , we can calculate if the plane will fail if the stress tensor σij at the interface between the object and the slope is known. Calculate σn and , τ as follows: Vector and Tensor Notation T = σ nˆ Summation Notation Ti = σ ijn j σ n = T ⋅ nˆ σ n = Ti n i τ = T − σ n nˆ τ = Ti − Tk n k n i This method is straightforward but cumbersome. A different approach involves rotating the coordinate system such that x1’ is along n̂ . In this case σn and τ are much easier to derive: σ n = σ '11 τ = σ '12 Mohr’s circle may be derived in two or three dimensions. This lecture explains the derivation in two dimensions because it is more straightforward and the results are easier to graph and understand. The derivation assumes that x1, x2, and x3 are principle directions. Consider the following figure in which the xi coordinate system is rotated clockwise about the x3 axis to xi’: x1 θ x1' φ11 Figure 5.2 x2' x2 φ22 φ21 Figure 5.2 Figure by MIT OCW. The rotation matrix is: ⎡α 11 α 12 α 13 ⎤ ⎡cos θ α ij = ⎢⎢α 21 α 22 α 23 ⎥⎥ = ⎢⎢ sin θ ⎢⎣α 31 α 32 α 33 ⎥⎦ ⎢⎣ 0 − sin θ cos θ 0 0⎤ 0⎥⎥ 1⎥⎦ The stress tensor σ in the xi coordinate system is transformed to σ ' in the xi’ coordinate system by the following equation: σ ' = ασα T ⎡cosθ −sin θ ⎢ σ ' = ⎢ sin θ cosθ ⎢⎣ 0 0 0⎤⎡σ 11 0 0 ⎤⎡ cosθ ⎥⎢ ⎥⎢ 0⎥⎢ 0 σ 22 0 ⎥⎢−sin θ 1⎥⎦⎢⎣ 0 0 σ 33 ⎥⎦⎢⎣ 0 ⎡σ 11 cosθ −σ 22 sin θ 0 ⎤⎡ cosθ ⎢ ⎥⎢ 0 ⎥⎢−sin θ = ⎢ σ 11 sin θ σ 22 cosθ ⎢⎣ 0 0 σ 33 ⎥⎦⎢⎣ 0 sin θ cosθ 0 sin θ cosθ 0 0⎤ ⎥ 0⎥ 1⎥⎦ 0⎤ ⎥ 0⎥ 1⎥⎦ ⎡σ 11 cos2 θ + σ 22 sin 2 θ (σ 11 − σ 22 )sin θ cosθ 0 ⎤ ⎢ ⎥ 2 2 = ⎢ (σ 11 − σ 22 )sin θ cosθ σ 11 sin θ + σ 22 cos θ 0 ⎥ ⎢ 0 0 σ 33 ⎥⎦ ⎣ Use the double-angle identities for sine and cosine to simplify the expressions for the normal stress σ11’ and the shear stress σ12’ become in the new coordinate system: sin 2θ = 2 sin θ cos θ cos 2θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ σ 11 ' = σ n = σ 11 + σ 22 + σ 11 − σ 22 2 2 σ − σ 22 σ 12 ' = τ = 11 sin 2θ 2 cos 2θ The expression for the normal stress and shear stress can be shown graphically in shear space: τ Allowable tractions 2θ σ2 σ1 σn τ Figure 5.3 Figure by MIT OCW. This figure is called Mohr’s circle. Mohr’s circle plots in stress space. Admonton’s law may also be plotted in stress space as a line with slope fs. When this line and Mohr’s circle intersect, the criterion for failure across a plane is met. A note on signs. As derived, assumed σ 1 > σ 2 . Simplest case: consider θ = 45 o σ1 n σ2 x2' 45 o x1 θ = P v x1 x2 x2 Figure 5.4 Figure 5.4 Figure by MIT OCW. P x1' Consider several cases: τ 1 0 2 σ2 σ1 σn 4 3 Figure 5.5 Figure by MIT OCW. 0) 2θ = 60o 1) 2θ = 90o 2) 2θ = 180o 3) 2θ = 270o 4) 2θ = 300o Back to the landslide: Consider a common experiment in soil mechanics or rock mechanics in which scientists apply a uniaxial stress σ2 to a cylindrical sample confined by a uniform stress σ1. x2 x1 x3 σ2 Figure 5.6 σ1 σ1 Figure 5.6 Figure by MIT OCW. Continue increasing σ2 until failure. n̂ of failure plane typically at θ ; 30o from σ1. Seal Rubber jacket Specimen Figure 5.7 Confining pressure Figure 5.7. Triaxial test deformation apparatus Figure 5.7 Figure by MIT OCW. 55 σ3 = 5000 p.s.i. 50 35 30 25 25 Stress: k.p.s.i. 30 20 15 10 5 0 0 1 40 2 3 4 Millistrains 5 35 Stress: k.p.s.i. 35 Stress: k.p.s.i. 40 σ3 = 0 45 σ3 = 1000 p.s.i. 20 30 25 20 15 15 10 10 5 5 0 0 1 2 3 4 Millistrains 5 0 0 1 2 3 4 5 Millistrains 6 7 Figure 5.8. Stress-strain curves for Rand quartzite at various confining pressures. Figure 5.8 1) Shear fracture. 2) Extension fracture. 3) Multiple shear fractures. Figure 5.9 4) Extension fracture produced by line loads. 5) Longitudinal splitting in uniaxial compression. Figure 5.9 Figures by MIT OCW. 9 8 1650 7 Strain (%) 6 685 5 500 4 3 3260 845 235 2 Figure 5.10 1 0 2490 0 0 1000 2000 3000 4000 5000 σ1- σ3 (bars) Figure 5.10. Stress-strain curves for Carrara marble at various confining pressures. The numbers on the curves are confining pressures in bars. Figure 5.10 15 Strain (%) Figure 5.11 800 oC 10 500 oC 300 oC 5 25 oC 0 0 5 10 15 20 σ1- σ3 (k.bars) Figure 5.11. Stress-strain curves for granite at a confining pressure of 5 kilobars and various temperatures. Figure 5.11 Figures by MIT OCW. Analyze this using a Mohr circle diagram: τ | τ| = µ| σn| + | σ0| | σ0| σ2 σ2 σn σ1 Figure 5.12 Figure 5.12 Figure by MIT OCW. Mohr circle tangent to failure curve ⇒ sample breaks. τ τ φ = tan-1µ 2θ σ2 φ σn Figure 5.13 σ1 Compressive stress negative Compressive stress positive Figure 5.13 Figure by MIT OCW. σn µ is the coefficient of friction. φ = tan −1 µ ; 2θ = 90 o − φ σ0 Figure 5.14 2θ = 90o Figure 5.14 Figure by MIT OCW. µ 0.0 0.6 1.0 lim µ →∞ θ 45° 30° 23° 0° Since most rocks have a coefficient of friction of about 0.6, the normal vector to the failure plane is typically 30°from the direction of the least compressive stress. Another way of saying this is that the failure plane is 30°from the direction of the most compressive stress. Styles of Faulting Faults are large-scale failure planes. Since the normals to failure planes are in the plane containing the least compressive and the most compressive stresses and are typically at 30o from the direction of the least compressive stress, different styles of faulting can be used to infer the directions of σ1, σ2, and σ3. The following diagram shows the deviatoric stresses associated with thrust faults, normal faults, and strike-slip faults. Total Stress Figure 5.15 Deviatoric x β y β Or Deviatoric Total β Or Or "Left Lateral" Figure 5.15 Figure by MIT OCW. "Right Lateral"