MIT - 16.20 Fall, 2002 Unit 7 Transformations and Other Coordinate Systems Readings : R 2-4, 2-5, 2-7, 2-9 BMP 5.6, 5.7, 5.14, 6.4, 6.8, 6.9, 6.11 T & G 13, Ch. 7 (74 - 83) On “other ” coordinate systems: T & G 27, 54, 55, 60, 61 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 As we ’ ve previously noted, we may often want to describe a structure in various axis systems. This involve s … Transformations (Axis, Deflection, Stress, Strain, Elasticity Tensors) e.g., loading axes <--> material principal axis Figure 7.1 Unidirectional Composite with Fibers at an Angle fibers Know stresses along loading axes, but want to know stresses (or whatever) in axis system referenced to the fiber. Paul A. Lagace ? 2001 Unit 7 - p . 2 MIT - 16.20 Fall, 2002 Problem : get expressions for (whatever) in one axis system in terms of (whatever) in another axis system (Review from Unified) Recall : nothing is inherently changing, we just describe a body from a different reference. Use ~ (tilde) to indicate transformed axis system. Figure 7.2 General rotation of 3-D rectangular axis system (still rectangular cartesian ) Paul A. Lagace ? 2001 Unit 7 - p . 3 MIT - 16.20 Fall, 2002 “Define ” this transformation via d irection cosines ~ ~ l mn = cosine of angle from y m to y n Notes : by convention, angle is measured positive counterclockwise (+ CCW) (not needed for cosine) ~ ~ since cos is an even function l mn = l nm cos ( θ ) = cos (- θ ) (reverse direction) ~ ~ But l mn ≠ l mn angle differs by 2 θ ! The order of a tensor governs the transformation needed. An n th order tensor requires an n th order transformation (can prove by showing link of order of tensor to axis system via governing equations). Paul A. Lagace ? 2001 Unit 7 - p . 4 MIT - 16.20 Thus: Quantity Transformation Equation ? Stress σ mn = l mp l nq σ pq ? ? Strain ε? = l ? l ? ε mn mp nq pq Axis x ? m = l mp x p ? Displacement u ? m = l mp u p ? Fall, 2002 Physical Basis equilibrium geometry geometry geometry Elasticity Tensor E ? mnpq = ll ns l pt l qu E rstu Hooke ’ s law ? mr ? ? ? In many cases, we deal with 2-D cases (replace the latin subscripts by greek subscripts) e.g., σ ? αβ = ll ? αθ βτ σ θτ ? (These are written out for 2-D in the handout). Paul A. Lagace ? 2001 Unit 7 - p . 5 MIT - 16.20 Fall, 2002 An important way to illustrate transformation of stress and strain in 2-D is via Moh r ’ s circle (recall from Unified). This was actually used B.C. (before calculators). It is a geometrical representation of the transformation. (See handout). (you will get to work with this in a problem set). Also recall … (Three) Important Aspects Associated with Stress/Strain Transformations 1. Principal Stresses / Strains (Axes) : there is a set of axes into which any state of stress / strain can be resolved such that there are no shear stresses / strains --> σ ij depend on applied loads --> ε ij depend on applied loads and material response Thus, note : For general materials … axes for principal strain ≠ axes for principal stress Generally: (have nothing to do with) material principal axes ≠ principal axes of stress / strain Paul A. Lagace ? 2001 Unit 7 - p . 6 MIT - 16.20 Fall, 2002 Find via roots of equation: σ 11 ? τ σ 12 σ 13 σ 12 σ 22 ? τ σ 23 = 0 σ 13 σ 23 σ 33 ? τ eigenvalues : σ I , σ II , σ III (same for strain) 2. Invariants : certain combinations of stresses / strains are invariant with respect to the axis system. Most important: Σ (extensional stresses / strains) = Invariant very useful in back-of-envelope / “quick check” calculations 3. Extreme shear stresses / strains: (in 3-D) there are three planes along which the shear stresses / strains are maximized. These values are often used in failure analysis (recall Tresca condition from Unified). These planes are oriented at 45 ° to the planes defined by the principal axes of stress / strain (use rotation to find these) Paul A. Lagace ? 2001 Unit 7 - p . 7 MIT - 16.20 Fall, 2002 Not only do we sometimes want to change the orientation of the axes we use to describe a body, but we find it more convenient to describe a body in a coordinate system other than rectangular cartesia n . Thus, conside r … Other Coordinate Systems The “ easiest ” case is Cylindrical (or Polar in 2-D) coordinates Figure 7.3 Loaded disk Figure 7.4 Stress around a hole Paul A. Lagace ? 2001 Unit 7 - p . 8 MIT - 16.20 Fall, 2002 Figure 7.5 Shaft Define the point p by a different set of coordinates other than y 1 , y 2 , y 3 Figure 7.6 Polar coordinate representation Volume = rd θ drdz Paul A. Lagace ? 2001 Unit 7 - p . 9 MIT - 16.20 Fall, 2002 Use θ , r, z where: y 1 = r cos θ y 2 = r sin θ y 3 = z are the “ mapping ” functions Now the way we describe stresses, etc. change … --> Differential element is now different Rectangular cartesian Figure 7.7 Differential element in rectangular cartesian system Volume = dy 1 dy 2 dy 3 Paul A. Lagace ? 2001 Unit 7 - p . 1 0 MIT - 16.20 Fall, 2002 Cylindrical Figure 7.8 Differential element in cylindrical system Volume = rd θ drdz Generally : dy 1 --> dr dy 2 --> rd θ dy 3 --> dz (get from “mapping ” functions) Paul A. Lagace ? 2001 Unit 7 - p . 1 1 MIT - 16.20 Fall, 2002 --> stresses follow same rules: σ mn direction face but now we deal with r, θ , z faces and directions Figure 7.9 Representation of stresses on rectangular cartesian and cylindrical differential elements equilibrium considerations (or your “mapping ” functions on equations in rectangular cartesian coordinates) yield: Paul A. Lagace ? 2001 Unit 7 - p . 1 2 MIT - 16.20 Fall, 2002 Equilibrium Equations (in cylindrical coordinates) r : ? σ rr + 1 ? σ θ r + ? σ zr + σ rr ? σ θ θ + f r = 0 ? r r ? θ ? z r ? σ r θ + 1 ? σ θ θ + ? σ z θ + 2 σ r θ + f θ = 0 θ : ? r r ? θ ? z r z : ? σ rz + 1 ? σ θ z + ? σ zz + σ rz + f z = 0 ? r r ? θ ? z r Body forces = f r , f θ , f z can do a similar manipulation for the Strain - Displacement Equations ? u ε = r rr ? r ε θθ = 1 ? u θ + u r r ? θ r ? u 3 ε = zz ? z Paul A. Lagace ? 2001 Unit 7 - p . 1 3 MIT - 16.20 Fall, 2002 (engineering shear strains) ε r θ = ? u θ + 1 ? u r ? u θ ? r r ? θ r ε θ z = 1 ? u 3 + ? u θ r ? θ ? z ε zr = ? u r + ? u 3 ? z ? r Stress - Strain Equations become, however, more complicated and not easy to “map ” into another coordinate systems. Why? Unless the material is isotropic, the properties change with direction (if the material principal axes are rectangular orthogonal). E ? mnpq = ll ? ? qu E rstu ? mr ns l pt l ? So, for cylindrical coordinates: E mnpq ( θ ) a function of θ Paul A. Lagace ? 2001 Unit 7 - p . 1 4 MIT - 16.20 Fall, 2002 Think of it this way: In cylindrical (polar 2-D) coordinates, we have rotated a “ local rectangular cartesian system ” . So we use the E ? mnpq transformation with the angle θ to find the elasticity tensor values and then the “local ” engineering constants. Recall / note : a material that is orthotropic may pick up additional coupling terms in this rotation and “ appear ” anisotropic in that local coordinate system. In the isotropic case, can write: 1 ε rr = E [ σ rr ? ν ( σ θθ + σ zz ) ] 1 ε θθ = E [ σ θθ ? ν ( σ rr + σ zz ) ] 1 ε zz = E [ σ zz ? ν ( σ rr + σ θθ ) ] Paul A. Lagace ? 2001 Unit 7 - p . 1 5 MIT - 16.20 Fall, 2002 21 + ν ) σ r θ ε r θ = ( E ( ε θ z = 21 + ν ) σ θ z E 21 + ν ) σ ε = ( zr zr E More generally, we can express transformation to any … General Curvilinear Coordinates (including locally non-rectangular Cartesian systems) functional forms: Fy 1 1 (, y 2 , y 3 ) = ξ 2 (, Fyy 2 , y 3 ) = η 1 3 (, Fyy 2 , y 3 ) = ζ 1 Use these to “transform” governing equations from basic rectangular case Paul A. Lagace ? 2001 Unit 7 - p . 1 6 MIT - 16.20 Fall, 2002 For cylindrical case: ξ = r η = θ ζ = z actual “mapping ” functions: Fy 1 1 (, y 2 , y 3 ) = y 2 2 y 1 2 + 2 (, Fyy 2 , y 3 ) = t an -1 ( y 2 / y 1 ) 1 Fyy 2 , y 3 ) = y 3 3 (, 1 Other cases Let ’s next consider some general solution approaches Paul A. Lagace ? 2001 Unit 7 - p . 1 7