GG303 Lecture 18 9/4/01 1 TENSOR TRANSFORMATION OF STRESSES Transformation of stresses between planes of arbitrar orientation In the 2-D eample of lecture 16, the normal and shear stresses (tractions) were found on one arbitraril oriented plane in the n,s reference frame. Here we consider two perpendicular (but otherwise arbitraril oriented) planes, one perpendicular to the -ais, and the other perpendicular to the -ais and seek to find the normal and shear stress acting on them. In lecture 16 we considered the case where no shear stresses acted on the planes perpendicular to the - and -aes. We address here the more general case where shear stresses might eist, but still appl the strateg of lecture 16. The ke concept is that the total value of each stress component in one reference frame is the sum of the weighted contributions from all the components in another frame (see Figures 18.1 18.4). We start with a two-dimensional eample: = w () 1 w () 2 w () 3 w ( 4) All four stress components are needed to completel describe stresses at a point in the, reference must be known. The w terms are the weighting factors. In order for the equation to be dimensionall consistent, all the weighting factors must be dimensionless. Figures 18.1 18.4 show that the weighting factors describe as dimensionless ratios how a force in one direction projects into another direction (here the direction), and how the ratio of an area with a normal in one direction projects to an area with a normal in another direction (here the direction). Algebraicall, we can write this as F A F F A F F A F F A F F = A A F A A F A A F A A F A Each term in parentheses weights the following, stress component to give a weighted contribution to the stress term in the, reference frame. Each ratio in parentheses equals the direction cosine between the reference frame aes of the numerator and denominator, so: = a a a a a a a a Stephen Martel 18-1 Universit of Hawaii
GG303 Lecture 18 9/4/01 2 Similarl, each of the other three 2-D stress components in the, reference frame also involve all four stress components in the, reference frame: A F A F A F A F F = A A F A A F A = a a a a a a a a F A F F A F F A F F A F F = A = a a a a a a a a F A F F A F F A F F A F F = A = a a a a a a a a The shorthand tensor notation for the four equations above is a ij = a ik jl kl ; i,j,k,l = (,) or (1,2) The eact same notation applies for 3-D! a ij = a ik jl kl ; i,j,k,l = (,,z) or (1,2,3) The same epressions result from multipling the following matrices. In 2-D 11 12 11 12 11 12 11 21 = a a a a 21 22 a a a a 21 22 21 22 12 22 In 3-D 11 12 13 11 12 1 3 11 12 13 = 11 21 31 21 22 23 21 22 23 21 22 23 12 22 32 31 32 3 3 31 32 33 31 32 33 13 23 33 In general a ij a T = [ ] [ ] [ ] Works for either 2-D or 3-D! Stephen Martel 18-2 Universit of Hawaii
GG303 Lecture 18 9/4/01 3 Advantages of Mohr circle approach over tensor/matri approach 1 Gives a geometric meaning to stress relationships. 2 Can do stress rotation problems in our head. Advantages of tensor or matri approach over Mohr circle approach 1 The phsical underpinning behind how stresses transform is eplicit; it is not obvious with a Mohr circle construction. First, the notion that all members of a stress tensor are involved in the transformation is more straightforward than with a Mohr circle. Second, the two rotation terms ai k and a j l reflect (a) the rotation of the area that the stress components act on, and (b) the rotation of the direction that the components act in, so the tendenc to incorrectl consider stress as a force is reduced; the tensor qualit of stresses is more apparent. 2 The double angle epressions in Mohr circle, which can be difficult to remember and work with, are not present here. 3 Easier to address with a computer or a calculator. 4 Can appl just as eas to 3-D as 2-D; far more useful for 3-D problems than Mohr diagrams. 5 Powerful methods of linear algebra eist for finding the magnitudes of the principal stresses ("eigenvalues") and the direction of the principal stresses ("eigenvectors), so the underling nature of the stress field is easier to identif; this is especiall important in 3- D. 6 The shear stress convention with tensors is logical. Stephen Martel 18-3 Universit of Hawaii
GG303 Lecture 18 9/4/01 4 Fig. 18.1 Contribution of to θ θ A A θ θ cos θ = a cos θ = a What does on face A of area A contribute to on face A of area A? Start with the definition of stress: (1) = (1) / A. The unknown quantities (1) and A must be found from the known quantities and θ. To do this we first find the force (1) associated with : (1) = A The component of (1) that acts along the -direction is (1) cos θ. (1) = (1) cos θ As can be seen from the diagram atop the page A = A cos θ, so A = A /cos θ (1) θ (1) So the contribution of to is: (1) = (1) / A = (1) cos θ / (A /cos θ ) = ( (1) / A ) cos θ cos θ (1) = a a (1) = A F (1) Contribution of to A A F A Stephen Martel 18-4 Universit of Hawaii
GG303 Lecture 18 9/4/01 5 Fig. 18.2 Contribution of to A θ A θ θ θ cos θ = a cos θ = a What does on face A of area A contribute to on face A of area A? Start with the definition of stress: (2) = F (2) / A. The unknown quantities F (1) and A must be found from the known quantities and θ. To do this we first find the force F (2) associated with : (2) = A (2) The component of (2) that acts along the -direction is (2) cos θ. (2) = (2) cos θ As can be seen from the diagram atop the page A = A cos θ, so A = A /cos θ θ (2) So the contribution of to is: (2) = (2) / A = cos θ / (A /cos θ ) = ( / A ) cos θ cos θ (2) = a a (2) = A F (2) Contribution of to A A F A Stephen Martel 18-5 Universit of Hawaii
GG303 Lecture 18 9/4/01 6 Fig. 18.3 Contribution of to θ θ A A θ θ cos θ = a cos θ = a What does on face A of area A contribute to on face A of area A? Start with the definition of stress: (3) = (3) / A. The unknown quantities (3) and A must be found from the known quantities and θ. To do this we first find the force (3) associated with : (3) = A The component of (3) that acts along the -direction is (3) cos θ. (3) = (3) cos θ As can be seen from the diagram atop the page A = A cos θ, so A = A /cos θ (3) θ (3) So the contribution of to is: (3) = (3) / A = (3) cos θ / (A /cos θ ) = ( (3) / A ) cos θ cos θ (3) = a a (3) = A F (3) Contribution of to A A F A Stephen Martel 18-6 Universit of Hawaii
GG303 Lecture 18 9/4/01 7 Fig. 18.4 Contribution of to A A θ θ θ θ cos θ = a cos θ = a What does on face A of area A contribute to on face A of area A? Start with the definition of stress: = F / A. The unknown quantities F and A must be found from the known quantities YY and θ. To do this we first find the force F associated with : = A The component of that acts along the -direction is cos θ. = cos θ As can be seen from the diagram atop the page A = A cos θ, so A = A /cos θ θ So the contribution of to is: = / A = cos θ / (A /cos θ ) = ( / A ) cos θ cos θ = a a = A F Contribution of to A A F A Stephen Martel 18-7 Universit of Hawaii