MSQ3 - 3 Layer Upon Layer Upon Layer A Determining Angle...

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3 Layer Upon Layer Upon Layer! A] Determining Angle α Assumptions Hookean ( = E σε ) Orthotropic (Makes matrix nicer, stress smearing, more on this in part B) Symmetry (looks like it is, and it has to be, more on this in part C) Solution Approach Values are given for principle direction variables, which I will use with the equations below to work towards and eventually setting equal to the testing moduli of the whole (all layers) annulus fibrosus. Following along my train of thought should be clear, and my motives behind what I calculate should be evident. Equations
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I must break this down into “Front Layer” and “2 nd Layer.” At the very end I will take into account that there are 5 “front layers” and 5 “2 nd layers,” but to get to that point I must start with single 2-D layers. Also, for sake of not having messy subscripts (you’ll see further on), instead of referring to the 2 nd layer as “2 nd layer”, from this point on I will refer to it as “back layer,” even though I know the true back layer is actually the same as the front layer.
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Front Layer Known = E1 120 MPa = E2 20 MPa =. ν12 4 = =. ν21 ν12E2E1 067 = . = . G12 E22 6 7 69 Solving for Q's = - = . Q11 E11 ν12ν21 123 31 = - = = . Q12 ν12E21 ν12ν21 Q21 8 22 = - = . Q22 E21 ν12ν21 20 55 = = . Q66 G12 7 69 = . ( ( )) + . + * . ( ( )) ( ( )) + . ( ( )) Q11 123 31 cos α 4 28 22 2 7 69 sin α 2 cos α 2 20 55 sin α 4 = . + . + * . ( ( )) ( ( )) + . ( +( ( )) ) Q12 123 31 20 55 4 7 69 sin α 2 cos α 2 8 22 sinα4 cos α 4 = . ( ( )) + . + * . ( ( )) ( ( )) + . ( ( )) Q22 123 31 sin α 4 28 22 2 7 69 sin α 2 cos α 2 20 55 cos α 4 = . - . - * . ( ( )) + . - . + * . ( ( )) ( ) Q16 123 31 8 22 2 7 69sinα cos α 3 8 22 20 55 2 7 69 sin α 3 cosα
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= . - . - * . ( ( )) +( . - . + * . ) ( ( )) Q26 123 31 8 22 2 7 69 sin α 3cosα 8 22 20 55 2 7 69 sinα cos α 3 =( . + . - * . - * . )( ( )) ( ( )) + . ( +( ( )) ) Q66 123 31 20 55 2 8 22 2 7 69 sin α 2 cos α 2 7 69 sinα4 cos α 4 Back Layer Known Take note that the axis is the same, however E1 now equals the front side’s E2 due to its inversion (back layer is just a horizontally-flipped version of the front layer).
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