MSQ2 - 2 Working Off Axis! A] Finding Tangent Moduli In...

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2 Working Off Axis! A] Finding Tangent Moduli In order to solve this problem, I need to use Hookean assumptions while treating the left sample as having constitutive relationships in parallel and the right one having them in series. Known , = ± Efiber parallel 330 59 MPa , = ± Efiber series 50 12 MPa =. νfiber 75 =. νmatrix 25 = Efiber 30Ematrix Left ACL: Parallel = = E σε σ Eε = → = = ε uniform ε εmatrix εfiber = = + σ additive σ σmatrix σfiber = + ± =. +. E Efiberνfiber Ematrixνmatrix 330 59MPa 7530Ematrix 25Ematrix , = . ←→ . Ematrix parallel Range of 12 033MPA 17 278MPa , = ←→ Efiber parallel Range of 361MPA 518MPa Right ACL: Series = = E σε ε σE = → = = σ uniform ε σmatrix σfiber = = + ε additive ε εmatrix εfiber =. +. = 1E 7530Em 25Em 150
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, = . ←→ . Ematrix series Range of 10 45MPA 17 05MPa , = . ←→ . Efiber series Range of 313 5MPA 511 5MPa B]Breaking Poisson’s Ratio Poisson’s ratio is the ratio of the contraction strain (squeezing/pulling sides, horizontal) to the axial strain (tension/compression of ends, vertical). In other words, it is the ratio of how much the specimen changes width (horizontal) over how much it changes length (vertical). In the parallel test, the Poisson’s ratio would primarily be controlled by the fibers. This is because they are taking most of the load (reminds me of stress-shielding), and any “necking” that would occur would be due to the fibers allowing it to do so. In the series case however, the Poisson’s ratio would primarily be controlled by the matrix.
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This note was uploaded on 03/25/2010 for the course ABBE BE430 taught by Professor Weiner during the Spring '10 term at Rose-Hulman.

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MSQ2 - 2 Working Off Axis! A] Finding Tangent Moduli In...

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