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 stressshielding), 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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 Spring '10
 WEINER
 Tensile strength, 25em, POISSON’S RATIO, 7530Em

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