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In a unidirectional continuous-fiber composite under tension PARALLEL to fibers, the longitudinal Young's modulus is given by the rule of mixtures: E_L = V_f·E_f + V_m·E_m. Why does this work for parallel loading but NOT for transverse loading?
A{'text': 'Transverse loading is always 0', 'label': 'C'}
B{'text': 'Under parallel loading both phases share the same STRAIN (iso-strain); under transverse loading they share the same STRESS (iso-stress) — yielding a different (Reuss-like) formula', 'label': 'B'}
C{'text': 'Both formulas are identical', 'label': 'D'}
D{'text': 'Strain in fiber ≠ strain in matrix only transversely', 'label': 'A'}
Answer & Solution
Correct answer: B. {'text': 'Under parallel loading both phases share the same STRAIN (iso-strain); under transverse loading they share the same STRESS (iso-stress) — yielding a different (Reuss-like) formula', 'label': 'B'}
Iso-strain (Voigt) parallel: E_L = V_f E_f + V_m E_m. Iso-stress (Reuss) transverse: 1/E_T = V_f/E_f + V_m/E_m. The transverse modulus is much lower because the soft matrix dominates the series.
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