Answer:
[tex]E = 12 * 10^{6} Psi[/tex]
Explanation:
Modulus of Elasticity perpendicular to the fibre,
[tex]\frac{1}{E} = \frac{B_{B} }{E_{B} } + \frac{B_{Al} }{E_{Al} }[/tex]
[tex]B_{B} = 20% = 0.2\\[/tex]
[tex]B_{Al} = 80% = 0.8[/tex]
[tex]E_{B} =55 * 10^{6} Psi[/tex]
[tex]E_{Al} =10 * 10^{6} Psi[/tex]
[tex]\frac{1}{E} = \frac{0.2 }{55 * 10^{6} } + \frac{0.8 }{10 *10^{6} }[/tex]
[tex]\frac{1}{E} = 8.36 * 10^{-8} \\E = \frac{1}{8.36 * 10^{-8} }\\E = 12 * 10^{6} Psi[/tex]
