Answer: (a). Ec(μ) = 165.6 GPa
(b). Ec(∝) = 83.09 GPa
Explanation:
this is quite straightforward, so we will go step by step.
from the data we have that,
Moduli of elasticity of the metal -(Em) is 60 Gpa
Moduli of elasticity of oxide is (Ep) is 380 Gpa
volume Vp = 33% = 0.33
(a). To solve the upper bound-modulus of the elasticity is calculate thus;
Ec (μ) = EmVm + EpVp ----------------(1)
where E rep the modulus of elasticity
v rep the volume fraction
c rep the composite
Vm = 100% - Vp
Vm = 100% - 33% = 67%
Vm = 0.67
substituting the valus of Em, Vm, Ep, Vp from equation (1) we have;
Ec(μ) = (60×0.67) + (380×0.33)
Ec(μ) = 40.2 + 125.4 = 165.6 GPa
Ec(μ) = 165.6 GPa
(b). The lower bound modulus of elasticity can be calculated thus;
Ec(∝) = EmVp / EpVm + EmVp -------------- (2)
substituting values Em,Vm,Ep,Vp.
Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)
Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa
Ec(∝) = 83.09 GPa
cheers i hope this helps!!!!
