Question

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elasticity of the metal and oxide are, respectively, 57 GPa and 390 GPa, what is the (a) upper-bound, and (b) lower-bound modulus of elasticity values (in GPa) for a composite that has a composition of 33 vol% of oxide particles.

Answer 1

a) the modulus of elasticity upper-bound is 166.89 GPa

b) the modulus of elasticity lower-bound is 79.36 Gpa

Step-by-step explanation:

For a two-phase composite, modulus of elasticity upper-bound expression is as follows;

Ec(U) = EmVm + EpVp

where Em is the modulus of elasticity of matrix, Ep is the modulus of elasticity of patriciate phase, Ec is the modulus of elasticity of composite, Vm is the volume fraction of matrix and Vp is the volume fraction of composite.

Also for a two-phase composite, modulus of elasticity lower-bound expression is as follows;

Ec(L) = (EmEp) / ( VmEp + VpEm)

a)

Now lets consider the expression of rule of mixtures for upper-bound and calculate the modulus of elasticity upper-bound.

Ec(U) = EmVm + EpVp --------- equ 1

Vm + Vp = 1

given that Vp = 33 vol% = 0.33, we substitute

Vm + 0.33 = 1

Vm = 0.67

so from our equation 1

we substitute our given data; Em = 57 Gpa, Ep = 390 Gpa, Vm = 0.67, Vp = 0.33

Ec(U) = EmVm + EpVp

Ec(U) = ( 57 × 0.67) + ( 390 × 0.33)

Ec(U) = 38.19 + 128.7

Ec(U) = 166.89 GPa

Therefore the modulus of elasticity upper-bound is 166.89 GPa

b)

Now lets consider the expression of rule of mixtures for lower-bound and calculate the modulus of elasticity upper-bound.

Ec(L) = (EmEp) / ( VmEp + VpEm)

we substitute our values

Ec(L) = (57 × 390) / ( (0.67 × 390) + (0.33 × 57)

Ec(L) = 22230 / ( 261.3 + 18.81)

Ec(L) = 22230 / 280.11

Ec(L) = 79.36 Gpa

Therefore the modulus of elasticity lower-bound is 79.36 Gpa