Suppose that you have an Al_75 Mg_25 (i.e., at.%) alloy at 300 degree C. What is

Question

Suppose that you have an Al_75 Mg_25 (i.e., at.%) alloy at 300 degree C. What is the weight fraction of each phase? A certain "rule of mixtures" in materials science states that, for some mixture of two phases alpha and beta, the overall alloy modulus E is given by E = E_alpha f)alpha + E_beta f_beta, where E_alpha and E_beta are the moduli of the individual phases and f_alpha and f_beta are the volume fractions of each phase. What is the modulus E of the alloy in part (a)? Data: E_AI = 70 GPa, E_AI3Mg2 = 140 GPa, rho_AI = 2.7 g/cm^3, and rho_AI3Mg2 = 4.0 g/cm^3. (As a simplification, assume that the density and modulus of the a phase is that same as that of pure Al.)

Explanation / Answer

a.

mass fraction of Al = weight of aluminium / total weight of the compound = 27*75/(27*75+24.305*25) = 0.7691

mass fraction of Mg =weight of magnesium / total weight of the compound = 24.305*25/(27*75+24.305*25) = 0.2308

b.

vf1=1/(1+(rho1/rho2)*(1/wf1-1)) = 1/(1+0.675*0.3) = 0.8316
vf2=1/(1+(rho2/rho1)*(1/wf2-1))=1/(1+1.481*3.332) = 0.1684

where, vf1=phase 1 volume fraction
vf2=phase 2 volume fraction
wf1=phase 1 mass fraction
wf2=phase 2 mass fraction
rho1=phase 1 density
rho2=phase 2 density

E = 70*0.8316 + 140*0.1684 = 81.788 GPa

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