The yield strength for an alloy that has an average grain diameter of 4.5 x 10-2 mm is 109 MPa. At a grain diameter of 6.8 x 10-3 mm, the yield strength increases to 263 MPa. At what grain diameter, in mm, will the yield strength be 224 MPa? d = Enter your answer in accordance to the question statement mm

Respuesta :

First we start withe the Hall-Pethch relation,

So,

[tex]\sigma_{y1} = \sigma_{0}+\frac{k_y}{\sqrt{d_1}}[/tex]

Where,

[tex]\sigma_{y1} [/tex] = Initial yield Strenght

[tex]d_1 =[/tex] Initial grain size of the alloy

Substituting,

[tex]109MPa = \sigma_0 + \frac{k_y}{\sqrt{4.5*10^-2mm}}[/tex]

We make the same relationship but now for the yield strength of 224MPa, so,

[tex]263MPa = \sigma_0 + \frac{k_y}{\sqrt{6.8*10^-3mm}}[/tex]

Solving,

[tex]\sigma_0 = 11.0655 MPa[/tex]

[tex]K_y= 20.7751 MPa.mm^{1/2}[/tex]

Hall-Petch relation is for the constant

[tex]\sigma_y = 11.0655MPa +\frac{20.7751MPa.mm^{1/2}}{\sqrt{d}}[/tex]

At end we make the sustitution for 224Mpa, so

[tex]\sigma_y=224MPa[/tex]

[tex]224MPa = 11.0655MPa +\frac{20.7751MPa-mm^{1/2}}{\sqrt{d}}[/tex]

[tex]d=\sqrt{0.0975657mm}[/tex]

[tex]d= 0.9877mm[/tex]