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The lower yield point for an iron that has an average grain diameter of 1 × 10–2 mm is 230 MPa. At a grain diameter of 6 × 10–3 mm, the yield point increases to 275 MPa. At what grain diameter will the lower yield point be 310 MPa? Express the answer in mm to three significant figures

Answer :

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Answer:

[tex]d=4.32*10^{-3}mm^{-2}[/tex]

Explanation:

We need to find the constant for the particular material given by the yield strenght equation,

That is,

[tex]\sigma_y =\sigma_0+k_yd^{1/2}[/tex]

Where

[tex]\sigma_y=[/tex]The yield strenght

[tex]d=[/tex] Average grain diameter

[tex]k_y =[/tex]constant for the particular material

Our values are,

[tex]d=1*10^{-2}mm[/tex]

[tex]\sigma_y=230Mpa[/tex]

Substituting for [tex]d=10^{-2}[/tex] mm and 230Mpa for [tex]\sigma_y[/tex],

[tex]230 = \sigma_0 +(1*10^{-2})^{-1/2}k_y[/tex]

[tex]230 = \sigma_0+10k_y[/tex] (1)

Substituting for [tex]d=6*10^{-3}mm[/tex] and 275Mpa for [tex]\sigma_y[/tex],

[tex]275 = \sigma_0 +(6*10^{-23})^{-1/2}k_y[/tex]

[tex]275 = \sigma_0+12.9k_y[/tex] (2)

Solving the two values (1) and (2) we have,

[tex]k_y=15.5Mpa (mm)^{1/2}[/tex]

[tex]\sigma_0 = 75Mpa[/tex]

Substituting now for 310Mpa calculate 310Mpa

[tex]\sigma_y =\sigma_0+k_yd^{1/2}[/tex]

[tex]310 =\75+15.5d^{1/2}[/tex]

Solving for d,

[tex]d=4.32*10^{-3}mm^{-2}[/tex]

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