Question

The yield strength for an alloy that has an average grain diameter of 5.7 × 10^-2 mm is 140 MPa. At a grain diameter of 7.3 × 10^-3 mm, the yield strength increases to 277 MPa. At what grain diameter, in mm, will the yield strength be 228 MPa?

d =

mm

Answer 1

Answer:

The Relation is,

$\sigma _y=\sigma _0+K_y\sqrt{\frac{1}{D}}$

For 1st yield strength

1 140 = 00 + hy 5.7 * 10-2

-------(1)

140 = 0o+K,(4.18)

$277=\sigma _0+K_y\sqrt{\frac{1}{7.3*10^-^3}}$

$277=\sigma _0+K_y(11.7)$----(2)

By solving both above equations

140-K_y(4.18)=277-K_y(11.7)

K_y(11.7-4.18)=277-140

K_y=137/7.52=18.21

By the substituting K_y value in equation 1

$140=\sigma _0+18.21(4.18)$

$\sigma _0=140-76.1178=63.88$

$228=63.88+18.21\sqrt{\frac{1}{D}}$

$\sqrt{\frac{1}{D}}=(228-63.88)/18.21=9.01$

Apply square on both sides

$1/D=9.01^2=81.18$

$\therefore$D=1/81.18=12.3*10^-3mm.