Question

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 1

`d=4.32*10^(-3)mm^(-2)`

Step-by-step explanation:

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

That is,

`\sigma_y =\sigma_0+k_yd^(1/2)`

Where

`\sigma_y=`The yield strenght

`d=` Average grain diameter

`k_y =`constant for the particular material

Our values are,

`d=1*10^(-2)mm`

`\sigma_y=230Mpa`

Substituting for
`d=10^(-2)` mm and 230Mpa for
`\sigma_y`,

`230 = \sigma_0 +(1*10^(-2))^(-1/2)k_y`

`230 = \sigma_0+10k_y` (1)

Substituting for
`d=6*10^(-3)mm` and 275Mpa for
`\sigma_y`,

`275 = \sigma_0 +(6*10^(-23))^(-1/2)k_y`

`275 = \sigma_0+12.9k_y` (2)

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

`k_y=15.5Mpa (mm)^(1/2)`

`\sigma_0 = 75Mpa`

Substituting now for 310Mpa calculate 310Mpa

`\sigma_y =\sigma_0+k_yd^(1/2)`

`310 =\75+15.5d^(1/2)`

Solving for d,

`d=4.32*10^(-3)mm^(-2)`