Solving for Hall Petch Unknowns

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The discussion centers on applying the Hall-Petch equation to determine the grain diameter at which the lower yield point of iron is 205 MPa. Given the yield points at two different grain diameters, participants explore how to solve for the unknowns in the equations. The equations provided are in MPa and mm, leading to a discussion about unit consistency. A suggested method for solving the equations involves substitution to isolate one variable, allowing for the calculation of the constants involved. Ultimately, the goal is to find the specific grain diameter corresponding to the desired yield point.
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The lower yield point for an iron that has an average grain diameter of 5x10^-2mm is 135 MPa. At a grain diameter of 8x10^-3, the yield point increases to 260 MPa. At what grain diameter will the lower yield point be 205 MPa?

Homework Equations


Hall-Petch Equation:
σy = σo + ky(d^-(1/2))

The Attempt at a Solution



135 MPa = σo + ky(22.36)
260 MPa = σo + ky(89.44)

I know this is going to sound silly, but I don't know to solve for two unknowns. If someone could just show me how to solve for one of them, I could solve for the other and figure the rest out.
 
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cperez said:
The lower yield point for an iron that has an average grain diameter of 5x10^-2mm is 135 MPa. At a grain diameter of 8x10^-3, the yield point increases to 260 MPa. At what grain diameter will the lower yield point be 205 MPa?



Homework Equations


Hall-Petch Equation:
σy = σo + ky(d^-(1/2))

The Attempt at a Solution



135 MPa = σo + ky(22.36)
260 MPa = σo + ky(89.44)

I know this is going to sound silly, but I don't know to solve for two unknowns. If someone could just show me how to solve for one of them, I could solve for the other and figure the rest out.

I'm not familiar with the physics of the problem, but I should be able to help you solve equations. What units are your last two equations in?
 
It's an engineering material science class problem.
The two equations are in MPa and mm.. so I think σy is in MPa/mm
 
cperez said:
It's an engineering material science class problem.
The two equations are in MPa and mm.. so I think σy is in MPa/mm

I believe those are mixed units. Don't you usually work in the mksA unit system?
 
I don't believe so.
 
135 MPa = σo + ky(22.36)
260 MPa = σo + ky(89.44)

I know this is going to sound silly, but I don't know to solve for two unknowns. If someone could just show me how to solve for one of them, I could solve for the other and figure the rest out.

One way you can do this is substitution. Isolate one of the variables and plug it into the other equation. In this instance σo seems like a good choice.

Using equation one:
σo = 135MPa - ky(22.36)

Plugging into equation two:
260Mpa = 135MPa - ky(22.36) + ky(89.44)

Solve for ky, and use that to get the other constant.
 

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