Solving for Hall Petch Unknowns

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Discussion Overview

The discussion revolves around solving for unknown parameters in the Hall-Petch equation related to the yield point of iron at varying grain diameters. Participants are attempting to determine the grain diameter corresponding to a specific yield point while navigating the complexities of the equations involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the Hall-Petch equation and provides specific yield point values for different grain diameters, seeking assistance in solving for two unknowns.
  • Another participant questions the units used in the equations, suggesting that they may be mixed and inquiring about the standard unit system typically employed.
  • A third participant suggests a method of substitution to isolate one variable, proposing to solve for σo first and then use that to find ky.

Areas of Agreement / Disagreement

There is no consensus on the correct approach to solving the equations, as participants express differing views on the units and methods to isolate variables. The discussion remains unresolved regarding the best path forward.

Contextual Notes

Participants express uncertainty about the units of measurement and the implications of using mixed units in their calculations. The discussion includes unresolved mathematical steps and assumptions regarding the constants in the Hall-Petch equation.

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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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