Solving for Hall Petch Unknowns

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In summary, the yield point for an iron with an average grain diameter of 5x10^-2mm is 135 MPa. As the grain diameter decreases to 8x10^-3mm, the yield point increases to 260 MPa. To find the grain diameter at which the yield point is 205 MPa, the Hall-Petch equation can be used by isolating one of the variables and substituting it into the other equation.
  • #1
cperez
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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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  • #2
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?
 
  • #3
It's an engineering material science class problem.
The two equations are in MPa and mm.. so I think σy is in MPa/mm
 
  • #4
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?
 
  • #5
I don't believe so.
 
  • #6
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.
 

Related to Solving for Hall Petch Unknowns

1. What is Hall-Petch equation and how is it used in materials science?

The Hall-Petch equation is a mathematical model that describes the relationship between the grain size of a material and its strength. It is used in materials science to predict the mechanical properties of polycrystalline materials based on their microstructure.

2. What are the unknowns in the Hall-Petch equation?

The unknowns in the Hall-Petch equation are the grain size, the yield strength, and the Hall-Petch coefficient. These parameters can be experimentally determined or calculated using other equations.

3. How do you solve for unknowns in the Hall-Petch equation?

To solve for unknowns in the Hall-Petch equation, you will need to have at least two known values. These can be obtained through experimental data or through other equations. Once you have at least two known values, you can rearrange the equation to solve for the unknown parameter.

4. What are the limitations of the Hall-Petch equation?

The Hall-Petch equation is limited in its applicability to certain materials and conditions. It is most accurate for materials with small grains and at low temperatures. It also does not take into account other factors that can affect the strength of a material, such as impurities or defects.

5. How can the Hall-Petch equation be used in practical applications?

The Hall-Petch equation can be used in practical applications to predict the strength of materials based on their grain size. This can be useful in designing and selecting materials for specific applications, as well as in understanding the effects of processing and heat treatment on material properties.

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