Fracture Toughness - Help with Units

[vtmike]

Hi,

Has anyone worked with or seen the units kg/mm^1.5 for fracture toughness before? I am trying to convert these to Mpa.sqrt(m) which is typically what we use for all our calculations.

Is anyone familiar with a conversion from kg/mm^1.5 to Mpa.sqrt(m)?



Any help would be appreciated!
Thanks,
Mike

 

[micalbrch]

Do you perhaps mean kg/mm^0.5 (root)?

 

[cloa]

Kic=MPA/sqrt(m)=MN/m2/sqrt(m)=MN/m^1.5=10^6*kg*(acc)/m^1.5
acc in m^2/s^2.

The acceleration division is probably just by gravity. Its just application of a fixed weight to open the crack which is actually silly as there is always stress concentration (simple geometry before considering crack effects) so the applied load in the testing method is not same as that experienced at the crack tip.

 

[metengr]

Please look here for information

http://books.google.com/books?id=LC...e&q=fracture toughness units of kg/mm&f=false

 

[cloa]

I can see that the units are kg/mm^3/2 and conversion ratios but they don't cover anything about how they get the conversions is pretty poor.

 

[Maui]

vtmike, the units you provided of kg/mm^1.5 are not units of fracture toughness. They were quoted incorrectly.

Maui

http://www.EngineeringMetallurgy.com

 

[mrfailure]

Actually, these units make sense to me although they are not common. Another way to express the units vtmike presented is: (kg/mm²)*mm^1/2. These are indeed valid units for fracture toughness.

To convert, I get x*9.807 MPa/(kg/mm²)*(1m/1000mm)^1/2, or

MPa-m^1/2 = 0.3101 kg/mm^1.5

Aaron Tanzer

http://www.lehightesting.com

 

[Maui]

The common units of fracture toughness involve (stress)*[(length)^1/2]. Stress, when written out as a force per unit area, contains units of seconds squared in the denomintor. What vtmike provided does not contain units of seconds squared in the denominmator, so they can't be valid units for fracture toughness.

Maui

http://www.EngineeringMetallurgy.com

 

[Maui]

To clarify my previous post for those of you who may be unclear on this, using units of Pascals for stress,

Pa = N/(m^2) = [kg*m/(s^2)]/(m^2) = kg/[m*(s^2)]

So for the units of fracture toughness we have,

Pa*(m^1/2) = kg/[(m^1/2)*(s^2)]

This contains units of seconds squared in the denominator.

Maui

http://www.EngineeringMetallurgy.com

 

[mrfailure]

Maui,

I'm confused here. Stress units such as pascals/megapascals, or psi/ksi do not contain a time component (seconds) in their denominators. Where does this come from?

Aaron Tanzer

http://www.lehightesting.com

 

[Maui]

Mrfailure, stress is force per unit area. Force can be calculated from Newton's law as

F = ma

where m is mass and a is acceleration. So the units of force are kg*[m/(s^2)]. This is where the time component originates from.

Maui

http://www.EngineeringMetallurgy.com

 

[mrfailure]

Thanks. Now I understand. All calculations are in earth gravity. Conversions I presented in the equations are based on kg (force) conversions on earth, which already account for the time component.

Aaron Tanzer

http://www.lehightesting.com

 

[RPstress]

Not sure where Maui and mrfailure are going with this. Thought they'd already got there!

Clearly it's a stress of kg/mm^2 (not an uncommon unit, although assumes the mass is in an unspecified gravity field, assume 1 g here on Earth). kg/mm^2 * 9.80665 is MPa (1 kg weighs 9.80665 N down here so kg/mm^2 are smaller numbers than MPa).

Stress times root(a) with a in mm is a valid stress intensity (once you've accounted for the acceleration in the conversion of kg/mm^2 to MPa). Converting the length from mm to m needs multipying the mm^0.5 by 1/1000^0.5. Total conversion is then kg/mm^2*mm^0.5 * 9.80665 / 31.623.

kg/mm^1.5 * 0.31011 = MPa.root(m). As stated above before things got unclear.

 

[Maui]

I think that there is some confusion here about what appropriate units are in describing this type of problem. According to the Systeme International D'unites (SI system), the kilogram is defined as a fundamental unit of mass, not force. Whether the fracture toughness measurement is made here on the earth or on the international space station, the units that are used to describe it should still be the same. The way that I decribed it is consistent with the requirements of the SI system. To refer to the kilogram as a fundamental unit of force is not consistent with its accepted definition.

Maui

http://www.EngineeringMetallurgy.com

 

[cloa]

This is example of engineering units not scientific- some old engineering units were quite dodgy as equivalents. They tested the fracture toughness using additions of dead weight in the olden days or perhaps in the third world. Not a modern practice. People still refer to things weighing kilograms.

 

[Maui]

No, it isn't. This is a question regarding one of the fundamental accepted definitions in physics. To portray such a thing as "old hat" is simply wrong. Using the incorrect units can lead to catastrophic consequences. The Mars Climate Orbiter that crashed years ago during it's desent to the planet's surface is a prime example.

http://en.wikipedia.org/wiki/Mars_Climate_Orbiter

Maui

http://www.EngineeringMetallurgy.com