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I am doing a Finite Element analysis of a continuum. In my problem definition I use these units: mm for length, kPa for applied stress, and MPa for material stifness (Young's modulus).

With these units, I get my output in: kPa for stress and micron for displacement, which is what I prefer.

Now I have to add Mass Density information to my problem. To benchmark my setup, I am trying to reproduce an old textbook example, in which the Mass Density is given in units of: $$ \\lb-sec^2 / in^4\, $$ What would be the equivalent of this in my adopted SI unit system.

My impulse is to use:

$$ \\kN / mm^3\, $$

Any suggestions or advice will be appreciated!

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  • $\begingroup$ Those units are that way to turn pounds into mass-pounds by dividing by 32 in/sec^2. There is no equivalent in SI becuse kg is mass. Just use regular mass and length units. $\endgroup$ – Tiger Guy Mar 29 at 6:21
  • $\begingroup$ Thanks @TigerGuy - but I need to know how to convert this. I have a value of 2.45E-4 lb-s^2/in^4 that I need to convert to my units. I have no idea what the material is, so I dont know what density to assume. I need to know how to convert it to kN/mm^3 or something like that. $\endgroup$ – Fritz45 Mar 29 at 6:50
  • $\begingroup$ based on my answer, i think 2.45E-4 lb-s^2/in^4 is equal to 2.619e-8 kN/mm^3, does it looks an acceptable value in you finite element model, or it is far away from what you expect? thanks $\endgroup$ – epsi1on Mar 29 at 12:53
  • $\begingroup$ @fritz - it's just density, kg per cubic meter, or whatever units you are working in. If you need to convert the textbook example, convert the given density by multiplying it by 32.2 to give it in pounds, then convert to the units you want. $\endgroup$ – Tiger Guy Mar 29 at 20:47
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First Part

Simplifying the $$ 1 \ lbf = 1 \ slug .ft/sec^2 $$ then $$ 1 \ lbf \ sec^2 / in^4 = 1 \ (slug .ft/sec^2 ). (sec^2/in^4) = 12 \ slug /in^3 $$

$$ 12 \ slug / in^3 * (14.6 kg / 1\ slug)*(in/25.4 \ mm)^3 = 0.0107 \ kg/mm^3 $$ Note that that each slug equals to 14.6 kg so (1 slug / 14.6 kg) equals to 1.0 and any expression multiplied with 1.0 will still matematically equal to original expression.

then

$$ 1.0 \ kg/mm^3 = 93.53 \ lbf \ sec^2 / in^4 $$

we name above as eq1


Second Part

Changing shape of the target formula:

$$ 1 \ kN / mm^3 = 1 \ kN / mm^3 * (1000 \ N/1 \ kN) * (1 \ kgf / 10 \ N) = 100 kgf/mm^3 $$

again note that

$$ (1000 \ N/1 \ kN) = (1 \ kgf / 10 \ N) = 1.0 $$

Also i took $ kgf = 10 N $ which is approximate, correct amount is $ kgf = 9.81 N $ finally $$ kN / mm^3 = 100 \ kgf/mm^3 $$

or

$$ 1.0 \ kgf/mm^3 = 0.01 \ kN / mm^3 $$

we name above as eq2


Third part

kg and kgf units are not consistent units on the paper, but looks equal on non scientific conversations. For example someone can say this chair mass is 10kg or it's weight is 10kgf, both refers to same chair with same mass and weight. so if we conventionally assume: 1 kgf = 1 kg, eq1 and eq2 can be combined like this:

$$ kgf/mm^3 = 0.01 \ kN / mm^3 = 93.53 \ lbf \ sec^2 / in^4 $$

Final part

so

$$ 1.0 \ kN / mm^3 = 9353.0 \ lbf \ sec^2 / in^4 $$

or

$$ 1.0 \ lbf \ sec^2 / in^4 = (0.00010691) \ kN / mm^3 $$

Which are the answers


Notes

I am not sure about the assumption of 1 kgf = 1 kg in third part, but the rest should not have problem. Anyways this answer could be considered as a suggestion.

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    $\begingroup$ 1 kgf = 1g N = 9.81 N. Unlike the British/US system of units 1 kgf does not equal 1 kg. The Metric system is much more straight forward. It doesn't mix mass & force. There is no equivalent to slugs in the Metric system. Also, in your 2nd & 3rd parts you have a capital K for kilo, it must be lower case k. $\endgroup$ – Fred Mar 29 at 9:34
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    $\begingroup$ @Fred thanks i've applied 9.81N and lowercase k for kilo. but my search says the slug and kilogram are unit for measuring same quantity so directly convertible. $\endgroup$ – epsi1on Mar 29 at 12:41
  • $\begingroup$ Many thanks @epsi1on for your very thorough answer. The part I was uncertain with, and which Google did not help much, was your part 1, which has clarified some units for me. Thanks again. with regards to your question above - I have put the value you suggest in my FE problem but get a numeric overflow. However, I am using an explicit dynamic analysis which is very sensitive to time step size. I suspect I will need to re-look at the size of my mesh etc. Thanks again for your help! $\endgroup$ – Fritz45 Mar 30 at 20:24
  • $\begingroup$ @Fritz45 you are welcome, your amount 2.45E-4 lb-s^2/in^4 is approx equal to 2.62 grf/cm^3 (gram force per cubic cm) or 2.62 Tonf/m^3 (Ton force per cubic meter) based on my calculation $\endgroup$ – epsi1on Mar 31 at 5:55

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