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Convert Metric Units to English Units

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SunshineTown

SunshineTown

Structural
Oct 22, 2019
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I have this formula from ACI 318-02 for the basic concrete breakout strength Nb = k*(fc^0.5)*(hef^1.5). I am thinking how this formula will led to a result in unit of force (kip)?
This question arises when I tried to use this formula as a reference to double check the concrete breakout strength using the standard in my region which is in English units and the 'timing' of converting the units leads to different results!
For example, say if I have k=24, fc=25MPa and hef=100mm, if I put these values into the formula without converting to metric ones then I will have the result of 120kN (which is 26.98 kips).
However, if I convert the units of fc and hef to psi and inch and then put it in the formula then the result will be 11.2879 which is 50.2kN. Can anyone explain which way is the correct one and why? I am confused...
 
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You should get a metric version of ACI318!

In the 2018M version, the k factor is shown as 24 or 17 for imperial units and 10 or 7 for SI/Metric. Presumably it is because the formula is not dimensionally correct, with (N/mm2^0.5)*(mm^1.5) not resulting in KN or kips!
 
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your formula is located in Appendix D parag D.5.2.2 and the explanation probably in Appendix F
equivalence_mlbwk0.jpg
 
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Can anyone explain how this formula was developed? I would expect it to be something like tension strength of concrete times the area of potential failure plane, which is factored hef^2. I know it is based on numerous tests but what is the theory behind this formula? How do you understand this formula is a simple way?
 
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1) The SQRT(f'c) factor is a normalized strength term used to match empirical tests to best-fit equations.
2) The k-term is a curve fitting term based on empirical tests. See Eligehausen, Mallee, Silva "Anchorage in Concrete Construction" for a deep dive.

If you are switching between imperial and metric, make sure the equation is right (the non-dimensional coefficients will change) and apply conversion to each unit, not just the result.
 
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Dik - your example gives the same wrong result as quoted in the OP.

If you want to use the k from the imperial code with metric input you need to assign the correct units to the k factor, or use the k appropriate to your input units, i.e. use the metric version of the code and make sure your input is in the same units as used in the code formula; stress in MPa and length in mm with results in N.

Doug Jenkins
Interactive Design Services
 
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Sorry Doug... I don't know what the 'right results' are.

The method was given as an example of how to solve it, and it's just a matter of changing the units to imperial or metric, and whether feet or cubits, and seeing what the end result is... assuming you have an idea of what it should be.

What units was the original formula presented in?

I would assume non-dimensional units, else the k value would have a really wierd set of units due to the exponents. I'm not familiar with the formula or the background.

Dik
 
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Dik said:
I would assume non-dimensional units, else the k value would have a really wierd set of units due to the exponents.
Click to expand...

That's the point; the k value does have weird units, so if you don't use the units assumed in the code you will get the wrong answer.

As stated in the OP, using k= 24 with N and mm input gives 120 kN, but with kip and inch input you get 11.29 kip = 50.2 kN.

Doug Jenkins
Interactive Design Services
 
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IDS said:
That's the point; the k value does have weird units
Click to expand...
Agreed. Though plenty of times the units for such coefficients are not given. Which IMO is terrible physics and pretty poor engineering. Though it is certainly not uncommon in many codes and engineering formula. Unit consistancy should always be maintained not only is it mathermatically correct, but it allows important checks and balances in calculations.

The formula here is a weird one and it does grate on me. But if it works empiracly then fine, but at making the units clear makes everybody's life easier.
(Also if you include the units in the coefficent it becomes clear that it needs to be converted to match.)

/END RANT
 
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I'll disagree here, since such equations typically make it very clear what the units for each input is, and what the units of the output is. The combination results in the units for the constant. The simplest approach, if you don't want to figure out a new constant is simply to convert all the inputs to the unit system of the original equation.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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IRStuff - I agree that sticking with the code units is the simplest and safest way to proceed, but that does require that the units are clearly stated in the code. Unfortunately that is not always the case. In this case for instance using the metric ACI 318 I didn't see anything to indicate that input should be in MPa and mm, and output is in N.

Doug Jenkins
Interactive Design Services
 
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Obviously, I've no dog in this hunt, but this describes the imperial units for 318-02, so presumably there's a corresponding table 1 in the metric version of 318-02

[edit] the imperial version of 318-02 Appendix E states "specified compressive strength of concrete, psi" and "effective anchor embedment depth, in." so presumably, any SI version would have similar statements

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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