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Hi, college student new to Mathcad here. I'm having trouble with the units of a concrete shear formula. I put units to the unit weight correction factor, lambda, but that didn't help.
Help would be much appreciated!
Solved! Go to Solution.
As Valery already pointed out - look at your definition of lambda - is this really what you intended?
Sorry, but what is it
A very neat work-around for an empirical formula that takes the square root of a stress.
As Valery already pointed out - look at your definition of lambda - is this really what you intended?
No, within the set of real numbers the square root is defined as being positive (or zero).
Only in the set of complex numbers we have two results (or n for an n-th root).
Confused with
?
Luc
Please attach your worksheet. You may have to ZIP it first. A serious bug in this forum often prevents from attaching a mathcad file. In Windows Explorer click with the right-mouse button on the mathcad file and select 'Send to' -> 'Compressed (zipped) folder'. The resulting .zip file can be attached.
Success!
Luc
My first guess was that you had a unit mis-match,especially after reading Luc's rant about lambda, but you don't.
Prime is notorious for getting the "labels" wrong. It's possible (in Prime) to have two things look exactly or nearly the same and be different things; you can have a variable m and it can be six meters long and write m := 6 m. If the first is labeled "variable" or "-", and the second is "unit" Prime will happily continue. But if you created something in the red rectangle with one label, and in the rectangle it has a different label, Prime will refuse to continue.
Check that.
Thank you all for your quick replies. I appreciate it very much!
if not wrong, You refer this.
The problem that you can't solve it because shear stress. just put psi because it from testing and result with square root but it keeping still Stress unit .
Hope it useful.
Yes, as Fred already wrote its an interesting workaround to be able to use an empiric formula where f is expected in psi and the square root of it is taken.
Guess most of us would have done it the "usual" way: remove the unit by dividing by the asked for unit, do the calculation (root) and then add the proper unit.