Is it possible in MathCAD to solve for a variable as I have shown in the snip? I've tried different ways but cannot get it to work. But I am also very new to this program.
I want to solve for ΦVn. I know my limiting capacity and the geometry of the weld to obtain the weld properties.
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And here's a snippet of your snippet, with 5 remarks:
(1) and (2) You've used an inline evaluation, that is, you put an evaluation ( an = ) at the end of the definition of fx. Well, phiHn1 is unknown, so you cannot evaluate this definition of fx, You should minimise the use of inline evaluations. Yes, they do work often, but more often they cause errors. If you want to know that value of a defined variable, evaluate it separately.
(3) phiHn1 is not part of the definition of phiHn. The normal way is to define fx and fy as functions of phiHn1, if you want to solve for that phiHn1.. You are using a symbolic evaluation here (the arrow, with the symbolic command above it, 'solve' this time.
You have to be(come) aware that the symbolic processor is not aware of units, it doesn't 'know' them and treats them as symbols. If you are lucky they cancel out. Having said that, there is a setting (under 'Calculation'=> 'Calculation options') to make the symbolic processor aware of units, but it doesn't always work.
Since you are after a numerical answer, it's better to use a solve block, or the root function. See my attached sheet. Look up the help regarding the root function to find out its two modes of operation.
(4) define Fx, fy and fz as functions of phiVn to prevent these errors. But note that the symbolic processor is OK with unknown variables.
(5) similar as above. But with fx, fy and fz known to the symbolic processor, it will solve this root. The result is 13.3, but without units.
Success!
Luc
And here's a snippet of your snippet, with 5 remarks:
(1) and (2) You've used an inline evaluation, that is, you put an evaluation ( an = ) at the end of the definition of fx. Well, phiHn1 is unknown, so you cannot evaluate this definition of fx, You should minimise the use of inline evaluations. Yes, they do work often, but more often they cause errors. If you want to know that value of a defined variable, evaluate it separately.
(3) phiHn1 is not part of the definition of phiHn. The normal way is to define fx and fy as functions of phiHn1, if you want to solve for that phiHn1.. You are using a symbolic evaluation here (the arrow, with the symbolic command above it, 'solve' this time.
You have to be(come) aware that the symbolic processor is not aware of units, it doesn't 'know' them and treats them as symbols. If you are lucky they cancel out. Having said that, there is a setting (under 'Calculation'=> 'Calculation options') to make the symbolic processor aware of units, but it doesn't always work.
Since you are after a numerical answer, it's better to use a solve block, or the root function. See my attached sheet. Look up the help regarding the root function to find out its two modes of operation.
(4) define Fx, fy and fz as functions of phiVn to prevent these errors. But note that the symbolic processor is OK with unknown variables.
(5) similar as above. But with fx, fy and fz known to the symbolic processor, it will solve this root. The result is 13.3, but without units.
Success!
Luc
