You've created a discontinuous function, which at no point does it actually equal zero. You could possibly create something that might detect the sign change.

TTFN
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I Know this maybe sound silly but this is my First try in programing something like this, so, Im I doing this the right way or the wrong way?
IRatuff I see that you noticed that Ive created an discontinous function is this the wrong way?
WOuld you do it someother way?
thank you

This is math issue, not a programming issue. A discontinuous function cannot have a value in the middle of the discontinuity, by definition. Since your function cannot possibly have a value where V(x)=0, there is no way to program to solve for solution that cannot exist. An obvious approach would be to make the function piecewise linear.

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Im not shure that I undestand you.
Please see file attached.
If I define a picewise linear function like in the file attached, I still have an discontinuity for the part were the graph is vertical.
So how to find these two values for x were the graph intesects with x=0?
Please help.

• http://files.engineering.com/getfile.aspx?folder=64c2cc12-6ac3-40df-8989-90

Maybe this would work for you.

• http://files.engineering.com/getfile.aspx?folder=e84524f1-7461-4ab7-8ba4-76

hmmmm...can you explain what you have done.
My goal would be that Mathcad spits 2 numbers out....in this case -1.5 and +1

Then maybe this? Although, I have no idea how you came up with -1.5 and +1 as your solutions.

• http://files.engineering.com/getfile.aspx?folder=4b928693-f3d6-4724-a8da-73

IRstuff heres an simple egsample.
http://s000.tinyupload.com/index.php?file_id=02292...
YOu can see that V(x) cuts the abscisa at 2 loactions.
I Know how to find one of them, but can Mathcad spit out both values if the egsist?
Thank you.

@Occupant
YOuve uploaded 2 same files...

Yeah, I thought Eng-Tips would update automatically, but apparently it doesn't.

• http://files.engineering.com/getfile.aspx?folder=fd1f4de5-fc1e-4c63-a24f-5e

Again, it's a math problem. When you specify 5 at x=1, and -x for x>1, you've created a discontinuity, for which no value of x results in f(x)=0 in that interval. By defining a piecewise linear transition, as shown in the attached file, a root can be found, and you can make it arbitrarily close to x=1


http://files.engineering.com/getfile.aspx?folder=a...

TTFN
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IRStuff
So basicly you defined that vertical portion with an equation BUT it has an very,very small slope?

To answer my own previous question.
Seem to work for this kind of problem

http://s000.tinyupload.com/index.php?file_id=04732...

Thank you.

One question,
In previous egsamples i have a function V(x)=....
I can get an value of V for any given x that I type in th brackets.
Is it possible vice versa?
I wanta to know at wich x would V= defined value?
So bascily I would need to First type eg. V=20kN and then Mathcad would give value for x at wich V=20kN
Possible?

You can look up "solve block" or "given-find" it would look something like this:

x:=1 [initial guess for solve block]

given [required for solve block]

V(x) <CTRL>= xx [press the <CTRL> key and then the "=", resulting in a bolded equal sign, unless you've dinked with the fonts]

x := find(x) [or] find(x) = [the former assigns the found value of x to x, so that it can be used in subsequent calculation; the latter simply displays the found value, but x remains at the initial guess value]

TTFN
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IRStuff,
one more question,see file attached.
http://s000.tinyupload.com/index.php?file_id=56723...
Is there anyway that I can condition the statement so that it looks for values belowe 4m and second statement for values above 4m?

Anyone?

Two things you can try:

> You need to change the guess value of x to be >4 -- for the second set of Finds
> Change find to minerr -- sometimes this works, and sometimes not - it worked a couple times on your sheet when I tried it, and failed most of the other times. The algorithms are all somewhat dependent on having continuous functions, so the discontinuous functions are very challenging

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There is a homework forum hosted by engineering.com: http://www.engineering.com/AskForum/aff/32.aspx

"Change find to minerr"

Is this spelt corectly?
I get an error every time when I try this.
So the solution woul dbe to convert this discountnous function into what you suggested "very steeep slope"wich would make it continous,

re: minerr, as I said, it was hit or miss as to whether minerr would work.

TTFN
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There is a homework forum hosted by engineering.com: http://www.engineering.com/AskForum/aff/32.aspx

Getting back to your beam4.xmcd file: The guess value for "x" that you put in has to be in the range of available solutions, therefore, x>4m will not give you a solution with x:=0 but it will with x:=6. Of course, V(x)=-20Kn will not work at all, there is no V(x)=-20Kn with x>4m

@irstuff
" You need to change the guess value of x to be >4 -- for the second set of Finds"
this works fine at my place
thanx!