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Finding max bending moment 1

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mar2805

mar2805

Structural
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Hi folks.
See file attached.
Its an simply supported beam with an concentrated force.
I cant program Mathcad to find the x value were shear force diagram intesects with x axis.
I know at wich value is this going to happen for this loading scenario but how to tell mathcad to slove this?
I tired using the root function but no luck.
Please help.
Thank you
 
 http://files.engineering.com/getfile.aspx?folder=ca47f6c5-6181-4ca7-ad2a-227d397b5d0e&file=Beam.xmcd
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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
faq731-376
7ofakss

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Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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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
 
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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.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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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-8989-90699e64443c&file=pice.xmcd
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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
 
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@Occupant
YOuve uploaded 2 same files...
 
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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




TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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IRStuff
So basicly you defined that vertical portion with an equation BUT it has an very,very small slope?
 
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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?
 
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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
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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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

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:
 
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"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,
 
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