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DB071-Visitor
1-Visitor
Jul 8, 2014 4:00 PM UTC
Solved

M14 Solve Block issue

Hello,

I am trying to use a solve block to find values for two variables but am running into difficulties. It seems that I am always getting an error "variable is undefined". After searching the forum it seems this is the standard error from a solve block and doesn't really give any clue as to what I'm doing wrong.

I originally suspected that I was having issues since I was making use of functions with conditional statements, but after experimenting and removing the statements I am still not getting any results. Manually, I can come close to getting a solution.

Please see attached sheet. Thanks.

RichardJAnswer
19-Tanzanite
Jul 8, 2014 8:34 PM UTC

I don't think it's possible to handle the mixed units the way you are trying to do it. It tries to substitute unitless values into the equations, and just comes up with other unit errors. You need to assign the result to a vector with two variable names, and then evaluate each variable separately.

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13 replies

RichardJ19-Tanzanite
19-Tanzanite
Jul 8, 2014 6:45 PM UTC

Sigma.n2 was not defined as a function of x. Also, x has no defined value. You also have two equations and one unknown, so there is no exact solution. That means you must use minerr, not Find.

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Fred_Kohlhepp23-Emerald I
23-Emerald I
Jul 8, 2014 6:51 PM UTC

Your force and moment are both linear functions of slope and will never satisfy both conditions of the solve block at the same slope.

It's not clear to me which two variables you're trying to solve for.

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DB071-VisitorAuthor
1-Visitor
Jul 8, 2014 7:46 PM UTC

Richard and Fred, thanks for your responses.

I will attempt to clarify what is going on in the worksheet. This is used to calculate the bearing stresses under a footing. The sheet is a subset of a larger, more complicated one.

sigma is the value of the stress at a specific location under the footing. The footing starts at x=0 and goes to x=L.incl (the length of the footing). The width of the footing, B(x), will vary along the length so it is also a function of x.

For some distance, a crack will form from x=0 to some distance (x=L.crack), where the stress must be equal to 0, but the length of that crack is unknown.

Essentially I am trying to describe a triangle with a base length of (L.incl-L.crack) and a maximum height of (slope*(L.incl-L.crack)). At intermediate values of x between L.crack and L.incl, the base of the triangle to the left of x is (x-L.crack), and the height is (slope*(x-L.crack)). The value of sigma at x=0 should be 0, at x=L.crack should be 0, and at x=L.incl some number which I am trying to find.

So, the way it is set up, sigma is a function of x, with the slope being a constant. I need to find both the length of the crack (L.crack, to find the length of the base of the triangle), and the slope of the top of the triangle (to find the height of the triangle).

So, two unknowns, L.crack and slope. Two conditions: Area under sigma is = F.perp, and the moment under sigma is equal to M.

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Fred_Kohlhepp23-Emerald I
23-Emerald I
Jul 8, 2014 8:14 PM UTC

Doesn't the length of the crack change the CGx?

Set up as you describe, it still won't solve.

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RichardJ19-TanzaniteAnswer
19-Tanzanite
Jul 8, 2014 8:34 PM UTC

I don't think it's possible to handle the mixed units the way you are trying to do it. It tries to substitute unitless values into the equations, and just comes up with other unit errors. You need to assign the result to a vector with two variable names, and then evaluate each variable separately.

Translate
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