Just looking, but I think your integral
is all constant terms that could be pulled out of the integral. the integral then evaluates to h_i and the summation should evaluate.
"Cntrl+shift+^" is the vectorization operator (matrices/tables\operators) and results in the arrow over an expression, meaning, "do this operation on each element of the vectors separately." It results in a vector result.
Why did you think I hadn't used it?
That is also the reason I had to add elements to your vectors for z and h--all of the vectors in an expression have to be the same length or vectorization will crash.
You're trying for a symbolic solution.
Mathcad is primarily a numeric program.
Try writing Fn(P) = SI(P, 6 ft)+SC(P,6ft)-1in,
Then investigate the "root" function.
But start with this:
you have
which (I think) can be rewritten as:
by taking all of the "constants" (that don't depend on x and y) out of the integration. (note bB = B)
Then evaluate:
And your entire second term would disappear.
Your expression blow up (division by zero) at v = 0.5
Can Poisson's ratio go that high?
Well, DUH!
You have a function 
Whether you evaluate for the elements of a vector, or over a range variable it's going to give you the same answer. The expression that circled red on the second sheet did so because you were dividing by zero.
You asked about the slope function, which does not work on range variables.
Congratulations! I believe you've found a bug!
You're right!
And a good reason not to do this:
Well, maybe it does.
I built a vector the usual way (using indices. Then I plotted all three, then I tried vectorized the range vector:
Fred Kohlhepp wrote:
Well, maybe it does.
I built a vector the usual way (using indices. Then I plotted all three, then I tried vectorized the range vector:
It has nothing to do with the way, the vector is created. It works also perfectly using the inline evaluation trick.
The problem was just the missing vectorization, which should always be done in cases like this, just to be on the safe side.
