topic Re: Mathcad usage in Mathcad

Mathcad usage

NM_10165576 — Thu, 12 Jun 2025 06:24:59 GMT

Hello everyone,
eff.K shown in picture is radius of curvature along beam length and X is the array for sectional points.
I have to double integrate eff.K to get deflection. How can i do it in Mathcad? Attached Mathcad 10 sheet.

Re: Mathcad usage

terryhendicott — Thu, 12 Jun 2025 07:54:36 GMT

Hi,

two steps.

1) turn the effK into a continuous function using the interp() function.

2) make a function to do the double integration.

Then plot the result

Re: Mathcad usage

AlanStevens — Thu, 12 Jun 2025 08:20:36 GMT

Here's a simple numerical integration:

Re: Mathcad usage

AlanStevens — Thu, 12 Jun 2025 08:22:23 GMT

Terry, shouldn't your outer integration be with respect to x?

Re: Mathcad usage

terryhendicott — Thu, 12 Jun 2025 08:41:44 GMT

Hi Alan,

If the second integral is wrt x it would be an area integration. It should be two integrals wrt length along the beam?

To be meaningful for deflection of the beam the slope at the ends of the beam needs to be included?

Cheers

Terry

Re: Mathcad usage

NM_10165576 — Thu, 12 Jun 2025 08:56:04 GMT

Thank you.
please send me the mathcad file.

Re: Mathcad usage

NM_10165576 — Thu, 12 Jun 2025 08:59:48 GMT

Thank you for the answer.
But the curve doesn't look like a beam deflected shape for simply supported beam. Shouldn't it be like parabola?

Re: Mathcad usage

terryhendicott — Thu, 12 Jun 2025 09:58:29 GMT

Hi,

But the curve doesn't look like a beam deflected shape for simply supported beam. Shouldn't it be like parabola?

Correct. Allowance needs to be made for the slope of the simply supported beam at the supports. Not enough information is given in the post to calculate this.

Cheers

Terry

Re: Mathcad usage

NM_10165576 — Thu, 12 Jun 2025 10:37:49 GMT

beam length is 5.46m and the supports are at 0.1m from left and 0.1m from right. Calculate for this new data, if possible.Deflected curve looks like in the image below. Attached the mathcad sheet.

Re: Mathcad usage

Werner_E — Thu, 12 Jun 2025 14:14:46 GMT

I have no expertise whatsoever in your field of work but the double integral as you had set it up

would mean that the evaluation of the inner definite integral would yield an expression in x. Variable y is gone and the subsequent integration wrt y would mean just a multiplication by y.

Given the limits 0 to x of the outer integral this would mean that you could replace the outer integration just by a multiplication with x.

So we could spare the outer integral speeding up the calculations. But is this really as it should be???

Here are is your formula and the equivalent one (with just the multiplication by x) compared with Alan's suggestion, which sure looks more like a parabola:

I have no knowledge about beams and deflection but I also guessed that either the variable of integration of the outer integral should not be y and/or that one of the integral limits should be different.

PS: While mathematically not correct in notation, the following would yield the very same result as my delta3 above and is maybe what you had in mind?:

Re: Mathcad usage

terryhendicott — Thu, 12 Jun 2025 15:00:31 GMT

Hi,

See enclosed file.

Re: Mathcad usage

NM_10165576 — Fri, 13 Jun 2025 03:38:52 GMT

This works. Thank you😊