Fixed end moments for Polygonal Loading 1

[JoeH78]

hi all,

What's the fixed end moments for polygonal load on a simple FEM rod element, additionally what if the load is offseted from "I" end or "J" end of rod element?

 

[IDS]

Joe - I think there may be a problem with the new Excel file formats here. It is saved as an .xlsb file, not a zip file.

I will re-save it in a zipped file, and also include an xls (2003) version, so the contents of the new zip file should be:

Macaulay.xls
Macaulay.xlsb

When downloaded and unzipped there should be a module in the VBE project listing called mMacaulay (it was mMacauley in the original version, I have just corrected the spelling).

Please let me know if the new file downloads OK.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[RFreund]

ishvaaag - very impressive and interesting. Thanks.

EIT

 

[JoeH78]

Thanks IDS, now it works

So as a result,
1. As long as the loading function remains first degree polyinomal then method described by Macauley can be used to calculate fixed-end moments. (With modifying the source code to accept more than 3 inputs for each two loading type)

2. If loading function is more complex, then curve estimation should be used by any means of numerical analyse methods (e.g. Fourier Series, finite difference, least square etcc...) to get the load function and to find fixed end moments

fixed end moment for the beam would be the integral of the fixed end moments for concentrated loads along the beam where P = w*dx.

mentioned method should be used.



What actually you mean with that P=w*dx?
(Centroid of concantrated loads)*(distance to end) = M ???

 

[BAretired]

If you can express the loading function as a function of x, then w = f(x) where w is the uniform load at point x and x is the distance from the left support. You can then take elements of load, i.e. w*dx and treat them like many concentrated loads.

The fixed end moments for a concentrated load are known, so you simply sum the fixed end moments for each element of load along the beam. Those two sums will be the fixed end moments due to the variable load.

BA

 

[BAretired]

For one trapezoidal load, you should be able to generate an exact solution for fixed end moments using area moment principles. The variables would be w1, w2, a, b and L where w1 and w2 are the uniformly varying load at each end, a is the distance from left support to w1, b is the length of load and L is the span of the beam.

For several trapezoidal loads, you would simply sum the results.



BA

 

[IDS]

1. As long as the loading function remains first degree polyinomal then method described by Macauley can be used to calculate fixed-end moments. (With modifying the source code to accept more than 3 inputs for each two loading type)

The function will accept any number of trapezoidal or point loads (up to about 1 million!). You just select a bigger range.

2. If loading function is more complex, then curve estimation should be used by any means of numerical analyse methods (e.g. Fourier Series, finite difference, least square etcc...) to get the load function and to find fixed end moments

If required it would be straightforward to use the same method for any degree of polynomial, but this would require some coding. Alternatively you can approximate any loading with a number of short trapezoidal loads, or point loads if you prefer.

For one trapezoidal load, you should be able to generate an exact solution for fixed end moments using area moment principles. The variables would be w1, w2, a, b and L where w1 and w2 are the uniformly varying load at each end, a is the distance from left support to w1, b is the length of load and L is the span of the beam.

For several trapezoidal loads, you would simply sum the results.

That's exactly what the spreadsheet does.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

Okay Doug, I haven't looked at your spread sheet, so I didn't know that.

BA

 

[paddingtongreen]

I didn't want to interfere with the relevant answers to the question but I wanted to relay the excitement I felt when I reasoned that each type of distributed load could be considered as a separate beam supporting only that load, the simple bending moments etc. worked out, with the reactions landing on the main beam. The main beam bending moments and shears are the sum of the moments due to the point loads plus the sub-beam values immediately above the point in question.

Not much use today, but it was in the pre-computer days, although I suppose it could still be useful in a spreadsheet

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

I agree, paddingtongreen that your method works and may be an easier way to see the problem. But, in essence it is no different than the moment/area method.

BA