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Fixed-end moments for Trapezoidal load on part of span?

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TarikHKJ

Structural
May 8, 2016
79
Hi,

Trying to solve an indeterminate beam using moment distribution displacement method, I used the general fixed-end moment equations, but one span has a trapezoidal load on part of the span which does not have a fixed-end moment equation,

here is the problem:
How to calculate the fixed-end moments for that span?


Many Thanks!
 
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A fixed-fixed beam with a triangular load had end moments of -wl^2/20 on the more heavily loaded end and -wl^2/30 on the less heavily loaded end. A fixed-fixed beam with a uniform load has end moments of -wl^2/12 on each end. Combine them and take it from there.

EDIT: Whoops, I now see you wrote that it has a trapezoidal load on only part of a span so the above does not apply.
 
You can also do the conjugate beam method solve for indeterminate beams.
 
@Archie your equations would be correct if the uniform/inclined loads are on the whole span, but in this case the trapezoidal load is on part of the span as you can see in the image referenced.


Kindest regards!
 
is this a student problem ?

I'd solve the span as a doubly cantilevered beam with a partial span loading. Not That hard ... 2 unknowns, unit force method would probably be my choice.

another day in paradise, or is paradise one day closer ?
 
@cal91 is it possible to use conjugate method to find fixed-end moments and continue on with the displacement method?

Kindest regards!
 
if this is a "real" problem, how accurate do you need, Need, to be ? replace with a UDL, or a point load at centroid.

if doing mdm, is there a way to derive the influence and carry-over coefficients ? ie mdm initally locks each support (fixed) and then releases each in turn, redistributes, and repeats

another day in paradise, or is paradise one day closer ?
 
Oh. My. Goodness. KootK Christmas in August. I've been waiting for this question for seventeen years. I had to develop general equations fixed end moments and shears for this condition for some 2D software that I wrote way back in the late 90's. It uses a generalized version of the triangular decomposition method that Archie suggested. I know, the equations look a little crazy. I vetted them extensively against hand calculations and commercial software however. They're correct, I personally guarantee it.

The equations below calculate the fixed end moments and shears for the left and right ends of the member under consideration. s1, s2, and s3 are the lengths of the beam in front of the load, under the load, and behind the load respectively. w1 and w2 are the load intensities at the beginning and end of the trapezoidal loading. I've also included some write up for the triangular decomposition method. There a,b,c = s1,s2,s3. I think you'll get the idea. I developed equations for outputting shear, moment, and deflection too if you need those. The deflection equations are wild. I had to use MathCAD to keep the algebra straight.

AF1_mzym01.png

AF2_gtiz8z.png


I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
TarikHKJ said:
@cal91 is it possible to use conjugate method to find fixed-end moments and continue on with the displacement method?

Yes! For fixed fixed, you will have two unknowns that you will need to solve for (do this by solving two equilibrium equations on the conjugate beam).

Once you solve for the moments, you can continue on to find the displacements.
 
@cal91 doesn't a fixed ended beam have 3 redundants not 2?

 
KootK, that's some piece of work! Congrats for pulling it off.
 
TarikHKJ said:
@cal91 doesn't a fixed ended beam have 3 redundants not 2?

It would be if both ends are considered fixed horizontally, but one end can be free to move horizontally while fixed against rotation and vertical displacement.

To solve the problem without KootK's equations, I would calculate the rotations at each end due to the trapezoidal loading on a simple span, then calculate the moments required to reduce both rotations to zero.


BA
 
Archie said:
KootK, that's some piece of work! Congrats for pulling it off.

Thanks. Intended for automation of course. I've often wondered how commercial packages handle this. Crazy equations but they cover uniform, triangular, trapezoidal, increasing, decreasing, and negative loads all in one go. Basically, any kind of standard distributed load. Very efficient in a programmatic sense.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
TarikHKJ said:

Not quite. That only works because the load is uniform and symmetric. In your case, the load is variable.

In the general case, using the Conjugate Beam Method, do the following:

1. Draw the Bending Moment diagram for the real beam assuming a simple span and the actual loading.
2. Load the Conjugate Beam with a fictitious load corresponding to M/EI of the real beam.
3. Calculate the shear at each end of the conjugate beam (which is also the rotation of the real beam).
4. Calculate the end moments required to bring the end rotations to zero.

If that sounds like too much work, you could use a reasonable approximation of the actual load.

BA
 
My ConBeamU spreadsheet has functions that will return end actions for any combination of partial trapezoidal loads, with any end conditions (fixed, free or spring restraints).

Download ConBeamU

Screenshot attached

On the FEA sheet (Fixed End Actions, not Finite Element Analysis) there are some examples compared with standard formulas, including Example 6 (starting row 131) which is a partial triangular load, using a formula from the Reinforced Concrete Designers Handbook by Reynolds and Steedman. The formulas are a bit simpler than Kootk's, but of course you would need to combine two to get trapezoidal loading.

The FEAU function will return fixed end moments and forces, and REAU will work with spring restraints. They both have versions without the U, that assume consistent units. The functions don't use standard formulas, they use Macaulay's method to calculate slopes and deflections for a cantilever, then solve to find end actions for the specified end conditions.

The functions work for beams with any number of segments with different section properties, and the spreadsheet also has functions for continuous beams, with any number of supports and segments, using the same approach.

Search the blog for Macaulay and/or ConbeamU for more details.

Doug Jenkins
Interactive Design Services
 
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