## Fixed end moments

## Fixed end moments

(OP)

Does anyone perhaps have formulas for the fixed end forces and moments for beams fixed at both ends with uniformly distributed loads and variably distributed loads where the loads do not start and end at the supports.

Thanks

Thanks

## RE: Fixed end moments

## RE: Fixed end moments

Thank you for your detailed reply which I wish to pursue further.

I have Roark and noticed that his loading goes up to the support, but I could not find the solution for stopping short of the support. Maybe you can indicate that to me.

What is TK and how do I get to that infomation ?

My prime interest in this is that I have written a program in VB6 which solves space and other frames, but I need to include a routine whereby a user can insert any type of load and the routine would then break it down into fixed end actions which I can then handle at the nodes.

I have tried to use the McCaulay theorem to deduce a standard formula, but the results I get from this does not correspond with known values for special cases. Maybe I do not remember the McCaulay theorem so well.

Thanks

P.K. Odendaal

pk@odendaal.com

## RE: Fixed end moments

## RE: Fixed end moments

Regarding Roark & Young on TK, see http://www.roarksformulas.com/ryupgrade.html

There is something called "Superposition capability" and "Superposition Wizard", which may do the trick - but you will have to check. From memory, I think this program used to cost between $500 and $1000 dollars.

## RE: Fixed end moments

You already know the fixed end moments for a single point load located at some location on a beam. It's a simple formula and is shown in most analysis text. A uniform load is nothing more than a "whole lot" of these single point loads all placed next to each other along a beams length.

You can use basic integral calculus to sum up the results of a bunch of these point loads ( size of P*dx each ) along the length of your beam in question.

For standard prismatic beams, I think you'll like the formulas you come up with better than the ones you'll find in Roark's. And... you get the satisfaction of having developed them yourself.

Dan

www.dtware.com

## RE: Fixed end moments

Thanks for the response.

I have tried using integration to get to a formula, but instead of starting from first principles (which I have long forgotten)I thought I might get by using McCaulays Theorem which is in effect in an already partially integrated from, but I just cannot get the correct answer.

The theorem looks like :

EId2y/dx2=Mx=Ma+Ra.x-w[x-a]^2/2

I have searched the web for his theorem, but could not find anything - although I have used his theorem many times in my career.

I agree Roarks formula is elaborate and involves deduction and adding various loads just to arrive at one loadcase.

P.K. Odendaal

pk@odendaal.com

## RE: Fixed end moments

The method I suggested is a little simpler ( understanding wise anyway ). Given a beam of length L. Put a point load, P, on it at a distance "a" from the left end. Let "b" equal the distance from the load to the right end.

The fixed end moment at the left is.. Ml= Pab^2/L^2. The fixed end moment at the right is.. Mr = Pa^2b/L^2. These formulas are in most books and can also be derived fairly easily. I like to use the conjugate beam method myself to derive them.

NOW... take the same beam and put a uniform load on it of "w" ( units of force/length ).

Pick some point in the beam that is "x" away from the left end of the beam. Take a thin slice of this load that is dx wide. This slice has a total load on the beam of "w*dx".

Due to just this one little slice of the load... the small moment at the left end of the beam ( from the point load formula above ) is...

d(Ml) = (w*dx)*x*(L-x)^2/L^2

Now.. just integrate:

Ml = Integral(from x=a to x=b) of d(Ml)

where a = distance to the start of the uniform load

where b = distance to the end of the uniform load

If your loading is not uniform... you have to derive a formula for how "w" changes as a function of "x". That will make your integration a little more difficult, however, intirely doable. ie.. you don't have to be a Calculus whizz. What's tough is cleaning up the algebra so that you have clean final equations. The actual integration is generally quite simple.

I didn't get this method from any book anywhere and I don't know if anyone else has ever done similar derivations. Several years back... I wanted to do the same thing you are currently doing. So.. I had to come up with the same formulas you are now seeking ( and in a way that made sense to me ). I used this technique to derive several formulas. I even derived the formula for the fixed end forces due to distributed moment loadings. I did extensive checking and verification on the formulas I derived and they all passed all the test.

Feel free to send me email directly if you wish.

Dan

dan@dtware.com

www.dtware.com