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Fixed end moments
- Thread starter odendaal
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EnglishMuffin
Mechanical
Roark's "Formulas for Stress and Strain" has formulas for beams with distributed loads of various types, some of which stop short of one end support. In the text, it is explained how to use the "three moment equation" to make use of these basic cases to solve more complex cases, where for example the distributed loads stop short of both ends, or there are point loads in addition to the distributed loads. You can't necessarily expect to find a ready made formula for every possible case - you often have to use superposition or force and slope matching to solve specific problems. But you might be lucky and find a ready made complete solution for your problem somewhere. There is a TK solver application program available from UTS called "Roark & Young on TK" , which has every formula in the book programmed and ready to use - I've never used it, but maybe theye have catered for these combination type solutions, which could save some work.
- Thread starter
- #3
EnglishMuffin
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
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
Any good text on energy theorems will help out here. Virtual work is the staple for this type of derivation, since you're interested in a response to a unit displacement or rotation. Take a look back in your old structures text, like McCormac.
EnglishMuffin
Mechanical
In my 5th edition, there is an example in section 7.3 of a case where the distributed load does not fully extend to either end. By studying this, which I confess I have not in detail, you should be able to solve your problem. In all these sort of situations, you can combine the basic cases by matching the slopes and deflections at the junction between them, thus enabling combinations of any degree of complexity to be solved, although it can be tedious.
Regarding Roark & Young on TK, see 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.
Regarding Roark & Young on TK, see 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.
You can derive such formulas "relatively easily" yourself ( as long as your beams cross section is prismatic ). I've done several myself assuming that varying loads vary in a linear fashion. You can also do this for non-linearly varying loads. The algebra is just a bit stickier.
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
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
- Thread starter
- #7
Hi Dan
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
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
I did a web search. It seems that what you are referring to is McCaulay Notation. It appears at a glance to be just a way to referring to the differential equations that can describe the load/deflection of an element. If I'm not mistaken... it merely allows you to write one equation that is understood differently "left" of a load point and "right" of the same load point. In the grand scheme of things though... I don't think it is really what you "Need"... though... it will do the trick if you want to use it.
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
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
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