Understanding Carry-Over Factor in Beam Analysis

[fonseh]

Homework Statement

I don't understand why for the first case , the 0.4 and 0.6 is used ? for the second question , we can see that in the first case , the moment from BC is carried over to CB ( highlighted part - line 4 ) ... Why for the second case , the moment from BC (6325.8) isn't carried over to CB ?

Homework Equations

The Attempt at a Solution

I think for the second part , the moment from BC (6325.8) should be carried over to CB (6325.8/2 )

 

[CivilSigma]

For Question #2, you carry over the moment from CB to BC and BC to CB because joints B and C are part of the member BC.

You can carry over the 6325.8 , but on the next itteration you will have to do: Moment to distribute = (-1) * DF * Previous moment . So you would have: 6325.8 + (-6325.8) which will add up to 0. Try continuing the iterations a bit, you will ultimately converge to the final solution shown.

For Question #1, it has to do with calculating the stiffness factors and then the distribution factors.

 

[fonseh]

You can carry over the 6325.8 , but on the next itteration you will have to do: Moment to distribute = (-1) * DF * Previous moment . So you would have: 6325.8 + (-6325.8) which will add up to 0. Try continuing the iterations a bit, you will ultimately converge to the final solution shown.

Why when the moment from BC is carried over to CB , shouldn't it become 0.5(6325.8 ) ? Juts like the moment BA( 5647.2) are being carried over to CB , become ( 2823.6) ??

 

[CivilSigma]

Yes (forgot about that) , but then after that you will have -2823.6 N*m and adding it up with 2823.6 = 0

 

[fonseh]

-2823.6 N*m and adding it up with 2823.6 = 0

Where is it ? It's not shown in the calculation ... the book only showed total moment 2823.6 ... , no -2823.6 N*m whatsover ... Or the book is wrong ?

 

[CivilSigma]

Where is it ? It's not shown in the calculation ... the book only showed total moment 2823.6 ... , no -2823.6 N*m whatsover ... Or the book is wrong ?

Sorry, ignore the last post (I thought DF=1).

For member AB, DF=0, so what ever you carry over, you will always have 0 in that column since you will need to do : DF * Moment * (-1)

 

[fonseh]

Sorry, ignore the last post (I thought DF=1).

For member AB, DF=0, so what ever you carry over, you will always have 0 in that column since you will need to do : DF * Moment * (-1)

So , do you mean the book is wrong ? the total moment about A should be = 0 ? not 2823.6Nm ??

 

[CivilSigma]

No, the book is correct... I was talking about your "next iterations" . Try it out, in that column you will have 2823.6 , then 0 then 0 then 0 ... when you finally take the sum , you will get a final moment on member AB of 2826.6 Nm

 

[fonseh]

Sorry, ignore the last post (I thought DF=1).

For member AB, DF=0, so what ever you carry over, you will always have 0 in that column since you will need to do : DF * Moment * (-1)

Ok , back to the problem , why when the moment from BC is carried over to CB , shouldn't it become 0.5(6325.8 ) ?

 

[CivilSigma]

Yes it will be.

Then when you continue the iteration, what will you get as a moment to distribute? You will have to distribute -3162.9 Nm . So in the column you now have : 3162.9 + -3162.9 =0.

Also, keep in mind that a pin/roller end cannot take a moment. So if you end up with anything other than 0, your solution is wrong.

You should set this problem up in excel to see for your self what values you will get if you continue iterating.

Maybe this can help more: https://en.wikipedia.org/wiki/Moment_distribution_method

 

[fonseh]

Yes it will be.

Then when you continue the iteration, what will you get as a moment to distribute? You will have to distribute -3162.9 Nm . So in the column you now have : 3162.9 + -3162.9 =0.

Also, keep in mind that a pin/roller end cannot take a moment. So if you end up with anything other than 0, your solution is wrong.

You should set this problem up in excel to see for your self what values you will get if you continue iterating.

Maybe this can help more: https://en.wikipedia.org/wiki/Moment_distribution_method

Why this is required ? Can you explain about in ( without assuming that the total moment about C = 0 ) ?

 

[CivilSigma]

Why this is required ? Can you explain about in ( without assuming that the total moment about C = 0 ) ?

I'm not assuming anything... Ok let's walk through this:

1. We carried over 3162.9 to member CB
2. Now we want to fill out a new row, let's start with column CB
3. To get the Moment to Distribute: FEM Dist. = CO * (-1) * DF = 3162.9 * (-1) * (1) = -3162.9 Nm

Also,

Why this is required ?

this was a suggestion for you to continue itterating

 

[fonseh]

I'm not assuming anything... Ok let's walk through this:

1. We carried over 3162.9 to member CB
2. Now we want to fill out a new row, let's start with column CB
3. To get the Moment to Distribute: FEM Dist. = CO * (-1) * DF = 3162.9 * (-1) * (1) = -3162.9 Nm

Also,

this was a suggestion for you to continue itterating

Well , i ended up getting this one ... I found that my moment about A is no longer 2823.6... I just did it halfway , for line 10 , the moment -3176.4 is carried over to BC become -1588.2 (This is similar in the first example , where the moment CB is carried over to BC )

If it's wrong , why can't i do in this way ?

 

[fonseh]

I'm not assuming anything... Ok let's walk through this:

1. We carried over 3162.9 to member CB
2. Now we want to fill out a new row, let's start with column CB
3. To get the Moment to Distribute: FEM Dist. = CO * (-1) * DF = 3162.9 * (-1) * (1) = -3162.9 Nm

Also,

this was a suggestion for you to continue itterating

why there's no need to carry over the moment from CB (-3176.4) to BC become -1588.2 ??

 

[CivilSigma]

Hey Fonseh ,

I read over my notes, and looked over the example again.

I forgot that when we have a pin end (So pin C), the moment distribution method can be simplified by the following two steps:

1. Stiffness factor is taken to be K = 3 EI/L
2. Carry over factor to that member is 0 , not 1/2

And when you do so and try to continue iterating you will find that all your new Dist and CO moments are 0.

 

[fonseh]

Hey Fonseh ,

I read over my notes, and looked over the example again.

I forgot that when we have a pin end (So pin C), the moment distribution method can be simplified by the following two steps:

1. Stiffness factor is taken to be K = 3 EI/L
2. Carry over factor to that member is 0 , not 1/2

And when you do so and try to continue iterating you will find that all your new Dist and CO moments are 0.

why in example in 607.jpg , the moment carried over from BC to CB or vice versa is half ? Can you explain the differences for both cases?

 

[CivilSigma]

In figure 607, the author solved the example without using the simplifications mentioned in post #15, he used a stiffness factor of 4 EI/L for member CB and when you use that stiffness value, you must use a carry over factor of 1/2. So the exercise took longer to solve.

 

[fonseh]

In figure 607, the author solved the example without using the simplifications mentioned in post #15, he used a stiffness factor of 4 EI/L for member CB and when you use that stiffness value, you must use a carry over factor of 1/2. So the exercise took longer to solve.

We already know that for far end pinned / roller supported , the Stiffness factor is taken to be K = 3 EI/L , why the author use K = 4 EI/L in 607.jpg and use carry over factor of 0.5 ?
Is it wrong to do so ?

 

[CivilSigma]

The formula K = 4 EI/ L is derived from a beam that is assumed to be pin at one end and fixed at the other (pg : 509 of the textbook your using), and with this assumption, the carry over factor is calculated to be 1/2.

When we use K=4 EI/ L for the pin end support as the author did , we are basically assuming it to be fixed (when in reality it is not) AND after iterating through the long process we see that the actual moment at the assumed fixed member (which is an actual pin) is 0. So what does that mean? Well if M=0, then we only have internal axial and shear forces which are reminiscent to a pin/roller support which is the true configuration of the given problem. So, everything works out in the end.

So, it is not wrong to do so. You will get the same answer if you use K= 4 EI/L AND CO=1/2 or K=3 EI/L AND CO = 0 for the far pinned end.

 

[fonseh]

The formula K = 4 EI/ L is derived from a beam that is assumed to be pin at one end and fixed at the other (pg : 509 of the textbook your using), and with this assumption, the carry over factor is calculated to be 1/2.

When we use K=4 EI/ L for the pin end support as the author did , we are basically assuming it to be fixed (when in reality it is not) AND after iterating through the long process we see that the actual moment at the assumed fixed member (which is an actual pin) is 0. So what does that mean? Well if M=0, then we only have internal axial and shear forces which are reminiscent to a pin/roller support which is the true configuration of the given problem. So, everything works out in the end.

So, it is not wrong to do so. You will get the same answer if you use K= 4 EI/L AND CO=1/2 or K=3 EI/L AND CO = 0 for the far pinned end.

So , do you mean Since i know it's pinned end support , using K=3 EI/L AND CO = 0 will bring me to the final ans faster ?

 

[CivilSigma]

So , do you mean Since i know it's pinned end support , using K=3 EI/L AND CO = 0 will bring me to the final ans faster ?

Yes!

Edit: also a nice trick to see if your solution is correct is to see whether the calculated moment is 0 at end pins/rollers

 

[fonseh]

The formula K = 4 EI/ L is derived from a beam that is assumed to be pin at one end and fixed at the other (pg : 509 of the textbook your using), and with this assumption, the carry over factor is calculated to be 1/2.

When we use K=4 EI/ L for the pin end support as the author did , we are basically assuming it to be fixed (when in reality it is not) AND after iterating through the long process we see that the actual moment at the assumed fixed member (which is an actual pin) is 0. So what does that mean? Well if M=0, then we only have internal axial and shear forces which are reminiscent to a pin/roller support which is the true configuration of the given problem. So, everything works out in the end.

So, it is not wrong to do so. You will get the same answer if you use K= 4 EI/L AND CO=1/2 or K=3 EI/L AND CO = 0 for the far pinned end.

hi , can you help me in this question ? https://www.physicsforums.com/threads/carry-over-factor-in-beam-confusion.910499/