# Sign of moment in buckling of column

## Homework Statement

Can someone explain why the M is assigned to be anticlockwise here ?

## The Attempt at a Solution

When i assign it as clockwise , i will get -P(δ -v) , which is different from the author ... Can i do so ? Why ?

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for the second example here , i can understand that M = -Pv , since M+Pv = 0 at either end when it's in equlibrium. ( I have showed in the working)

P/s : I know the the sign convention of the bending moment of beam is positive when the beam upwards as shown ...

For the first example in post#1 , i gt M+P(∂-v) = 0 , so M = - P(∂-v) .
I am not sure whether is my concept correct or not . My working is in the 3rd photo here

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Is my concept wrong ? can someone explain it ?

PhanthomJay
Homework Helper
Gold Member
Draw a FBD appropriately of an arbitrary cut section of the top part of the deflected column and determine the direction and magnitude of the moment of the applied force P. What must be the direction and magnitude of the internal moment at the cut?

The moment M has to be counter clock wise at the base because from the given figure the applied force P will try and rotate the column clockwise

fonseh
The moment M has to be counter clock wise at the base because from the given figure the applied force P will try and rotate the column clockwise
How about the case in post#2 ? Why
M = -Pv ? For this case , why the M at the bottom = clockwise? since to balance the anticlockwise moment produced by P , the Moment has to be clockwise to counter the effect , am i right ? Just like the case in post#1 ...

How about the case in post#2 ? Why
M = -Pv ? For this case , why the M at the bottom = clockwise? since to balance the anticlockwise moment produced by P , the Moment has to be clockwise to counter the effect , am i right ? Just like the case in post#1 ...
Yes, that's correct. So, note that in post#2 , the assumed direction of the moment in figure (b) is incorrect as it is drawn counter clockwise.

fonseh
Yes, that's correct. So, note that in post#2 , the assumed direction of the moment in figure (b) is incorrect as it is drawn counter clockwise.
So , in notes 2 , the assumed direction of the moment should be in clockwise direction . So , M = Pv ?
But , i checked out so many books and so may links , they still give M = -Pv , so are they all wrong ?

So , in notes 2 , the assumed direction of the moment should be in clockwise direction . So , M = Pv ?
But , i checked out so many books and so may links , they still give M = -Pv , so are they all wrong ?

No , M=-Pv is correct for the assumed internal moments (positive is counter clockwise) as seen in figure (b). The assumed internal moment is counter clockwise in direction and the force P also rotates the column in a counter clockwise motion. So, ΣMoments = 0 , M+Pv=0 , M=-Pv.

Obviously, since the column is assumed to be in equilibrium, the internal moment will actually be clockwise. This is exactly what the equation tells us, M= - Pv, the negative sign means that the assumed internal moment is equal to PV but is clock wise. But the example doesn't explicitly state this. For mathematical consistency we assign clockwise moments a negative sign, and counter clock wise moments a positive sign.

fonseh
No , M=-Pv is correct for the assumed internal moments (positive is counter clockwise) as seen in figure (b). The assumed internal moment is counter clockwise in direction and the force P also rotates the column in a counter clockwise motion. So, ΣMoments = 0 , M+Pv=0 , M=-Pv.

Obviously, since the column is assumed to be in equilibrium, the internal moment will actually be clockwise. This is exactly what the equation tells us, M= - Pv, the negative sign means that the assumed internal moment is equal to PV but is clock wise. But the example doesn't explicitly state this. For mathematical consistency we assign clockwise moments a negative sign, and counter clock wise moments a positive sign.
Do you mean for all the cases in post 1 and post 2 , the author assumed the internal moment is in counter clockwise direction ?

Do you mean for all the cases in post 1 and post 2 , the author assumed the internal moment is in counter clockwise direction ?
Yes, but in case 1 it is not an internal moment but a reaction moment

fonseh
Do you mean for all the cases in post 1 and post 2 , the author assumed the internal moment is in counter clockwise direction ?
It seems that the theory that the author assumed that the moment is counterclckwise all the time is incorrect here . I have 2 example below (from another book) . We can see that in the firstexample ( photo 1 and photo 2) book , the assumed internal moment is clockwise , for another case(photo 3 and photo4) , we can see that the reaction moment is clockwise

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You can assume any direction to be positive, but for that particular problem, you need to be consistent and carry out the same assumption.

fonseh
You can assume any direction to be positive, but for that particular problem, you need to be consistent and carry out the same assumption.
please refer to the edited post in post #12 .

You can assume any direction to be positive, but for that particular problem, you need to be consistent and carry out the same assumption.
It seems that the author of the example in the book that i uploaded doesnt follow the rules , or i missed out something ? can you explain it ?

You can assume any direction to be positive, but for that particular problem, you need to be consistent and carry out the same assumption.
I also found that the assumed internal moment always has the same direction with the applied force moment in both example , is this true ? why we need to make that assumption ?

Assuming a direction is all but a perspective for positive and negative, have a read about the conventional sign convention:
https://en.wikipedia.org/wiki/Shear_and_moment_diagram#Convention

Also,
I also found that the assumed internal moment always has the same direction with the applied force moment in both example
This is not true, it depends on the situation of analysis. In post #12, question 4.4, the internal moment is drawn counter clockwise and the force P rotates the column clockwise.

fonseh
Assuming a direction is all but a perspective for positive and negative, have a read about the conventional sign convention:
https://en.wikipedia.org/wiki/Shear_and_moment_diagram#Convention

Also,

This is not true, it depends on the situation of analysis. In post #12, question 4.4, the internal moment is drawn counter clockwise and the force P rotates the column clockwise.
Well , i agree that in 4.4 the assumed moment is counterclockwise , because the moment will turn the beam in U shape (smile curve) . But , i dont understand why in 4.3 , the assumed moment is clockwise , why not anticlockwise ? Because when the assumed moment is anticlockwise , it will bend the beam in U shape

It doesn't really matter what direction you assume the moment is as long as you adopt a consistent general sign convention: Assume that all counter clockwise moments are'+' and all clock wise moments are '-'.

Say in our FBD, we drew the unknown internal moment ( call it 'M') counter clockwise, then taking sum of the moments and using the above statement:
(+M) + (-PV) = 0 , then M=PV. How do we interpret this result? It means that the internal moment is of counter cockwise orientation and of magnitude PV.

Say in our FBD, we drew the unknown internal moment ( call it 'M') clockwise, then taking sum of the moments and using the above statement:
(-M) + (-PV) = 0 , then M=-PV. How do we interpret this result? It means that the internal moment is not of clockwise direction (because our answer is negative) , it is of counter clock wise direction and has magnitude PV.

Have a watch:

So , shouldn,t the assumed moment is anticlockwise at the beam to enable it to have a curve U shape ? Just like the case below

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Say in our FBD, we drew the unknown internal moment ( call it 'M') counter clockwise, then taking sum of the moments and using the above statement:
(+M) + (-PV) = 0 , then M=PV. How do we interpret this result? It means that the internal moment is of counter cockwise orientation and of magnitude PV.
You are referring to which case ? photo in 483 or photo in 485 ? I'm confused

It doesn't really matter what direction you assume the moment is as long as you adopt a consistent general sign convention: Assume that all counter clockwise moments are'+' and all clock wise moments are '-'.

Say in our FBD, we drew the unknown internal moment ( call it 'M') counter clockwise, then taking sum of the moments and using the above statement:
(+M) + (-PV) = 0 , then M=PV. How do we interpret this result? It means that the internal moment is of counter cockwise orientation and of magnitude PV.

Say in our FBD, we drew the unknown internal moment ( call it 'M') clockwise, then taking sum of the moments and using the above statement:
(-M) + (-PV) = 0 , then M=-PV. How do we interpret this result? It means that the internal moment is not of clockwise direction (because our answer is negative) , it is of counter clock wise direction and has magnitude PV.

Have a watch:
Ya, i understand hat . But , when i try out M = Py , where EI(d2y.dx2) = Py , i gt different things with the derived equation , why cant i do so ?

PhanthomJay
Homework Helper
Gold Member
The negative sign in the "-Py" term is correct, although when you assume the correct direction of the internal moment for the 'pinned-pinned' case, you end up with "+Py", and get the wrong signage in the differential equation . This paradox has confused me for years. I believe the paradox is resolved when you consider that the internal moment is , when written in terms of the curvature - stiffness relationship where the absolute value of M = EI(d^2y/dx^2), that actually M= - EI(d^2y/dx^2), because the curve is concave with a negative curvature. Unlike the cantilever case in you first post, where M is shown correctly and the curvature is convex or positive.

fonseh
, because the curve is concave with a negative curvature
How do you know that ? When we view it from different sides , it can be negative and positive curvature , right ?

The negative sign in the "-Py" term is correct, although when you assume the correct direction of the internal moment for the 'pinned-pinned' case, you end up with "+Py", and get the wrong signage in the differential equation . This paradox has confused me for years. I believe the paradox is resolved when you consider that the internal moment is , when written in terms of the curvature - stiffness relationship where the absolute value of M = EI(d^2y/dx^2), that actually M= - EI(d^2y/dx^2), because the curve is concave with a negative curvature. Unlike the cantilever case in you first post, where M is shown correctly and the curvature is convex or positive.
Do you mean the author suume that the right is positive axis ? How about the case where the beam is curve to the left ( the max/min point located at the left) ? if this case , M= Py ?

When we apply the axial forces at the end , the beam can be deflected to both right and to the left , right ? just like the below , but , how do we determine that the beam is deflected to which direction ?

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