# Twisting at shear center / centroid

• fonseh
In summary: O. Since we just defined O as a distance e from the wall we need to specify another coordinate to show where the point O is.In summary, the author discusses the concept of shear center and its relation to the twisting and deflection of a beam. When a force is applied through the centroid along an axis of symmetry, the beam only deflects downward. However, when the force is applied through the centroid but there is no axis of symmetry, an imbalance in shear flow distribution causes a net moment and results in twisting and deflecting of the beam. To prevent this, the force can be moved towards the side with more shear distribution. The shear center, which is located at a distance from the wall defined by the
fonseh
In the notes , the author stated that when the force is applied through the centorid of cross section , the channel will bend and twist.
but , on the second page , the author stated that the shear center lies on an axis of symmetry of member's cross sectional area...
So, i am confused whether the member will twist or not when the force P is applied thru the centroid or shear center ...

Why the shear center is located at O , which is located outside of the C channel ?

In the notes , it's also stated that to prevent twisting , the P should be applied at O , does it mean we have to apply the force thru the centoid and also at the O at the same time to prevent twisting ? I'm confused
Is there anything wrong with the notes ?

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When you apply a force on the beam through the centroid along an axis of symmetry for the cross section ( see fig 7.25a) then the beam just deflects downward. This is to be expected and seems reasonable.

Now let's apply that same force the way it is shown in Fig 7.24a, that is through the centroid, but this time there is not an axis of symmetry for the cross section (by definition of the centroid there will still be an equal amount of the beam on either side of the imaginary line drawn through the cross section by this force).
However, if you look at 7.24b we see that even though the beam material is split evenly by area with this force, the geometries are different and we see this causes more shear flow distribution on one side of the cross section.

This imbalance of shear flow distribution will cause a net moment and results in the member twisting as well as deflecting.

NOTE: if we go back and look at the first case where we had a force through the centroid BUT it is symmetrical that shear flow distribution will be equal and no net moment will occur, hence no twisting just deflection.

Well now that we know if the shear distribution is not equally balanced on both sides we get a net moment and twisting occurs.
So it makes sense that if we move that force over a bit toward the side with more shear distribution then we could rebalance them and then we would not have a moment, hence no twisting.

Looking at that shear distribution in fig 7.24b if you wanted to draw a line through it an have the same area on both sides (which balances the moments) that line ends up outside the C channel, might not always be the case but for this example it is.

So that's why it ends up outside the C channel, because along that axis the shear distributions are equal on each side and no net moment occurs. This distance is defined by the ecentricity e .

When the Book says "The shear center will always lie on an axis of symmetry they are just talking about that point O. Since we just defined O as a distance e from the wall we need to specify another coordinate to show where the point O is.

That is a confusing topic and that book does not do a great job at explaining it, hope this helps.

fonseh
pistolpete said:
see fig 7.25a
i assume you say 7.23a ?

pistolpete said:
However, if you look at 7.24b we see that even though the beam material is split evenly by area with this force, the geometries are different and we see this causes more shear flow distribution on one side of the cross section.

This imbalance of shear flow distribution will cause a net moment and results in the member twisting as well as deflecting.
do you mean the beam in 7.24b undergo deflection and also twisting ?

pistolpete said:
When you apply a force on the beam through the centroid along an axis of symmetry for the cross section ( see fig 7.25a) then the beam just deflects downward. This is to be expected and seems reasonable.

Now let's apply that same force the way it is shown in Fig 7.24a, that is through the centroid, but this time there is not an axis of symmetry for the cross section (by definition of the centroid there will still be an equal amount of the beam on either side of the imaginary line drawn through the cross section by this force).
However, if you look at 7.24b we see that even though the beam material is split evenly by area with this force, the geometries are different and we see this causes more shear flow distribution on one side of the cross section.

This imbalance of shear flow distribution will cause a net moment and results in the member twisting as well as deflecting.

NOTE: if we go back and look at the first case where we had a force through the centroid BUT it is symmetrical that shear flow distribution will be equal and no net moment will occur, hence no twisting just deflection.

Well now that we know if the shear distribution is not equally balanced on both sides we get a net moment and twisting occurs.
So it makes sense that if we move that force over a bit toward the side with more shear distribution then we could rebalance them and then we would not have a moment, hence no twisting.

Looking at that shear distribution in fig 7.24b if you wanted to draw a line through it an have the same area on both sides (which balances the moments) that line ends up outside the C channel, might not always be the case but for this example it is.

So that's why it ends up outside the C channel, because along that axis the shear distributions are equal on each side and no net moment occurs. This distance is defined by the ecentricity e .

When the Book says "The shear center will always lie on an axis of symmetry they are just talking about that point O. Since we just defined O as a distance e from the wall we need to specify another coordinate to show where the point O is.

That is a confusing topic and that book does not do a great job at explaining it, hope this helps.
I'm curious how is the force applied directly at the centroid so that no twisting occur and only deflection occur ? in figure 7.23a , we can only apply the force on top of the c beam , right? how it is possible to apply the force directly at the centoid ?

pistolpete said:
When you apply a force on the beam through the centroid along an axis of symmetry for the cross section ( see fig 7.25a) then the beam just deflects downward. This is to be expected and seems reasonable.

Now let's apply that same force the way it is shown in Fig 7.24a, that is through the centroid, but this time there is not an axis of symmetry for the cross section (by definition of the centroid there will still be an equal amount of the beam on either side of the imaginary line drawn through the cross section by this force).
However, if you look at 7.24b we see that even though the beam material is split evenly by area with this force, the geometries are different and we see this causes more shear flow distribution on one side of the cross section.

This imbalance of shear flow distribution will cause a net moment and results in the member twisting as well as deflecting.

NOTE: if we go back and look at the first case where we had a force through the centroid BUT it is symmetrical that shear flow distribution will be equal and no net moment will occur, hence no twisting just deflection.

Well now that we know if the shear distribution is not equally balanced on both sides we get a net moment and twisting occurs.
So it makes sense that if we move that force over a bit toward the side with more shear distribution then we could rebalance them and then we would not have a moment, hence no twisting.

Looking at that shear distribution in fig 7.24b if you wanted to draw a line through it an have the same area on both sides (which balances the moments) that line ends up outside the C channel, might not always be the case but for this example it is.

So that's why it ends up outside the C channel, because along that axis the shear distributions are equal on each side and no net moment occurs. This distance is defined by the ecentricity e .

When the Book says "The shear center will always lie on an axis of symmetry they are just talking about that point O. Since we just defined O as a distance e from the wall we need to specify another coordinate to show where the point O is.

That is a confusing topic and that book does not do a great job at explaining it, hope this helps.
Bump

I aplogize for the confusion, I have that book but a different edition so I had the wring numbers for the figures I referenced, I went through and changed them to what they should be for the pages you posted.

pistolpete said:
When you apply a force on the beam through the centroid along an axis of symmetry for the cross section ( see fig 7.24a) then the beam just deflects downward. This is to be expected and seems reasonable.

Now let's apply that same force the way it is shown in Fig 7.23a, that is through the centroid, but this time there is not an axis of symmetry for the cross section (by definition of the centroid there will still be an equal amount of the beam on either side of the imaginary line drawn through the cross section by this force).
However, if you look at 7.23b we see that even though the beam material is split evenly by area with this force, the geometries are different and we see this causes more shear flow distribution on one side of the cross section.

This imbalance of shear flow distribution will cause a net moment and results in the member twisting as well as deflecting.

NOTE: if we go back and look at the first case where we had a force through the centroid BUT it is symmetrical that shear flow distribution will be equal and no net moment will occur, hence no twisting just deflection.

Well now that we know if the shear distribution is not equally balanced on both sides we get a net moment and twisting occurs.
So it makes sense that if we move that force over a bit toward the side with more shear distribution then we could rebalance them and then we would not have a moment, hence no twisting.

Looking at that shear distribution in fig 7.23b if you wanted to draw a line through it an have the same area on both sides (which balances the moments) that line ends up outside the C channel, might not always be the case but for this example it is.

So that's why it ends up outside the C channel, because along that axis the shear distributions are equal on each side and no net moment occurs. This distance is defined by the ecentricity e .

When the Book says "The shear center will always lie on an axis of symmetry they are just talking about that point O. Since we just defined O as a distance e from the wall we need to specify another coordinate to show where the point O is.

That is a confusing topic and that book does not do a great job at explaining it, hope this helps.

fonseh
fonseh said:
do you mean the beam in 7.24b undergo deflection and also twisting ?

I assume this question was due to my error on the figure references, so no this beam would just deflect downward. This is because there would be no net moment due to shear distribution (same on each side, but opposite so net moment is 0).

fonseh said:
I'm curious how is the force applied directly at the centroid so that no twisting occur and only deflection occur ? in figure 7.23a , we can only apply the force on top of the c beam , right? how it is possible to apply the force directly at the centoid ?

Yes, if we are thinking about an actual beam we would apply that force to the top through the centroid, and as the beam deflected and twisted that point where we applied the load would not line up with the centroidal axis of the beam anymore.

SO the book is basically just saying to arbitarilly consider a loading through the centroid to use for analysis of what is happening.

Does that make sense?

To try to summarize the concept, if a load is applied along centroidal axis the shear distribtions will create moments (for symmetrical these will be equal and opposite and no net moment => No twisting occurs, just deflection). If the cross section of the beam is not symmetrical the moments may not be equal and a net moment occurs =>twisting and deflection of the beam occurs.

So if we do not want twisitng we need to get that moment to be equal and opposite in magnitude, this is achieved by offsetting the loading, the distance we need to do so is achieved by finding e.

fonseh
pistolpete said:
Yes, if we are thinking about an actual beam we would apply that force to the top through the centroid, and as the beam deflected and twisted that point where we applied the load would not line up with the centroidal axis of the beam anymore.

SO the book is basically just saying to arbitarilly consider a loading through the centroid to use for analysis of what is happening.

Does that make sense?

To try to summarize the concept, if a load is applied along centroidal axis the shear distribtions will create moments (for symmetrical these will be equal and opposite and no net moment => No twisting occurs, just deflection). If the cross section of the beam is not symmetrical the moments may not be equal and a net moment occurs =>twisting and deflection of the beam occurs.

So if we do not want twisitng we need to get that moment to be equal and opposite in magnitude, this is achieved by offsetting the loading, the distance we need to do so is achieved by finding e.
In 7-23 a , the author apply the force thru the centroid , why twisting will also occur ? The shear center in the beam also equals to the centroid of the beam , right ?