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Center of Gravity
- Thread starter lippie
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EnglishMuffin
Mechanical
What have externally applied loads got to do with the center of gravity ? To all intents and purposes, center of gravity = center of mass which depends only on the density.
I must agree with EnglishMuffin. Read STAAD's COG command and understand it! COG in STAAD commands returns the location of the COG for your structure model! It has nothing to do with the applied loads.
Now I think may be what you are looking for is the COG of a group of loads applied to a structure. To answer that, you can do it by hand. I do not think STAAD would do that for you.
As always, please understand the software that you are using before use it!
Good Luck.
Now I think may be what you are looking for is the COG of a group of loads applied to a structure. To answer that, you can do it by hand. I do not think STAAD would do that for you.
As always, please understand the software that you are using before use it!
Good Luck.
EnglishMuffin
Mechanical
Lutfi : Sorry, but there is no such thing as "the C.O.G. of a group of loads".
SlideRuleEra
Structural
Like EnglishMuffin, I am not aware of a "C.O.G. of a group of loads" but you could have a single equivalent resultant force from a group of loads.
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EnglishMuffin
Mechanical
If a body is in tranlational and rotational equilibrium under exactly three forces, there is a unique point through which they all must pass. However, this is not referred to as the "Center of Gravity", although it could happen to coincide with the COG in special cases. For any other number of forces acting on a body, in general all you can say is that they can be reduced to a single force (the resultant) at some arbitrary point, together with a couple.
EnglishMuffin
Mechanical
Sorry - missed your post - was posting at the same time.
The center of gravity, for all practical purposes, is the same as the center of mass.
The center of mass can be defined as follows :
"That point in a body about which the sum of the mass moments of all the individual masses constituting the body is zero".
So if there are j individual point masses denoted by m(j), and the vector position from an arbitrary origin be denoted r(j), the center of mass is located r from the origin, where :
r* Sum(m(j)) = Sum(m(j)*r(j))
So in the most general case, you would have to find it using calculus or numerical methods. However, if you have a body composed of masses with simple individual shapes having known centers of mass, you can find it quite simply.
The center of gravity, for all practical purposes, is the same as the center of mass.
The center of mass can be defined as follows :
"That point in a body about which the sum of the mass moments of all the individual masses constituting the body is zero".
So if there are j individual point masses denoted by m(j), and the vector position from an arbitrary origin be denoted r(j), the center of mass is located r from the origin, where :
r* Sum(m(j)) = Sum(m(j)*r(j))
So in the most general case, you would have to find it using calculus or numerical methods. However, if you have a body composed of masses with simple individual shapes having known centers of mass, you can find it quite simply.
EnglishMuffin
Mechanical
By the way, there seems to be a lot of confusion about this rather basic topic, and there is presently another thread which someone has started on the subject : thread727-68451. This thread also seems to contain some rather incoherent statements (at least in my opinion).
EnglishMuffin
Mechanical
By the way AGAIN :
Although I don't know anything about STAAD, I can't believe that STAAD's COG command won't give you the center of gravity, if that is indeed what you are trying to find.
Although I don't know anything about STAAD, I can't believe that STAAD's COG command won't give you the center of gravity, if that is indeed what you are trying to find.
What I was talking about is to equivalent concentrated load (ECL) to replace several concentrated and even partial uniform loads.
Sorry for the confusion. STAAD will print the CG of the structure just as fleisher stated. But that will give the center of gravity for the frame/structure.
Sorry for not being more clear in my response.
Sorry for the confusion. STAAD will print the CG of the structure just as fleisher stated. But that will give the center of gravity for the frame/structure.
Sorry for not being more clear in my response.
I am not sure what Lippie is after.
By definition COG has nothing to do with force. It is simply the centre of mass. The formula put forward by EnglishMuffin adequately defines it. And if Lippie wants it for a 3D structure it is simply a matter going through the 3 axes. I have done a fair share of it with turbine supporting structures. There is a such thing called Empirical rules in CP 2012 where COG of the combined foundation and the machine has to be within the centroid of the bearing surface by a specified tolerance in order to achieve uniform settlements. In today's computer age the calculation of a global COG can be easily performed by breaking down a complicated structure into a series of brick and cylindrical elements with known local COG.
I believe the confusion here could have been the use of terms. May be the word "resultant" is better for describing the summation of a series of forces and the COG for forces be replaced with loading centroid.
To illustrate my point, I would suggest the COG of a cantilever, with uniform section and loaded with a point load at the free end, is at its mid-span. The centriod of the external load application is at the free end and the centroid of the internal reaction is at the fixed end.
For any force system there should be a centroid for the external loads and another centroid for the inernal reactions. The two hold each other in equilibrium, sometimes with the presence of couples or moments. That is my experience from analysing complicated reinforced concrete sections (may contain additional material like carbon fibre laminates) subjected to biaxial bending.
By definition COG has nothing to do with force. It is simply the centre of mass. The formula put forward by EnglishMuffin adequately defines it. And if Lippie wants it for a 3D structure it is simply a matter going through the 3 axes. I have done a fair share of it with turbine supporting structures. There is a such thing called Empirical rules in CP 2012 where COG of the combined foundation and the machine has to be within the centroid of the bearing surface by a specified tolerance in order to achieve uniform settlements. In today's computer age the calculation of a global COG can be easily performed by breaking down a complicated structure into a series of brick and cylindrical elements with known local COG.
I believe the confusion here could have been the use of terms. May be the word "resultant" is better for describing the summation of a series of forces and the COG for forces be replaced with loading centroid.
To illustrate my point, I would suggest the COG of a cantilever, with uniform section and loaded with a point load at the free end, is at its mid-span. The centriod of the external load application is at the free end and the centroid of the internal reaction is at the fixed end.
For any force system there should be a centroid for the external loads and another centroid for the inernal reactions. The two hold each other in equilibrium, sometimes with the presence of couples or moments. That is my experience from analysing complicated reinforced concrete sections (may contain additional material like carbon fibre laminates) subjected to biaxial bending.
ibrahimdemir
Mechanical
Lippie,
Sometimes, you may simulate forces by using lumped masses at their effective locations in the model. So, all the masses will be in the model and there will not be a problem getting COG through the analysis software.
If you modelled some masses as forces introduced onto the structural members and trying to find out where the COG is, you need to work out matematically as EnglishMuffin suggested.
If you are trying to check the stability of the structure and trying to find where the equivalent external force is acting, you can still find it matematically by replacing Sum(m(j)) by Sum(F(j)) in the same example.
Hope this helps.
Ibrahim Demir
Sometimes, you may simulate forces by using lumped masses at their effective locations in the model. So, all the masses will be in the model and there will not be a problem getting COG through the analysis software.
If you modelled some masses as forces introduced onto the structural members and trying to find out where the COG is, you need to work out matematically as EnglishMuffin suggested.
If you are trying to check the stability of the structure and trying to find where the equivalent external force is acting, you can still find it matematically by replacing Sum(m(j)) by Sum(F(j)) in the same example.
Hope this helps.
Ibrahim Demir
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