How to calculate the center of gravity of a steel pipe with a weight to the side (middle)? 1

[StathPol12]

Hi guys,

I have a steel pipe with the dimensions shown below:

Length of the pipe = 6.4 metres.
External diameter = 2.48 metres.
Internal diameter = 2.46 metres.
Self-weight of the steel pipe = 6766.85 kg (uniform steel pipe).

A steel door of 120 kg is bolted to the right side of the steel pipe, in the middle. Can you please help me with calculating the centre of gravity?

Please see attached image.

 

[Luceid]

Responding here rather than the other post, I would delete the second post in the structural engineering forum.

I like to break up a problem like this into a group of 2D views so we can take it one step at a time! Especially in this case, since there is a fair bit of symmetry. Let's look at it in the direction of the pipe axes.
Overall:

Looking down the Y-Axis

Looking down the X-Axis

Looking down the Z-Axis

For each of these, you can start to get a sense of symmetry, if the shape is symmetric across its axis, you can visually get a sense of where you expect the centroid to be. You can get to the same conclusions by using the equations in my last post. I'll give you one example, you give me the other two.

To find the "Z" coordinate of the center of gravity:

W1 = 6766 kg (Pipe Weight)
Z1 = 0m (Distance from Pipe Centroid to origin, in the z-direction)

W2 = 120 kg (Door Weight)
Z2 = 0m (Distance from Door Centroid to origin, in the z-direction)

Sum of Wi * zi = 6766*0 + 120*0 = 0
Sum of Wi = 6766+120 = 6886

Wi*zi / Wi = 0, so your z-coordinate of the origin is 0.

Try the same equations with x and y and let me know how it works out. This will give you a final centroid coordinate of (x,y,0).

 

[waross]

Getting close.

Wi*zi / Wi = 0, so your z-coordinate of the origin is 0.

Valid for the two end sections but not for the section where the door is mounted.
Overlooked, the section of the pipe where the undimensioned door is mounted.
Missing information:
How much of the pipe was removed to create the doorway?
Calculate the CoG of the removed section and reverse the sign for that section of pipe.
Now calculate the CoG of the door and the door surround. The 120 kg will be acting at the CoG of the door assembly.
If the door is 2 meters wide, it may be possible to locate it at such a radius that it doesn't change the CoG of the pipe.
Tweaking the width of the door around 2 meters will probably allow you to mount it so that the CoG is not changed.
(If that is any advantage.)

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[rb1957]

the point is to set up a co-ordinate system (like Luceid shows), or you could use the mid-length of the pipe as your origin (so that the co-ordinate of the pipe CG is (0,0,0) .... this helps later.
then to define the CG of each part in that co-ordinate system (as waross notes, you haven't specified the zCG of the "door") ... "middle" may define the lengthwise position, but how about the vertical or the lateral (the Y direction) ? As this is school work, the idea may be to assume something ... state your assumptions !

then it is (for X_CG ... sum the (Weight*X_CG) for each part (so that if the origin is at the CG of the pipe this is zero for the pipe) and divide by the sum of the weights of the parts

that is X_CG = weight_door*X_CG_door/(weight_pipe + weight_door)

so you need to define the position of the door.
also, define if the pipe is cut away behind the door (which seems reasonable) ... if it is the easy way is to count this as negative weight and look up in a "strength of materials" text the CG for a sector (the piece of pipe you've cut away).

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[StathPol12]

So, the centre of gravity in the x direction is

SFpipe = 6766 kg
SFdoor = 120 kg
Centre of gravity at x direction: ((6766 x 1.24) + (120 x 2.48)) / (6766 + 120) = 1.26 metres (From Point A). Is this correct?

 

[rb1957]

1) why put your origin at point A, when the center of the circle is more convenient ?
2) you are assuming that there is no hole in the pipe under the door ... you should state this.
3) it is much better practice to show the axes going thru the origin, rather than some random point. The Y-axis should be up through point A.
4)"SF" ? Shear Force ? Weight is a much better term, when calculating a CG.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[dik]

I'd have used the centre of the steel pipe, but it doesn't matter where you reference the origin. It's the methodology... and weight is generally in Newtons, not kg... Is there any material that connects the door to the pipe... that would add to the calculations.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[StathPol12]

Thanks for the feedback. Just posting Luceid's diagrams but with axis x(horizontal), y (vertical), z (depth). Just still working to find the COG for each (x,y,z coordinate).

Self weight of pipe (SFpipe) = 6766 kg
Self weight of door (SFdoor)= 120 kg

"Y" coordinate of the center of gravity:

SFpipe = 6766 kg (Pipe Weight)
a1 -> Distance of the pipe centroid to origin (assume that point A is the origin),in the y-direction = 0m

SFdoor = 120 kg (Door Weight)
a2 -> Distance of the door centroid to origin (assume that point A is the origin),in the y-direction = 0m

Sum = (SFpipe * a1) + (SFdoor * a2) = (6766*0 + 120*0) = 0 kgm
Sum of self-weights = 6766+120 = 6886 kg
Sum / Sum of self-weights = 0, y-coordinate of the origin is 0.

"X" coordinate of the center of gravity


SFpipe = 6766 kg (Pipe Weight)
b1 -> Distance of the pipe centroid to origin (assume that point A is the origin),in the x-direction = 0m

SFdoor = 120 kg (Door Weight)
b2 -> Distance of the door centroid to origin (assume that point A is the origin),in the x-direction = 1.24m

Sum = (SFpipe * b1) + (SFdoor * b2) = (6766*0 + 120*1.24) = 148.8 kgm
Sum of self-weights = 6766+120 = 6886 kg
Sum / Sum of self-weights = 148.8/6886 = 0.021 metres to the right from point A (I feel my calculation is wrong).

 

[dik]

Does anyone have a weight and centroid for the attached?

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[Luceid]

Waross, I suppose you may be right, I was assuming the problem was simpler than it may be. I had taken the door as a simple attachment, with no missing pipe, because from the information given there’s no way to tell how much is removed.

 

[Luceid]

Stahl, if you don’t need to account for the missing pipe, your calc seems to be in the right ballpark. I would expect the centroid to shift slightly to the right since the door creates an unsymmetric section, and the assembly is slightly heavier on the right of the x direction. I would also expect it to be not very far to the right, since the door weight is small compared to the pipe weight. So I would review the comments others have made about the possible missing pipe material but otherwise you are getting the hang of it.

 

[dik]

I missed a part...

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[rb1957]

so the piece of pipe removed for the transverse pipe ?

1st principles ... integration of d_theta ?

nah, too much like work ... model it in CAD ...

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Snickster]

If you sum weights about the center of the pipe the C.G, would be located to the right of the center by:

(120 x 1.24)/(6766.85 + 120) = 0.0216 meters to right of centerline.

 

[StathPol12]

Thanks guys,

I have taken all the comments and created a word file with the details of the pipe and the centre of gravity for the x and y coordinates. Still trying to work out the centre of gravity for z direction . Can you please have a look if you have a few minutes free time and let me know what you think?

 

[rb1957]

why wouldn't the z_CG be zero ? assuming the door is centered (it's not dimensioned so it's open for assumption).

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[3DDave]

I am more surprised the CAD system doesn't do the original calculation.

 

[StathPol12]

How do we work out the load distribution if the steel pipe (with the door) is lifted by 4 chains (i.e. the point loads acting on the beam)? Assume that there are 4 lifting attachment points on the steel pipe where they will be used to mount the chains.

We know that the combined self-weight of steel pipe (66.4 kN) and steel door (1.2 kN) is 67.6 kN). And we know that the center of gravity due to that door is:

x - coordinate = 0.022 meters
y - coordinate = 0 meters
z - coordinate = 0 meters


If the there was no steel door, then, we would divide 67.6 kN / 4 and each chain would take a quarter of the load.

 

[Snickster]

Couldn't you just look at it as a 2-dimensional problem in the Y-X plane and sum forces and moments = 0.

 

[StathPol12]

We could if we didn't have the extra weight...How do I put the shift of the CoG to the equation? If this is going to be lifted by 4 chains, then it would be the total weight divided by 4..but how do we consider the shift of CoG 0.023 m to the right?