×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

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

 Forum Search FAQs Links MVPs

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

(OP)
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.

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

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).

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

Getting close.

#### Quote:

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!

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

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.

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

(OP)
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?

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

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.

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

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

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

(OP)
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). :(

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

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

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

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.

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

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.

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

I missed a part...

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

-Dik

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

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.

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

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.

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

(OP)
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?

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

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.

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

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

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

(OP)
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.

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

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

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

(OP)
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?

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

The width of the chain anchor points relative to the pipe diameter is significant.
You are showing the chains at an angle.
The angle is significant.

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

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

Given the geometry I think it isn't redundant. Cutting any of the four chains/cables would allow the assembly to move to a new equilibrium position.

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

Depending on the anchorage location, cutting a chain can become unstable. You cannot push on a chain, if I recall. I always like saddles, myself.

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

-Dik

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

you know the weight of the total thing (pipe and door) and you know the CG of the total thing ... put the weight at the CG and calculate from there.

you can analyze like a simply supported beam ... read up on that. Basically look at the XZ plane ... weight down, and two vertical reactions (two because the two points at each end line up, yes?). Your CG is at the mid-span yes? if the lift points are equally distant from the mid-pt (from the CG) then each end will support 1/2 the weight. But how to distribute this between the two lifts (at either end).

look at the pipe in the YZ plane. Again the weight is down and in this view you see two lifts (for the same reason as before, yes?) but here the CG is slightly to one side (there is a small Y value to the total CG). So the XZ solution told you that the weight at the ends is 1/2 the weight) so now you can solve the vertical reactions for the load (weight) not being exactly in the middle.

So now you know the vertical load in each lift. But your lifts don't "look" vertical. If the lifts are angled find the length of the lift and it's vertical component, then the load in the inclined lift = (lift_length/lift_vertical_component)*vertical_load. yes?

And Greg has a point, you should state your assumptions (in order to solve the indeterminate problem). And the key assumption is that each support has the same stiffness (and you should understand why).

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

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

(OP)
rb1957 The chains will be positioned vertically, I did a small error on the previous drawing - just spent some time to correct it and here how it looks. I also did a small error in the weight of the pipe and corrected it. We have the following:

We know that:

- Weight (self-weight) of the pipe is 64.3 kN (6550 kg)
- Weight (self-weight) of the door is 1.2 kN (120 kg)
- Total self-weight of pipe and door is 65.5 kN (ignore the saddles for now).

From the calculations, we also know that the center of gravity due to the door attachment will shift 0.023 m to the right from the center point of the pipe (If we assume the datum is the center of the pipe only).

I am trying to work out the load each chain will take in order to put them as point loads on the steel beams. We know that the chains to the side where the door is located will take slightly higher point loads than the other side

What confuses and cannot work out the load in each chain, is how to consider the shift of the center of gravity (0.023m) due to the door. Lifting points are at quarter points (1.6 m from each end of the 6.4 m pipe).

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

think about it ... I'm not going to do it for you.

I've said each end (with two chains) reacts 1/2 the load ... and you understand why ?
Look at the structure in the YZ plane ... things appear, at least, to be symmetric.

So consider each end in the XY plane, with two chains and 1/2 the load, as a simply supported beam; look this up if you don't know what that means for the analysis.
The load is positioned at the X CG, slightly off the mid-point between the two lifts.
I may have confused you as I had mentally up the Z axis up, X along, Y lateral (to the Left, why?)

and what are the two grey things, joining the lifts ...how much do they weigh ? ('cause the chains supports their weight too)

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

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

(OP)
I took the XY plane put the total load of 65.5 kN (sum of pipe and door) on the shifted center of gravity point as shown above. Then, I solved it as a simply supported beam.

If this is correct, then

16.0715 kN on the each of the two chains located to the side of the pipe without the door
16.6785 kN on the each of the two chains located to the side of the pipe with the door

(16.0715 x 2) + (16.6785 x 2)= 65.5 kN..no?

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

that seems reasonable.

what about the grey things (joining the lifts) ?

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

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

#### Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!