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# centre of area/centre of mass

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1. I am trying to locate the centroid of the area (or centre of mass) of a segment of an arch ring. I have calculated the centroid shown by approximately breaking the curve into equivalent triangles and squares and applying the parallel axis theorem, however am sure the results have significant error. I Know the radius and can calculate the formula of the curve if this helps.

Any help would be much appreciated...

p.s. only interested in finding the horizontal distance (i.e. x-axis)
Attached Images

2. (Original post by andymorl)
I am trying to locate the centroid of the area (or centre of mass) of a segment of an arch ring. I have calculated the centroid shown by approximately breaking the curve into equivalent triangles and squares and applying the parallel axis theorem, however am sure the results have significant error. I Know the radius and can calculate the formula of the curve if this helps.

Any help would be much appreciated...

p.s. only interested in finding the horizontal distance (i.e. x-axis)
Do we know the width of the arch?

And does it have constant density - e.g. stone or cast metal - or does it vary by the width - e.g. wood that's been bent - or something else entirely.

Knowing these, then working in polar coordinates, we can get an exact CofM.
3. Thanks for the reply, the bridge is assumed to be a constant material through out. Please find attached a more detail picture of what I am trying to achieve. The left hand top picture is the full half of the span of the bridge (not interested in the other half). It is cut vertically through the crown section (top section), the inner radius is given and the thickness of the arch is 1.5m at the crown and increases to 2m at the end. Further the arch is subdivided into six sections parallel to the vertical through the crown at an angle of 4.6 degree.

I am trying to find the centre of mass/centre of area of each of these segments as the arch reduced or increased by one segment from the crown to the end of one half of the arch.

Hope my drawing makes sense....
Attached Images

4. p.s.

These are the approximate locations of the centroids I have calculated
5. (Original post by andymorl)
Thanks for the reply, the bridge is assumed to be a constant material through out. Please find attached a more detail picture of what I am trying to achieve. The left hand top picture is the full half of the span of the bridge (not interested in the other half). It is cut vertically through the crown section (top section), the inner radius is given and the thickness of the arch is 1.5m at the crown and increases to 2m at the end. Further the arch is subdivided into six sections parallel to the vertical through the crown at an angle of 4.6 degree.

I am trying to find the centre of mass/centre of area of each of these segments as the arch reduced or increased by one segment from the crown to the end of one half of the arch.

Hope my drawing makes sense....

Problem just got a lot more complicated.

I was going to use rotational symmetry, but that's no longer possible as each section's width varies.

Would it be safe to assume that the width increases linearly as a function of the angle. With each section having a different starting width, and different amount of increase.

And the radius of curavature of the inner/lower edge remains the same?
6. (Original post by ghostwalker)
Problem just got a lot more complicated.

I was going to use rotational symmetry, but that's no longer possible as each section's width varies.

Would it be save to assume that the width increase linearly as a function of the angle. With each section having a different starting width, and different amount of increase.

And the radius of curavature of the inner/lower edge remains the same?
The width increase isn't linear
7. Yes that assumption is fine. I am also interested in the solution of the same thickness as well as the varying thickness of arch given.

Thank you again
8. linear increase from 1.5m at the crown to 2m at the end is fine forget the other dimensions in between as the solution dose not need to be that exact

9. I have found the centroid with <7% error according to the literature by applying parallel axis theorem, but this process takes a long time and I need something straightforward that can be iterated in excel as part of an analysis package I am creating.

Anything<5% is more than acceptable
10. (Original post by andymorl)
linear increase from 1.5m at the crown to 2m at the end is fine forget the other dimensions in between as the solution dose not need to be that exact

It is quite a significant difference.
Top section's width changes by 10 mm.
End section's width changes by 150 mm.

Anyhow, this is now going to take longer than I have time for at the moment - I'll get back to you this evening.
11. Thanks for the reply, appreciated
12. (Original post by andymorl)
Thanks for the reply, appreciated
Okay, lets consider one section:

We need the area of the whole section:

We're going to need the radius of the outer rim as a function of theta, which will be:

Arranging we get:

Calling these constants m and c, we now have

Our integral now becomes:

Now we need the total moment in the x-axis direction.

Same integral as before, except this time we want to be integrating , rather than just "r".

If we divide this by our area, we end up with the position of the CofM along the x-axis for this one section.

I've worked with area, rather than mass, assuming a density of "1" - doesn't effect anything.

Note theta will need to be in radians.

...And breathe.

If you can sort this (integrals), then you can adapt it for multiple sections, or treating the whole span as one section, or even it being circular.
13. (Original post by ULTRALIGHT BEAM)
The width increase isn't linear
Aye, but OP is happy to work with that.

Aside from the final section, it's actually quadratic.
14. Thanks for the reply again, I am a bit out of practice with polar coordinates at the moment but will look over this. In the mean time I have applied another method with is giving reasonable results <1.5% error from the worked example I am studying. I can calculate the equation of the curve, which I have done in excel for the polynomial which is attached (just done this for speed).

Then integrated as shown in the other attachment, is this correct too, my maths is a bit lame atm
Attached Images

15. That is a different section I am working on btw
16. this is the other example I mentioned with the coordinated of the arch, and a comparison of the integration method I used to find the centroid vs. figures given in the literature.

It literature is from the 1800's with no explanation (apparently a graphical method was used to find the centroid)
Attached Images

17. (Original post by andymorl)
Thanks for the reply again, I am a bit out of practice with polar coordinates at the moment but will look over this. In the mean time I have applied another method with is giving reasonable results <1.5% error from the worked example I am studying. I can calculate the equation of the curve, which I have done in excel for the polynomial which is attached (just done this for speed).

Then integrated as shown in the other attachment, is this correct too, my maths is a bit lame atm
The formulae look right, but I don't understand why you're only working to the middle of a section - that would surely miss out the final half section.

You separate it into two parts:
First part from x=0 to where the lowest point on the lower curve of the final section and a
Second from there to the extreme right of the section, using an equation for the straight line of the end as the lower curve.

Though you may have done that already.

PS: If your lower curve is the arc of a circle, you should be able to get an exact equation for that, rather than a polynomial approximation.
18. ah right my maths is ok, but I am no mathematician. I was working to the centre of each section as I will be taking moments about the centre line. This is assuming the self weight of the arch is acting vertically as it is static
19. (Original post by andymorl)
ah right my maths is ok, but I am no mathematician. I was working to the centre of each section as I will be taking moments about the centre line. This is assuming the self weight of the arch is acting vertically as it is static
If your integrals only go as far as the centre of a section, then you're finding the CofM for the portion from the midline of the arch to the centre of the section, but not including the part of the section after there.

PS: Quote if you want a reply - I'll notice it quicker. Though I'm just going off now for the night.
20. thanks

Yea I know the formula of the inner curve but it is a function of the radius i.e. x^2+y^2=r^2. I just thought it would be easier to integrate the polynomial in terms of x and y, I know my limits lol

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