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Hi I read this response but it isn't exactly my situation I have a circle where the axis of rotation is on the c axis like this for a segment equal to the green. enter image description here

Specifically, I have a round steel plate on a concrete base attached with anchors. There a moment across the center (out of plane) I am trying to determine the resisting moment of the concrete surface.

I am very much not sure what the integration result would be.

https://en.wikipedia.org/wiki/Circular_segment

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  • $\begingroup$ So you want the second moment of area of the green bit about the C axis? I must admit I completely fail to understand the actual loading scenario. $\endgroup$ Sep 21, 2023 at 21:30
  • $\begingroup$ yes, I have a circular base plate on concrete. I have anchors all around and it is subject to a moment. My objective is to find the max tension in the anchors but to do so I need to find the concrete area resistance (the green zone) and iterate on what "d" is to balance. $\endgroup$ Sep 22, 2023 at 12:09
  • $\begingroup$ I suspect you don't want the moment of inertia as such. So far as I can tell you have a disc restrained at its circumference, and a moment applied along the dashed line. The disc lies above (and presumably in contact with) the concrete underneath. A decent diagram of the actual situation including the location of the anchors would be vastly helpful. If my description is correct then you have a plate deformation problem, it is non linear, and so you will have to be very clever, make some assumptions or use FEA. $\endgroup$ Sep 22, 2023 at 21:09

2 Answers 2

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I think I found my answer, thanks to https://structx.com/Shape_Formulas_003.html and the Parallel axis theorem https://en.wikipedia.org/wiki/Second_moment_of_area.

$I_x$ around the center of the circle is

$I_x=\frac{r^4}{8}(\theta-\sin{\theta}+2\sin{\theta}\sin^2{\frac{\theta}{2}})$

From the Theorem

$I_{x'}=I_x-Ad^2$

Where $d$ is the distance between the center of the circle and the line

$A$ is the area $A=\frac{r^2}{2}(\theta-\sin{\theta})$

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  • $\begingroup$ it's not correct. Ix'=Ix+A d^2 . A is the area, d is the distance between the two axes! $\endgroup$
    – kamran
    Sep 22, 2023 at 20:45
  • $\begingroup$ You are right, I will correct $\endgroup$ Sep 23, 2023 at 22:48
  • $\begingroup$ Thank you so much! For others: This solution is also applicable for a moment of inertia around the centroid of the segment, you just need to set the distance d as the distance of the centroid from the circle center. $\endgroup$ Nov 18, 2023 at 23:49
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The formula you can use is:

$$I=\frac{1}{8}R^4(\theta- sin\theta cos\theta) $$

Where:

  • I is the second area moment of the circular segment about the chord.

  • R is the radius of the circle.

  • θ is the angle (in radians) subtended by the circular segment at the center of the circle.

Maximum stress to the concrete happens at the top of the segment H=R and is

$$\sigma=\frac{M*C}{I}$$

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  • $\begingroup$ Hi. Are you sure? It give me very large "I". Also to find the Resisting moment, I use M = I/h $\endgroup$ Sep 22, 2023 at 12:11
  • $\begingroup$ It gives a large I because it's about the cord not the centroid of the segment! if in fact you need the I about the centroid it will be much smaller. $\endgroup$
    – kamran
    Sep 22, 2023 at 19:28

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