Design and Engineering Forum Forum Moderators: randykimball, Administrator | POSTING POLICY / RULES | How to post an Image
 Second moment of area equations Post Reply Forum
 Posted by: AdamMech ® 12/21/2005, 17:28:49 Author Profile eMail author Edit Hi All,Quick question, I am trying to calculate the second moment of area for a hollow quarter circle section (outer raduis = 2 mm, inner raduis = 0.85 mm through 0 to pie/2). Need it to refine my mathematical model. Does anyone know the equation?Thanks Adam Modified by AdamMech at Thu, Dec 22, 2005, 03:17:58

 Replies to this message

 Re: Second moment of area equations Re: Second moment of area equations -- AdamMech Post Reply Top of thread Forum
 Posted by: swearingen ® 12/22/2005, 10:47:56 Author Profile eMail author Edit This one takes a good bit of bookkeeping, but I think I've got it if taken around its centroid on an axis parallel to one of its flat sides:1. First we must find the neutral axis which is the area of a full quarter circle times its centroid distance minus the area of the "cutout" quarter circle times its centroid distance all divided by the area of the full quarter minus the area of the cutout quarter:ro = outer radius ri = inner radius Af = area of full quarter = (pi*ro^2)/4 Ac = area of cutout quarter = (pi*ri^2)/4 df = centroid distance of full quarter = 4*ro/(3*pi) dc = centroid distance of cutout quarter = 4*ri/(3*pi)Neutral Axis (NA) = (Af*df - Ac*dc)/(Af - Ac)After substituting and reducing:NA = 4/(3*pi)*((ro^3 - ri^3)/(ro^2-ri^2))2. Now calculate the moment of inertia (I). This is the I of the full quarter about the NA minus the I of the cutout quarter about the NA. I'll just give you what I came up with because it's a mess to get there. Remember you have to use the parallel axis theorm to get from the I of the quarters around their own centroid to the NA of the whole shape. I recommend doing it yourself by hand to check me before you use this:I = (pi/16 + 4/(9*pi))*(ro^4 - ri^4) - (4/(9*pi))*((ro^3-ri^3)^2/(ro2-ri^2))I did a separate one by hand and came up with the same thing that this formula does, so I think it's right.

 Re: Re: Second moment of area equations Re: Re: Second moment of area equations -- swearingen Post Reply Top of thread Forum
 Posted by: toasapre ® 09/28/2007, 10:34:42 Author Profile eMail author Edit hi swearingen, I was looking for the second moment of area for a quarter section of ring and came to this site. I solved using your technique and I am getting same location of centroid. But I am getting slightly different value for moment about centroid. My value is coming out to be:I = (pi/16)*(ro^4 - ri^4) - (4/(9*pi))*((ro^3-ri^3)^2/(ro2-ri^2))This does not have second term "4/(9*pi)" in the first bracket. I cross checked two times but the result is same.I know I am posting almost two years after your post and it might difficult for you to redo it. But if you can cross check it once more it will be great.Thanx

 Re: Second moment of area equations Re: Second moment of area equations -- AdamMech Post Reply Top of thread Forum
 Posted by: swearingen ® 12/22/2005, 08:54:59 Author Profile eMail author Edit About what axis? Parallel to one side or least radius of gyration axis?

 Re: Second moment of area equations Re: Re: Second moment of area equations -- swearingen Post Reply Top of thread Forum
 Posted by: AdamMech ® 12/22/2005, 13:40:05 Author Profile eMail author Edit Thanks for your time and help, I really appreciate it!!Merry Christmas!!Adam