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Area Moment of Inertia | |||
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Posted by: DavidH ® 12/08/2006, 21:25:55 Author Profile eMail author Edit |
I have got a bit of a problem understanding the calculation of this. Can somebody give me some help. |
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Re: Moment of Inertia | |||
Re: Moment of Inertia -- DavidH | Post Reply | Top of thread | Forum |
Posted by: hjens56 ® 12/12/2006, 15:15:38 Author Profile eMail author Edit |
Need a little more info about what you don't understand. |
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Re: Re: Moment of Inertia -- hjens56 | Post Reply | Top of thread | Forum |
Posted by: DavidH ® 12/12/2006, 22:06:18 Author Profile eMail author Edit |
Well if for instance we use the wide flange charts from AISC for a W21 x 62 beam, 21"x62lb the Inertia for that is 1330 and when I use the formula for that beam 20.99 d, 8.24 w, .615tf, & .400tw than I come up with a different figure using the formula bd^3 - h^3(b-t) / 12. How come? |
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Re: Re: Re: Moment of Inertia -- DavidH | Post Reply | Top of thread | Forum |
Posted by: Kelly Bramble ® 12/13/2006, 10:27:43 Author Profile eMail author Edit |
A W flange has radii at all surface intersections and corners I beleive. So, manually calculating is only a estimation (though probably good enough). If you realy use the AISC W21 x 62 beam (specify in your material block) I would use the AISC number. Also, be sure your are using the same Ixx, Iyy, or Izz. Related: |
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Re: Re: Re: Moment of Inertia -- DavidH | Post Reply | Top of thread | Forum |
Posted by: hjens56 ® 12/13/2006, 04:30:43 Author Profile eMail author Edit |
First of all, you need to specify which moment of inertia for the beam cross-section you are calculating: about the horizontal or vertical axis. From the numbers you gave, I assume you are trying to get the moment of inertia about the horizontal axis (Ixx). Second, I don't recognize the formula you gave. It doesn't even have the same variables as you listed for the beam dimensions. I am going to also assume that it is incorrect. The first step to calculating the moment is to divide the beam cross-section into 3 rectangles: one tall and skinny rectangle for the web section, and two rectangles for the top and bottom flange sections. Ixx for the beam can be calculated by summing the moments of the simpler shapes about the horizontal axis of the center of the beam. To do this you first calculate the moments for each of the simple shapes about their individual centers. The formula for these calculations for the rectangular cross-sections is (1/12)*b*h^3 where b is the length of the horizontal side of the rectangle and h is the length of the vertical side of the rectangle. Once you have the individual moments of inertia for each of the simple rectangles about their own centers, you must find the moments of inertia of the shapes about the center of the beam section. For the web section rectangle, this is easy since the two centers are the same and no calculation is needed. For the other two rectangles, the moment of the rectangles about the center of the beam is given by the formula: Ixx'=Ixx + A*d^2 where Ixx' is the moment about the center of the beam, Ixx is the moment about the rectangle's own center that you calculated earlier, A is the area of the rectangle, and d is the distance between the center of the rectangle and the center of the beam. Once you have the moments for each of the rectangles about the center of the beam, you can simply add them together to get the overall moment for the beam. If you follow this method correctly with the numbers you gave for the W21 beam, you will get approximately 1300 in^4 for a moment of inertia about the horizontal axis. I hope this helps. Modified by hjens56 at Wed, Dec 13, 2006, 04:34:03 |
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Re: Re: Re: Re: Moment of Inertia -- hjens56 | Post Reply | Top of thread | Forum |
Posted by: DavidH ® 12/13/2006, 17:37:48 Author Profile eMail author Edit |
Yes you got my question. What helps me alot to understand the calc. is when you spell out the actual numbers in your illustration of a formula - then I can follow it good. Like I get a little mixed up when it comes to what the difference is between Ixx' and Ixx. So if you would use the beam I gave dimensions for and then type out the way you came to 1300 I would appreciate that. |
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Re: Re: Re: Re: Re: Moment of Inertia -- DavidH | Post Reply | Top of thread | Forum |
Posted by: DavidH ® 12/13/2006, 18:37:01 Author Profile eMail author Edit |
I just got now!! Thanks for your help. |
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