Second Moment of Area Homework: Answer = 351.88 x 10^6

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Discussion Overview

The discussion revolves around the calculation of the second moment of area (Iy) for a given shape, with a specific answer provided (351.88 x 10^6). The scope includes homework-related queries, clarification of concepts, and technical calculations.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant states they divided the shape into two sections to find the second moment of area for each and then added them together.
  • Another participant suggests that the integration limits for the second part of the calculation should be adjusted and emphasizes the importance of calculating the centroid before applying the parallel axis theorem.
  • A participant expresses confusion regarding the distinction between the second moment of area and moment of inertia, questioning whether their calculations were correct.
  • One participant clarifies that the second moment of area is sometimes referred to as the area moment of inertia and confirms that the initial calculation was correct for the left-hand edge, but suggests that the problem likely requires the second moment of area about the centroid.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the second moment of area, with some suggesting adjustments to the method and others confirming the initial calculations. The discussion remains unresolved regarding the correct interpretation of the problem and the final answer.

Contextual Notes

There are limitations in the clarity of the problem statement, particularly regarding the axis about which the second moment of area should be calculated. Additionally, there is ambiguity in the definitions and distinctions between second moment of area and moment of inertia as discussed by participants.

Who May Find This Useful

This discussion may be useful for students or individuals studying structural engineering or mechanics, particularly those grappling with concepts related to moments of inertia and second moments of area in their homework or practical applications.

v_pino
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Homework Statement


I have to find the second moment of area, Iy.
The answer is 351.88 x 10^6

Homework Equations






The Attempt at a Solution


I divided the shape into 2 sections and found the second moment of area for each of them then added the two. I have attached my work but is rather poor quality. I'll explain my working more if it is unclear.

Thanks :D
 

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It looks like using calculus you are attempting to calculate the moment of inertia about the right edge of the left wide flange; if so, your second part integration limits seem like they ought to be from 0 to .3 rather than .5 to .35. But in any case, unless the problem specifically states, you are trying to find the second moment of area about the centroid of the overall section. So you must first calculate the centroid using the moment area method, then calculate the momemt of inertia about the centroid using the parallel axis theorem. Are you familiar with both? You don't need calculus if you know that the moment fo inertia of a rectangle about its own centroid is bh^3/12. I try to steer away from using calculus as much as possible..it's a great learning tool, but it can throw you off if not used properly.
 
Hi thanks for the reply!

I am a little confused with the difference between second moment of area and moment of inertia.

Isn't the second moment of area = integral of y^2 dA

and the moment of inertia = integral of y^2 dm

So in my solution for my problem, didn't I calculate the moment of area instead of inertia?

thanks
 
v_pino: Second moment of area, integral[(y^2)dA], is sometimes called area moment of inertia. Mass moment of inertia is integral[(y^2)dm]. You got the correct answer for second moment of area about the left-hand edge of your cross section. Nice work. But as PhanthomJay mentioned, the question is probably asking you to compute the second moment of area about a vertical axis passing through the centroid of the cross section. However, when you do this, the answer is not 351.88e6 mm^4.
 

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