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Engineering Mechanics - Moments of Inertia - Discussion

@ : Home > Engineering Mechanics > Moments of Inertia > General Questions - Discussion

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"Do not wait for leaders; do it alone, person to person."
- Mother Teresa
4. 

The irregular area has a moment of inertia about the AA axis of 35 (106) mm4. If the total area is 12.0(103) mm2, determine the moment of inertia if the area about the BB axis. The DD axis passes through the centroid C of the area.

[A]. IBB = 5.00(106 ) mm4
[B]. IBB = 17.00(106 ) mm4
[C]. IBB = 16.80(106 ) mm4
[D]. IBB = 55.4(106 ) mm4

Answer: Option E

Explanation:

No answer description available for this question.


Sujin said: (Mon, Mar 26, 2012 02:34:37 PM)    
 
Ibb=Idd+A(r*r)
Iaa=Idd+A(x*x)
r=10mm,x=40mm
Ibb-Iaa=A(r*r-x*x)
Ibb=Iaa+A(r*r-x*x)
Ibb=[35+12(100-1600)]*100000
Ibb=17*10^6mm^4

Aniruddha said: (Tue, Jun 5, 2012 04:23:51 PM)    
 
First find Idd ?.
Idd = Iaa-(area*Distance betn DD and AA^sq.)
= Iaa-(Area*40^2)
= (35X10^6) - ((12X10^3)X40^2)
= 15.8 x 10^6
Now find Ibb
= Idd + Area X h^2
= (15.8x10^6) + (12x10^3)(10^2)
ANS = 17 X 10^6

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