Engineering Mechanics - Moments of Inertia - Discussion
Discussion Forum : Moments of Inertia - General Questions (Q.No. 4)
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.
Discussion:
4 comments Page 1 of 1.
RUPESH said:
8 years ago
Ibb = Idd+12*10^3*100.
= (35*10^6)-(12*10^3*1600)+12*10^3*100.
= (35*10^6)-(12*10^3*1600)+12*10^3*100.
Gowthamy said:
1 decade ago
Ibb = Idd+12*10^3*100.
= (35*10^6)-(12*10^3*1600)+12*10^3*100.
= (35*10^6)-(12*10^3*1600)+12*10^3*100.
Aniruddha said:
1 decade ago
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
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
Sujin said:
1 decade ago
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
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
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