Engineering Mechanics - Moments of Inertia - Discussion
Discussion Forum : Moments of Inertia - General Questions (Q.No. 2)
2.
Determine the inertia of the parabolic area about the x axis.
Discussion:
4 comments Page 1 of 1.
Vio said:
1 decade ago
dIx = y^2.dA.
Where dA = x.dy.
40.y = x^2.
=> x=square root of(40.y).
Then, susbstitute the above values in main equation,
dIx = y^2.over root of(40.y).dy.
Then take integral on both side from 0 to 10.
Then you get Ix=11,430 in^4.
Where dA = x.dy.
40.y = x^2.
=> x=square root of(40.y).
Then, susbstitute the above values in main equation,
dIx = y^2.over root of(40.y).dy.
Then take integral on both side from 0 to 10.
Then you get Ix=11,430 in^4.
Aksharkumar said:
1 decade ago
Determine the moment of inertia for section about horizontal and vertical centroidal 120.
(1)
G.VASU FROM IIIT(RKVALLEY) said:
1 decade ago
dIx=y^2.dA
where dA=x.dy
40.y=x^2
=>x=square root of(40.y)
then, susbstitute the above values in main equation
dIx=y^2.over root of(40.y).dy
then take integral on both side from 0 to 10.
then you get Ix=11,430 in^4.
where dA=x.dy
40.y=x^2
=>x=square root of(40.y)
then, susbstitute the above values in main equation
dIx=y^2.over root of(40.y).dy
then take integral on both side from 0 to 10.
then you get Ix=11,430 in^4.
Dhinesh said:
1 decade ago
Give answer descripition.
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