Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 13)
13.
Moment of inertia of a circular section about an axis perpendicular to the section is
Discussion:
30 comments Page 3 of 3.
Prakash patel said:
10 years ago
It's perpendicular so,
πd^4/64 divided by d/2.
It's true and πd^3/32.
πd^4/64 divided by d/2.
It's true and πd^3/32.
Sridhar.repaka said:
1 decade ago
Its not n it is "PI".
Prashant kadam said:
1 decade ago
ixx+iyy = izz.
So ixx and iyy is nd4/64+nd4/64 = nd4/32.
So ixx and iyy is nd4/64+nd4/64 = nd4/32.
Shibunath said:
1 decade ago
According to perpendicular axis theorem.
Ixx+Iyy= πd raise to 4/64+πd raise to 4/64 = πd raise to 4/32.
Ixx+Iyy= πd raise to 4/64+πd raise to 4/64 = πd raise to 4/32.
Joel Kumar said:
1 decade ago
What is 'n' here?
Jay said:
1 decade ago
M.O.I means Moment of Inertia.
Sabari skcet said:
1 decade ago
What is M.O.I?
Prathap said:
1 decade ago
Can anyone explain clearly how nd^4/32?
HITESH said:
1 decade ago
But axis perpendicular to section is longitudinal axis and about which M.O.I of circle is n(d^4)/32.
Abhishek Tiwari said:
1 decade ago
BY PERPENDICULAR AXIS THEOREM,
Izz = Ixx+Iyy.
So about x&y nd4/64.
Izz = nd4/64+nd4/64.
= nd4/32.
Izz = Ixx+Iyy.
So about x&y nd4/64.
Izz = nd4/64+nd4/64.
= nd4/32.
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