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 1 of 3.
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.
Noor ahmed said:
1 decade ago
About x axis pid^4/64 similarly about why axis. Perpendicular means about z axis. Sum of two axes (perpendicular axes theorem).
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.
DEVENDRA MISHRA said:
9 years ago
Area moment of inertia.
Ix = nd4/64.
Iy = nd4/64.
The axis perpendicular to the section.
Iz = nd4/32.
Ix = nd4/64.
Iy = nd4/64.
The axis perpendicular to the section.
Iz = nd4/32.
Kuldeep Maher said:
5 years ago
Axis perpendicular to the cross-section would be longitudinal axis so the answer should be π*d^4/64.
(1)
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.
Fazeel said:
3 years ago
Izz = Ixx+Iyy.
Izz= π/64 x d^4 + π/64 x d^4,
Izz =(π/64 +π/64) x d^4,
Izz= π/32 x d^4.
Izz= π/64 x d^4 + π/64 x d^4,
Izz =(π/64 +π/64) x d^4,
Izz= π/32 x d^4.
(10)
Edward said:
8 years ago
m.o.i of circular lamina= pi *r power4/2.
then pi/2*(d/2)power4.
then pi/32*d power4.
then pi/2*(d/2)power4.
then pi/32*d power4.
NITISH KUMAR said:
9 years ago
By perpendicular theorem Ixx + Iyy = Ixx.
φd^4/64 + φd^4/64 = φd^4/32.
φd^4/64 + φd^4/64 = φd^4/32.
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.
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