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 2 of 3.
Vikash said:
8 years ago
Explain in step by step.
VINOD KUMAR said:
8 years ago
I can understand this problem. Please explain this clearly.
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.
Sunil said:
8 years ago
I can't understand this problem can anyone help me?
Mohan said:
8 years ago
I can't understand. Can anyone help me?
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).
Anuraag said:
9 years ago
But, the moment of inertia has "M". Where is it?
I= MK^2. Where is M?
I= MK^2. Where is M?
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.
Rashed said:
9 years ago
Thank you all. the Answer is right.
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.
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