Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 16)
16.
Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is
Discussion:
16 comments Page 1 of 2.
Fawi said:
5 years ago
CG point 1,2.
So CG from the base: 1/3.
So, 1/12 * 1/3 = 1/36.
So CG from the base: 1/3.
So, 1/12 * 1/3 = 1/36.
Srinidhi said:
5 years ago
Can someone explain in detail?
Ganipalli Israel said:
5 years ago
Parallel axis:
I=Icg+A*H^2
I=BH^3/36 + 1/2BH*H/3^2.
I=BH^3/36+ BH^3/18,
I=BH^3 +2BH^3/36,
I=BH^3(1+2)/36,
I=BH^3(3)/36,
I=BH^3/12.
I=Icg+A*H^2
I=BH^3/36 + 1/2BH*H/3^2.
I=BH^3/36+ BH^3/18,
I=BH^3 +2BH^3/36,
I=BH^3(1+2)/36,
I=BH^3(3)/36,
I=BH^3/12.
(1)
Kavvala Laxman said:
6 years ago
I can't understand the solution please explain in detail.
(2)
Ajeya said:
7 years ago
How bh^3/36? Please explain me to understand it.
Rohan said:
7 years ago
What is the moment of inertia of triangle at its apex?
Dinesh said:
7 years ago
The Center of gravity lies at 1/3 rd of height from base.
Shehab zakaria said:
8 years ago
Why 1/3?
Jasir said:
8 years ago
Please can anyone explain more about the moment of inertia rule?
Mahmood Ahmad said:
8 years ago
It is BH3/12.
(1)
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