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.
Bina Mistry said:
1 decade ago
The moment of inertia of a triangular section of height h about an axis passing through its C.G and parallel to its base is given as,
I = bh3/36.
The moment of inertia of a triangular section of height h about its base is given as,
I = bh3/12.
I = bh3/36.
The moment of inertia of a triangular section of height h about its base is given as,
I = bh3/12.
(1)
Priyesh said:
1 decade ago
I = bh3/12 for triangular section oh height h about its base, and centroid of triangular section is given by 1/3.
So I = bh3/12*1/3. Which is bh3/36?
So I = bh3/12*1/3. Which is bh3/36?
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)
Jasir said:
8 years ago
Please can anyone explain more about the moment of inertia rule?
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.
Dinesh said:
7 years ago
The Center of gravity lies at 1/3 rd of height from base.
Kavvala Laxman said:
6 years ago
I can't understand the solution please explain in detail.
(2)
Rohan said:
7 years ago
What is the moment of inertia of triangle at its apex?
Ajeya said:
7 years ago
How bh^3/36? Please explain me to understand it.
Charul said:
1 decade ago
Want to know the solution of this question.
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