Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 8)
8.
The moment of inertia of a square of side a about its diagonal is
a2/8
a3/12
a4/12
a4/16
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Partha said:   5 years ago
It's about a diagonal axis. It divides the Sqr in two triangles. So h=√3a/2, b=a. By using the formula for triangle I=bh^3/12.

So, I= √3a^4/32. As on diagonal, there are two triangles, So, the net moment of Inertia, I=√3a^4/16. So, I think the given options are wrong.

SOURABH SWARNKAR said:   5 years ago
a4/24 + a4/24 = a4/12 is answer.

Ashutosh Agrawal said:   7 years ago
Perpendicular axis theorem MIz=MIx+MIy=MI about one diagonal + MI about another diagonal. That is 2*MIx=2*MI about one diagonal.

Because it is square MIx=MId.
Therefore MI about diagonal is b^4/12.
(1)

SAMIM MOLLA said:   8 years ago
Diagonal of a square is sqrt of 2 *a ( assume one side) height is a/sqrt of 2. Since the square is consist of two triangles. So MI of the one triangle is 1/12 bh^3..here b= sqrt of 2 *a and h= a/sqrt of 2..

So MI OF THE SQUARE ABOUT ITS DIAGONAL IS TWICE THE MI OF ONE TRIANGLE, IT MEANS 1/12 *√ 2* a * (a/√ 2)^3 * 2= a^4/12.

Sonu said:   8 years ago
The Correct answer is a4/12.

L. Gopi said:   8 years ago
Correct answer is a^4/24. Since it is about the diagonal.

S.Bera said:   8 years ago
But here give about its diagonal.

Ashish said:   9 years ago
You are Correct @Shariful Alam.

Md.shariful alam said:   1 decade ago
We know Moment of inertia of squqre bar = bh3/12.

Here side are equal then b=a & h=a.

So M.I = a4/12.

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