Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 43)
43.
The moment of inertia of the shaded portion of the area shown in below figure about the X-axis, is
Discussion:
6 comments Page 1 of 1.
Ravina said:
3 years ago
Please solve in detail.
Tabarak said:
3 years ago
Triangle moment of Inertia = (B*h^3)/12 = (8*8^3)/12.
Semicircle = (π*R^4)/8 = (π*4^4)/8.
Circle = (π*R^4)/4=(π*2^4)/4.
Semicircle = (π*R^4)/8 = (π*4^4)/8.
Circle = (π*R^4)/4=(π*2^4)/4.
(2)
Nadar khan said:
4 years ago
Explain the answer in Detail, please.
(1)
Yash said:
5 years ago
Please explain in detail.
Rutuja misal said:
7 years ago
Please explain in detail.
Sachin said:
8 years ago
Correct answer is 441.86 cm^4.
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