Question

A hollow circular section has an external diameter of 8 cm and an internal diameter of 6 cm. Moment of inertia about the horizontal x axis passing through its center is

Answer 1

Answer:

.A hollow C.I. pipe with external diameter 100 mm and thickness of metal 10 mm is used as a strut. Calculate the moment of inertia and radius of gyration about its diameter.

Step-by-step explanation:

Answer 2

Answer:

Moment of Inertia passing through its centre is  137.45 $cm^{4}$

Step-by-step explanation:

Given:

A hollow circular section has an external diameter of 8 cm

A hollow circular section has an external diameter of 6 cm

To find: Moment of inertia passing through its centre.

Formula used: $I = \frac{\pi }{64} (d_{0} ^{4} - d_{i} ^{4} )$

Solution:

Moment of inertia:

Moment of inertia is the sum of the product of mass of each particle with the square of its distance from the axis of the rotation.

Moment of inertia for hollow circular cross section = $\frac{\pi }{64} (d_{0} ^{4} - d_{i} ^{4} )$

Where $d_{0}$ = outer diameter of cross section

$d_{i}$ = inner diameter of cross section

Calculation:

$d_{0}$ = 8cm

$d_{i}$ = 6 cm

I = $\frac{\pi }{64} (8^{4} - 6^{4} )$

I = $\frac{\pi }{64} (4096 - 1296)$

I = $\frac{\pi }{64} (2800)$

I = $3.14 *(43.75)$

I = 137.45 $cm^{4}$

Final answer:

Moment of Inertia passing through its centre is  137.45 $cm^{4}$

SPJ3