Calculate area moment of inertia for a circular cross-section with 3 mm diameter:


28 mm4


254 mm4


7 mm4


4 mm4


81 mm4



Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.


52 cm4


72 cm4


32 cm4


24 cm4


2 cm4

Answers

Answer 1

Answer:

Part 1

4 mm⁴

Part 2

32 cm³

Explanation:

Part 1

The diameter of the circular cross section, d = 3 mm

The area moment of inertia of a circle, [tex]I_C[/tex], is given as follows;

[tex]I_x = I_y = \dfrac{\pi \cdot d^4 }{64}[/tex]

Where;

d = The diameter of the circle

Therefore, the area moment of inertia of the given circular cross section, with d = 3 mm, is found as follows;

[tex]I_x = I_y = \dfrac{\pi \times (3 \, mm)^4 }{64} \approx 4 \, mm^4[/tex]

Part 2

The minimum area moment of inertia for a rectangular cross-section is given as follows;

[tex]I_x = \dfrac{1}{12} \cdot b \cdot h^3[/tex]

[tex]I_y = \dfrac{1}{12} \cdot h \cdot b^3[/tex]

The minimum moment of inertia, for the rectangular cross-section is given by placing, the height, h = The short side length and calculating for, Iₓ, and vice versa

Therefore, for the question, where, h = 4 cm, and b = 6 cm, we have;

[tex]I_x = \dfrac{1}{12} \times 6 \, cm \times (4 \, cm)^3 = 32 \, cm^4[/tex]


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