Respuesta :
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]
