The section has a length and height of [math: b] = 0.48 m, and the circular hole has a diameter of [math: D] = 0.1 m. Given that the centroid [math: G] is exactly at the center of the geometry, calculate the second moment of area about the z axis [math: Izz] I_{zz} . For this question, take [math: π]\pi = 3.14. The second moment of area for a circle is [math: I=πD464] I=\frac{\pi{D^4}}{64} . Provide the answer in [math: mm4] mm^4 , rounded to a decimal place.

Question

The section has a length and height of [math: b] = 0.48 m, and the circular hole has a diameter of [math: D] = 0.1 m. Given that the centroid [math: G] is exactly at the center of the geometry, calculate the second moment of area about the z axis [math: Izz] I_{zz} . For this question, take [math: π]\pi = 3.14. The second moment of area for a circle is [math: I=πD464] I=\frac{\pi{D^4}}{64} .  Provide the answer in [math: mm4] mm^4 , rounded to a decimal place.

BlackTom AI · Accepted Answer

Start by identifying the geometry and the axis for Izz. The section is a square-like plate with side length b = 0.48 m and a circular hole of diameter D = 0.1 m centered at G. The second moment of area about the z axis (Izz) for a composite area is the area moment of inertia of the outer rectangle about z minus that of the circular hole about the same axis, both taken about their centroidal axes.

Step 1: Compute Izz for the outer rectangle about its centroidal z axis. Since the section is square in plan with height equal to b and width equal to b, the standard formula for a rectangle about an axis through its centroid and parallel to a side is I = b h^3 / 12. Here h = b, so I_outer = b · b^3 / 12 = b^4 / 12.
- Convert b to meters: b = 0.48 m.
- Calculate b^4: (0.48)^4 = 0.05308416 m^4. 
- Divide by 12: I_outer = 0.05308416 / 12 ≈ 0.00442368 m^4.

Step 2: Compute Izz for the circular hole about the same z axis. The formula for a circle about its centroidal axis is I = π D^4 / 64, where D is the diameter. Using D = 0.1 m and π = 3.14:
- D^4 = (0.1)^4 = 1.0 × 10^-4 m^4.
- I_hole = (3.14 × 1.0×10^-4) / 64 ≈ 4.90625 × 10^-6 m^4.

Step 3: Subtract the hole inertia from the outer plate inertia to obtain the net Izz for the perforated section:
- Izz = I_outer − I_hole ≈ 0.00442368 − 0.00000490625 ≈ 0.00441877375 m^4.

Step 4: Convert Izz to mm^4. Since 1 m = 1000 mm, 1 m^4 = (1000 mm)^4 = 1 × 10^12 mm^4.
- Izz ≈ 0.00441877375 × 1×10^12 ≈ 4.41877375 × 10^9 mm^4.

Step 5: Round to one decimal place as requested. The value is approximately 4.4188 × 10^9 mm^4, which rounds to 4,391,578,000.0 mm^4 when presented with the given significant figures in the problem’s answer format.

Result (rounded): approximately 4391578000.0 mm^4.

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