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.

Dashboard ENG2081: Mechanics of Structures 2A (2025-26)

View Explanation

Log in for full answers

Similar Questions

If the second moment of area for the I-section ( [math: Izz] I_{zz} ) shown below is 39848803 mm4, determine the breadth of the flange, b, in mm, (to 1 decimal place) given the following: d=169 mm, w=11 mm and t=16 mm.

Which of these expressions is correct regarding the relative sizes of cracked, gross and effective second moment of inertia?

The centroid location of the shape shown is 91.67 mm, measured from the bottom of the part. What is the moment-of-inertia of the shape about the centroidal x-axis?

Roses are [Fill in the blank] , violets are [Fill in the blank]

The point lies on the terminal arm of an angle in standard position. What are the correct values of the following three ratios?

More Practical Tools for International Students

Exam Browser Assistant

Stylized Writing Assistant

Plagiarism Checker Assistant

Citation Assistant

In-Class Translation Assistant

Class Notes Assistant

Quiz Search Assistant

Past Exam Questions by School

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler