Area properties – rectangular section about centroidal X–X axis Compute the second moment of area (moment of inertia) about the X–X axis for a rectangle that is 3 cm wide (b = 3 cm) and 4 cm deep (h = 4 cm), taking X–X through the centroid and parallel to the base.

Difficulty: Easy

Correct Answer: 16 cm^4

Explanation:


Introduction / Context:
The second moment of area (area moment of inertia) of a plane section is a geometric property used in bending, deflection, and stability calculations. For standard shapes, compact formulas exist. Here we determine I about the centroidal horizontal axis X–X for a rectangular section.



Given Data / Assumptions:

  • Width b = 3 cm (horizontal).
  • Depth/height h = 4 cm (vertical).
  • X–X axis passes through the centroid and is parallel to the base (horizontal axis).
  • Use standard formula I_x = b * h^3 / 12.


Concept / Approach:
For a rectangle, the centroidal second moment about the axis parallel to its base is I_x = b * h^3 / 12. This comes from integrating y^2 dA over the area. Units are in length^4.



Step-by-Step Solution:

Write the formula: I_x = b * h^3 / 12.Substitute b = 3 cm, h = 4 cm.Compute h^3 = 4^3 = 64 cm^3.Compute b * h^3 = 3 * 64 = 192 cm^4.Divide by 12: I_x = 192 / 12 = 16 cm^4.


Verification / Alternative check:
Dimensional check: input lengths in cm yield I in cm^4. Also note symmetry: increasing depth strongly increases I because of the h^3 factor, consistent with 16 cm^4 being larger than values using smaller h.



Why Other Options Are Wrong:

  • 9 cm^4 or 12 cm^4: arise from using b^3 h / 12 or arithmetic slips.
  • 20 cm^4: not supported by the formula; likely from averaging or misusing 1/10 instead of 1/12.


Common Pitfalls:
Confusing axes: I_x uses h^3, whereas I_y about the vertical centroidal axis would be h * b^3 / 12. Always align the cubic dimension with the axis perpendicular to bending.



Final Answer:
16 cm^4

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