Area Moments — Triangle About Different Parallel Axes For a triangular section of base b and height h, the moment of inertia about an axis through the vertex and parallel to the base is how many times the moment of inertia about the centroidal axis parallel to the base?

Difficulty: Medium

Correct Answer: Nine times

Explanation:


Introduction / Context:
Parallel-axis transformations are common in structural analysis. For triangles, knowing the relationship between inertia about centroidal and non-centroidal axes helps in composite-section calculations.


Given Data / Assumptions:

  • Triangle with base b and height h.
  • Centroidal axis parallel to base with known inertia I_G = (b h^3) / 36.
  • Vertex axis: a line through the vertex, parallel to the base.


Concept / Approach:
Use the parallel-axis theorem: I_vertex = I_G + A d^2, where A is the area and d is the perpendicular distance between the two parallel axes. For a triangle, the centroid lies at h/3 from the base and at 2h/3 from the opposite vertex.


Step-by-Step Solution:

Area: A = (1/2) b h. Centroid-to-vertex distance: d = 2h / 3. Centroidal inertia: I_G = (b h^3) / 36. Apply parallel-axis: I_vertex = I_G + A d^2 = (b h^3)/36 + (1/2 b h) * (2h/3)^2. Compute increment: (1/2 b h) * (4h^2/9) = (2/9) b h^3 = (8/36) b h^3. Sum: I_vertex = (1/36 + 8/36) b h^3 = (9/36) b h^3 = (1/4) b h^3. Ratio: I_vertex / I_G = (1/4) / (1/36) = 9.


Verification / Alternative check:
Numeric example with b = h = 1 yields I_G = 1/36 and I_vertex = 1/4; the ratio is indeed 9.


Why Other Options Are Wrong:
Two, four, and six times do not satisfy the exact parallel-axis calculation for a triangle.


Common Pitfalls:
Using h/3 instead of 2h/3 for the centroid-to-vertex distance; mixing up which axis is being referenced.


Final Answer:
Nine times.

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