Second moment of area of a triangle about its base For a triangular cross-section with base b and height h, what is the moment of inertia (second moment of area) about its base line?

Introduction / Context:
The second moment of area (moment of inertia) is a geometric property used to predict a member’s resistance to bending. For standard shapes like triangles, rectangles, and circles, closed-form expressions enable quick section property calculations used in beam design and deflection analysis.

Given Data / Assumptions:

• Cross-section is a triangle with base b and height h.
• Axis of interest is along the base line of the triangle.
• Material is homogeneous; the property is purely geometric.

Concept / Approach:
For a triangle, the centroidal moment of inertia about an axis parallel to the base is I_centroid = (b * h^3) / 36. Using the parallel-axis theorem to shift from centroid to the base line a distance h/3, the base-axis moment of inertia becomes I_base = I_centroid + A * (h/3)^2, where A = (1/2) * b * h. This yields I_base = (b * h^3) / 12.

Step-by-Step Solution:
Compute area: A = 0.5 * b * h.Use centroidal value: I_centroid = (b * h^3) / 36.Apply parallel-axis theorem: I_base = I_centroid + A * (h/3)^2.I_base = (b * h^3) / 36 + (0.5 * b * h) * (h^2 / 9) = (b * h^3) / 12.

Verification / Alternative check:
Dimension check: I has units of length^4, satisfied by (b * h^3). Also, compared to the centroidal value, the base-axis inertia must be larger, and 1/12 is larger than 1/36 by a factor of 3, consistent with the shift.

Why Other Options Are Wrong:
(b * h^3)/36: This is the centroidal inertia, not about the base.Terms with b^3 * h or with h^2 indicate wrong axis or incorrect application of the theorem.

Common Pitfalls:

• Confusing centroidal and base-line axes.
• Applying the parallel-axis theorem with an incorrect distance (not using h/3).

Final Answer:
I_base = (b * h^3) / 12