Area moments – triangular section about its base\nThe second moment of area (area moment of inertia) of a triangular section with base b and height h about its base is:

Difficulty: Easy

Correct Answer: b h^3 / 12

Explanation:


Introduction / Context:
Area moments of inertia are central to bending stress and deflection analyses. For common shapes, standard formulas are used to speed design and checks.



Given Data / Assumptions:

  • Plane area is a triangle of base b and height h.
  • Axis of interest is along the base line of the triangle (in-plane axis through the base).
  • Small-deflection, linear elasticity context (typical beam theory usage).


Concept / Approach:
The area moment of inertia about the base for a triangle is derived by integrating y^2 dA from the base toward the apex, using a linear variation of width with height.



Step-by-Step Solution:

Let the triangle have base along y = 0 and apex at y = h.Width at a distance y from base varies linearly: w(y) = (b/h) (h − y).Differential area: dA = w(y) dy.Area moment about base: I_base = ∫ y^2 dA = ∫_0^h y^2 w(y) dy = ∫_0^h y^2 (b/h)(h − y) dy.Compute: (b/h) ∫_0^h (h y^2 − y^3) dy = (b/h) [ h * (h^3 / 3) − (h^4 / 4) ] = (b/h) (h^4/3 − h^4/4) = (b/h) (h^4/12) = b h^3 / 12.


Verification / Alternative check:
Parallel-axis transformation from centroidal axis (I_cg = b h^3 / 36) to base: I_base = I_cg + A d^2 with d = h/3 and A = b h / 2. This also yields b h^3 / 12.



Why Other Options Are Wrong:

  • b h^3 / 4 and b h^3 / 8 are too large; they correspond to different shapes/axes.
  • b h^3 / 36 is the centroidal moment for a triangle about a base-parallel centroidal axis, not about the base itself.


Common Pitfalls:
Mixing centroidal and base axes or forgetting linear width variation.


Final Answer:
b h^3 / 12

More Questions from Engineering Mechanics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion