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

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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

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

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