Cantilever beam with linearly varying load\n\nA cantilever beam of length l carries a triangular distributed load that varies from zero at the free end to w per unit length at the fixed end. Identify the correct nature of the bending moment diagram (BMD) for this loading.

Introduction / Context:
This question checks understanding of how distributed load shapes translate into shear force and bending moment diagrams in strength of materials. Recognizing whether the bending moment diagram (BMD) is linear, parabolic, or cubic is critical for sizing beams and predicting maximum stresses and deflections.

Given Data / Assumptions:

• Cantilever beam of length l, fixed at one end and free at the other.
• Transverse load varies linearly along the span: zero at the free end, w per unit length at the fixed end.
• Small-deflection beam theory; sign conventions standard.

Concept / Approach:
For any beam: load intensity w(x) integrates to shear V(x), and shear integrates to bending moment M(x). If w(x) is uniform (constant), V is linear and M is parabolic. If w(x) is linear (triangular), then V is quadratic in x and M is cubic in x.

Step-by-Step Solution:
Let w(x) = k x (with x measured from the free end toward the fixed end).Shear force: V(x) = ∫ w(x) dx = ∫ k x dx = (k/2) x^2 + C → quadratic.Bending moment: M(x) = ∫ V(x) dx = ∫ (k/2) x^2 dx = (k/6) x^3 + C′ → cubic.Therefore the BMD is a cubic curve (not a parabola).

Verification / Alternative check:
Dimensional progression rule: increase the polynomial order by one upon integration. Triangular load (order 1) → shear (order 2) → moment (order 3). This quick rule confirms the cubic nature.

Why Other Options Are Wrong:
Agree: incorrect because it calls the BMD parabolic. The BMD is cubic. The options about dependence on elastic modulus or shear shape are distractors; the curve family depends directly on the load distribution and basic equilibrium relations.

Common Pitfalls:
Confusing the UDL case (parabolic M) with triangular loading (cubic M); mixing the coordinate origin; forgetting that shear is the integral of load and moment is the integral of shear.

Final Answer:
Disagree