Elongation of a conical bar due to its own weight: A conical bar of length l has base diameter d at the fixed upper end and tapers linearly to a point at the free lower end. Weight per unit volume is w. Find the total elongation under self-weight (assume linear elasticity with modulus E).

Introduction:
Members with variable cross-section exhibit nonuniform stress under self-weight. A conical bar fixed at the large end and tapering to zero at the free tip has a neat closed-form elongation independent of base diameter.

Given Data / Assumptions:

• Length l; base diameter d at the fixed top; linear taper to zero at bottom.
• Specific weight (weight per unit volume) w; Young’s modulus E.
• Small strains; linear elasticity.

Concept / Approach:
At a section x measured from the fixed end, area A(x) decreases linearly, and the axial force equals the weight of the segment below. Strain is σ/E, and total elongation is the integral of strain over the length.

Step-by-Step Solution:

Verification / Alternative check:
The result is independent of d because both P(x) and A(x) scale with A0 and cancel. Dimensional check: w has units of force/volume; multiplying by l^2/E gives length.

Why Other Options Are Wrong:

• wl^2/(2E) or /(3E): overestimates elongation; incorrect integration factors.
• wl/(6E): missing one power of l from the integral.
• wl^3/(6Ed^2): introduces d dependance that cancels in a true conical taper to a point.

Common Pitfalls:
Setting up A(x) incorrectly; forgetting that the line of action for self-weight is the weight of the portion below the section; missing the linear taper square in area.

Final Answer: