Cantilever with triangular (linearly varying) load: A cantilever beam of length l carries a load varying from zero at the free end to w per unit length at the fixed end. What is the shear force at the fixed end?

Introduction / Context:
Linearly varying distributed loads are common in structural members. For cantilevers with a triangular load intensity that is zero at the tip and maximum at the fixed end, the reaction shear and moment at the fixed support are obtained from resultants of the equivalent load diagram.

Given Data / Assumptions:

• Cantilever length = l.
• Load intensity varies linearly from 0 (free end) to w (fixed end).
• Static equilibrium; prismatic member.

Concept / Approach:
The total load equals the area under the load-intensity diagram. For a triangle of base l and height w, the resultant load is (1/2) * w * l, acting at the centroid of the triangle located at a distance l/3 from the high-intensity end (i.e., from the fixed end) or 2l/3 from the free end. The reaction shear at the fixed end equals this resultant (taking the cantilever as a free body).

Step-by-Step Solution:

Verification / Alternative check:
Integrate the load intensity q(x) = (w / l) * x measured from the free end to the fixed end: ∫_0^l q(x) dx = ∫_0^l (w/l) x dx = (w/l) * (l^2 / 2) = w l / 2, confirming the resultant magnitude equals the support shear.

Why Other Options Are Wrong:

• zero: contradicts presence of a distributed load.
• w l / 4: underestimates by half.
• w l and 2 w l: overestimates the triangular resultant.

Common Pitfalls:
Placing the resultant at l/3 from the free end (incorrect); confusing triangular versus uniform loads (for uniform, resultant would be w l with different symbols).

Final Answer:
w l / 2