Cantilever — Uniform Distributed Load

Calculate support reaction, shear force, and bending moment for a cantilever beam under a full-span UDL

Inputs

w

UDL Intensity

N/mm

l

Span

Formula Interpretation

Support Reaction

R = wl \quad \text{(N)}

The UDL of intensity w (N/mm) over full span l is equivalent to a single concentrated force W = wl (N) at the midpoint. The fixed support must carry this entire load upward.

Shear Force

F_x = wx \quad (0 \le x \le l)

The shear force increases linearly from zero at the free end (x = 0) to the maximum value wl at the fixed support (x = l). The SFD is a straight line (first-degree polynomial).

Bending Moment

M_x = \dfrac{wx^2}{2} \qquad M_{\max} = \dfrac{wl^2}{2}

The bending moment increases parabolically from zero at the free end to Mmax = wl²/2 at the fixed end. The BMD is a second-degree (parabolic) curve. The fixed end is the critical section.

Knowledge Points

UDL Equivalence

A uniformly distributed load w (N/mm) over length l can be replaced by a single equivalent concentrated load W = wl (N) acting at the centroid of the loaded region — the midpoint l/2 for a uniform load. This simplifies reaction calculations.

Linear Shear Force Diagram

Unlike the constant SFD under a concentrated end load, the UDL produces a linearly varying shear force. It is zero at the free end and reaches its maximum magnitude wl at the fixed support. The SFD is a straight inclined line.

Parabolic Bending Moment Diagram

The bending moment under a UDL follows a parabolic (second-order) distribution. This is because bending moment is the integral of shear force: integrating a linear function gives a quadratic. The BMD is concave-up, with the vertex at the free end.

Worked Example

A cantilever beam of span l = 1500 mm carries a UDL of w = 3 N/mm (3 kN/m) over its full length. Find the support reaction, maximum shear force, and maximum bending moment. Also find the shear force and bending moment at x = 750 mm from the free end.

Knowns

• Span: l = 1500 mm
• UDL intensity: w = 3 N/mm (= 3 kN/m)
• Section position: x = 750 mm (from free end B)

Solution

Step 1 — Support Reaction (Formula ①)

R = wl = 3 \times 1500 = 4500 \text{ N}

Step 2 — Shear Force at fixed end A (x = l)

F_{\max} = wl = 3 \times 1500 = 4500 \text{ N}

Step 3 — Shear Force at x = 750 mm

F_x = wx = 3 \times 750 = 2250 \text{ N}

Step 4 — Max Bending Moment at fixed end A (Formula ③)

M_{\max} = \dfrac{wl^2}{2} = \dfrac{3 \times 1500^2}{2} = 3{,}375{,}000\,\text{N·mm} = 3375\,\text{N·m}

Step 5 — Bending Moment at x = 750 mm

M_x = \dfrac{wx^2}{2} = \dfrac{3 \times 750^2}{2} = 843{,}750\,\text{N·mm} = 843.75\,\text{N·m}

Result: R = 4500 N; F_max = 4500 N (at A); Mmax = 3375 N·m (at A); F(750mm) = 2250 N; M(750mm) = 843.75 N·m.

Extended Knowledge

•A canopy extending from a building wall acts as a cantilever. Snow accumulation creates a nearly uniform distributed load. Engineers calculate the bending moment at the wall connection (fixed end) to size the structural section.
•A vertical retaining wall embedded in the ground experiences lateral soil or water pressure that varies with depth (triangular, not uniform). Cantilever UDL analysis is the starting point; more complex distributions require numerical methods.
•Every horizontal beam carries its own weight as a UDL equal to its mass per unit length times g. Self-weight is often the dominant load in long-span cantilevers and must be included in all structural calculations.
•A sign post or mast treated as a vertical cantilever with wind pressure as a uniform lateral load. The bending moment at the base equals wl²/2, which determines the required section modulus and anchor bolt design.
•The parabolic BMD shows that the moment increases rapidly near the fixed end. A tapered cross-section — larger at the root and smaller at the tip — matches the moment distribution, saving material and reducing self-weight.