Part I: Statics · Topic 4 of 8 · 21 min read

Distributed Loads, Shear & Bending

Connect the load diagram, internal actions and structural response.

What you'll learn

Find equivalent forces for uniform, triangular and varying loads.
Calculate external reactions before taking sections.
Develop piecewise shear-force and bending-moment equations.
Sketch SFDs and BMDs with correct slopes, jumps and boundary values.

Three parts stacked and lined up vertically. Top: a horizontal beam with a pin support on the left and a roller on the right, with evenly spaced downward arrows across the full span. Middle: a shear diagram in which a straight line falls from a positive value at the left end, through zero at midspan, to an equal negative value at the right end. Bottom: a bending moment diagram with one smooth parabola, zero at both ends and highest at midspan. Labels: uniform downward load w (kN/m); V(x); M(x). Centred caption below: V decreases linearly; M is parabolic

Understand the concepts

External reactions and internal resultants are different

Reactions are external support forces. Shear $V(x)$ and bending $M(x)$ are internal effects exposed by cutting the beam at location $x$ . A single load resultant can solve reactions, but does not replace the distributed load when finding internal-action functions.

State the section-cut convention

For the LEFT isolated beam segment in these notes, positive $V$ acts downward at the cut and positive sagging acts CCW at the cut. This produces and for downward-positive load intensity .

Try it yourself

Move the load along the beam and watch the two reactions trade off.

Assumptions and sign conventions

Straight Euler-Bernoulli-style beam idealisation; static planar loading.
Small deformation is assumed for equilibrium geometry; shear deformation and deflection are outside these introductory examples.
Downward distributed-load intensity $w(x)$ is positive in the differential relations shown.
Use N–m or kN–m consistently; do not mix mm with m without conversion.

Equations that matter

R=\int_a^b w(x)\,dx,\qquad \bar x=\frac{\int_a^b xw(x)\,dx}{R}

Step-by-step method

Draw all support reactions and compute them using whole-beam equilibrium.
Define the x-origin, loading regions and the positive internal sign convention.
Make cuts in every interval separated by point loads, couples or loading changes.
Apply section equilibrium (or integrate load and shear) to obtain $V(x)$ and $M(x)$ .
Apply point-force jumps to V and couple jumps to M where needed.
Check support-end boundary conditions, moments at cuts and the location of maximum .

Worked example: Uniformly loaded simply supported beam

A $6.0\,\mathrm m$ simply supported beam carries a uniform $2.0\,\mathrm{kN/m}$ downward load along its full span. Find reactions, $V(x)$ , $M(x)$ and maximum positive bending moment.

Common mistakes

Using the load resultant at midspan to compute every internal shear instead of retaining the actual load over each section.
Misplacing a triangular load centroid.
Drawing a straight moment line under uniform loading (it must be parabolic).
Confusing the location of zero shear with a zero bending moment.

Check yourself

What does a downward point load do to a shear-force diagram?
Where is the centroid of a triangular load?
Why can a bending moment be nonzero when the shear force is zero?

Ready to try it yourself?

Try Practice Questions (4)

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Distributed Loads, Shear & Bending

What you'll learn

Understand the concepts

External reactions and internal resultants are different

State the section-cut convention

Try it yourself

Assumptions and sign conventions

Equations that matter

Step-by-step method

Worked example: Uniformly loaded simply supported beam

Common mistakes

Check yourself

Load intensity controls shear slope

Shear controls moment slope

Boundary values and extrema are clues

Step 1: Compute the external resultant

Step 2: Cut at an arbitrary section

Step 3: Find the bending moment

Step 4: Locate the maximum