Determine the shear force and bending moment at points along a simply supported beam under point and distributed loads, and relate maximum bending moment to the risk of structural failure

A QCE Engineering Unit 3 answer on beam bending. Covers shear force and bending moment, how to find them at a section, the meaning of maximum bending moment, and how bending stress and deflection link to material choice, with worked numbers.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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What this dot point is asking

QCAA wants you to look inside a loaded beam and quantify the two internal effects that bending produces: shear force and bending moment. You need to find their values at points along the beam, identify where the bending moment is greatest, and explain why that location is where the beam is most likely to fail. This extends the equilibrium and stress work to the everyday structural element of a floor joist, lintel or bridge girder.

The answer

Shear force and bending moment

When a beam bends, each internal cross-section carries two internal reactions:

Shear force $V$ : the net force acting transverse to the beam axis, tending to slide one part of the beam past the other.
Bending moment $M$ : the net turning effect about the section, tending to curve the beam.

To find them at a chosen point, make an imaginary cut there and consider the equilibrium of the part to one side. The shear force is the algebraic sum of the vertical forces on that part; the bending moment is the algebraic sum of their moments taken about the cut:

V = \sum F_{\text{(one side)}} \qquad M = \sum (F \times d)_{\text{(one side)}}

Sign conventions vary, so the key is consistency: sagging (concave up) bending is conventionally positive.

Where the beam is most stressed

Bending moment varies along the beam, and the section carrying the largest bending moment is the most highly stressed. The bending stress at the outer fibre of a beam is:

\sigma = \frac{M y}{I}

where $y$ is the distance from the neutral axis to the outer fibre and $I$ is the second moment of area of the cross-section. Because $\sigma$ is proportional to $M$ , the point of maximum bending moment is the point of maximum bending stress, and therefore the most likely failure location. A deeper section (larger $I$ ) or a higher-strength material reduces the stress.

Standard results worth knowing

For a simply supported beam of span $L$ :

Central point load $W$ : maximum bending moment $M_{max} = \dfrac{WL}{4}$ at midspan.
Uniformly distributed load of total weight $W$ : maximum bending moment $M_{max} = \dfrac{WL}{8}$ at midspan.

Deflection (how far the beam sags) depends on the load, the span cubed or to the fourth power, and on $EI$ , the product of Young's modulus and second moment of area. A serviceable beam must satisfy both a strength limit (stress below the allowable) and a deflection limit.

Bending moment under a central load

Find the maximum bending moment and reactions

A simply supported beam spans $L = 4.0\,\text{m}$ between supports A and B. A single point load $W = 6000\,\text{N}$ acts at midspan. Self-weight is neglected. Find the support reactions and the maximum bending moment.

Reactions. By symmetry the load splits equally:

R_A = R_B = \frac{W}{2} = \frac{6000}{2} = 3000\,\text{N}

Bending moment at midspan. Cut at midspan and take moments of the left-hand part about the cut. Only $R_A$ acts on that part, at a distance of $2.0\,\text{m}$ :

M_{max} = R_A \times \frac{L}{2} = 3000 \times 2.0 = 6000\,\text{N m}

Check against the standard formula:

M_{max} = \frac{WL}{4} = \frac{6000 \times 4.0}{4} = 6000\,\text{N m} \checkmark

The maximum bending moment of $6000\,\text{N m}$ occurs at midspan, directly under the load, which is where this beam would first fail in bending.

Why this matters for civil structures

Every floor, roof and bridge deck is a beam problem. Engineers locate the maximum bending moment, compute the bending stress, and pick a section deep enough to keep that stress safe while also limiting deflection. This is why structural beams are deep rather than wide and why an I-section concentrates material at the top and bottom, far from the neutral axis where it does the most good.