Skip to main content

Beams, shear force and bending moments for QCE Engineering Unit 3

Syllabus dot point

“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”

QCEEngineeringUnit 3: Civil structures6 min read

Quick answer

A loaded beam carries internal shear force VV (the net transverse force on a cut section) and internal bending moment MM (the net turning effect on that section). To find them at any point, cut the beam there and apply equilibrium to one side. Shear force equals the sum of vertical forces on that side; bending moment equals the sum of their moments about the cut. The maximum bending moment marks the most highly stressed section, where bending stress is greatest and failure is most likely. For a central point load WW on a simply supported span LL, the maximum bending moment is Mmax=WL/4M_{max} = WL/4 at midspan. Larger bending moments demand a deeper section or a stiffer material.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

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 VV: the net force acting transverse to the beam axis, tending to slide one part of the beam past the other.
  • Bending moment MM: 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=∑F(one side)M=∑(F×d)(one side)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:

σ=MyI\sigma = \frac{M y}{I}

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

Standard results worth knowing

For a simply supported beam of span LL:

  • Central point load WW: maximum bending moment Mmax=WL4M_{max} = \dfrac{WL}{4} at midspan.
  • Uniformly distributed load of total weight WW: maximum bending moment Mmax=WL8M_{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 EIEI, 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 mL = 4.0\,\text{m} between supports A and B. A single point load W=6000 NW = 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:

RA=RB=W2=60002=3000 NR_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 RAR_A acts on that part, at a distance of 2.0 m2.0\,\text{m}:

Mmax=RA×L2=3000×2.0=6000 N mM_{max} = R_A \times \frac{L}{2} = 3000 \times 2.0 = 6000\,\text{N m}

Check against the standard formula:

Mmax=WL4=6000×4.04=6000 N m✓M_{max} = \frac{WL}{4} = \frac{6000 \times 4.0}{4} = 6000\,\text{N m} \checkmark

The maximum bending moment of 6000 N m6000\,\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.

Common errors
Confusing shear force with bending moment
Shear is a force (newtons); bending moment is a turning effect (newton metres). They are different quantities with different units and different peak locations.
Forgetting the distributed-load case
A uniformly distributed load gives WL/8WL/8, not WL/4WL/4. Using the point-load formula for a distributed load overestimates the moment.
Including forces from the wrong side
When you cut the beam, use the forces on one side only. Mixing both sides double-counts the load.
Sizing on strength alone
A beam can be strong enough yet sag too far. Both the bending-stress limit and the deflection limit must be checked.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks
A simply supported beam of span 6.0 m6.0\ \text{m} carries a single point load of 24 kN24\ \text{kN} at 2.0 m2.0\ \text{m} from the left support A. Determine the support reactions, then determine the bending moment directly under the load.
Show worked answer →

A 7 mark determine question rewards correct reactions then the moment at the load.

Reactions by taking moments about A: RB×6.0=24×2.0R_B \times 6.0 = 24 \times 2.0, so RB=48/6.0=8.0 kNR_B = 48/6.0 = 8.0\ \text{kN}. Vertical equilibrium: RA=24−8.0=16 kNR_A = 24 - 8.0 = 16\ \text{kN}.

Bending moment under the load (using the left segment): M=RA×2.0=16×2.0=32 kN mM = R_A \times 2.0 = 16 \times 2.0 = 32\ \text{kN m}.

Markers reward the moment equation about a support, both reactions summing to the load, and the maximum bending moment of 32 kN m32\ \text{kN m} taken under the point load.

Practice4 marks
Justify why the maximum bending moment, rather than the maximum shear force, is usually the critical quantity when checking a simply supported beam against bending failure.
Show worked answer →

A 4 mark justify answer needs the link between bending moment and bending stress.

Bending stress is proportional to bending moment, σ=My/I\sigma = My/I, so the point of maximum bending moment carries the largest tensile and compressive fibre stresses, which is where bending failure (yielding or fracture of the extreme fibres) initiates. Shear force governs a different failure mode and for typical slender beams the bending stress reaches the material limit first.

Markers reward tying maximum moment to maximum bending stress via σ=My/I\sigma = My/I and concluding it is the critical location for bending failure.

Practise this

Sources & how we know this

ExamExplained