A simply supported beam of span 4.0\ m carries a central point load of 600\ N. Find the maximum bending moment. [2 marks]

SQA SQA AH style-style practice: A simply supported beam of span 4.0\ m carries a single central point load of 800\ N. Calculate the reactions, the maximum shear force and the maximum bending moment.

By symmetry the two reactions are equal. Reactions: R_A = R_B = 8002 = 400\ N. Maximum shear force equals the reaction, 400\ N (constant either side of the load, changing sign under it). Maximum bending moment is at the centre, under the load. M_max = R_A × L2 = 400 × 2.0 = 800\ N m. Markers reward equal reactions of 400\ N, a maximum shear of 400\ N, and a central maximum bending moment of 800\ N m. For a central point load M_max = WL/4 = 800 × 4/4 = 800\ N m, the same answer.

ScotlandEngineering ScienceSyllabus dot point

How do we find the shear force and bending moment along a loaded beam, and why does beam shape matter?

The analysis of loaded beams: shear force and bending moment, their diagrams along the span, and the role of the second moment of area in resisting bending.

An SQA Advanced Higher Engineering Science answer on loaded beams, covering shear force and bending moment, how to draw shear force and bending moment diagrams along a span, and the role of the second moment of area in resisting bending.

Generated by Claude Opus 4.816 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A beam carrying transverse loads experiences two internal effects. The shear force at a section is the net transverse force to one side of it; the bending moment is the net moment to one side. Drawn along the span, the shear force diagram steps down at each downward load, and the bending moment diagram rises to a maximum where the shear force passes through zero. For a simply supported beam with a central point load $W$ and span $L$ , the maximum bending moment is $\dfrac{WL}{4}$ at mid-span; for a uniformly distributed load $w$ per unit length it is $\dfrac{wL^2}{8}$ . How well a beam resists bending depends on its second moment of area $I$ : placing material far from the neutral axis (as in an I-section) gives a large $I$ , so the beam is stiffer and stronger for the same amount of material.

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What this key area is asking
Shear force and bending moment
Shear force and bending moment diagrams
The second moment of area
Examples in context
Try this

What this key area is asking

The SQA wants you to analyse a loaded beam: to find the shear force and bending moment at points along its span, to draw shear force and bending moment diagrams, and to explain how the second moment of area of the cross-section governs its resistance to bending. This extends the equilibrium of structures into how a beam carries load along its length.

Shear force and bending moment

To find them at any point, make an imaginary cut and consider the forces on one side: add up the vertical forces for the shear force, and take moments about the cut for the bending moment. A beam must resist both: shear across the section and bending along its length, which is why both are calculated.

Shear force and bending moment diagrams

A shear force diagram plots the shear force along the span. Starting from a support reaction, it stays constant between loads and steps down by each downward point load (and rises at an upward reaction); under a uniformly distributed load it slopes. A bending moment diagram plots the bending moment along the span. It is zero at the simply supported ends and rises to a maximum where the shear force is zero (changes sign).

For the two standard cases the maximum bending moment is worth remembering:

M_{max} = \frac{WL}{4} \ \text{(central point load }W\text{)}, \qquad M_{max} = \frac{wL^2}{8} \ \text{(UDL }w\text{ per unit length)}.

The second moment of area

The second moment of area is the shape factor in bending: a larger $I$ means a stiffer, stronger beam for the same material. This is why beams are made deep rather than wide, and why an I-section puts most of its material in the top and bottom flanges, far from the neutral axis, with only a thin web between. The same mass of steel resists far more bending as an I-beam than as a square bar.

Reactions, shear and bending for a central point load

A simply supported beam spans $6.0\ \text{m}$ and carries a $1200\ \text{N}$ point load at its centre. Find the reactions, the maximum shear force and the maximum bending moment.

step 1 Find the reactions

By symmetry the load splits equally: $R_A = R_B = \dfrac{1200}{2} = 600\ \text{N}$ .

step 2 Find the maximum shear force

Either side of the centre the shear force equals the reaction, so the maximum shear is $600\ \text{N}$ (it changes sign under the load).

step 3 Find the maximum bending moment

At mid-span $M_{max} = \dfrac{WL}{4} = \dfrac{1200 \times 6.0}{4} = 1800\ \text{N m}$ , the peak that the section must be sized to resist.

Examples in context

A floor joist is sized so its bending moment under the floor load stays within safe limits, which is why joists are deep and set on edge, not flat. A steel I-beam in a building frame uses its large second moment of area to span far with little material. A diving board is a cantilever whose bending moment is greatest at the fixed end, so it is thickest there. A bridge girder is checked at the section of maximum bending moment. In each case the shear force and bending moment diagrams locate the critical section, and the second moment of area sets how much the chosen shape can resist.

Try this

Q1. State what the shear force at a section of a beam represents. [1 mark]

Cue. The net transverse (vertical) force acting to one side of the section.

Q2. A simply supported beam of span $4.0\ \text{m}$ carries a central point load of $600\ \text{N}$ . Find the maximum bending moment. [2 marks]

Cue. $M_{max} = \dfrac{WL}{4} = \dfrac{600 \times 4.0}{4} = 600\ \text{N m}$ .

Q3. State why an I-section resists bending well for its weight. [2 marks]

Cue. It places most of its material in the flanges, far from the neutral axis, giving a large second moment of area.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style5 marksA simply supported beam of span

4.0\ \text{m}

carries a single central point load of

800\ \text{N}

. Calculate the reactions, the maximum shear force and the maximum bending moment.

Show worked answer →

By symmetry the two reactions are equal.

Reactions: $R_A = R_B = \dfrac{800}{2} = 400\ \text{N}$ .

Maximum shear force equals the reaction, $400\ \text{N}$ (constant either side of the load, changing sign under it).

Maximum bending moment is at the centre, under the load.

$M_{max} = R_A \times \dfrac{L}{2} = 400 \times 2.0 = 800\ \text{N m}$ .

Markers reward equal reactions of $400\ \text{N}$ , a maximum shear of $400\ \text{N}$ , and a central maximum bending moment of $800\ \text{N m}$ . For a central point load $M_{max} = WL/4 = 800 \times 4/4 = 800\ \text{N m}$ , the same answer.

SQA AH style4 marksExplain why an I-section beam is more efficient at resisting bending than a solid rectangular bar of the same cross-sectional area, referring to the second moment of area.

Show worked answer →

Bending resistance depends on the second moment of area, which grows rapidly as material is placed further from the neutral axis.

An I-section concentrates most of its material in the flanges, far from the neutral axis, giving a large second moment of area.

A solid bar of the same area has material spread evenly, much of it near the neutral axis where it does little to resist bending.

So the I-section has a larger second moment of area for the same amount of material, making it stiffer and stronger in bending, and lighter for a given strength.

Markers reward linking bending resistance to the second moment of area, explaining that the I-section places material far from the neutral axis (in the flanges), and concluding it resists bending better for the same area.

Related dot points

Sources & how we know this

Advanced Higher Engineering Science Course Specification (C823 77) — SQA (2019)
Advanced Higher Engineering Science Data Booklet — SQA (2019)