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Structures & Materials

Why structures stand up — forces and free body diagrams, the triangle of forces, moments and beam reactions, materials selection, stress and strain.

Topic 7 booklet — PDF coming soon Open NoStressSim Data booklet (formulae) Go to self-check
Revise the topicForces, beams & materials Worked examplesMoments, reactions, stress & strain Test myselfQuiz & confidence check

Exam = tested in the written paper most years. Moments, beam reactions and stress/strain calculations appear every year.

Sections:
Concept 1

Forces and free body diagrams Exam

A force is a push or a pull. Every force has a size (in newtons, N) and a direction. Engineers must understand the forces on a structure to make sure it is strong, stable and safe.

ForceDescription
Weightforce pulling an object towards the centre of the Earth
Applied forcea push or pull from another object
Reaction forcethe force a support pushes back with, to keep the structure in place
Tensiona pulling force in a member, e.g. a rope or cable
Compressiona squeezing force in a member, e.g. a column or strut

Free body diagrams

A free body diagram shows all the forces acting on an object as labelled arrows in the correct direction. Draw it first — it makes sure no force is missed before you calculate.

Free body diagram — simply-supported beam with a central load 600 N (load) RA RB
Loads act down (red); the supports push back up with reactions (green). The reactions must add up to the total downward load.
▶ Simulate it 1

Free body diagram of a loaded beam

  1. Create Beam: add a support at each end and a point load in the middle. Press SOLVE.
  2. Predict first: are the two reactions equal? Why? Compare with the sim, then Export the image as your free body diagram.
  3. Change one thing: move the load nearer one support. Predict which reaction grows, then check.
Open NoStressSim

Tension or compression?

For each structural member, decide which force it carries, then check.

Concept 2

Triangle of forces Exam

When three forces act on an object in equilibrium, the three force vectors form a closed triangle when drawn head to tail. Find unknown forces by accurate scale drawing.

National 5 note: at N5 you find unknown forces by drawing an accurate scale diagram, not by resolving forces. Resolution (sin/cos components) is Higher Engineering Science only.

The scale-drawing method

  1. Identify the three forces acting on the object.
  2. Choose a sensible scale (for example, 1 cm = 100 N).
  3. Draw the first force as an arrow of the correct length and direction.
  4. Draw the second force head to tail, in the correct direction.
  5. Draw the third force from the head of the second back to the tail of the first — it should close the triangle.
  6. Measure the unknown side and convert back using your scale.
Worked example — the hanging sign
Scale drawing
A sign weighing 600 N is held by two cables meeting at a point: one at 30° above the horizontal, the other at 45°. Find the tension in each cable by scale drawing.
  • Scale: 1 cm = 100 N, so the weight is a 6 cm vertical line.
  • From the bottom of the weight line, draw a line at 30° to the horizontal; from the top, a line at 45°.
  • Where the two cable lines meet closes the triangle. Measure each side and convert.
  • Answers: cable at 30° ≈ 440 N; cable at 45° ≈ 540 N (±20 N for drawing accuracy).
▶ Simulate it 2

Triangle of forces — Static Node mode

  1. Static Node mode: build the sign example — a 600 N weight held by cables at 30° and 45°. Do the scale drawing by hand first and predict each tension.
  2. SOLVE and open Show Working (ΣFx = 0, ΣFy = 0). Did your drawing match within about ±20 N?
  3. Change one thing: make both cables steeper (60°/60°). Predict whether the tensions rise or fall, then check. Why do shallow cables carry more tension?
Open NoStressSim
Concept 3

Moments and the principle of moments Exam

The moment of a force is its turning effect about a pivot. When an object is balanced, the principle of moments applies: total clockwise moments = total anticlockwise moments.

Moment
M = F x
force × perpendicular distance · Nm, N, m
Principle of moments
ΣCWM = ΣACWM
clockwise = anticlockwise, about the same pivot
Worked example — moment of a force (substitute & solve)
Substitute & solve
A force of 60 N acts 0.4 m from a pivot. Calculate the moment.
M=F x
M=60 × 0.4
M=24 Nm
Worked example — balance the see-saw (rearrange)
Rearrange — numbers in first
A 200 N child sits 1.5 m to the left of the pivot of a see-saw. Where must a 300 N child sit on the right to balance it?
ΣACWM=ΣCWMprinciple of moments
200 × 1.5=300 × dnumbers in first
d=300 ÷ 300now rearrange
d=1.0 m
▶ Simulate it 3

Moments and the principle of moments

  1. Build a beam pivoted at the centre. Place a 200 N load 1.5 m to the left and a 300 N load on the right. Predict (clockwise = anticlockwise) where the 300 N load balances it, before moving it.
  2. Position it and SOLVE / Show Working — check ΣCWM = ΣACWM.
  3. Change one thing: double the right-hand load. Predict the new balance distance, then test.
Open NoStressSim
Concept 4

Simply-supported beams Exam

A simply-supported beam rests on two supports. To find the reaction at each support, take moments about one support — this removes one unknown reaction from the calculation. Then check: the two reactions must add up to the total downward load.

Worked example 1 — one load
Take moments about A
A 4 m beam rests on supports A and B at each end. A 600 N load is placed 1 m from support A. Find the reaction at each support.
ΣCWM=ΣACWM (about A)R_A drops out
600 × 1=RB × 4
RB=600 ÷ 4 = 150 N
RA=600 − 150 = 450 Nreactions sum to the load
Worked example 2 — two loads
Take moments about A
A 6 m beam rests on supports A and B at each end. A 400 N load is 2 m from A and a 600 N load is 5 m from A. Find the reaction at each support.
400 × 2 + 600 × 5=RB × 6moments about A
3800=6 RB
RB=3800 ÷ 6 = 633 N
RA=1000 − 633 = 367 Nsum check: 1000 N

🏗️ Beam-reactions explorer — live

A 2000 N person stands on a 4 m walkway beam supported at each end (this is Design Brief A from your booklet). Drag their position and watch the two reactions — taking moments about A gives RB = 2000 × x ÷ 4.

—RA (N)
—RB (N)

Challenge: stand the person where support B carries exactly 1500 N.

▶ Simulate it 4

Simply-supported beam reactions

  1. Build Worked Example 2: a 6 m beam, supports at each end, a 400 N load 2 m from A and a 600 N load 5 m from A. Predict both reactions by taking moments about A.
  2. SOLVE and open Show Working — compare your moments line with the sim's ΣCWM = ΣACWM. Do the two reactions add up to 1000 N?
  3. Change one thing: slide the 600 N load to the far support. Predict, then check. Export the Simulation Report PDF as a model worked solution.
Open NoStressSim
◆ Design brief B

Two-cable sign mount

  1. Design a two-cable mount for a 500 N sign so that neither cable tension exceeds 400 N. Using Static Node mode, try different cable angles and watch the tensions.
  2. Record a pair of angles that works, and explain what makes the tensions rise or fall as you change the angles.
Open NoStressSim
Assignment link. Export the Simulation Report (it carries your name and SCN, the diagram and the full ΣCWM = ΣACWM working) and write one sentence justifying your design against the brief — evidence you can use in the Booklet 8 assignment.
Concept 5

Materials selection Exam

To justify a material choice you must name a property and say why that property matters for the application — "aluminium because it is light" scores less than "aluminium because its low density makes the laptop light to carry".

FamilyExamplesTypical use
Ferrous metalsmild steel, cast iron, stainless steelframes, beams, tools
Non-ferrous metalsaluminium, copper, brasswiring, lightweight structures, fittings
Polymers (plastics)PVC, ABS, polythene, nyloncasings, containers, gears
Ceramicsporcelain, glasselectrical insulators, oven parts
CompositesGRP (fibreglass), CFRP (carbon fibre)boats, sports equipment, racing cars
Smart materialsshape memory alloys, thermochromic filmspecialist applications, sensors
PropertyMeaning
Strengthhow much force a material can take before failing
Stiffnesshow much a material resists changing shape
Hardnessresistance to scratching or denting
Toughnessability to absorb energy without breaking (resists impact)
Ductilitycan be drawn into wires; deforms without breaking
Malleabilitycan be hammered or rolled into thin sheets
Elasticityreturns to its original shape after a force is removed
Conductivityallows heat or electricity to flow through it
Corrosion resistanceresists rust and chemical attack
Densitymass per unit volume; high density = heavy material

Justify the choice

Choose the property an engineer would rely on for each application, then check. (Task 2 in your booklet.)

Concept 6

Stress and strain Exam

Stress is the force acting on each unit of cross-sectional area — high stress means a small area carrying a large force, making failure more likely. Strain is the change in length divided by the original length — it has no units, because it is a ratio.

⚠ Units — read this first

  • Work in newtons (N) and square millimetres (mm²) — stress comes out in N mm⁻². Cross-sectional areas of cables and bars are almost always quoted in mm², so there is no need to convert to metres.
  • 1 N mm⁻² = 1 000 000 Pa = 1 MPa. Never divide newtons by mm² and call the answer N/m².
  • For strain, use the same length unit for Δl and l (e.g. change both to metres) before dividing.
Stress
σ = FA
force ÷ cross-sectional area · N mm⁻²
Strain
ε = Δll
change in length ÷ original length · no units
Worked example — stress (substitute & solve)
Substitute & solve
A steel cable carries a load of 5000 N. The cross-sectional area of the cable is 100 mm². Calculate the stress.
σ=F ÷ A
σ=5000 ÷ 100
σ=50 N mm⁻²
Worked example — strain with a unit conversion
Substitute & solve — convert first
A wire has an original length of 2 m. When loaded, it stretches by 4 mm. Calculate the strain.
Convert first
Δl = 4 mm = 0.004 m mm → m
ε=Δl ÷ l
ε=0.004 ÷ 2
ε=0.002no units — a ratio
Worked example — stress and strain together
Multi-step
A bar of cross-sectional area 150 mm² is 3 m long and carries a load of 4500 N. It stretches by 6 mm. Calculate the stress and the strain.
σ=4500 ÷ 150 = 30 N mm⁻²stress first
ε=0.006 ÷ 3 = 0.0026 mm → 0.006 m

Common mistakes — watch out for these in the exam

  • Mixing units in stress — dividing newtons by mm² and writing N/m². Work in N and mm² (N mm⁻²) throughout.
  • Forgetting that strain has no units (it is a ratio).
  • Triangle of forces: not drawing the vectors head-to-tail, or choosing a scale too small to read accurately.
  • Taking moments: leaving out a load, or measuring a distance from the wrong support.
  • Justifying a material by naming a property but not saying why it matters for the application.
Booklet check

Check your booklet work

Try each task in your booklet first, then open the matching answer. Scale-drawing answers are a target value ± a tolerance for drawing accuracy.

Try This — final answers

Try This — Triangle of forces §2
  1. Cable at 40° ≈ 205 N; cable at 60° ≈ 310 N (±20 N)
  2. Strut (60°) ≈ 580 N in compression; horizontal cable ≈ 290 N (±20 N)
  3. Cable at 35° ≈ 400 N; cable at 55° ≈ 575 N (±20 N)
Try This — Moments & balanced beams §3–4

Moments:

  1. M = 80 × 0.5 = 40 Nm
  2. M = 40 × 0.25 = 10 Nm
  3. M = 12 × 0.6 = 7.2 Nm

Balanced beams:

  1. 250 × 2 = 500 × d → d = 1.0 m
  2. 400 × 1.8 = 240 × d → d = 3.0 m
  3. 600 × 1.2 = 300 × d → d = 2.4 m
Try This — Simply-supported beams §5
  1. Moments about A: 900 × 2 = RB × 3 → RB = 600 N; RA = 900 − 600 = 300 N
  2. Moments about A: 500 × 1 + 700 × 4 = RB × 5 → RB = 660 N; RA = 1200 − 660 = 540 N
  3. Moments about A: 1000 × 3 = RB × 4 → RB = 750 N; RA = 1000 − 750 = 250 N
Try This — Stress & strain §7–8

Stress:

  1. σ = 4000 ÷ 500 = 8 N mm⁻²
  2. σ = 2000 ÷ 50 = 40 N mm⁻²
  3. σ = 800 ÷ 80 = 10 N mm⁻²

Strain:

  1. ε = 0.003 ÷ 1.5 = 0.002
  2. ε = 0.001 ÷ 2 = 0.0005
  3. ε = 0.01 ÷ 5 = 0.002

Section practice — final answers

Practice — Forces and free body diagrams §1
  1. Newton (N).
  2. A pulling (stretching) force.
  3. Tension stretches / pulls a material apart; compression squashes / pushes it together.
  4. Weight (the force of gravity).
  5. The reaction force.
  6. FBD: load (weight) arrow down at the centre; an equal upward reaction arrow at each support.
  7. FBD: weight arrow down from the sign; a tension arrow up along each of the two cables.
  8. Compression.
  9. Tension.
  10. It shows every force and its direction so none is missed — the reactions/moments are then set up correctly.
Practice — Triangle of forces §2
  1. The three forces are in equilibrium (they balance — no net force).
  2. Higher.
  3. e.g. 1 cm = 100 N (the 800 N force is then 8 cm long).
  4. Cable at 45° ≈ 155 N; cable at 60° ≈ 220 N (±20 N)
  5. Cable at 30° ≈ 440 N; cable at 45° ≈ 540 N (±20 N)
  6. Strut (60°) ≈ 580 N compression; horizontal cable ≈ 290 N (±20 N)
  7. A larger scale (longer lines) is read more accurately; too small a scale magnifies drawing errors.
  8. A closed triangle means the vectors return head-to-tail to the start — they sum to zero, so the object is in equilibrium.
  9. 4.5 × 100 = 450 N.
  10. e.g. not drawing head-to-tail, or the wrong angle — use a sharp pencil, a protractor and an accurate scale.
Practice — Moments and beams §3–5
  1. M = F x (moment = force × perpendicular distance).
  2. For a balanced object, total clockwise moments = total anticlockwise moments about the same pivot.
  3. M = 40 × 0.25 = 10 Nm
  4. M = 120 × 0.6 = 72 Nm
  5. 350 × 1.2 = 420 × d → d = 1.0 m
  6. Moments about A: 800 × 1 = RB × 4 → RB = 200 N; RA = 600 N
  7. Moments about A: 1000 × 2 + 600 × 5 = RB × 6 → RB = 833 N; RA = 767 N
  8. Central load → RA = RB = 600 N
  9. Moments about A: 600 × 1 + 400 × 2 = RB × 3 → RB = 467 N; RA = 533 N
  10. e.g. forgetting a load or using the wrong distance — take moments about a support to remove one reaction, and measure every distance from that pivot.
Practice — Materials selection §6
  1. Copper — high electrical conductivity.
  2. The ability to absorb energy / resist fracture from an impact without breaking.
  3. It can be drawn out (stretched permanently) into a wire without breaking.
  4. It returns to its original shape once the load is removed.
  5. Aluminium: low density (light to carry) + good corrosion resistance + adequate strength/stiffness — property + why.
  6. Any two, e.g. strength (carry wind/loads) and corrosion resistance/durability (survive weather).
  7. Toughened glass / glass-ceramic — heat resistance (and transparency).
  8. Hardened (carbon) steel — hardness and toughness.
  9. CFRP: very high strength-to-weight ratio (strong yet light) and high stiffness — two properties justified.
  10. Any two: strong, cheap, readily available, easily welded/joined.
Practice — Stress and strain §7–8
  1. σ = F ÷ A.
  2. ε = Δl ÷ l.
  3. N mm⁻² (newtons per square millimetre).
  4. It is a ratio of two lengths, so the units cancel.
  5. σ = 30 000 ÷ 20 000 = 1.5 N mm⁻²
  6. σ = 6000 ÷ 200 = 30 N mm⁻²
  7. σ = 1500 ÷ 25 = 60 N mm⁻²
  8. ε = 0.001 ÷ 2 = 0.0005
  9. ε = 0.008 ÷ 4 = 0.002
  10. Stress: σ = 4500 ÷ 150 = 30 N mm⁻²; strain: ε = 0.006 ÷ 3 = 0.002
Check yourself

Check yourself

Mixed multiple choice

One question from every part of the topic. Choose an answer for each, then mark them.

Rate your confidence — the booklet's success criteria

These are the success criteria from your booklet. Red = not yet, Amber = getting there, Green = confident. Saved on this device.

Ready for exam questions? Open the Past Paper Finder and look for the structures and materials questions.

Sources & credits: The Topic 7 booklet © R Stewart, 2026. NoStressSim is a free simulator by R Stewart. The Past Paper Finder is compiled by Mr McDonald, 2024; past-paper questions © Qualifications Scotland (SQA). The N4/N5 data booklet is reproduced for educational use, © Qualifications Scotland (SQA).