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Stress, strain and Young's modulus: HSC Engineering Studies Civil Structures

Syllabus dot point

“Define and calculate stress, strain and Young's modulus, interpret stress-strain curves for ductile and brittle materials, and apply the relationships to typical civil engineering materials”

7 min read

Quick answer

Stress is force on area, strain is change in length on original length, and Young's modulus is the elastic-region slope of stress over strain. Ductile materials yield then harden; brittle materials snap. Structural steel sits at E=200E = 200 GPa with yield around 250 MPa, well above the working stresses used in design.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions
  4. Practice questions
In plain English

Imagine pulling on a long spring versus pulling on a dry stick of spaghetti. The spring stretches smoothly and springs back if you let go gently, but pull it too far and it stays stretched out of shape, then eventually snaps after a lot of visible stretching, that is a ductile material like mild steel. The spaghetti barely stretches at all; it feels rigid right up until the instant it snaps cleanly in two, with almost no warning, that is a brittle material like cast iron or concrete in tension. Stress is just how hard you are pulling relative to how thick the thing is, strain is how much it stretches relative to its original length, and Young's modulus is a number that tells you how "spring-like" versus "spaghetti-like" a material is in the stretchy, spring-back part of its behaviour, before anything permanent happens. Civil engineers care about both: a stiff, spring-like material resists sagging, but its strength (how hard you can pull before it yields or snaps) is a separate question entirely.

What this dot point is asking

NESA wants you to define stress, strain and Young's modulus, calculate them from a load case, and read a stress-strain curve to identify yield point, ultimate tensile strength and the elastic and plastic regions. You must also recognise the differences between ductile materials (mild steel) and brittle materials (cast iron, concrete in tension).

The answer

Definitions

Stress is force per unit cross-sectional area:

σ=FA\sigma = \frac{F}{A}

with units of pascals (Pa) or megapascals (MPa).

Strain is the fractional change in length:

ε=ΔLL\varepsilon = \frac{\Delta L}{L}

Strain is dimensionless.

Young's modulus (the modulus of elasticity) is the slope of the elastic part of the stress-strain curve:

E=σεE = \frac{\sigma}{\varepsilon}

Typical values: structural steel E=200E = 200 GPa, aluminium alloys E=70E = 70 GPa, concrete E=25E = 25 to 35 GPa, timber E=10E = 10 GPa.

The three defining formulas

σ=FAε=ΔLLE=σε\sigma = \frac{F}{A} \qquad \varepsilon = \frac{\Delta L}{L} \qquad E = \frac{\sigma}{\varepsilon}

  • Stress σ\sigma: force per unit area, in Pa or MPa.
  • Strain ε\varepsilon: change in length over original length, dimensionless.
  • Young's modulus EE: the slope of the elastic (straight-line) region only, a measure of stiffness, not strength.
  • Combine all three into a single working formula for extension: ΔL=FLAE\Delta L = \dfrac{FL}{AE}.

Young's modulus compared across four structural materials A bar chart comparing typical Young's modulus values in gigapascals for four civil engineering materials: structural steel at 200 GPa, aluminium alloy at 70 GPa, concrete at 30 GPa, and timber at 10 GPa. Structural steel is by far the stiffest, roughly twenty times stiffer than timber. 200 GPa 70 GPa 30 GPa 10 GPa Steel Aluminium Concrete Timber Higher bar = stiffer material (less elastic deflection for the same stress).

Stress-strain curves

Ductile materials (mild steel, structural steel grades like 250 and 350) show a linear elastic region up to the yield point, a yield plateau, then strain hardening up to the ultimate tensile stress, then necking and fracture. Typical yield stress for grade 250 structural steel is 250 MPa; ultimate tensile strength is around 410 MPa.

Brittle materials (cast iron, concrete in tension, glass) show an almost linear curve with little or no plastic deformation before fracture. They fail suddenly.

The area under the stress-strain curve is the energy absorbed per unit volume before failure (toughness). Ductile materials have high toughness; brittle materials have low toughness.

Key points on a stress-strain curve

Reading a tensile-test curve is a common HSC task. The features you must label, in order of increasing strain, are:

  • Limit of proportionality. The highest stress at which stress is exactly proportional to strain (Hooke's law holds). The curve is straight up to this point.
  • Elastic limit. The highest stress from which the material returns to its original length when unloaded. Beyond it, deformation is permanent.
  • Yield point (upper and lower). In mild steel the curve drops slightly then runs almost flat. This yield plateau is where the material deforms plastically at roughly constant stress. The lower yield stress is taken as fyf_y for design.
  • Ultimate tensile strength (UTS). The peak stress on the curve. Beyond it the specimen necks (cross-section reduces locally).
  • Fracture point. Where the specimen breaks. Because engineering stress uses the original area, the plotted fracture stress sits below the UTS even though the true stress is rising.

Two derived properties come straight off the curve. Ductility is the total plastic strain at fracture (high for mild steel, near zero for cast iron). Toughness is the area under the whole curve, the energy absorbed per unit volume before fracture.

Annotated engineering stress-strain curve for mild steel A plot of engineering stress against strain for a mild steel tensile test specimen. The curve rises in a straight line through the elastic region to the limit of proportionality, continues to the elastic limit and yield point at about 250 megapascals, runs along a near-flat yield plateau, rises again through strain hardening to the ultimate tensile strength peak of about 410 megapascals, then falls to a lower fracture stress as the specimen necks. Marked points sit directly on the drawn curve at the yield point, the ultimate tensile strength, and the fracture point. 410 300 200 100 0 Yield point about 250 MPa UTS about 410 MPa Fracture necked section Elastic region (straight) then yield plateau then strain hardening Strain (increasing to the right, not to scale near fracture) Stress (MPa) Illustrative ExamExplained curve for grade 250 mild steel; the flat yield plateau and long strain-hardening run mark it as ductile.

Hooke's law and the elastic relationship

In the elastic region the three quantities link directly. Rearranging E=σ/εE = \sigma / \varepsilon and substituting the definitions gives a single working equation for extension:

ΔL=FLAE\Delta L = \frac{F L}{A E}

This is the form to reach for when a question gives load, geometry and modulus and asks for deflection or extension. It shows extension is proportional to load and length, and inversely proportional to area and stiffness.

Application in civil structures

Civil engineers use allowable stress well below the yield stress, dividing by a factor of safety of 1.5 to 3. The allowable (working) stress is σallow=fy/FoS\sigma_{\text{allow}} = f_y / FoS. A member is acceptable when its working stress stays below this value; the actual factor of safety is the yield stress divided by the working stress. Concrete is strong in compression (typical 32 MPa) but weak in tension (about 3 MPa), which is why it is reinforced with steel.

Stiffness versus strength is a distinction examiners probe. Young's modulus measures stiffness (resistance to elastic deflection); yield and ultimate stress measure strength (resistance to permanent deformation and failure). A material can be stiff but brittle (cast iron) or strong but relatively flexible. Selecting a beam to limit deflection is a stiffness problem governed by EE; selecting one to avoid yielding is a strength problem governed by fyf_y. The Sydney Harbour Bridge arch, for example, needed both high stiffness to limit sag and high strength to carry the compression in its members.

Working through a stress-strain calculation

Work in a consistent order every time: (1) convert every given quantity to base SI units (mm2^2 to m2^2, GPa to Pa, mm to m), (2) find stress from σ=F/A\sigma = F/A, (3) find strain from ε=σ/E\varepsilon = \sigma/E (or go straight to extension via ΔL=FL/(AE)\Delta L = FL/(AE)), (4) if a factor of safety is asked for, compare against the ALLOWABLE stress fy/FoSf_y/FoS, not the raw yield stress. Writing the unit on every line catches most arithmetic slips before they reach the final answer.

Worked example

An aluminium strut of area 400 mm2400 \text{ mm}^2 and length 1.51.5 m carries 28 kN compression. E=70E = 70 GPa.

Step 1: Calculate the stress

σ=28×103/(400×10−6)=70\sigma = 28 \times 10^3 / (400 \times 10^{-6}) = 70 MPa.

Step 2: Calculate the strain

ε=70×106/(70×109)=1.0×10−3\varepsilon = 70 \times 10^6 / (70 \times 10^9) = 1.0 \times 10^{-3}.

Step 3: Calculate the change in length

ΔL=1.0×10−3×1.5=1.5\Delta L = 1.0 \times 10^{-3} \times 1.5 = 1.5 mm of shortening.

Common traps
Mixing units
Convert mm2^2 to m2^2 (divide by 10610^6), and GPa to Pa (multiply by 10910^9). Or work entirely in MPa and mm consistently.
Using diameter for area
Cross-sectional area of a circular bar is A=πd2/4A = \pi d^2 / 4, not πd2\pi d^2.
Treating concrete as ductile
Concrete is brittle in tension. That is the entire reason for reinforced concrete: steel takes the tensile stresses.
Reading Young's modulus from the wrong part of the curve
EE is the slope of the elastic (initial linear) region only. Past yield the relationship is no longer linear.

Exam-style questions

Questions in the style of NESA 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.

HSC 2021-style2 marks
Define mechanical strain and state why it is a dimensionless quantity.
Show worked answer →

Strain is the ratio of the change in length of a member to its original length, ε=ΔL/L\varepsilon = \Delta L / L. Because it is a length divided by a length, the units cancel, so strain has no units (it is dimensionless), often expressed as microstrain (×10−6\times 10^{-6}) or as a percentage. Markers award one mark for the correct definition or formula and one mark for explaining that dividing length by length leaves no units.

HSC 2021-style4 marks
A steel tie bar of cross-sectional area 250 mm^2 and original length 2.0 m carries a tensile load of 50 kN. Young's modulus for the steel is 200 GPa. Calculate the stress, strain and extension of the bar.
Show worked answer →

Convert units. A=250 mm2=250×10−6 m2A = 250 \text{ mm}^2 = 250 \times 10^{-6} \text{ m}^2. E=200 GPa=200×109 PaE = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa}.

Stress.

σ=FA=50×103250×10−6=2.0×108 Pa=200 MPa\sigma = \frac{F}{A} = \frac{50 \times 10^3}{250 \times 10^{-6}} = 2.0 \times 10^8 \text{ Pa} = 200 \text{ MPa}

Strain.

ε=σE=2.0×108200×109=1.0×10−3\varepsilon = \frac{\sigma}{E} = \frac{2.0 \times 10^8}{200 \times 10^9} = 1.0 \times 10^{-3}

Extension.

ΔL=ε×L=1.0×10−3×2.0=2.0×10−3 m=2.0 mm\Delta L = \varepsilon \times L = 1.0 \times 10^{-3} \times 2.0 = 2.0 \times 10^{-3} \text{ m} = 2.0 \text{ mm}

Markers reward (1) consistent SI units throughout, (2) the formula for stress used first, then Young's modulus to find strain, then the strain definition to find extension, and (3) units on every line. Stating that 200 MPa is well below the yield stress of structural steel (about 250 MPa) earns the engineering-reasoning mark.

HSC 2023-style6 marks
A circular aluminium tie rod of diameter 20 mm and length 3.0 m carries a tensile load of 40 kN. Young's modulus is 70 GPa and the yield stress is 250 MPa. Determine the tensile stress, the extension under load, and assess whether the rod is safe using a factor of safety of 2.5.
Show worked answer →

Cross-sectional area.

A=πd24=π(0.020)24=3.14×10−4 m2A = \frac{\pi d^2}{4} = \frac{\pi (0.020)^2}{4} = 3.14 \times 10^{-4} \text{ m}^2

Stress.

σ=FA=40×1033.14×10−4=1.27×108 Pa=127 MPa\sigma = \frac{F}{A} = \frac{40 \times 10^3}{3.14 \times 10^{-4}} = 1.27 \times 10^8 \text{ Pa} = 127 \text{ MPa}

Strain and extension.

ε=σE=127×10670×109=1.82×10−3\varepsilon = \frac{\sigma}{E} = \frac{127 \times 10^6}{70 \times 10^9} = 1.82 \times 10^{-3}

ΔL=εL=1.82×10−3×3.0=5.45×10−3 m=5.45 mm\Delta L = \varepsilon L = 1.82 \times 10^{-3} \times 3.0 = 5.45 \times 10^{-3} \text{ m} = 5.45 \text{ mm}

Safety assessment. The allowable stress is σallow=fy/FoS=250/2.5=100\sigma_{\text{allow}} = f_y / FoS = 250 / 2.5 = 100 MPa. The working stress (127 MPa) exceeds the allowable stress, so the rod is NOT safe at a factor of safety of 2.5. The actual factor of safety is 250/127=1.97250 / 127 = 1.97. To make it safe the designer should increase the diameter or use a higher-strength alloy. Markers reward the circular area formula, the stress and extension working, and the explicit comparison of working stress against allowable stress to reach a justified safe/unsafe verdict.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
A steel bracket bolt has a cross-sectional area of 78.5 mm^2 and carries a tensile load of 12 kN. Calculate the stress in the bolt in MPa.
Show worked solution →

Convert area to consistent units: A=78.5 mm2=78.5×10−6 m2A = 78.5 \text{ mm}^2 = 78.5 \times 10^{-6} \text{ m}^2.

σ=FA=12×10378.5×10−6=1.53×108 Pa=153 MPa\sigma = \frac{F}{A} = \frac{12 \times 10^3}{78.5 \times 10^{-6}} = 1.53 \times 10^8 \text{ Pa} = 153 \text{ MPa}

Marking criteria: 1 mark for correct unit conversion, 1 mark for correctly applying σ=F/A\sigma = F/A, 1 mark for the correct final value with MPa units.

foundation3 marks
A timber post 2.4 m long shortens by 1.8 mm under load. Calculate the strain, and state why strain has no units.
Show worked solution →

ε=ΔLL=1.8×10−32.4=7.5×10−4\varepsilon = \frac{\Delta L}{L} = \frac{1.8 \times 10^{-3}}{2.4} = 7.5 \times 10^{-4}

Strain has no units because it is a length divided by a length, so the metre units cancel completely.

Marking criteria: 1 mark for correct unit conversion of the extension, 1 mark for the correct strain value, 1 mark for explaining why the units cancel.

core4 marks
The table below gives Young's modulus for four materials used in civil structures. | Material | E (GPa) | |---|---| | Structural steel | 200 | | Aluminium alloy | 70 | | Concrete | 30 | | Timber | 10 | Two beams of identical length, cross-section and load are made from steel and timber. Using the table, explain which beam deflects more, and by approximately what factor.
Show worked solution →

Deflection under a given load is inversely proportional to Young's modulus for members of identical geometry and loading (from ΔL=FL/(AE)\Delta L = FL/(AE), and the equivalent bending-deflection relationship also scales as 1/E1/E). The timber beam (E=10E = 10 GPa) deflects more than the steel beam (E=200E = 200 GPa).

deflectiontimberdeflectionsteel≈EsteelEtimber=20010=20\frac{\text{deflection}_{\text{timber}}}{\text{deflection}_{\text{steel}}} \approx \frac{E_{\text{steel}}}{E_{\text{timber}}} = \frac{200}{10} = 20

The timber beam deflects roughly 20 times as much as the steel beam.

Marking criteria: 1 mark for correctly reading both E values from the table, 1 mark for stating deflection is inversely proportional to E, 1 mark for identifying timber as the more-deflecting beam, 1 mark for the correct approximate factor of 20.

core5 marks
A structural steel tie of circular cross-section, diameter 16 mm and length 2.5 m, carries a tensile load of 35 kN. Young's modulus is 200 GPa and the yield stress is 250 MPa. Calculate the stress and the extension, and state the actual factor of safety against yielding.
Show worked solution →
Area
A=πd24=π(0.016)24=2.01×10−4 m2A = \dfrac{\pi d^2}{4} = \dfrac{\pi (0.016)^2}{4} = 2.01 \times 10^{-4} \text{ m}^2
Stress
σ=FA=35×1032.01×10−4=1.74×108 Pa=174 MPa\sigma = \dfrac{F}{A} = \dfrac{35 \times 10^3}{2.01 \times 10^{-4}} = 1.74 \times 10^8 \text{ Pa} = 174 \text{ MPa}
Strain and extension
ε=σE=174×106200×109=8.7×10−4\varepsilon = \dfrac{\sigma}{E} = \dfrac{174 \times 10^6}{200 \times 10^9} = 8.7 \times 10^{-4}

ΔL=εL=8.7×10−4×2.5=2.18×10−3 m=2.18 mm\Delta L = \varepsilon L = 8.7 \times 10^{-4} \times 2.5 = 2.18 \times 10^{-3} \text{ m} = 2.18 \text{ mm}

Factor of safety. FoS=fy/σ=250/174=1.44FoS = f_y / \sigma = 250 / 174 = 1.44.

Marking criteria: 1 mark for the circular area formula, 1 mark for correct stress, 1 mark for correct strain and extension, 1 mark for the correct factor of safety, 1 mark for consistent SI units throughout.

core4 marks
Sketch and describe, in words, the key difference between the stress-strain curve of a ductile material (mild steel) and a brittle material (cast iron) up to the point of fracture.
Show worked solution →

A ductile material such as mild steel shows a linear elastic region up to the limit of proportionality, followed by a yield plateau where the material deforms plastically at roughly constant stress, then strain hardening up to the ultimate tensile strength, then necking and fracture at a large total strain. A brittle material such as cast iron shows an almost linear curve with no discernible yield plateau or strain-hardening region; it fractures suddenly at a low strain shortly after (or at) its elastic limit, with far less area under the curve.

Marking criteria: 1 mark for correctly describing the ductile curve's stages (elastic, yield, hardening, necking/fracture), 1 mark for correctly describing the brittle curve (linear, sudden fracture, little plastic strain), 1 mark for contrasting the total strain at fracture, 1 mark for linking the smaller area under the brittle curve to lower toughness.

exam6 marks
A circular steel rod of diameter 25 mm and length 4.0 m carries a tensile load of 70 kN. Young's modulus is 200 GPa, the yield stress is 250 MPa, and the required factor of safety is 2.0. Determine the working stress, the extension under load, and assess whether the rod satisfies the required factor of safety. If it does not, recommend a suitable minimum diameter.
Show worked solution →
Area
A=πd24=π(0.025)24=4.91×10−4 m2A = \dfrac{\pi d^2}{4} = \dfrac{\pi (0.025)^2}{4} = 4.91 \times 10^{-4} \text{ m}^2
Working stress
σ=FA=70×1034.91×10−4=1.43×108 Pa=143 MPa\sigma = \dfrac{F}{A} = \dfrac{70 \times 10^3}{4.91 \times 10^{-4}} = 1.43 \times 10^8 \text{ Pa} = 143 \text{ MPa}
Extension
ε=σ/E=(143×106)/(200×109)=7.14×10−4\varepsilon = \sigma/E = (143 \times 10^6)/(200 \times 10^9) = 7.14 \times 10^{-4}

ΔL=εL=7.14×10−4×4.0=2.85×10−3 m=2.85 mm\Delta L = \varepsilon L = 7.14 \times 10^{-4} \times 4.0 = 2.85 \times 10^{-3} \text{ m} = 2.85 \text{ mm}

Allowable stress
σallow=fy/FoS=250/2.0=125 MPa\sigma_{\text{allow}} = f_y/FoS = 250/2.0 = 125 \text{ MPa}.
Assessment
The working stress (143 MPa) exceeds the allowable stress (125 MPa), so the rod does NOT satisfy the required factor of safety (actual FoS=250/143=1.75FoS = 250/143 = 1.75).
Recommended diameter
Require σ≤125\sigma \leq 125 MPa: A≥F/σallow=(70×103)/(125×106)=5.60×10−4 m2A \geq F/\sigma_{\text{allow}} = (70 \times 10^3)/(125 \times 10^6) = 5.60 \times 10^{-4} \text{ m}^2.

d≥4Aπ=4×5.60×10−4π=0.0267 m=26.7 mmd \geq \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 5.60 \times 10^{-4}}{\pi}} = 0.0267 \text{ m} = 26.7 \text{ mm}

A minimum diameter of about 27 mm (the next standard size up) is recommended.

Marking criteria: 1 mark for area, 1 mark for working stress, 1 mark for extension, 1 mark for allowable stress, 1 mark for a correctly justified fail verdict with actual factor of safety, 1 mark for the correct recommended minimum diameter.

exam7 marks
Explain, with reference to Young's modulus and yield stress, why a civil engineer selecting a floor beam to limit deflection is solving a different problem to one selecting a beam to prevent failure. Illustrate your answer with reference to a named Australian structure.
Show worked solution →

This is a 7-mark EXPLAIN: markers reward a clear conceptual distinction supported by a correctly used example, not just two separate definitions.

The stiffness problem (deflection)
Deflection under service load is governed by the member's stiffness, controlled by Young's modulus EE together with its geometry (via ΔL=FL/(AE)\Delta L = FL/(AE) for axial members, and the analogous bending-deflection relationship for beams). A stiffer material or a larger/deeper section reduces deflection for the same load, independent of how close the material is to yielding. This matters because excessive deflection can cause a serviceability failure, such as cracked finishes or a floor that feels "springy", well before the material comes close to yielding.
The strength problem (failure)
Preventing failure is governed by the member's strength, controlled by the yield stress fyf_y (or ultimate stress) and the applied working stress, checked against a factor of safety: σallow=fy/FoS\sigma_{\text{allow}} = f_y/FoS. A member can be adequately stiff (small deflection) yet still be at risk of yielding if its cross-section is too small for the load, and vice versa: a very strong but slender member can be strong enough to avoid yielding yet deflect excessively in service.
Worked distinction
For example, a steel beam with E=200E = 200 GPa and fy=250f_y = 250 MPa could easily satisfy a strength check (working stress well below the 250/FoS allowable limit) while still deflecting beyond an acceptable serviceability limit for a long, lightly-loaded span, because deflection depends on EE and geometry, not on how far the stress sits below yield.
Australian example
The Sydney Harbour Bridge arch was designed to satisfy both requirements simultaneously: sufficient stiffness (high EE, large steel sections) to limit sag and vibration under traffic and wind load, and sufficient strength (steel grade and section size selected against yield stress with a factor of safety) to carry the enormous compressive forces in the arch without permanent deformation or failure. Meeting one requirement does not automatically satisfy the other, which is why both are checked independently in design.

Marking criteria: 2 marks for correctly explaining the stiffness/deflection problem via Young's modulus, 2 marks for correctly explaining the strength/failure problem via yield stress and factor of safety, 1 mark for explicitly contrasting that satisfying one does not guarantee the other, 2 marks for a correctly used, relevant Australian structural example.

Practise this

Sources & how we know this

Based on the official NESA syllabus. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

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