Tensile Stress, Tensile Strain, Young Modulus Experiment & Brittle vs Ductile Graphs
Solid mechanics distinguishes between the extrinsic stiffness of a particular spring or wire (the spring constant k in Hooke's Law, F = kΔL) and the intrinsic stiffness of a material independent of its dimensions: the Young Modulus, E = Tensile Stress / Tensile Strain = (FL) / (AΔL). While the area under a Force-Extension (F-ΔL) graph gives the total elastic strain energy stored (E_p = ½FΔL = ½k(ΔL)²), the area under a Stress-Strain (σ-ε) graph gives the elastic strain energy stored per unit volume (½σε). In the laboratory, Young Modulus is determined using a long, thin metal wire to maximise extension ΔL while measuring diameter d at multiple positions with a micrometer screw gauge. Examiners frequently test springs in series and parallel, plastic deformation unloading lines, and comparing brittle (glass), ductile (copper), and polymeric (rubber) stress-strain graphs.
- Hooke's Law (F = kΔL) & Springs in Series vs Parallel
- Hooke's Law states that the tensile or compressive force F needed to extend or compress a spring is directly proportional to its extension ΔL (F = kΔL), up to the limit of proportionality (where the spring constant k = F / ΔL has units N m⁻¹). For two identical springs of constant k: 1. In Parallel (sharing the load equally, half the extension): effective spring constant k_parallel = k₁ + k₂ = 2k (stiffer). 2. In Series (both experience the full load F, double the total extension): 1 / k_series = 1 / k₁ + 1 / k₂, so k_series = ½k (less stiff).
- Elastic Strain Energy (E = ½FΔL = ½k(ΔL)²) & Loading-Unloading Hysteresis
- For any wire or spring obeying Hooke's Law, the work done in stretching it is stored as elastic strain energy, equal to the triangular area under the Force-Extension graph: E = ½FΔL = ½k(ΔL)². If a wire is stretched beyond its elastic limit (undergoing plastic deformation where atomic planes slip permanently), the unloading line is parallel to the loading line but returns to a permanent non-zero extension; the area trapped between the loading and unloading curves equals the energy dissipated as internal thermal energy (heat).
- Tensile Stress (σ), Tensile Strain (ε) & Breaking Stress (UTS)
- 1. Tensile Stress (σ): force applied per unit cross-sectional area perpendicular to the force: σ = F / A (where A = πr² = ¼πd² for a cylindrical wire; units: Pa or N m⁻²). 2. Tensile Strain (ε): ratio of extension to original length: ε = ΔL / L (dimensionless ratio, no units or expressed as a percentage). 3. Main Tensile Stress (UTS): the maximum stress a material can withstand while being stretched before necking and breaking.
Tensile Stress, Tensile Strain, Young Modulus Experiment & Brittle vs Ductile Graphs: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Hooke's Law (F = kΔL) & Springs in Series vs Parallel
- Hooke's Law states that the tensile or compressive force F needed to extend or compress a spring is directly proportional to its extension ΔL (F = kΔL), up to the limit of proportionality (where the spring constant k = F / ΔL has units N m⁻¹). For two identical springs of constant k: 1. In Parallel (sharing the load equally, half the extension): effective spring constant k_parallel = k₁ + k₂ = 2k (stiffer). 2. In Series (both experience the full load F, double the total extension): 1 / k_series = 1 / k₁ + 1 / k₂, so k_series = ½k (less stiff).
- Elastic Strain Energy (E = ½FΔL = ½k(ΔL)²) & Loading-Unloading Hysteresis
- For any wire or spring obeying Hooke's Law, the work done in stretching it is stored as elastic strain energy, equal to the triangular area under the Force-Extension graph: E = ½FΔL = ½k(ΔL)². If a wire is stretched beyond its elastic limit (undergoing plastic deformation where atomic planes slip permanently), the unloading line is parallel to the loading line but returns to a permanent non-zero extension; the area trapped between the loading and unloading curves equals the energy dissipated as internal thermal energy (heat).
- Tensile Stress (σ), Tensile Strain (ε) & Breaking Stress (UTS)
- 1. Tensile Stress (σ): force applied per unit cross-sectional area perpendicular to the force: σ = F / A (where A = πr² = ¼πd² for a cylindrical wire; units: Pa or N m⁻²). 2. Tensile Strain (ε): ratio of extension to original length: ε = ΔL / L (dimensionless ratio, no units or expressed as a percentage). 3. Main Tensile Stress (UTS): the maximum stress a material can withstand while being stretched before necking and breaking.
- The Young Modulus: E = σ / ε = (FL) / (AΔL)
- Within the limit of proportionality, tensile stress is directly proportional to tensile strain; the constant of proportionality is the Young Modulus: E = Tensile Stress / Tensile Strain = (F / A) ÷ (ΔL / L) = (FL) / (AΔL) (units: Pa or N m⁻², typically GPa = 10⁹ Pa for metals). Note: The gradient of a Stress-Strain graph equals E; the area under the linear region of a Stress-Strain graph equals the elastic strain energy stored per unit volume (Energy / Volume = ½σε in J m⁻³).
- Experimental Determination of the Young Modulus of a Wire
- A long (L > 2 m), thin (d < 0.5 mm) metal wire is clamped horizontally over a pulley (or vertically in Searle's apparatus alongside a reference wire to cancel thermal expansion and support sag): 1. Measure original length L from clamp to fiducial marker with a metre rule. 2. Measure diameter d in at least 5 different places and at 90° orientations using a micrometer screw gauge (to check uniformity and calculate mean A = ¼πd²). 3. Add slotted masses of weight F = mg in steps, reading extension ΔL of the marker against a fixed vernier scale, then unload masses to verify the elastic limit was not exceeded. 4. Plot a graph of Force F (y-axis) against Extension ΔL (x-axis): Gradient = F / ΔL, so E = Gradient × (L / A).
- Brittle, Ductile & Polymeric Stress-Strain Curves
- 1. Brittle material (e.g. glass, cast iron, ceramic): obeys Hooke's law (straight line of steep gradient, high E) right up to the breaking point where it snaps suddenly via crack propagation with zero plastic deformation. 2. Ductile material (e.g. copper, mild steel): obeys Hooke's law initially up to the limit of proportionality and elastic limit, reaches a yield point, and then undergoes extensive plastic deformation (necking and permanent stretching as planes of metal ions slide over one another) before fracture. 3. Polymeric material (e.g. rubber): does not obey Hooke's law; curves smoothly as tangled long-chain polymer molecules uncoil (easy to stretch initially, then stiffens as covalent bonds are stretched); returns to original length (elastic) but unloading curve lies below loading curve (elastic hysteresis loop; area inside loop = thermal energy heating the rubber).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Distinguish between the limit of proportionality and the elastic limit of a metal wire.
Two identical springs, each of spring constant k = 40 N m⁻¹, are connected (a) in parallel and (b) in series to support a 20 N weight. Calculate the effective spring constant and extension in each case.
What do the gradient and the area under the linear region of a Stress-Strain (σ vs ε) graph represent?
Why is a very long, very thin wire used in the laboratory experiment to measure the Young Modulus of steel or copper?
Why does the uncertainty in the diameter d of a wire contribute twice as much to the percentage uncertainty in the Young Modulus as the uncertainty in length L or extension ΔL?
Why does stretching and releasing a rubber band repeatedly make the rubber band feel warm?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A steel wire of length 2.40 m and diameter 0.64 mm is stretched by a tensile load of 45.0 N. Given that the Young Modulus of steel is 2.10 × 10¹¹ Pa and the wire obeys Hooke's law, calculate: (a) the tensile stress in the wire, (b) the extension of the wire in mm, and (c) the elastic strain energy stored in the wire. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate radius r = 0.32 × 10⁻³ m and cross-sectional area A = πr² = π × (0.32 × 10⁻³)² = 3.217 × 10⁻⁷ m² [1 mark]. 2. (a) Tensile stress σ = F ÷ A = 45.0 N ÷ (3.217 × 10⁻⁷ m²) = 1.40 × 10⁸ Pa (140 MPa) [1 mark]. 3. (b) From E = σ ÷ ε, calculate tensile strain ε = σ ÷ E = (1.40 × 10⁸) ÷ (2.10 × 10¹¹) = 6.662 × 10⁻⁴ [1 mark]. 4. Calculate extension ΔL = ε × L = (6.662 × 10⁻⁴) × 2.40 m = 1.60 × 10⁻³ m = 1.60 mm [1 mark]. 5. (c) Elastic strain energy E_p = ½FΔL = 0.5 × 45.0 N × (1.60 × 10⁻³ m) = 0.0360 J (3.60 × 10⁻² J) [1 mark].
- Q2
A brass rod of length L and diameter d is welded end-to-end in series with a steel rod of length 2L and diameter 2d, and the composite rod is subjected to a tensile force F. Given that the Young Modulus of steel is twice that of brass (E_steel = 2 E_brass), calculate the ratio of the extension of the brass rod to the extension of the steel rod (ΔL_brass / ΔL_steel). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Because the two rods are connected in series, both rods experience the exact same tensile force F [1 mark]. 2. Since cross-sectional area A = ¼πd² is proportional to d², doubling the diameter from d (brass) to 2d (steel) makes the cross-sectional area of the steel rod 4 times larger: A_steel = 4 A_brass [1 mark]. 3. Rearrange the Young Modulus formula for extension: ΔL = (FL) ÷ (EA) [1 mark]. 4. Substitute the ratios for steel relative to brass: ΔL_steel = [F × (2L)] ÷ [(2 E_brass) × (4 A_brass)] = (2 ÷ 8) × [(FL) ÷ (E_brass A_brass)] = ¼ ΔL_brass; therefore ΔL_brass / ΔL_steel = 4 (or 4 : 1) [1 mark].
- Q3
Compare and explain the mechanical behaviour of glass, copper, and polythene when each is subjected to an increasing tensile stress until it breaks. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Glass is a brittle material: its stress-strain graph is a steep straight line (obeying Hooke's law with a high Young Modulus) right up to its breaking stress, where it fractures suddenly with no plastic deformation [1 mark]. 2. Copper is a ductile metallic material: it obeys Hooke's law initially up to its elastic limit, and then beyond its yield point it undergoes large permanent plastic deformation [1 mark]. 3. In copper, plastic deformation occurs because layers/planes of positive metal ions slide over one another (dislocation motion) and form new metallic bonds in new positions, causing the wire to neck and thin before breaking [1 mark]. 4. Polythene is a ductile polymeric material that does not obey Hooke's law; under small stress, long tangled polymer chains untangle and align parallel to the force, producing a very large strain (extension) before breaking [1 mark]. 5. When unloaded before breaking, copper and polythene retain a permanent extension (irreversible plastic deformation), whereas glass prior to fracture is purely elastic [1 mark].
- Q4
A vertical spring of spring constant k = 250 N m⁻¹ is compressed by 0.120 m and used to launch a toy rocket of mass 0.080 kg vertically upwards into the air. Calculate the elastic strain energy stored in the compressed spring, the launch speed of the rocket as it leaves the spring, and the maximum height reached above its launch point (neglecting air resistance, g = 9.81 m s⁻²). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Elastic strain energy stored in the spring: E_e = ½k(ΔL)² = 0.5 × 250 × (0.120)² = 1.80 J [1 mark]. 2. Assuming negligible height change during decompression (or measuring from compressed position), equate E_e to kinetic energy ½mv²: 1.80 = 0.5 × 0.080 × v² → v² = 45.0 → v = 6.71 m s⁻¹ [1 mark]. 3. At the maximum height h, all 1.80 J of elastic strain energy has been converted into gravitational potential energy (E_p = mgh) [1 mark]. 4. Solve for h: h = E_e ÷ (mg) = 1.80 ÷ (0.080 × 9.81) = 1.80 ÷ 0.7848 = 2.29 m [1 mark].
Common Exam Questions (4)
When a mass m is hung gently on an unstretched vertical spring and comes to rest at equilibrium extension ΔL, why is the elastic energy stored ½mgΔL while the gravitational potential energy lost by the mass is mgΔL?
The mass loses gravitational potential energy ΔE_p = mgΔL, while the spring gains elastic strain energy E_e = ½FΔL = ½mgΔL. The other half of the energy (½mgΔL) was either transferred to the person's hand supporting the mass as it was lowered gently, or - if dropped suddenly, caused the mass to oscillate up and down until air resistance and internal friction damped the oscillations as heat.
Does a thicker wire of the same metal have a larger Young Modulus than a thinner wire?
No. The Young Modulus (E) is a material property that depends only on the chemical identity and temperature of the material (e.g. steel is always ~210 GPa regardless of wire thickness or length). A thicker wire has a larger spring constant k (k = EA / L, stiffer object), not a larger Young Modulus.
What safety precaution is essential when carrying out the Young Modulus wire experiment?
Always wear safety goggles (eye protection) because the thin wire is under high tension and can whip back sharply into the eyes if it snaps, and place a padded box or sand tray on the floor directly beneath the hanging slotted masses to protect feet if the wire breaks.
What is the difference between elastic deformation and plastic deformation at the atomic level?
In elastic deformation, atoms are pulled slightly further apart than their equilibrium separation against interatomic electrostatic forces, and spring back to their exact original neighbours when the force is removed. In plastic deformation, planes of atoms slip permanently past one another (via crystal dislocations) and form bonds with new neighbours, so the shape change is irreversible.
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