Stress-strain behaviour and mechanical properties
Reading the tensile stress-strain curve: elastic constants, yield and 0.2 % proof stress, UTS and necking, ductility measures, resilience and toughness, true stress-strain and the Hollomon law.
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Why it matters
The tensile test is the single most important source of design data: Young's modulus for deflection, yield strength for allowable stress, ultimate strength and elongation for forming and for judging ductility. Reading the stress-strain curve correctly, including the difference between engineering and true values, is needed in machine design, metal forming and every materials-selection decision.
Key ideas
The tensile test. A standard specimen of original gauge length L0 and cross-section A0 is pulled at a slow, constant rate; load F and elongation ΔL are recorded. Engineering stress and strain are based on the original dimensions, so they describe the specimen, not the instantaneous state of the material.
Regions of a ductile metal's curve.
- Linear elastic: stress proportional to strain (Hooke's law) up to the proportional limit. Unloading returns the specimen to zero strain. The elastic limit is the highest stress with no permanent set; it is very close to the proportional limit.
- Yielding: plastic flow starts. Some low-carbon steels show a sharp upper yield point, a drop to a lower yield point and a flat yield-point elongation (Lüders bands), caused by carbon and nitrogen atoms pinning dislocations. Metals without a distinct yield point are given a 0.2 % offset (proof) stress: draw a line parallel to the elastic slope from strain 0.002 and read where it meets the curve.
- Uniform plastic deformation with strain hardening: the load keeps rising because the metal hardens faster than its area shrinks.
- Ultimate tensile strength (UTS): the maximum engineering stress, F_max/A0. Beyond it, deformation localises in a neck.
- Necking to fracture: the load falls on the engineering curve even though the true stress in the neck keeps rising.
Elastic properties. E (Young's modulus) measures stiffness and depends on bonding, so heat treatment and alloying hardly change it (all steels are about 200–210 GPa). Poisson's ratio ν = −lateral strain / axial strain, about 0.3 for metals; G = E / [2(1 + ν)] for isotropic materials.
Ductility. Percent elongation (depends on gauge length, so always quote it) and percent reduction in area (independent of gauge length, a better measure of the neck's ductility). A material with less than about 5 % elongation is usually called brittle.
Energy measures. Resilience is the elastic strain energy per unit volume up to yield (area under the elastic part). Tensile toughness is the total area under the curve to fracture; it needs both strength and ductility.
True stress and true strain. True stress σT = F/Ai uses the instantaneous area; true (logarithmic) strain εT = ln(Li/L0) is additive over successive steps. Assuming constant volume (plastic flow is volume-conserving) and uniform deformation (valid only up to UTS), they convert from engineering values. Beyond UTS, use the measured neck area.
Strain hardening. In the uniform plastic region many metals follow Hollomon's law σT = K·εTⁿ. The exponent n (0.1–0.5) measures how strongly the metal hardens; necking starts when εT = n (Considère criterion), so high-n metals stretch-form well.
Brittle and polymeric behaviour. Ceramics and grey cast iron fracture in the elastic region with almost no plasticity and are much stronger in compression than tension. Polymers are viscoelastic: their curves depend strongly on strain rate and temperature.
Effect of temperature and rate. Raising temperature lowers E, yield strength and UTS and usually raises ductility; raising strain rate does the reverse. BCC steels become brittle at low temperature (ductile-to-brittle transition).
Formulas
σ = F / A0 and ε = ΔL / L0
Engineering stress (Pa) and strain (dimensionless); F in N, A0 in m², L0 in m.
σ = E·ε
Hooke's law, linear elastic region only; E in Pa.
ν = −ε_lateral / ε_axial and G = E / (2(1 + ν))
Poisson's ratio (dimensionless); shear modulus G in Pa, isotropic materials.
%EL = (Lf − L0) / L0 × 100 and %RA = (A0 − Af) / A0 × 100
Lf, Af = gauge length and minimum area after fracture.
Ur = σy² / (2E)
Modulus of resilience (J/m³), assumes linear elasticity up to yield σy.
σT = σ·(1 + ε) and εT = ln(1 + ε)
True stress (Pa) and true strain; valid up to the onset of necking.
σT = K·εTⁿ
Hollomon law; K = strength coefficient (Pa), n = strain-hardening exponent. At necking, εT = n.
Worked examples
Example 1 (standard): reducing tensile-test data. Given: steel specimen, d0 = 12.5 mm, L0 = 50 mm; load at yield 40 kN; maximum load 62 kN; after fracture Lf = 62 mm and neck diameter df = 9.0 mm; E = 200 GPa.
- A0 = π/4 × 12.5² = 122.7 mm².
- σy = 40 000 / 122.7 = 326 MPa.
- UTS = 62 000 / 122.7 = 505 MPa.
- %EL = (62 − 50)/50 × 100 = 24 %.
- Af = π/4 × 9.0² = 63.6 mm²; %RA = (122.7 − 63.6)/122.7 × 100 = 48.2 %.
- Ur = σy²/(2E) = 326²/(2 × 200 000) = 0.266 MPa = 0.266 MJ/m³.
Example 2 (GATE level): true values at necking and the Hollomon law. Given: an annealed metal reaches its UTS of 400 MPa at an engineering strain of 0.20. Assume Hollomon behaviour.
- True stress at UTS: σT = 400 × (1 + 0.20) = 480 MPa.
- True strain at UTS: εT = ln 1.20 = 0.182.
- Since necking starts at εT = n, n = 0.182.
- K = σT / εTⁿ = 480 / 0.182^0.182 = 480 / 0.733 = 655 MPa.
Example 3 (elastic): a 10 mm diameter, 2 m long steel rod carries 20 kN (E = 200 GPa, ν = 0.3).
- σ = 20 000 / (π/4 × 10²) = 254.6 MPa.
- ε = 254.6 / 200 000 = 1.273 × 10⁻³; ΔL = 1.273 × 10⁻³ × 2000 mm = 2.55 mm.
- Diameter change = −ν·ε·d = −0.3 × 1.273 × 10⁻³ × 10 = −0.0038 mm.
Common mistakes
- Mixing MPa, N/mm² and Pa: 1 MPa = 1 N/mm², which makes hand calculations in mm easy.
- Applying the true-stress conversions after necking has started.
- Quoting %EL without the gauge length, or comparing elongations from different gauge lengths.
- Thinking heat treatment raises E. It raises yield strength and hardness, not stiffness.
- Reading the 0.2 % offset as 0.2 (instead of 0.002) strain.
- Assuming the drop in the engineering curve after UTS means the material is getting weaker; it is the neck's area that is falling.
For GATE PI
Expect direct numericals on stress, strain, elongation, Hooke's law and Poisson's ratio; conversions between engineering and true stress/strain; the necking criterion εT = n with Hollomon's law (a favourite in metal-forming questions); and resilience or toughness as area under the curve. Conceptual one-markers ask which property measures stiffness, ductility or toughness and what the 0.2 % proof stress is.
Quick check
- A specimen of 100 mm gauge length breaks at 122 mm. What is %EL?
- Engineering strain is 0.10. What is the true strain?
- Which property is unchanged by quenching and tempering a steel: yield strength, hardness or Young's modulus?
- For a Hollomon material with n = 0.25, at what true strain does necking begin?
- E = 200 GPa and ν = 0.25. What is G?
Answers: 1. 22 %; 2. 0.0953; 3. Young's modulus; 4. 0.25; 5. 80 GPa.
Interview questions
All Engineering Materials interview questionsTry answering each one aloud before you open it.
1.What is stress in the context of engineering materials?Concept
Stress is the internal resistance offered by a material to an external force. It is defined as the force applied per unit area and is measured in Pascals (Pa) or N/m². Stress can be tensile, compressive, or shear, depending on the nature of the force applied.
2.Explain the concept of strain in materials.Concept
Strain is the deformation of a body relative to its original size. Normal (engineering) strain is the change in length divided by the original length, ε = ΔL/L0, and shear strain is the change in an originally right angle, in radians; both are dimensionless. True strain, εT = ln(L/L0), uses the current length and is additive over successive deformation steps, which is why it is used in metal forming. Strain is partly elastic (recovered on unloading) and partly plastic (permanent) once the yield point is passed.
3.What is the stress-strain curve and why is it important?Concept
The stress-strain curve is a graphical representation of the relationship between stress and strain for a material. It is important because it provides valuable information about the material's mechanical properties, such as elasticity, yield strength, ultimate tensile strength, and ductility. Engineers use this curve to predict how materials will behave under different types of loads.
4.Define Young's Modulus and its significance.Concept
Young's Modulus, also known as the modulus of elasticity, is a measure of the stiffness of a material. It is defined as the ratio of tensile stress to tensile strain in the linear elastic region of the stress-strain curve. It is significant because it helps in determining how much a material will deform under a given load, which is crucial for design and analysis.
5.Why is steel commonly used in construction for its mechanical properties?Application
Steel is commonly used in construction due to its high tensile strength, ductility, and toughness. It can withstand significant stress without permanent deformation, making it ideal for structures that need to support heavy loads. Additionally, steel's ability to be easily welded and formed into various shapes adds to its versatility in construction applications.
6.What happens to a material when it is loaded beyond its yield strength?Application
When a material is loaded beyond its yield strength, it undergoes plastic deformation, meaning it will not return to its original shape when the load is removed. This permanent deformation occurs because the material's internal structure has been altered, and it can lead to failure if the load continues to increase.
7.How does temperature affect the stress-strain behaviour of materials?Application
Temperature can significantly affect the stress-strain behaviour of materials. Generally, as temperature increases, materials tend to become more ductile and less brittle, which can lower their yield strength and ultimate tensile strength. Conversely, at lower temperatures, materials may become more brittle and prone to fracture under stress.
8.Calculate the stress experienced by a rod with a cross-sectional area of 0.005 m² subjected to a force of 1000 N.Numerical
Stress (σ) is calculated using the formula σ = F / A, where F is the force applied, and A is the cross-sectional area. Here, F = 1000 N and A = 0.005 m². Therefore, σ = 1000 N / 0.005 m² = 200,000 N/m² or 200 kPa.
9.A material has a Young's Modulus of 200 GPa. If a tensile stress of 50 MPa is applied, what is the resulting strain?Numerical
Strain (ε) is calculated using the formula ε = σ / E, where σ is the stress and E is Young's Modulus. Here, σ = 50 MPa = 50 x 10⁶ N/m² and E = 200 GPa = 200 x 10⁹ N/m². Therefore, ε = (50 x 10⁶) / (200 x 10⁹) = 0.00025 or 250 microstrain.
10.Explain the difference between elastic and plastic deformation.Concept
Elastic deformation is reversible, meaning the material returns to its original shape when the applied stress is removed. It occurs when the stress is within the elastic limit of the material. Plastic deformation, on the other hand, is permanent and occurs when the material is stressed beyond its yield strength, causing a permanent change in shape.
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