Strength of materials: stress and strain

In short

Stress is the internal resisting force per unit area set up in a body when external loads act on it; strain is the deformation it produces per unit original dimension. Together they decide whether a shaft, bolt, beam or bracket will carry its load safely and by how much it will stretch, which is the starting point of every design calculation in mechanical engineering. The topic carries direct marks in every university Strength of Materials paper and 4-8 marks in GATE ME every year, and the stress-strain curve, elastic constants and factor of safety are the most commonly asked opening questions in design and manufacturing interviews (L&T, Tata Motors, BHEL, Godrej, Thermax, Cummins, ISRO/DRDO). Material grades and allowable stresses come from Indian standards such as IS 2062 for structural steel and IS 800 for its design.

Explained

Stress: definition, types and sign convention

• Stress σ = P/A, force per unit area, SI unit N/m² = pascal; engineering practice uses N/mm² = MPa (1 MPa = 10⁶ Pa; 1 kgf/mm² ≈ 9.81 MPa).
• Normal (direct) stress acts perpendicular to the section: tensile (+, elongates) or compressive (−, shortens).
• Shear stress τ = F/A acts parallel to the section - as in a rivet, bolt, key or punched sheet.
• Single shear: one cross-section resists (τ = P/A). Double shear: two sections resist (τ = P/2A) - a bolt in a double-cover butt joint.
• Bearing (crushing) stress σ_b = P/(d·t) between a rivet shank and the plate it presses against; checked separately in riveted joints.
• Engineering (nominal) stress uses the original area A₀; true stress uses the instantaneous area A. True stress always rises until fracture; engineering stress falls after the ultimate point because the neck is thinning.
• Simple stress assumes the load passes through the centroid and is uniformly distributed over the section - valid away from holes, notches and points of application (Saint-Venant's principle).
• Stress concentration: near a hole or fillet the local stress is Kt times the nominal value (Kt ≈ 3 for a small circular hole in a wide plate in tension). Ductile materials yield locally and redistribute under static load; brittle materials and fatigue loading do not forgive it.

Strain and deformation

• Linear (normal) strain ε = δ/L, change in length per unit original length - dimensionless, often quoted in microstrain (1 µε = 10⁻⁶) or percent.
• Shear strain φ (or γ) = lateral displacement / distance between the planes = angular distortion in radians.
• Lateral strain accompanies axial strain; Poisson's ratio µ (or ν) = −lateral strain / axial strain. Typical values: steel 0.27-0.30, cast iron 0.21-0.26, aluminium 0.33, copper 0.34, concrete 0.1-0.2, rubber ≈ 0.5, cork ≈ 0. Thermodynamic limits: −1 < µ < 0.5.
• Volumetric strain ε_v = δV/V. For a bar under uniaxial stress ε_v = ε(1 − 2µ); for three mutually perpendicular stresses ε_v = (σx + σy + σz)(1 − 2µ)/E. At µ = 0.5 the volume does not change - the material is incompressible (rubber, and plastic deformation of metals).
• True strain ε_t = ln(L/L₀) = ln(1 + ε); used in metal forming because strains add.
• Elongation of a bar of varying section: integrate δ = ∫ P dx /(A(x)E) - sum stepwise for a stepped bar, use the closed form for a taper.

Stress-strain curve and material behaviour

• Mild steel in tension (ductile): O-A straight line to the proportional limit (Hooke's law holds); elastic limit just above it (load removed, no permanent set); upper and lower yield points where the specimen yields at nearly constant load (Lüders bands appear); strain hardening to the ultimate tensile strength (the highest engineering stress); then necking and fracture at the breaking stress, with a cup-and-cone fracture.
• Ductility measures: percentage elongation (gauge length matters - typically 5.65√S₀ or 50 mm) and percentage reduction in area. IS 2062 E250 requires about 23% minimum elongation on a 5.65√S₀ gauge length; check the current edition for the exact grade values.
• Brittle materials (cast iron, glass, concrete) show no yield point, little plastic strain and fail suddenly; cast iron is far stronger in compression than in tension, so grey cast iron components are designed to be loaded in compression.
• Non-ferrous metals and high-strength steels have no distinct yield point, so proof stress is quoted - commonly 0.2% proof stress, found by drawing a line parallel to the elastic line from 0.002 strain.
• Resilience and toughness: area under the curve up to the elastic limit is the modulus of resilience (energy absorbed elastically per unit volume); total area to fracture is the modulus of toughness.
• Related terms: hardness (resistance to indentation), creep (slow deformation under sustained load at high temperature), fatigue (failure under repeated load below the yield stress), elastic hysteresis.
• Indian test standards: IS 1608 (tensile testing of metallic materials, aligned with ISO 6892), IS 1599 (bend test), IS 1757/IS 1499 (Charpy V-notch and U-notch impact), IS 1598 (Izod impact).

Hooke's law, the four elastic constants and their relations

• Hooke's law: within the proportional limit, stress ∝ strain. σ = Eε for direct stress, τ = Gφ for shear.
• Young's modulus E = σ/ε - the slope of the elastic line. Typical: mild steel 200-210 GPa, cast iron 100-120 GPa, aluminium alloys 70 GPa, copper 110 GPa, concrete 25-35 GPa, timber 10-15 GPa.
• Modulus of rigidity (shear modulus) G = τ/φ - about 77-80 GPa for steel.
• Bulk modulus K = σ_hydrostatic / volumetric strain - about 160 GPa for steel.
• Poisson's ratio µ as defined above. Only two of E, G, K, µ are independent for an isotropic material; the other two follow from E = 2G(1 + µ) = 3K(1 − 2µ).
• Isotropic / homogeneous / orthotropic: properties the same in all directions / same at every point / different along three axes (timber, FRP laminates).
• E is a measure of stiffness, not strength: a high-strength alloy steel and mild steel have nearly the same E, so a tie rod of either stretches by the same amount under the same working load.

Factor of safety and allowable stress

• Factor of safety FOS = failure stress / working (allowable) stress. Ductile materials take yield stress as the failure stress; brittle materials take ultimate stress.
• Typical values: 1.5-2 for ductile material under static load with well-known loads, 3-4 for brittle or shock-loaded parts, 5-10 for uncertain loads, hoisting ropes and lifting tackle (statutory values for lifting appliances are prescribed under the State factory rules — framed under the Factories Act 1948 and continued under the OSH Code, 2020, in force from 21 November 2025 — and IS 3938; verify the current state rules).
• IS 800:2007 moved Indian steel design from working-stress to limit state design: partial safety factor on material γm = 1.10 for yielding and 1.25 for ultimate/rupture, with load factors (e.g. 1.5 on dead and imposed load) on the other side.
• For machine design, IS 2062 gives structural steel grades (E250 ⇒ f_y ≈ 250 MPa, f_u ≈ 410 MPa; also E300, E350, E410, E450), and design data books quote allowable stresses as f_y/FOS.

Thermal stress and composite (compound) bars

• A free bar heated through ΔT expands by δ = αLΔT with no stress. Stress arises only when the expansion is restrained.
• Fully restrained bar: σ = EαΔT (compressive on heating, tensile on cooling) - independent of length and area.
• Partially restrained (yield of supports = a): σ = E(αLΔT − a)/L.
• α values (per °C): mild steel ≈ 12 × 10⁻⁶, cast iron ≈ 11 × 10⁻⁶, aluminium ≈ 23 × 10⁻⁶, brass ≈ 19 × 10⁻⁶, copper ≈ 17 × 10⁻⁶, concrete ≈ 10-12 × 10⁻⁶ (close to steel - which is why RCC works).
• Composite bar under load (two materials sharing one load, same length): equilibrium P = P₁ + P₂ and compatibility ε₁ = ε₂ ⇒ σ₁/E₁ = σ₂/E₂. The stiffer material carries the larger share of load.
• Composite bar under temperature change: the material with the larger α is compressed and the other is stretched; equilibrium gives equal and opposite forces, compatibility gives α₁LΔT − δ₁ = α₂LΔT + δ₂.
• Practical consequences: expansion gaps in rails and bridges, bellows in piping, creep/thermal allowance in boiler supports, hot-riveting shrinkage stress, shrink-fitting a gear ring onto a hub.
• Residual stresses from welding, quenching, casting and cold working exist with no external load; relieved by stress-relief annealing (typically 600-650 °C for steel), shot peening deliberately puts compressive residual stress at the surface.

Key points

• Stress - internal resistance per unit area, σ = P/A, in N/mm² (MPa); tension positive, compression negative.
• Strain - deformation per unit original length, dimensionless; shear strain is an angle in radians.
• Shear vs bearing - τ = P/A (single) or P/2A (double); crushing checked as P/(d·t).
• Hooke's law - σ = Eε only up to the proportional limit, not up to the yield point.
• Young's modulus E - stiffness, slope of the elastic line; ≈ 200 GPa for all steels regardless of strength.
• Poisson's ratio µ - ratio of lateral to axial strain, 0 to 0.5 for real isotropic materials; 0.3 for steel.
• Elastic constant relations - E = 2G(1 + µ) = 3K(1 − 2µ); only two are independent.
• Volumetric strain - ε_v = ε(1 − 2µ) uniaxially; zero when µ = 0.5.
• Proportional limit, elastic limit, yield point - three different points, in that order on the curve.
• Ultimate stress - peak of the engineering stress-strain curve; breaking stress is lower because of necking.
• Proof stress - 0.2% offset strength, used where there is no distinct yield point (aluminium, high-tensile steel).
• Ductile vs brittle - large plastic strain and necking vs sudden fracture; cast iron is weak in tension, strong in compression.
• True stress/strain - σ_t = σ(1 + ε), ε_t = ln(1 + ε); needed in forming calculations.
• Factor of safety - yield stress / working stress for ductile parts; ultimate stress / working stress for brittle ones.
• Limit state design (IS 800:2007) - γm = 1.10 on yielding, 1.25 on rupture, replacing working-stress factors.
• Thermal stress - σ = EαΔT when fully restrained; depends on restraint, not on length or area.
• Composite bar - equal strain, so load splits in proportion to AE; equal and opposite forces under temperature change.
• Stress concentration - Kt ≈ 3 at a small circular hole; critical for brittle materials and fatigue, not static ductile design.

Formulas and examples

Example 1 - stress, strain and elongation of a stepped bar. A steel bar carries 40 kN in tension. Portion AB is 500 mm long, 20 mm diameter; portion BC is 800 mm long, 30 mm diameter. E = 200 GPa. Find the stresses and the total extension.

• A_AB = π/4 × 20² = 314.16 mm²; A_BC = π/4 × 30² = 706.86 mm².
• σ_AB = 40,000/314.16 = 127.3 MPa; σ_BC = 40,000/706.86 = 56.6 MPa.
• δ_AB = 40,000 × 500/(314.16 × 200,000) = 20,000,000/62,832,000 = 0.3183 mm.
• δ_BC = 40,000 × 800/(706.86 × 200,000) = 32,000,000/141,372,000 = 0.2264 mm.
• Total δ = 0.545 mm. Strain in AB = 0.3183/500 = 6.37 × 10⁻⁴ (637 µε).

Example 2 - elastic constants from a tension test. A 50 mm diameter, 300 mm long steel rod carries 300 kN. Its length increases by 0.219 mm and its diameter decreases by 0.0091 mm. Find E, µ, G and K.

• A = π/4 × 50² = 1,963.5 mm²; σ = 300,000/1,963.5 = 152.8 MPa.
• ε = 0.219/300 = 7.30 × 10⁻⁴ ⇒ E = 152.8/7.30 × 10⁻⁴ = 209,300 N/mm² ≈ 209 GPa.
• Lateral strain = 0.0091/50 = 1.82 × 10⁻⁴ ⇒ µ = 1.82/7.30 = 0.249 (≈ 0.25).
• G = E/[2(1 + µ)] = 209,300/2.498 = 83,800 N/mm² ≈ 84 GPa.
• K = E/[3(1 − 2µ)] = 209,300/(3 × 0.502) = 139,000 N/mm² ≈ 139 GPa.
• Volumetric strain = ε(1 − 2µ) = 7.30 × 10⁻⁴ × 0.502 = 3.66 × 10⁻⁴; ΔV = 3.66 × 10⁻⁴ × (1,963.5 × 300) = 216 mm³ increase.

Example 3 - thermal stress in a restrained rail. A 12 m rail of mild steel is laid at 24 °C with no expansion gap and rigidly held. α = 12 × 10⁻⁶/°C, E = 200 GPa. (a) Stress when it reaches 54 °C. (b) Stress if a 2 mm gap is left.

• ΔT = 30 °C. Free expansion = 12 × 10⁻⁶ × 12,000 × 30 = 4.32 mm.
• (a) Fully restrained: σ = EαΔT = 200,000 × 12 × 10⁻⁶ × 30 = 72 MPa compressive (note: independent of the 12 m length).
• (b) Restrained expansion = 4.32 − 2 = 2.32 mm ⇒ ε = 2.32/12,000 = 1.933 × 10⁻⁴ ⇒ σ = 200,000 × 1.933 × 10⁻⁴ = 38.7 MPa compressive.
• Force on each support if A = 6,000 mm²: 38.7 × 6,000 = 232 kN.

Example 4 - composite bar (steel tube and copper core) under load. A copper rod 40 mm diameter is enclosed in a steel tube of 50 mm outside and 40 mm inside diameter, both 400 mm long and rigidly joined at the ends. An axial compressive load of 120 kN is applied to the assembly. E_steel = 200 GPa, E_copper = 100 GPa. Find the stress in each and the shortening.

• A_copper = π/4 × 40² = 1,256.6 mm²; A_steel = π/4 (50² − 40²) = π/4 × 900 = 706.86 mm².
• Equal strain ⇒ σ_s/200 = σ_c/100 ⇒ σ_s = 2σ_c.
• Equilibrium: σ_s × 706.86 + σ_c × 1,256.6 = 120,000 ⇒ σ_c(1,413.7 + 1,256.6) = 120,000 ⇒ σ_c = 120,000/2,670.3 = 44.94 MPa; σ_s = 89.88 MPa (both compressive).
• Strain = 44.94/100,000 = 4.494 × 10⁻⁴ ⇒ shortening = 4.494 × 10⁻⁴ × 400 = 0.180 mm.
• Load shared: steel 89.88 × 706.86 = 63.5 kN (53%), copper 44.94 × 1,256.6 = 56.5 kN (47%).

Example 5 - design using factor of safety and double shear. A tie rod of IS 2062 E250 steel (f_y = 250 MPa) must carry 90 kN with FOS 2.5. It is attached by a pin in double shear; allowable shear stress for the pin is 100 MPa.

• Allowable tensile stress = 250/2.5 = 100 MPa ⇒ required area = 90,000/100 = 900 mm² ⇒ d = √(4 × 900/π) = √1,145.9 = 33.9 mm ⇒ use 36 mm diameter (A = 1,017.9 mm², actual σ = 88.4 MPa).
• Pin in double shear: required area per section = 90,000/(2 × 100) = 450 mm² ⇒ d = √(4 × 450/π) = √572.96 = 23.9 mm ⇒ use a 25 mm pin (τ = 90,000/(2 × 490.9) = 91.7 MPa, safe).
• Extension of a 2 m rod: δ = 90,000 × 2,000/(1,017.9 × 200,000) = 0.884 mm.

Where it is used

• Machine design and product engineering: sizing shafts, bolts, keys, tie rods, springs, lifting hooks and pressure-vessel plate - every calculation starts from σ = P/A and an allowable stress from IS 2062 / design data books. Roles at L&T, Thermax, Godrej & Boyce, Kirloskar, Tata Motors.
• Structural and fabrication engineering: limit state design of steel structures to IS 800:2007, welded and bolted joint checks (shear, bearing, tension), plate girders in BHEL and L&T Heavy Engineering workshops.
• Materials testing and QA labs: universal testing machines running tensile tests to IS 1608, hardness and impact tests, mill test certificates for incoming steel; NABL-accredited labs, NTPC/Railways vendor inspection.
• Experimental stress analysis and instrumentation: strain gauges and Wheatstone bridges in load cells, torque sensors and weighbridges - everything measured is strain, converted to stress through E.
• Thermal and piping design: expansion loops, bellows, slip joints and support design for boiler and refinery piping where EαΔT stresses dominate; railway rail welding and destressing by Indian Railways.
• FEA and CAD roles: ANSYS/Abaqus/SolidWorks Simulation engineers at Tata Technologies, KPIT, Infosys Engineering Services interpret von Mises stress plots - which are meaningless without an understanding of simple stress, strain and yield.

Interview questions

• Q: What is the difference between stress and pressure? A: Pressure is an externally applied force per unit area acting normal to a surface, usually from a fluid; stress is the internal resisting force per unit area developed inside the material, and it can be normal or shear, tensile or compressive.
• Q: Does Hooke's law hold up to the yield point? A: No - it holds only up to the proportional limit, which lies slightly below the elastic limit and the yield point; beyond it the curve is no longer straight even though the deformation may still be elastic.
• Q: Two rods, one mild steel and one high-strength alloy steel, of the same size carry the same working load. Which stretches more? A: Neither - both stretch almost equally, because extension depends on E, which is about 200 GPa for both; the alloy steel only allows a higher load before yielding.
• Q: Why is engineering stress lower than true stress after necking? A: Engineering stress divides the load by the original area while the actual area has reduced sharply at the neck, so dividing by the real area gives a higher, still-rising true stress.
• Q: What is the theoretical range of Poisson's ratio and why can it not exceed 0.5? A: For isotropic materials −1 < µ < 0.5; at µ = 0.5 volumetric strain ε(1 − 2µ) becomes zero, so a larger value would mean the volume increases under hydrostatic compression, which is physically impossible.
• Q: A bar is heated but free to expand - what is the stress in it? A: Zero. Thermal stress needs restraint; a free bar simply elongates by αLΔT.
• Q: Does thermal stress in a fully restrained bar depend on its length or cross-section? A: No - σ = EαΔT contains neither; length and area affect the force and the free expansion, not the stress.
• Q: In a composite bar of steel and copper under one axial load, which carries more stress? A: The strains are equal, so stress is proportional to E - the steel, with twice copper's modulus, carries twice the stress and a load share proportional to AE.
• Q: How many independent elastic constants does an isotropic material have? A: Two. E, G, K and µ are linked by E = 2G(1 + µ) = 3K(1 − 2µ), so fixing any two fixes the rest.
• Q: Why is cast iron used for machine beds and brackets but not for tie rods? A: It is brittle with low tensile strength but high compressive strength and good damping and castability, so it suits compression-loaded, vibration-damping bodies, not members in tension.
• Q: What is 0.2% proof stress and when is it used? A: The stress at which 0.2% permanent strain remains, found by an offset line parallel to the elastic region; it replaces yield stress for materials with no distinct yield point, such as aluminium alloys and high-tensile steel.
• Q: On what basis do you pick the factor of safety for a ductile part under static load? A: Yield stress divided by working stress, typically 1.5-2 when loads and material are well known, raised to 3-10 for shock, fatigue, brittle material, lifting tackle or uncertain loading - and governed by the relevant code (IS 800 limit state factors for steel structures).

Common mistakes

• Using the deformed or reduced area instead of the original area for engineering stress, or mixing up true and engineering values in the same problem.
• Mixing units - substituting E in GPa with P in N and A in mm² (E must be 200,000 N/mm² for a result in MPa), or forgetting that 1 GPa = 1,000 MPa.
• Assuming Hooke's law and δ = PL/AE apply beyond the proportional limit, or to a bar whose section varies continuously without integrating.
• Taking τ = P/A for a bolt that is actually in double shear (halving the stress is correct there), or forgetting the separate bearing/crushing check.
• Writing thermal stress as EαΔT for a bar that is only partly restrained, or applying it to a free bar and getting a non-zero answer.
• In composite bars, equating stresses instead of strains - the condition is σ₁/E₁ = σ₂/E₂ when the lengths and extensions are equal.
• Treating E as a measure of strength and quoting a higher E for high-strength steel, or confusing stiffness (E) with hardness and toughness.

Related topics

Principal stresses and theories of failure · Strain energy and impact loading · Bending, torsion and columns · Thin and thick cylinders · Deflection of beams · Engineering materials and heat treatment