Stress, Strain and Safety Factors
Loads and stress types, Hooke’s law, elastic modulus, yield and ultimate strength, safety factors and a method for a strength check.
Loads, internal forces and stress
A load is an external force or moment on a part. Inside the material the load is carried by internal forces distributed over the cross-section; the intensity of that distribution, force per unit area, is the stress, the measure that compares with the strength of the material. A part fails when the stress at some point exceeds what the material can take, so stress, not force, is what matters: the same 10 kN pull is harmless on a thick bar and fatal on a thin wire.
σ = F ÷ A (normal stress) τ = V ÷ A (average shear stress) 1 MPa = 1 N/mm²
Normal stress σ acts perpendicular to the cut face (tension is positive, compression negative); shear stress τ acts along it. The area A is the cross-section that carries the force: the net area, after subtracting holes. For a member in pure tension or compression, with the force through the centroid, the stress is uniform and σ = F ÷ A.
Strain and Hooke’s law
Strain ε is the stretch per unit length, ε = ΔL ÷ L (a pure number, often quoted in microstrain, 10-6). In the elastic range stress and strain are proportional: σ = E ε (Hooke’s law), where E is the modulus of elasticity (Young’s modulus), a measure of the stiffness of the material. Combining the two gives the elongation of a bar: δ = F L ÷ (A E).
Worked example: a steel tie rod
- A round steel rod of diameter 12 mm carries 10 kN in tension. The area is A = π × 12² ÷ 4 = 113.1 mm², so σ = 10 000 ÷ 113.1 = 88.4 MPa.
- With E = 200 GPa the strain is ε = 88.4 ÷ 200 000 = 442 µε, and over 1 000 mm the rod stretches δ = ε L = 0.442 mm: stiff metals stretch very little.
- The yield strength of the steel is 350 MPa, so the safety factor against yielding is n = 350 ÷ 88.4 = 4.0.
- The stiffness of the rod as a spring is k = A E ÷ L = 22 619 N/mm (23 kN/mm).
The stress–strain curve
| Point | Meaning | Use |
|---|---|---|
| Elastic region | The straight line: the part springs back when unloaded | Design for stresses in this region |
| Yield strength Sy | The stress where permanent (plastic) deformation begins; for materials without a clear yield point, the stress at 0.2 percent permanent strain | The usual limit for ductile metals under static load |
| Ultimate tensile strength Su | The highest stress the material reaches before necking | Used for brittle materials and in fatigue and fracture |
| Elongation | Strain at fracture: ductility | Over 5 percent is ductile; cast iron and ceramics are brittle |
| Toughness | The area under the curve: energy absorbed before fracture | Impact and shock loads |
Ductile materials (mild steel, aluminum, copper) give a warning by stretching and deforming; brittle materials (cast iron, glass, ceramics, some plastics) break suddenly, and are much weaker in tension than in compression. A part designed against yielding is safe from permanent set; against fracture needs the ultimate strength and the notch toughness. Material data are in Material Properties.
Elastic moduli and Poisson’s ratio
| Material | Elastic modulus E (GPa) | Poisson’s ratio ν | Shear modulus G (GPa) |
|---|---|---|---|
| Carbon and alloy steel | 200 to 210 | 0.29 to 0.30 | 78 to 82 |
| Stainless steel | 190 to 200 | 0.30 | 75 to 77 |
| Aluminum alloys | 69 to 72 | 0.33 | 26 to 27 |
| Titanium (Ti-6Al-4V) | 110 to 115 | 0.32 | 42 |
| Copper, brass, bronze | 100 to 125 | 0.33 to 0.35 | 37 to 45 |
| Cast iron | 95 to 150 | 0.26 | 40 to 60 |
| Glass-filled nylon, PETG, ABS | 2 to 8 | 0.35 to 0.40 | 0.7 to 3 |
| Concrete | 25 to 40 | 0.15 to 0.20 | 10 to 17 |
| Wood (along the grain) | 9 to 16 | 0.3 to 0.5 | 0.6 to 1 |
When a bar is stretched it also gets thinner: the lateral strain is −ν times the axial strain (Poisson’s ratio ν). For an isotropic material the three constants are linked: G = E ÷ (2 (1 + ν)); for steel with ν = 0.30 that is G = 200 ÷ (2 × 1.30) = 76.9 GPa. Note that E is almost the same for all steels: a stronger steel is not stiffer, only stronger (so a higher-strength alloy deflects the same as mild steel for the same size). Aluminum is one third as stiff as steel and one third the weight, so the stiffness per weight is similar.
Shear and bearing stress
Pins, bolts, rivets, keys and welds carry load in shear: τ = F ÷ Ashear, with a pin in double shear (clevis) having two shear planes, so A = 2 × (π d² ÷ 4). Where the pin presses on the hole there is also a bearing stress σb = F ÷ (d t) on the projected area of the hole. The yield in shear of a ductile metal is about 0.577 of its tensile yield (von Mises) or 0.5 (Tresca).
Worked example: a clevis pin in double shear
- A 10 mm pin (yield 400 MPa) carries 8 kN in double shear: τ = 8 000 ÷ (2 × 78.5) = 50.9 MPa. The shear yield strength is 0.577 × 400 = 231 MPa, so n = 231 ÷ 50.9 = 4.5.
- The bearing stress on two 6 mm plates is σb = 8 000 ÷ (10 × 2 × 6) = 67 MPa: fine for steel plate but close to the limit for aluminum or thin plate (check the plate for tear-out at the edge, at least 1.5 to 2 pin diameters from the center of the hole).
Thermal stress
A part that is heated and free to expand grows by ΔL = α L ΔT without stress (α is the coefficient of thermal expansion: about 12 × 10-6 per K for steel, 23 × 10-6 for aluminum). If the expansion is prevented the part is stressed as if it had been compressed back: σ = E α ΔT. Compare Material Properties for the table of expansion coefficients.
Worked example: a rod held between rigid walls
A steel bar rises 50 K between rigid walls. Free expansion would be 12.00×10-6 × 50 = 0.00060 (0.60 mm per meter). Prevented, the stress is σ = 200 000 × 12.00×10-6 × 50 = 120 MPa in compression: a third of the yield strength of mild steel from a 50 K temperature rise. Allow expansion with a sliding joint, a gap or a flexible element (as in a pipe loop), or the structure will buckle or the supports yield.
Combined stresses and principal stresses
At a point in a part there may be normal stresses σx and σy and a shear stress τxy at once. On some plane through the point the shear vanishes and the normal stresses are at their maximum and minimum: the principal stresses. They are σ1,2 = (σx + σy) ÷ 2 ± √(((σx − σy) ÷ 2)² + τxy²), and the maximum shear stress is τmax = √(((σx − σy) ÷ 2)² + τxy²) (the radius of Mohr’s circle). The principal angle is tan 2θp = 2 τxy ÷ (σx − σy). A ductile material is checked with the von Mises stress (see Torsion, Shear and Combined Loads).
Worked example: principal stresses
- A point has σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. The average is 50 MPa and the radius R = √(30² + 30²) = 42.4 MPa.
- σ1 = 50 + 42.4 = 92.4 MPa, σ2 = 50 − 42.4 = 7.6 MPa, and τmax = 42.4 MPa. The principal planes are rotated by θp = 22.5° from the x axis.
- The von Mises equivalent stress in plane stress is σ′ = √(σ1² − σ1 σ2 + σ2²) = 88.9 MPa: compare this with the yield strength, not the larger principal stress.
Safety factors and allowable stress
The safety factor is the ratio of the strength to the stress: n = S ÷ σ, where S is the yield strength (for ductile metals, static load), the ultimate strength (brittle materials), the endurance limit (fatigue) or the critical load (buckling). The allowable stress is the strength divided by the chosen factor. The factor covers everything that cannot be known exactly: the real loads, the real material strength (scatter, defects, heat treatment), the accuracy of the calculation and the manufacturing, the consequence of failure and any wear and corrosion.
| Situation | Typical safety factor on yield (or on the relevant strength) |
|---|---|
| Well-known loads, controlled materials, low consequence | 1.25 to 1.5 (aerospace, where weight is critical and testing is extensive) |
| Ordinary machine design, static, known loads | 1.5 to 2.5 |
| Uncertain loads, shock, average materials | 2.5 to 4 |
| Brittle materials, on the ultimate strength | 3 to 5 or more |
| Lifting equipment, hoists, people at risk | 4 to 8 or as the code states (cables and slings often 5) |
| Pressure vessels and structures | As required by the design code (ASME, EN 13445, AISC) |
- Load factors versus material factors: modern codes (LRFD, Eurocodes) apply factors to the loads and to the strength separately; the simple safety factor combines them.
- Do not over-trust the number: a factor of 3 on a wrongly estimated load can be less safe than a factor of 1.5 on a measured one. Measure or bound the loads (see Force, Load Cells and Vibration) and check the worst case.
- Weak links: the part is only as strong as its weakest section, joint or fastener: check all of them.
- Consider failure modes beyond yield: buckling (Columns and Buckling), fatigue (Fatigue and Stress Concentration), fracture, creep at high temperature, corrosion, and excessive deflection (Beam Bending and Deflection).
Units and conversions
| Quantity | SI unit | Conversions |
|---|---|---|
| Force | newton (N) | 1 kN = 224.8 lbf; 1 kgf = 9.807 N |
| Stress, pressure | pascal (Pa); MPa = N/mm² | 1 MPa = 145.04 psi; 1 ksi = 6.895 MPa |
| Modulus | GPa | 1 GPa = 145 038 psi = 145 ksi; steel 200 GPa ≈ 29 000 ksi |
| Moment, torque | N·m | 1 N·m = 8.851 lbf·in = 0.7376 lbf·ft |
| Second moment of area | mm⁴ (cm⁴, m⁴) | 1 in⁴ = 416 231 mm⁴ = 41.62 cm⁴ |
| Length | m, mm | 1 in = 25.4 mm |
Keep the units consistent: with force in N and length in mm the stress comes out in N/mm² = MPa, the modulus in MPa (200 000 for steel) and the second moment of area in mm⁴. Mixing m and mm is the commonest cause of a result wrong by a factor of 1 000 or 1 000 000 (see Shafts, Keys and Couplings for an example worked in N and mm).
Related
- Structures and Strength of Materials: the overview of this guide.
- Beam Bending and Deflection and Torsion, Shear and Combined Loads: the main loading cases.
- Material Properties: properties of the materials.
- Fatigue and Stress Concentration: when the load repeats.
- Worked Design Checks: complete checks from loads to safety factor.
FAQ
What is the difference between stress and strain?
Stress is the force per unit area inside the material (MPa); strain is the relative deformation (a ratio). They are related through the modulus of elasticity: σ = E ε in the elastic range.
What does yield strength mean?
The stress at which the material starts to deform permanently. Below it the part springs back; above it it keeps some deformation. For a design against yielding, keep the stress below the yield strength divided by the safety factor.
Why does steel strength not change its stiffness?
The modulus of elasticity is set by the atomic bonds in iron, almost the same for all steels. Alloying and heat treatment change the yield and ultimate strength, but not the stiffness, so a deflection-limited part gains nothing from a stronger steel.
What safety factor should I use?
For ordinary machine parts 1.5 to 3 on yield, depending on how well the loads are known and what a failure would cost. For lifting and people-carrying equipment follow the applicable code, which usually demands much higher values.
References
- Beer, F. P., Johnston, E. R. and DeWolf, J. T., Mechanics of Materials, McGraw-Hill: stress and strain, Hooke’s law, Mohr’s circle.
- Shigley, J. E. and Mischke, C. R., Mechanical Engineering Design, McGraw-Hill: stress, strength and the design factor.
- ASM Handbook, vol. 1 and ASM Metals Handbook for material properties; ASTM E8 (tension testing of metals).
Last reviewed: October 2026. Values are typical; use the material supplier’s certificate and the applicable design code.