Stress-strain relationships, deformation mechanisms, and strengthening
Understanding mechanical behavior begins with two fundamental concepts: stress (the intensity of force) and strain (the measure of deformation). These quantities allow us to describe material response in a way that is independent of specimen size.
Engineering stress and strain are defined using the original specimen dimensions, which makes them easy to calculate from test data:
Engineering Stress:
$$\sigma = \frac{F}{A_0}$$Engineering Strain:
$$\varepsilon = \frac{\Delta L}{L_0} = \frac{L - L_0}{L_0}$$Where: $F$ = applied force, $A_0$ = original cross-sectional area, $L_0$ = original length, $L$ = current length
True stress and strain account for the changing dimensions of the specimen during deformation. They are more physically meaningful, especially at large strains:
True Stress:
$$\sigma_T = \frac{F}{A} = \sigma(1 + \varepsilon)$$True Strain:
$$\varepsilon_T = \ln\left(\frac{L}{L_0}\right) = \ln(1 + \varepsilon)$$Note: The relationships $\sigma_T = \sigma(1 + \varepsilon)$ and $\varepsilon_T = \ln(1 + \varepsilon)$ assume constant volume during plastic deformation.
At small strains (less than about 1%), engineering and true values are nearly identical. At large strains, the differences become significant. True stress continues to increase even after the engineering stress reaches a maximum.
In general, stress is a tensor quantity that describes the state of stress at a point. For a three-dimensional stress state, nine components are needed (though symmetry reduces this to six independent components):
$$\sigma_{ij} = \begin{bmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{bmatrix}$$The diagonal components ($\sigma_{xx}$, $\sigma_{yy}$, $\sigma_{zz}$) are normal stresses, while the off-diagonal components are shear stresses. For equilibrium, $\tau_{xy} = \tau_{yx}$, and similarly for other pairs.
Elastic deformation is reversible: when the load is removed, the material returns to its original shape. At the atomic level, elastic deformation corresponds to the stretching of atomic bonds without atoms changing their neighbors.
For most materials at small strains, stress is proportional to strain. This linear relationship is known as Hooke's Law:
Hooke's Law (Uniaxial):
$$\sigma = E\varepsilon$$Where: $E$ = Young's modulus (elastic modulus), with units of Pa or GPa
Young's modulus is a measure of stiffness. Materials with strong atomic bonds (like ceramics and covalent solids) have high moduli, while materials with weak bonds (like polymers) have low moduli.
| Material | Young's Modulus (GPa) |
|---|---|
| Diamond | 1000 |
| Tungsten | 400 |
| Steel | 200 |
| Aluminum | 70 |
| Magnesium | 45 |
| Polystyrene | 3 |
| Rubber | 0.01 |
When a material is stretched in one direction, it typically contracts in the perpendicular directions. Poisson's ratio quantifies this effect:
$$\nu = -\frac{\varepsilon_{lateral}}{\varepsilon_{axial}}$$For most metals, $\nu \approx 0.3$. For rubber and other nearly incompressible materials, $\nu \approx 0.5$. Cork has $\nu \approx 0$, which is why it works well as a bottle stopper.
For shear deformation, the analogous relationship is:
$$\tau = G\gamma$$where $G$ is the shear modulus and $\gamma$ is the shear strain. For isotropic materials, the elastic constants are related:
$$G = \frac{E}{2(1 + \nu)}$$A steel wire 2 m long with a diameter of 1 mm supports a mass of 10 kg. Calculate the elastic elongation. Take $E = 200$ GPa.
Solution:
Cross-sectional area: $A = \pi r^2 = \pi (0.0005)^2 = 7.85 \times 10^{-7}$ m²
Force: $F = mg = 10 \times 9.8 = 98$ N
Stress: $\sigma = F/A = 98 / (7.85 \times 10^{-7}) = 125$ MPa
Strain: $\varepsilon = \sigma/E = (125 \times 10^6) / (200 \times 10^9) = 6.25 \times 10^{-4}$
Elongation: $\Delta L = \varepsilon \cdot L_0 = (6.25 \times 10^{-4})(2) = 1.25$ mm
The tensile test is the most common mechanical test. A specimen with a known geometry is pulled at a controlled rate while measuring force and elongation. The resulting stress-strain curve reveals many important properties.
A typical stress-strain curve for a ductile metal shows several distinct regions:
Yield Strength ($\sigma_y$): Stress at which plastic deformation begins. For materials without a distinct yield point, the 0.2% offset method is used: draw a line parallel to the elastic region offset by 0.2% strain, and the yield strength is where this line intersects the curve.
Ultimate Tensile Strength (UTS): Maximum engineering stress on the curve.
Ductility: Measured as percent elongation or percent reduction in area:
$$\% EL = \frac{L_f - L_0}{L_0} \times 100$$ $$\% RA = \frac{A_0 - A_f}{A_0} \times 100$$Resilience is the ability to absorb energy during elastic deformation. The modulus of resilience is the area under the stress-strain curve up to the yield point:
$$U_r = \frac{\sigma_y^2}{2E}$$Toughness is the ability to absorb energy before fracture. It equals the total area under the stress-strain curve and depends on both strength and ductility. A material can be strong but not tough (like glass) or tough but not very strong (like rubber).
Material A has $\sigma_y$ = 500 MPa, UTS = 600 MPa, and 15% elongation. Material B has $\sigma_y$ = 300 MPa, UTS = 400 MPa, and 40% elongation. Both have $E$ = 200 GPa. Which is tougher?
Solution:
Approximate toughness as the area under the curve. For a rough estimate, use the average stress times the strain to failure:
Material A: $U \approx \frac{(500 + 600)}{2} \times 0.15 = 82.5$ MJ/m³
Material B: $U \approx \frac{(300 + 400)}{2} \times 0.40 = 140$ MJ/m³
Despite being weaker, Material B is tougher due to its greater ductility.
Plastic deformation is permanent deformation that occurs when the stress exceeds the yield strength. In crystalline materials, plastic deformation primarily occurs by the movement of dislocations along specific crystallographic planes and directions.
A slip system consists of a slip plane (typically the most densely packed plane) and a slip direction (typically the most densely packed direction). The number and geometry of slip systems strongly influence the ductility of a material.
| Crystal Structure | Slip Plane | Slip Direction | Number of Systems |
|---|---|---|---|
| FCC | {111} | <110> | 12 |
| BCC | {110}, {112}, {123} | <111> | 48 |
| HCP | (0001) | <1120> | 3 |
FCC metals (Cu, Al, Au, Ni) are generally ductile because they have 12 slip systems, all with favorable geometry. HCP metals (Mg, Zn, Ti) tend to be less ductile because their primary slip systems are limited.
Slip begins when the shear stress on a slip system reaches a critical value called the critical resolved shear stress (CRSS), $\tau_{CRSS}$. For a single crystal under uniaxial tension, the resolved shear stress is given by Schmid's Law:
Schmid's Law:
$$\tau_R = \sigma \cos\phi \cos\lambda$$Where: $\phi$ = angle between stress axis and slip plane normal, $\lambda$ = angle between stress axis and slip direction
Schmid Factor: $m = \cos\phi \cos\lambda$ (maximum value = 0.5)
Yield occurs when $\tau_R = \tau_{CRSS}$, so the yield stress is:
$$\sigma_y = \frac{\tau_{CRSS}}{\cos\phi \cos\lambda} = \frac{\tau_{CRSS}}{m}$$The slip system with the largest Schmid factor will yield first.
An FCC single crystal is oriented such that the tensile axis makes angles of 35° with both the slip plane normal and the slip direction for the most favorably oriented slip system. If $\tau_{CRSS}$ = 3 MPa, what is the tensile yield stress?
Solution:
Schmid factor: $m = \cos(35°)\cos(35°) = (0.819)(0.819) = 0.671$
Yield stress: $\sigma_y = \tau_{CRSS}/m = 3/0.671 = 4.5$ MPa
Since plastic deformation occurs by dislocation motion, anything that impedes dislocation motion will increase the strength of a material. There are four main strengthening mechanisms in metals.
Grain boundaries act as obstacles to dislocation motion because dislocations cannot easily cross from one grain to another (different slip plane orientations). Smaller grains mean more grain boundaries and higher strength. The Hall-Petch equation describes this relationship:
Hall-Petch Equation:
$$\sigma_y = \sigma_0 + k_y d^{-1/2}$$Where: $\sigma_0$ = friction stress (lattice resistance), $k_y$ = Hall-Petch slope (material constant), $d$ = average grain diameter
This relationship holds over a wide range of grain sizes, from millimeters down to about 20 nm. Below about 10-20 nm, other mechanisms (like grain boundary sliding) become important and the Hall-Petch relationship breaks down.
Impurity atoms in a metal lattice create local stress fields that interact with dislocations, making their motion more difficult. The strengthening effect depends on the size and concentration of the solute atoms:
$$\Delta\sigma \propto c^{1/2}$$where $c$ is the solute concentration. Larger size differences between solute and solvent atoms produce greater strengthening.
As a metal is plastically deformed, the dislocation density increases dramatically. These dislocations interact with each other, creating tangles that impede further dislocation motion. The flow stress increases approximately as:
$$\sigma \propto \rho^{1/2}$$where $\rho$ is the dislocation density. Strain hardening is described by:
$$\sigma_T = K\varepsilon_T^n$$where $K$ is the strength coefficient and $n$ is the strain hardening exponent (typically 0.1 to 0.5 for metals).
Fine, hard precipitate particles dispersed in the matrix can dramatically increase strength. Dislocations must either cut through the particles (for small, coherent precipitates) or bow around them (for larger, incoherent precipitates). The strength contribution depends on particle size, spacing, and volume fraction.
For the Orowan bowing mechanism:
$$\Delta\tau = \frac{Gb}{\lambda}$$where $G$ is the shear modulus, $b$ is the Burgers vector, and $\lambda$ is the interparticle spacing.
A low-carbon steel has $\sigma_0$ = 70 MPa and $k_y$ = 0.74 MPa·m$^{1/2}$. Calculate the yield strength for grain sizes of 100 µm and 10 µm.
Solution:
For $d$ = 100 µm = $10^{-4}$ m:
$$\sigma_y = 70 + 0.74(10^{-4})^{-1/2} = 70 + 0.74(100) = 144 \text{ MPa}$$For $d$ = 10 µm = $10^{-5}$ m:
$$\sigma_y = 70 + 0.74(10^{-5})^{-1/2} = 70 + 0.74(316) = 304 \text{ MPa}$$Reducing the grain size by a factor of 10 more than doubles the yield strength.
Hardness is a measure of a material's resistance to localized plastic deformation (typically indentation). Hardness tests are quick, inexpensive, and usually non-destructive, making them valuable for quality control and material characterization.
Rockwell Hardness: Uses different indenter shapes and loads for different materials. The hardness number is calculated from the depth of indentation. Common scales include HRC (diamond cone, 150 kg load) for hard steels and HRB (1/16" ball, 100 kg load) for softer metals.
Brinell Hardness (HB): A hardened steel or tungsten carbide ball (usually 10 mm diameter) is pressed into the surface under a known load (typically 500 or 3000 kg). The Brinell number is:
$$HB = \frac{2P}{\pi D(D - \sqrt{D^2 - d^2})}$$where $P$ is the load, $D$ is the ball diameter, and $d$ is the impression diameter.
Vickers Hardness (HV): Uses a diamond pyramid indenter with a 136° angle. The Vickers number is based on the diagonal length of the impression and can be used across a wide range of hardnesses:
$$HV = \frac{1.854P}{d^2}$$For many metals, there is an approximate relationship between hardness and tensile strength:
Hardness-Tensile Strength Correlations:
For steels: $UTS \text{ (MPa)} \approx 3.45 \times HB$
For copper alloys: $UTS \text{ (MPa)} \approx 3.0 \times HB$
These are approximate relationships useful for estimation. Actual values may vary by 10-20%.
| Test | Indenter | Load Range | Best For |
|---|---|---|---|
| Rockwell C | Diamond cone | 150 kg | Hard steels, tool steels |
| Rockwell B | 1/16" steel ball | 100 kg | Soft steels, brass, aluminum |
| Brinell | 10 mm ball | 500-3000 kg | Cast iron, forgings |
| Vickers | Diamond pyramid | 1-120 kg | All materials, thin sections |
| Knoop | Elongated diamond | 0.025-5 kg | Brittle materials, thin coatings |
Fracture is the separation of a material into two or more pieces under stress. Understanding fracture is critical for preventing catastrophic failures in engineering structures.
Ductile fracture involves significant plastic deformation before failure. The fracture surface typically shows a cup-and-cone appearance in tensile specimens, with evidence of void formation and coalescence. Ductile fracture absorbs considerable energy and usually gives warning before final failure.
Brittle fracture occurs with little or no plastic deformation. The fracture surface is relatively flat and may show cleavage facets or a granular appearance. Brittle fracture is dangerous because it occurs suddenly without warning and at stress levels below the yield strength.
Griffith recognized that real materials contain flaws (cracks, inclusions, pores) that concentrate stress. For a brittle material with a sharp crack, the stress required for fracture is:
Griffith Equation:
$$\sigma_c = \sqrt{\frac{2E\gamma_s}{\pi a}}$$Where: $E$ = Young's modulus, $\gamma_s$ = surface energy, $a$ = half-crack length
This equation shows that fracture stress decreases as crack size increases. For ductile materials, the equation is modified to include the plastic work at the crack tip.
Fracture toughness ($K_{IC}$) is a material property that describes resistance to crack propagation. It has units of MPa·m$^{1/2}$ and is measured using specimens with pre-existing cracks. Fracture occurs when:
$$K = Y\sigma\sqrt{\pi a} \geq K_{IC}$$where $Y$ is a geometric factor (typically 1.0-1.2 for common geometries). Materials with high $K_{IC}$ can tolerate larger cracks at a given stress level.
| Material | $K_{IC}$ (MPa·m$^{1/2}$) |
|---|---|
| Aluminum alloys | 20-40 |
| Steel (low alloy) | 50-150 |
| Titanium alloys | 50-100 |
| Alumina (Al₂O₃) | 3-5 |
| Silicon carbide | 3-4 |
| Glass | 0.7-0.8 |
A steel plate with $K_{IC}$ = 80 MPa·m$^{1/2}$ is subjected to a stress of 200 MPa. What is the maximum crack size that can be tolerated? Assume $Y$ = 1.0.
Solution:
At the critical condition: $K_{IC} = Y\sigma\sqrt{\pi a_c}$
Solving for the critical crack half-length:
$$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{Y\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{80}{1.0 \times 200}\right)^2$$ $$a_c = \frac{1}{\pi}(0.4)^2 = 0.051 \text{ m} = 51 \text{ mm}$$The full crack length (2$a$) is about 100 mm. This seems large, but lower-toughness materials would have much smaller critical crack sizes.
Fatigue is the progressive damage and failure of a material under cyclic loading. Fatigue failures are particularly dangerous because they occur at stress levels well below the yield strength and often without obvious warning. It is estimated that fatigue accounts for 90% of all mechanical failures in service.
The relationship between stress amplitude and the number of cycles to failure is shown on an S-N curve (also called a Wöhler curve). For ferrous metals and titanium alloys, there is often a fatigue limit (endurance limit) below which fatigue failure does not occur regardless of the number of cycles. For most non-ferrous metals, there is no true fatigue limit and failure eventually occurs at any stress amplitude.
The stress amplitude is defined as:
$$\sigma_a = \frac{\sigma_{max} - \sigma_{min}}{2}$$And the mean stress:
$$\sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2}$$Once a fatigue crack initiates (typically at a stress concentration), it grows incrementally with each load cycle. The rate of crack growth is described by the Paris Law:
Paris Law:
$$\frac{da}{dN} = C(\Delta K)^m$$Where: $da/dN$ = crack growth rate per cycle, $\Delta K = K_{max} - K_{min}$ = stress intensity range, $C$ and $m$ = material constants (typically $m$ = 2-4)
The Paris Law applies in the "steady-state" region of crack growth. Below a threshold $\Delta K_{th}$, cracks do not propagate. Above a critical value, rapid unstable fracture occurs.
A component has an initial crack of 2 mm and will fail when the crack reaches 20 mm. If $C = 2 \times 10^{-11}$ m/cycle (with $\Delta K$ in MPa·m$^{1/2}$), $m = 3$, and $\Delta K = 20$ MPa·m$^{1/2}$ (constant), estimate the fatigue life.
Solution:
Integrating the Paris Law for constant $\Delta K$:
$$N = \int_{a_i}^{a_f} \frac{da}{C(\Delta K)^m} = \frac{a_f - a_i}{C(\Delta K)^m}$$ $$N = \frac{0.020 - 0.002}{(2 \times 10^{-11})(20)^3}$$ $$N = \frac{0.018}{(2 \times 10^{-11})(8000)} = \frac{0.018}{1.6 \times 10^{-7}}$$ $$N = 112,500 \text{ cycles}$$Creep is the time-dependent plastic deformation of materials under constant stress. It becomes significant at temperatures above about 0.4 $T_m$ (where $T_m$ is the absolute melting temperature). Creep is a major concern for high-temperature applications such as jet engines, power plants, and furnace components.
A typical creep curve (strain vs. time at constant stress and temperature) shows three stages:
For design purposes, the secondary creep rate is most important. It follows a power-law relationship:
Power Law Creep:
$$\dot{\varepsilon}_s = A\sigma^n \exp\left(-\frac{Q_c}{RT}\right)$$Where: $\dot{\varepsilon}_s$ = steady-state creep rate, $A$ = material constant, $n$ = stress exponent (typically 3-8), $Q_c$ = activation energy for creep
Different mechanisms dominate at different stress and temperature regimes:
The Larson-Miller parameter (LMP) allows comparison of creep data at different temperatures and times. It is based on the observation that temperature and time can be combined into a single parameter:
Larson-Miller Parameter:
$$LMP = T(C + \log t_r)$$Where: $T$ = temperature (K or °R), $t_r$ = rupture time (hours), $C$ = material constant (typically 20 for many alloys)
Plotting stress vs. LMP gives a master curve that can be used to predict rupture life at different temperature-time combinations.
A turbine blade material fails after 10,000 hours at 800°C under a certain stress. Using the Larson-Miller parameter with $C$ = 20, estimate the rupture time at 850°C under the same stress.
Solution:
At constant stress, LMP is constant. First calculate LMP at 800°C (1073 K):
$$LMP = 1073(20 + \log(10000)) = 1073(20 + 4) = 25,752$$At 850°C (1123 K), solve for $t_r$:
$$25,752 = 1123(20 + \log(t_r))$$ $$20 + \log(t_r) = 22.93$$ $$\log(t_r) = 2.93$$ $$t_r = 851 \text{ hours}$$The 50°C temperature increase reduces the rupture life by more than a factor of 10. This illustrates why temperature control is critical in high-temperature applications.
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