Materials Science Cheatsheet

Cheatsheet Content

1. Material Selection Process a) Procedure for Material Selection Analyze Application Requirements: Define function, environment, lifespan, aesthetics, and cost. Identify Property Requirements: Translate application needs into specific material properties (mechanical, physical, chemical, electrical). Screening Materials: Use databases or charts (e.g., Ashby charts) to narrow down material classes based on primary properties. Detailed Evaluation: Consider secondary properties and processing characteristics. Evaluate cost-effectiveness (material cost, processing cost, maintenance). Assess environmental impact and sustainability. Final Selection & Testing: Choose the best candidate(s) and conduct prototypes or simulations. b) Crucial Criteria for Material Selection Mechanical Properties: Strength (yield, tensile), hardness, ductility, toughness, fatigue, creep. Physical Properties: Density, melting point, thermal conductivity, electrical conductivity, coefficient of thermal expansion. Chemical Properties: Corrosion resistance, oxidation resistance, biodegradability. Processing Characteristics: Formability, machinability, weldability, castability, heat treatability. Cost: Raw material cost, manufacturing cost, assembly cost, maintenance cost. Availability & Sustainability: Supply chain, recyclability, environmental impact. 2. Edge Dislocation Alignment When two edge dislocations of opposite sign (one with an extra half-plane from the top, one from the bottom) and separated by several atomic distances become aligned on the same slip plane, they effectively "annihilate" each other. The extra half-plane associated with one dislocation fills the void created by the missing half-plane of the other. The resulting defect is a perfect lattice or a region of significantly reduced strain energy, as the lattice distortion caused by the dislocations is eliminated or greatly minimized. _ _ _ _ _ _ _ _ _ _ _ _ | | | _ _ _ _ _ _ _ _ | <-- Extra half-plane | | | | | | | | | | | | | | | | | |_ _ _ _ _ _ _ _| | | | | | | | | | | | |_ _ _ _ _ _ _ _ _ _ _ _| ^ ^ | | Positive Edge Dislocation _ _ _ _ _ _ _ _ _ _ _ _ | | | | | | | | | | | _ _ _ _ _ _ _ _ | <-- Missing half-plane | | | | | | | | | | | | | | | | | |_ _ _ _ _ _ _ _| | | | |_ _ _ _ _ _ _ _ _ _ _ _| ^ ^ | | Negative Edge Dislocation (opposite sign) When these align, they cancel out, reducing lattice strain. 3. Dislocation Motion vs. Shear Stress Direction Dislocation Type Relationship between Applied Shear Stress Direction and Dislocation Line Motion Edge Dislocation The dislocation line moves perpendicular to the dislocation line itself and parallel to the Burgers vector, in the direction of the applied shear stress. Screw Dislocation The dislocation line moves parallel to the dislocation line itself and perpendicular to the Burgers vector, in the direction of the applied shear stress. Mixed Dislocation The dislocation line motion has components both perpendicular and parallel to the dislocation line, influenced by both edge and screw components, generally in the direction of the resolved shear stress. 4. Small-Angle vs. High-Angle Grain Boundaries Small-angle grain boundaries (SAGBs) are less effective in interfering with slip than high-angle grain boundaries (HAGBs) because: Lattice Misorientation: SAGBs have a small misorientation angle ($ < 10^\circ - 15^\circ $) between adjacent grains. This means the crystal lattice across the boundary is only slightly disrupted. Dislocation Structure: SAGBs can often be described as an array of dislocations. The stress fields of these dislocations are less severe and more localized than the highly disordered structure of HAGBs. Slip Transmission: Due to the small misorientation and less severe lattice distortion, it is easier for dislocations to pass through or be transmitted across SAGBs. The resistance to slip is minimal compared to HAGBs where a significant change in slip plane orientation occurs. Energy: SAGBs have lower interfacial energy compared to HAGBs. HAGBs present a much larger crystallographic discontinuity, requiring more energy for dislocations to cross, thus acting as stronger barriers to slip. 5. Brittleness of HCP Metals HCP (Hexagonal Close-Packed) metals are typically more brittle than FCC (Face-Centered Cubic) and BCC (Body-Centered Cubic) metals due to: Limited Slip Systems: HCP crystal structures have fewer independent slip systems than FCC and BCC. For a material to be ductile, it generally requires at least five independent slip systems to accommodate arbitrary changes in shape by plastic deformation (von Mises criterion). Primary Slip System: The primary slip system in most HCP metals is on the basal plane ($\{0001\}\langle 11\bar{2}0 \rangle$). This offers only 3 independent slip systems. Activation of Non-Basal Slip: Other slip systems (e.g., prismatic or pyramidal) often require significantly higher critical resolved shear stress (CRSS) to activate, especially at room temperature. This makes plastic deformation more difficult. Twinning: When slip is restricted, HCP metals tend to deform by twinning, which is a more limited mode of plastic deformation and often leads to fracture. FCC (12 slip systems) and BCC (many slip systems, though not all active at low temperatures) have ample slip systems, allowing for significant plastic deformation before fracture, hence their higher ductility. 6. Hall-Petch Equation and Yield Point The Hall-Petch equation describes the relationship between yield strength and grain size: $$ \sigma_y = \sigma_0 + k_y d^{-1/2} $$ Where: $\sigma_y$ is the yield strength $\sigma_0$ is the friction stress (a material constant) $k_y$ is the Hall-Petch constant (a material constant) $d$ is the average grain diameter Given data for iron: Case 1: $d_1 = 1 \times 10^{-2}$ mm, $\sigma_{y1} = 230$ MPa Case 2: $d_2 = 6 \times 10^{-3}$ mm, $\sigma_{y2} = 275$ MPa We have two equations: $230 = \sigma_0 + k_y (1 \times 10^{-2})^{-1/2}$ $275 = \sigma_0 + k_y (6 \times 10^{-3})^{-1/2}$ Calculate inverse square roots: $(1 \times 10^{-2})^{-1/2} = (1/100)^{-1/2} = 10$ $(6 \times 10^{-3})^{-1/2} = (0.006)^{-1/2} \approx 12.91$ So, the equations become: $230 = \sigma_0 + 10 k_y$ $275 = \sigma_0 + 12.91 k_y$ Subtract (1) from (2): $275 - 230 = (12.91 - 10) k_y$ $45 = 2.91 k_y$ $k_y = 45 / 2.91 \approx 15.46$ MPa $\cdot$ mm$^{1/2}$ Substitute $k_y$ into (1) to find $\sigma_0$: $230 = \sigma_0 + 10 \times 15.46$ $230 = \sigma_0 + 154.6$ $\sigma_0 = 230 - 154.6 = 75.4$ MPa Now, we want to find the grain diameter $d_3$ when $\sigma_{y3} = 310$ MPa: $310 = \sigma_0 + k_y d_3^{-1/2}$ $310 = 75.4 + 15.46 d_3^{-1/2}$ $310 - 75.4 = 15.46 d_3^{-1/2}$ $234.6 = 15.46 d_3^{-1/2}$ $d_3^{-1/2} = 234.6 / 15.46 \approx 15.17$ $d_3 = (15.17)^{-2}$ $d_3 \approx 0.00435$ mm The lower yield point will be 310 MPa at a grain diameter of approximately $4.35 \times 10^{-3}$ mm. 7. Interstitial Impurity Atom near Edge Dislocation Location An interstitial impurity atom (e.g., carbon in iron) would be expected to situate itself in the compressive strain region just below the extra half-plane of an edge dislocation, or more precisely, in the region where the lattice is "stretched" or in tension, but the most stable position is in the dilated region of the lattice. For an edge dislocation, this is typically the region below the extra half-plane , where the atomic spacing is slightly larger than in the perfect lattice. It would sit in the "open" region or the dilated part of the lattice. Explanation in terms of Lattice Strains An edge dislocation introduces both compressive and tensile lattice strains into the crystal lattice: Compressive Strain: Above the slip plane, where the extra half-plane is inserted, atoms are pushed closer together, causing compressive stress. Tensile Strain: Below the slip plane, where atoms are pulled apart to accommodate the extra half-plane above, there is a region of tensile stress (dilated region). Interstitial impurity atoms are typically smaller atoms that fit into the spaces between the host atoms. When such an impurity atom is introduced into a crystal, it creates local lattice strains. To minimize the overall strain energy of the system, the interstitial atom will migrate to a region where it can relieve existing lattice strains. Specifically, an interstitial atom, being generally larger than the interstitial site it occupies in a perfect lattice, will tend to create a local tensile strain. Therefore, it will preferentially occupy a position in the crystal where there is already a local tensile (dilational) strain. For an edge dislocation, this is the region just below the extra half-plane (the "open" side). By moving to this dilated region, the interstitial atom effectively "fills" the expanded space, thereby reducing the tensile strain locally and lowering the total strain energy of the system. Conversely, a substitutional impurity atom that is larger than the host atom would prefer the tensile region, while a smaller substitutional atom would prefer the compressive region. --> Extra Half-Plane Compressive Strain Tensile Strain (Dilated) Interstitial Impurity Atom