🇮🇳 GATE Metallurgical Engineering · subject
GATE Metallurgical Engineering Mechanical Metallurgy Syllabus
The Mechanical Metallurgy syllabus for GATE Metallurgical Engineering in full: 6 chapters, 21 topics and 2 sub-topics, every one listed below, plus 49 Mechanical Metallurgy flashcards with answers.
Mechanical Metallurgy syllabus: all 6 chapters
Expand any chapter to see its topics and sub-topics.
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Strain Tensor and Stress Tensor
2 topics- Representation by Mohr’s Circle
- Elasticity, Stiffness, and Compliance Tensor
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Dislocation Theory
5 topics- Edge, Screw, and Mixed Dislocations
- Source and Multiplication of Dislocations
- Stress Fields Around Dislocations
- Partial Dislocations
- Dislocation Interactions and Reactions
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Strengthening Mechanisms
5 topics- Work/Strain Hardening
- Strengthening due to Grain Boundaries
- Strengthening due to Solid Solution
- Strengthening due to Precipitation
- Strengthening due to Dispersion
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Fracture Behaviour
5 topics- Griffith Theory
- Linear Elastic Fracture Mechanics
- Fracture Toughness
- Fractography
- Ductile to Brittle Transition
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Fatigue
2 topics- Cyclic Stress Strain Behaviour
- Low Cycle Fatigue
- High Cycle Fatigue
- Crack Growth
- Cyclic Stress Strain Behaviour
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Mechanisms of High Temperature Deformation and Failure
2 topics- Creep and Stress Rupture
- Stress Exponent and Activation Energy
GATE Metallurgical Engineering Mechanical Metallurgy flashcards — 49 questions with answers
What does a Mohr's circle graphically represent, and what are the coordinates of its center and radius for a 2D stress state with normal stresses $\sigma_x,\sigma_y$ and shear stress $\tau_{xy}$?
Mohr's circle is a graphical representation of the transformation of stress (or strain) on differently oriented planes. Center: $\left(\frac{\sigma_x+\sigma_y}{2},\,0\right)$; Radius: $R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}$.
Using Mohr's circle, give the formulas for the principal stresses and the maximum in-plane shear stress.
Principal stresses: $\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}$. Maximum in-plane shear stress: $\tau_{max}=R=\frac{\sigma_1-\sigma_2}{2}$.
On a Mohr's circle, how does the angle between physical planes relate to the angle subtended at the circle's center?
Angles on Mohr's circle are doubled: a rotation of $\theta$ between physical planes corresponds to an arc of $2\theta$ on the circle. Principal planes (where $\tau=0$) are $90^{\circ}$ apart physically but $180^{\circ}$ apart on the circle.
State Hooke's law in generalized tensor form and define the stiffness and compliance tensors.
$\sigma_{ij}=C_{ijkl}\,\varepsilon_{kl}$ and $\varepsilon_{ij}=S_{ijkl}\,\sigma_{kl}$. $C_{ijkl}$ is the fourth-rank elastic stiffness tensor and $S_{ijkl}$ is the compliance tensor; they are inverses of one another ($C=S^{-1}$).
How many independent elastic constants does the stiffness tensor have in the general anisotropic case, and how many for a cubic and an isotropic material?
General anisotropic (triclinic): 21 independent constants. Cubic: 3 ($C_{11},C_{12},C_{44}$). Isotropic: 2 (e.g., $E$ and $\nu$, or the two Lamé constants $\lambda$ and $\mu$).
For an isotropic material, relate the shear modulus $G$, Young's modulus $E$, bulk modulus $K$ and Poisson's ratio $\nu$.
$G=\frac{E}{2(1+\nu)}$ and $K=\frac{E}{3(1-2\nu)}$. The Zener anisotropy ratio for cubic crystals is $A=\frac{2C_{44}}{C_{11}-C_{12}}$, equal to 1 for an isotropic material.
Define the Burgers vector orientation relative to the dislocation line for edge and screw dislocations.
Edge dislocation: Burgers vector $\vec{b}$ is perpendicular to the dislocation line ($\vec{b}\perp\vec{\xi}$). Screw dislocation: $\vec{b}$ is parallel to the line ($\vec{b}\parallel\vec{\xi}$). A mixed dislocation has $\vec{b}$ at an arbitrary angle and contains both components.
How do edge and screw dislocations differ in their freedom to glide and cross-slip?
An edge dislocation has a uniquely defined slip plane (containing both $\vec{b}$ and $\vec{\xi}$) and can only climb out of it (non-conservative). A screw dislocation has no unique slip plane, so it can cross-slip onto any plane containing $\vec{b}$.
What is the magnitude of the Burgers vector for a perfect dislocation in an FCC crystal of lattice parameter $a$?
Perfect FCC dislocation: $\vec{b}=\frac{a}{2}\langle110\rangle$, with magnitude $|\vec{b}|=\frac{a}{2}\sqrt{2}=\frac{a}{\sqrt{2}}$.
Describe how a Frank-Read source multiplies dislocations and give its critical activation stress.
A dislocation segment pinned at two points of length $L$ bows out under shear stress; when it reaches a semicircle it becomes unstable, expands, and pinches off a loop while regenerating the segment, repeating to multiply dislocations. Critical stress: $\tau_c=\frac{Gb}{L}$.
Define the dislocation density and give the Taylor (forest) hardening relation between flow stress and dislocation density.
Dislocation density $\rho$ = total dislocation line length per unit volume (units $\text{m}^{-2}$). Taylor relation: $\tau=\tau_0+\alpha G b\sqrt{\rho}$, where $\alpha\approx0.3$–$0.5$.
Give the stress field of a screw dislocation along the $z$-axis.
A screw dislocation has only shear components: $\tau_{\theta z}=\frac{Gb}{2\pi r}$ (in cylindrical coordinates), or $\tau_{xz}=-\frac{Gb}{2\pi}\frac{y}{x^{2}+y^{2}}$, $\tau_{yz}=\frac{Gb}{2\pi}\frac{x}{x^{2}+y^{2}}$. The stress decays as $1/r$ and is axisymmetric.
Write the elastic strain energy per unit length stored in a dislocation and explain the cut-off radii.
$E\approx\frac{Gb^{2}}{4\pi}\ln\!\left(\frac{R}{r_0}\right)$ for a screw (and $\frac{Gb^{2}}{4\pi(1-\nu)}\ln\frac{R}{r_0}$ for an edge). Often simplified to $E\approx\tfrac{1}{2}Gb^{2}$. $r_0$ is the core radius (continuum elasticity invalid inside) and $R$ is the outer cut-off (crystal size or spacing).
State the line-tension approximation for a dislocation and the stress to bend it to radius $R$.
Line tension $T\approx\frac{1}{2}Gb^{2}$ (energy per unit length). The shear stress needed to curve a dislocation to radius of curvature $R$ is $\tau=\frac{T}{bR}=\frac{Gb}{2R}$.
What is a partial dislocation, and write the dissociation reaction of a perfect FCC dislocation into Shockley partials.
A partial dislocation has a Burgers vector that is not a full lattice translation, bounding a stacking fault. FCC dissociation: $\frac{a}{2}[10\bar{1}]\rightarrow\frac{a}{6}[11\bar{2}]+\frac{a}{6}[2\bar{1}\bar{1}]$.
Why does a perfect dislocation dissociate into partials, and what sets the equilibrium separation distance?
It dissociates because the energy criterion $b_1^{2}>b_2^{2}+b_3^{2}$ (Frank's rule: $\frac{a^{2}}{2}>\frac{a^{2}}{6}+\frac{a^{2}}{6}$) is satisfied, lowering energy. Equilibrium separation $d$ is set by the balance of the partials' elastic repulsion against the stacking-fault energy $\gamma$: $d\approx\frac{G b_p^{2}}{2\pi\gamma}$.
How does stacking fault energy (SFE) affect the separation of partials and the ease of cross-slip?
Low SFE gives wide separation of partials (wide stacking-fault ribbons), making cross-slip and climb difficult and favoring planar slip and twinning. High SFE gives narrow/constricted partials, easy cross-slip, and wavy slip.
State Frank's rule for whether a dislocation reaction is energetically favorable.
A reaction $\vec{b}_1\rightarrow\vec{b}_2+\vec{b}_3$ is favorable if it lowers the total energy, i.e. $|\vec{b}_1|^{2}>|\vec{b}_2|^{2}+|\vec{b}_3|^{2}$ (since energy $\propto b^{2}$).
What is a Lomer-Cottrell lock and why does it impede slip?
A Lomer-Cottrell lock forms when two dislocations on intersecting {111} planes react to give a sessile (stair-rod) partial $\frac{a}{6}\langle110\rangle$ lying along the line of intersection. Because its Burgers vector lies in neither slip plane, it cannot glide, acting as a barrier that contributes to work hardening.
Give the Peach-Koehler equation for the force per unit length on a dislocation.
$\vec{F}=(\boldsymbol{\sigma}\cdot\vec{b})\times\vec{\xi}$, where $\boldsymbol{\sigma}$ is the stress tensor, $\vec{b}$ the Burgers vector, and $\vec{\xi}$ the unit line vector. For a simple glide force: $F=\tau b$ per unit length.
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Write the force per unit length between two parallel edge dislocations (glide component) and state the sign convention.
Glide force: $F_x=\frac{Gb^{2}}{2\pi(1-\nu)}\frac{x(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}$. Like-sign edges repel when on the same side; they form a stable vertical wall (low-angle tilt boundary) when stacked one above another at $90^{\circ}$.
What is work (strain) hardening, and define the strain-hardening exponent in the Hollomon equation.
Work hardening is the increase in flow stress with plastic strain due to dislocation multiplication and interaction. Hollomon equation: $\sigma=K\varepsilon^{n}$, where $n$ is the strain-hardening exponent ($0\leq n\leq1$) and $K$ is the strength coefficient.
In a true stress-strain curve obeying $\sigma=K\varepsilon^{n}$, at what true strain does necking (the onset of plastic instability) begin?
Necking begins at the maximum-load (Considère) condition $\frac{d\sigma}{d\varepsilon}=\sigma$, which gives the true uniform strain $\varepsilon_u=n$ (the strain-hardening exponent).
What are the three stages of work hardening in single crystals?
Stage I: easy glide (low hardening, single slip system active). Stage II: linear hardening (high rate, multiple slip and dislocation forest interactions). Stage III: parabolic/dynamic recovery (rate decreases as cross-slip relieves stresses).
State the Hall-Petch relation for grain-boundary strengthening and identify each term.
$\sigma_y=\sigma_0+k_y\,d^{-1/2}$, where $\sigma_y$ is the yield strength, $\sigma_0$ the friction stress (lattice resistance), $k_y$ the Hall-Petch (locking) coefficient, and $d$ the average grain diameter.
Physically, why do grain boundaries strengthen a polycrystal (basis of Hall-Petch)?
Grain boundaries act as barriers to dislocation motion because slip planes are discontinuous across them and misorientation requires nucleating slip in the neighboring grain. Dislocations pile up at boundaries; finer grains mean shorter pile-ups, higher stress concentration needed to propagate slip, hence higher yield strength.
What is the inverse Hall-Petch effect?
At very small grain sizes (typically below about $10$–$20$ nm), the strength decreases with further grain refinement instead of increasing, because deformation shifts from dislocation-mediated slip to grain-boundary sliding/diffusion (Coble-type) accommodation.
How does solid-solution strengthening scale with solute concentration, and what factors control its magnitude?
Strengthening typically scales as $\Delta\tau\propto G\,\varepsilon^{3/2}c^{1/2}$ (often written $\Delta\sigma\propto c^{1/2}$ or $c^{2/3}$). It depends on the size misfit $\varepsilon_b=\frac{1}{b}\frac{db}{dc}$ and the modulus misfit $\varepsilon_G=\frac{1}{G}\frac{dG}{dc}$.
Compare substitutional and interstitial solid-solution strengthening and their lattice strain fields.
Substitutional solutes produce mainly symmetric (spherical) distortions and interact chiefly with edge dislocations. Interstitial solutes (e.g., C, N in BCC Fe) produce asymmetric (tetragonal) distortions that interact with both edge and screw dislocations, making them far more potent strengtheners per atom.
Distinguish the two mechanisms by which dislocations overcome precipitates: shearing vs. Orowan looping.
Shearing (cutting) occurs for small, coherent, soft precipitates; strength increases with particle size. Orowan looping (bypassing) occurs for large, hard, or incoherent precipitates; the dislocation bows between and leaves loops, and strength decreases with increasing particle spacing. Peak strength occurs at the crossover (critical) radius.
Write the Orowan equation for the increase in shear stress due to bypassing of impenetrable particles spaced $L$ apart.
$\Delta\tau=\frac{Gb}{L}$ (more completely $\Delta\tau\approx\frac{Gb}{L-2r}$, where $L$ is the inter-particle spacing and $r$ the particle radius). Strength is inversely proportional to particle spacing.
How does precipitation strengthening differ from dispersion strengthening in terms of particle origin and stability?
Precipitation (age) hardening uses coherent/semi-coherent particles formed by solid-state precipitation from a supersaturated solution; they are thermodynamically metastable and can over-age/coarsen and dissolve at high temperature. Dispersion strengthening uses incoherent, thermally stable, insoluble particles (e.g., oxides like $\ce{Y2O3}$, $\ce{ThO2}$) added externally, retaining strength at high temperature.
Sketch the sequence of GP zones and precipitates in age-hardening of Al-Cu, and where peak hardness occurs.
Sequence: supersaturated solid solution $\rightarrow$ GP zones $\rightarrow$ $\theta''$ (coherent) $\rightarrow$ $\theta'$ (semi-coherent) $\rightarrow$ $\theta$ ($\ce{CuAl2}$, incoherent equilibrium). Peak hardness corresponds to the $\theta''/\theta'$ stage; over-aging (coarse incoherent $\theta$) reduces hardness.
State Griffith's energy criterion for brittle fracture and the resulting critical fracture stress.
Crack propagates when the elastic strain energy released equals the surface energy created. Critical stress: $\sigma_f=\sqrt{\frac{2E\gamma_s}{\pi a}}$ (plane stress), where $\gamma_s$ is the surface energy and $2a$ the central crack length.
How did Orowan modify the Griffith criterion for crystalline (metallic) materials?
Orowan added the plastic work term $\gamma_p$ to the surface energy: $\sigma_f=\sqrt{\frac{2E(\gamma_s+\gamma_p)}{\pi a}}$. Since $\gamma_p\gg\gamma_s$ for metals, this accounts for the much higher fracture stress of metals than predicted by pure surface energy.
Define the stress intensity factor $K$ and write its general form for mode I.
$K$ characterizes the magnitude of the elastic crack-tip stress field. Mode I: $K_I=Y\sigma\sqrt{\pi a}$, where $Y$ is a dimensionless geometry factor, $\sigma$ the applied stress, and $a$ the crack length. The crack-tip stresses scale as $\sigma_{ij}=\frac{K_I}{\sqrt{2\pi r}}f_{ij}(\theta)$.
Name the three loading modes in fracture mechanics.
Mode I: opening (tensile, crack faces pulled apart normal to crack plane). Mode II: in-plane sliding (shear normal to crack front). Mode III: out-of-plane tearing (anti-plane shear parallel to crack front).
Relate the strain energy release rate $G$ to the stress intensity factor $K_I$.
$G=\frac{K_I^{2}}{E}$ (plane stress) or $G=\frac{K_I^{2}(1-\nu^{2})}{E}$ (plane strain). $G$ is the elastic energy released per unit area of crack advance; fracture occurs when $G\geq G_c$.
Define fracture toughness $K_{IC}$ and the condition for it to be a valid material property.
$K_{IC}$ is the critical mode-I stress intensity factor (plane-strain fracture toughness) at which a crack propagates unstably: $K_{IC}=Y\sigma_f\sqrt{\pi a}$. Validity (plane strain, small-scale yielding) requires thickness and crack length $\geq2.5\left(\frac{K_{IC}}{\sigma_y}\right)^{2}$.
Estimate the plastic zone size at a crack tip under plane stress and plane strain.
Plane stress: $r_p=\frac{1}{2\pi}\left(\frac{K_I}{\sigma_y}\right)^{2}$. Plane strain: $r_p=\frac{1}{6\pi}\left(\frac{K_I}{\sigma_y}\right)^{2}$ (about one-third the plane-stress value due to triaxial constraint).
How does specimen thickness affect fracture toughness and the fracture appearance?
Thin sections are in plane stress, giving higher (thickness-dependent) toughness with slant/shear (45 degree) fracture lips. As thickness increases, constraint rises toward plane strain, toughness falls to the minimum plateau value $K_{IC}$ and fracture becomes flat/square. $K_{IC}$ is the conservative, thickness-independent property.
In fractography, what surface features distinguish ductile, brittle cleavage, and fatigue fractures?
Ductile (microvoid coalescence): dimples (cup-and-cone macroscopically). Brittle transgranular cleavage: flat facets with river patterns and feathers; intergranular brittle fracture shows faceted grain boundaries. Fatigue: macroscopic beach marks and microscopic striations, each striation representing one load cycle.
What is the difference between beach marks and striations in fatigue fractography?
Beach marks (clamshell marks) are macroscopic features visible to the eye, formed by changes in loading or environment over many cycles (marking crack-front rest positions). Striations are microscopic ridges (seen under SEM/TEM), each typically corresponding to a single load cycle of crack advance.
What does the ductile-to-brittle transition temperature (DBTT) represent, and which crystal structures show it?
The DBTT is the temperature below which a material's fracture mode changes from ductile (high energy) to brittle (low energy), measured by a sharp drop in impact toughness (e.g., Charpy). BCC metals (e.g., ferritic steels) and HCP metals show a pronounced DBTT; FCC metals generally do not and remain ductile to very low temperatures.
List factors that raise the ductile-to-brittle transition temperature (make a steel more brittle).
Higher strain rate, triaxial stress state / notches (constraint), larger grain size, increased carbon content, interstitials (P, N), and irradiation embrittlement all raise the DBTT. Grain refinement, Mn, and Ni lower it (improve low-temperature toughness).
Why do BCC metals exhibit a DBTT while FCC metals usually do not?
In BCC metals the yield/flow stress is strongly temperature-dependent (high Peierls stress for screw dislocations); at low temperature yield stress exceeds the cleavage stress, so fracture occurs before yielding. FCC metals have low, temperature-insensitive Peierls stress and abundant slip systems, so they yield before cleaving at all temperatures.
Describe a hysteresis loop in cyclic stress-strain behaviour and what its area represents.
Under cyclic loading, plotting stress versus strain gives a closed hysteresis loop; its width is the plastic strain range $\Delta\varepsilon_p$ and its height the stress range $\Delta\sigma$. The enclosed area equals the plastic strain energy dissipated per cycle.
Distinguish cyclic hardening from cyclic softening and how to predict which occurs.
Cyclic hardening: stress amplitude rises with cycles (typical of initially soft/annealed metals). Cyclic softening: stress amplitude falls (typical of cold-worked/hard metals). Rule of thumb: if $\frac{\sigma_{UTS}}{\sigma_y}>1.4$ the material cyclically hardens; if $<1.2$ it cyclically softens.
State the Coffin-Manson and Basquin relations describing strain-life fatigue.
Coffin-Manson (plastic, low-cycle): $\frac{\Delta\varepsilon_p}{2}=\varepsilon_f'(2N_f)^{c}$. Basquin (elastic, high-cycle): $\frac{\Delta\varepsilon_e}{2}=\frac{\sigma_f'}{E}(2N_f)^{b}$. Total: $\frac{\Delta\varepsilon}{2}=\frac{\sigma_f'}{E}(2N_f)^{b}+\varepsilon_f'(2N_f)^{c}$.
GATE Metallurgical Engineering Mechanical Metallurgy FAQ
What is in the GATE Metallurgical Engineering Mechanical Metallurgy syllabus?
GATE Metallurgical Engineering Mechanical Metallurgy is split into 6 chapters: Strain Tensor and Stress Tensor, Dislocation Theory, Strengthening Mechanisms, Fracture Behaviour, Fatigue and Mechanisms of High Temperature Deformation and Failure. Together they contain 21 topics and 2 sub-topics.
How many chapters are in GATE Metallurgical Engineering Mechanical Metallurgy?
6 chapters, containing 21 topics and 2 sub-topics. The largest chapter is Dislocation Theory, with 5 topics.
Are there flashcards for GATE Metallurgical Engineering Mechanical Metallurgy?
Yes — a 49-card Mechanical Metallurgy deck, printed in full in the flashcards section of this page and available in the Examius app with spaced repetition. The free plan covers one exam of your choice, including its flashcards; Examius Pro, a one-time purchase, unlocks every other exam and deck.