GATE MT · Chapter-wise

Solid Mechanics & Mechanical Properties

Stress, strain, fracture, fatigue, creep · PYQs 1990–2026 with answers & solutions

Solid Mechanics & Mechanical Properties

Stress, strain, fracture, fatigue, creep

221 questions

GATE 2026 · Q16

In a face-centered cubic metal, Shockley partial is:

MCQ1M

Perfect and mobile dislocation

Perfect and immobile dislocation

Imperfect and immobile dislocation

Imperfect and mobile dislocation

Solution

Shockley partial: b = a/6⟨112⟩ — NOT a lattice vector (imperfect), glissile on {111} plane (mobile). Answer: D

GATE 2026 · Q17

Deformation mechanism map is used for determining which property?

MCQ1M

Fatigue strength

Creep rate

Tensile strength

Impact toughness

Solution

Ashby deformation mechanism maps show dominant creep mechanisms and strain rates vs. T/T_m. Answer: B

GATE 2026 · Q18

Which dislocation dissociation reaction is feasible in FCC metals?

MCQ1M

\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{2}1]+\frac{a}{6}[\bar{1}\bar{1}2]\)

\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[112]+\frac{a}{6}[21\bar{1}]\)

\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{1}2]+\frac{a}{6}[\bar{1}\bar{2}\bar{1}]\)

\(\frac{a}{2}[0\bar{1}1]\rightarrow\frac{a}{6}[1\bar{2}1]+\frac{a}{6}[2\bar{1}\bar{1}]\)

Solution

Check A: vectors add to \(\frac{a}{6}[0\bar{3}3]=\frac{a}{2}[0\bar{1}1]\) \(\checkmark\). Energy decreases: \(a^2/2 > a^2/6+a^2/6\) \(\checkmark\). Answer: A

GATE 2026 · Q37

Match: P) Paris Law, Q) Schmid Factor, R) Larson-Miller Parameter, S) Portevin-Le Chatelier Effect
with 1) Creep, 2) Fatigue, 3) Dynamic Strain Aging, 4) Critical Resolved Shear Stress

MCQ2M

P-3, Q-1, R-2, S-4

P-2, Q-4, R-1, S-3

P-2, Q-4, R-3, S-1

P-1, Q-3, R-2, S-4

Solution

Paris Law:Fatigue(2), Schmid:CRSS(4), Larson-Miller:Creep(1), PLC:DSA(3). Answer: B

GATE 2025 · Q60

Nabarro–Herring creep in polycrystalline Ni. \(\dot{\varepsilon}=10^{-8}\) s\(^{-1}\) at \(\sigma=10\) MPa. What stress gives \(\dot{\varepsilon}=10^{-9}\) s\(^{-1}\)? (integer MPa)

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Solution

N–H creep: \(n=1\Rightarrow\dot{\varepsilon}\propto\sigma\). \(\sigma_2=\sigma_1\times(\dot{\varepsilon}_2/\dot{\varepsilon}_1)=10\times10^{-1}=\mathbf{1}\) MPa.

GATE 2025 · Q62

A cylindrical specimen is plastically tensioned to 10% uniform elongation. Final gage-section area = 20 mm\(^2\). Initial gage-section area is ______ mm\(^2\) (integer).

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Solution

Volume conserved: \(A_iL_i=A_fL_f\). \(L_f=1.1L_i\). \(A_i=A_f\times(L_f/L_i)=20\times1.1=\mathbf{22}\) mm\(^2\).

GATE 2025 · Q64

Al alloy billet (300 mm dia.) hot extruded to 75 mm dia. at \(\dot{\varepsilon}=10\) s\(^{-1}\). Flow stress \(\sigma=10(\dot{\varepsilon})^{0.3}\) MPa. Ideal plastic work of deformation per unit volume is ______ × 10\(^6\) J m\(^{-3}\) (1 decimal place).

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Solution

\(\sigma=10\times10^{0.3}=19.95\) MPa. True strain: \(\varepsilon=2\ln(300/75)=2\ln4=2.773\). Ideal work \(=\sigma\varepsilon=19.95\times2.773=\mathbf{55.3}\) MJ m\(^{-3}\).

GATE 2024 · Q16

Match the laws with corresponding material properties:
Column I: (P) Hooke's law, (Q) Fick's law, (R) Fourier's law, (S) Darcy's law
Column II: (1) Thermal conductivity, (2) Young's modulus, (3) Permeability, (4) Diffusivity

MCQ1M

P–2, Q–1, R–4, S–3

P–4, Q–3, R–1, S–2

P–2, Q–4, R–1, S–3

P–4, Q–3, R–2, S–1

Solution

Hooke's law → Young's modulus (2), Fick's law → Diffusivity (4), Fourier's law → Thermal conductivity (1), Darcy's law → Permeability (3). Answer: C

GATE 2024 · Q25

Match the concepts (Column I) with phenomena (Column II):
P. Peierls-Nabarro stress Q. Cottrell's atmosphere R. Paris law S. Considère's criterion
1. Yield point phenomenon 2. Fatigue 3. Dislocation glide 4. Onset of necking

MCQ1M

P–1, Q–2, R–3, S–4

P–4, Q–1, R–2, S–3

P–3, Q–1, R–2, S–4

P–3, Q–4, R–2, S–1

Solution

FCC slip systems: {111} planes (4) × <110> directions (3 per plane) = 12 total. Answer: C

GATE 2024 · Q34

A single crystal: slip plane normal at 60° to tensile axis; slip direction at 45° to tensile axis; critical resolved shear stress = 2 MPa. The tensile stress at which plastic deformation commences is ________ MPa. (Round off to one decimal place)

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Solution

Schmid factor: \(m=\cos\phi\cos\lambda=\cos 60°\times\cos 45°=0.5\times0.707=0.354\). \(\sigma_{yield}=\tau_{CRSS}/m=2/0.354\approx5.66\) MPa. Answer: 5.5–5.8 MPa

GATE 2024 · Q45

Which one of the following graphs represents Griffith's criterion for the growth of a crack in a brittle isotropic infinitely large plate with a center crack? (\(\Delta SE\) = strain energy released; \(\Gamma_s\) = total surface energy; \(a_c\) = critical crack length)
GATE 2024 Q45 figure

MCQ2M

Option A (see figure)

Option B (see figure)

Option C (see figure)

Option D (see figure)

Solution

Griffith's criterion for crack growth: energy vs crack length curves showing characteristic minimum. Answer: D

GATE 2024 · Q48

A creep test of a pure polycrystalline metal is performed in tension and the creep strain rate decreases during the primary stage. The creep mechanism is dislocation-climb-controlled. The observed decrease in creep strain rate is/are due to

MSQ2M

an increase in dislocation density.

grain growth.

a decrease in the dislocation density.

an increase in the cross-sectional area of the sample.

Solution

During primary creep with dislocation-climb mechanism, dislocations multiply and pile up, increasing dislocation density. This raises the back-stress on gliding dislocations and reduces the creep strain rate. Answer: A

GATE 2024 · Q54

A steel bar under fatigue loading has tensile mean stress. Ultimate tensile strength = 1000 MPa, fatigue limit under fully reversed loading = 250 MPa. Using the Goodman relationship, the fatigue limit for a mean stress of 100 MPa is ________ MPa. (Round off to the nearest integer)

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Solution

Goodman relation: σ_a/250 + 100/1000 = 1, giving σ_a = 225 MPa. Answer: 225

GATE 2024 · Q63

A large rectangular component undergoes fully-reversed cyclic loading with fatigue crack from the outer surface. Stress amplitude \(\sigma_A\) = 100 MPa, \(K_{1C}\) = 50 MPa·m½, geometric factor \(\alpha\) = 1.12. The crack length at which the component will fail catastrophically is ________ mm. (Round off to one decimal place)

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Solution

Griffith: K_IC = ασ_A√(πa). Solving: a = [K_IC/(ασ_A)]²/π = 63.4 mm. Answer: 62.5-64.5

GATE 2023 · Q19

Diamond has low

MCQ1M

electrical conductivity

modulus of elasticity

hardness

thermal conductivity

Solution

Diamond has high hardness, high elastic modulus, and high thermal conductivity, but low electrical conductivity. Answer: A

GATE 2023 · Q22

The mechanism of creep for a single crystal as depicted in the schematic is
GATE 2023 Q22 figure

MCQ1M

Nabarro-Herring creep

Grain boundary sliding

Dislocation creep

Coble creep

Solution

The shown single-crystal creep schematic corresponds to vacancy diffusion through the lattice, i.e. Nabarro-Herring creep. Answer: A

GATE 2023 · Q27

Which of the following is/are responsible for reducing the high cycle fatigue life
of a component?

MSQ1M

increasing the mean stress at constant amplitude

increasing the surface roughness

employing shot peening

absence of sharp corners in the component

Solution

Higher mean stress and rougher surfaces reduce high-cycle fatigue life; shot peening and avoiding sharp corners improve it. Correct options: A, B

GATE 2023 · Q63

Strain hardening behavior of an alloy is given by sigma = 1100 epsilon0.3 , where sigma and epsilon
are true stress and true strain, respectively. The alloy is cold drawn to an unknown
amount of strain, followed by tensile testing. If the tensile test showed
10 % reduction in area at maximum load, then the unknown amount of strain from
prior cold work is ____________ (round off to 2 decimal places).

NAT2M

Solution

At maximum load Considere's criterion gives remaining uniform true strain equal to n = 0.3. Convert 10% area reduction to true strain and subtract from n to obtain the prior cold-work strain. Answer range: 0.18 to 0.21

GATE 2023 · Q64

A specimen containing maximum initial surface crack of size 1.5 mm is subjected to cyclic loading with sigma_max = 300 MPa and sigma_min = 0 MPa. Assuming specimen geometric factor of 1, and referring to the given figure, the crack growth rate in micrometre cycle^-1 is ____________ (round off to nearest integer).
Given: N = number of cycles; a = crack length; R = stress ratio; Delta K = stress intensity range.
GATE 2023 Q64 figure

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Solution

Compute Delta K = Y Delta sigma sqrt(pi a) with R = 0 and read the crack-growth rate from the supplied plot. Answer range: 7 to 15 micrometre/cycle

GATE 2022 · Q26

For a material that undergoes strain hardening, necking instability occurs during
tensile testing when ___________
Given: \(\sigma\) = true stress and \(\epsilon\) = true strain.

MCQ1M

\(\dfrac{d\sigma}{d\epsilon} = 0\)

\(\dfrac{d\sigma}{d\epsilon} = \epsilon\)

\(\dfrac{d\sigma}{d\epsilon} = \sigma\)

\(\dfrac{d\sigma}{d\epsilon} = \infty\)

Solution

Considere criterion for necking in true stress-true strain form is d sigma/d epsilon = sigma. Answer: C

GATE 2022 · Q29

With reference to the stress intensity factor, find the correct match of
nomenclature (Column A) with the mode of deformation applied to the crack
(Column B).
Column A Column B
(P) Mode I (X) Forward shear mode
(Q) Mode II (Y) Parallel shear mode
(R) Mode III (Z) Crack opening mode

MCQ1M

P - Z, Q - Y, R - X

P - Z, Q - X, R - Y

P - Y, Q - X, R - Z

P - Y, Q - Z, R - X

Solution

Mode I is opening, Mode II forward/in-plane shear, and Mode III parallel/tearing shear. Answer: B

GATE 2022 · Q41

Match the phenomena (Column I) with the descriptions (Column II)
Column I Column II
(P) Cottrell atmosphere (1) Decrease in yield stress when loading
direction is reversed
(Q) Suzuki interaction (2) Stress assisted diffusion of vacancies
resulting in plastic deformation in a
polycrystalline material
(R) Bauschinger effect (3) Lü ders bands
(S) Nabarro-Herring creep (4) Segregation of solutes to the stacking fault

MCQ2M

P - 1, Q - 2, R - 3, S - 4

P - 1, Q - 2, R - 4, S - 3

P - 3, Q - 4, R - 1, S - 2

P - 3, Q - 1, R - 4, S - 2

Solution

Cottrell atmosphere causes Luders bands; Suzuki interaction is solute segregation to stacking faults; Bauschinger effect is reverse loading yield drop; Nabarro-Herring is vacancy diffusion creep. Answer: C

GATE 2022 · Q46

From high temperature tensile testing, the flow stress (measured at the same
value of strain) of an alloy was found to be 50 MPa at a strain rate of 0.1 s⁻¹ and 70 MPa at a strain rate of 10 s⁻¹. The strain rate sensitivity parameter is
_______ (round off to 3 decimal places).

NAT2M

Solution

Use sigma = C strain_rate^m, so m = ln(70/50)/ln(10/0.1) ≈ 0.073. Answer range: 0.069 to 0.075

GATE 2022 · Q52

High cycle fatigue data for an alloy at various alternating stresses is given in the figure/table. A specimen is subjected sequentially to: first 5000 cycles at sigma_a = 400 MPa, then 25000 cycles at sigma_a = 300 MPa, and finally cycles at sigma_a = 500 MPa. Assuming Miner's law is obeyed, the number of cycles to failure at the final applied stress of 500 MPa is ___________.
GATE 2022 Q52 figure

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Solution

Miner's damage: 5000/10000 + 25000/100000 = 0.75, leaving 0.25 life at 500 MPa where Nf=1000. Remaining cycles = 250. Answer: 250

GATE 2021 · Q17

For uniaxial tensile stress-strain behaviour of polycrystalline aluminium,
which one of the following statements is FALSE?
: ; : ; do
) At the ultimate tensile stress point on the true stress - strain curve, aE =0
@

MCQ1M

Option A (see MT2021.pdf)

Option B (see MT2021.pdf)

| Resilience is the area under the elastic region of the engineering stress - straincurve.

| Maximum true stress does not correspond to the maximum load.los |

Solution

At UTS on the TRUE stress-strain curve, dσ/dε = σ (Considère criterion), NOT dσ/dε = 0. Option B claims dσ/dε = 0 on the true curve, which is false. Answer: B

GATE 2021 · Q18

Which one of the following is FALSE for creep deformation?
The minimum creep rate is obtained in the primary stage (stage I).
Creep resistance decreases with decrease in grain size.
Coble creep occurs via grain boundary diffusion.
Nabarro-Herring creep occurs via lattice diffusion.
Organising Institute - IIT Bombay

MCQ1M

Option A (see MT2021.pdf)

Option B (see MT2021.pdf)

Option C (see MT2021.pdf)

Option D (see MT2021.pdf)

Solution

Minimum creep rate (steady-state) occurs in Stage II (secondary creep), NOT Stage I (primary creep). The statement that minimum creep rate is in stage I is FALSE. Answer: A

GATE 2021 · Q28

A body is subjected to a state of stress given by the following stress tensor:
50 0 0
0 200 O | MPa.
0 0 100
If yielding is predicted by the Tresca Criterion, the uniaxial tensile yield
stress (in MPa) of the body should be less than or equal to:
(round off to nearest integer).

NAT1M

Solution

Tresca criterion: τ_max = (σ_max − σ_min)/2 = (200 − 50)/2 = 75 MPa. Yielding occurs when τ_max = σ_y/2, so σ_y = 150 MPa. Answer: 150 to 150

GATE 2021 · Q58

A metal plate is in a state of plane strain (¢,, = 0) with o,, = o,,#0 and
Ty = Tz = Tyz = 0. If the Poisson's ratio is 0.3, the ratio, o,,/0,, is
(round off to 1 decimal place).
(e748 <a 2
Organising Institute - IT Bombay
80

NAT2M

Solution

Plane strain: ε_zz=0. From Hooke's law: σ_zz = ν(σ_xx+σ_yy). With σ_xx=σ_yy: σ_zz/σ_xx = 2ν = 2×0.3 = 0.6. Answer: 0.6 to 0.6

GATE 2021 · Q59

An infinite metal plate has a central through-thickness crack of length a
mm. The maximum applied stress (in MPa) that the plate can sustain in
mode I is: (round off to nearest integer).
Assume: Linear elastic fracture mechanics is valid
Given: Fracture toughness, K;c= 20 MPa m"?

NAT2M

Solution

LEFM: K_Ic = σ√(πa) where a is half-crack length. With K_Ic=20 MPa√m and crack half-length a≈0.04 m: σ = 20/√(π×0.04) ≈ 100 MPa. Answer: 98 to 102

GATE 2020 · Q12

The number of independent elastic constants of an isotropic material is:

MCQ1M

Solution

An isotropic material has the same properties in all directions; only 2 independent elastic constants are needed (e.g., E and ν, or λ and μ). Answer: B

GATE 2020 · Q13

A slip system consists of a slip plane and a slip direction. Which one of the following is NOT a valid slip system in a FCC copper crystal?

MCQ1M

\((111)[\bar{1}\bar{1}0]\)

\((\bar{1}11)[011]\)

\((1\bar{1}1)[10\bar{1}]\)

\((11\bar{1})[101]\)

Solution

In FCC, valid slip systems are {111}⟨110⟩. For option B: (1̄11)[011] — check if [011] lies in (1̄11): dot product = 0×(−1)+1×1+1×1 = 2 ≠ 0, so [011] is NOT in (1̄11) — invalid slip system. Answer: B

GATE 2020 · Q18

For a material to exhibit superplasticity, one of the requirements is:

MCQ1M

Coarse-grained microstructure

High strain-rate sensitivity

Low strain-hardening exponent

High modulus of elasticity

Solution

Superplasticity requires high strain-rate sensitivity (m ≈ 0.5), fine and stable grain size, and deformation near 0.5T_m. High strain-rate sensitivity prevents necking and enables large elongations. Answer: B

GATE 2020 · Q26

The indenter used in Rockwell hardness measurements on C scale is

MCQ1M

diamond cone

10 mm steel ball

diamond pyramid

1/16-in. steel ball

Solution

Rockwell C scale uses a 120° diamond cone (Brale indenter) with a 150 kgf load, suitable for hard materials. Answer: A

GATE 2020 · Q47

Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].
Assertion [a]: Low-alloy steels used for medium-temperature creep resistance often have additions of strong carbide-forming elements.
Reason [r]: During creep deformation, the particles with higher misfit with the matrix, lose coherency.

MCQ2M

Both [a] and [r] are true and [r] is the correct reason for [a].

Both [a] and [r] are true but [r] is not the correct reason for [a].

Both [a] and [r] are false.

[a] is true but [r] is false.

Solution

[a] is true: carbide formers (Mo, Cr, V) pin dislocations during creep. [r] is also true but is not the reason — carbides resist coarsening at high temperature, which is the actual mechanism. Answer: B

GATE 2020 · Q61

The steady state creep rate of a material increases by a factor of 20 when the temperature is increased from 890 K to 980 K. The creep rate at a temperature of __________ K (round off to the nearest integer) will be 5 times the creep rate at 890 K.

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Solution

ε̇ ∝ exp(−Q/RT). From 890→980 K, rate increases 20×: Q/R = ln20/(1/890 − 1/980) ≈ 327,500 K. For 5× increase: 1/T = 1/890 − ln5/327500 ≈ 1/936 K. Answer: 933 to 939

GATE 2020 · Q62

Crack growth is being continuously measured in a test specimen subjected to constant amplitude cyclic stress with a mean stress of zero. The crack growth rate is related to the stress intensity range, ΔK as \(\dfrac{da}{dN} \propto (\Delta K)^3\), where \(a\) is the crack length and \(N\) is the number of cycles. When the crack length increases by a factor of two, the crack growth rate will increase by a factor of __________ (round off to one decimal place).

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Solution

ΔK ∝ σ√(πa), so ΔK scales as √a. da/dN ∝ (ΔK)³ ∝ a^(3/2). If a doubles: rate increases by 2^(3/2) = 2√2 ≈ 2.83. Answer: 2.6 to 3.0

GATE 2019 · Q26

During low strain rate (≤ 0.1 per second) deformation of a metal at room temperature, the one that deforms by twinning mode is _______________.

MCQ1M

Solution

Mg (HCP) has limited slip systems and deforms primarily by twinning at low strain rates. Answer: B

GATE 2019 · Q27

In a tensile creep test of a metal, Nabarro-Herring mechanism is favored over Coble mechanism for _________________.

MCQ1M

larger grain size and lower temperature

smaller grain size and higher temperature

larger grain size and higher temperature

smaller grain size and lower temperature

Solution

Nabarro-Herring (lattice diffusion, \(\propto d^{-2}\)) dominates at larger grain size and higher temperature; Coble (grain boundary diffusion) dominates at smaller grain size and lower temperature. Answer: C

GATE 2019 · Q28

Beach marks are commonly observed on the fractured surfaces of metals after a ________.

MCQ1M

Creep test

Fatigue test

Impact test

Compression test

Solution

Beach marks (macroscopic striations) are a classic fracture surface feature of fatigue failure — they mark crack front positions during intermittent crack growth. Answer: B

GATE 2019 · Q29

The length of internal cracks in two samples of the same glass is \(c_1 = 0.5\) mm and \(c_2 = 2\) mm. The ratio \(\left(\dfrac{\sigma_1}{\sigma_2}\right)\) of the fracture strength of the two samples is ________________.

MCQ1M

0.5

1.0

2.0

4.0

Solution

Griffith: \(\sigma \propto c^{-1/2}\). So \(\dfrac{\sigma_1}{\sigma_2} = \sqrt{\dfrac{c_2}{c_1}} = \sqrt{\dfrac{2}{0.5}} = \sqrt{4} = 2\). Answer: C

GATE 2019 · Q42

An aluminium single crystal is loaded in tension along \([1\bar{1}0]\) axis. Among the following slip systems, the one that will be activated first is__________________.

MCQ2M

\((1\bar{1}1)[0\bar{1}1]\)

\((\bar{1}11)[011]\)

\((\bar{1}\bar{1}1)[1\bar{1}0]\)

\((\bar{1}\bar{1}1)[101]\)

Solution

Highest Schmid factor for loading along \([1\bar{1}0]\) corresponds to slip system \((1\bar{1}1)[0\bar{1}1]\). Answer: A

GATE 2019 · Q59

A material made of alternating layers of metals A and B is loaded parallel to the layers (isostress condition). If the volume % of B is 25%, the elastic modulus (in GPa, rounded off to one decimal place) of the material is __________________.
Given: Elastic moduli of A and B are 200 GPa and 100 GPa respectively. GATE 2019 Q59 figure

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Solution

Loading is parallel to layers → isostrain (rule of mixtures): \(E_c = V_A E_A + V_B E_B = 0.75\times200 + 0.25\times100 = 150 + 25 = 175\) GPa. However the figure shows perpendicular loading (isostress): \(1/E_c = V_A/E_A + V_B/E_B = 0.75/200 + 0.25/100\). \(1/E_c = 0.00625\). \(E_c = 160\) GPa. Answer: 155.0 to 165.0

GATE 2019 · Q60

The S-N curve for a steel shows an endurance limit (stress amplitude) of 300 MPa. If the stress ratio \(\sigma_{min}/\sigma_{max} = -0.8\), the maximum stress (in MPa, rounded off to nearest integer) that the steel can withstand for infinite fatigue life is ____________. GATE 2019 Q60 figure

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Solution

Stress amplitude \(\sigma_a = \frac{\sigma_{max}-\sigma_{min}}{2}\). With \(\sigma_{min} = -0.8\sigma_{max}\): \(\sigma_a = \frac{\sigma_{max}(1+0.8)}{2} = 0.9\sigma_{max}\). Setting \(\sigma_a = 300\): \(\sigma_{max} = 300/0.9 \approx 333\) MPa. Answer: 330 to 335

GATE 2019 · Q61

True stress–true strain behavior of a metal is given by \(\sigma = 1750\,\varepsilon^{0.37}\) where \(\sigma\) is in MPa. The true stress at necking (in MPa, rounded off to nearest integer) is ___________________.

NAT2M

Solution

Necking occurs when \(\varepsilon = n = 0.37\). \(\sigma = 1750\times(0.37)^{0.37} = 1750\times0.693 \approx 1213\) MPa. Answer: 1160 to 1260

GATE 2018 · Q20

The c/a ratio of Zn (hcp) is 1.856. Slip at room temperature occurs most easily on which of
the following slip systems in Zn:

Note: In hcp metals, the ideal c/a ratio is 1.633.

MCQ1M

{11̅00} 〈112̅0〉

{11̅00} 〈0002〉

{0001} 〈112̅0〉

{101̅1̅} 〈112̅3〉

Solution

In Zn with c/a = 1.856 > ideal 1.633, the basal plane (0001) is the most closely packed and (0001)⟨112̄0⟩ is the primary slip system. Answer: C

GATE 2018 · Q37

Determine the correctness (or otherwise) of the following Assertion [A] and the Reason [R]

Assertion [A]: For a material exhibiting Coble creep, a reduction in grain size results in a
significant increase in creep rate
Reason [R]: Grain boundaries act as a barrier to motion of dislocations

MCQ2M

Both [A] and [R] are true and [R] is the correct reason for [A]

Both [A] and [R] are true, but [R] is not the correct reason for [A]

Both [A] and [R] are false

[A] is true but [R] is false

Solution

[A] is true: Coble creep (grain boundary diffusion) ∝ d⁻³, so finer grains greatly increase rate. [R] is also true, but the reason for [A] is grain boundary diffusion paths, not dislocation barriers. Answer: B

GATE 2018 · Q39

A glass fibre of 5 micron diameter is subjected to a tensile stress of 20 MPa. The surface
energy and elastic modulus of this material are 0.3 J·m⁻² and 70 GPa, respectively. Pick the
correct answer based on the information provided above:
Note: The glass fibre contains a population of flaws of different lengths.

MCQ2M

The fibre will undergo brittle fracture

The fibre will undergo plastic deformation, but not fracture

The fibre will undergo elastic deformation, but not fracture

The fibre will undergo buckling

Solution

Griffith critical flaw size: a_c = 2Eγ_s/(πσ²) = 2×70×10⁹×0.3/(π×(20×10⁶)²) ≈ 33 μm. The largest flaw in a 5 μm fibre is < 2.5 μm < 33 μm, so no fracture — only elastic deformation. Answer: C

GATE 2018 · Q51

A continuous and aligned carbon-fiber composite consists of 25 vol.% of fibers in an epoxy matrix. The Young’s modulus of fiber and matrix, respectively are \(E_f = 250\,\mathrm{GPa}\) and \(E_m = 2.5\,\mathrm{GPa}\).

If the composite is subjected to longitudinal loading (iso-strain condition and assuming elastic response), the fraction of load borne by the reinforcement is ________ (to two decimal places)

NAT2M

Solution

Iso-strain: load fraction on fibre = V_fE_f/(V_fE_f+V_mE_m) = 62.5/(62.5+1.875) ≈ 0.971. Answer: 0.96 to 0.98

GATE 2018 · Q56

A single crystal of aluminium is subjected to 10 MPa tensile stress along the [321]
crystallographic direction. The resolved shear stress on the (111̅) [101] slip system is
______ (in MPa to two decimal places)

NAT2M

Solution

Schmid factor τ = σ·cosλ·cosφ. [321]/(111̅)[101]: cosφ = 4/√42 ≈ 0.617, cosλ = 4/√28 ≈ 0.756. τ = 10×0.617×0.756 ≈ 4.66 MPa. Answer: 4.5 to 4.8

GATE 2018 · Q65

The ideal plastic work involved in extruding a cylindrical billet of length 100 mm, from an
initial diameter of 20 mm to a final diameter of 16 mm is __________ (in J to one decimal
place).

The flow stress in compression is 40 MPa, and remains constant throughout the process.

NAT2M

Solution

W = V·σ_f·ε. V = π×10²×100 = 31416 mm³. ε = 2ln(20/16) = 0.446. W = 31416×40×0.446 ≈ 561 J. Answer: 555.5 to 565.5

GATE 2017 · Q26

A brittle material (Young’s modulus = 60 GPa and surface energy = 0.5 J.m²) has a surface crack of length 2 μm. The fracture strength (in MPa) of this material is ___

NAT1M

Solution

Griffith: σ = √(2Eγ/(πa)) = √(2×60×10&sup9;×0.5/(π×10⁻&sup6;)) ≈ 138 MPa. Answer: 95.00 to 125.00

GATE 2017 · Q27

Both creep resistance and tensile strength of a metal can be enhanced by

MCQ1M

increase in the grain size

decrease in the grain size

addition of dispersoids

annealing

Solution

Dispersoids pin grain boundaries (creep resistance) and block dislocations (strength). Fine grains hurt creep. Answer: C

GATE 2017 · Q28

Stress required to operate a Frank-Read source of length L is approximately given by:

MCQ1M

Gb/L

Gb²/L

Gb²/L²

Gb²/2L²

Solution

Frank-Read source stress τ ≈ Gb/L. Answer: A

GATE 2017 · Q58

A steel component is subjected to fatigue: σ_max = 200 MPa, σ_min = 0. Initial crack length = 1 mm. Crack propagation: da/dN = 10⁻¹²(ΔK)³, where a in meters, ΔK in MPa√m. The crack length (in m) after one million cycles is ___

NAT2M

Solution

Integration of Paris law with ΔK = Δσ√(πa). After 10⁶ cycles, a ≈ 0.012 m. Answer: 0.009 to 0.015

GATE 2017 · Q61

A single crystal of an FCC metal is subjected to tensile stress along [110]. Which slip system will be activated?

MCQ2M

a/2[1̅10](111)

a/2[011](11̅1)

a/2[01̅1](1̅11)

a/2[110](1̅1̅1)

Solution

The slip system with highest Schmid factor is activated. For [110] loading, a/2[011](11̅1) has the highest resolved shear stress. Answer: B

GATE 2016 · Q29

Creep resistance decreases due to:

MCQ1M

Small grain size

Fine dispersoid size

Low stacking fault energy

High melting point

Solution

Small grains promote grain boundary sliding at high temperatures, reducing creep resistance. Answer: A

GATE 2016 · Q57

Fatigue S-N plot for an aluminium alloy. Piston rod subjected to (i) 1000 cycles at 420 MPa, then (ii) 1000 cycles at 300 MPa. Using Miner’s rule, the remaining fatigue life (cycles) at 250 MPa is ___
GATE 2016 Q57 figure

NAT2M

Solution

By Miner’s rule: Σ(n_i/N_i) = 1. From S-N curve, compute remaining life ≈ 2640 cycles. Answer range: 2630 to 2650

GATE 2016 · Q64

Liquid phase sintered SiC-Ni composite: γ_ss = 0.80 J/m², γ_sl = 0.43 J/m². SiC grain size = 20 μm. Average intergranular neck size (μm):

MCQ2M

3.03

4.28

9.16

18.32

Solution

cos(φ/2) = γ_ss/(2γ_sl) = 0.80/0.86 = 0.93. Neck size calculated from dihedral angle and grain size. Answer: C

GATE 2015 · Q35

In epoxies, creep resistance is enhanced by:

MCQ1M

increasing bulkiness of side groups

increasing cross-link density

addition of plasticizers

annealing

Solution

Increasing cross-link density restricts chain mobility, thereby enhancing creep resistance. Answer: B

Metallurgical Engineering — Q.36 to Q.65 (2 Marks Each)

GATE 2015 · Q61

Match the fracture surface features with fracture types: P. Striations Q. Dimples/microvoids R. Flat facets with river markings S. Jagged grain-like features — 1. Intergranular 2. Cleavage 3. Ductile 4. Fatigue

MCQ2M

P-1, Q-2, R-3, S-4

P-1, Q-3, R-2, S-4

P-4, Q-3, R-2, S-1

P-2, Q-1, R-4, S-3

Solution

Striations → Fatigue (4), Dimples → Ductile (3), River markings → Cleavage (2), Jagged → Intergranular (1). Answer: C

GATE 2015 · Q62

Match: P. Hall-Petch Q. Nabarro-Herring R. Lomer-Cottrell S. Frank-Read — 1. Dislocation reaction product 2. Diffusional creep 3. Dislocation source 4. Grain boundary strengthening

MCQ2M

P-1, Q-2, R-3, S-4

P-1, Q-2, R-4, S-3

P-4, Q-2, R-1, S-3

P-4, Q-1, R-2, S-3

Solution

Hall-Petch → Grain boundary strengthening (4), Nabarro-Herring → Diffusional creep (2), Lomer-Cottrell → Dislocation reaction product/lock (1), Frank-Read → Dislocation source (3). Answer: C

GATE 2015 · Q64

Match: P. Creep resistance Q. Modulus enhancement R. Superplasticity S. Increased strength — 1. Fine-grained two-phase alloy 2. Single crystal 3. Coherent precipitates 4. Glass fibres in epoxy

MCQ2M

P-2, Q-4, R-1, S-3

P-1, Q-2, R-3, S-4

P-2, Q-4, R-3, S-3

P-1, Q-4, R-2, S-3

Solution

Creep resistance → Single crystal (2), Modulus enhancement → Glass fibres in epoxy/composite (4), Superplasticity → Fine-grained two-phase (1), Increased strength → Coherent precipitates (3). Answer: A

GATE 2015 · Q65

The fracture stress of a brittle material is 300 MPa at a surface energy γ_s = 0.9 J/m². If γ_s is reduced to 0.1 J/m², the fracture stress (in MPa) is ___

NAT2M

Solution

Griffith criterion: σ ∝ √γ_s. σ = 300 × √(0.1/0.9) = 300 × 1/3 = 100 MPa. Answer range: 98 to 102

Load	Elongation
45 kN	1 mm
75 kN	4.4 mm

Load

Elongation

45 kN

1 mm

75 kN

4.4 mm