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GATE Mechanical: Strength of Materials, previous-year questions solved

Real Strength of Materials questions from the last four GATE Mechanical papers, each with a step-by-step solution that matches the official answer key. Every answer also says where the concept turns up in actual engineering work: FEA, pressure vessels, shafts, welds and springs. That's the part coaching notes skip, and the part interviews ask about.

Questions from official GATE ME papers (IIT Kharagpur 2022, IISc Bengaluru 2024, IIT Roorkee 2025, IIT Guwahati 2026); solutions by Skill-Lync. · Updated 15 Nov 2026 · 26 questions

GATE 2022 · ME Set-2 · Q.16MCQ · 1 markStress, strain and material testing

How is ultimate tensile strength defined from a stress-strain test?

Which one of the following is the definition of ultimate tensile strength (UTS) obtained from a stress-strain test on a metal specimen?
  1. (A)Stress value where the stress-strain curve transitions from elastic to plastic behavior
  2. (B)The maximum load attained divided by the original cross-sectional area
  3. (C)The maximum load attained divided by the corresponding instantaneous cross-sectional area
  4. (D)Stress where the specimen fractures

Answer: (B) Official key: B

The standard (engineering) stress-strain curve divides the load by the original cross-sectional area A0. UTS is the peak of that curve: the maximum load the specimen carries, divided by A0.

  • (A) describes the yield point, not UTS.
  • (C) is the true stress at maximum load (instantaneous area), which is higher than UTS because the specimen has already thinned.
  • (D) is the fracture stress; after necking the engineering stress falls, so fracture comes below UTS.
Where this shows up in industry

Material datasheets and the allowables you type into a CAE material card quote engineering UTS. If you build a true stress-strain curve for a nonlinear crash or forming model, you convert from this engineering value first.

GATE 2025 · ME · Q.58NAT · 2 marksStress, strain and material testing

How do you find the strain energy density from a bilinear stress-strain curve?

An isotropic brittle material is tested in the universal testing machine. The stress-strain diagram for the material shows a bi-linear elastic behavior as shown in the figure given below. The strain energy density is ________ MJ m−3 (rounded off to 2 decimal places).
Bilinear elastic stress-strain curve: straight line to 100 MPa at strain 0.01, then to 120 MPa at strain 0.0151001200.010.015σ (MPa)εO

Answer: 1.05 Official key: 1.00 to 1.10

Strain energy density is the area under the stress-strain curve up to the final strain.

  1. Triangle from 0 to ε = 0.01: ½ × 100 MPa × 0.01 = 0.50 MJ/m³.
  2. Trapezium from ε = 0.01 to 0.015: ½ × (100 + 120) MPa × 0.005 = 0.55 MJ/m³.
  3. Total = 0.50 + 0.55 = 1.05 MJ/m³ (1 MPa × strain = 1 MJ/m³).
Where this shows up in industry

The same area under the curve is the energy-absorption number crash engineers look at when they choose a material for a crush can or bumper beam. In FEA it is the internal energy the solver reports per unit volume.

GATE 2022 · ME Set-1 · Q.26MSQ · 1 markStress, strain and material testing

Does a steel rod's volume increase in tension, and does hydrostatic pressure cause shear stress?

Assuming the material considered in each statement is homogeneous, isotropic, linear elastic, and the deformations are in the elastic range, which one or more of the following statement(s) is/are TRUE?
  1. (A)A body subjected to hydrostatic pressure has no shear stress.
  2. (B)If a long solid steel rod is subjected to tensile load, then its volume increases.
  3. (C)Maximum shear stress theory is suitable for failure analysis of brittle materials.
  4. (D)If a portion of a beam has zero shear force, then the corresponding portion of the elastic curve of the beam is always straight.

Answer: (A) and (B) Official key: A, B

  • (A) True. Under hydrostatic pressure all three normal stresses are equal, so every plane is a principal plane and the shear stress on any plane is zero (the Mohr's circle shrinks to a point).
  • (B) True. Volumetric strain in uniaxial tension is εv = ε(1 − 2ν). For steel ν ≈ 0.3 < 0.5, so εv > 0 and the volume increases.
  • (C) False. Maximum shear stress (Tresca) theory is for ductile materials; brittle materials are checked with maximum normal stress (Rankine) or Mohr-Coulomb.
  • (D) False. Zero shear force means a constant bending moment, and curvature = M/EI. The elastic curve is straight only if that constant moment is also zero.
Where this shows up in industry

Statement (C) is a real design choice: picking the wrong failure criterion for cast iron or glass-filled plastic parts gives unsafe results. Statement (D) is why a beam in pure bending still curves even though its shear force diagram is empty.

GATE 2025 · ME · Q.28MCQ · 1 markMohr's circle and stress transformation

What does Mohr's circle look like for a plate under hydrostatic pressure?

A metallic square plate is subjected to a uniform hydrostatic pressure (P). Choose the correct Mohr’s circle representing the state of stress at any point in the plate from the options given below.
For the Mohr’s circle, normal stress is positive towards right and shear stress is positive in the upward direction.
  1. (A)Mohr's circle optionστOP
  2. (B)Mohr's circle optionστO−P
  3. (C)Mohr's circle optionστO−PP
  4. (D)Mohr's circle optionστO−P0

Answer: (B) Official key: B

Hydrostatic pressure is compressive, so σx = σy = −P and τxy = 0.

Mohr's circle has centre (σx + σy)/2 = −P and radius √[((σx − σy)/2)² + τ²] = 0. A circle of zero radius is a single point at σ = −P on the negative normal-stress axis: option (B).

(A) would be hydrostatic tension. (C) is pure shear. (D) is uniaxial compression.

Where this shows up in industry

Deep-sea housings and parts in hydraulic presses see near-hydrostatic stress. Because there's no shear, von Mises stress is zero there, which is why FEA shows a pressurised solid block as 'safe' even at very high pressure.

GATE 2022 · ME Set-2 · Q.38MCQ · 2 marksMohr's circle and stress transformation

How do you add two uniaxial stresses applied in different directions?

A linear elastic structure under plane stress condition is subjected to two sets of loading, I and II. The resulting states of stress at a point corresponding to these two loadings are as shown in the figure below. If these two sets of loading are applied simultaneously, then the net normal component of stress σxx is ________.
Loading I: uniaxial stress sigma along x. Loading II: uniaxial stress sigma along an axis at 45 degrees to xLoading - IσσxyLoading - IIσσxx'y'45°
  1. (A)3σ/2
  2. (B)σ(1 + 1/√2)
  3. (C)σ/2
  4. (D)σ(1 − 1/√2)

Answer: (A) 3σ/2 Official key: A

The structure is linear elastic, so stresses from the two loadings can be added once they're in the same x-y axes.

  1. Loading I: σxx = σ.
  2. Loading II is uniaxial σ along an axis at 45° to x. Transforming to x: σxx = σ cos²45° = σ/2.
  3. Total: σxx = σ + σ/2 = 3σ/2.
Where this shows up in industry

This is exactly what an FEA post-processor does when it combines load cases: it transforms each result to a common coordinate system and adds them. It's only valid while the model stays linear.

GATE 2026 · ME · Q.52NAT · 2 marksMohr's circle and stress transformation

What are the normal and shear stresses on a 45° weld in a plate under biaxial stress?

A welded square plate of 1 m × 1 m is subjected to biaxial stress of magnitude 6.5 MPa and 25 MPa, as shown in the figure below. The ratio of the normal stress acting in the perpendicular direction of the weld to the shear stress of the weld is ________ (rounded off to 2 decimal places).
Square plate with a weld along the diagonal; 25 MPa tension top and bottom, 6.5 MPa compression on the sidesWeld25 MPa25 MPa6.5 MPa6.5 MPa

Answer: 0.59 Official key: −0.61 to −0.57 OR 0.57 to 0.61

From the figure, σy = +25 MPa (tension, arrows pointing out) and σx = −6.5 MPa (compression, arrows pointing in), with no shear on the plate edges.

  1. On a plane at 45°: σn = (σx + σy)/2 = (−6.5 + 25)/2 = 9.25 MPa.
  2. τ = (σy − σx)/2 = (25 + 6.5)/2 = 15.75 MPa (the 45° plane carries the maximum in-plane shear).
  3. Ratio = 9.25 / 15.75 = 0.59.
Where this shows up in industry

Weld codes check normal and shear stress on the weld throat separately, which is this exact transformation. In FEA you'd read the same values by defining a local coordinate system along the weld line.

GATE 2024 · ME · Q.57NAT · 2 marksMohr's circle and stress transformation

What is the normal stress on an inclined weld in a thin pressure vessel?

The figure shows a thin cylindrical pressure vessel constructed by welding plates together along a line that makes an angle α = 60° with the horizontal. The closed vessel has a wall thickness of 10 mm and diameter of 2 m. When subjected to an internal pressure of 200 kPa, the magnitude of the normal stress acting on the weld is _________ MPa (rounded off to 1 decimal place).
Closed thin cylindrical vessel with a weld line inclined at angle alpha to the horizontalαWeld line

Answer: 12.5 Official key: 12.3 to 12.7

  1. Hoop stress σh = pd/2t = 0.2 × 2000 / (2 × 10) = 20 MPa (circumferential).
  2. Longitudinal stress σl = pd/4t = 10 MPa (along the axis, horizontal in the figure).
  3. The weld line makes α = 60° with the axis, so the normal to the weld makes 60° with the hoop direction. Transforming: σn = σl sin²α + σh cos²α.
  4. σn = 10 × sin²60° + 20 × cos²60° = 7.5 + 5 = 12.5 MPa.
Where this shows up in industry

Spiral-welded pipes and LPG cylinders put the weld at an angle on purpose: the seam sees less than the full hoop stress. Pressure-vessel codes use the same transformation to set the weld efficiency.

GATE 2025 · ME · Q.56NAT · 2 marksMohr's circle and stress transformation

How do you find the thickness strain under plane stress?

The state of stress at a point is shown in the figure given below. Under plane stress assumption, the normal strain along the thickness direction (εzz) is ____________ (rounded off to 2 decimal places).
Use the Young’s Modulus E = 200 GPa and Poisson’s ratio ν = 0.27.
Plane stress element: 350 (tension) along X, 150 (compression) along Y, shear 50; values as printed in the paper350 GPa350 GPa150 GPa150 GPa50 GPa50 GPaXYZ

Answer: −0.27 Official key: −0.29 to −0.25 OR 0.25 to 0.29

Under plane stress σzz = 0, but the thickness still changes because of Poisson's effect. Shear stress doesn't produce normal strain.

  1. From the figure: σxx = +350 (tension), σyy = −150 (compression).
  2. εzz = −(ν/E)(σxx + σyy) = −(0.27/200)(350 − 150) = −0.27.

Note: the official figure labels the stresses in GPa, so the strain works out to −0.27 as printed, and the key accepts it. With the more realistic MPa the same working gives −0.00027.

Where this shows up in industry

Sheet-metal forming simulations and thin-shell FEA elements use exactly this: σzz = 0, and the thickness change is computed from in-plane stresses. That thinning is what a forming engineer checks for splits.

GATE 2022 · ME Set-1 · Q.52NAT · 2 marksThin cylinders

How much does the wall thickness of a thin cylinder change under internal pressure?

A thin-walled cylindrical pressure vessel has mean wall thickness of t and nominal radius of r. The Poisson’s ratio of the wall material is 1/3. When it was subjected to some internal pressure, its nominal perimeter in the cylindrical portion increased by 0.1% and the corresponding wall thickness became t̄. The corresponding change in the wall thickness of the cylindrical portion, i.e. 100 × (t̄ − t)/t, is ________% (round off to 3 decimal places).

Answer: −0.060 Official key: −0.062 to −0.058

  1. Perimeter strain = hoop strain: εh = (σh/E)(1 − ν/2) = 0.001, with σl = σh/2.
  2. σh/E = 0.001 / (1 − 1/6) = 0.0012.
  3. Radial (thickness) stress is negligible in a thin wall, so εt = −ν(σh + σl)/E = −(1/3)(1.5)(0.0012) = −0.0006.
  4. Change in thickness = −0.060% (the wall gets thinner).
Where this shows up in industry

Ultrasonic wall-thickness checks on pressurised tanks and pipelines have to allow for this elastic thinning. It's small, but it's the same Poisson coupling that makes shell elements report thickness change.

GATE 2026 · ME · Q.51NAT · 2 marksThin cylinders

What internal pressure can a thin cylinder take if the surface strain is limited?

A cylindrical pressure vessel made of steel has diameter of 3 m and wall thickness of 15 mm. For steel, Young’s modulus and Poisson’s ratio are 210 GPa and 0.3, respectively. The cylinder is designed such that the allowable normal strain at the outer cylindrical surface is equal to 0.00034. The permissible pressure in the tank is ________ kPa (rounded off to 1 decimal place).

Answer: 840.0 Official key: 838.0 to 842.0

  1. σh = pd/2t, σl = pd/4t. The larger surface strain is the hoop strain.
  2. εh = (pd/2tE)(1 − ν/2) = p × 3000 / (2 × 15 × 210,000) × 0.85.
  3. Set εh = 0.00034: p = 0.00034 × 2 × 15 × 210,000 / (3000 × 0.85) = 0.84 MPa = 840.0 kPa.
Where this shows up in industry

Strain-gauge acceptance tests on pressure vessels work this way: a gauge glued in the hoop direction during the hydro test confirms the shell is behaving as designed.

GATE 2024 · ME · Q.41MSQ · 2 marksShear force and bending moment

How do you tell if a beam is statically indeterminate?

Which of the following beam(s) is/are statically indeterminate?
  1. (A)Beam optionP1P2
  2. (B)Beam optionM
  3. (C)Beam optionP1P2
  4. (D)Beam optionMP

Answer: (C) and (D) Official key: C; D

A planar beam has 3 equilibrium equations. Count the support reactions: a pin gives 2, a roller 1, a fixed support 3.

  • (A) pin + roller = 3 reactions: determinate (the overhang doesn't matter).
  • (B) pin + roller = 3: determinate (an applied moment is a load, not a support).
  • (C) pin + 2 rollers = 4: indeterminate (degree 1).
  • (D) fixed + roller = 4: indeterminate (a propped cantilever).
Where this shows up in industry

Real structures are usually indeterminate: a chassis rail on several mounts, a shaft on three bearings. That's why they're solved with FEA (stiffness method) rather than equilibrium alone.

GATE 2024 · ME · Q.53NAT · 2 marksShear force and bending moment

What is the maximum bending moment in a simply supported beam with a triangular load?

A horizontal beam of length 1200 mm is pinned at the left end and is resting on a roller at the other end as shown in the figure. A linearly varying distributed load is applied on the beam. The magnitude of maximum bending moment acting on the beam is _______ N.m. (round off to 1 decimal place)
Beam of length 1200 mm, pinned at the left and on a roller at the right, with a load rising linearly from zero to 100 N/m100 N/m1200 mm

Answer: 9.2 Official key: 9.0 to 9.5

  1. Total load W = ½ × 100 × 1.2 = 60 N, acting at 2L/3 from the left. Reactions: RA = W/3 = 20 N, RB = 40 N.
  2. Load intensity at x: w(x) = 100x/1.2. Shear V(x) = 20 − 100x²/(2 × 1.2). V = 0 at x = L/√3 = 0.6928 m.
  3. M(x) = 20x − 100x³/(6 × 1.2). At x = 0.6928 m: Mmax = w0L²/(9√3) = 100 × 1.44 / 15.59 = 9.2 N·m.
Where this shows up in industry

Hydrostatic load on a tank wall or soil pressure on a retaining wall is triangular, and the maximum moment is not at mid-span. Designers place stiffeners where V = 0, the same point found here.

GATE 2025 · ME · Q.57NAT · 2 marksShear force and bending moment

What is the bending moment in a beam loaded by a uniformly distributed moment?

A simply supported beam of length 1 m is subjected to a uniformly distributed bending moment of 1 N m per m throughout the length as shown in the figure given below. The bending moment at the mid-point of the beam is_____N m (rounded off to the nearest integer).
Simply supported beam AB of length 1 m carrying a uniformly distributed bending moment of 1 N m per mAB1 N m per m1 m

Answer: 0 Official key: 0 to 0

  1. Total applied moment = 1 N·m/m × 1 m = 1 N·m. The supports balance it with a couple: RA = −RB, |R| = 1 N.
  2. Moments about B: RA × 1 + m × 1 = 0, so RA = −m.
  3. At any section x: M(x) = RAx + m·x = −mx + mx = 0.
  4. The bending moment is zero everywhere, so at mid-span it's 0 N·m. The beam carries only a constant shear force of 1 N.
Where this shows up in industry

This looks like a trick question, but it's a useful check: loads that only add moment can come back entirely as shear, e.g. a beam with many small brackets each applying a twisting couple.

GATE 2026 · ME · Q.40MCQ · 2 marksShear force and bending moment

How do you draw the SFD and BMD of a cantilever with a point load and an end moment?

A cantilever beam of length L is subjected to a concentrated load P and a moment M as shown in the figure below. Neglecting the mass of the beam, which one of the following options is the correct representation of shear force diagram (SFD) and bending moment diagram (BMD)?
Cantilever of length L, fixed at the left, with load P at L/2 and moment M = PL/2 at the free endPM = PL/2L/2L/2
  1. (A)Shear force and bending moment diagram optionPL/2L/2SFDPL/2L/2L/2BMD
  2. (B)Shear force and bending moment diagram optionPL/2L/2SFDPLPL/2L/2L/2BMD
  3. (C)Shear force and bending moment diagram optionPL/2L/2SFDPL/2L/2L/2BMD
  4. (D)Shear force and bending moment diagram optionPL/2L/2SFDPLPL/2L/2L/2BMD

Answer: (A) Official key: A

  1. Shear force: cutting anywhere between the free end and the load, the only load to the right is the moment M, so V = 0 there. Between the support and L/2, V = P. So the SFD is a block of height P over the left half only (rules out C and D).
  2. Bending moment, measured from the free end: for the outer half, M(x) = PL/2 (constant, from the end moment).
  3. For the inner half the point load acts against the end moment: M = PL/2 − P(x − L/2). At the fixed end (x = L) this is PL/2 − PL/2 = 0.
  4. So the BMD rises from 0 at the support to PL/2 at mid-span and stays flat to the free end: option (A).
Where this shows up in industry

The fixed-end moment being zero means the root weld or bolt joint carries only shear here. Getting the direction of the applied moment right is the same check you do when you apply moments in an FEA load case.

GATE 2026 · ME · Q.41MCQ · 2 marksShear force and bending moment

How do you find support reactions in a compound beam with an internal hinge?

Two structural members are connected by a hinge and they are externally loaded as shown in the figure below. A and B are roller supports, and C is a pin support. Neglecting the mass of the members, the magnitude of the vertical reaction force at C in kN is
Two members joined by a hinge: roller A, 10 kN load, hinge, roller B, 25 kN load, pin CABCHinge10 kN25 kN0.5 m0.5 m0.2 m0.2 m0.6 m
  1. (A)5
  2. (B)10
  3. (C)20
  4. (D)15

Answer: (A) 5 Official key: A

  1. The hinge carries no moment. Left member (A to hinge): moments about the hinge: RA × 1.0 = 10 × 0.5, so RA = 5 kN, and the hinge passes 10 − 5 = 5 kN down onto the right member.
  2. Right member (hinge at 1.0 m, B at 1.2 m, 25 kN at 1.4 m, C at 2.0 m). Moments about B: 5 × 0.2 − 25 × 0.2 + RC × 0.8 = 0.
  3. RC = (5 − 1)/0.8 = 5 kN, option (A). (Check: RB = 5 + 25 − 5 = 25 kN.)
Where this shows up in industry

Gerber beams in bridges and long conveyor frames use internal hinges so that settlement of one support doesn't load the whole structure. The 'cut at the hinge, M = 0' step is the same one used to split a model into free bodies.

GATE 2026 · ME · Q.49NAT · 2 marksBending and shear stress

How do you find the load from the shear stress at the neutral axis of a rectangular beam?

A simply supported beam is subjected to an external point load P as shown in the figure below. The beam has a rectangular cross-section of 20 mm × 45 mm. A is a pin support and B is a roller support. The shear stress developed at point C, lying on the neutral axis of the beam, is 3 MPa. Neglecting the mass of the beam, the magnitude of the applied load is _______ kN (rounded off to 1 decimal place).
Simply supported beam of span 5 m: pin A, point C on the neutral axis 1 m from A, load P 2 m from A, roller BABPoint CP1 m1 m3 m

Answer: 3.0 Official key: 2.9 to 3.1

  1. Span 5 m, load 2 m from A: RA = P × 3/5 = 0.6P. Point C (1 m from A) is left of the load, so the shear force there is V = 0.6P.
  2. For a rectangle, the maximum shear stress (at the neutral axis) is τ = 1.5 V/A = 1.5 × 0.6P / (20 × 45).
  3. 3 = 0.9P / 900, so P = 3000 N = 3.0 kN.
Where this shows up in industry

Short, deep members (brackets, gear teeth, lugs) fail in shear before bending. The 1.5 factor is why FEA shows peak shear at mid-depth, where bending stress is zero.

GATE 2026 · ME · Q.42MCQ · 2 marksDeflection, stiffness and energy methods

How do you find the stiffness and natural frequency of an overhanging beam with a tip mass?

A 5 kg mass is suspended at the free end of an overhanging massless beam, having a pin support and a roller support, as shown in the figure below. Young’s modulus of the material of the beam is 200 GPa and area moment of inertia of the beam is 10−8 m4. The natural frequency of the beam in rad/s is
Overhanging beam: pin support, roller support 1 m away, 5 kg mass at the free end 2 m beyond the roller5 kg1 m2 m
  1. (A)10
  2. (B)10/3
  3. (C)5
  4. (D)20/3

Answer: (A) 10 Official key: A

  1. Tip deflection of an overhang a beyond a span l, load P at the tip: δ = Pa²(l + a)/(3EI).
  2. With a = 2 m, l = 1 m: δ = P × 4 × 3 / (3EI) = 4P/EI. EI = 200×109 × 10−8 = 2000 N·m², so k = EI/4 = 500 N/m.
  3. ωn = √(k/m) = √(500/5) = 10 rad/s, option (A).
Where this shows up in industry

Every modal analysis is this calculation scaled up: stiffness from the structure, mass from the parts. Mounting a heavy component on a long overhang drops its natural frequency into the range of engine or road excitation.

GATE 2022 · ME Set-2 · Q.51NAT · 2 marksDeflection, stiffness and energy methods

How do you find string tensions on a rigid beam supported by elastic strings?

A rigid beam AD of length 3a = 6 m is hinged at frictionless pin joint A and supported by two strings as shown in the figure. String BC passes over two small frictionless pulleys of negligible radius. All the strings are made of the same material and have equal cross-sectional area. A force F = 9 kN is applied at C and the resulting stresses in the strings are within linear elastic limit. The self-weight of the beam is negligible with respect to the applied load. Assuming small deflections, the tension developed in the string at C is _________ kN (round off to 2 decimal places).
Rigid beam AD hinged at A; string from B over two pulleys down to C; separate string at D; force F at C; spacings aABCDFRigid beamaaaa

Answer: 1.50 Official key: 1.48 to 1.52

The problem is statically indeterminate, so add a compatibility condition.

  1. Let the beam rotate by a small angle θ about A. Downward movements: B = aθ, C = 2aθ, D = 3aθ.
  2. String BC is one string (same tension T1 at B and C) of length a + a + a = 3a; it stretches by aθ + 2aθ = 3aθ, so its strain is θ. String D has length a and stretches 3aθ: strain 3θ.
  3. Same E and A: T1 = EAθ, TD = 3EAθ = 3T1.
  4. Moments about A: T1(a) + T1(2a) + TD(3a) = F(2a), so 3T1 + 9T1 = 2 × 9 and T1 = 1.50 kN.
Where this shows up in industry

Load sharing between bolts, cables or mounts of different lengths follows this rule: the stiffer path (here the short string D) takes more load. It's why FEA bolt-load results rarely split evenly.

GATE 2022 · ME Set-1 · Q.37MCQ · 2 marksDeflection, stiffness and energy methods

How do you use Castigliano's theorem to find the reaction on an L-shaped frame?

An L-shaped elastic member ABC with slender arms AB and BC of uniform cross-section is clamped at end A and connected to a pin at end C. The pin remains in continuous contact with and is constrained to move in a smooth horizontal slot. The section modulus of the member is same in both the arms. The end C is subjected to a horizontal force P and all the deflections are in the plane of the figure. Given the length AB is 4a and length BC is a, the magnitude and direction of the normal force on the pin from the slot, respectively, are
L-shaped member: AB of length 4a clamped at A, BC of length a hanging down from B, end C pinned in a smooth horizontal slot, horizontal force P at CAB4aaPC
  1. (A)3P/8, and downwards
  2. (B)5P/8, and upwards
  3. (C)P/4, and downwards
  4. (D)3P/4, and upwards

Answer: (A) 3P/8, downwards Official key: A

  1. The slot stops vertical movement at C, so it applies an unknown vertical force R (take upward as positive). Use Castigliano: ∂U/∂R = 0 (bending energy only, slender arms).
  2. Arm BC (distance s up from C): M = Ps, which doesn't depend on R.
  3. Arm AB (distance u from B towards A): M = Ru + Pa.
  4. δv = (1/EI)∫04a(Ru + Pa)u du = (1/EI)[R(64a³/3) + Pa(8a²)] = 0.
  5. R = −8P × 3/64 = −3P/8. The negative sign means it acts downward: 3P/8, downwards, option (A).
Where this shows up in industry

Energy methods are how hand calculations handle frames that equilibrium can't solve, and they're the basis of the finite element method itself: minimising strain energy gives the stiffness equations.

GATE 2022 · ME Set-2 · Q.50NAT · 2 marksTorsion and springs

How is shear stress shared between the core and rim of a composite shaft in torsion?

A shaft of length L is made of two materials, one in the inner core and the other in the outer rim, and the two are perfectly joined together (no slip at the interface) along the entire length of the shaft. The diameter of the inner core is di and the external diameter of the rim is do, as shown in the figure. The modulus of rigidity of the core and rim materials are Gi and Go, respectively. It is given that do = 2di and Gi = 3Go. When the shaft is twisted by application of a torque along the shaft axis, the maximum shear stress developed in the outer rim and the inner core turn out to be τo and τi, respectively. All the deformations are in the elastic range and stress-strain relations are linear. Then the ratio τi/τo is ______ (round off to 2 decimal places).
Shaft cross-section: inner core of diameter d_i perfectly bonded inside an outer rim of diameter d_odido

Answer: 1.50 Official key: 1.48 to 1.52

  1. Bonded together, core and rim twist by the same angle per unit length θ, so shear strain γ = rθ is continuous.
  2. Stress τ = Grθ. The maximum in each material is at its outer radius: τi = Gi(di/2)θ, τo = Go(do/2)θ = Godiθ.
  3. τi/τo = 3Go(di/2) / (Godi) = 1.50.

The stiffer core carries the higher peak stress even though it sits closer to the axis.

Where this shows up in industry

Composite drive shafts (carbon tube over a steel or aluminium core) and bimetal shafts are designed this way. The stress jump at the interface is the location that FEA and fatigue checks focus on.

GATE 2026 · ME · Q.19MCQ · 1 markTorsion and springs

What is the maximum shear stress in a closely coiled helical spring?

A closely coiled helical compression spring of mean coil diameter D and wire diameter d, is loaded by an axial force F.
The maximum shear stress developed in the wire is
  1. (A)8FD/(πd³) + 4F/(πd²)
  2. (B)8FD/(πd³) + 2F/(πd²)
  3. (C)16FD/(πd³) + 4F/(πd²)
  4. (D)32FD/(πd³) + 4F/(πd²)

Answer: (A) Official key: A

  1. At any wire section the force F produces a torque T = F × D/2 and a direct shear force F.
  2. Torsional shear: τt = 16T/(πd³) = 16(FD/2)/(πd³) = 8FD/(πd³).
  3. Direct shear: τd = F/(πd²/4) = 4F/(πd²). Both add at the inner side of the coil.
  4. τmax = 8FD/(πd³) + 4F/(πd²): option (A).

(The Wahl factor would also add curvature; this question asks for the basic direct + torsional sum.)

Where this shows up in industry

Valve springs, suspension coils and clutch springs are sized with this formula plus the Wahl correction. Spring fatigue cracks start on the inside of the coil, where both terms add.

GATE 2022 · ME Set-1 · Q.16MCQ · 1 markColumns and buckling

What is the buckling load of a beam held by an inclined cable?

A uniform light slender beam AB of section modulus EI is pinned by a frictionless joint A to the ground and supported by a light inextensible cable CB to hang a weight W as shown. If the maximum value of W to avoid buckling of the beam AB is obtained as βπ²EI, where π is the ratio of circumference to diameter of a circle, then the value of β is
Beam AB pinned at A to a wall, held at B by a cable to C on the wall at 30 degrees, weight W hanging at BCAB30°W2.5 mg
  1. (A)0.0924 m−2
  2. (B)0.0713 m−2
  3. (C)0.1261 m−2
  4. (D)0.1417 m−2

Answer: (A) 0.0924 m−2 Official key: A

  1. Joint B: cable tension T at 30°. Vertical: T sin30° = W, so T = 2W. Horizontal: compression in AB = T cos30° = √3 W.
  2. AB is pinned at both ends (pin at A, cable joint at B): Pcr = π²EI/L², L = 2.5 m.
  3. √3 W = π²EI/6.25, so W = π²EI/(6.25√3) and β = 1/10.83 = 0.0924 m−2, option (A).
Where this shows up in industry

Jib cranes, boom arms and truss members in compression are checked like this. The geometry (cable angle) decides how much of the hung load becomes column compression.

GATE 2022 · ME Set-2 · Q.18MCQ · 1 markCombined stresses and yield

How do you apply the distortion energy (von Mises) theory to a plane stress state?

A structural member under loading has a uniform state of plane stress which in usual notations is given by σx = 3P, σy = −2P and τxy = √2 P, where P > 0. The yield strength of the material is 350 MPa. If the member is designed using the maximum distortion energy theory, then the value of P at which yielding starts (according to the maximum distortion energy theory) is
  1. (A)70 MPa
  2. (B)90 MPa
  3. (C)120 MPa
  4. (D)75 MPa

Answer: (A) 70 MPa Official key: A

  1. Von Mises for plane stress: σvm² = σx² + σy² − σxσy + 3τxy².
  2. = 9P² + 4P² + 6P² + 6P² = 25P², so σvm = 5P.
  3. Yield when 5P = 350 MPa: P = 70 MPa, option (A).
Where this shows up in industry

The 'von Mises stress' plot in every FEA tool is this formula applied at each node. For ductile metals it's the number you compare against yield strength.

GATE 2026 · ME · Q.34NAT · 1 markCombined stresses and yield

What is the permissible stress for a biaxial state using distortion energy theory?

A steel plate having yield strength of 550 MPa, is subjected to biaxial state of stress as σx = σ and σy = −2σ. Using distortion energy theory and considering the factor of safety as 2, the permissible value of σ that can be applied to the plate is ________ MPa (rounded off to 1 decimal place).

Answer: 103.9 Official key: 102.0 to 105.0

  1. σvm = √(σx² + σy² − σxσy) = √(σ² + 4σ² + 2σ²) = √7 σ.
  2. Allowable = 550/2 = 275 MPa. √7 σ = 275, so σ = 103.9 MPa.

Opposite-sign stresses are the worst case: the −σxσy term adds instead of subtracting.

Where this shows up in industry

Plates under tension one way and compression the other (shear webs, torsion boxes) reach yield much earlier than either stress alone suggests. FEA von Mises plots pick this up automatically.

GATE 2024 · ME · Q.56NAT · 2 marksCombined stresses and yield

What is the factor of safety of a shaft under axial force and torque by distortion energy theory?

A solid massless cylindrical member of 50 mm diameter is rigidly attached at one end, and is subjected to an axial force P = 100 kN and a torque T = 600 N.m at the other end as shown. Assume that the axis of the cylinder is normal to the support. Considering distortion energy theory with allowable yield stress as 300 MPa, the factor of safety in the design is _____ (rounded off to 1 decimal place).
Solid cylinder of 50 mm diameter fixed to a wall at one end, with axial force P and torque T at the free endTP

Answer: 4.5 Official key: 4.3 to 4.7

  1. Axial stress σ = P/A = 100,000 / (π × 50²/4) = 50.9 MPa.
  2. Torsional shear at the surface τ = 16T/(πd³) = 16 × 600,000 / (π × 50³) = 24.4 MPa.
  3. σvm = √(σ² + 3τ²) = √(2594 + 1793) = 66.2 MPa.
  4. FoS = 300 / 66.2 = 4.5.
Where this shows up in industry

Drive shafts with axial thrust (helical gears, propeller shafts) are checked with exactly this combination before fatigue factors are added.

GATE 2022 · ME Set-2 · Q.53NAT · 2 marksCombined stresses and yield

How do you size a shaft for combined bending and torsion using maximum shear stress theory?

A shaft AC rotating at a constant speed carries a thin pulley of radius r = 0.4 m at the end C which drives a belt. A motor is coupled at the end A of the shaft such that it applies a torque Mz about the shaft axis without causing any bending moment. The shaft is mounted on narrow frictionless bearings at A and B where AB = BC = L = 0.5 m. The taut and slack side tensions of the belt are T1 = 300 N and T2 = 100 N, respectively. The allowable shear stress for the shaft material is 80 MPa. The self-weights of the pulley and the shaft are negligible. Use the value of π available in the on-screen virtual calculator. Neglecting shock and fatigue loading and assuming maximum shear stress theory, the minimum required shaft diameter is _______ mm (round off to 2 decimal places).
Shaft AC on bearings at A and B with AB = BC = L; motor torque Mz at A; pulley of radius r at C with belt tensions T1 and T2 both downwardABMzCT1T2rLLzxy

Answer: 23.94 Official key: 23.60 to 24.20

  1. Both belt tensions act downward at C: total 400 N on an overhang of L = 0.5 m beyond bearing B. Maximum bending moment (at B): M = 400 × 0.5 = 200 N·m.
  2. Torque from the belt: T = (T1 − T2)r = 200 × 0.4 = 80 N·m (between A and C).
  3. Maximum shear stress theory: τmax = 16√(M² + T²)/(πd³) = 16 × 215.4 / (πd³).
  4. Set equal to 80 MPa: d³ = 16 × 215.4 / (π × 80×106) = 1.371×10−5 m³, d = 23.94 mm.
Where this shows up in industry

This is the first-pass sizing for any belt- or gear-driven shaft before bearing loads, keyways and fatigue are added, the same numbers you'd later confirm with an FEA of the shaft.

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