UPSC Mechanical Engineering 2026 — Paper I

(a) Compute the three reaction components (two force components and the fixing couple) at the built-in support O of a bent cantilever beam, including the self-weight of the uniform beam (50 kg/m). 'Compute the reactions' means give Ox, Oy and M_O with correct signs/senses. (b) A statically indeterminate stepped bar fixed at both ends carries two intermediate axial loads. (i) Find the two support reactions R_A and R_D; (ii) find the compressive axial force F_BC in the middle segment. Both sub-parts must be answered. (c) Three demands: explain WHY grain size matters in polycrystalline metals, explain HOW it is measured (ASTM E112 methods), and then solve (i) the ASTM grain-size number for 75 grains/in^2 at 100x and (ii) the grains/in^2 the same specimen would show at 60x. (d) From the data 'power component = 0.94 x normal thrust' (i.e. cos(phi) = 0.94, so pressure angle phi = 20°) and stub-tooth addendum = 0.85 m, find for BOTH gear ratios G = 1 and G = 2.5 the minimum number of teeth on the pinion to avoid interference/undercutting AND the length of the path of contact expressed in terms of the module m. (e) For a solid circular shaft of d = 60 mm carrying an axial thrust P together with a torque T = 2 kN.m, find the largest permissible P so that the maximum (principal) normal stress does not exceed sigma_all = 220 MPa. This is a combined axial-plus-torsion principal-stress problem.

• Self-weight 0.4905 kN/m applied as 1.177 kN on the horizontal part and 0.589 kN on the inclined part at their midpoints
• Resolution of the 1.4 kN load perpendicular to the inclined member into 0.70 kN and 1.21 kN components
• Force balance: O_x = 0.70 kN (left), O_y = 5.98 kN (up)
• Moment balance with the 4 kN·m couple giving M_O ≈ 9.12 kN·m (counter-clockwise)
• Equilibrium plus compatibility (zero total elongation) for a statically indeterminate bar
• Axial forces expressed in one unknown, with L₁/A₁ and L₂/A₂ used correctly
• R_A = 10.5 kN and R_D = 2.0 kN with senses
• F_BC = 15.0 kN compressive
• Importance: Hall–Petch strengthening, toughness, creep and other properties
• Estimation methods of ASTM E112: comparison, planimetric and intercept
• ASTM number n = 1 + log₂ 75 ≈ 7.2
• N₆₀ = 75 × (100/60)² ≈ 208 grains/in²
• Pressure angle from cos φ = 0.94, φ ≈ 20°
• Minimum teeth formula with a_w = 0.85 m: 11 teeth for G = 1 and 32 for G = 2.5
• Path of approach and recess from addendum and pitch radii
• Path of contact ≈ 3.62 m for G = 1 and ≈ 4.46 m for G = 2.5
• Torsional shear stress τ = 16T/πd³ = 47.16 MPa
• Principal stress σ_max = σ/2 + √((σ/2)² + τ²) with σ = P/A
• Solving σ = (σ_all² − τ²)/σ_all = 209.9 MPa
• P = σA ≈ 593 kN

(a) Define heat treatment, state its purposes, and explain the main heat-treatment processes (annealing, normalising, hardening/quenching, tempering, austempering/martempering, case hardening) by locating them on the Fe-Fe3C equilibrium diagram and naming the microstructural phases produced. (b) For a thin steel tube in pure torsion (mean diameter 100 mm, thickness 3 mm), calculate the transmissible torque for a factor of safety of 2.25 under three separate failure theories: (i) maximum shear stress (Tresca/Guest), (ii) maximum total strain energy (Haigh/Beltrami), and (iii) maximum shear strain energy (von Mises/Distortion energy). All three must be evaluated. (c) A flywheel problem with a harmonic turning-moment diagram T = 8000 + 1000 sin 2theta - 2000 cos 2theta over a 180° cycle. Four sub-parts must all be answered: (i) T_max, T_min, T_mean plus a labelled T-theta sketch; (ii) power developed; (iii) total percentage fluctuation of speed; (iv) the new percentage fluctuation if k falls from 750 mm to 600 mm, and by how much it rises.

• Definition and purposes of heat treatment
• Critical lines A₁, A₃, A_cm and 0.8% C eutectoid on the Fe–Fe₃C diagram
• Annealing and normalising with temperature ranges and products (coarse and fine pearlite)
• Hardening to martensite, tempering, austempering/martempering (bainite) and case hardening
• Thin-tube shear stress τ = T/(2A_m t) with A_m = πd²/4 and principal stresses ±τ
• Tresca: τ = 50 N/mm², T ≈ 2.36 kN·m
• Haigh: 2.6τ² = 100², T ≈ 2.92 kN·m
• von Mises: 3τ² = 100², T ≈ 2.72 kN·m
• T_max = 10 236 N·m, T_min = 5764 N·m, T_mean = 8000 N·m, with labelled T–θ diagram
• Power = 8000 × 31.416 ≈ 251.3 kW
• ΔE = 2236 J, I = 281.25 kg·m², C_s ≈ 0.806%
• New C_s ≈ 1.259%, an increase of about 0.453 percentage points

(a(i)) Rectilinear kinematics with variable acceleration a = (2t - 9) m/s^2. Integrate twice with the given initial conditions and report, at t = 9 s, (I) the position s, (II) the TOTAL DISTANCE travelled (not displacement), and (III) the velocity. (a(ii)) For the cylindrical part of a thin-walled pressure vessel (r = 0.6 m, t = 18 mm, p = 2.8 MPa, E = 200 GPa, nu = 0.30) compute (I) circumferential and longitudinal stresses, (II) the maximum in-plane AND out-of-plane shear stresses, and (III) circumferential and longitudinal strains. All three sub-parts carry marks. (b(i)) An inverse plane-stress problem: given sigma_x = 110 MPa and the normal stress on two inclined planes (37 MPa tension at theta = 32°, 12 MPa compression at theta = 48°), find the two unknowns sigma_y and tau_xy by setting up and solving two simultaneous transformation equations. (b(ii)) Derive the damped natural frequency of a single-degree-of-freedom rotational system: a rigid massless bar pivoted at O, with a spring k at distance a on one side, a viscous damper c at distance b on the other side, and a concentrated mass m at distance l. The answer is a symbolic expression, since no numerical data are given. (c) For a continuous unidirectional glass-epoxy composite (E_f = 72.5 GPa, E_m = 3.5 GPa, V_f = 35%), compute (i) the longitudinal and transverse moduli using the rule of mixtures and the inverse rule of mixtures, and (ii) the strain in the fibres and in the matrix when a stress of 50 MPa acts on a 250 mm^2 cross-section.

• v = t² − 9t + 10 and s = t³/3 − 4.5t² + 10t + 1 with the initial conditions
• Position s(9) = −30.5 m and velocity v(9) = 10 m/s
• Turning points from v = 0 at t = 1.298 s and 7.702 s
• Total distance = 56.0 m, distinct from displacement −31.5 m
• Hoop stress 93.3 MPa and longitudinal stress 46.7 MPa
• Maximum in-plane shear 23.3 MPa
• Maximum out-of-plane shear 46.7 MPa
• Strains ε_c ≈ 3.97 × 10⁻⁴ and ε_L ≈ 9.33 × 10⁻⁵
• Transformation equation σ_θ = (σx+σy)/2 + (σx−σy)/2 cos 2θ + τ sin 2θ written for both angles
• Two simultaneous equations set up with correct trigonometric values
• σ_y ≈ −60.7 MPa
• τ_xy ≈ −27.9 MPa
• Equation of motion m l² θ̈ + c b² θ̇ + k a² θ = 0
• Equivalent inertia, damping and stiffness identified
• Natural frequency ω_n = (a/l)√(k/m) and ζ = c b²/(2 a l √(k m))
• ω_d = ω_n√(1 − ζ²) = √(k a²/(m l²) − c² b⁴/(4 m² l⁴))
• E_L = 27.65 GPa by the rule of mixtures
• E_T = 5.25 GPa by the inverse rule
• Load = 12.5 kN and equal strain ε = 1.81 × 10⁻³ in fibre and matrix along the fibres
• Fibre and matrix stresses 131.1 and 6.33 MPa with loads adding to 12.5 kN

(a) A simply supported beam of span L carries a triangular distributed load of peak intensity w0 at A falling linearly to zero at B. Using the DOUBLE-INTEGRATION method only, obtain (i) the equation of the elastic curve, (ii) the slope at A, and (iii) the deflection at midspan. The method is prescribed, so a moment-area or Macaulay short cut earns reduced credit. (b) Balancing of several masses in different planes. (i) Find the magnitude and angular position of two balance masses placed at radius 700 mm in planes 100 mm either side of the middle eccentric's plane; (ii) if one balance mass is moved to a radius of 50 mm, find its new magnitude and state whether the other mass or any position changes. (c) A friction problem with two bodies: a uniform bar pinned at B and resting on a block of mass m0 on the floor. Derive a GENERAL expression for the horizontal force P needed to start the block moving, for P acting towards the right and for P acting towards the left, then evaluate both for theta = 30°, m = m0 = 3 kg, mu_s = 0.60, mu_k = 0.50.

• Reactions R_A = w₀L/3 and R_B = w₀L/6, and bending moment M(x) = w₀Lx/3 − w₀x²/2 + w₀x³/(6L)
• EI y″ = M integrated twice with constants from y(0) = y(L) = 0
• Elastic curve y = (w₀/(360EIL))(20L²x³ − 15Lx⁴ + 3x⁵ − 8L⁴x), C₁ = −w₀L³/(45)
• θ_A = w₀L³/(45EI) and y(L/2) = 5w₀L⁴/(768EI), both downward
• Force polygon Σ mr = 0 and couple polygon Σ mrz = 0 about the central plane
• (mr)_A = (mr)_B = 10 kg·mm at 240° and 0° respectively
• Balancing masses 14.3 g each at 700 mm
• Shifted mass = 10/50 = 0.2 kg (200 g), with no change to the other mass or the angular positions
• Moment equilibrium of the bar about B with normal force N and friction ±μ_k N at A
• N = mg/[2(1 ∓ μ_k tan θ)] for the two directions
• P = (μ_k + μ_s)N + μ_s m₀ g
• P_A ≈ 40.4 N (right) and P_B ≈ 30.2 N (left)

(a) Compute the machining (turning) time for a single straight-turning pass that reduces a 60 mm bar to 50 mm diameter over 120 mm length at V = 120 m/min and f = 0.3 mm/rev. The chain required is V -> N -> feed rate -> time. (b) A break-even/cost-comparison problem with three explicit sub-parts: (i) the break-even volume of each process separately, (ii) the cross-over (indifference) volume between X and Y, and (iii) at a demand of 35,000 units, the profit of each process, a recommendation, and the margin of safety percentage under the recommended process. (c) Compute the power density of a laser beam that delivers 5 J in 0.01 s over 0.5 mm^2. A short calculation, so credit also attaches to correct unit handling, sensible unit conversions and a brief comment on why power density governs the LBM mechanism. (d) Statistical process control: (i) compute the upper and lower control limits for both the X-bar chart and the R chart from X-bar-bar = 48.6 mm, R-bar = 4.2 mm and the given constants; (ii) from a separate attribute inspection of 20 lots of 200 units with 184 defectives, compute p-bar and the UCL and LCL of a p-chart. (e) Two demands: (a) discuss the factors that affect surface finish in machining, and (b) list the profile-finishing techniques used in modern manufacturing. Both the discussion and the list carry marks.

• N = 1000V/(πD) = 636.6 rpm
• Feed rate = fN ≈ 191 mm/min
• t = L/(fN) ≈ 0.628 min (37.7 s)
• Break-even X = 17,455 units and Y = 20,834 units
• Cross-over at 31,765 units from equating total costs
• Profits ₹9,65,000 (X) and ₹10,20,000 (Y) at 35,000 units
• Recommend Y, margin of safety about 40.5%
• Power = E/t = 500 W
• Area converted to 0.5 × 10⁻⁶ m²
• Power density = 10⁹ W/m² (1000 W/mm² or 10⁵ W/cm²)
• Comment on why power density governs material removal
• X̄ chart limits 51.02 and 46.18 mm about 48.6 mm
• R chart UCL 8.88 mm, LCL 0
• p̄ = 184/4000 = 0.046
• p-chart UCL 0.0904 and LCL 0.0016
• Cutting parameters: feed, speed (built-up edge) and depth
• Tool factors: nose radius, geometry, wear
• Workpiece and machine factors: material, rigidity, vibration
• List of finishing techniques: honing, lapping, superfinishing, polishing, burnishing and advanced processes such as abrasive flow, magnetic abrasive and magnetorheological finishing

(a) Flat rolling calculation with three sub-parts: (i) the projected arc/contact length, (ii) the rolling (separating) force using the average flow stress, and (iii) a check on whether the biting (self-gripping) condition is satisfied for mu = 0.25. (b) A process-selection question for biomedical implant machining demanding excellent surface integrity with NO thermal damage. Four explicit sub-parts: (i) suitability of USM, LBM and EDM; (ii) surface defects produced by each; (iii) the best process with justification; (iv) post-processing needed for the selected process. (c) A 5-job, 4-machine job shop. (i) Define the critical ratio, compute it for each job at t = 0, rank by urgency, name the most critical job and explain in two sentences why CR beats due date or processing time alone. (ii) Simulate SPT dispatching at every machine, draw the complete Gantt chart, and report C_j, F_j, L_j, T_j per job plus makespan, mean flow time, total tardiness, maximum tardiness and number of tardy jobs.

• Contact length L = √(RΔh) = 63.2 mm
• Rolling force F = Y L w ≈ 4.74 MN
• Biting condition μ ≥ tan α with α = 9.07°, tan α ≈ 0.16
• Conclusion that biting is satisfied (μ = 0.25 > 0.16, Δh_max = 25 mm)
• Suitability of the three processes: mechanical USM against thermal LBM and EDM
• Surface defects: recast layer, heat-affected zone and micro-cracks in LBM and EDM, embedded abrasive and chipping in USM
• USM selected as best, justified by no thermal damage, no recast layer and compressive residual stress
• Post-processing of USM: ultrasonic cleaning, polishing, passivation and inspection or sterilisation
• CR = (due date − now)/remaining work; values 1.333, 1.250, 1.500, 1.667, 1.600 with J₂ most critical
• SPT dispatching applied at every machine, including tie decisions at t = 23 and t = 33
• Gantt chart with completion times J₁ = 36, J₂ = 22, J₃ = 33, J₄ = 40, J₅ = 25
• Makespan 40, mean flow time 31.2, total tardiness 16, maximum tardiness 8, three tardy jobs

(a(i)) Orthogonal (Merchant) cutting analysis from Fc = 1500 N, Ft = 900 N, alpha = 10°, r = 0.4. Four quantities must be found: (I) shear angle phi, (II) friction angle beta, (III) coefficient of friction mu, (IV) shear force Fs from Merchant's circle relations. (a(ii)) Limit-gauge design for a hole specified as 25 +/- 0.01 mm using Taylor's principle, with gauge tolerance = 10% of work tolerance and a 10% wear allowance on the GO gauge only. Give the GO and NO-GO plug gauge sizes. (b) Two demands: define/explain the roles of manufacturing flexibility, lean practices and agility in an organisation; then explain HOW the three are complementary (not competing) approaches. The second half is the analytical core and carries the heavier weight. (c) Work measurement: (i) explain the conceptual basis of predetermined time standards (PTS) and identify the fundamental assumption that lets basic-motion times be pre-assigned independently of the task; (ii) give TWO specific situations where PTS is preferable to stopwatch time study for capacity planning and TWO where it is inappropriate.

• Shear angle φ = 22.9° from tan φ = r cos α/(1 − r sin α)
• Friction force 1146.8 N and normal force 1320.9 N resolved on the tool face
• β ≈ 41.0° and μ ≈ 0.868
• Shear force F_s ≈ 1030 N from F_c cos φ − F_t sin φ
• Work limits 25.01 and 24.99 mm, tolerance 0.02 mm
• Gauge tolerance 0.002 mm and wear allowance 0.002 mm on the GO gauge only
• GO gauge 24.992 to 24.994 mm
• NO-GO gauge 25.008 to 25.010 mm
• Flexibility defined as adaptability to variety at low cost and time
• Lean practices as waste removal with JIT, kaizen and pull
• Agility as strategic response to unpredictable change
• Complementarity: lean needs flexibility, agility rests on both, a decoupling point resolves conflicts
• Tasks broken into basic motions with tabulated times
• Assumption: motion time depends only on the motion, not on the task, and times are additive at a normal pace
• Two situations favouring PTS, such as a line that does not yet exist and high-volume short-cycle assembly
• Two inappropriate situations, such as long-cycle job-shop or mental or machine-paced work

(a) Limits-and-fits calculation for a 60 mm shaft in a hole with hole tolerance 0.030 mm, shaft tolerance 0.020 mm and a required minimum clearance of 0.010 mm. Three sub-parts: (i) maximum clearance, (ii) hole and shaft limits, (iii) allowance. (b) Facility layout for a firm making both standardised and customised products. (i) Compare product layout with process layout analytically on FIVE named criteria: material flow, flexibility, cost, demand variability and machine utilisation. (ii) Explain cellular layout and analyse both its benefits and the challenges of implementing it. (c) Continuous-review (Q, R) inventory system under stochastic demand. Four sub-parts: (i) EOQ; (ii) mean and standard deviation of demand during the lead time; (iii) safety stock and reorder point at a 95% service level (z = 1.645); (iv) total annual inventory cost (ordering + cycle holding + safety-stock holding) and the inflation of holding cost caused by the safety stock.

• Hole limits 60.000 and 60.030 mm (hole basis)
• Shaft limits 59.970 and 59.990 mm
• Maximum clearance 0.060 mm
• Allowance 0.010 mm
• Product layout versus process layout compared on all five named criteria
• Recommendation matching layout to standardised and customised products
• Cellular layout concept: part families, group technology, machine cells
• Cell advantages (set-up, WIP, lead time, quality) and challenges (family formation, duplication, inflexibility, re-layout cost)
• EOQ = 462 units
• Lead-time demand mean 96 and standard deviation 60 units
• Safety stock ≈ 99 units and reorder point ≈ 195 units
• Total cost ≈ ₹10 090 and holding cost inflated by 42.7% because of the safety stock