Mechanics of Materials is a formula-driven, high-scoring subject in Part B2 (Mechanical) of the GATE Robotics and Automation (RA) 2027 paper. Once you can compute a stress, apply Hooke’s law, read a Mohr’s circle, and use the torsion and beam-bending equations, most questions become one- or two-step calculations. This guide reviews stress and strain, principal stresses, torsion, beam bending and column buckling, and then works through five GATE-style problems in full.
TABLE OF CONTENTS
Stress, strain & Hooke’s law
Normal stress is force per unit area, σ = P/A, and normal strain is the fractional change in length, ε = δ/L. Within the elastic limit they are related by Hooke’s law, σ = Eε, where E is Young’s modulus. Combining these gives the axial deformation formula δ = PL/(AE), one of the most-used results in the subject. The three elastic constants E, G (shear modulus) and ν (Poisson’s ratio) are linked by E = 2G(1 + ν).
Principal stresses & Mohr’s circle
Under combined loading, the principal stresses are the maximum and minimum normal stresses, acting on planes free of shear. They and the maximum in-plane shear stress come from the plane-stress transformation, which Mohr’s circle represents graphically: the centre sits at the average normal stress and the radius equals the maximum shear stress.
Torsion of shafts
For a circular shaft carrying torque T, the torsion equation links shear stress, torque and twist: T/J = τ/r = Gθ/L. The shear stress is maximum at the outer radius. For a solid shaft the polar moment of area is J = πd⁴/32, which gives the handy result τmax = 16T/(πd³). The angle of twist is θ = TL/(GJ).
Beam bending (SFD/BMD)
Transverse loads produce a shear force and bending moment that vary along the beam — drawn as the SFD and BMD. The flexure formula M/I = σ/y = E/R relates bending moment to bending stress. Two standard simply-supported results are worth memorising: a central point load W gives Mmax = WL/4, and a uniformly distributed load w gives Mmax = wL²/8 with maximum shear wL/2 at the supports.
Figure 1. Simply supported beam under a UDL: maximum bending moment wL²/8 at mid-span.
Columns & Euler buckling
Slender columns fail by buckling before reaching the yield stress. The Euler critical load is Pcr = π²EI/Le², where Le is the effective length set by the end conditions: Le = L for pinned–pinned, 2L for one end fixed and one free, and 0.5L for both ends fixed. Buckling depends on the least moment of inertia of the cross-section.
Worked examples (GATE-style)
Example 1 — axial stress and elongation
A steel bar carries an axial tensile load of 50 kN. Its length is 2 m, cross-sectional area 500 mm², and E = 200 GPa. Find the axial stress and the elongation.
Solution. σ = P/A = 50×10³ / (500×10−6) = 100 MPa.
δ = PL/(AE) = (50×10³×2) / (500×10−6×200×10&sup9;) = 10⁵/10⁸ = 1 mm.
Example 2 — principal stresses (Mohr’s circle)
At a point, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. Find the principal stresses and the maximum shear stress.
Solution. Centre = (80+20)/2 = 50 MPa. Radius = √[((80−20)/2)² + 30²] = √(30² + 30²) = √1800 = 42.43 MPa.
σ1 = 50 + 42.43 = 92.43 MPa; σ2 = 50 − 42.43 = 7.57 MPa.
τmax = radius = 42.43 MPa.
Example 3 — torsional shear stress
A solid circular shaft of 50 mm diameter transmits a torque of 1 kN·m. Find the maximum shear stress.
Solution. For a solid shaft, τmax = 16T/(πd³) = 16×1000 / (π×0.05³).
= 16000 / (π×1.25×10−4) = 16000 / (3.927×10−4) = 40.74 MPa.
Example 4 — simply supported beam under UDL
A simply supported beam of span 4 m carries a uniformly distributed load of 10 kN/m. Find the maximum bending moment and maximum shear force.
Solution. Mmax = wL²/8 = 10×4²/8 = 160/8 = 20 kN·m (at mid-span).
Maximum shear force = wL/2 = 10×4/2 = 20 kN (at the supports).
Example 5 — Euler buckling load
A pinned–pinned column is 3 m long, with E = 200 GPa and least moment of inertia I = 4×10⁶ mm⁴. Find the Euler critical buckling load.
Solution. For pinned ends Le = L = 3 m, and I = 4×10⁶ mm⁴ = 4×10−6 m⁴.
Pcr = π²EI/L² = π²×200×10⁹×4×10−6 / 3² = π²×800000/9.
= 7.896×10⁶/9 = 877.3 kN.
Key formulas
FORMULA SHEET
Axial: σ = P/A, δ = PL/(AE), E = 2G(1+ν)
Principal: σ1,2 = (σx+σy)/2 ± √[((σx−σy)/2)²+τxy²]
Torsion: T/J = τ/r = Gθ/L, τmax = 16T/(πd³)
Bending: M/I = σ/y = E/R
SS beam UDL: Mmax = wL²/8, Vmax = wL/2
Euler: Pcr = π²EI/Le²
Common mistakes to avoid
- Unit slips. Convert mm to m and kN to N consistently — a 10⁶ or 10⁹ error is the most common way to lose a Mechanics of Materials mark.
- Using diameter for radius in the torsion formula, or forgetting that τmax = 16T/(πd³) already uses diameter.
- Wrong effective length in Euler buckling — fixed–free is 2L, not L, which changes Pcr by a factor of four.
- Taking the larger moment of inertia: buckling always uses the least I of the section.
- Confusing WL/4 (point load) with wL²/8 (UDL) for the maximum bending moment of a simply supported beam.
GATE ROBOTICS & AUTOMATION 2027
Master Mechanics of Materials for GATE RA
Get in-depth lectures on stress–strain, Mohr’s circle, torsion, beams and columns, with PYQs and full-length tests, taught by Piyush Wairale (IIT Madras).
Frequently asked questions
What is the difference between stress and strain?
Stress is the internal resisting force per unit area (units: Pa or MPa), while strain is the resulting fractional deformation (dimensionless). Within the elastic limit they are proportional through Young’s modulus, σ = Eε.
Why does buckling use the least moment of inertia?
A column buckles about the axis of least bending stiffness, which corresponds to the smallest moment of inertia of the cross-section. Using any larger value would overestimate the critical load and be unsafe.
Where is the bending moment maximum in a simply supported beam?
For a central point load or a symmetric uniformly distributed load, the maximum bending moment occurs at mid-span — WL/4 for a central point load and wL²/8 for a UDL. The shear force is maximum at the supports.
This solved-problems guide is part of the complete GATE RA 2027 Syllabus overview. For the full topic breakdown, see the Mechanics of Materials syllabus guide.
Recent Post
The complete GATE General Aptitude 2027 guide: official GA syllabus, exam pattern, section-wise weightage, a 4-week preparation plan and links to all 16 free topic-wise guides for Verbal, Quantitative, Analytical and Spatial Aptitude, by Piyush Wairale (IIT Madras).
Paper folding, cutting, pattern completion, figure series and embedded figures for GATE Spatial Aptitude 2027: the unfold-in-reverse method with the 2^n hole rule, sketch-and-mirror for cuts, rotation and count tracking for series, 8 solved PYQs, three diagrams, practice set and rule sheet by Piyush Wairale (IIT Madras).
Dice, cube nets, painted cubes, cube counting, assembling and grouping for GATE Spatial Aptitude 2027: the two-views rule, the 11 nets of a cube, the 8 / 12(n-2) / 6(n-2)^2 / (n-2)^3 formulas, Euler's formula, four diagrams, 8 solved PYQs and GATE-style questions, a practice set and a rule sheet by Piyush Wairale (IIT Madras).
Master 2D transformations for GATE Spatial Aptitude 2027: rotation by 90, 180 and 270 degrees, mirror and water images, symmetric letters and digits, the clock mirror rule, combined transformations and shape-matching MCQs, with 8 solved questions, four diagrams, a practice set and a one-page rule sheet by Piyush Wairale (IIT Madras).
Solve Puzzles, Seating Arrangement, Blood Relations and Direction Sense for GATE General Aptitude 2027: circular and linear arrangement method, family-tree decoding, coordinate method for directions, diagrams, 8 solved PYQs, practice set and rule sheet by Piyush Wairale (IIT Madras).
Crack Analogy, Series and Numerical Relations for GATE General Aptitude 2027: number and letter series methods, word and number analogies, coding-decoding, grid and operator puzzles, 8 solved PYQs, practice set and rule sheet by Piyush Wairale (IIT Madras).
Learn Daily, Wherever You Are
Free lectures, exam updates, PYQ discussions, and job alerts — delivered through our YouTube channel and Telegram communities.
