Engineering Mechanics and Mechanics of Materials

GATE ME syllabus for engineering mechanics and mechanics of materials, with equilibrium, friction, work-energy and impulse-momentum, stress and strain, Mohr's circle, bending, torsion, deflection, columns, strain energy and thermal stress, with solved examples.

Syllabus

Engineering mechanics: free-body diagrams and equilibrium, friction, trusses and frames, virtual work, kinematics and dynamics of rigid bodies in plane motion, impulse and momentum, energy methods.
Mechanics of materials: stress and strain, elastic constants, Mohr's circle, thin cylinders, SFD and BMD, bending and shear stresses, shear centre, deflection of beams, torsion, Euler columns, energy methods, thermal stresses, strain gauges, material testing.

Key concepts and formulas

Dynamics.

Work-energy: work done = change in kinetic energy. Rolling disc: KE = ½mv² + ½Iω², with v = rω.
Impulse-momentum: F Δt = m Δv. Coefficient of restitution e = (relative velocity of separation)/(relative velocity of approach).
Moments of inertia about the centre: rod mL²/12, solid disc mr²/2, ring mr², solid sphere (2/5)mr². Parallel axis: I = I_G + md².

Friction. F ≤ μN; angle of friction tan φ = μ. A block on an incline slides when tan θ > μ. Belt friction: T₁/T₂ = e^μθ.

Stress and strain. σ = Eε, E = 2G(1 + ν) = 3K(1 − 2ν). Volumetric strain of a rod: ε_v = ε(1 − 2ν). Thermal stress in a fully restrained bar: σ = EαΔT.

Mohr's circle. Centre (σ_x + σ_y)/2, radius √(((σ_x − σ_y)/2)² + τ_xy²) = maximum in-plane shear stress. Principal planes are at 2θ_p = tan⁻¹(2τ_xy/(σ_x − σ_y)).

Bending, torsion and deflection. M/I = σ/y = E/R; T/J = τ/r = Gθ/L. Power transmitted P = 2πNT/60 (N in rpm). Hollow shaft J = π(D⁴ − d⁴)/32. Deflections: cantilever with end load PL³/(3EI); simply supported beam with central load PL³/(48EI) and with UDL 5wL⁴/(384EI).

Columns. Euler load P_cr = π²EI/L_{e</sub²; slenderness ratio L_e/k with I = Ak².}

Strain energy. Axial U = P²L/(2AE); resilience per unit volume σ²/(2E). A load dropped suddenly (h = 0) produces twice the stress of the same load applied gradually.

Springs. Series: 1/k = 1/k₁ + 1/k₂; parallel: k = k₁ + k₂.

Worked example

A solid shaft transmits 50 kW at 300 rpm, with an allowable shear stress of 60 MPa. Find the torque and the minimum diameter. Then find the principal stresses for σ_x = 100 MPa, σ_y = 40 MPa and τ_xy = 40 MPa.

Python

Open in playground

import math

P, N, tau = 50e3, 300, 60e6
T = P * 60 / (2 * math.pi * N)
d = (16 * T / (math.pi * tau)) ** (1 / 3)
print(f"T = {T:.2f} N m, d_min = {d * 1000:.2f} mm")
sx, sy, t = 100, 40, 40
c, r = (sx + sy) / 2, math.hypot((sx - sy) / 2, t)
print(f"sigma1 = {c + r:.1f} MPa, sigma2 = {c - r:.1f} MPa, tau_max(in-plane) = {r:.1f} MPa")
print(f"thermal stress, steel, 50 K: {200e9 * 12e-6 * 50 / 1e6:.0f} MPa")

Output

T = 1591.55 N m, d_min = 51.31 mm
sigma1 = 120.0 MPa, sigma2 = 20.0 MPa, tau_max(in-plane) = 50.0 MPa
thermal stress, steel, 50 K: 120 MPa