Engineering Mechanics and Solid Mechanics (GATE CE)

GATE CE syllabus for engineering mechanics and solid mechanics, with the key formulas for equilibrium, SFD and BMD, stress and strain, elastic constants, Mohr's circle, bending, torsion and deflection, and solved examples.

Syllabus

Engineering mechanics: force systems, free-body diagrams, equilibrium, friction, centroid and moment of inertia, kinematics and dynamics, virtual work.
Solid mechanics: shear force and bending moment diagrams, stress and strain, Hooke's law, elastic constants, principal stresses and Mohr's circle, theories of failure, bending and shear stresses, torsion, thin cylinders, deflection of beams, buckling of columns.

Key concepts and formulas

Equilibrium. In two dimensions, ΣF_x = 0, ΣF_y = 0 and ΣM = 0. Limiting friction F = μN.

Shear force and bending moment. dM/dx = V and dV/dx = −w (load intensity). The bending moment is maximum where the shear force is zero.

Simply supported beam, span L	Max bending moment	Max deflection
Central point load P	PL/4	PL³/(48EI)
Uniformly distributed load w	wL²/8	5wL⁴/(384EI)

Cantilever, length L	Max bending moment	Max deflection
Point load P at the free end	PL	PL³/(3EI)
Uniformly distributed load w	wL²/2	wL⁴/(8EI)

Stress, strain and elastic constants. σ = P/A, ε = δL/L, σ = Eε, δL = PL/(AE). The elastic constants are related by E = 2G(1 + μ) = 3K(1 − 2μ), where μ is Poisson's ratio.

Principal stresses for a plane stress state (σ_x, σ_y, τ_xy):

σ_1,2 = (σ_x + σ_y)/2 ± √(((σ_x − σ_y)/2)² + τ_xy²), and the maximum in-plane shear stress is the radius of Mohr's circle.

Bending equation: M/I = σ/y = E/R. The section modulus is Z = I/y_max. For a rectangle b × d: I = bd³/12, Z = bd²/6. The maximum shear stress in a rectangular section is 1.5 times the average.

Torsion: T/J = τ/r = Gθ/L. For a solid shaft of diameter d: J = πd⁴/32, so τ_max = 16T/(πd³).

Thin cylinder (pressure p, diameter d, thickness t): hoop stress pd/(2t), longitudinal stress pd/(4t).

Euler buckling load: P_cr = π²EI/L_{e</sub², with effective length L_e = L (both ends pinned), 2L (one end fixed, one free), L/√2 (fixed-pinned) and L/2 (both ends fixed).}

Worked example

A simply supported steel beam of span 6 m carries a UDL of 20 kN/m. Its rectangular section is 200 mm wide and 400 mm deep, and E = 200 GPa. Find the maximum bending moment, the maximum bending stress and the maximum deflection.

Python

Open in playground

import math

L, w = 6.0, 20e3          # m, N/m
b, d, E = 0.2, 0.4, 200e9
I = b * d ** 3 / 12
M = w * L ** 2 / 8
sigma = M * (d / 2) / I
delta = 5 * w * L ** 4 / (384 * E * I)
print(f"I = {I:.6e} m^4")
print(f"M_max = {M / 1e3:.1f} kN m")
print(f"sigma_max = {sigma / 1e6:.3f} MPa")
print(f"deflection = {delta * 1000:.3f} mm")

# Principal stresses for sigma_x = 80, sigma_y = -20, tau_xy = 30 (MPa)
sx, sy, t = 80, -20, 30
c, r = (sx + sy) / 2, math.hypot((sx - sy) / 2, t)
print(f"sigma1 = {c + r:.2f} MPa, sigma2 = {c - r:.2f} MPa, tau_max = {r:.2f} MPa")

Output

I = 1.066667e-03 m^4
M_max = 90.0 kN m
sigma_max = 16.875 MPa
deflection = 1.582 mm
sigma1 = 88.31 MPa, sigma2 = -28.31 MPa, tau_max = 58.31 MPa