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Mechanical Engineering Essentials, Summaries of Analytical mechanics

University of Liverpool OnlineAnalytical mechanics

Key formulas and step-by-step solved problems in core mechanical engineering topics for quick learning, revision, and academic support.

Typology: Summaries

2022/2023

Available from 07/12/2025

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Mechanical Engineering Formulas & Solved Problems
1. Engineering Mathematics
Formulas:
- Derivative: d/dx(x^n) = nx^(n-1)
- Integral: Integralx^n dx = x^(n+1)/(n+1) + C
- Unit Conversion: 1 inch = 25.4 mm, 1 ft = 0.3048 m
Solved Problems:
Q1: Differentiate f(x) = 3x^4 + 2x^2 - 5x + 7
A1: f'(x) = 12x^3 + 4x - 5
Q2: Convert 10 inches to mm
A2: 10 * 25.4 = 254 mm
2. Statics
Formulas:
- Resultant Force: F = sqrt(Fx^2 + Fy^2)
- Moment: M = F * d
- Equilibrium: SumFx = 0, SumFy = 0, SumM = 0
Solved Problem:
Q: A 200 N force is applied 0.5 m from a pivot. Find the moment.
A: M = 200 * 0.5 = 100 Nm
3. Dynamics
Formulas:
- Newton's Second Law: F = ma
- Kinematic Equations:
pf3
pf4

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Download Mechanical Engineering Essentials and more Summaries Analytical mechanics in PDF only on Docsity!

1. Engineering Mathematics

Formulas:

  • Derivative: d/dx(x^n) = nx^(n-1)
  • Integral: Integralx^n dx = x^(n+1)/(n+1) + C
  • Unit Conversion: 1 inch = 25.4 mm, 1 ft = 0.3048 m

Solved Problems: Q1: Differentiate f(x) = 3x^4 + 2x^2 - 5x + 7 A1: f'(x) = 12x^3 + 4x - 5

Q2: Convert 10 inches to mm A2: 10 * 25.4 = 254 mm

2. Statics

Formulas:

  • Resultant Force: F = sqrt(Fx^2 + Fy^2)
  • Moment: M = F * d
  • Equilibrium: SumFx = 0, SumFy = 0, SumM = 0

Solved Problem: Q: A 200 N force is applied 0.5 m from a pivot. Find the moment. A: M = 200 * 0.5 = 100 Nm

3. Dynamics

Formulas:

  • Newton's Second Law: F = ma
  • Kinematic Equations:

v = u + at s = ut + 0.5at^ v^2 = u^2 + 2as

Solved Problem: Q: A 10 kg object accelerates at 2 m/s^2. Find the force. A: F = 10 * 2 = 20 N

4. Thermodynamics

Formulas:

  • First Law: Q = DeltaU + W
  • Heat Transfer: Q = mcDeltaT

Solved Problem: Q: Find heat to raise 2 kg water from 20 degreesC to 100 degreesC (c = 4186 J/kg degreesC) A: Q = 2 * 4186 * (100 - 20) = 669,760 J

5. Fluid Mechanics

Formulas:

  • Bernoulli's Equation: P + 0.5rhov^2 + rhogh = constant
  • Flow Rate: Q = A * v

Solved Problem: Q: Find flow rate through a 0.1 m^2 pipe at 3 m/s. A: Q = 0.1 * 3 = 0.3 m^3/s

6. Mechanics of Materials

Formulas:

  • Gear Ratio: GR = Teeth of Driven / Teeth of Driver
  • Efficiency: Eff = (Output Power / Input Power) * 100

Solved Problem: Q: Gear A has 20 teeth, Gear B has 40. Find gear ratio. A: GR = 40 / 20 = 2

10. Common Engineering Units

Length:

  • 1 inch = 25.4 mm
  • 1 ft = 0.3048 m

Force:

  • 1 N = 0.2248 lbf
  • 1 lbf = 4.448 N

Pressure:

  • 1 atm = 101.3 kPa
  • 1 bar = 100,000 Pa
  • 1 psi = 6,894.76 Pa

Energy:

  • 1 J = 1 Nm
  • 1 cal = 4.184 J