Mechanics of Materials

Mechanics of Materials (also called Strength of Materials) studies how solid materials deform and fail under various loads. This is essential for designing safe structures and machines.

Stress and Strain

Stress (σ)

Stress is internal force per unit area within a material:

σ = F/A

Where:
- σ = stress (Pa or N/m², often MPa)
- F = force (N)
- A = cross-sectional area (m²)

Types of Stress:

Strain (ε)

Strain is deformation relative to original size (dimensionless):

ε = ΔL/L₀

Where:
- ε = strain (no units, often given as %)
- ΔL = change in length (m)
- L₀ = original length (m)

Example: Stretching a Wire

A 2 m steel wire with cross-section 5 mm² is pulled with 1000 N. It stretches by 2 mm. Find stress and strain.

Stress:
σ = F/A = 1000 N / (5 × 10⁻⁶ m²) = 200 × 10⁶ Pa = 200 MPa

Strain:
ε = ΔL/L₀ = 0.002 m / 2 m = 0.001 = 0.1%

Stress-Strain Relationship

Hooke's Law (Elastic Region)

For small deformations, stress is proportional to strain:

σ = Eε

Where:
- E = Young's Modulus (Modulus of Elasticity) (Pa)
- σ = stress (Pa)
- ε = strain (dimensionless)

Rearranging for elongation:

ΔL = (FL₀)/(AE)

Young's Modulus Values

Higher E means stiffer material (less deformation under load).

Stress-Strain Curve

Stress (σ)
  ↑
  |         Fracture ×
  |                ,'
  |              ,'  Necking
  |            ,'
  |    Yield ×'______ Plastic Region
  |         /'
  |       /'
  |     /'
  |   /' Elastic Region (Hooke's Law)
  | /'
  |/________________________→ Strain (ε)
   0

Key Points:

1. Elastic Region: Material returns to original shape when load removed
2. Yield Point: Permanent deformation begins
3. Plastic Region: Material permanently deformed
4. Ultimate Strength: Maximum stress material can withstand
5. Fracture: Material breaks

Material Properties

Examples:

• Ductile: Steel, aluminum (bend before breaking)
• Brittle: Glass, concrete, cast iron (break suddenly)

Types of Loading

Axial Loading (Tension/Compression)

     ← F        F →
    =========|=========
    
σ = F/A
ΔL = (FL)/(AE)

Shear Stress

    ________
    |  τ →  |
    |_______|
       ↑ F

τ = F/A

Where:
- τ = shear stress (Pa)
- F = shear force (N)
- A = area parallel to force (m²)

Shear Strain:

γ = Gθ

Where:
- γ = shear strain
- G = shear modulus
- θ = angular deformation

Shear Modulus:

G = E / [2(1 + ν)]

Where ν = Poisson's ratio ≈ 0.3 for metals

Example: Bolt in Shear

A bolt (diameter 10 mm) connects two plates. Shear force is 5000 N. Find shear stress.

A = π d²/4 = π(0.01)²/4 = 7.85 × 10⁻⁵ m²
τ = F/A = 5000 / (7.85 × 10⁻⁵) = 63.7 × 10⁶ Pa = 63.7 MPa

If bolt is in double shear (shear plane crosses bolt twice), divide stress by 2.

Beam Bending

When a beam bends, one side is in tension, the other in compression.

Bending Stress

σ = My/I

Where:
- M = bending moment (N⋅m)
- y = distance from neutral axis (m)
- I = second moment of area (m⁴)

Maximum stress occurs at outer fibers (largest y).

Second Moment of Area (I)

Measure of resistance to bending. Larger I means less bending.

Example: Rectangular Beam

A steel beam (100 mm wide × 200 mm tall) has bending moment M = 10 kN⋅m. Find maximum stress.

b = 0.1 m, h = 0.2 m
I = bh³/12 = 0.1(0.2)³/12 = 6.67 × 10⁻⁵ m⁴

Maximum stress at top/bottom (y = h/2 = 0.1 m):
σ_max = My/I = (10,000 × 0.1) / (6.67 × 10⁻⁵)
σ_max = 15 × 10⁶ Pa = 15 MPa

Section Modulus

For convenience, define section modulus:

S = I/c

Where c = distance to outer fiber (usually h/2)

Then: σ_max = M/S

Torsion (Twisting)

Torsional Shear Stress

For circular shafts:

τ = Tr/J

Where:
- T = torque (N⋅m)
- r = radius from center (m)
- J = polar moment of inertia (m⁴)

Maximum stress at outer surface (r = R).

Polar Moment of Inertia

Angle of Twist

φ = TL/(GJ)

Where:
- φ = angle of twist (radians)
- T = torque (N⋅m)
- L = length (m)
- G = shear modulus (Pa)
- J = polar moment of inertia (m⁴)

Example: Drive Shaft

A solid steel shaft (diameter 50 mm, length 1 m) transmits 1000 N⋅m torque. Find maximum shear stress and angle of twist.

R = 0.025 m
J = πR⁴/2 = π(0.025)⁴/2 = 6.14 × 10⁻⁷ m⁴

Maximum shear stress:
τ_max = TR/J = 1000 × 0.025 / (6.14 × 10⁻⁷)
τ_max = 40.7 × 10⁶ Pa = 40.7 MPa

Angle of twist (G = 80 GPa for steel):
φ = TL/(GJ) = 1000 × 1 / (80 × 10⁹ × 6.14 × 10⁻⁷)
φ = 0.0203 rad = 1.16°

Deflection of Beams

How much does a beam bend under load?

Common Beam Deflections

Simply supported beam, center load P:

    P
    ↓
  ===*===
  ↑     ↑
  
  L
  
δ_max = PL³/(48EI)  (at center)

Simply supported beam, uniform load w:

↓↓↓↓↓↓↓
=========
↑       ↑

δ_max = 5wL⁴/(384EI)  (at center)

Cantilever beam, end load P:

    P
    ↓
┤====*

L

δ_max = PL³/(3EI)  (at end)

Example: Beam Deflection

A steel beam (L = 4 m, I = 1 × 10⁻⁵ m⁴, E = 200 GPa) supports 5000 N at center. Find deflection.

δ = PL³/(48EI)
δ = 5000 × 4³ / (48 × 200 × 10⁹ × 1 × 10⁻⁵)
δ = 320,000 / (96 × 10⁶)
δ = 0.00333 m = 3.33 mm

Combined Loading

Real components often experience multiple types of stress simultaneously.

Principal Stresses

When normal and shear stresses act together, use Mohr's circle or formulas:

σ_1, σ_2 = (σ_x + σ_y)/2 ± √[((σ_x - σ_y)/2)² + τ_xy²]

Where:
- σ_1, σ_2 = principal stresses (max and min)
- σ_x, σ_y = normal stresses
- τ_xy = shear stress

Failure Theories

How do we predict failure under combined loading?

Maximum Shear Stress Theory (Tresca):

Failure when: τ_max = σ_y / 2

Distortion Energy Theory (von Mises):

Failure when: σ_equivalent = √(σ₁² - σ₁σ₂ + σ₂²) ≥ σ_y

Von Mises theory is more accurate for ductile materials.

Factor of Safety

Design with margin above expected loads:

Factor of Safety = Strength / Applied Stress

FOS = σ_failure / σ_actual

Typical values:

• Aerospace: 1.5-2.0 (weight-critical)
• Automotive: 2.0-3.0
• Buildings: 3.0-4.0
• Pressure vessels: 4.0-8.0
• Unknown loads/brittle: 5.0+

Example: Safety Factor

A steel rod (σ_y = 250 MPa) experiences stress of 100 MPa. What is the factor of safety?

FOS = σ_y / σ_actual = 250 / 100 = 2.5

This means the rod can handle 2.5× the current load before yielding.

Practice Problems

Problem 1: Stress and Strain

A 1 m aluminum bar (E = 70 GPa, A = 100 mm²) is stretched by 1 mm. Find the force required.

<details> <summary>Solution</summary>

ε = ΔL/L₀ = 0.001/1 = 0.001
σ = Eε = 70 × 10⁹ × 0.001 = 70 × 10⁶ Pa = 70 MPa
F = σA = 70 × 10⁶ × 100 × 10⁻⁶ = 7000 N = 7 kN

</details>

Problem 2: Beam Bending

A beam (I = 5 × 10⁻⁶ m⁴) has maximum bending moment 2000 N⋅m. Height is 200 mm. Find maximum stress.

<details> <summary>Solution</summary>

y_max = h/2 = 0.1 m
σ_max = My/I = 2000 × 0.1 / (5 × 10⁻⁶) = 40 × 10⁶ Pa = 40 MPa

</details>

Problem 3: Torsion

A hollow shaft (outer diameter 40 mm, inner diameter 30 mm) transmits 500 N⋅m. Find maximum shear stress.

<details> <summary>Solution</summary>

R = 0.02 m, r = 0.015 m
J = π(R⁴ - r⁴)/2 = π(0.02⁴ - 0.015⁴)/2 = 1.23 × 10⁻⁷ m⁴
τ_max = TR/J = 500 × 0.02 / (1.23 × 10⁻⁷) = 81.3 MPa

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Key Takeaways

✓ Stress: Force per unit area (σ = F/A)
✓ Strain: Deformation ratio (ε = ΔL/L)
✓ Hooke's Law: σ = Eε (elastic region)
✓ Bending: σ = My/I (maximum at outer fibers)
✓ Torsion: τ = Tr/J (maximum at outer surface)
✓ Factor of Safety: Design margin above expected loads

Next Steps

With solid mechanics covered, explore thermal systems:

• Chapter 05: Thermodynamics: energy, heat, and power cycles
• Practice more stress analysis problems
• Learn finite element analysis (FEA) for complex geometries

Pro Tip: Always check units! MPa and Pa differ by 10⁶. Missing this causes big errors!