ANALYSIS OF BARS OF COMPOSITE SECTIONS

ANALYSIS OF BARS OF COMPOSITE SECTIONS

A composite bar is made up of two or more bars of different materials, of equal lengths, fixed rigidly so that they behave as one single unit. When such a bar is subjected to axial tensile or compressive load, the load is shared between each material according to its stiffness.

Important Points

  1. Strain (extension per unit length) in each bar is the same, because bars are connected rigidly.
  2. Total load on the composite bar = Sum of loads carried by each material.

Consider a Composite Bar

Let the bar consist of two materials:

  • L = Length of both bars
  • A₁, A₂ = Cross-sectional areas of bar 1 and bar 2
  • E₁, E₂ = Young’s Modulus of materials 1 and 2
  • P = Total load
  • P₁, P₂ = Load shared by bar 1 and bar 2
  • σ₁, σ₂ = Stress in bar 1 and bar 2

Total Load

P = P₁ + P₂     ...(i)

Stress in Bar 1

σ₁ = P₁ / A₁

Stress in Bar 2

σ₂ = P₂ / A₂

Strain Condition

Because bars deform equally:

σ₁ / E₁ = σ₂ / E₂

Modular Ratio

Modular Ratio = E₁ / E₂

Load Sharing in Steel Rod and Copper Tube

A steel rod (diameter 3 cm) is centrally enclosed in a hollow copper tube. Both are of equal length (15 cm) and subjected to an axial pull of 45,000 N. Young’s modulus: Eₛ = 2.1 × 10⁵ N/mm², E꜀ = 1.1 × 10⁵ N/mm².

Step 1: Find Areas

Steel rod area:

Aₛ = π/4 × (30)² = 706.86 mm²

Copper tube area:

External diameter = 50 mm Internal diameter = 40 mm

A꜀ = π/4 × (50² − 40²) = 706.86 mm²

Interestingly, both areas are equal.

Step 2: Use Strain Equality

σₛ / Eₛ = σ꜀ / E꜀

Substitute values:

σ꜀ = (E꜀ / Eₛ) × σₛ = (1.1 × 10⁵ / 2.1 × 10⁵) × σₛ

σ꜀ = 0.523 σₛ

Step 3: Total Load Condition

P = Load on steel + Load on copper

45,000 = σₛ × 706.86 + σ꜀ × 706.86

Substitute σ꜀ = 0.523σₛ:

45,000 = σₛ × 706.86 (1 + 0.523)

45,000 = σₛ × 706.86 × 1.523

σₛ = 41.77 N/mm²

σ꜀ = 21.88 N/mm²

Step 4: Load Carried by Each Material

Load on steel:

Pₛ = σₛ × Aₛ = 41.77 × 706.86 = 29,525.5 N

Load on copper:

P꜀ = 45,000 − 29,525.5 = 15,474.5 N

✔ Final Answers: Stress in steel = 41.77 N/mm² Stress in copper = 21.88 N/mm² Load carried by steel = 29.53 kN Load carried by copper = 15.47 kN

Problem 1.20 – Composite Tube (Steel + Brass)

A compound tube consists of:

  • Steel tube: 140 mm internal diameter, 160 mm external
  • Brass tube: 160 mm internal, 180 mm external

Axial load = 900 kN Young’s modulus: Eₛ = 2 × 10⁵ N/mm² Eᵦ = 1 × 10⁵ N/mm²

Step 1: Areas

Steel tube area:

Aₛ = π/4 × (160² – 140²) = 4712.4 mm²

Brass tube area:

Aᵦ = π/4 × (180² – 160²) = 5340.7 mm²

Step 2: Strain Equality

σₛ / Eₛ = σᵦ / Eᵦ

σᵦ = 2 σₛ

Step 3: Total Load Equation

900,000 = σₛAₛ + σᵦAᵦ

900,000 = σₛ × 4712.4 + 2σₛ × 5340.7

900,000 = σₛ (4712.4 + 10,681.4)

900,000 = σₛ (14,765.5)

σₛ = 60.95 N/mm²

σᵦ = 121.9 N/mm²

Step 4: Individual Loads

Load on brass:

Pᵦ = σᵦAᵦ = 121.9 × 5340.7 = 651,110 N = 651.1 kN

Load on steel:

Pₛ = 900 − 651.1 = 248.9 kN

Extension of Composite Tube

Extension, dL = (σₛ × L) / Eₛ

= (60.95 × 140) / (2 × 10⁵)

= 0.0427 mm

✔ Final Answers: Stress in steel = 60.95 N/mm² Stress in brass = 121.9 N/mm² Load carried by brass = 651.1 kN Load carried by steel = 248.9 kN Extension = 0.0427 mm

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