Question

A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 0.520 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 5.00×104 N at its ends. Find the length L of the nickel rod if the elongations of the two rods are equal. What is the stress in the brass rod? What is the stress in the nickel rod? What is the strain in the brass rod? What is the strain in the nickel rod?

Answer 1

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂   ⇒  P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒  L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒  L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒  L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁    ⇒   ε₁ = σ₁ / E₁

⇒   ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂    ⇒   ε₂ = σ₂ / E₂

⇒   ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875