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
