Question

A steel rod 0.450 m long and an aluminum rod 0.250 m long, both with the same diameter, are placed end to end between rigid supports with no initial stress in the rods. The temperature of the rods is now raised by 60.0 ∘C.

What is the stress in each rod? (Hint: The length of the combined rod remains the same, but the lengths of the individual rods do not.)

Psteel,Paluminium = ?

Answer 1

First of all, write down the Bulk modulus of elasticity for steel and aluminium.

Please consult your text book to find out these data.

These are -

E₁ = 200 GPa for steel
E₂ = 69 GPa for aluminum

Also -

Coefficient of linear thermal expansion -

α₁ = 12×10⁻⁶ °C⁻¹ for steel

α₂ = 23×10⁻⁶ °C⁻¹ for aluminum

Each rod would undergo a tensile strain due to temperature of ∆T = 60.0 deg C:

ε₁ = α₁∙∆T = 12×10⁻⁶ °C⁻¹ ∙ 60°C = 7.2×10⁻⁴

ε₂ = α₂∙∆T = 23×10⁻⁶ °C⁻¹ ∙ 60°C = 1.38×10⁻³

Therefore, the total elongation of the composite rod would be,

∆L = L₁∙ε₁ + L₂∙ε₂ = 0.45 m∙72×10⁻⁵ + 0.25∙138×10⁻⁵ = 6.69 ×10⁻⁴ m

Since the rods are fitted in rigid gap we need compressive strain to keep the total length constant. The strain of reach rod is given by:

ε₁' = σ₁/E₁

ε₂' = σ₂/E₂

As mentioned in the problem, the total length of the composite rod keeps constant, these strains might differ from the thermal strain due to the different Young's modules (the steel rod is harder to compress).

In other words the length of the rods might change but the total length does not.

Since total length of the rod is constant, the magnitude of the total thermal expansion of two rods equals the magnitude of the mechanical compression, means,

∆L = ∆L' = L₁∙ε₁' + L₂∙ε₂'

In statical equilibrium the force acting along the composite rode are constant. Since both rods have the same diameter the stress is also the same:

σ₁ = σ₂

=> ε₁'∙E₁ = ε₂'∙E₂

=> ε₂ = ε₁'∙E₁/E₂

Therefore,

∆L = L₁∙ε₁' + L₂∙ε₁'∙E₁/E₂

=> ε₁' = ∆L /( L₁ + L₂∙E₁/E₂)

put the values -

ε₁' = 6.69×10⁻⁴m /( 0.45 m + 0.25 m∙200GPa/69GPa)

= 5.69 × 10⁻⁴

So the stress in each of the rod will be -

σ = σ₁ = 5.69×10⁻⁴ ∙ 200GPa = 113.8 MPa

Therefore, our answers are -

Psteel = Paluminium = 113.8 MPa (Answer)