Answer:
ΔL = 1.68 mm
σ = 84 MPa
Explanation:
Thermal expansion is:
ΔL = α ΔT L
Thermal stress is:
σ = α ΔT E
Given:
α = 1.2×10⁻⁵ /°C
E = 1.0×10⁵ MPa
ΔT = 80°C − 10°C = 70°C
L = 2 m
ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)
ΔL = 0.00168 m
ΔL = 1.68 mm
σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)
σ = 84 MPa
The expansion of the rod is 0.00168 meters and the stress induced is 84MPa
Linear thermal expansion, ΔL = α(ΔT)L
Length of rod, L = 2m
Change in temperature, ΔT = (T2 - T1) = (80°C - 10°C) = 70°C
Coefficient of expansion, α = 0.000012
Plugging values into the formula :
ΔL = 0.000012 × 70 × 2 = 0.00168 m
The stress induced, σ; if expansion is prevented is defined as : α(ΔT)E ; where E = modulus of elasticity :
σ = 0.000012 × 70 × 1.0 x 10^5
σ = 0.000012 × 70 × 100000
σ = 84 MPa
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