ΔL = 1.504 – 1.500 = 0.004 m ΔT = 85 – 15 = 70°C α = ΔL / (L₀ ΔT) = 0.004 / (1.500 × 70) = 3.81 × 10⁻⁵ /°C
ΔT = 30°C Strain if free = α ΔT = (2.3e-5)(30) = 6.9e-4 But walls prevent expansion → compressive stress: Stress = Y × (α ΔT) = (7.0e10)(6.9e-4) = 4.83 × 10⁷ Pa linear thermal expansion problems and solutions pdf
✅ Key takeaway: Expansion is small but critical in precision work. Question: A metal bar is 1.500 m at 15°C and 1.504 m at 85°C. Find α. ΔL = 1
✅ Compare to known values: This is likely aluminum or brass. Question: An aluminum rod (α = 2.3 × 10⁻⁵ /°C, Y = 7.0 × 10¹⁰ Pa) is 1.000 m at 20°C. It is fixed between two rigid walls. If the temperature rises to 50°C, what is the thermal stress? ✅ Compare to known values: This is likely
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This article serves as a preview to that guide. Below, we explain the core formula, walk through three common types of problems, and tell you how to download a comprehensive PDF with 20+ solved problems. The change in length of a solid due to a temperature change is given by: