Question
A steel rod of length 1.000 m at 20°C is heated to 120°C. Find:
(a) the new length, (b) the strain produced if the rod is clamped at both ends and not allowed to expand, and (c) the thermal stress, given αsteel=1.2×10−5 /°C and Ysteel=2×1011 Pa.
Solution — Step by Step
L′=L0(1+αΔT)
With ΔT=120−20=100°C:
L′=1.000×(1+1.2×10−5×100)=1.000×(1+1.2×10−3)
L′=1.0012 m
If clamped, the rod would have expanded by ΔL=L′−L0=1.2×10−3 m, but it can’t. The clamps effectively “compress” the rod back, producing a compressive strain:
strain=L0ΔL=αΔT=1.2×10−5×100=1.2×10−3
Stress = Y× strain:
σ=2×1011×1.2×10−3=2.4×108 Pa
This is 240 MPa — a serious load!
Final answers: L′=1.0012 m, strain =1.2×10−3, stress =240 MPa.
Why This Works
Thermal stress is the elastic stress that develops in a constrained body when the temperature changes. The body wants to expand but can’t, so it experiences compression equivalent to forcing it back to its original length.
Notice the stress formula σ=YαΔT depends only on material properties and ΔT — not on the length. This is why long railway rails buckle just as easily as short ones if expansion gaps are missing.
Alternative Method
Direct formula: F=YAαΔT for the constraint force on a clamped rod of cross-section A. Stress is F/A=YαΔT. We dodged needing to know A.
Common Mistake
Forgetting that ΔT=T2−T1 in Celsius (or Kelvin — the size of a degree is the same). Some students convert to absolute Kelvin (e.g., 293→393, still ΔT=100) and confuse themselves into multiplying by 293. For changes in temperature, °C and K give identical numbers.