Thermal Stress, Fixed-End Bar

δ_{T} = α Δ T L δ_{mech} = \frac{P L}{A E} = - δ_{T} (fully fixed)

Free thermal expansion is α ΔT L. If a bar is fully restrained, mechanical shortening cancels δT, so σ = E α ΔT (compression on heating for a fixed bar). Statically indeterminate: compatibility + equilibrium.

$α$ (1/°C): Coefficient of thermal expansion
$Δ T$: Temperature change
$P$: Mechanical force from restraint

Step-by-step solved example

Steel α=12×10⁻⁶ /°C, E=200 GPa, ΔT=+50°C, fully fixed. σ?

Fixed-end bar: thermal expansion is cancelled by mechanical shortening.

1. Free δ
Would expand αΔTL.
2. Stress
σ=EαΔT=200e3 MPa ×12e-6×50=120 MPa compression.
$σ = 120 MPa$

Answer: 120 MPa compression

10 practice questions

0/10 correct

1.A free bar (no restraint) has thermal stress

2.Fully fixed heating produces

3.σ = E α ΔT assumes

4.If the gap is larger than free expansion, stress is

5.Cooling a fully fixed bar produces

6.Two bars in series with different α: you must use

7.α of steel is about

8.Concrete α is

9.Redundant support reactions from ΔT are found by

10.Units: E in MPa, α in 1/°C, ΔT in °C → σ in