CivilSolve

Statics & Mechanics of Materials

Thermal Stress (Constrained Bar)

A bar restrained between rigid supports and subjected to a temperature change: free expansion δT = α·ΔT·L, the compressive (or tensile, on cooling) stress σ = E·α·ΔT when fully restrained, the reduced stress when an initial gap absorbs part of the expansion, and the restraint force when the area is given. Use for: 'the rail/bar is fixed at both ends and heated', gap-closing problems, restraint force questions. Not for: free (unrestrained) thermal elongation — that's axial-member.

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Inputs

Load a sample problem:
alphaThermal expansion coefficient

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

fully restrained steel bar heated 40 °C

Given

length
2 m
modulus
200 GPa
alpha
value 0.000012per degC
temperature change
40 degC

Assumptions

  • Rigid (unyielding) supports, linear-elastic material, uniform temperature change along the bar.

Solution steps

  1. Free thermal expansion

    Unrestrained, the bar would change length in proportion to α, ΔT and L.

    δT = α · ΔT · L

    δT = 0.96 mm

    = 0.96 mm

  2. Stress from the suppressed strain

    The supports suppress the entire expansion — the bar carries the strain as compression.

    σ = E·(δ_blocked)/L, = E·α·ΔT when there is no gap

    suppressed δ = 0.96 mm; σ = 96 MPa (compression)

    = 96 MPa

Results

Free thermal expansion δT

0.00096m

Thermal stress σ (compression)

96MPa

Where this answer was checked
source
σ = E·α·ΔT (NCEES FE Reference Handbook) — no gap, SI
verified by
hand-recomputed
derivation
σ = 200e9·12e-6·40 = 96 MPa compression. Free δ = 12e-6·40·2 = 0.96 mm.

Example 2

gap absorbs part of the expansion

Given

length
1 m
modulus
200 GPa
alpha
value 0.000012per degC
temperature change
50 degC
gap
0.2 mm

Assumptions

  • Rigid (unyielding) supports, linear-elastic material, uniform temperature change along the bar.

Solution steps

  1. Free thermal expansion

    Unrestrained, the bar would change length in proportion to α, ΔT and L.

    δT = α · ΔT · L

    δT = 0.6 mm

    = 0.6 mm

  2. Stress from the suppressed strain

    Only the expansion beyond the gap is suppressed; the restraint pushes back on that portion.

    σ = E·(δ_blocked)/L, = E·α·ΔT when there is no gap

    suppressed δ = 0.4 mm; σ = 80 MPa (compression)

    = 80 MPa

Results

Free thermal expansion δT

0.0006m

Thermal stress σ (compression)

80MPa

Where this answer was checked
source
σ = E·(δT − gap)/L — 1 m bar, 0.2 mm gap, ΔT = 50 °C
verified by
hand-recomputed
derivation
δT = 12e-6·50·1 = 0.6 mm; blocked = 0.4 mm; σ = 200e9·0.4e-3/1 = 80 MPa.

Example 3

US: restrained bar heated 100 °F with per-°F alpha

Given

length
10 ft
modulus
29000 ksi
alpha
value 0.0000065per degF
temperature change
100 degF

Assumptions

  • Rigid (unyielding) supports, linear-elastic material, uniform temperature change along the bar.

Solution steps

  1. Free thermal expansion

    Unrestrained, the bar would change length in proportion to α, ΔT and L.

    δT = α · ΔT · L

    δT = 0.078 in

    = 0.078 in

  2. Stress from the suppressed strain

    The supports suppress the entire expansion — the bar carries the strain as compression.

    σ = E·(δ_blocked)/L, = E·α·ΔT when there is no gap

    suppressed δ = 0.078 in; σ = 18.85 ksi (compression)

    = 18.85 ksi

Results

Free thermal expansion δT

0.0065ft

Thermal stress σ (compression)

18.85ksi

Where this answer was checked
source
σ = E·α·ΔT in US units — E = 29000 ksi, α = 6.5e-6/°F
verified by
hand-recomputed
derivation
σ = 29000·6.5e-6·100 = 18.85 ksi (unit-consistent in ksi directly; SI cross-check 129.97 MPa / 6.8948 = 18.85).