Heating a free bar causes no stress, it just gets longer. Stress appears only when expansion is blocked. This calculator solves the fully constrained bar equation sigma = -E * alpha * dT in all four directions, computes Timoshenko bimetallic strip curvature and cantilever tip deflection, and includes a spot-checked E / alpha table for common metals.
Scope boundary: this tool covers blocked expansion (stress and bending). The free length change of an unconstrained part, dL = alpha * L * dT, belongs to our thermal expansion tool. Related stress tools: principal stress states, stress concentration factors, thin-wall hoop stress, and shear stress.
Constrained bar solver
sigma = -E * alpha * dT (tensile positive: heating a clamped bar gives negative sigma, i.e. compression)
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Bimetallic strip (Timoshenko, 1925)
kappa = 6*(a2-a1)*(1+m)^2*dT / (h*(3*(1+m)^2 + (1+m*n)*(m^2 + 1/(m*n)))) m = t1/t2, n = E1/E2, h = t1+t2, rho = 1/kappa. kappa > 0 bends toward layer 1.
Layer 1 (top)
Layer 2 (bottom)
Loading
-1/m curvature
Material data: Young's modulus and thermal expansion
Typical room-temperature values. Spot-checked against engineering references (matweb, amesweb, textbook tables), October 2026. Real alloys vary with temper and temperature: use certified datasheets for final design.
Material
Category
E (GPa)
alpha (10^-6/K)
Note
Carbon steel (typical)
Steel
200
12
Textbook value; AISI 1020 measures 11.7
Stainless steel 304
Steel
193
17.2
Austenitic; high expansion vs carbon steel
Aluminium 6061-T6
Aluminium
68.9
23.6
AA typical, 20-100 C
Copper C11000
Copper alloy
120
17
Electrolytic tough pitch
Brass C36000
Copper alloy
97
20.5
Free-cutting brass
Invar 36
Nickel alloy
141
1.2
Fe-36Ni; near-zero expansion
Worked examples
Steel bar, fully clamped, heated 35 K
E = 200 GPa, alpha = 12 x 10-6/K, dT = +35 K. sigma = -(200e9)(12e-6)(35) = -84 MPa: 84 MPa compressive (12183.17 psi). A 100 K rise in the same bar gives 240 MPa, past the yield of mild steel.
Copper bar, fully clamped, heated 50 K
Copper C11000: E = 120 GPa, alpha = 17.0 x 10-6/K. sigma = -(120e9)(17.0e-6)(50) = -102 MPa (102 MPa compressive). Copper's higher alpha outweighs its lower modulus versus steel.
Aluminium bar, fully clamped, cooled 20 K
Aluminium 6061-T6: E = 68.9 GPa, alpha = 23.6 x 10-6/K, dT = -20 K. sigma = +32.5208 MPa: 32.5208 MPa tensile.
Bimetallic strip: brass on steel, heated 100 K
Brass C36000 (1 mm, 97 GPa, 20.5e-6/K) on steel (1 mm, 200 GPa, 12.0e-6/K): m = 1, n = 0.485, h = 2 mm. kappa = -0.61643 1/m, radius 1.62224 m; a 200 mm cantilever deflects 12.313 mm at the tip, bowing toward the steel side.
Thermal stress is the internal stress that appears when a material's thermal expansion or contraction is blocked. Heating alone causes no stress: a free bar simply gets longer by dL = alpha*L*dT. Stress only develops when something prevents that length change, for example walls clamping both ends of the bar, and the material must then deform elastically to fit the constraint.
What is the formula for thermal stress in a fully constrained bar?
For a bar whose total strain is forced to zero, sigma = -E*alpha*dT, where E is Young's modulus, alpha the coefficient of linear thermal expansion, and dT the temperature change. The minus sign is the tensile-positive convention: heating (dT > 0) produces compression. Example: a steel bar with E = 200 GPa and alpha = 12e-6/K heated by 35 K develops sigma = -84 MPa, i.e. 84 MPa of compressive stress.
Is thermal stress tensile or compressive?
It depends on the direction of the temperature change. Heating a constrained bar pushes it against the walls, so the stress is compressive (negative in the tensile-positive convention). Cooling a constrained bar pulls it against the walls, so the stress is tensile (positive). Example: an aluminium 6061 bar (E = 68.9 GPa, alpha = 23.6e-6/K) cooled by 20 K develops +32.5208 MPa, i.e. 32.5208 MPa tensile.
Worked example: steel bar, temperature rise 35 K
Take a carbon steel bar clamped between rigid walls: E = 200 GPa, alpha = 12 x 10^-6/K, dT = +35 K. sigma = -E*alpha*dT = -(200e9)(12e-6)(35) = -84,000,000 Pa = -84 MPa. The magnitude is 84 MPa compressive, which equals 12183.17 psi. Because the stress scales linearly, a 100 K rise in the same bar gives 240 MPa, enough to yield ordinary structural steel.
Worked example: copper bar, temperature rise 50 K
Copper C11000 has E = 120 GPa and alpha = 17.0 x 10^-6/K. Fully constrained and heated by 50 K: sigma = -(120e9)(17.0e-6)(50) = -102,000,000 Pa = -102 MPa, i.e. 102 MPa compressive. Copper's higher expansion coefficient outweighs its lower modulus, so it develops more thermal stress than steel for the same temperature change.
How do I convert thermal stress between MPa, psi and ksi?
Stress converts like any pressure: 1 MPa = 145.0377 psi and 1 GPa = 145.0377 ksi (from the exact definition 1 psi = 6894.757293168 Pa). So 84 MPa = 12183.17 psi, and 1 ksi = 6.89476 MPa. For the expansion coefficient, 12 x 10^-6/K equals 6.6667 x 10^-6 per degree F, and a 63 F temperature difference equals 35 K.
What is the Timoshenko bimetallic strip formula?
Timoshenko (1925) gives the curvature of a bonded two-layer strip as kappa = 6*(a2-a1)*(1+m)^2*dT / (h*(3*(1+m)^2 + (1+m*n)*(m^2 + 1/(m*n)))), where m = t1/t2 is the thickness ratio, n = E1/E2 the modulus ratio, h = t1+t2 the total thickness, and a1, a2 the expansion coefficients. The radius is rho = 1/kappa. When both layers have equal thickness and modulus, this collapses to the familiar kappa = 3*(a2-a1)*dT/(2*h).
Worked example: brass on steel bimetallic strip
Layer 1 brass C36000 (t = 1 mm, E = 97 GPa, alpha = 20.5e-6/K) bonded to layer 2 steel (t = 1 mm, E = 200 GPa, alpha = 12.0e-6/K), heated by 100 K. Thickness ratio m = 1, modulus ratio n = 0.485, total thickness h = 2 mm. Timoshenko gives kappa = -0.61643 per metre, so the radius of curvature is 1.62224 m. A 200 mm cantilever of this strip deflects 12.313 mm at the tip.
Which way does a bimetallic strip bend when heated?
The strip bends away from the layer that wants to expand more, i.e. it curves toward the lower-expansion layer. In this calculator's convention kappa is positive, and the strip bends toward layer 1, exactly when (a2-a1)*dT > 0. Cooling reverses the curvature. A brass-on-steel strip heated up therefore bows toward the steel side, which is the principle behind bimetallic thermostats and thermal actuators.
What are typical E and alpha values for common metals?
Typical room-temperature values used by this calculator: carbon steel E = 200 GPa, alpha = 12.0e-6/K; stainless steel 304 E = 193 GPa, alpha = 17.2e-6/K; aluminium 6061-T6 E = 68.9 GPa, alpha = 23.6e-6/K; copper C11000 E = 120 GPa, alpha = 17.0e-6/K; brass C36000 E = 97 GPa, alpha = 20.5e-6/K; Invar 36 E = 141 GPa, alpha = 1.2e-6/K. Real alloys vary with temper and temperature, so treat these as starting points, not certified data.
What is the difference between free thermal expansion and thermal stress?
Free thermal expansion (dL = alpha*L*dT, no stress) happens when the part can grow; constrained expansion produces stress instead. If you need the length change of an unconstrained part, use a thermal expansion calculator; this tool covers the blocked-expansion case: the stress in a clamped bar and the curvature of a bimetallic strip. Real assemblies are often partially constrained, which needs a compatibility analysis beyond either simple formula.
What are the limits of these thermal stress calculations?
The formulas assume linear elasticity, a uniform temperature change, constant E and alpha, and small deflections. They stop being valid once the stress passes the material's yield strength (for example, the 240 MPa from a 100 K rise would permanently deform mild steel), when alpha varies strongly with temperature, or under thermal shock with steep gradients. They are screening calculations, not a substitute for qualified mechanical design or pressure-vessel codes.
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