Axial & thermal stress calculator
Calculate normal stress, strain and elongation of an axially loaded bar as well as the thermal stress under restrained expansion. Choose the mode, enter the cross-section as area or diameter, set the material with its expansion coefficient – the calculator performs the safety check against yield strength with traffic-light rating, live with every input.
Axial & thermal stress calculator
Model: uniaxial prismatic tension/compression bar, linear-elastic per Hooke, uniform temperature over the cross-section. Elastic stress check with a global safety factor against yield. No buckling, notch, fatigue or creep check. Sizing tool for machine and fixture design.
Results
Calculating …
Formulas and fundamentals
The bar loaded axially in tension or compression follows Hooke's law: the normal stress is uniformly distributed over the cross-section, sigma = F/A. This gives the strain eps = sigma/E = F/(E·A) and the elongation dL = F·L/(E·A) = eps·L. Tension is counted positive, compression negative. The available safety factor against yield is S = Rp/|sigma|; it is checked against the required factor S_req and shown as a traffic light.
A temperature change dT expands the free bar by dL_th = alpha·L·dT without any stress. If this expansion is fully restrained, for instance by two rigid supports, the thermal stress sigma_th = -E·alpha·dT builds up. The negative sign means a temperature rise (dT > 0) produces compression and cooling produces tension. Notably the thermal stress depends neither on length nor on cross-section, only on elastic modulus, expansion coefficient and temperature difference.
If force and restrained thermal expansion act simultaneously, both contributions superimpose: sigma_ges = F/A - E·alpha·dT. A tensile force can thus be partly or fully compensated by heating, or cooling adds to the tensile load. The calculator reports the contributions separately and the governing total stress, checked against yield strength.
Worked example
A tension bar of A = 100 mm² in S235 steel (Rp = 235 N/mm²) carries a force of 10,000 N over a length of 1000 mm. The normal stress is sigma = 10,000/100 = 100 N/mm², the strain eps = 100/210,000 = 4.76·10⁻⁴ and the elongation dL = 0.476 mm. The safety factor against yield is S = 235/100 = 2.35 – the check passes with S_req = 1.5.
If the same bar is heated by dT = 50 K instead of loaded mechanically and its expansion is prevented by rigid supports (alpha = 12·10⁻⁶ 1/K), a compressive thermal stress of sigma_th = 210,000 · 12·10⁻⁶ · 50 = 126 N/mm² arises. Free to expand, the bar would have grown by dL_th = 12·10⁻⁶ · 1000 · 50 = 0.6 mm.
Acting together, sigma_ges = 100 − 126 = −26 N/mm²: the tensile force is overcompensated by heating and the bar is under slight net compression. The example shows why thermal stresses must not be neglected in machine design – they can considerably exceed the mechanical load.
Frequently asked questions
Why does the thermal stress not depend on length?
Because the free thermal strain eps_th = alpha·dT and the mechanical strain needed to restrain it must exactly cancel. The resulting stress sigma_th = E·alpha·dT contains neither length nor cross-section. Only the free expansion dL_th = alpha·L·dT grows with length, not the stress under restrained expansion.
When does a thermal stress arise at all?
Only when the thermal expansion is restrained. A free bar expands without stress. A stress builds up only when supports, adjacent parts or a temperature gradient prevent the expansion fully or partly. In practice the real case usually lies between free and fully restrained.
What do the signs of force and temperature mean?
Tension is positive, compression negative. A positive force produces tension. Under restrained expansion a temperature rise (dT > 0) produces compression, i.e. a negative stress, while cooling produces tension. The governing total stress is the signed sum of both contributions.
Does the calculation also apply to compression members?
For stress and shortening, yes. Under compression, however, buckling must additionally be checked; it depends on length, support conditions and second moment of area and is not included here. The calculator flags compression loading. Use the buckling calculator for the stability check.
Which expansion coefficient should be used?
The mean linear coefficient of thermal expansion in the temperature range considered. For structural steel it is about 12·10⁻⁶ 1/K, for austenitic stainless steel around 16·10⁻⁶ 1/K and for aluminium about 23·10⁻⁶ 1/K. Selecting a material sets suitable reference values that can be overridden.
Is this a complete strength verification?
No. It is an elastic stress check with a global safety factor against yield for the uniaxial case. Notch effects, multiaxial stress states, fatigue, buckling and creep at high temperatures are not included and must be verified separately where relevant.
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