Library / Mechanical / MEC-81
Mechanical
Normal strain from three-dimensional stress and temperature change
Strain in the x direction in an isotropic elastic solid under normal stresses in all three directions, plus the free thermal expansion.
- Ref
- MEC-81
- Rev
- v1
- Field
- Mechanical
- Result
- εx [µε]
For an isotropic linear elastic material, the normal strain in the x direction is εx = [σx − ν(σy + σz)]/E + αΔT. The first part is the generalised Hooke’s law: stress in x stretches the material and stress in y and z shrinks it through the Poisson effect. The second part is thermal expansion. The result is in microstrain (µε): 1 µε is 10⁻⁶. Enter tensile stresses as positive and compressive as negative; shear stress does not affect normal strain.
Strain in the x direction in an isotropic elastic solid under normal stresses in all three directions, plus the free thermal expansion. The other two normal strains follow by cycling x, y and z.
Calculator
- Mechanical strain (stress part)425 µε
- Thermal strain, αΔT240 µε
- Stress that would cancel the thermal strain if the member were fully restrained, EαΔT48 MPa
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- With σx = 100 MPa, σy = 50 MPa and a 20 K temperature rise, the strain in x is 665 µε: 425 µε from the stresses and 240 µε from heating.
- The y stress of 50 MPa reduces the x strain by 75 µε through the Poisson effect: ignoring it would overstate the strain by 17.6 %.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (µε) | Local step | Change in result |
|---|---|---|---|---|---|---|
| E Young’s modulus | 1 to 500 GPa | 410 | 85,240 | 84,830 | +10% | −5.81 % |
| σx Normal stress in x (tension positive) | −300 to 500 MPa | −1,335 | 2,665 | 4,000 | +16 MPa | +12 % |
| ΔT Temperature change | −100 to 200 K | −775 | 2,825 | 3,600 | +6 K | +10.8 % |
| σy Normal stress in y (tension positive) | −300 to 500 MPa | −10 | 1,190 | 1,200 | +16 MPa | −3.61 % |
| σz Normal stress in z (tension positive) | −300 to 500 MPa | −85 | 1,115 | 1,200 | +16 MPa | −3.61 % |
| α Thermal expansion coefficient | 0 to 30 10⁻⁶/K | 425 | 1,025 | 600 | +10% | +3.61 % |
| ν Poisson’s ratio | 0 to 0.5 | 615 | 740 | 125 | +10% | −1.13 % |
Sweep of Normal stress in y (tension positive)
| Normal stress in y (tension positive) (MPa) | Normal strain in x (µε) |
|---|---|
| −300 | 1,190 |
| −200 | 1,040 |
| −100 | 890 |
| 0 | 740 |
| 100 | 590 |
| 200 | 440 |
| 300 | 290 |
| 400 | 140 |
| 500 | −10 |
Sweep one variable
Variables and how each one contributes
Normal strain in x
What the equation gives you. Everything else feeds it.
Normal stress in x (tension positive)
The normal stress in the x direction, in MPa (tension positive). Raises the strain in direct proportion: each extra MPa adds 1000/E microstrain, with E in GPa. Raising it by 10 MPa (to 110 MPa) moves the result by +7.52 %, from 665 to 715 µε.
Normal stress in y (tension positive)
The normal stress in the y direction, in MPa. Lowers the x strain through the Poisson effect: a tensile y stress narrows the material in x. Raising it by 10 MPa (to 60 MPa) moves the result by −2.26 %, from 665 to 650 µε.
Normal stress in z (tension positive)
The normal stress in the z direction, in MPa. It is zero in plane stress. Lowers the x strain through the Poisson effect, in the same way as σy. Raising it by 10 MPa (to 10 MPa) moves the result by −2.26 %, from 665 to 650 µε.
Young’s modulus
Young’s modulus of the material, in GPa (about 200 for steel, 70 for aluminium). Lowers the mechanical part of the strain, because a stiffer material strains less for the same stress; the thermal part is not affected. Raising it by 10 GPa (to 210 GPa) moves the result by −3.04 %, from 665 to 644.8 µε.
Poisson’s ratio
Poisson’s ratio of the material (about 0.3 for steel). Lowers the strain in x while the sum of σy and σz is positive, because a larger Poisson effect shrinks the material more. Raising it by 0.05 (to 0.35) moves the result by −1.88 %, from 665 to 652.5 µε.
Thermal expansion coefficient
The coefficient of thermal expansion, in 10⁻⁶ per kelvin (about 12 for steel, 23 for aluminium). Raises the thermal part of the strain in direct proportion when ΔT is positive. Raising it by 1 (to 13) moves the result by +3.01 %, from 665 to 685 µε.
Temperature change
The temperature change from the stress-free state, in kelvin. Positive means heating. Raises the strain in direct proportion to the temperature rise, through thermal expansion. Raising it by 5 K (to 25 K) moves the result by +9.02 %, from 665 to 725 µε.
Worked example with the values above
- Mechanical part: [σx − ν(σy + σz)]/E = [100 − 0.3 × 50] / 200,000 MPa = 425 µε.
- Thermal part: αΔT = 12 × 20 = 240 µε.
- εx = 425 + 240 = 665 µε.
Assumptions and limits
- A linear elastic, homogeneous and isotropic material, with strains small.
- Stresses in MPa with moduli in GPa, giving strain in microstrain.
- Thermal expansion free of any constraint except the stress components entered.
Common mistakes
- Leaving out the Poisson contribution of the other two stresses.
- Using the plane-stress shortcut when the through-thickness stress is not zero.
- Mixing MPa and GPa without converting.
Statistics and uncertainty
Strain-gauge and temperature errors add directly: a 2 K error in ΔT with α = 12 × 10⁻⁶/K causes 24 µε of error, which may be as large as the stress effect being measured, so temperature compensation matters.
Where it is used
- Interpreting strain-gauge readings in a pressurised or heated structure.
- Checking the strain in a barrel, hull or pipe under combined pressure and temperature.
Maths you need
- Stress and strain: force per area and relative change in length
- Elastic modulus: stress divided by strain in the elastic range