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Normal (Axial) Stress

Also known as axial stress · P over A · tensile stress

σ=PA\sigma = \frac{P}{A}

Worked example: 50 kN on 500 mm^2 → 100 MPa — press Try an example to run it live, then adjust anything.

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Normal (Axial) Stress explained

PPAσ

Force alone tells you nothing about whether a part will survive — a 50 kN pull is nothing to a bridge chord and fatal to a coat hanger. What matters is the force spread over the material actually resisting it, and that intensity is stress. Claude-Louis Navier formalised it in his 1826 Leçons, the book that turned the craft of building into the discipline of strength of materials. Work an example: a 50 kN load on a rod of 500 mm² cross-section gives σ = 50 000 N ÷ 0.0005 m² = 100 000 000 Pa = 100 MPa, comfortably inside structural steel's 250 MPa yield.

The classic trap is using the gross area when a hole has been drilled through it. A 1/2-inch bolt hole through a 3-inch by 1/4-inch bar removes a sixth of the section, and the stress at that net section is a sixth higher — this is why steel design codes make you check gross yielding and net-section rupture separately. Threaded rod is the same story: a 1/2-13 rod has 0.196 in² of shank but only 0.1419 in² of tensile stress area, and the threads are where it breaks.

Normal (Axial) Stress formula

σ=PA\sigma = \frac{P}{A}
Where
  • σ\sigma= Normal stress (kPa)
  • PP= Axial force (N)
  • AA= Cross-sectional area (m²)

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Part of this formula set

Stress, strain and the elastic constants

9 formulasNormal and shear stress, strain, axial deformation, and the four elastic constants E, G, K and ν that link them.