Stress, Strain & Elongation Calculator
Axial stress, strain, extension and safety factor from load and section.
Uniform axial load on a prismatic member. It does not check buckling, which governs any slender part in compression and will fail it far below the stress shown here.
How it works
Stress is force divided by area, strain is the fraction the piece stretches, and below yield the two are proportional. That proportionality constant is Young's modulus, and Hooke's law is the whole of elastic behaviour in one line: σ = F ÷ A, ε = σ ÷ E, ΔL = ε × L.
The thing worth internalising is that E is a property of the element, not of the alloy or its heat treatment. Every steel — mild, tool, spring, stainless — has E within a few percent of 200 GPa. Heat treating a bolt from Grade 2 to Grade 8 triples what it carries before yielding and changes its stiffness by nothing at all. If a part is too bendy, a stronger material will not help; only more material, or better-placed material, will.
Aluminium is a third the stiffness of steel and about a third the density, which is why an aluminium part matched to a steel one for stiffness ends up substantially lighter — you spend the weight saving on making it thicker, and come out ahead because stiffness rises faster than thickness does.
Safety factor here is against yield, the point of permanent deformation. Ductile metals still have useful margin beyond it before fracture; brittle ones — castings, hardened tool steel, ceramics — do not, and should be worked to a larger factor for that reason.
Common questions
My part is in compression — is this still right?
Only if it is short and stout. Anything slender buckles long before it reaches yield stress, and buckling depends on length and end fixity rather than on strength. A 1 m steel rod 10 mm across buckles at roughly 1.5 kN while this would tell you it yields at 19.6.
Why doesn't a stronger steel make it stiffer?
Because stiffness comes from Young's modulus, and E is essentially identical across all steels. Strength and stiffness are independent. To deflect less you need more section or a shorter span, not a better alloy.
What safety factor is appropriate?
1.5 to 2 for a well-understood static load in a ductile metal, 3 to 4 where the load is uncertain or the material brittle, 5 to 8 for anything lifting over people. Fatigue is a separate calculation and governs anything cyclically loaded.
Where do stress concentrations come in?
They do not, here. A hole, a notch, a sharp internal corner or a thread root can multiply local stress by two or three. This gives the nominal stress; anything with a feature in the loaded path needs that multiplied.