How this instrument works
Elongation is the actual length a straight member gains when an axial force pulls along it. The formula ΔL = FL ⁄ (AE) builds it in two steps that are easy to separate: F ⁄ A is the stress running through the cross-section, and dividing that stress by the elastic modulus E converts it into strain, the fractional stretch per unit length. Multiply that strain by the original length L and the fraction becomes an actual distance — millimetres, not a ratio.
Each variable earns its place on a particular side of the line. Force and length sit on top because a harder pull or a longer starting member both produce more total stretch in direct proportion — double either one and the elongation doubles with it. Area and modulus sit underneath because they resist stretching: a fatter cross-section spreads the same force thinner, and a stiffer material — one with a higher E — simply refuses to strain as much for a given stress. Steel's modulus near 200 GPa is roughly three times aluminum's, which is why a steel tie-rod sized for the same load stretches a third as far.
The formula only holds below the material's proportional limit, the stress at which the stress-strain curve stops being a straight line. Push a mild-steel rod past roughly 250 MPa and the relationship breaks — the rod starts to yield, part of the stretch stops reversing when the load is removed, and ΔL = FL ⁄ (AE) understates what actually happens. It also assumes a uniform prismatic section carrying pure axial load; a rod with a hole, a taper, or any bending mixed in needs a more careful calculation than this single line provides.
- Enter the Applied force (F) — the steady axial pull or push along the member's centerline, in newtons or pound-force.
- Enter the Original length (L), the unstretched length of the rod or cable, in metres, feet, or inches.
- Enter the Cross-sectional area (A) — the member's cut area perpendicular to the load, typically in mm² or in².
- Enter the Elastic modulus (E), the material's Young's modulus — about 200 GPa for steel, 69 GPa for aluminum.
- Read the Elongation (dL), the length gained under load, in millimetres or inches.
Worked example — a steel tie-rod under 5 kN
Take a 2 m steel tie-rod with a 100 mm² cross-section, roughly the footprint of a large coin, carrying a steady 5,000 N pull — close to the weight of a small car hanging from it. Its elastic modulus is 200 GPa, ordinary mill-certificate steel. Converting the cross-section to base units first, 100 mm² is 0.0001 m², and 200 GPa is 200,000,000,000 Pa. The formula then reads ΔL = FL ⁄ (AE) = (5000 × 2) ⁄ (0.0001 × 200,000,000,000) = 10,000 ⁄ 20,000,000 = 0.0005 m.
That is 0.5 mm of stretch on a 2 m rod — a strain of 0.00025, well inside structural steel's elastic range, which typically extends to about 0.001 before yielding begins. A load-cell calibration technician runs this exact reasoning backwards: clamp the rod, measure the 0.5 mm stretch with a dial indicator, and knowing the rod's geometry and modulus, back-calculate the 5,000 N force that produced it.
Questions
Does the same formula work for a compressive load?
Yes, with a sign flip. ΔL = FL ⁄ (AE) applies equally to a compressive force, returning a negative ΔL that represents shortening rather than stretch. The physics is the same axial relationship, but slender members under compression can buckle sideways well before the material yields, a failure mode this formula does not predict.
What happens once the load passes the elastic limit?
The formula stops matching reality. ΔL = FL ⁄ (AE) assumes stress stays proportional to strain, which holds only up to a material's proportional limit — roughly 250 MPa for mild steel. Beyond that the rod yields, part of the deformation becomes permanent, and the true stretch grows faster than this straight-line formula predicts.
Why does area sit under the line instead of over it?
A thicker member spreads the same force across more material, which lowers the stress it carries — stress is force divided by area. Since strain, and therefore elongation, follows stress, doubling A halves the stress and halves the resulting stretch, which is why A belongs in the denominator alongside E.
How does this relate to a spring's F = kx?
It is the same law with the stiffness spelled out. Rearranging gives F = (AE ⁄ L) × ΔL, matching F = kx with an effective spring constant k = AE ⁄ L. A rod's stiffness comes from its cross-section, length, and modulus the way a coil spring's comes from its wire diameter and coil count.
Does the cross-section's shape matter, or only its area?
Only the area matters for elongation — a 100 mm² square bar and a 100 mm² round bar stretch by the same amount under the same axial load. Shape becomes important for other properties, such as bending stiffness or buckling resistance, which depend on how that area is arranged around the centroid, not just its total value.
What elastic modulus should I use for common materials?
Structural steel is about 200 GPa, aluminum alloys about 69 GPa, and copper around 110–130 GPa, while engineering plastics like nylon sit near 2–4 GPa, meaning roughly a hundred times more stretch for an identical load and geometry. Concrete varies widely, typically 20–40 GPa depending on mix and curing age.