Stress, strain and constitutive behaviour

Define what is divided by what

For our uniform axial specimen, engineering normal stress σ is axial force F divided by original cross-sectional area A0. Engineering strain ε is extension δ divided by original gauge length L0. Normal describes an action perpendicular to the section; shear describes an action tangent to it. A load in newtons is not itself a stress.

Stress has units of force per area. Using N and mm² gives N/mm², which equals MPa. Strain is dimensionless because its numerator and denominator are both lengths. A strain of 0.001 means one part in a thousand, or 0.1%, not 0.001 mm. Preserve the gauge length and area definitions when comparing records.

Connect stress with strain within a range

A constitutive relation describes material response under specified conditions. The simplest axial linear elastic relation is σ=Eε, where E is Young’s modulus. E has stress units and is the slope of the stress–strain line. Combining the definitions gives specimen stiffness F/δ=EA0/L0. This stiffness depends on both material and geometry.

Elastic means recoverable on unloading; linear means proportional over the considered range. They are different ideas: a nonlinear response can still be recoverable. Permanent strain after unloading indicates an inelastic contribution. The first straight segment alone cannot establish the yield point, fracture strain, cyclic life or response at a different temperature or loading rate.

Averages require a geometry assumption

The uniform-bar calculation approximates a region far from grips, holes and abrupt section changes. Real stress and strain can vary from point to point. Dividing the total force by area gives an average normal stress for the section, not every local peak. A narrow notch or eccentric pull can make a simple average insufficient.

A measurement of overall machine movement can also include grip slip and fixture deformation. To infer material strain, identify the gauge region and how its extension was determined. A plausible straight line can arise from a consistent setup while still giving the wrong material modulus if area, length or displacement attribution is wrong.

Convert a constructed coupon record

Use original area 20 mm² and gauge length 100 mm. Forces 0, 500 and 1000 N correspond to extensions 0, 0.05 and 0.10 mm. At 500 N,σ=500/20=25 MPa and ε=0.05/100=0.0005. At 1000 N, the values are 50 MPa and 0.001. Both give E=σ/ε=50000 MPa, or 50 GPa, within the supplied linear assumption.

The specimen stiffness is 1000/0.10=10000 N/mm. Doubling its original length while keeping area and material unchanged would halve this stiffness in the same model. It would not halve E. No unloading or failure data are supplied, so the plotted record should end at the last stated point rather than imply an unmeasured continuation.

Try the reasoning

1. What is the difference between 0.001 strain and 0.001 mm extension?

   Show the answer and reasoning

   Strain is a length ratio; extension is a length. A gauge length is needed to convert between them.

2. Would twice the specimen length change E in this ideal model?

   Show the answer and reasoning

   No. It changes structural stiffness and extension, but E remains the supplied material parameter.

3. Does this line identify a yield strength?

   Show the answer and reasoning

   No. The record contains no transition or unloading evidence from which to establish yield.

What to carry forward

Keep force, stress, extension, strain, material modulus and specimen stiffness distinct.