Plastic deformation and yield behaviour

Unloading reveals a permanent part

Plastic deformation remains after unloading. In a simple metal tensile test, the total loaded strain can combine elastic and plastic contributions. The onset can be gradual, so a specified offset proof stress provides an operational comparison rather than a perfectly sharp universal boundary.

Imagine two marks initially 40.0 mm apart on a small metal tensile coupon. Under load they move to 40.240 mm apart. After the force is removed, suppose they settle at 40.144 mm, with no further time-dependent recovery in this example. The loaded extension was 0.240 mm; 0.096 mm recovered, while 0.144 mm remained. The specimen did not recover all its extension, but it did not retain all of it either.

Use engineering strain ε = (L − L₀)/L₀, where L₀ is the original gauge length and L is the present separation. The original cross-section A₀ defines engineering stress σ = F/A₀. Here A₀ = 8.00 mm²; a force of 1920 N gives 240 N/mm² = 240 MPa. Deformation and relative strain explains the gauge and percent conversion. Elastic response and elastic modulus provides the reversible-slope preparation.

For the constructed response below, the uniform gauge is in small uniaxial tension at fixed temperature, loading direction and prescribed strain rate. The elastic unloading slope is E = 100 GPa = 100000 MPa. Strain splits as ε = ε_e + ε_p, where ε_e is the recoverable elastic part and ε_p the retained plastic part in this model. Since ε_e = σ/E, rearranging gives ε_p = ε − σ/E. Insert stress and modulus in matching units.

At the loaded state A, ε = 0.240/40.0 = 0.00600. Its elastic part is 240/100000 = 0.00240 and its plastic part is 0.00600 − 0.00240 = 0.00360. Thus elastic recovery is 40.0 × 0.00240 = 0.0960 mm, leaving 40.0 × 0.00360 = 0.144 mm. At zero stress the model has ε_e = 0, not ε = 0.

The calculation stipulates immediate linear elastic unloading, no reverse plasticity and no further evolution while unloaded. It excludes creep, delayed recovery, thermal strain, damage, grip slip and necking. A measured residual at one recovery time is not automatically permanent; those possibilities need observations. A nonlinear loading curve alone also does not prove plasticity, because a nonlinear elastic material can recover along a curved path.

A microscopic slip can leave a macroscopic history

In a crystalline metal, slip is relative shear displacement of material on opposite sides of a crystal plane. A dislocation is a line defect separating regions that have and have not yet slipped. As it moves, nearby atoms change neighbours and the slipped region grows. The rearrangement can proceed locally rather than requiring a whole plane to move at once. Defects, boundaries and interfaces develops that moving-line picture.

When a dislocation crosses a crystal and leaves its surface, the relative slip can remain. Removing the external load permits elastic lattice distortion to recover; it need not reverse every earlier slip. Many such events, across suitably oriented systems and grains, can contribute to a tensile shape change. Motion within the slip plane is called glide. This is a representative route, not every plastic mechanism in every material. A noncrystalline polymer has no identical lattice-slip description.

Obstacles and interactions can make further glide more difficult. Continued plastic deformation may build a dislocation arrangement requiring greater stress for further flow: work hardening. This changes resistance to plastic flow, rather than necessarily changing the initial elastic modulus by the same factor. Why processing history matters follows this state change through deformation and annealing; a material name alone does not specify it.

Use the original invented loading record in the table. Join the points by the stipulated straight segments: the initial branch reaches 150 MPa at ε = 0.00150 with slope 100000 MPa. From there to ε = 0.00600, the rising branch has slope 20000 MPa. This deliberately simple curve has a sharp model knee; it does not claim that real yielding is always sharp or that these are values for a named alloy.

The second slope is the slope of the total loading response, not a new elastic modulus. For example, from ε = 0.00300 to 0.00500 stress rises by 40 MPa. Elastic strain rises by only 40/100000 = 0.000400; the remaining 0.00200 − 0.000400 = 0.00160 is an increase in the model’s plastic strain.

Unload from A, at ε = 0.00600 and σ = 240 MPa. The fixed plastic part is then 0.00360, so any state on this unloading segment follows ε = 0.00360 + σ/100000. At 120 MPa the strain is 0.00480; at zero stress it is 0.00360. Neither point lies on the original loading curve.

For this model only, tensile reloading retraces that elastic segment up to the previous maximum. At 180 MPa it gives ε = 0.00360 + 0.00180 = 0.00540, whereas first loading reached 180 MPa at ε = 0.00300. Same current stress does not mean same strain or internal state. No rule for reverse loading, repeated cycling or continuation beyond the previous maximum is supplied.

The offset is a definition

For a 0.2% offset construction, draw a line parallel to the initial elastic slope beginning at strain 0.002. Its intersection with the curve defines that proof stress. Do not confuse 0.2% plastic offset with a total strain of 0.2% at the intersection.

An offset specifies a reproducible macroscopic comparison when the onset is hard to pinpoint. Write the 0.2% strain offset as ε₀ = 0.002, since 0.2/100 = 0.002. A line with the elastic slope E but shifted right by ε₀ has equation σ_off = E(ε − ε₀). Its zero-stress intercept is ε₀. The offset is horizontal strain displacement, not 0.2% of the stress or of the modulus.

Use the actual constructed record, rather than guessing a proof stress. At ε = 0.00300 the curve is 180 MPa and the offset line is 100000(0.00300 − 0.00200) = 100 MPa, so the curve is 80 MPa above the line. At ε = 0.00500 the curve is 220 MPa but the line is 300 MPa, so the difference is −80 MPa. An intersection lies within that stipulated straight interval.

The curve’s interval slope is (220 − 180)/(0.00500 − 0.00300) = 20000 MPa. Within it, σ_curve = 180 + 20000(ε − 0.00300). Equate this to 100000(ε − 0.00200): 120 + 20000ε = 100000ε − 200, with stresses in MPa. Thus 320 = 80000ε and ε = 0.00400. Substitution gives σ = 200 MPa. The intersection was calculated from the supplied piecewise-linear model, not inferred solely from a label.

At this intersection the total strain is 0.00400 = 0.400%, not 0.200%. In the stipulated unloading model, σ/E = 0.00200 is recovered and ε_p = 0.00200 remains. Over 40.0 mm those parts are 0.0800 mm each. The worked case below checks the same interpretation from a supplied proof stress.

The model first departs from elastic loading at 150 MPa, whereas its 0.2% proof stress is 200 MPa. Proof stress is therefore not the first plastic event even in this simple construction. In a real polycrystal, grains and local regions need not begin slipping together; finite resolution and the chosen criterion matter. A different stated offset can give a different comparison value.

Keep the graphical definition distinct from its plastic-strain interpretation. If actual unloading is nonlinear, has a different slope or includes time-dependent recovery, the offset-line intersection alone does not prove the final residual strain is exactly 0.002. Establish the relevant unloading/recovery behaviour instead of applying this ideal metal model to every specimen.

Carry the endpoint, state and measurement into the comparison

A proof-stress report needs the chosen offset and strain scale, original area, specimen gauge and strain measurement, material state, direction, temperature and rate. The line’s slope must come from a justified elastic region, rather than a line fitted through significant plastic flow or machine movement. A percent axis must be converted consistently: a horizontal reading of 0.4% means ε = 0.004.

The two figures describe the same invented record for different questions. Loading–unloading asks what recovers and what remains. The offset construction asks which point satisfies a chosen comparison rule. Changing from an initial loading curve to an unloading line, or changing the offset, changes the question even when every axis still says stress and strain.

The 240 MPa at A is only the largest stress in this short supplied record. It has not been established as ultimate tensile strength or fracture stress. Tensile curves, strength and ductility develops those later endpoints. A proof stress likewise does not establish toughness, fatigue life or the load a component can safely support; geometry, local stress and other failure modes require their own evidence.

Rate, direction and temperature effects extends the operating conditions. Keep them with the result rather than transferring this record to a hot, rapidly loaded, differently oriented or previously deformed state. For a real recovery experiment, retain the force and gauge records, measurement corrections and elapsed times so apparent residual extension can be distinguished from slip, drift or delayed recovery.

Worked examples

Try the reasoning

1. Does a proof stress mark the first atomic slip event?

   Show the answer and reasoning

   Not necessarily.

   It is a specified macroscopic measurement convention. The constructed record first leaves its elastic branch at 150 MPa but reaches the 0.2% proof intersection at 200 MPa. This numerical difference makes the distinction explicit.

2. What strain remains after ideal elastic unloading?

   Show the answer and reasoning

   The plastic part.

   The elastic part is recovered. For state A, subtract σ/E = 0.00240 from ε = 0.00600 to obtain ε_p = 0.00360. That statement relies on the specified linear elastic unloading and absent delayed evolution.

3. A different uniform coupon has L₀ = 80.0 mm and E = 200 GPa. At a supplied engineering stress of 300 MPa its gauge length is 80.320 mm. Assuming the same type of ideal elastic unloading with no time-dependent recovery, find loaded strain, recovered extension and residual extension. Does this supply its offset proof stress?

   Show the answer and reasoning

   Loaded strain is 0.00400; recovered extension is 0.120 mm and residual extension is 0.200 mm. A chosen offset proof stress has not been determined.

   The extension is 0.320 mm and ε = 0.320/80.0 = 0.00400. Elastic strain is 300/200000 = 0.00150; plastic strain is 0.00250. Multiplying each by 80.0 mm gives 0.120 and 0.200 mm, summing to 0.320 mm. This is one loaded state and an unloading model, not a full curve or a specified 0.2% intersection.

4. Keep the constructed loading curve and E = 100000 MPa, but specify a 0.1% rather than 0.2% offset. The rising branch is σ = 120 + 20000ε MPa for 0.00150 ≤ ε ≤ 0.00600. Find the new intersection and check that it lies in the branch. Did the physical curve change?

   Show the answer and reasoning

   ε = 0.00275 and σ = 175 MPa. The same physical curve is being read with a different comparison criterion.

   A 0.1% offset is 0.001, so equate 120 + 20000ε = 100000(ε − 0.001). This gives 220 = 80000ε, hence 0.00275. Substitution gives 175 MPa in both expressions. The strain lies between 0.00150 and 0.00600; its ideal residual is 0.001. Do not compare 175 and 200 MPa as if they used the same proof definition.

5. The invented specimen first reaches 180 MPa at strain 0.00300. It is then taken to A (240 MPa, 0.00600), unloaded to zero and reloaded elastically to 180 MPa, staying below its previous maximum. Find the new strain and compare gauge extensions over the original 40.0 mm.

   Show the answer and reasoning

   On reloading ε = 0.00540 and extension is 0.216 mm, compared with 0.120 mm on first loading. The difference is 0.0960 mm despite the same current stress.

   After A the fixed plastic part is 0.00360. At 180 MPa the elastic part is 0.00180, giving total 0.00540. On first loading the plastic part at 180 MPa was 0.00300 − 0.00180 = 0.00120. Its later increase explains the extra extension. This is the stipulated tensile history model, not a prediction for reversed loading or arbitrary cyclic evolution.

6. Between Y (150 MPa, ε = 0.00150) and C (180 MPa, ε = 0.00300), calculate the elastic and plastic strain increments using E = 100000 MPa. Is the total interval slope of 20000 MPa evidence that E became 20 GPa?

   Show the answer and reasoning

   The total increment is 0.00150, elastic increment 0.000300 and plastic increment 0.00120. The 20 GPa total loading slope is not a changed elastic modulus in this model.

   Δσ = 30 MPa, so Δε_e = 30/100000 = 0.000300. Subtract from Δε = 0.00300 − 0.00150 = 0.00150 to get Δε_p = 0.00120. The total slope is 30/0.00150 = 20000 MPa, while unloading is independently stipulated to retain slope E = 100000 MPa.

7. A different time-dependent sample has measured residual strain 0.002 immediately after unloading, falling to 0.0005 after one hour. A report calls the entire immediate 0.002 permanent plastic strain and uses the ideal model to guarantee it will never change. What does the record actually establish?

   Show the answer and reasoning

   The reported residual decreases by 0.0015. The readings do not justify treating the immediate 0.002, or the remaining 0.0005, as an established final permanent value.

   Residual strain describes the reading at a stated recovery time. The reported change is 0.002 − 0.0005 = 0.0015. If validated gauge measurements represent actual continued recovery at the same conditions, the immediate residual was not all permanent and the no-delayed-recovery idealisation does not apply. Check drift and relevant measurement controls; one hour of observations cannot guarantee that no further change will occur.

What to carry forward

• State the yield or proof-stress definition.
• Separate total, elastic and plastic strain.
• A loading curve and a current stress do not erase prior plastic deformation; follow the actual loading–unloading path.
• The total postyield slope need not equal the elastic unloading slope.
• An offset proof stress is found by a specified graphical construction. Its residual-strain interpretation requires the appropriate unloading model.
• Keep specimen strain, state, direction, rate, temperature and recovery time with the result; no component safety or lifetime follows from proof stress alone.