Module 16 · Lesson 16.1
When the restoring force remembers
Hysteresis, the bilinear model, tangent stiffness and the iteration inside every nonlinear time step.
Why this matters
Everything up to here has assumed one thing about the structure: that the restoring force depends only on the current displacement, through fs = ku. That single assumption is what makes superposition work, and superposition is what makes modal analysis, response spectra and every closed-form solution in this course possible.
A structure designed to survive a severe earthquake is not designed to stay elastic. It is designed to yield in chosen places, and once it yields, fs = ku is false. The force now depends on where the structure has BEEN — on whether this is the first excursion or the fifth, and on which direction it came from.
Everything built on superposition falls with it. This lesson is about what replaces it.
By the end of this lesson you should be able to
- State the bilinear hysteretic model and what each parameter controls
- Distinguish tangent, secant and initial stiffness
- Explain the iteration inside a nonlinear time step
- Describe how yielding changes the effective period during the response
What you should already know
- Newmark integration (Module 8)
- The equation of motion and the meaning of each term (Module 3)
- Energy in a vibrating system (Module 7)
The picture before the equations
Push a steel brace. It stretches elastically, and if you let go it returns to where it started — the force–displacement path retraces itself and no energy is left behind.
Push it far enough and it yields. Now let go. It comes back along a line PARALLEL to the original elastic line, but it does not return to zero displacement: it stops short, permanently longer than it was. Push it the other way and it goes elastic again until it yields in compression.
Draw force against displacement for a full cycle of that and you get a closed loop rather than a line. The loop encloses an area, and that area is energy — energy that went into permanently deforming the steel and is not coming back. This is hysteresis, and the loop is the single most informative picture in earthquake engineering.
The bilinear model
The simplest model that captures it needs three numbers:
Initial stiffness k — the elastic slope, the same k as in every earlier module.
Yield force fy — where the slope changes. The yield DISPLACEMENT follows: uy = fy/k.
Post-yield stiffness ratio α — the yielded slope as a fraction of the elastic one. α = 0 is elastic–perfectly plastic, a flat plateau. Real steel members strain-harden, so α ≈ 0.02 to 0.05 is more realistic, and a small positive α also makes the numerics much better behaved.
The hardening rule decides what happens on reversal. This course uses KINEMATIC hardening: the elastic range keeps its full width 2fy and slides along with the plastic deformation. It is the standard first model for steel, and it means a structure that has yielded in one direction reaches yield SOONER on the way back.
Three stiffnesses, and which is which
They get confused constantly, and they answer different questions.
Initial stiffness k — the slope before any yielding. Sets the elastic period.
Tangent stiffness kt — the slope at the current point. It is k while elastic and αk while yielding, and it flips between them within a single time step. This is what the SOLVER uses.
Secant stiffness ksec — the slope of the straight line from the origin to the current point. It is what an EQUIVALENT LINEAR system would have, and it is the basis of every displacement-based design method. It is never used inside the time-stepping.
Tangent stiffness is for solving the step. Secant stiffness is for replacing the system with a linear one. Using one where the other belongs is a common and expensive error.
Why a nonlinear step needs iteration
In Module 8 a Newmark step was a single linear solve, because the effective stiffness was known before the step began.
It no longer is. The restoring force at the end of the step depends on the displacement at the end of the step, which is what the step is trying to find. The two are circular, so the step is solved iteratively:
- 1.Guess the displacement increment using the current tangent stiffness.
- 2.Compute the restoring force that increment actually produces, following the hysteresis rule.
- 3.Compare it with what equilibrium demands. The difference is the residual.
- 4.Correct the displacement by residual ÷ current tangent stiffness, and go back to step 2.
This is Newton–Raphson, and it usually converges in two or three iterations. It stops converging where the tangent stiffness is nearly zero — which is exactly what an elastic–perfectly plastic model gives you, and one of the practical reasons to carry a small positive α.
If a nonlinear analysis reports non-converged steps, the results after the first one are not equilibrium solutions. They are whatever the solver was holding when it gave up. This is not a warning to acknowledge and move past.
The period lengthens during the response
Here is the physical consequence that matters most. When the system yields, the tangent stiffness drops from k to αk. The instantaneous period is set by that stiffness, so the structure's period LENGTHENS the moment it yields, and it lengthens more the further it goes.
For most ground motions, moving to a longer period moves down the acceleration branch of the spectrum. A yielding structure attracts less force than an elastic one partly because it has stopped being the structure that was being shaken. It softened, and the earthquake was no longer tuned to it.
That is not the whole story — energy dissipation is the other half — but it is the half that is usually missed.
Try it
Yielding, hysteresis and what it buys
Cycle a yielding element and read the loop, then put the same element under an earthquake.
1.0 means the structure is designed to stay elastic.
0 gives elastic–perfectly plastic.
Record
- Inelastic response
- Elastic response
Show the numbers behind this plot
- Yield force fy
- 0.66 kN
- Yield displacement uy
- 10.7 mm
- Loop area, prescribed cycle
- 0.0821 kJ
- Compare 4fy(u₀ − uy)
- 0.0847 kJ
- Equivalent viscous damping at this ductility
- 48%
- Peak elastic force demand
- 2.64 kN
- Peak inelastic force
- 0.752 kN
- Ductility demand μ
- 5.66
- Residual displacement
- 24.3 mm
- Peak elastic displacement
- 42.8 mm
- Peak inelastic displacement
- 60.6 mm
- Effective period at peak ductility
- 1.78 s
- Newton iterations that failed
- 0
Approximate — hardening enlarges the loop.
Lengthened from 0.80 s as the structure softened.
Strength reduced to 25% of the elastic demand, and the displacement is 1.42× the elastic value — close to the equal-displacement rule, which predicts a ratio near 1 for structures of this period.
What the numbers behind the rules look like
- Equal-displacement rule: R = μ, so a ductility of 5.7 would justify a strength reduction of 5.7.
- Equal-energy rule: R = √(2μ − 1) = 3.21 — more demanding, and the one that applies at short period.
- Set α to zero and watch the residual displacement grow. Post-yield stiffness is what pulls a structure back towards its original position; without it, drifts accumulate in one direction.
- Neither rule is a derivation. Both are observations fitted to large numbers of analyses, and they are the reasoning behind the behaviour factors in seismic codes.
What this shows: A hysteresis loop's AREA is energy dissipated per cycle. That is the whole reason a ductile structure survives forces several times its own strength — it converts the earthquake's energy into plastic work instead of storing it.
From first principles
Response of a bilinear hysteretic system
We want to show: Find the restoring force of a yielding member from its displacement history alone, and from that the energy one full cycle dissipates.
An elastic spring has no memory: tell it the displacement and it tells you the force. A yielding member has memory, and the memory is exactly one number — how much permanent, plastic stretch it has accumulated so far. Everything else follows from that one number. Give the model a plastic displacement up that slides as the member yields, and measure the elastic stretch from there rather than from the origin, and the whole of hysteresis is just Hooke's law applied to the RIGHT quantity. The loop, the parallel unloading, the early yield on reversal and the enclosed area all drop out of that single idea.
Worked example
Reading a bilinear system before running anything
Given
- A single-storey steel frame: lateral stiffness k = 12 MN/m
- Yield force fy = 480 kN (governed by plastic hinging in the beams)
- Post-yield stiffness ratio α = 0.05
- Seismic mass 240 tonnes
- A nonlinear analysis reports a peak displacement of 120 mm
Find
The yield displacement, the ductility demand, the effective period at peak, and the energy dissipated per cycle at that amplitude.
Assumptions
- Bilinear kinematic hardening; no strength or stiffness degradation
Predict first
A bilinear system yields during an earthquake. What happens to its instantaneous period at that moment?
Practice
A bilinear system has k = 12 MN/m and fy = 480 kN. What is its yield displacement, in millimetres?
Practice
The same system reaches a peak displacement of 120 mm. What is the displacement ductility demand?
Practice
Using the equal-energy rule, what strength reduction factor R corresponds to a ductility of 4?
Practice
For the system in question 1 (k = 12 MN/m, fy = 480 kN) with α = 0.05 and μ = 3, what is the secant stiffness at peak, in MN/m?
Practice
An elastic–perfectly plastic system has fy = 480 kN and uy = 40 mm and is cycled to μ = 3. How much energy does one full symmetric cycle dissipate, in kJ?
Check yourself
Which stiffness does a nonlinear solver use inside its Newton iteration?
Check yourself
Why can a response spectrum not be used directly on a structure that yields?
Summary
- A yielding structure's restoring force depends on its history, so fs = ku no longer holds
- The bilinear model needs three numbers: k, fy and α
- Kinematic hardening slides the elastic range without changing its width
- Tangent stiffness solves the step; secant stiffness replaces the system with a linear one
- Every step needs Newton iteration, and non-converged steps invalidate everything after them
- Yielding lengthens the instantaneous period, which usually reduces the demand
- Loop area is dissipated energy: 4fy·uy(μ − 1) for a perfectly plastic system
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint