Structural Dynamics / Nonlinear structural dynamics
When the restoring force remembers
Hysteresis, the bilinear model and equilibrium iteration in implicit nonlinear integration.
Lesson review: expert verification remains open
This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.
Checks needed for project use
- A dynamics reviewer must confirm excitation models, damping, modal assumptions and any comfort or serviceability interpretation. Small displacement alone is not an acceptance criterion.
- Independent validation of the associated interactive implementation, input domain and project applicability remains required; this review checks lesson context and selected calculations, not design certification.
- Independent nonlinear engineering review needed for constitutive calibration, cyclic degradation, P–Δ effects, member/system ductility and post-event residual definition. R/q factors cannot be selected directly from a member ductility number.
These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.
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Read the lesson register and outstanding review notesWhy this matters
Linear modal superposition assumes a linear restoring-force relation. If a structure yields, restoring force depends on its loading history and independent linear modes no longer reproduce the full response.
Some seismic systems are detailed to develop controlled inelastic deformation; others use different response strategies. This lesson models yielding with a bilinear hysteretic element and solves its equation of motion through time.
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)
Force depends on loading history
An ideal elastic element unloads along its loading line. A yielding element can retain plastic displacement after the force is removed. A stable cyclic force–displacement loop encloses an area equal to the energy dissipated over that cycle. Real braces may also buckle, degrade or pinch; the bilinear model here omits those effects.
The bilinear model
Initial stiffness k sets the elastic slope. Yield force fy gives uy = fy/k. The post-yield ratio α sets the yielded tangent αk; α = 0 gives an elastic–perfectly plastic plateau. Select α from an appropriate material or member model, rather than choosing a positive value to improve convergence.
The model uses linear kinematic hardening. Its back force q = H up shifts an elastic range of width 2fy, with H = αk/(1−α) for 0 ≤ α < 1. The α = 1 limit is handled separately as elastic response.
Tangent and secant stiffness
The initial stiffness describes the elastic response. Tangent stiffness is the local slope used by a consistent Newton iteration. Secant stiffness joins the chosen reference point to the current force–displacement point; an equivalent-linear method may use it to define an effective period. A secant approximation is not the consistent Newton derivative, although other iterative schemes can use approximate stiffnesses.
Solving an implicit nonlinear step
A Newmark step seeks a displacement that satisfies dynamic equilibrium. The solver predicts displacement, updates the constitutive state, calculates the force residual and corrects the displacement. Its effective tangent is kt + γc/(βΔt) + m/(βΔt²). Damping and inertia therefore contribute even when the material tangent is zero.
Investigate non-convergence using residuals, time-step sensitivity, the constitutive update and possible physical instability. Explicit or event-based methods use different procedures; nonlinear response does not universally require Newton iteration.
Period measures during yielding
For 0 < α < 1, the local tangent period on a yielded branch is 2π√(m/(αk)). In this bilinear model it stays constant along that branch and returns to the elastic value on unloading. For α = 0 there is no finite tangent period on the plateau. These local measures do not define a single period for the complete nonlinear response.
A secant period at a specified backbone amplitude can increase with that amplitude. Whether the corresponding spectral demand rises or falls depends on the input and spectrum. Hysteretic energy dissipation must also be included when interpreting the nonlinear response.
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 sets the yield force to the elastic force demand for this record.
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
- EPP equivalent damping reference
- 48%
- Peak elastic force demand
- 2.64 kN
- Peak inelastic force
- 0.752 kN
- Ductility demand μ
- 5.66
- End-of-record 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
Elastic–perfectly plastic reference; hardening reduces the stable loop area at fixed amplitude.
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 . Compare this response with the equal-displacement approximation; its accuracy depends on the period and excitation.
What the numbers behind the rules look like
- Equal-displacement rule: R = μ, so a ductility of 5.7 corresponds to a nominal reduction factor of 5.7.
- Equal-energy rule: R = √(2μ − 1) = 3.21 . This is an idealised energy comparison; it is not a general short-period design rule.
- Vary α and compare the response histories. Kinematic hardening changes the restoring force but does not guarantee self-centring or a monotonic change in residual displacement.
- These idealised response approximations are useful comparisons. Code behaviour factors also depend on the structural system, ductility, overstrength and the applicable design provisions.
What this shows: A stable closed hysteresis loop measures energy dissipated over a cycle. Compare this idealised behaviour with the changing-amplitude response to the selected record; member capacity and cyclic degradation need separate assessment.
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.
The model separates total displacement into elastic and plastic parts. The elastic part gives restoring force, while plastic displacement determines the translated yield surface through q = H up. These state variables describe loading and unloading under the stated linear kinematic-hardening assumptions.
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 with 0 < α < 1 yields. How does its local tangent period compare with its elastic period?
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 material stiffness defines a consistent, updated Newton tangent for the bilinear model?
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
- Implicit nonlinear integration requires an equilibrium solution; other methods use different procedures. Unresolved steps require investigation
- Positive post-yield softening lengthens a local tangent period; spectral demand still depends on the input
- Loop area is dissipated energy: 4fy·uy(μ − 1) for a perfectly plastic system