Module 18 · Lesson 18.1
How a viscous damper works
Force from velocity, the phase shift that makes it almost free, and the exponent that keeps it from becoming expensive.
Why this matters
Damping has appeared in every module as a number, ζ, taken from a table. For an ordinary building 2% to 5% is a reasonable estimate of what friction in connections, cracking and non-structural elements happen to provide. It is not designed; it is observed.
A supplemental damper changes that. It is a manufactured component with a stated force law, placed where the engineer chooses, and it can raise the effective damping to 20% or 30%. Damping stops being a property you estimate and becomes one you specify.
The consequences are more interesting than simply 'less response', and they turn on one fact: a viscous damper's force depends on velocity, and velocity is ninety degrees out of phase with displacement.
By the end of this lesson you should be able to
- State the damper force law and its phase relationship to displacement
- Explain why damper force adds little to peak column force
- Say what the velocity exponent does and why it is set below 1
- Compute energy dissipated per cycle
What you should already know
- Viscous damping in the equation of motion (Module 3)
- Energy dissipated per cycle (Module 7)
- Harmonic response and phase (Module 6)
The force law
A fluid viscous damper is a piston forced through an orifice. Its force is
F = C |v|^α sign(v)
where v is the velocity ACROSS the damper — the rate of change of the relative displacement of its two ends — C is the damping coefficient, and α is the velocity exponent.
Two things follow immediately, and both matter more than they look.
The force depends on velocity, not displacement. A damper does not resist being somewhere; it resists getting there quickly. Hold it at a large displacement and it carries no force at all.
The force acts along the damper's own axis, which for a diagonal brace is not the direction the storey is moving.
The phase shift, and why it is close to free
In harmonic motion at frequency ω:
u(t) = u₀ sin ωt, so v(t) = u₀ω cos ωt
The velocity is ninety degrees ahead of the displacement. When displacement is at its peak, velocity is zero. When displacement is zero, velocity is at its peak.
So a viscous damper delivers its maximum force at the instant the structure passes through its undeformed position — the instant when the columns and braces are carrying their LEAST force.
The damper force and the frame force peak at different times. Adding dampers therefore adds far less to the peak column force than their rated capacity suggests, and often reduces it, because the whole response has become smaller.
This is the property that makes viscous damping attractive compared with adding stiffness. A brace resists displacement, so its force peaks with the frame's and everything downstream — connections, columns, foundations — grows. A damper's peak arrives when the frame is momentarily unloaded.
It is not exactly free. Real dampers have some stiffness, the response is not perfectly harmonic, and the phasing is not exactly ninety degrees. But the effect is real and large, and it is why the columns in a damped frame are usually checked for the combination at the instant of peak damper force as well as at the instant of peak drift.
The velocity exponent
α = 1 is a linear damper. Force proportional to velocity, the classical viscous dashpot, and the only case for which a modal damping ratio can be written exactly.
Seismic dampers are usually made with α ≈ 0.3 to 0.5, and the reason is a force limit.
Consider a damper with C = 900 kN·(s/m)^α. At a storey velocity of 0.35 m/s a linear damper delivers 315 kN. Double the velocity and it delivers 630 kN — twice as much, in strict proportion.
With α = 0.4 the same coefficient gives 591 kN at 0.35 m/s and 780 kN at 0.70 m/s. Doubling the velocity raises the force by only 32%.
That is what the exponent is for. Design events are moderate; the maximum credible event is fast. A linear damper sized for the design event delivers an enormous force in the rare one, and that force has to be carried by the brace, the connection, the beam and the column. A nonlinear damper delivers nearly the same force in the design event and refuses to escalate in the rare one. The exponent caps the force the device can impose on the structure around it.
The cost is analytical: a nonlinear damper has no exact modal damping ratio, so the modal machinery of Module 12 stops applying and the analysis must be a nonlinear time history.
Energy per cycle
For a linear damper in harmonic motion of amplitude u₀ at frequency ω, the energy dissipated per cycle is
ED = π C ω u₀²
Three dependencies worth carrying: linear in C, so doubling the damper doubles the dissipation; linear in ω, so a damper is more effective at higher frequency; and quadratic in u₀, so amplitude matters far more than anything else.
That last one is the practical lever. A damper placed where the structure moves a lot does much more than the same damper placed where it moves a little — and 'a lot' means relative movement ACROSS the damper, not absolute movement of the floor it sits on.
Try it
A manufactured viscous damper
Force from velocity, energy from the loop, and what the velocity exponent is really for.
1 is linear. 0.3–0.5 is typical for seismic dampers, and it caps the force in a rare event.
- α = 0.50
- α = 1 (linear)
- Displacement (mm)
- Damper force (kN ÷ 10)
- Peak relative velocity
- 0.188 m/s
- Peak damper force (axial)
- 868 kN
- Brace efficiency cos²θ
- 0.750
- Effective horizontal force
- 752 kN
- Damper stroke required
- 52 mm
- Energy dissipated per cycle
- 91.1 kJ
- Equivalent modal damping (linear damper)
- 2.7%
- Peak response without dampers
- 90 mm
- Peak response with dampers
- 46.2 mm
- Reduction
- 49%
Shown for α = 1; a nonlinear damper has no exact modal damping ratio.
With α = 0.50, doubling the velocity raises the force by only 41%. That is the point of a nonlinear damper: it delivers nearly full force at moderate velocity and refuses to deliver an enormous force in a rare, fast event — which protects the braces, the connections and the frame behind them.
Sizing and placement
- Added damping goes with the SQUARE of the relative modal displacement across the damper. A damper in a storey where the mode barely drifts does almost nothing, however large it is.
- Dampers add damping without adding stiffness, so they do not change the period and do not attract more force to the structure — unlike a brace, which does both.
- Damper force and structural force peak at different instants, because one depends on velocity and the other on displacement. Adding their peaks arithmetically over-designs the connection.
- The stroke is a hard limit. A damper that bottoms out becomes a rigid strut, and the force that follows is not a damping force at all.
What this shows: A viscous damper's force depends on VELOCITY, not displacement, so it peaks where the displacement is zero — which is why it adds damping without adding stiffness, and why it does not change the structure's period.
Worked example
Sizing a damper and checking what it does to the frame
Given
- A steel frame storey with a peak relative velocity of 0.35 m/s in the design event
- In the maximum credible event the peak relative velocity reaches 0.70 m/s
- Two candidate dampers, both with C = 900 kN·(s/m)^α: one with α = 1.0, one with α = 0.4
- Damper braced at 38° to the horizontal
Find
The force each delivers in both events, and what the brace and connection must be designed for.
Assumptions
- Velocity along the damper axis taken as the storey velocity times cos θ; the axial force is resolved back through cos θ
- Rate effects and damper stiffness neglected, as is usual at this stage of sizing
Predict first
A frame is retrofitted with viscous dampers. At the instant of peak inter-storey drift, what is the damper force?
Practice
A linear damper has C = 900 kN·s/m. What force does it deliver at a relative velocity of 0.35 m/s, in kN?
Practice
A damper with C = 900 kN·(s/m)0.4 and α = 0.4 is driven at 0.35 m/s. What force does it deliver, in kN?
Practice
The same α = 0.4 damper is driven at 0.70 m/s. What force does it deliver, in kN?
Practice
A damper is braced at 38° to the horizontal. What fraction of a horizontal damper's effectiveness does it retain?
Practice
A linear damper with C = 2.0 MN·s/m is cycled at 30 mm amplitude at ω = 4.19 rad/s. How much energy does it dissipate per cycle, in kJ?
Check yourself
Why are seismic viscous dampers usually made with a velocity exponent below 1?
Worked example
Why a viscous damper adds no stiffness
Given
- A linear viscous damper with coefficient c, in a structure oscillating harmonically
Find
The phase of the damper force relative to the displacement, and why it matters
Summary
- Damper force depends on velocity: F = C|v|^α sign(v)
- Velocity leads displacement by 90°, so damper force peaks where frame force is least
- Dampers therefore add little to peak column force
- α ≈ 0.3–0.5 caps the force in rare fast events, at the cost of modal analysis
- A brace at θ loses cos θ twice, so effectiveness goes as cos²θ
- ED = πCωu₀² for a linear damper — amplitude matters most, and it enters squared
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