Structural dynamics · steady-state response

Damping Under Repeated Loading

Compare transient and steady response when energy is dissipated.

Approx. 32 min

Learning objectives

By the end of this lesson

Define steady-state response in physical engineering terms.
Use damped magnification with consistent quantities and units.
Change model inputs and explain the resulting trend.
Separate an educational model result from a code-based safety decision.

Before you calculate

What this means in the real world

What the formula is for

Damping limits the resonance peak and changes the phase relationship between load and response.

R = 1 / √[(1−r²)² + (2ζr)²]

Where an engineer uses it

Mechanical plant supported on a roof: This application turns steady-state response into a real modelling decision.

Combine measured operating frequency with an equivalent modal damping estimate.

What a wrong value can do

A wrong unit, sign, scale, boundary condition, or property can produce a believable number with the wrong physical meaning. It may understate demand, exaggerate it, or lead to the wrong failure check.

Definition

steady-state response

The repeating part of forced response that remains after the transient contribution has decayed.

Why it mattersIt provides the central modelling idea for damping under repeated loading and connects observed motion to a solvable engineering representation.

Build the physical picture

Compare transient and steady response when energy is dissipated. Start by identifying the system boundary, response coordinates, applied action, and the properties that store or dissipate energy. A useful model is the simplest one that preserves the behaviour needed for the engineering question.

Formula

Damped magnification

R = 1 / √[(1−r²)² + (2ζr)²]

Damping limits the resonance peak and changes the phase relationship between load and response.

u: response displacement or displacement vectorm
t: times
ω: angular frequency where usedrad/s

Force-balance lab

Explore steady-state response

Change the available model inputs, watch the motion or force relationship, and describe the trend before reading the numerical result.

Displacement, u 20 mmVelocity, u̇ 0.12 m/sAcceleration, ü 0.60 m/s²Educational force limit 3.0 kN

Illustrative linear SDOF: m = 1000 kg, k = 20 kN/m, c = 5 kN·s/m

/\/\/\/

Fᵢ 0.60 kN ←→ Fₛ + Fᴅ

Stiffness forceFₛ = ku = 0.40 kN

Damping forceFᴅ = cu̇ = 0.60 kN

Inertia forceFᵢ = mü = 0.60 kN

Educational demand/limit ratio0.53

Calculated component-force sum is within the limit you entered.

Not a structural safety verdict. The entered limit is only a learning comparison. Real safety requires code-defined load combinations, strength and serviceability checks, geometry, materials, detailing, and professional review.

Real-world application

Mechanical plant supported on a roof

This application turns steady-state response into a real modelling decision.

Modelling move: Combine measured operating frequency with an equivalent modal damping estimate.
Engineering question: Would added damping materially reduce the steady vibration?

Calculate, compare, interpret

Editable formula question

Explore velocity-dependent resistance: Damping Under Repeated Loading

Change damping and velocity, then calculate the viscous damping-force magnitude.

Fᴅ = c u̇

Damping coefficient c = 500 kN·s/mMin 1 · Textbook example 500 · Max 5000 kN·s/mVelocity u̇ = 0.2 m/sMin 0.01 · Textbook example 0.2 · Max 2 m/s

Substitute your current valuesFᴅ = 500 × 0.2Answer format: enter the numerical result only in kN. Example: 12, not “answer = 12”.

Your calculated answer

Round only at the final step. The accepted tolerance for this exercise is ±0.02.

The displayed minimum, maximum, and textbook values define this learning exercise only. They are not universal design limits.

Editable formula question

Compare a model demand with your entered criterion

Enter a demand and a comparison value chosen for this exercise, then interpret the ratio.

utilization indicator = model demand / entered criterion

Model demand D = 4 unitsMin 0.1 · Textbook example 4 · Max 20 unitsEntered criterion L = 5 unitsMin 0.1 · Textbook example 5 · Max 20 units

Substitute your current valuesratio = 4 ÷ 5Answer format: enter the numerical result only (no unit for this dimensionless result). Example: 12, not “answer = 12”.

Your calculated answer

—

Round only at the final step. The accepted tolerance for this exercise is ±0.01.

The displayed minimum, maximum, and textbook values define this learning exercise only. They are not universal design limits.