Foundations of structural motion · Force balance

The Forces That Govern Motion

Experiment with the restoring, dissipating, and inertial actions that shape structural movement.

Approx. 36 min

Learning objectives

By the end of this lesson

Relate stiffness force to displacement, damping force to velocity, and inertia force to acceleration.
Read a free-body diagram for a linear SDOF system.
Combine individual elements into an equivalent dynamic model.
Recognise when gravity cancels through static equilibrium and when it affects stability.

Before you calculate

What this means in the real world

What the formula is for

At every instant, the applied force is balanced by inertia, damping, and stiffness terms. The response coordinate and its derivatives all depend on time.

m ü(t) + c u̇(t) + k u(t) = p(t)

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

Equation of motion

A time-dependent force balance that relates applied loading to inertia, damping, and stiffness actions in the idealized system.

Why it mattersOnce the equation is formulated with a consistent sign convention and initial conditions, an analytical or numerical method can be chosen to obtain response.

Three responses, three resisting forces

A moving structure develops restoring and resisting actions. In the linear idealization, deformation produces stiffness force, velocity produces damping force, and acceleration produces inertia force. The equation of motion is a force balance written for a system whose state changes with time.

Formula

Linear SDOF equation of motion

m ü(t) + c u̇(t) + k u(t) = p(t)

At every instant, the applied force is balanced by inertia, damping, and stiffness terms. The response coordinate and its derivatives all depend on time.

m ü(t): inertia termN
c u̇(t): damping termN
k u(t): stiffness termN
p(t): externally applied time-varying forceN

Force-balance lab

Change the dynamic state

Adjust displacement, velocity, and acceleration to see which component force changes. Values use a transparent illustrative linear SDOF model.

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.

Read the force relationships

The stiffness force is Fₛ = ku, the viscous damping force is Fᴅ = cu̇, and the inertia force is Fᵢ = mü. These terms are assembled with the applied loading using a consistent sign convention. D’Alembert’s principle provides an equivalent equilibrium view by placing inertia force opposite the acceleration.

Relative and total motion

For base excitation such as an earthquake, deformation and damping depend on motion relative to the moving base, while inertia depends on total acceleration. Keeping this distinction clear is essential when formulating the equation of motion.

Editable formula question

Assemble the component-force demand

Choose the instantaneous mass, motion, stiffness, and damping inputs, then calculate mü + cu̇ + ku.

p(t) = mü + cu̇ + ku

Mass m = 1200 kgMin 100 · Textbook example 1200 · Max 10000 kgAcceleration ü = 0.5 m/s²Min 0 · Textbook example 0.5 · Max 3 m/s²Damping coefficient c = 3000 N·s/mMin 0 · Textbook example 3000 · Max 20000 N·s/mVelocity u̇ = 0.1 m/sMin 0 · Textbook example 0.1 · Max 1 m/sStiffness k = 20000 N/mMin 1000 · Textbook example 20000 · Max 100000 N/mDisplacement u = 20 mmMin 0 · Textbook example 20 · Max 100 mm

Substitute your current valuesp = (1200×0.5) + (3000×0.1) + (20000×0.02)Answer 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 demand with an entered limit

Choose an educational model demand and comparison limit, then calculate their ratio.

ratio = demand / entered limit

Model demand D = 1.3 kNMin 0.1 · Textbook example 1.3 · Max 10 kNEntered comparison limit L = 1.5 kNMin 0.1 · Textbook example 1.5 · Max 10 kN

Substitute your current valuesratio = 1.3 ÷ 1.5Answer format: enter the numerical result only (no unit for this dimensionless result). Example: 12, not “answer = 12”.

Your calculated answer

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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.