Linear force-excited model — one degree of freedom
Rotating-Unbalance SDOF Force-Transmission Check
Calculate peak 1× excitation force,steady-state displacement and transmitted support force for one documented equivalent mass–spring–damper mode.
Peak-amplitude SDOF result
Implemented steady-state model
ω=2πn/60 ; F₀=Uω²
ωₙ=√(K/m) ; r=ω/ωₙ ; ζ=c/(2mωₙ)
D=√[(1−r²)²+(2ζr)²]
X=(F₀/K)/D
ტფ=√[1+(2ζr)²]/D ; Fტ=F₀Tფ
m is the combined participating mass in kg,K is the equivalent stiffness in N/m,U is converted from g·mm to kg·m by 10⁻⁶,n is rpm,ω and ωₙ are rad/s,X is the peak displacement amplitude in metres,and F₀ and Fტ are peak force amplitudes in newtons. Force reduction is (1−Tფ)×100%; a negative value explicitly means amplification.
Source and dimensional check
MIT OpenCourseWare,Rotating Imbalance derives the harmonic force components with amplitude meω². MIT OpenCourseWare,Vibration Isolation derives the force-excited SDOF response and transmitted-force ratio. Units check:F₀=(kg·m)/s²=N;F₀/K=N/(N/m)=m.
Interpretation limits
The result is not a balance tolerance,a vibration-severity value,a foundation design or a safe/unsafe verdict. Enter only equivalent mass,stiffness,damping and unbalance supported by project data. Total phase-coherent U means the vector-equivalent unbalance producing the modeled 1× force component; unrelated rotors or phases cannot simply be added as scalars. Peak amplitudes must not be compared directly with RMS or peak-to-peak limits.
Ideal undamped resonance
When ζ=0 and r=1,the steady-state transfer denominator is zero. The ideal linear model has no finite harmonic steady-state response. The calculator reports that singularity explicitly; real systems require measured/modelled damping and transient analysis near resonance.
